Exploring the Mountain

Time-independent Schrodinger equation 02

2014-07-23T02:39:00.000-07:00

— 2. The infinite square well —

Imagine a situation where a projectile in confined to a one dimensional movement and bounces off two infinitely rigid walls while conserving kinetic energy.

This situation can be modeled by the following potential:

$ \displaystyle f(x) = \begin{cases} 0 \quad 0\leq x \leq a\\ \infty \quad \mathrm{otherwise} \end{cases}$

Classically speaking the description is basically what we said in our initial paragraph and on what follows we'll derive the Quantum Mechanical description of the physics that result from this potential.

Outside the potential well it is $ {\psi=0}$ since the wave function cannot exist outside the well.

Inside the well the potential is $ {0}$, hence the particle is a free particle and its wave equation is

$ \displaystyle -\frac{\hbar ^2}{2m}\frac{d^2 \psi}{dx^2}=E \psi $

Since $ {E>0}$ we can define a new quantity $ {k}$

$ \displaystyle k=\frac{\sqrt{2mE}}{\hbar}$

and rewrite the wave equation in the following way

$ \displaystyle \frac{d^2 \psi}{dx^2}=-k^2\psi$

The previous equation is the equation of a harmonic oscillator whose solution in known to be of the form

$ \displaystyle \psi=A\sin kx+B\cos kx $

Were $ {A}$ and $ {B}$ are arbitrary constants whose value will be defined by normalization and boundary conditions.

The boundary conditions that have to be respected are

$ {\psi(0)=A\sin k0+B\cos k0=0 }$
$ {\psi(a)=A\sin ka+B\cos ka=0 }$

since one has to have continuity of the wave function and outside the potential well the wave function is vanishing.

The first condition implies $ {B=0}$ and hence the wave function simply is

$ \displaystyle \psi=A\sin kx$

For the second condition it is $ {A\sin ka=0}$. This implies that either $ {A=0}$ or $ {\sin ka=0}$. The first possibility can be discarded since with $ {A=B=0}$ one would have $ {\psi (x)=0}$ and that solution has no physical interest whatsoever. Hence one is left with $ {\sin ka=0}$.

The previous condition implies that

$ \displaystyle ka=0,\pm\pi,\pm 2\pi,\pm 3\pi,\cdots$

Since $ {k=0}$ again leads us to $ {\psi (x)=0}$ we'll discard this value.

The parity of the sine function ($ {\sin (-x)=-\sin x}$) allows one to absorb the sign of the negative solutions into the constant $ {A}$ (which remains undetermined) and we are left with

$ \displaystyle k_n=\frac{n\pi}{a} $

where $ {n}$ runs from $ {1}$ to infinity.

Since the second boundary condition determines the allowed values of $ {k}$ it also determines the allowed values of the energy of the system.

$ \displaystyle E_n=\frac{\hbar ^2 k_n^2}{2m}=\frac{n^2\pi^2\hbar^2}{2ma^2} $

Even though in a classical context a particle trapped in a infinite square well can have any value for its energy that doesn't happen in the quantum mechanical context. The allowed values for the energy arise from the fact that we imposed boundary conditions in a differential equation. Even in classical contexts one is sure to face quantization conditions as long as one imposes boundary conditions in differential equations.

The wave functions are

$ \displaystyle \psi_n(x)=A\sin\left(\frac{n\pi}{a}x\right) $

At this point one just have to find the value of $ {A}$ in order to determine the time-independent wave function.

In order to determine $ {A}$ one must normalize the wave function:

$ {\begin{aligned} 1&=\int_0^a|A|^2\sin^2(kx)dx\\ &=|A|^2\dfrac{a}{2} \end{aligned}}$

Hence

$ \displaystyle A=\sqrt{\frac{2}{a}}$

Where we have chosen the positive real root because the phase of $ {A}$ has no physical significance.

Finally one can write the wave functions that represent the particle inside of the infinite square well:

$ \displaystyle \psi_n(x)=\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right) $

Like we said previously the time-independent Schroedinger equation has an infinite set of solutions. As we can see in the expression for $ {E}$ the energy of the particle increases as $ {n}$ increases. For that reason the state $ {n=1}$ is said to be the ground state while the other values of $ {n}$ are said to be excited states.

— 2.0.1. Properties of infinite square well solutions —

The solutions we just found to the infinite square well have some interesting properties. As an example we'll sketch the first four solutions to the time-independent Schroedinger equation (here we set $ {a=1}$).

The wave functions are alternatively even and odd relative to the center of square well.
The wave functions of successive energy states have one more node than the wave function that precedes it. $ {\psi_11}$ has $ {0}$ nodes, $ {\psi_2}$ has one node, $ {psi_3}$ has two nodes, and so on and so forth.
The wave functions are orthonormal

$ \displaystyle \int\psi_m^*(x)\psi_n(x)dx =\delta_{mn}$
The set of wave functions is complete.

$ \displaystyle f(x)=\sum_{n=1}^{\infty}c_n\psi_n(x)=\sqrt{\frac{2}{a}}\sum_{n=1}^{\infty}c_n\sin\left(\frac{n\pi}{a}x\right) $

To evaluate the coefficients $ {c_n}$ one uses the expression $ {\int \psi_m^*(x)f(x)dx}$. The proof is
$ {\begin{aligned} \int \psi_m^*(x)f(x)dx&=\sum_{n=1}^{\infty}c_n\int \psi_m^*(x)\psi_n(x)dx\\ &=\sum_{n=1}^{\infty}c_n\delta_{mn}\\ &=c_m \end{aligned}}$

The first property is valid for all symmetric potentials. The second property is always valid. Orthonormality also is universally valid. The property of completeness is the more subtle one, but for all practical purposes we can consider that for every potential we'll encounter that the set of solutions is complete.

— 2.1. Time-dependent solutions —

The stationary states for the infinite square well are

$ \displaystyle \Psi_n(x,t)=\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right)e^{-in^2\pi^2\hbar/(2ma^2)t} \ \ \ \ \ (1)$

For our job to be done we need to show that general solutions to the time-dependent Schroedinger equation can be written as linear combinations of stationary states.

In order to do that one must first write the general solution for $ {t=0}$

$ \displaystyle \Psi(x,0)=\sum_{n=1}^\infty c_n\psi _n(x) $

Since $ {\psi _n}$ form a complete set we know that $ {\Psi (x,0)}$ can be written in that way. Using the orthonormality condition we know that the coefficients are:

$ \displaystyle c_n=\sqrt{\frac{2}{a}}\int_0^a \sin\left(\frac{n\pi}{a}x\right)\Psi(x,0)dx$

With this we can write

$ \displaystyle \Psi(x,t)=\sum_{n=1}^\infty c_n\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right)e^{-in^2\pi^2\hbar/(2ma^2)t} $

which is the most general solution to the infinite square well potential.

Time-independent Schrodinger equation 01

2014-05-31T22:55:00.000-07:00

— 1. Stationary states —

In the previous posts we've normalized wave functions, we've calculated expectation values of momenta and positions but never at any point we've made a quite logical question:

How does one calculate the wave function in the first place?

The answer to that question obviously is:

You have to solve the Schroedinger equation.

The Schroedinger equation is

$ \displaystyle i\hbar\frac{\partial \Psi(x,t)}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi(x,t)}{\partial x^2}+V\Psi(x,t)$

Which is partial differential equation of second order. Partial differential equations are very hard to solve whereas ordinary differential equations are easily solved.
The trick is to to turn this partial differential equation into ordinary differential equation.
To do such a thing we'll employ the separation of variables technique.
We'll assume that $ {\Psi(x,t)}$ ca be written as the product of two functions. One of the functions is a function of the position alone whereas the other function is solely a function of $ {t}$.

$ \displaystyle \Psi(x,t)=\psi(x)\varphi(t)$

This restriction might seem as overly restrictive to the class of solutions of the Schroedinger Equations, but in this case appearances are deceiving. As we'll see later on more generalized solutions of the Schroedinger Equation can be constructed with separable solutions.
Calculating the appropriate derivatives for $ {\Psi(x,t)}$ yields:

$ \displaystyle \frac{\partial \Psi}{\partial t}=\psi\frac{d\varphi}{dt} $

and

$ \displaystyle \frac{\partial^2 \Psi}{\partial x^2} = \frac{d^2 \psi}{d x^2}\varphi $

Substituting the previous equations into the Schroedinger equation results is:

$ \displaystyle i\hbar\psi\frac{d\varphi}{dt}=-\frac{\hbar^2}{2m}\frac{d \psi^2}{d x^2}\varphi+V\psi\varphi$

Dividing the previous equality by $ {\psi\varphi}$

$ \displaystyle i\hbar\frac{1}{\varphi}\frac{d\varphi}{dt}=-\frac{\hbar^2}{2m}\frac{1}{\psi}\frac{d \psi^2}{d x^2}+V$

Now in the previous equality the left-hand side is a function of $ {t}$ while the right-hand side is a function of $ {x}$ (remember that by hypothesis $ {V}$ isn't a function of $ {t}$).
These two facts make the equality expressed in the last equation require a very fine balance. For instance if one were to vary $ {x}$ without varying $ {t}$ then the right-hand side would change while left-hand side would remain the same spoiling our equality. Evidently such a thing can't happen! Te only way for all equality to hold is that both sides of the equation are in fact constant. That way there's no more funny business of changing one side while the other remains constant.
For reasons that will become obvious in the course of this post we'll denote this constant (the so-called separation constant) by $ {E}$.

$ \displaystyle i\hbar\frac{1}{\varphi}\frac{d\varphi}{dt}=E \Leftrightarrow \frac{d\varphi}{dt}=-\frac{i E}{\hbar}\varphi$

and for the second equation

$ \displaystyle -\frac{\hbar^2}{2m}\frac{1}{\psi}\frac{d^2 \psi}{d x^2}+V=E \Leftrightarrow -\frac{\hbar^2}{2m}\frac{d^2 \psi}{d x^2}+V\psi=E\psi$

The first equation of this group is ready to be solved and a solution is

$ \displaystyle \varphi=e^{-i\frac{E}{\hbar}t} $

The second equation, the so-called time-independent Schroedinger equation can only be solved when a potential is specified.
As we can see the method of separable solutions had lived to my promise. With it we were able to produce two ordinary equations which can in principle be solved. In fact one of the equations is already solved.
At this point we'll state a few characteristics of separable solutions in order to better understand their importance (of one these characteristics was already hinted before):

Stationary states The wave function is

$ \displaystyle \Psi(x,t)=\psi(x)e^{-i\frac{E}{\hbar}t}$
and it is obvious that it depends on $ {t}$. On the other hand the probability density doesn't depend on $ {t}$. This result can easily be proven with the implicit assumption that $ {E}$ is real (in a later exercise we'll see why $ {E}$ has to be real).

$ \displaystyle \Psi(x,t)^*\Psi(x,t)=\psi^*(x)e^{i\frac{E}{\hbar}t}\psi(x)e^{-i\frac{E}{\hbar}t}=|\psi(x)|^2 $
If we were interested in calculating the expectation value of any dynamical variable we would see that those values are constant in time.

$ \displaystyle <Q(x,p)>=\int\Psi^*Q\left( x,\frac{\hbar}{i}\frac{\partial}{\partial x} \right)\Psi\, dx $
In particular $ {<x>}$ is constant in time and as a consequence $ {=0}$.
Definite total energy As we saw in classical mechanics the Hamiltonian of a particle is

$ \displaystyle H(x,p)=\frac{p^2}{2m}+V(x) $
Doing the appropriate substitutions the corresponding quantum mechanical operator is (in quantum mechanics operators are denoted by a hat):

$ \displaystyle \hat{H}=-\frac{\hbar^2}{2m}\frac{d^2}{d x^2}+V $
Hence the time-independent Schroedinger equation can be written in the following form:

$ \displaystyle \hat{H}\psi=E\psi$
The expectation value of the Hamiltonian is

$ \displaystyle <\hat{H}>=\int\psi ^*\hat{H}\psi\, dx=E\int|\psi|^2\, dx=E $
It also is

$ \displaystyle \hat{H}^2\psi=\hat{H}(\hat{H}\psi)=\hat{H}(E\psi)=E\hat{H}\psi=EE\psi=E^2\psi $
Hence

$ \displaystyle <\hat{H}^2>=\int\psi ^*\hat{H}^2\psi\, dx=E^2\int|\psi|^2\, dx=E^2 $
So that the variance is

$ \displaystyle \sigma_{\hat{H}}^2=<\hat{H}^2>-<\hat{H}>^2=E^2-E^2=0$
In conclusion in a stationary state every energy measurement is certain to return the value $ {E}$ since that the energy distribution has value $ {E}$.
Linear combination The general solution of the Schroedinger equation is a linear combination of separable solutions.
We'll see in future examples and exercises that the time-independent Schroedinger equation holds an infinite number of solutions. Each of these different wave functions is associated with a different separation constant. Which is to say that for each allowed energy level there is a different wave function.
It so happens that for the time-dependent Schroedinger equation any linear combination of a solution is itself a solution. After finding the separable solutions the task is to construct a more general solution of the form

$ \displaystyle \Psi(x,t)=\sum_{n=1}^{+\infty}c_n\psi_n(x)e^{-i\frac{E_n}{\hbar}t}=\sum_{n=1}^{+\infty}c_n\Psi_n(x,t) $
The point is that every solution of the time-dependent Schroedinger equation can be written like this with the initial conditions of the problem being being studied fixing the values of the constants $ {c_n}$.
I understand that all of this may be a little too abstract so we'll solve a few exercises to make it more palatable.

As an example we'll calculate the time evolution of a particle that starts out in a linear combination of two stationary states:

$ \displaystyle \Psi(x,0)=c_1\psi_1(x)+c_2\psi_2(x)$

For the sake of our discussion let's take $ {c_n}$ and $ {\psi_n}$ to be real.
Hence the time evolution of the particle is simply:

$ \displaystyle \Psi(x,t)=c_1\psi_1(x)e^{-i\frac{E_1}{\hbar}t}+c_2\psi_2(x)e^{-i\frac{E_2}{\hbar}t} $

For the probability density it is
$ {\begin{aligned} |\Psi(x,t)|^2 &= \left( c_1\psi_1(x)e^{i\frac{E_1}{\hbar}t}+c_2\psi_2(x)e^{i\frac{E_2}{\hbar}t} \right) \left( c_1\psi_1(x)e^{-i\frac{E_1}{\hbar}t}+c_2\psi_2(x)e^{-i\frac{E_2}{\hbar}t} \right)\\ &= c_1^2\psi_1^2+c_2^2\psi_2^2+2c_1c_2\psi_1\psi_2\cos\left[ \dfrac{E_2-E_1}{\hbar}t \right] \end{aligned}}$
As we can see even though $ {\psi_1}$ and $ {\psi_2}$ are stationary states and hence their probability density is constant the probability density of the final wave function oscillates sinusoidally with angular frequency $ {(E_2-E_1)/t}$.
Prove that for for normalizable solutions the separation constant $ {E}$ must be real.
Let us write $ {E}$ as
$ {E=E_0+i\Gamma}$
Then the wave equation is of the form

$ \displaystyle \Psi(x,t)=\psi(x)e^{-i\frac{E_0}{\hbar}t}e^{\frac{\Gamma}{\hbar}t} $

$ {\begin{aligned} 1 &= \int_{-\infty}^{+\infty}|\Psi(x,t)|^2\, dx \\ &= \int_{-\infty}^{+\infty} \psi(x,t)^*\psi(x,t)e^{-i\frac{E_0}{\hbar}t}e^{i\frac{E_0}{\hbar}t}e^{\frac{\Gamma}{\hbar}t}e^{\frac{\Gamma}{\hbar}t}\, dx \\ &= e^{\frac{2\Gamma}{\hbar}t}\int_{-\infty}^{+\infty}|\psi(x,t)|^2\, dx \end{aligned}}$
The final expression has to be equal to $ {1}$ to all values of $ {t}$. The only way for that to happen is that we set $ {\Gamma=0}$. Thus $ {E}$ is real.
Show that the time-independent wave function can always be taken to be a real valued function.
We know that $ {\psi(x)}$ is a solution of

$ \displaystyle -\frac{\hbar^2}{2m}\frac{d^2 \psi}{d x^2}+V\psi=E\psi $

Taking the complex conjugate of the previous equation

$ \displaystyle -\frac{\hbar^2}{2m}\frac{d^2 \psi^*}{d x^2}+V\psi^*=E\psi^* $

Hence $ {\psi^*}$ is also a solution of the time-independent Schroedinger equation.
Our next result will be to show that if $ {\psi_1}$ and $ {\psi_2}$ are solutions of the time-independent Schroedinger equation with energy $ {E}$ then their linear combination also is a solution to the time-independent Schroedinger equation with energy $ {E}$.
Let's take

$ \displaystyle \psi_3=c_1\psi_1+c_2\psi_2$

as the linear combination.
$ {\begin{aligned} -\frac{\hbar^2}{2m}\frac{d^2 \psi_3}{d x^2}+V\psi_3 &= -\frac{\hbar^2}{2m}\left( c_1\dfrac{\partial ^2\psi_1}{\partial x^2}+c_2\dfrac{\partial ^2\psi_2}{\partial x^2} \right)+ V(c_1\psi_1+c_2\psi_2)\\ &= c_1\left( -\frac{\hbar^2}{2m}\dfrac{\partial ^2\psi_1}{\partial x^2}+V\psi_1 \right)+c_2\left( -\frac{\hbar^2}{2m}\dfrac{\partial ^2\psi_2}{\partial x^2}+V\psi_2 \right)\\ &= c_1E\psi_1 + c_2E\psi_2\\ &= E(c_1\psi_1+c_2\psi_2)\\ &= E\psi_3 \end{aligned}}$
After proving this result it is obvious that $ {\psi+\psi^*}$ and that $ {i(\psi-\psi^*)}$ are solutions to the time-independent Schroedinger equation. Apart from being solutions to the time-independent Schroedinger equation it is also evident from their construction that these functions are real functions. Since they have same value for the $ {E}$ as $ {\psi}$ we can use either one of them as a solution to the time-independent Schroedinger equation
If $ {V(x)}$ is an even function than $ {\psi(x)}$ can always be taken to be either even or odd.
Since $ {V(x)}$ is even we know that $ {V(-x)=V(x)}$. Now we need to prove that if $ {\psi(x)}$ is a solution to the time-independent Schroedinger equation then $ {\psi(-x)}$ also is a solution.
Changing from $ {x}$ to $ {-x}$ in the time-independent Schroedinger equation

$ \displaystyle -\frac{\hbar^2}{2m}\frac{d^2 \psi(-x)}{d (-x)^2}+V(-x)\psi(-x)=E\psi(-x) $

in order to understand what's going on with the previous equation we need to simplify

$ \displaystyle \dfrac{d^2}{d (-x)^2}$

Let us introduce the variable $ {u}$ and define it as $ {u=-x}$. Then

$ \displaystyle \frac{d}{du}=\frac{dx}{du}\frac{d}{dx}=-\frac{d}{dx} $

And for the second derivative it is

$ \displaystyle \frac{d^2}{du^2}=\frac{dx}{du}\frac{d}{dx}\frac{dx}{du}\frac{d}{dx}=\left(-\frac{d}{dx}\right)\left(-\frac{d}{dx}\right)=\frac{d^2}{dx^2} $

In the last expression $ {u}$ is a dummy variable and thus can be substituted by any other symbol (also see this post Mathematical trick in Statistical Physics to see what kind of manipulations you can do with change of variables and derivatives). For convenience we'll change it back to $ {x}$ and it is

$ \displaystyle \dfrac{d^2}{d (-x)^2}=\dfrac{d^2}{d x^2}$

So that our initial expression becomes

$ \displaystyle -\frac{\hbar^2}{2m}\frac{d^2 \psi(-x)}{d x^2}+V(-x)\psi(-x)=E\psi(-x) $

Using the fact that $ {V(x)}$ is even it is

$ \displaystyle -\frac{\hbar^2}{2m}\frac{d^2 \psi(-x)}{d x^2}+V(x)\psi(-x)=E\psi(-x) $

Hence $ {\psi(-x)}$ is also a solution to the time-independent Schroedinger equation.
Since $ {\psi(x)}$ and $ {\psi(-x)}$ are solutions to the time-independent Schroedinger equation whenever $ {V(x)}$ is even we can construct even and odd functions that are solutions to the time-independent Schroedinger equation.
The even functions are constructed as

$ \displaystyle h(x)=\psi(x)+\psi(-x) $

and the odd functions are constructed as

$ \displaystyle g(x)=\psi(x)-\psi(-x) $

Since it is

$ \displaystyle \psi(x)=\frac{1}{2}(h(x)+g(x)) $

we have showed that any solution to the time-independent Schroedinger equation can be expressed as a linear combination of even and odd functions when the potential function is an even function.
Show that $ {E}$ must exceed the minimum value of $ {V(x)}$ for every normalizable solution to the time-independent Schroedinger equation.
Rewriting the time-independent Schroedinger equation in order of its second x derivative

$ \displaystyle \frac{d^2\psi}{dx^2}=\frac{2m}{\hbar ^2}(V(x)-E)\psi $

Let us proceed with a proof by contradiction and assume that we have $ {V_{\mathrm{min}}>E}$. Using the previous equation this implies that $ {\dfrac{d^2\psi}{dx^2}}$ and $ {\psi }$ have the same sign. This comes from the fact that $ {\frac{2m}{\hbar ^2}(V(x)-E)}$ is positive.
Let us suppose that $ {\psi}$ is always positive. Then $ {\dfrac{d^2\psi}{dx^2}}$ is also always positive. Hence $ {\psi}$ is concave up. In the first quadrant the graph of the function is shaped like

Since by hypothesis our function is normalizable it needs to go to $ {0}$ as $ {x\rightarrow -\infty}$. in order for the function to go to $ {0}$ the plot needs to do something like this

Such a behavior would imply that there is region of space where the function is positive and its second derivative is negative (in our example such a region is delimited in $ {-0.5\leq x \leq 0.5}$).
Such a behavior is in direct contradiction with the conclusion that $ {\psi}$ and $ {\dfrac{d^2\psi}{dx^2}}$ always have the same sign. Since this contradiction arose from the hypothesis that $ {V_{\mathrm{min}}>E}$ the logical conclusion is that $ {V_{\mathrm{min}}<E}$.
With $ {V_{\mathrm{min}}<E}$, $ {\psi}$ and $ {\dfrac{d^2\psi}{dx^2}}$ no longer need to have the same sign at all times. Hence $ {\psi}$ can turn over to $ {0}$ and $ {\psi}$ can go to $ {0}$.

The Wave Function 05

2014-05-29T10:28:00.000-07:00

— 1.6. The Uncertainty Principle —

Imagine that is holding a rope in one's hand at that the rope is tied at the end to brick wall. If one suddenly jerks the rope it would cause a pulse formation that would travel along the rope until hitting the brick wall. At every instant of time you could fairly reasonably ascribe a position to this wave pulse but on the other hand if you would be asked to calculate its wavelength you wouldn't know how to do it since this phenomenon isn't periodic.

Imagine now that, instead of just producing one jerk, you continuously wave the rope so that you end up producing a standing wave. In this case the wavelength is perfectly defined, since this a phenomenon that is periodic, but the wave position loses its meaning.

Quantum mechanics, as we'll see in later posts, asks for a particle description that is given in terms of wave packets. Roughly speaking, a wave packet is the result of summing an infinite number of waves (with different wave numbers and phases) that exhibit constructive interference in just a small region of space. An infinite number of waves with different momenta is needed to ensure constructive and destructive interference in the appropriate regions of space.

Hence we see that by summing more and more waves we are able to make the position of the particle more and more defined while simultaneously making its momentum less and less defined (remember that the waves that we are summing all have different momenta).

In a more formal language one would say that one is working in two different spaces. The position space and the momentum space. What we're seeing is that in the wave packet formalism it is impossible to have a phenomenon that is perfectly localized in both spaces at the same time.

More physically speaking this means that for a particle its position and momentum have an inherent spread. One can theoretically make the spread of one of the quantities as small as one wants but that would cause the spread in the other quantity to get larger and larger. That is to say the more localized a particle is the more its momentum is spread and the more precise a particle's momentum is the more fuzzy is its position.

This result is known as Heisenberg's uncertainty principle and one can make it a mathematically rigorous, but for now this handwaving argument is enough. With it we can already see that Quantum Mechanics needs a radical new way of confronting reality.

For now we'll just put this result in a quantitative footing and leave its proof for a later post

$ \displaystyle \sigma_x \sigma_p \geq \frac{\hbar}{2} \ \ \ \ \ (31)$

One can interpret the uncertainty principle in the language of measurements being made on an ensemble of identically prepared systems. Imagine that you prepare an ensemble whose position measurements are very defined. That is to say that every time you measure the position of a particle the results are very much alike. Well, in this case if you were to also measure the momentum of each particle you would see that the values of momentum you'd end up measuring would be wildly different.

On the other hand you could possibly want to have an ensemble of particles whose momentum measurements would end up with values that have small differences between them. In this case the price to pay would be that the positions of the particles would be scattered all over the place.

Evidently that between those two extremes there is a plethora of possible results. The only limitation that the uncertainty principle stipulates is that the product of the spreads of the two quantities has to be bigger than $ {\dfrac{\hbar}{2}}$.

Exercise 5 A particle of mass $ {m}$ is in the state

$ \displaystyle \Psi(x,t)=Ae^{-a\left[\dfrac{mx^2}{\hbar}+it\right]} \ \ \ \ \ (32)$

where $ {A}$ and $ {a}$ are positive constants.

Find A

To find the value of $ {A}$ one has to normalize the wave function

$ {\begin{aligned} 1 &= \int_{-\infty}^{+\infty} |\Psi(x,t)|^2\,dx\\ &= |A|^2\int_{-\infty}^{+\infty} e^{2a\dfrac{mx^2}{\hbar}}\, dx\\ &= |A|^2 \sqrt{\dfrac{\hbar\pi}{2am}} \end{aligned}}$

Thus

$ \displaystyle A=\sqrt[4]{\frac{2am}{\hbar\pi}} $

For what potential energy function $ {V(x)}$ does $ {\Psi}$ satisfy the Schroedinger equation?

The Schroedinger equation is

$ \displaystyle i\hbar\frac{\partial \Psi}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}+V\Psi \ \ \ \ \ (33)$

For the first term it follows

$ \displaystyle \frac{\partial \Psi}{\partial t}=-ia\Psi$

For the first $ {x}$ derivative it is

$ \displaystyle \frac{\partial \Psi}{\partial x}=-\frac{2amx}{\hbar}\Psi $

For the second order $ {x}$ derivative it is

$ {\begin{aligned} \frac{\partial ^2 \Psi}{\partial x^2} &= -\frac{2am}{\hbar}\Psi+ \dfrac{4a^2m^2x^2}{\hbar ^2}\Psi\\ &= -\dfrac{2am}{\hbar}\left( 1-\dfrac{2amx^2}{\hbar} \right)\Psi \end{aligned}}$

Replacing these expressions into the Schroedinger equation yields

$ {\begin{aligned} V\Psi &= i\hbar\dfrac{\partial \Psi}{\partial t}+\dfrac{\hbar ^2}{2m}\dfrac{\partial^2 \Psi}{\partial x^2}\\ &= a\hbar\Psi+\dfrac{\hbar ^2}{2m}\left[ -\dfrac{2am}{\hbar} \left( 1-\dfrac{2amx^2}{\hbar} \right)\Psi \right]\\ &= a\hbar\Psi-a\hbar\Psi+\hbar a\dfrac{2amx^2}{\hbar}\Psi\\ &= 2ma^2x^2\Psi \end{aligned}}$

Thus

$ \displaystyle V=2ma^2x^2 $

Calculate the expectation values of $ {x}$, $ {x^2}$, $ {p}$ and $ {p^2}$.

The expectation value of $ {x}$

$ \displaystyle <x>=|A|^2\int_{-\infty}^{+\infty}xe^{-2ax\frac{x^2}{\hbar}}\, dx=0$

The expectation value of $ {p}$

$ \displaystyle =m\frac{d<x>}{dt}=0 $

The expectation value of $ {x^2}$

$ {\begin{aligned} <x^2> &= |A|^2\int_{-\infty}^{+\infty}x^2e^{-2ax\frac{x^2}{\hbar}}\, dx\\ &= 2|A|^2\dfrac{1}{4(2m/\hbar)}\sqrt{\dfrac{\pi\hbar}{2am}}\\ &= \dfrac{\hbar}{4am} \end{aligned}}$

The expectation value of $ {p^2}$

$ {\begin{aligned} <p^2> &= \int_{-\infty}^{+\infty}\Psi ^* \left( \dfrac{\hbar}{i}\dfrac{\partial }{\partial x} \right)^2\Psi\, dx\\ &= -\hbar ^2\int_{-\infty}^{+\infty}\Psi ^* \dfrac{\partial ^2 \Psi}{\partial x^2}\, dx\\ &= -\hbar ^2\int_{-\infty}^{+\infty}\Psi ^* \left[ -\dfrac{2am}{\hbar} \left( 1-\dfrac{2amx^2}{\hbar} \right)\Psi \right]\, dx\\ &= 2am\hbar\int_{-\infty}^{+\infty}\Psi ^* \left( 1-\dfrac{2amx^2}{\hbar} \right)\Psi\, dx\\ &= 2am\hbar\left[ \int_{-\infty}^{+\infty}\Psi ^*\Psi\, dx -\dfrac{2am}{\hbar}\int_{-\infty}^{+\infty}\Psi ^* x^2 \Psi\, dx\right]\\ &= 2am\hbar\left[ 1-\dfrac{2am}{\hbar}<x^2> \right]\\ &= 2am\hbar\left[ 1-\dfrac{2am}{\hbar}\dfrac{\hbar}{4am}\right]\\ &=2am\hbar\left( 1-1/2 \right)\\ &=am\hbar \end{aligned}}$

Find $ {\sigma_x}$ and $ {\sigma_p}$. Is their product consistent with the uncertainty principle?

$ \displaystyle \sigma_x=\sqrt{<x^2>-<x>^2}=\sqrt{\dfrac{\hbar}{4am}} $

$ \displaystyle \sigma_p=\sqrt{<p^2>-^2}=\sqrt{am\hbar} $

And the product of the two previous quantities is

$ \displaystyle \sigma_x \sigma_p=\sqrt{\dfrac{\hbar}{4am}}\sqrt{am\hbar}=\frac{\hbar}{2} $

The product is consistent with the uncertainty principle.

The Wave Function 04

2014-05-24T12:07:00.000-07:00

— 1.5. Momentum and other Dynamical quantities —

Let us suppose that we have a particle that is described by the wave function $ {\Psi}$ then the expectation value of its position is (as we saw in The Wave Function 02):

$ \displaystyle <x>=\int_{-\infty}^{+\infty}x|\Psi(x,t)|^2\, dx $

Neophytes interpret the previous equations as if it was saying that the expectation value coincides with the average of various measurements of the position of a particle that is described by $ {\Psi}$. This interpretation is wrong since the first measurement will make the wave function collapse to the value that is actually obtained and if the following measurements of the position are done right away they'll just be of the same value of the first measurement.

Actually $ {<x>}$ is the average of position measurements of particles that are all described by the state $ {\Psi}$. That is to say that we have two ways of actually accomplishing what is implied by the previous interpretation of $ {<x>}$:

We have a single particle. Then after a position measurement is made we have to able to make the particle to return to its $ {\Psi}$ state before we make a new measurement.
We have a collection - a statistical ensemble is a more respectable name - of a great number of particles (in order for it to be statistically significant) and we arrange them all to be in state $ {\Psi}$. If we perform the measurement of the position of all this particles, then average of the measurements should be $ {<x>}$.

To put it more succinctly:

The expectation value is the average of repeated measurements on an ensemble of identically prepared systems.

Since $ {\Psi}$ is a time dependent mathematical object it is obvious that $ {<x>}$ also is a time dependent quantity:

$ {\begin{aligned} \dfrac{d<x>}{dt}&= \int_{-\infty}^{+\infty}x\dfrac{\partial}{\partial t}|\Psi|^2\, dx \\ &= \dfrac{i\hbar}{2m}\int_{-\infty}^{+\infty}x\dfrac{\partial}{\partial x}\left( \Psi^*\dfrac{\partial \Psi}{\partial x}-\dfrac{\partial \Psi^*}{\partial x}\Psi \right)\, dx \\ &= -\dfrac{i\hbar}{2m}\int_{-\infty}^{+\infty}\left( \Psi^*\dfrac{\partial \Psi}{\partial x}-\dfrac{\partial \Psi^*}{\partial x}\Psi \right)\,dx \\ &= -\dfrac{i\hbar}{m}\int_{-\infty}^{+\infty}\left( \Psi^*\dfrac{\partial \Psi}{\partial x}\right)\,dx \end{aligned}}$

where we have used integration by parts and the fact that the wave function has to be square integrable which is to say that the function is vanishingly small as $ {x}$ approaches infinity.

(Allow me to go on a tangent here but I just want to say that rigorously speaking the Hilbert space isn't the best mathematical space to construct the mathematical formalism of quantum mechanics. The problem with the Hilbert space approach to quantum mechanics is two fold:

the functions that are in Hilbert space are necessarily square integrable. The problem is that many times we need to calculate quantities that depend not on a given function but on its derivative (for example), but just because a function is square integrable it doesn't mean that its derivative also is. Hence we don't have any mathematical guarantee that most of the integrals that we are computing actually converge.
The second problem is that when we are dealing with continuous spectra (later on we'll see what this means) the eigenfunctions (we'll see what this means) are divergent

The proper way of doing quantum mechanics is by using rigged Hilbert spaces. A good first introduction to rigged Hilbert spaces and their use in Quantum Mechanics is given by Rafael de la Madrid in the article The role of the rigged Hilbert space in Quantum Mechanics )

The previous equation doesn't express the average velocity of a quantum particle. In our construction of quantum mechanic nothing allows us to talk about the velocity of particle. In fact we don't even know what the meaning of

velocity of a particle

is in quantum mechanics!

Since a particle doesn't have a definitive position prior to is measurement it also can't have a well defined velocity. Later on we'll see how how to construct the probability density for velocity in the state $ {\Psi}$.

For the purposes of the present section we'll just postulate that the expectation value of the velocity is equal to the time derivative of the expectation value of position.

$ \displaystyle <v>=\dfrac{d<x>}{dt} \ \ \ \ \ (24)$

As we saw in the lagrangian formalism and in the hamiltonian formalism posts of our blog it is more customary (since it is more powerful) to work with momentum instead of velocity. Since $ {p=mv}$ the relevant equation for momentum is;

$ \displaystyle =m\dfrac{d<x>}{dt}=-i\hbar\int_{-\infty}^{+\infty}\left( \Psi^*\dfrac{\partial \Psi}{\partial x}\right)\,dx \ \ \ \ \ (25)$

Since $ {x}$ represents the position operator operator we can say in an analogous way that

$ \displaystyle \frac{\hbar}{i}\frac{\partial}{\partial x}$

represents the momentum operator. A way to see why this definition makes sense is to rewrite the definition of the expectation value of the position

$ \displaystyle <x>=\int \Psi^* x \Psi \, dx$

and to rewrite equation 25 in a more compelling way

$ \displaystyle = \int \Psi^*\left( \frac{\hbar}{i}\frac{\partial}{\partial x} \right) \Psi \, dx$

After knowing how to calculate the expectation value of these two dynamical quantities the question now is how can one calculate the expectation value of other dynamical quantities of interest?

The thing is that all dynamical quantities can be expressed as functions of of $ {x}$ and $ {p}$. Taking this into account one just has to write the appropriate function of the quantity of interest in terms of $ {p}$ and $ {x}$ and then calculate the expectation value.

In a more formal (hence more respectable) way the equation for the expectation value of the dynamical quantity $ {Q=Q(x,p)}$ is

$ \displaystyle <Q(x,p)>=\int\Psi^*Q\left( x,\frac{\hbar}{i}\frac{\partial}{\partial x} \right)\Psi\, dx \ \ \ \ \ (26)$

As an example let us look into what would be the relevant expression for the kinetic energy the relevant definition can be found at Newtonian Mechanics 01. Henceforth we'll use $ {T}$ to denote the kinetic energy instead of $ {K}$ in order to use the same notation that is used in Introduction to Quantum Mechanics (2nd Edition).

$ \displaystyle T=\frac{1}{2}mv^2=\frac{p^2}{2m} $

Hence the expectation value is

$ \displaystyle <T>=-\frac{\hbar ^2}{2m}\int\Psi^*\frac{\partial ^2\Psi}{\partial x^2}\, dx \ \ \ \ \ (27)$

Exercise 3 Why can't you do integration by parts directly on

$ \displaystyle \frac{d<x>}{dt}=\int x\frac{\partial}{\partial t}|\Psi|^2 \, dx$

pull the time derivative over onto $ {x}$, note that $ {\partial x/\partial t=0}$ and conclude that $ {d<x>/dt=0}$?

Because integration by parts can only be used when the differentiation and integration are done with the same variable.

Exercise 4 Calculate

$ \displaystyle \frac{d}{dt}$

First lets us remember the the Schroedinger equation:

$ \displaystyle \frac{\partial \Psi}{\partial t}=\frac{i\hbar}{2m}\frac{\partial^2\Psi}{\partial x^2}-\frac{i}{\hbar}V\Psi \ \ \ \ \ (28)$

And its complex conjugate

$ \displaystyle \frac{\partial \Psi^*}{\partial t}=-\frac{i\hbar}{2m}\frac{\partial^2\Psi^*}{\partial x^2}+\frac{i}{\hbar}V\Psi^* \ \ \ \ \ (29)$

for the time evolution of the expectation value of momentum is

$ {\begin{aligned} \dfrac{d}{dt} &= \dfrac{d}{dt}\int\Psi ^* \dfrac{\hbar}{i}\dfrac{\partial \Psi}{\partial x}\, dx\\ &= \dfrac{\hbar}{i}\int \dfrac{\partial}{\partial t}\left( \Psi ^* \dfrac{\partial \Psi}{\partial x}\right)\, dx\\ &= \dfrac{\hbar}{i}\int\left( \dfrac{\partial \Psi^*}{\partial t}\dfrac{\partial \Psi}{\partial x}+\Psi^* \dfrac{\partial}{\partial x}\dfrac{\partial \Psi}{\partial t} \right) \, dx\\ &= \dfrac{\hbar}{i}\int \left[ \left( -\dfrac{i\hbar}{2m}\dfrac{\partial^2\Psi^*}{\partial x^2}+\dfrac{i}{\hbar}V\Psi^* \right)\dfrac{\partial \Psi}{\partial x} + \Psi^*\dfrac{\partial}{\partial x}\left( \dfrac{i\hbar}{2m}\dfrac{\partial^2\Psi}{\partial x^2}-\dfrac{i}{\hbar}V\Psi \right)\right]\, dx\\\ &= \dfrac{\hbar}{i}\int \left[ -\dfrac{i\hbar}{2m}\left(\dfrac{\partial^2\Psi^*}{\partial x^2}\dfrac{\partial\Psi}{\partial x}-\Psi^*\dfrac{\partial ^3 \Psi}{\partial x^3} \right)+\dfrac{i}{\hbar}\left( V\Psi ^*\dfrac{\partial\Psi}{\partial x}-\Psi ^*\dfrac{\partial (V\Psi)}{\partial x}\right)\right]\, dx \end{aligned}}$

First we'll calculate the first term of the integral (ignoring the constant factors) doing integration by parts (remember that the boundary terms are vanishing) two times

$ {\begin{aligned} \int \left(\dfrac{\partial^2\Psi^*}{\partial x^2}\dfrac{\partial\Psi}{\partial x}-\Psi^*\dfrac{\partial ^3 \Psi}{\partial x^3}\right)\, dx &= \left[ \dfrac{\partial \Psi^*}{\partial x^2} \dfrac{\partial \Psi}{\partial x}\right]-\int\dfrac{\partial \Psi^*}{\partial x}\dfrac{\partial ^2 \Psi}{\partial x^2}\, dx- \int \Psi^*\dfrac{\partial ^3 \Psi}{\partial x^3}\, dx \\ &=-\left[ \Psi ^*\dfrac{\partial ^2 \Psi}{\partial x^2} \right]+\int \Psi^*\dfrac{\partial ^3 \Psi}{\partial x^3}\, dx - \int \Psi^*\dfrac{\partial ^3 \Psi}{\partial x^3}\, dx \\ &= 0 \end{aligned}}$

Then we'll calculate the second term of the integral

$ {\begin{aligned} \int \left( V\Psi ^*\dfrac{\partial\Psi}{\partial x}-\Psi ^*\dfrac{\partial (V\Psi)}{\partial x} \right)\, dx &= \int \left( V\Psi ^*\dfrac{\partial\Psi}{\partial x}-\Psi ^* \dfrac{\partial V}{\partial x}\Psi-\Psi ^*V\dfrac{\partial \Psi}{\partial x} \right)\, dx\\ &= -\int\Psi ^* \dfrac{\partial V}{\partial x}\Psi\, dx\\ &=<-\dfrac{\partial V}{\partial x}> \end{aligned}}$

In conclusion it is

$ \displaystyle \frac{d}{dt}=<-\dfrac{\partial V}{\partial x}> \ \ \ \ \ (30)$

Hence the expectation value of the momentum operator obeys Newton's Second Axiom. The previous result can be generalized and its generalization is known in the Quantum Mechanics literature as Ehrenfest's theorem

The wave function 03

2014-05-13T03:40:00.000-07:00

— 1.4. Normalization —

The Scroedinger equation is a linear partial differential equation. As such, if $ {\Psi(x,t)}$ is a solution to it, then $ {A\Psi(x,t)}$ (where $ {A}$ is a complex constant) also is a solution.

Does this mean that a physical problem has an infinite number of solutions in Quantum Mechanics? It doesn't! The thing is that besides the The Scroedinger equation one also has condition 11 to take into account. Stating 11 for the wave function:

$ \displaystyle \int_{-\infty}^{+\infty}|\Psi (x,t)|^2dx=1 \ \ \ \ \ (15)$

The previous equations states the quite obvious fact that the particle under study has to be in some place at a given instant.

Since $ {A}$ was a complex constant the normalization condition fixes $ {A}$ in absolute value but can't tell us nothing regarding its phase. Apparently once again one is haunted with the perspective of having an infinite number of solutions to any given physical problem. The things is that this time the phase doesn't carry any physical significance (a fact that will be demonstrated later) and thus we actually have just one physical solution.

In the previous discussion one is obviously assuming that the wave function is normalizable. That is to say that the function doesn't blow up and vanishes quickly enough at infinity so that the integral being computed makes sense.

At this level it is customary to say that these wave functions don't represent physical states but that isn't exactly true. A wave function that isn't normalizable because integral is infinite might represent a beam of particles in a scattering experiment. The fact that the integral diverges to infinity can then be said to represent the fact that beam is composed by an infinite amount of particles.

While the identically null wave function represents the absence of particles.

A question that now arises has to do with the consistency of our normalization and this is a very sensible question. The point is that we normalize the Schroedinger equation for a given time instant, so how does one know that the normalization holds for other times?

Let us look into the time evolution of our normalization condition 15.

$ \displaystyle \frac{d}{dt}\int_{-\infty}^{+\infty}|\Psi (x,t)|^2dx=\int_{-\infty}^{+\infty}\frac{\partial}{\partial t}|\Psi (x,t)|^2dx \ \ \ \ \ (16)$

Calculating the derivative under the integral for the right hand side of the previous equation

The complex conjugate of the Schroedinger equation is

$ \displaystyle \frac{\partial \Psi^*(x,t)}{\partial t}=-\frac{i\hbar}{2m}\frac{\partial^2\Psi^*(x,t)}{\partial x^2}+\frac{i}{\hbar}V\Psi^*(x,t) \ \ \ \ \ (17)$

Hence for the derivative under the integral

$ {\begin{aligned} \frac{\partial}{\partial t}|\Psi (x,t)|^2&=\frac{\partial}{\partial t}(\Psi^* (x,t)\Psi (x,t))\\ &=\Psi^* (x,t)\frac{\partial\Psi (x,t)}{\partial t}+\frac{\partial \Psi^* (x,t)}{\partial t}\Psi (x,t)\\ &=\frac{i\hbar}{2m}\left( \Psi^*(x,t)\frac{\partial^2\Psi(x,t)}{\partial x^2}-\frac{\partial^2\Psi^*(x,t)}{\partial x^2}\Psi (x,t)\right)\\ &=\frac{\partial}{\partial x}\left[ \frac{i\hbar}{2m}\left( \Psi^*(x,t)\frac{\partial\Psi(x,t)}{\partial x}-\frac{\partial\Psi^*(x,t)}{\partial x}\Psi(x,t) \right) \right] \end{aligned}}$

Getting back to 16

$ \displaystyle \frac{d}{dt}\int_{-\infty}^{+\infty}|\Psi (x,t)|^2dx=\frac{i\hbar}{2m}\left[ \Psi^*(x,t)\frac{\partial\Psi(x,t)}{\partial x}-\frac{\partial\Psi^*(x,t)}{\partial x}\Psi(x,t) \right]_{-\infty}^{+\infty} \ \ \ \ \ (18)$

Since we're assuming that our wave function is normalizable the wave function (and its complex conjugate) must vanish for $ {+\infty}$ and $ {-\infty}$.

In conclusion

$ \displaystyle \frac{d}{dt}\int_{-\infty}^{+\infty}|\Psi (x,t)|^2dx=0$

Since the derivative vanishes one can conclude that the integral is constant.

In conclusion one can say that if one normalizes the wave equation for a given time interval it stays normalized for all time intervals.

Exercise 1 At time $ {t=0}$ a particle is represented by the wave function

$ \displaystyle \Psi(x,0)=\begin{cases} Ax/a & \text{if } 0\leq x\leq a\\ A(b-x)/(b-a) & \text{if } a\leq x\leq b \\ 0 & \text{otherwise}\end{cases} \ \ \ \ \ (19)$

where $ {A}$, $ {a}$ and $ {b}$ are constants.

Normalize $ {\Psi}$.
$ {\begin{aligned} 1&=\int_{-\infty}^{+\infty} |\Psi|^2\,dx\\ &=\int_0^a|\Psi|^2\,dx+\int_a^b|\Psi|^2\,dx\\ &=\dfrac{|A|^2}{a^2}\int_0^a|x^2\,dx+\dfrac{|A|^2}{(b-a)^2}\int_a^b(b-x)^2\,dx\\ &=\dfrac{|A|^2}{a^2}\left[ \dfrac{x^3}{3} \right]_0^a+\dfrac{|A|^2}{(b-a)^2}\left[ \dfrac{(b-x)^3}{3} \right]_a^b\\ &=\dfrac{|A|^2a}{3}+\dfrac{|A|^2}{(b-a)^2}\dfrac{(b-a)^3}{3}\\ &=\dfrac{|A|^2a}{3}+|A|^2\dfrac{b-a}{3}\\ &=\dfrac{b|A|^2}{3} \end{aligned}}$
Hence for $ {A}$ it is

$ \displaystyle A=\sqrt{\dfrac{3}{b}} $
Sketch $ {\Psi(x,0)}$
In $ {0\leq x \leq a}$ $ {\Psi(x,0)}$ is a strictly increasing function that goes from $ {0}$ to $ {A}$.
In $ {a \leq x \leq b}$ $ {\Psi(x,0)}$ is strictly decreasing function that goes from $ {A}$ to $ {0}$.
Hence the plot of $ {\Psi(x,0)}$ is (choosing the following values $ {a=1}$, $ {b=2}$ and $ {A=\sqrt{b}=\sqrt{2}}$):
Where is the particle most likely to be found at $ {t=0}$? Since $ {x=a}$ is maximum of the $ {\Psi}$ function the most likely value for the particle to be found is at $ {x=a}$.
What is the probability of finding the particle to the left of $ {a}$? Check the answers for $ {b=a}$ and $ {b=2a}$.
$ {\begin{aligned} P(x<a)&=\int_0^a|\Psi|^2\,dx\\ &=\dfrac{|A|^2}{a^2}\int_0^a x^2\,dx\\ &=\dfrac{|A|^2}{a^2}\left[ \dfrac{x^3}{3} \right]_0^a\\ &=\dfrac{|A|^2}{3}a\\ &=\dfrac{3}{3b}a\\ &=\dfrac{a}{b} \end{aligned}}$
At first let us look into the $ {b=a}$ limiting case. We can imagine that this is the end result of $ {b}$ getting nearer and nearer to $ {a}$. That is to say that the domain of the strictly decreasing part of $ {\Psi(x,0)}$ is getting shorter and shorter and when finally $ {b=a}$ $ {\Psi(x,0)}$ doesn't have a domain where its is strictly decreasing and $ {\Psi(x,0)}$ is defined by its strictly increasing and vanishing features (in the appropriate domains). That is to say that to the right of $ {a}$ the function is $ {0}$. Hence the probability of the particle being found to the left of $ {a}$ is $ {1}$.
From the previous calculation $ {P(x<a)_{b=a}=1}$ which is indeed the correct result.
The $ {b=2a}$ case can be analyzed in a different way. In this case:
- $ {x=a}$ is the half point of the domain of $ {\Psi(x,0)}$ where $ {\Psi(x,0)}$ is non vanishing (end points of the domain are excluded).
- $ {\Psi(x,0)}$ is strictly increasing in the first half of the domain ($ {0\leq x\leq a}$).
- $ {\Psi(x,0)}$ is strictly decreasing in the second half of the domain ($ {a\leq x\leq b}$).
- $ {\Psi(x,0)}$ is continuous.
Thus one can conclude that $ {\Psi(x,0)}$ is symmetric around $ {a}$ and consequently the probability of the particle being found to the left of $ {a}$ has to be $ {1/2}$.
From the previous calculation $ {P(x<a)_{b=2a}=1/2}$ which is indeed the correct result.
What is the expectation value of $ {x}$?
$ {\begin{aligned} <x>&= \int_a^b x|\Psi|^2\,dx\\ &=\dfrac{|A|^2}{a^2}\int_0^a x^3\,dx+\dfrac{|A|^2}{(b-a)^2}\int_a^b x(b-x)^2\,dx\\ &=\dfrac{|A|^2}{a^2}\left[ \dfrac{x^4}{4} \right]_0^a+\dfrac{|A|^2}{(b-a)^2}\left[ 1/2x^2b^2-2/3x^3b+x^4/4 \right]_a^b\\ &=\dfrac{2a+b}{4} \end{aligned}}$

Exercise 2 Consider the wave function

$ \displaystyle \Psi(x,t)=Ae^{-\lambda |x|}e^{-i\omega t} \ \ \ \ \ (20)$

where $ {A}$, $ {\lambda}$ and $ {\omega}$ are positive real constants.

Normalize $ {\Psi}$
$ {\begin{aligned} 1&=\int_{-\infty}^{+\infty} |\Psi|^2\,dx\\ &=\int_{-\infty}^{+\infty} |A|^2e^{-2\lambda |x|}\,dx\\ &=2|A|^2\int_0^{+\infty}e^{-2\lambda |x|}\,dx \\ &=2|A|^2\int_0^{+\infty}e^{-2\lambda x}\,dx \\ &=-\dfrac{|A|^2}{\lambda}\left[ e^{-2\lambda x} \right]_0^{+\infty}\\ &=\dfrac{|A|^2}{\lambda} \end{aligned}}$
Hence it is

$ \displaystyle A=\sqrt{\lambda} $
Determine $ {<x>}$ and $ {<x^2>}$
$ {\begin{aligned} <x>&=\int_{-\infty}^{+\infty} x|\Psi|^2\,dx\\ &=|A|^2\int_{-\infty}^{+\infty} xe^{-2\lambda |x|}\,dx\\ &=0 \end{aligned}}$
The integral is vanishing because we're calculating the integral of an odd function between symmetrical limits.
$ {\begin{aligned} <x^2>&=\int_{-\infty}^{+\infty} x^2|\Psi|^2\,dx\\ &=2\lambda\int_0^{+\infty} x^2e^{-2\lambda x}\,dx\\ &=2\lambda\int_0^{+\infty} \dfrac{1}{4}\dfrac{\partial^2}{\partial \lambda ^2}\left( e^{-2\lambda x} \right)\,dx\\ &=\dfrac{\lambda}{2} \dfrac{\partial^2}{\partial \lambda ^2}\int_0^{+\infty}e^{-2\lambda\,dx} x \,dx\\ &= \dfrac{\lambda}{2} \dfrac{\partial^2}{\partial \lambda ^2} \left[ -\dfrac{e^{-2\lambda\,dx}}{2\lambda} \right]_0^{+\infty}\\ &= \dfrac{\lambda}{2}\dfrac{\partial^2}{\partial \lambda ^2}\left(\dfrac{1}{2\lambda} \right)\\ &=\dfrac{\lambda}{2}\dfrac{\partial}{\partial \lambda}\left(-\dfrac{1}{\lambda ^2} \right)\\ &=\dfrac{\lambda}{2}\dfrac{1}{\lambda^3}\\ &= \dfrac{1}{2\lambda^2} \end{aligned}}$
Find the standard deviation of $ {x}$. Sketch the graph of $ {\Psi ^2}$. What is the probability that the particle will be found outside the range $ {[<x>-\sigma,<x>+\sigma]}$?

$ \displaystyle \sigma ^2=<x^2>-<x>^2=\frac{1}{2\lambda ^2}-0=\frac{1}{2\lambda ^2}$

Hence the standard deviation is

$ \displaystyle \sigma=\dfrac{\sqrt{2}}{2\lambda}$

The square of the wave function is proportional to $ {e^{-2\lambda |x|}}$. Dealing for piecewise definitions of the square of the wave function, its first derivative in order to $ {x}$ and its second derivative in order to $ {x}$

$ \displaystyle |\Psi|^2=\begin{cases} e^{2\lambda x} & \text{if } x < 0\\ e^{-2\lambda x} & \text{if } x \geq 0 \end{cases} \ \ \ \ \ (21)$

$ \displaystyle \dfrac{\partial}{\partial x}|\Psi|^2=\begin{cases} 2\lambda e^{2\lambda x} & \text{if } x < 0\\ -2\lambda e^{-2\lambda x} & \text{if } x \geq 0 \end{cases} \ \ \ \ \ (22)$

$ \displaystyle \dfrac{\partial ^2}{\partial x ^2}|\Psi|^2=\begin{cases} 4\lambda ^2 e^{2\lambda x} & \text{if } x < 0\\ 4\lambda ^2 e^{-2\lambda x} & \text{if } x \geq 0 \end{cases} \ \ \ \ \ (23)$

As we can see the first derivative of $ {|\Psi|^2}$ changes its sign on $ {0}$ from positive to negative. Hence it was strictly increasing before $ {0}$ and it is strictly decreasing after $ {0}$. Hence $ {0}$ is a maximum of $ {|\Psi|^2}$.
The second derivative is always positive so $ {|\Psi|^2}$ is always concave up (convex).
Hence its graphical representation is:

The probability that the particle is to be found outside the range $ {[<x>-\sigma, <x>+\sigma ]}$ is
$ {\begin{aligned} P(<x>-\sigma, <x>+\sigma)&= 2\int_\sigma^{+\infty}|\Psi|^2\,dx\\ &= 2\lambda\int_\sigma^{+\infty}e^{2\lambda x}\\ &= 2\lambda\left[ -\dfrac{e^{2\lambda x}}{2\lambda} \right]_\sigma^{+\infty}\\ &=\lambda \dfrac{e^{2\lambda x}}{2\lambda}\\ &=e^{-2\lambda\dfrac{\sqrt{2}}{2\lambda}}\\ &=e^{-\sqrt{2}} \end{aligned}}$

The Wave Function Exercises 01

2014-03-28T02:48:00.000-07:00

Exercise 1

$ {<j>^2=21^2=441}$
$ {<j^2>=1/N\sum j^2N(J)=\dfrac{6434}{14}=459.6}$
Calculating for each $ {\Delta j}$

$ {j}$ $ {\Delta j=j-<j>}$
14 $ {14-21=-7}$
15 $ {15-21=-6}$
16 $ {16-21=-5}$
22 $ {22-21=1}$
24 $ {24-21=3}$
25 $ {25-21=4}$

Hence for the variance it follows
$ {\sigma ^2=1/N\sum (\Delta j)^2N(j)=\dfrac{260}{14}=18.6}$
Hence the standard deviation is

$ \displaystyle \sigma =\sqrt{18.6}=4.3$
$ {\sigma^2=<j^2>-<j>^2=459.6-441=18.6}$
And for the standard deviation it is

$ \displaystyle \sigma =\sqrt{18.6}=4.3$

Which confirms the second equation for the standard deviation.

Exercise 2 Consider the first $ {25}$ digits in the decimal expansion of $ {\pi}$.

What is the probability of getting each of the 10 digits assuming that one selects a digit at random.

The first 25 digits of the decimal expansion of $ {\pi}$ are

$ \displaystyle \{3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9, 3, 2, 3, 8, 4, 6, 2, 6, 4, 3\}$

Hence for the digits it is

$ {N(0)=0}$	$ {P(0)=0}$
$ {N(1)=2}$	$ {P(1)=2/25}$
$ {N(2)=3}$	$ {P(2)=3/25}$
$ {N(3)=5}$	$ {P(3)=1/5}$
$ {N(4)=3}$	$ {P(4)=3/25}$
$ {N(5)=3}$	$ {P(5)=3/25}$
$ {N(6)=3}$	$ {P(6)=3/25}$
$ {N(7)=1}$	$ {P(7)=1/25}$
$ {N(8)=2}$	$ {P(8)=2/25}$
$ {N(9)=3}$	$ {P(9)=3/25}$

The most probable digit is $ {5}$. The median digit is $ {4}$. The average is $ {\sum P(i)N(i)=4.72}$.
$ {\sigma=2.47}$

Exercise 3 The needle on a broken car is free to swing, and bounces perfectly off the pins on either end, so that if you give it a flick it is equally likely to come to rest at any angle between $ {0}$ and $ {\pi}$.

Along the $ {\left[0,\pi\right]}$ interval the probability of the needle flicking an angle $ {d\theta}$ is $ {d\theta/\pi}$. Given the definition of probability density it is $ {\rho(\theta)=1/\pi}$.
Additionally the probability density also needs to be normalized.

$ \displaystyle \int_0^\pi \rho(\theta)d\theta=1\Leftrightarrow\int_0^\pi 1/\pi d\theta=1 $

which is trivially true.
The plot for the probability density is
Compute $ {\left\langle\theta \right\rangle}$, $ {\left\langle\theta^2 \right\rangle}$ and $ {\sigma}$.
$ {\begin{aligned} \left\langle\theta \right\rangle &= \int_0^\pi\frac{\theta}{\pi}d\theta\\ &= \frac{1}{\pi}\int_0^\pi\theta d\theta\\ &= \frac{1}{\pi} \left[ \frac{\theta^2}{2} \right]_0^\pi\\ &= \frac{\pi}{2} \end{aligned}}$
For $ {\left\langle\theta^2 \right\rangle}$ it is
$ {\begin{aligned} \left\langle\theta^2 \right\rangle &= \int_0^\pi\frac{\theta^2}{\pi}d\theta\\ &= \frac{1}{\pi} \left[ \frac{\theta^3}{3} \right]_0^\pi\\ &= \frac{\pi^2}{3} \end{aligned}}$
The variance is $ {\sigma^2=\left\langle\theta^2 \right\rangle-\left\langle\theta\right\rangle^2 =\dfrac{\pi^2}{3}-\dfrac{\pi^2}{4}=\dfrac{\pi^22}{12}}$.
And the standard deviation is $ {\sigma=\dfrac{\pi}{2\sqrt{3}}}$.
Compute $ {\left\langle\sin\theta\right\rangle}$, $ {\left\langle\cos\theta\right\rangle}$ and $ {\left\langle\cos^2\theta\right\rangle}$.
$ {\begin{aligned} \left\langle\sin\theta \right\rangle &= \int_0^\pi\frac{\sin\theta}{\pi}d\theta\\ &= \frac{1}{\pi}\int_0^\pi\sin\theta d\theta\\ &= \frac{1}{\pi} \left[ -\cos\theta \right]_0^\pi\\ &= \frac{2}{\pi} \end{aligned}}$
and
$ {\begin{aligned} \left\langle\cos\theta \right\rangle &= \int_0^\pi\frac{\cos\theta}{\pi}d\theta\\ &= \frac{1}{\pi}\int_0^\pi\cos\theta d\theta\\ &= \frac{1}{\pi} \left[ \sin\theta \right]_0^\pi\\ &= 0 \end{aligned}}$
We'll leave $ {\left\langle\cos\theta^2 \right\rangle}$ as an exercise for the reader. As a hint remember that $ {\cos^2\theta=\dfrac{1+\cos(2\theta)}{2}}$.

Exercise 4

In exercise $ {1.1}$ it was shown that the the probability density is

$ \displaystyle \rho(x)=\frac{1}{2\sqrt{hx}}$

Hence the mean value of $ {x}$ is
$ {\begin{aligned} \left\langle x \right\rangle &= \int_0^h\frac{x}{2\sqrt{hx}}dx\\ &= \frac{h}{3} \end{aligned}}$
For $ {\left\langle x^2 \right\rangle}$ it is
$ {\begin{aligned} \left\langle x^2 \right\rangle &= \int_0^h\frac{x}{2\sqrt{hx}}dx\\ &= \frac{1}{2\sqrt{h}}\int_0^h x^{3/2}dx\\ &= \frac{1}{2\sqrt{h}}\left[\frac{2}{5}x^{5/2} \right]_0^h\\ &= \frac{h^2}{5} \end{aligned}}$
Hence the variance is

$ \displaystyle \sigma^2=\left\langle x^2 \right\rangle-\left\langle x \right\rangle^2=\frac{h^2}{5}-\frac{h^2}{9}=\frac{4}{45}h^2 $

and the standard deviation is

$ \displaystyle \sigma=\frac{2h}{3\sqrt{5}} $
For the distance to the mean to be more than one standard deviation away from the average we have two alternatives. The first is the interval $ {\left[0,\left\langle x \right\rangle+\sigma\right]}$ and the second is $ {\left[\left\langle x \right\rangle+\sigma,h\right]}$.
Hence the total probability is the sum of these two probabilities.
Let $ {P_1}$ denote the probability of the first interval and $ {P_2}$ denote the probability of the second interval.
$ {\begin{aligned} P_1 &= \int_0^{\left\langle x \right\rangle-\sigma}\frac{1}{2\sqrt{hx}}dx\\ &= \frac{1}{2\sqrt{h}}\left[2x^{1/2} \right]_0^{\left\langle x \right\rangle-\sigma}\\ &= \frac{1}{\sqrt{h}}\sqrt{\frac{h}{3}-\frac{2h}{3\sqrt{5}}}\\ &=\sqrt{\frac{1}{3}-\frac{2}{3\sqrt{5}}} \end{aligned}}$
Now for the second interval it is
$ {\begin{aligned} P_2 &= \int_{\left\langle x \right\rangle+\sigma}^h\frac{1}{2\sqrt{hx}}dx\\ &= \ldots\\ &=1-\sqrt{\frac{1}{3}+\frac{2}{3\sqrt{5}}} \end{aligned}}$
Hence the total probability $ {P}$ is $ {P=P_1+P_2}$
$ {\begin{aligned} P&=P_1+P_2\\ &= \sqrt{\frac{1}{3}-\frac{2}{3\sqrt{5}}}+1-\sqrt{\frac{1}{3}+\frac{2}{3\sqrt{5}}}\\ &\approx 0.3929 \end{aligned}}$

Exercise 5 The probability density is $ {\rho(x)=Ae^{-\lambda(x-a)^2}}$

Determine $ {A}$.
Making the change of variable $ {u=x-a}$ ($ {dx=du}$) the normalization condition is
$ {\begin{aligned} 1 &= A\int_{-\infty}^\infty e^{-\lambda u^2}du\\ &= A\sqrt{\frac{\pi}{\lambda}} \end{aligned}}$
Hence for $ {A}$ it is

$ \displaystyle A=\sqrt{\frac{\lambda}{\pi}}$
Find $ {\left\langle x \right\rangle}$, $ {\left\langle x^2 \right\rangle}$ and $ {\sigma}$.
$ {\begin{aligned} \left\langle x \right\rangle &= \sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^\infty (u+a)e^{-\lambda u^2}du\\ &= \sqrt{\frac{\lambda}{\pi}}\left(\int_{-\infty}^\infty ue^{-\lambda u^2}du+a\int_{-\infty}^\infty e^{-\lambda u^2}du \right)\\ &=\sqrt{\frac{\lambda}{\pi}}\left( 0+a\sqrt{\frac{\pi}{\lambda}} \right)\\ &= a \end{aligned}}$
If you don't see why $ {\displaystyle\int_{-\infty}^\infty ue^{-\lambda u^2}du=0}$ check this post on my other blog.
For $ {\left\langle x^2 \right\rangle}$ it is
$ {\begin{aligned} \left\langle x^2 \right\rangle &= \sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^\infty (u+a)^2e^{-\lambda u^2}du\\ &= \sqrt{\frac{\lambda}{\pi}}\left(\int_{-\infty}^\infty u^2e^{-\lambda u^2}du+2a\int_{-\infty}^\infty u e^{-\lambda u^2}du+a^2\int_{-\infty}^\infty e^{-\lambda u^2}du \right) \end{aligned}}$
Now $ {\displaystyle 2a\int_{-\infty}^\infty u e^{-\lambda u^2}du=0}$ as in the previous calculation.
For the third term it is $ {\displaystyle a^2\int_{-\infty}^\infty e^{-\lambda u^2}du=a^2\sqrt{\frac{\pi}{\lambda}}}$.
The first integral is the hard one and a special technique can be employed to evaluate it.
$ {\begin{aligned} \int_{-\infty}^\infty u^2e^{-\lambda u^2}du &= \int_{-\infty}^\infty-\frac{d}{d\lambda}\left( e^{-\lambda u^2} \right)du\\ &= -\frac{d}{d\lambda}\int_{-\infty}^\infty e^{-\lambda u^2}du\\ &=-\frac{d}{d\lambda}\sqrt{\frac{\pi}{\lambda}}\\ &=\frac{1}{2}\sqrt{\frac{\pi}{\lambda^3}} \end{aligned}}$
Hence it is
$ {\begin{aligned} \left\langle x^2 \right\rangle &= \sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^\infty (u+a)^2e^{-\lambda u^2}du\\ &= \sqrt{\frac{\lambda}{\pi}}\left(\int_{-\infty}^\infty u^2e^{-\lambda u^2}du+2a\int_{-\infty}^\infty u e^{-\lambda u^2}du+a^2\int_{-\infty}^\infty e^{-\lambda u^2}du \right)\\ &= \sqrt{\frac{\lambda}{\pi}}\left( \frac{1}{2}\sqrt{\frac{\pi}{\lambda^3}}+0+a^2\sqrt{\frac{\pi}{\lambda}} \right)\\ &=a^2+\frac{1}{2\lambda} \end{aligned}}$
The variance is

$ \displaystyle \sigma^2=\left\langle x^2 \right\rangle-\left\langle x \right\rangle^2=\frac{1}{2\lambda}$

Hence the standard deviation is

$ \displaystyle \sigma=\frac{1}{\sqrt{2\lambda}} $

— Mathematica file —

The resolution of exercise 2 was done using some basic Mathematica code which I'll post here hoping that it can be helpful to the readers of this blog.

// N[Pi, 25]

piexpansion = IntegerDigits[3141592653589793238462643]

digitcount = {}

For[i = 0, i <= 9, i++, AppendTo[digitcount, Count[A, i]]]

digitcount

digitprobability = {}

For[i = 0, i <= 9, i++, AppendTo[digitprobability, Count[A, i]/25]]

digitprobability

digits = {}

For[i = 0, i <= 9, i++, AppendTo[digits, i]]

digits

j = N[digits.digitprobability]

digitssquared = {}

For[i = 0, i <= 9, i++, AppendTo[digitssquared, i^2]]

digitssquared

jsquared = N[digitssquared.digitprobability]

sigmasquared = jsquared - j^2

std = Sqrt[sigmasquared]

deviations = {}

deviations = piexpansion - j

deviationssquared = (piexpansion - j)^2

variance = Mean[deviationssquared]

standarddeviation = Sqrt[variance]

The Wave Function 02

2014-03-28T02:28:00.000-07:00

— 1.3. Probability —

In the previous post we were introduced to the Schrodinger equation (equation 1), stated Born's interpretation of what is the physical meaning of the wave function and took a little glimpse into some philosophical positions one might have regarding Quantum Mechanics.
Since probability plays such an essential role in Quantum Mechanics it seems that a brief revision of some of its concepts is in order so that we are sure that we have the tools that allows one to do Quantum Mechanics.

— 1.3.1. Discrete variables —

The example used in the book in order to expound on the terminology and concepts of probability is of a set that consists of 14 people in a class room:

one person has 14 years
one person has 15 years
three people have 16 years
two people 22 years
five people have 25 years

Let $ {N(j)}$ represent the number of people with age $ {j}$. Hence

$ {N(14)=1}$
$ {N(15)=1}$
$ {N(16)=3}$
$ {N(22)=2}$
$ {N(25)=5}$

One can represent the previous data points by use of a histogram:

The the total number of people in the room is given by

$ \displaystyle N=\sum_{j=0}^{\infty}N(j) \ \ \ \ \ (2)$

Adopting a frequentist definition of probability Griffiths then makes a number of definitions of probability concepts under the assumption that the phenomena at study are discrete ones.

Definition 1 The probability of an event $ {j}$, $ {P(j)}$ is proportional to the number elements that have the property $ {j}$ and inversely proportional to the total elements ($ {N}$) under study.

$ \displaystyle P(j)=\frac{N(j)}{N} \ \ \ \ \ (3)$

It is easy to see that from equation 3 together with equation 2 it follows

$ \displaystyle \sum_{j=0}^{\infty}P(j)=1 \ \ \ \ \ (4)$

After defining $ {P(j)}$ we can also define what is the most probable value for $ {j}$.

Definition 2 The most value for $ {j}$ is the one for which $ {P(j)}$ is a maximum.

Definition 3 The average value of $ {j}$ is given by

$ \displaystyle <j>=\sum_{j=0}^{\infty}jP(j) \ \ \ \ \ (5)$

But what if we are interested in computing the average value of $ {j^2}$? Then the appropriate expression must be

$ \displaystyle <j>=\sum_{j=0}^{\infty}j^2P(j) $

Hence one can write with full generality that average value for some function of $ {j}$, denoted by $ {f(j)}$ is given by

$ \displaystyle <f(j)>=\sum_{j=0}^{\infty}f(j)P(j) \ \ \ \ \ (6)$

After introducing the definition of maximum of a probability distribution it is time to introduce a couple of definitions that relate t the symmetry and spread of a distribution.

Definition 4 The median is the value of $ {j}$ for which the probability of having a larger value than $ {j}$ is the same as the probability of having a value with a smaller value than $ {j}$.

After seeing a definition that relates to the the symmetry of a distribution we'll introduce a definition that is an indication of its spread.
But first we'll look at two examples that will serve as a motivation for that:

and

Both histograms have the same median, the same average, the same most probable value and the same number of elements. Nevertheless it is visually obvious that the two histograms represent two different kinds of phenomena.
The first histogram represents a phenomenon whose values are sharply peaked about the average (central) value.
The second histogram on the other hand represents a phenomenon represents a more broad and flat distribution.
The existence of such a difference in two otherwise equal distributions introduces the necessity of introducing a measure of spread.
A first thought could be to define the difference about the average for each individual value

$ \displaystyle \Delta j=j-<j> $

This approach doesn't work since that for random distributions one would expect to find equally positive and negative values for $ {\Delta j}$.
One way to circumvent this issue would be to use $ {|\Delta j|}$, and even though this approach does work theoretically it has the problem of not using a differentiable function.
These two issues are avoided if one uses the squares of the deviations about the average.
The quantity of interest in called the variance of the distribution.

Definition 5 The variance of a distribution ,$ {\sigma ^2}$, that has an average value is given by the expression

$ \displaystyle \sigma ^2=<(\Delta j)^2> \ \ \ \ \ (7)$

Definition 6 The standard deviation, $ {\sigma}$, of a distribution is given by the square root of its variance.

For the variance it is

$ \displaystyle \sigma ^2=<j^2>-<j>^2 \ \ \ \ \ (8)$

Since by definition 5 the variance is manifestly non-negative then it is

$ \displaystyle <j^2> \geq <j>^2 \ \ \ \ \ (9)$

where equality only happens when the distribution is composed of equal elements and equal elements only.

— 1.3.2. Continuous variables —

Thus far we've always assumed that we are dealing with discrete variables. To generalize our definitions and results to continuous distributions.
One has to have the initial care to note that when dealing with phenomena that allow for a description that is continuous probabilities of finding a given value are vanishing, and that one should talk about the probability of a given interval.
With that in mind and assuming that the distributions are sufficiently well behaved one has that the probability of and event being between $ {x}$ and $ {x+d}$ is given by

$ \displaystyle \rho(x)dx \ \ \ \ \ (10)$

The quantity $ {\rho (x)}$ is the probability density.
The generalizations for the other results are:

$ \displaystyle \int_{-\infty}^{+\infty}\rho(x)dx=1 \ \ \ \ \ (11)$

$ \displaystyle <x>=\int_{-\infty}^{+\infty}x\rho(x)dx \ \ \ \ \ (12)$

$ \displaystyle <f(x)>=\int_{-\infty}^{+\infty}f(x)\rho(x)dx \ \ \ \ \ (13)$

$ \displaystyle \sigma ^2=<(\Delta x)^2>=<x^2>-<x>^2 \ \ \ \ \ (14)$

The Wave Function 01

2014-03-28T02:03:00.001-07:00

— 1. The Wave Function —

The purpose of this section is to introduce the wave function of Quantum Mechanics and explain its physical relevance and interpretation.

— 1.1. The Schroedinger Equation —

Classical Dynamics' goal is to derive the equation of motion, $ {x(t)}$, of a particle of mass $ {m}$. After finding $ {x(t)}$ all other dynamical quantities of interest can be computed from $ {x(t)}$.
Of course that the problem is how does one finds $ {x(t)}$? In classical mechanics this problem is solved by applying Newton's Second Axiom

$ \displaystyle F=\frac{dp}{dt} $

For conservative systems it is $ {F=-\dfrac{\partial V}{\partial x}}$ (previously we've used $ {U}$ to denote the potential energy but will now use $ {V}$ to accord to Griffith's notation).
Hence for classical mechanics one has

$ \displaystyle m\frac{d^2 x}{dt^2}=-\dfrac{\partial V}{\partial x} $

as the equation that determines $ {x(t)}$ (with the help of the suitable initial conditions).
Even though Griffith's only states the Newtonian formalism approach of Classical Dynamics we already know by Classical Physics that apart from Newtonian formalism one also has the Lagrangian formalism and Hamiltonian formalism as suitable alternatives (and most of time more appropriate alternatives) to Newtonian formalism as ways to derive the equation of motion.
As for Quantum Mechanics one has to resort the Schrodinger Equation in order to derive the equation of motion the specifies the Physical state of the particle in study.

$ \displaystyle i\hbar\frac{\partial \Psi}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}+V\Psi \ \ \ \ \ (1)$

— 1.2. The Statistical Interpretation —

Of course now the question is how one should interpret the wave function. Firstly its name itself should sound strange. A particle is something that is localized while a wave is something that occupies an extense region of space.
According to Born the wave function of a particle is related to the probability of it occupying a region of space.
The proper relationship is $ {|\Psi(x,t)|^2dx}$ is the density probability of finding the particle between $ {x}$ and $ {x+dx}$.
This interpretation of the wave function naturally introduces an indeterminacy to Quantum Theory, since one cannot predict with certainty the position of a particle when it is measured and only its probability.
The conundrum that now presents itself to us is: after measuring the position of a particle we know exactly where it is. But what about what happens before the act of measurement? Where was the particle before our instruments interacted with it and revealed is position to us?
These questions have three possible answers:

The realist position: A realist is a physicist that believes that the particle was at the position where it was measured. If this position is true it implies that Quantum Mechanics is an incomplete theory since it can't predict the exact position of a particle but only the probability of finding it in a given position.
The orthodox position: An orthodox quantum physicist is someone that believes that the particle had no definite position before being measured and that it is the act of measurement that forces the particle to occupy a position.
The agnostic position: An agnostic physicist is a physicist that thinks that he doesn't know the answer to this question and so refuses to answer it.

Until 1964 advocating one these three positions was acceptable. But on that year John Stewart Bell proved a theorem, On the Einstein Podolsky Rosen paradox, that showed that if the particle has a definite position before the act of measurement then it makes an observable difference on the results of some experiments (in due time we'll explain what we mean by this).
Hence the agnostic position was no longer a respectable stance to have and it was up to experiment to show if Nature was a realist or if Nature was an orthodox.
Nevertheless the disagreements of what exactly is the position of a particle when it isn't bening measured, all three groups of physicists agreed to what would be measured immediately after the first measurement of the particle's position. If at first one has $ {x}$ then the second measurement has to be $ {x}$ too.
In conclusion the wave function can evolve by two ways:

It evolves without any kind of discontinuity (unless the potential happens to be unbounded at a point) under the Schrodinger Equation.
It collapses suddenly to a single value due to the act of measurement.

The interested reader can also take a look at the following book from Bell: Speakable and Unspeakable in Quantum Mechanics

Hamiltonian formalism exercises

2014-03-02T05:59:00.004-08:00

Exercise 1 Choose a set of generalized coordinates that totally specify the mechanical state of the following systems:

A particle of mass $ {m}$ that moves along an ellipse. Let $ {x=a\cos\theta}$ and $ {y=b\sin\theta}$. Then the generalized coordinate is $ {\theta}$. Perhaps you might be surprised to find out that we need only a single coordinate to describe the motion of this particle since its motion is described on a plane which is a two dimensional entity. But the particle motion is restricted to be along the ellipse and that constraint decreases the particle's degrees of freedom to one instead of being two.
A cylinder that moves along an inclined plane. If the cylinder rotates we need $ {x}$, the distance travelled, and $ {\theta}$, the angle of rotation. If the cylinder doesn't rotate we only need $ {x}$.
The two masses on a double pendulum.

The generalized coordinates are $ {\theta_1}$ and $ {\theta_2}$.
Do you see why?

Exercise 2 Derive the transformation equations for the double pendulum.
It is $ {x_1=l_1\cos\theta_1}$, $ {x_2=l_1\cos\theta_1+l_2\cos\theta_2}$, $ {y_1=l_1\sin\theta_1}$ and $ {y_2=l_1\sin\theta_1+l_2\sin\theta_2}$

Exercise 3 Show that $ {\dfrac{\partial \dot{\vec{r}}_\nu}{\partial \dot{q}_\alpha}=\dfrac{\partial\vec{r}_\nu}{\partial q_\alpha}}$.
$ {\begin{aligned} \vec{r}_\nu&=\vec{r}_\nu(q_1,q_2,\cdots,q_n,t)\Rightarrow \\ \dot{\vec{r}_\nu}&=\dfrac{\partial\vec{r}_\nu }{\partial q_1}\dot{q}_1+\cdots+\dfrac{\partial\vec{r}_\nu }{\partial q_n}\dot{q}_n+\dfrac{\partial\vec{r}_\nu}{\partial t} \Rightarrow \\ \dfrac{\partial \dot{\vec{r}}_\nu}{\partial\dot{q}_\alpha}&=\dfrac{\partial \vec{r}_\nu}{\partial q_\alpha} \end{aligned}}$

Exercise 4 Consider a system of particles that experiences an increment $ {dq_j}$ on its generalized coordinates. Derive the following expression $ {dW=\displaystyle \sum_\alpha \Phi_\alpha dq_\alpha}$ for the total work done by the force and indicate the physical meaning of $ {\Phi_\alpha}$.
First note that

$ \displaystyle d\vec{r}_\nu =\sum_{\alpha=1}^n \dfrac{\partial \vec{r}_\nu}{\partial q_\alpha}dq_\alpha$

For $ {dW}$ it is
$ {\begin{aligned} dW &=\sum_{\nu=1}^N\vec{F}_\nu\cdot d\vec{r}_\nu\\ &=\sum_{\nu=1}^N\left( \sum_{\alpha=1}^n \vec{F}_\nu\cdot\dfrac{\partial \vec{r}_\nu}{\partial q_\alpha} \right) dq_\alpha\\ &= \sum_{\alpha=1}^n \Phi_\alpha dq_\alpha \end{aligned}}$
with $ {\displaystyle \Phi_\alpha=\sum_{\nu=1}^N \vec{F}_\nu\cdot\dfrac{\partial \vec{r}_\nu}{\partial q_\alpha}}$ being the generalized force.

Exercise 5 Show that $ {\Phi_\alpha=\dfrac{\partial W}{\partial q_\alpha}}$.
We have $ {\displaystyle dW=\sum_\alpha\frac{\partial W}{\partial q_\alpha}dq_\alpha}$ and $ {\displaystyle dW=\sum_\alpha\Phi_\alpha dq_\alpha}$. Hence $ {\displaystyle \sum_\alpha\left( \Phi_\alpha- \frac{\partial W}{\partial q_\alpha}\right)dq_\alpha=0}$. Since $ {dq_\alpha}$ are linearly independent it is $ {\Phi_\alpha=\dfrac{\partial W}{\partial q_\alpha}}$.

Exercise 6 Derive the lagrangian for a simple pendulum and obtain an equation to describe its motion.
The generalized coordinate for the simple pendulum is $ {\theta}$ and the transformation equations are $ {x=l\sin\theta}$ and $ {y=-l\cos\theta}$.
For the kinetic energy it is $ {K=1/2mv^2=1/2m(l\dot{\theta})^2=1/2ml^2\dot{\theta}^2}$.
for the potential $ {V=mgl(1-\cos\theta)}$.
Hence the Lagrangian is $ {L=K-V=1/2ml^2\dot{\theta}^2-mgl(1-\cos\theta)}$.
$ {\dfrac{\partial L}{\partial \theta}=-mgl\sin\theta}$ and $ {\dfrac{\partial L}{\partial \dot{\theta}}=ml^2\dot{\theta}}$.
Hence the Euler Lagrange equation is
$ {\begin{aligned} \frac{d}{dt}\dfrac{\partial L}{\partial \dot{\theta}}-\dfrac{\partial L}{\partial \theta}&=0\Rightarrow\\ ml^2\dot{\theta}+mgl\sin\theta&=0\Rightarrow\\ \ddot{\theta}+g/l\sin\theta=0 \end{aligned}}$

Exercise 7 Two particles of mass $ {m}$ are connected with each other and to two points $ {A}$ and $ {B}$ by springs with constant factor $ {k}$. The particles are free to slide along the direction of $ {A}$ and $ {B}$. Use the Euler-Lagrange equations to derive the equations of motion of the particles.
The kinetic energy is $ {K=1/2m\dot{x}^2_1+1/2m\dot{x}^2_2}$.
The potential energy is $ {V=1/2kx^2_1+1/2k(x_2-x_1)^2+1/2kx^2_2}$.
Hence the Lagrangian is $ {L=1/2m\dot{x}^2_1+1/2m\dot{x}^2_2-1/2kx^2_1-1/2k(x_2-x_1)^2-1/2kx^2_2}$.
The partial derivatives of the Lagrangian are:

$ {\dfrac{\partial L}{\partial x_1}=k(x_2-x_1)}$
$ {\dfrac{\partial L}{\partial x_2}=k(x_1-x_2)}$
$ {\dfrac{\partial L}{\partial \dot{x_1}}=m\dot{x}_1}$
$ {\dfrac{\partial L}{\partial \dot{x_2}}=m\dot{x}_2}$

And the Euler-Lagrange equations are:

$ {m\ddot{x}_1=k(x_2-x_1)}$
$ {m\ddot{x}_2=k(x_1-2x_2)}$

Exercise 8 A particle of mass $ {m}$ moves subject to a conservative force field. Use cylindrical coordinates to derive:

The Lagrangian. The kinetic energy is $ {K=1/2m(1/2m\dot{\rho}^2+\rho^2\dot{\phi}^2+\dot{z^2})}$. The potential is $ {V=V(\rho,\phi,z)}$.
The equations of motion.
- $ {m(\ddot{\rho}-\rho\dot{\phi}^2)=-\dfrac{\partial v}{\partial \rho}}$
- $ {m\dfrac{d}{dt}(\rho^2\dot{\phi}=-\dfrac{\partial V}{\partial \phi}}$
- $ {m\ddot{z}=-\dfrac{\partial V}{\partial z}}$

Exercise 9 A double pendulum oscillates on a vertical plane.

Calculate:

The Lagrangian. The transformation equations for the coordinates are
- $ {x_1=l_1\cos\theta_1}$
- $ {y_1=l_1\sin\theta_1}$
- $ {x_2=l_1\cos\theta_1+l_2\cos\theta_2}$
- $ {y_2=l_1\sin\theta_1+l_2\sin\theta_2}$
Applying $ {\dfrac{d}{dt}}$ to the previous equations
- $ {\dot{x}_1=-l_1\dot{\theta}_1\sin\theta_1}$
- $ {\dot{y}_1=l_1\dot{\theta}_1\cos\theta_1}$
- $ {\dot{x}_2=-l_1\dot{\theta}_1\sin\theta_1-l_2\dot{\theta}_2\sin\theta_2}$
- $ {\dot{y}_2=l_1\dot{\theta}_1\cos\theta_1+l_2\dot{\theta}_2\cos\theta_2}$
Hence the kinetic energy is

$ \displaystyle K=1/2m_1l^2_1\theta^2_1+1/2m\left[ l^2_1\dot{\theta}^2_1+l^2_2\dot{\theta}^2_2+2l_1l_2\dot{\theta}_1\dot{\theta}_2\cos(\theta_1-\theta_2) \right]$

And the potential is

$ \begin{aligned}\displaystyle V&=m_1g(l_1+l_2-l_1\cos\theta_1)\\ &+m_2g\left[l_1+l_2-(l_1\cos\theta_1+l_2\cos\theta_2)\right]\end{aligned}$

As always the Lagrangian is $ {L=K-V=\cdots}$
The equations of motion.
- ${\begin{aligned}\dfrac{\partial L}{\partial \theta_1}&=-m_2l_1l_2\dot{\theta}_1\dot{\theta}_2\sin(\theta_1-\theta_2)\\ &- m_1gl_1\sin\theta_1-m_2gl_1\sin\theta_1\end{aligned}}$
- $ {\dfrac{\partial L}{\partial \dot{\theta_1}}=m_1 l_1^2\dot{\theta}_1^2+m_2l_1^2\dot{\theta}_1+m_2l_1l_2\dot{\theta}_2\cos(\theta_1-\theta_2)}$
- $ {\dfrac{\partial L}{\partial \theta_2}=m_2l_1l_2\dot{\theta}_1\dot{\theta}_2\sin(\theta_1-\theta_2)-m_2gl_2\sin\theta_2}$
- $ {\dfrac{\partial L}{\partial \dot{\theta_2}}=m_2l_2^2\dot{\theta}_2^2+m_2l_1l_2\dot{\theta}_1\cos(\theta_1-\theta_2)}$
${\begin{aligned}-(m_1+m_2)gl\sin\theta_1&=(m_1+m_2)l_1\ddot{\theta}_1+ m_2l_1l_2\ddot{_2}\cos(\theta_1-\theta_2)\\ &+ m_2l_1l_2\dot{\theta}_2\sin(\theta_1-\theta_2)\end{aligned}}$
and
$ {\begin{aligned}-m_2gl_2\sin\theta_2 &=m_2l_2\ddot{\theta}_2+m_2l_1l_2\ddot{\theta}_1\cos(\theta_1-\theta_2)\\ &+ m_2l_1l_2\dot{\theta}_1^2\sin(\theta_1-\theta_2)\end{aligned}}$
Assume that $ {m_1=m_2=m}$ e $ {l_1=l_2=l}$ and write the equations of motion. Left as an exercise for the reader.
Write the previous equations in the limit of small oscillations. If $ {\theta\ll1}$ implies $ {\sin \theta\approx\theta}$ and $ {\cos \theta\approx1}$.
Hence the equations of motion are
$ {2l\ddot{\theta}_1+l\ddot{\theta}_2=-2g\theta_1}$
$ {l\ddot{\theta}_1+l\ddot{\theta}_2=-g\theta_2}$

Exercise 10 A particle moves along the plane $ {xy}$ subject to a central force that is a function of the distance between the particle and the origin.

Find the hamiltonian of the system. The generalized coordinates are $ {r}$ and $ {\theta}$. The potential is of the form $ {V=V(r)}$.
The Lagrangian is $ {L=1/2m(\dot{r}^2+r^2\dot{\theta}^2)-V(r)}$.
The conjugate momenta are:
- $ {p_r=\dfrac{\partial L}{\partial \dot{r}}=m\dot{r}}$
- $ {p_\theta=\dfrac{\partial L}{\partial \dot{\theta}}=mr^2\dot{\theta}}$
For the Hamiltonian it is
$ {\begin{aligned} H&=\displaystyle \sum_{\alpha=n}^np_\alpha\dot{q}_\alpha\\ &=p_r\dot{r}+p_\theta\dot{\theta}-(1/2m(\dot{r^2}+r^2\dot{\theta}^2)-V(r))\\ &=\dfrac{p_r^2}{2m}+\dfrac{p_theta^2}{2mr^2}+V(r) \end{aligned}}$
Write the equations of motion.
- $ {\dot{r}=\dfrac{\partial H}{\partial p_r}=\dfrac{p_r}{m}}$
- $ {\dot{\theta}=\dfrac{\partial H}{\partial p_\theta}=\dfrac{p_\theta}{mr^2}}$
- $ {\dot{p}_r=-\dfrac{\partial H}{\partial r}=\dfrac{p_\theta}{mr^3}-V'(r)}$
- $ {\dot{p}_\theta=-\dfrac{\partial H}{\partial \theta}=0)}$

Exercise 11 A particle describes a one dimensional motion subject to a force

$ \displaystyle F(x,t)= \frac{k}{x^2}e^{-t/\tau} $

where$ {k}$ and $ {\tau}$ are positive constants. Find the lagrangian and the hamiltonian.
Compare the hamiltonian with the total energy and discuss energy conservation for this system.
Since $ {F(x,t)= \frac{k}{x^2}e^{-t/\tau}}$ it follows $ {V=\dfrac{k}{x}e^{-t/\tau}}$.
For the kinetic energy it is $ {K=1/2m\dot{x}^2}$. Hence the lagrangian is

$ \displaystyle L=1/2m\dot{x}^2-\dfrac{k}{x}e^{-t/\tau}$

. Now $ {p_x=m\dot{x}\Rightarrow\dot{x}=\dfrac{p_x}{m}}$.
For the Hamiltonian it is $ {H=p_x\dot{x}-L=\dfrac{p_x^2}{2m}+\dfrac{k}{x}e^{-t/\tau}}$.
Since $ {\dfrac{\partial L}{\partial t}=0}$ the system isn't conservative. Since $ {\dfrac{\partial U}{\partial \dot{x}}=0}$ it is $ {H=E}$.

Exercise 12 Consider two functions of the generalized coordinates and the generalized momenta, $ {g(q_k,p_k)}$ and $ {h(q_k,p_k)}$. The Poisson brackets are defined as:

$ \displaystyle [g,h]=\sum_k \left(\frac{\partial g}{\partial q_k}\frac{\partial h}{\partial p_k}-\frac{\partial g}{\partial p_k}\frac{\partial h}{\partial q_k}\right) $

Show the following properties of the Poisson brackets:

$ {\dfrac{dg}{dt}=[g,H]+\dfrac{\partial g}{\partial t}}$. Left as an exercise for the reader.
$ {\dot{q}_j=[q_j,H]}$ e $ {\dot{p}_j=[p_j,H]}$. Left as an exercise for the reader.
$ {[p_k,p_j]=0}$ e $ {[q_k,q_j]=0}$. Left as an exercise for the reader.
$ {[q_k,p_j]=\delta_{ij}}$. Left as an exercise for the reader.
If the Poisson brackets between two functions is null the two functions are said to commute.
Show that if a function $ {f}$ doesn't depend explicitly on time and $ {[f,H]=0}$ the function is a constant of movement.

Newtonian formalism exercises

2014-03-02T05:55:00.001-08:00

Exercise 1 A particle of mass $ {m}$ moves in the plane $ {xy}$. Its position vector is $ {\vec{r}=a\cos \omega t \vec{i}+b\sin \omega t \vec{j}}$ with $ {a,b,\omega}$ positive constants and $ {a>b}$. Show that:

The particle's trajectory is an ellipse.
For $ {x=a\cos\omega\ t}$ and $ {y=b\sin\omega\ t}$ it is

$ \displaystyle \left( \frac{x}{a} \right)^2+\left( \frac{y}{b} \right)^2=\cos^2\omega t+\sin^2\omega t=1 $

Which is the equation of an ellipse
The particle is subject to a central force oriented to the origin
It is
$ {\begin{aligned} \vec{F} &= m\dfrac{d\vec{v}}{dt}\\ &= m\dfrac{d^2\vec{r}}{dt^2}\\ &= m\dfrac{d^2}{dt^2}a\cos \omega t \vec{i}+b\sin \omega t \vec{j}\\ &=-m\omega^2\vec{r} \end{aligned}}$
Which is a central force oriented to the origin (note the minus sign).

Exercise 2 A particle of constant mass $ {m}$ is subject to to a force $ {F}$. Suppose that for $ {t_1}$ e $ {t_2}$ the velocities are $ {\vec{v}_1}$ and $ {\vec{v}_2}$, respectively. Show that the work the force does on the particle equals its change in kinetic energy.

$ {\begin{aligned} W &= \int_{t_1}^{t_2}\vec{F}\cdot d\vec{r}\\ &= \int_{t_1}^{t_2}\vec{F}\cdot\dfrac{d\vec{r}}{dt}dt \\ &= \int_{t_1}^{t_2}\vec{F}\cdot\vec{v}dt\\ &= \int_{t_1}^{t_2}m\dfrac{d\vec{v}}{dt}\cdot\vec{v}dt\\ &=m\int_{t_1}^{t_2}\vec{v}\cdot d\vec{v}\\ &=\dfrac{1}{2} m\int_{t_1}^{t_2}d(\vec{v}\cdot\vec{v})\\ &=1/2m(v_2^2-v_1^2) \end{aligned}}$

Exercise 3 For the conditions of exercise 1 calculate:

The particle's kinetic energy in the positive semi-major axis and on the positive semi-minor axis.
First note that $ {\vec{v}=-\omega a\sin \omega t \vec{i}+\omega b \cos \omega t \vec{j}}$.
Now $ {K=1/2m(\omega^2a^2\sin^2\omega t+\omega^2b^2\cos^2\omega t)}$. Let $ {A}$ denote the semi-major axis extension and $ {B}$ the semi-minor axis extension.
Hence $ {K_A=1/2m\omega ^2b^2}$ and $ {K_B=1/2m\omega ^2a^2}$
The work done by the force field when it moves the particle from the positive semi-major axis to the positive semi-minor axis.
$ {\begin{aligned} W &= \int_A^B\vec{F}\cdot d\vec{r}\\ &= \int_A^B(-m\omega ^2\vec{r})\cdot d\vec{r}\\ &= -m\omega ^2\int_A^B\vec{r}\cdot d\vec{r}\\ &= -1/2m\omega ^2\int_A^B d(\vec{r}\cdot\vec{r})\\ &=1/2m\omega ^2(a^2-b^2) \end{aligned}}$
The total work done by the force field by moving the particle along the ellipse.
$ {W=0}$. Do you see why?

Exercise 4 Show that if a particle is subject to $ {\vec{F}}$ and $ {\vec{v}}$ is its instantaneous velocity then the instantaneous power is

$ \displaystyle \mathcal{P}=\vec{F}\cdot\vec{v} $

By definition it is $ {dW=\vec{F}\cdot d\vec{r}}$. Hence

$ {\begin{aligned} \mathcal{P}&=\dfrac{dW}{dt}\\ &= \vec{F}\cdot\dfrac{d\vec{r}}{dt}\\ &= \vec{F}\cdot\vec{v} \end{aligned}}$

Exercise 5 Show that the integral $ {\displaystyle \int _{P_1}^{P_2} \vec{F}\cdot d\vec{r}}$ is independent of a particle's trajectory if and only if $ { \displaystyle\oint\vec{F}\cdot d\vec{r}=0 }$.

Let $ {\Gamma=P_1AP_2BP_1}$ denote a closed curve and admit that $ {\displaystyle \int _{P_1}^{P_2} \vec{F}\cdot d\vec{r}}$ is path independent. Then

$ {\begin{aligned} \oint \vec{F}\cdot d\vec{r}&=\int_\Gamma\vec{F}\cdot d\vec{r} \\ &= \int_{P_1AP_2}\vec{F}\cdot d\vec{r}+\int_{P_2BP_1}\vec{F}\cdot d\vec{r}\\ &= \int_{P_1AP_2}\vec{F}\cdot d\vec{r}-\int_{P_1BP_2}\vec{F}\cdot d\vec{r}\\ &=0 \end{aligned}}$

Where the last equality follows from our assumption that $ {\displaystyle \int _{P_1}^{P_2} \vec{F}\cdot d\vec{r}}$ is path independent.

Suppose now that $ { \displaystyle\oint\vec{F}\cdot d\vec{r}=0 }$.

$ {\begin{aligned} \int_\Gamma\vec{F}\cdot d\vec{r}&=\int_{P_1AP_2}\vec{F}\cdot d\vec{r}+\int_{P_2BP_1}\vec{F}\cdot d\vec{r}\\ &= \int_{P_1AP_2}\vec{F}\cdot d\vec{r}-\int_{P_1BP_2}\vec{F}\cdot d\vec{r}\\ &=0 \end{aligned}}$

Where the last equality follows from our $ { \displaystyle\oint\vec{F}\cdot d\vec{r}=0 }$. Hence $ {\displaystyle\int_{P_1AP_2}\vec{F}\cdot d\vec{r}=\int_{P_1BP_2}\vec{F}\cdot d\vec{r}}$.

Exercise 6 A particle of mass $ {m}$ moves along $ {x}$ subject to to a conservative force field $ {V(x)}$. If the particle's positions are $ {x_1}$ and $ {x_2}$ on $ {t_1}$ and $ {t_2}$, respectively, show that, if $ {E}$ is the total energy then

$ \displaystyle t_2-t_1=\sqrt{\frac{m}{2}}\int_{x_1} ^{x_2}\frac{dx}{\sqrt{E-V(x)}} $

Write $ {1/2m\left( \dfrac{dx}{dt} \right)^2+V(x)=E}$ solve in order to $ {dt}$ and integrate.

Exercise 7 Consider a particle of mass $ {m}$ that moves vertically on a resistive medium where the retarding force is proportional to the particle's velocity. Consider that particle initially moves in the downward direction with an initial velocity $ {v_0}$ from an height $ {h}$. Derive the particle's equation of motion $ {z=z(t)}$.

It is $ {F=m\dfrac{dv}{dt}=-mgg-kmv}$ with $ {v<0}$ and $ {-kmv>0}$. Then solve in order to $ {dv}$ and integrate to find $ {v=\dfrac{dz}{dt}=-\dfrac{g}{k}+\dfrac{kv_0+g}{k}e^{-kt}}$ (the terminal velocity $ {v_t}$ is $ {v_t=\displaystyle \lim_{t \rightarrow +\infty} v=-g/k}$).

Solving in order to $ {dz}$ and integrating and it is $ {z=h-\dfrac{gt}{k}+\dfrac{kv_0+g}{k^2}(1-e^{-kt})}$.

Exercise 8 A particle is shot vertically on a region where exists a constant gravitational field with a constant initial velocity $ {v_0}$. Show that, in the presence of resistive force proportional to the square of the particle's instantaneous velocity, the velocity of the particle when it returns to its initial position is:

$ \displaystyle \frac{v_0 v_t}{\sqrt{v_0^2+v_t^2}} $

Where $ {v_t}$ denotes the particle's terminal velocity.

Quantum Mechanics Introduction

2014-03-02T05:45:00.001-08:00

— 1. Motivation —

Unlike we did so far in the previous posts first we'll begin to study Quantum Mechanics by mentioning some of the experiments that motivated the reformulation of Classical Physics into the body of knowledge that is Quantum Mechanics. Also we'll spend more time explaining our initial formulations.

— 1.1. New results, new conceptions —

Anyone that has ever realized an experiment before knows that if one wants to know anything about a system that is being studied one needs to interact with it. In a more formal language one should say: "The act of measuring a system introduces a disturbance in the system". Thus far we've used the concept of mechanical state on our analysis of physical systems. A more critical eye to the concept of reveals the following:

In principle the disturbance can, in some cases, be made as small as one wants. The fact that exists a limit in our measuring apparatus lies in the nature of the apparatus and not in the nature of the theory being used.
There exists some disturbances whose effect can't be neglected. Nevertheless one can always calculate exactly the value of the effect of the disturbance and compensate it in the value of the quantity that is being measured afterwards.

Thus one can say that our very successful Classical Theory of Physics is casual and deterministic. Despite its many successes our Classical Theory of Physics had a few dark clouds hovering over it:

The persistence of these experimental results and the failure of accommodating them in the classical world view indicated that a revision of concepts was needed:

Corpuscular entities were demonstrating wavelike behavior.
Wave entities were demonstrating pointlike behavior.
There exists a statistical nature in atomic and subatomic phenomena that seems to be an essential characteristic of Nature.
Nature's atomic nature (no pun intended) implies that the process of measure has to be thought over: some disturbances can't be made arbitrarily small since there exists a natural limit to how small a disturbance can be made.

— 1.2. Double-slit experiment —

In order to make our previous discussion more concrete we'll look into a particular experiment that makes it clear how disparate are the two world views that we've been discussing.

— 1.2.1. Double-slit and particles —

Suppose an experimental apparatus which consists of a wall with two opening and a second wall that will serve as target. A beam of particles is fired upon the first wall. Some of those particles will hit the wall and of some will pass through the slits and hit the second wall.

The particles that pass through slit $ {1}$ are responsible for the probability curve $ {P_1}$ while the particles that pass through slit $ {2}$ are responsible for the probability curve $ {P_2}$. The resulting probability curve, $ {P_{12}}$, results from the algebraic sum of the curves $ {P_1}$ and $ {P_2}$.

— 1.2.2. Double-slit and waves —

Now if we direct a beam of waves on the same experimental apparatus the pattern one observes in the second wall is:

Now what we'll study is the intensity curve. The intensity curve $ {I_1}$ is the result one would observe if only slit $ {1}$ was open, while the intensity curve $ {I_2}$ is what one would observe if only slit $ {2}$ was open. The resulting intensity is given by the mathematical expression $ {I_{12}=|h_1+h_2|^2= I_1+2I_1I_2 \cos \theta}$. The last term is responsible for the interaction between the waves that come from slit $ {1}$ with the waves that come from slit $ {2}$ which ultimately is what causes the interference pattern one observes in the second wall.

— 1.2.3. Double slit and electrons —

Now that we familiarized ourselves with the behavior of particles and waves under the double slit we'll study what one observes when we subject an electron beam through the same apparatus. Experimental results show conclusively that electrons are particles so we're expecting to see the same pattern that we saw for macroscopic particles. Nevertheless this is what Nature has for us!:

In the case of electrons Nature forces to think again in terms of curves related to wave phenomena. But unlike macroscopic wave phenomena this aren't curves of intensity but they are curves of probability. This introduces an apparent oxymoron since probability curves are a concept that is intrinsic to particles while an interference pattern is intrinsic to waves...

In order to explain what we're observing one has to assume that each probability curve $ {P_i}$ is associated to a probability amplitude $ {\phi_i}$. In order to calculate the probability one has to calculate the modulus squared of the probability amplitude, $ {P_i=\phi_i^2}$. Thus one has to first calculate the sum of probability amplitudes of an electron passing through slit $ {1}$ or passing through slit $ {2}$ and only then should we calculate the modulus squared of the probability amplitude of an electron passing through slit $ {1}$ or slit $ {2}$:

$ \displaystyle P_{12}=|\phi_1+\phi_2|^2$

— 2. Basic concepts and preliminary definitions —

After our initial discussion I suppose that you understand why physicists had the need to introduce a new paradigm that made sense of what's happening in the atomic and sub- atomic level. Now it's time to introduce our usual initial definitions.

Definition 1 An operator is a mathematical operations that carries a given function into another function.

As an example of an operator we have $ {\dfrac{d}{dx}}$ which a transforms a functions into its $ {x}$ derivative. As another example of an operator we have $ {2\times}$ that transforms a function into its double.

Definition 2 The linear momentum is represented by the operator

$ \displaystyle p=\dfrac{\hbar}{i}\dfrac{d}{dx} \ \ \ \ \ (1)$

Definition 3 The energy of a particle is represented by the operator:

$ \displaystyle E=i\hbar\dfrac{d}{dt} \ \ \ \ \ (2)$

Definition 4 For a free particle the following mathematical relationships are valid:$ {\begin{aligned} \label{eq:relacoesdebroglie} k &= \frac{\hbar}{p}\\ \omega &= \frac{E}{\hbar} \end{aligned}}$

— 3. Quantum Mechanics Axioms —

After this batch of initial definitions that allows us to know the entities that will take part in the construction of our quantum vision of the world it is time for us to define the rules that govern them

Axiom 1 The quantum state is defined by the specification of the relevant physical quantities and is represented by a complex function $ {\Psi(x,t)}$

Axiom 2 To every observable in Classical Physics ($ {A}$) there corresponds a linear, Hermitian operator ($ {\hat{A}}$) in quantum mechanics.

Axiom 3 Measurements made to an observable associated with the operator $ {\hat{A}}$ done on a quantum system specified by $ {\Psi(x,t)}$ will always return $ {a}$ where $ {a}$ is an eigenvalue of $ {a}$.

$ \displaystyle \hat{A}\Psi(x,t)=a\Psi(x,t) \ \ \ \ \ (3)$

Axiom 4 The probability that a particle is in the volume element $ {dx}$ is denoted by$ {P(x)dx}$ and is calculated by

$ \displaystyle P(x)dx=|\Psi(x,t)|^2dx \ \ \ \ \ (4)$

Axiom 5 The average value of a physical quantity $ {A}$, represented by $ {<A>}$, is

$ \displaystyle <A> = \int \Psi^*\hat{A}\Psi \ \ \ \ \ (5)$

Where the integral is calculated in the relevant volume region.

Axiom 6 The state of a quantum system evolves according to the Schrodinger Equation.

$ \displaystyle \hat{H}\Psi= i\hbar\frac{\partial \Psi}{\partial t} \ \ \ \ \ (6)$

Where$ {\hat{H}}$ is the hamiltonian operator and corresponds to the total energy of a quantum system.

Newtonian Mechanics 06

2014-02-28T00:07:00.000-08:00

— 7. Newton and Euler-Lagrange Equations —

We've seen in the previous examples that solving a problema while using the lagrangian formalism would lead us to the same equations of Newton's formalism.

In this section we'll show that both formulations are indeed equivalent for conservative systems.

We have $ {\dfrac{\partial L}{\partial x_i}-\dfrac{d}{dt}\dfrac{\partial L}{\partial \dot{x}_i}=0 }$ for $ {i=1,2,3}$. The previous equation can be written in the following form:

$ \displaystyle \dfrac{\partial (K-U)}{\partial x_i}-\dfrac{d}{dt}\dfrac{\partial (K-U)}{\partial \dot{x}_i}=0$

Since our analysis doesn't depend on the type of coordinates one uses we'll choose to use rectangular coordinates. Hence it is $ {K=K(\dot{x}_i}$ and $ {U=U(x)}$.

It is $ {\dfrac{\partial T}{\partial x_i}=0}$ and $ {\dfrac{\partial U}{\partial \dot{x}_i}}$.

Hence $ {-\dfrac{\partial U}{\partial \dot{x}_i}=\dfrac{d}{dt}\dfrac{\partial T}{\partial \dot{x}_i}}$.

Since for a conservative system it is $ {-\dfrac{\partial U}{\partial \dot{x}_i}=F_i}$ it follows that $ {F_i=\dfrac{d}{dt}\dfrac{\partial T}{\partial \dot{x}_i}}$

$ {\begin{aligned} \dfrac{d}{dt}\dfrac{\partial T}{\partial \dot{x}_i} &= \dfrac{d}{dt}\dfrac{\partial \sum_j(1/2m\dot{x}_j^2)}{\partial \dot{x}_i} \\ &= \dfrac{d}{dt}(m\dot{x}_i) \\ &= \dot{P}_i \end{aligned}}$

Finally it is $ {F_i=\dot{P}_i}$ which is Newton's Second Axiom. Since the dynamics of a particle are a result of this axiom the dynamics of a given particle have to be the same on both formulations of mechanics.

— 8. Symmetry considerations —

As you certainly noticed in the previous examples the absence of generalized coordinate on the lagrangian of the system implied the conservation of a momentum (angular or linear). These coordinates that don't appear on the lagrangian are called cyclic coordinates in the literature.

Obviously that the presence or absence of cyclical coordinates on a lagrangian depend on the choice of coordinate that one makes. But the fact that a moment is conserved cannot depend on the choice of the set of coordinates one makes.

Since the right choice of coordinates is linked to the symmetry that the system exhibits one can conclude that symmetry and conserved quantities are intrinsically connected.

In this section we'll understand why symmetry considerations are so important in contemporary Physics and what is the relationship between symmetry and conservation. If a system exhibits some kind of continuous symmetry this symmetry will always manifest in the form some conserved quantity.

The mathematical proof of this theorem (and its multiple generalizations) is Noether's theorem and I won't provide a proof of it here. Instead we'll look into the consequences of three types of continuous symmetry and I'll provide you links for you to study Noether's theorem:

— 8.1. Continuous symmetry for time translations —

As we saw in Newtonian Mechanics 01 a frame is said to be inertial if time is homogeneous. When one says that time is homogeneous one is saying that one can perform a continuous time translation (formally one says $ {t \rightarrow t+\delta t}$) and the characteristics of the mechanical system won't change.

Let $ {L}$ denote the lagrangian of an isolated system. Since the system is isolated its physical characteristics must remain unchanged for all times. This is equivalent to saying that the lagrangian can't depend on time ($ {\dfrac{\partial L}{\partial t}=0}$).

Hence the total derivative is just

$ \displaystyle \frac{dL}{dt}= \sum_j \frac{\partial L}{\partial q_j}\dot{q}_j+ \sum_j \frac{\partial L}{\partial \dot{q}_j}\ddot{q}_j$

Using Euler-Lagrange equations 13 for generalized coordinates it is

$ {\begin{aligned} \frac{dL}{dt} &= \sum_j \dot{q}_j\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_j}+ \sum_j \frac{\partial L}{\partial \dot{q}_j}\ddot{q}_j \Rightarrow \\ &\Rightarrow \frac{dL}{dt}-\sum_j\frac{d}{dt}\left( \dot{q}_j\frac{\partial L}{\partial \dot{q}_j} \right)= 0 \\ &\Rightarrow \frac{d}{dt} \left( L-\sum_j\dot{q}_j\frac{\partial L}{\partial \dot{q}_j}=0\right) \end{aligned}}$

In conclusion it is

$ \displaystyle L-\sum_j\dot{q}_j\dfrac{\partial L}{\partial \dot{q}_j}=-H \ \ \ \ \ (14)$

where $ {-H}$ (the $ {-}$ sign will be apparent later) is some constant.

Suppose that $ {U=U(x_{\alpha,i})}$ and $ {x_{\alpha,i}=x_{\alpha,i}(q_j)}$. Then it is $ {U=U(q_j)}$ and $ {\dfrac{\partial U}{\partial \dot{q}_j}=0}$.

Hence $ {\dfrac{\partial L}{\partial \dot{q}_j}=\dfrac{\partial (K-U)}{\partial \dot{q}_j}=\dfrac{\partial K}{\partial \dot{q}_j}}$.

Hence we can write equation 14 as $ {\displaystyle (K-U)-\sum_j\dot{q}_j\dfrac{\partial K}{\partial \dot{q}_j}=-H}$. From this it follows $ {K+U=H}$.

The function $ {H}$ is called the Hamiltonian and its definition is given by equation 14.
Furthermore one can identify the Hamiltonian with the total energy of the system if the following conditions are met:

The equations of coordinate transformations are time independent which implies that the kinetic energy is a quadratic homogeneous function of $ {\dot{q}_j}$
The potential energy is velocity independent so that the terms $ {\dfrac{\partial U}{\partial \dot{q}_j}}$ can be eliminated.

— 8.2. Continuous symmetry for space translations —

As we saw in Newtonian Mechanics 01 a frame is said to be inertial if space is homogeneous. When one says that space is homogeneous one is saying that the lagrangian is invariant under space translations.

Formally one says that $ {\delta L=0}$ for $ {\vec{r}_\alpha \rightarrow \vec{r}_\alpha+\delta\vec{r}}$.

Without loss of generality let us consider just one particle. Now $ {L=L(x_i),\dot{x_i}}$ and $ {\displaystyle \delta \vec{r} = \sum_i\delta x_i \vec{e}_i}$.

In this case the variation in $ {L}$ due to $ {\delta \vec{r}}$ is

$ \displaystyle \delta L = \sum_i \frac{\partial L}{\partial x_i}\delta x_i + \sum_i\frac{\partial L}{\partial \dot{x}_i}\delta \dot{x}_i=0$

Now $ {\delta x_i=\delta\dfrac{dx_i}{dt}=\frac{d}{dt}\delta x_i=0}$ so the expression for the variation in the lagrangian becomes

$ \displaystyle \delta L = \sum_i \frac{\partial L}{\partial x_i}\delta x_i=0$

For the previous expression to be identically $ {0}$ one has to have $ {\dfrac{\partial L}{\partial x_i}=0}$ since the $ {\delta x_i}$ are arbitrary variations.

According to Euler-Lagrange equations 13 one also has $ {\dfrac{\partial L}{\partial \dot{x}_i}=\mathrm{const}}$.

$ {\begin{aligned} \frac{\partial (K-U)}{\partial \dot{x}_i} &= \frac{\partial K}{\partial\dot{x}_i}\\ &= \frac{\partial}{\partial \dot{x}_i}\left( 1/2m\sum_j\dot{x}_j^2 \right) \\ &= m\dot{x}_i \\ &= P_i \end{aligned}}$

Hence the homogeneity of space to translations implies the conservation of linear momentum in an isolated system.

— 8.3. Continuous symmetry for space rotations —

As we saw in Newtonian Mechanics 01 a frame is said to be inertial if space is isotropic. When one says that space is isotropic one is saying that the lagrangian is invariant under space rotations.

Formally one says that $ {\delta L=0}$ for $ {\vec{r}_\alpha \rightarrow \vec{r}_\alpha+\delta\vec{r}}$ where $ {\delta\vec{r}=\delta \vec{\theta} \times \vec{r}}$.

First we have (we're considering one particle) $ {\delta\vec{\dot{r}}=\delta \vec{\theta} \times \vec{\dot{r}}}$

$ \displaystyle \delta L = \sum_i \frac{\partial L}{\partial x_i}\delta x_i + \sum_i\frac{\partial L}{\partial \dot{x}_i}\delta \dot{x}_i=0$

From $ {p_i=\dfrac{\partial L}{\partial \dot{x}_i}}$ and $ {\dot{p}_i=\dfrac{\partial L}{\partial x_i}}$ it follows

$ {\begin{aligned} \delta L &= \sum_i\dot{p}_i\delta x_i+ \sum_i p_i\delta\dot{x}_i\\ &= \dot{\vec{p}}\cdot\delta\vec{r}+ \vec{p}\cdot\delta\dot{\vec{r}} \\ &= \dot{\vec{p}}\cdot(\delta \vec{\theta} \times \vec{r})+ \vec{p}\cdot(\delta \vec{\theta} \times \dot{\vec{r}}) \\ &= \delta\vec{\theta}\cdot(\vec{r}\times\dot{\vec{p}}) + \delta\vec{\theta}\cdot(\dot{\vec{r}}\times\vec{p})\\ &= \delta\vec{\theta}\cdot (\vec{r}\times\dot{\vec{p}} + \dot{\vec{r}}\times\vec{p}) \end{aligned}}$

Since $ {\delta\vec{\theta}\cdot (\vec{r}\times\dot{\vec{p}} + \dot{\vec{r}}\times\vec{p})=\delta\vec{\theta}\cdot\dfrac{d}{dt}(\vec{r}\times\vec{p})}$ and $ {\delta L=0}$, it follows $ {\delta\vec{\theta}\cdot\dfrac{d}{dt}(\vec{r}\times\vec{p})=0}$.

Since $ {\delta\vec{\theta}}$ is an arbitrary vector it follows $ {\dfrac{d}{dt}(\vec{r}\times\vec{p})=0}$.

Hence $ {\vec{r}\times\vec{p}}$ is constant.

In conclusion one can say that space isotropy implies the conservation of angular momentum. Another important result is that whenever a mechanical system has a symmetry axis the angular momentum about that axis is a conserved quantity.

— 9. Hamiltonian Dynamics —

As was seen previously if the potential energy of a system doesn't depend on velocity then $ {p_i=\dfrac{\partial L}{\partial \dot{x}_i}}$. Consequently one can introduce the following definition:

Definition 7 In a system of generalized coordinates $ {q_j}$ the generalized momentum is given by the following expression

$ \displaystyle p_j=\frac{\partial L}{\partial \dot{q}_j} \ \ \ \ \ (15)$

As a consequence of the previous definition it is $ {\dot{p}_j=\frac{\partial L}{\partial q_j}}$.

And the Hamiltonian can be written as a Legendre Transformation of the Lagrangian

$ \displaystyle H=\sum_j p_j\dot{q}_j-L \ \ \ \ \ (16)$

Since $ {\dot{q}_j=\dot{q}_j(q_k,p_k,t)}$ equation 16 can be written

$ \displaystyle H(q_k,p_k,t)=\sum_j p_j\dot{q}_j-L(q_k,\dot{q}_k,t) \ \ \ \ \ (17)$

So we have $ {H=H(q_k,p_k,t)}$ and $ {L=L(q_k,\dot{q}_k,t)}$. Hence the differential f $ {H}$ is

$ \displaystyle dH=\sum_k\left( \frac{\partial H}{\partial q_k}dq_k+\frac{\partial H}{\partial p_k}dp_k \right) + \frac{\partial H}{\partial t}dt \ \ \ \ \ (18)$

Calculating $ {\dfrac{\partial H}{\partial q_k}}$ and $ {\frac{\partial H}{\partial p_k}}$ via 16 and substituting into 18 it is

$ \displaystyle dH=\sum_k (\dot{q}_kdp_k-\dot{p}_kdq_k)-\frac{\partial L}{\partial t}dt \ \ \ \ \ (19)$

Identifying the coefficients of $ {dq_k}$, $ {dt_k}$ and $ {dt}$ it follows:

$ \displaystyle \dot{q}_k=\frac{\partial H}{\partial p_k} \ \ \ \ \ (20)$

and

$ \displaystyle -\dot{p}=\frac{\partial H}{\partial q_k} \ \ \ \ \ (21)$

Which are called the canonical equations of motion. When one uses these equations to study the time evolution of a physical system one is using Hamiltonian Dynamics.

One has $ {-\dfrac{\partial L}{\partial t}=\dfrac{\partial H}{\partial t}}$. Furthermore one also have $ {\dfrac{dH}{dt}=\dfrac{\partial H}{\partial t}}$ which implies that if the Hamiltonian function doesn't depend explicitly on $ {t}$ then $ {H}$ is a conserved quantity.

Example 7 A particle of mass $ {m}$ moves on the surface of a cylinder subject to a force which points to the center of the cylinder (the origin of our frame) and is proportional to the distance between the particle and the origin.
Since $ {\vec{F}=-k\vec{r}}$ it follows that $ {U=1/2kr^2=1/2k(R^2+z^2)}$

For the velocity it is $ {v^2=\dot{r}^2+r^2\dot{\theta}^2+\dot{z}^2}$. Since $ {r=R}$ is a constant in our example $ {K=1/2m(R^2\dot{\theta}^2+\dot{z}^2)}$

Hence the Lagrangian is $ {L=1/2m(R^2\dot{\theta}^2+\dot{z}^2)-1/2k(R^2+z^2)}$. The generalized coordinates are $ {\theta}$ and $ {z}$ and the generalized momenta are

$ \displaystyle p_\theta=\frac{\partial L}{\partial \dot{\theta}}=mR^2\dot{\theta} $
and

$ \displaystyle p_z=\frac{\partial L}{\partial \dot{z}}=m\dot{z} $

Since this a conservative system and and transformation equations between coordinates systems don't depend on time, $ {H}$ is the total energy of the system and it is a function of $ {\theta}$, $ {p_\theta}$, $ {z}$ and $ {p_z}$. But $ {\theta}$ doesn't appear in the Lagrangian (thus it is a cyclic coordinate).

$ \displaystyle H(z,p_\theta,p_z)=K+U= \frac{p_\theta^2}{2mR^2}+\frac{p_z^2}{2m} +1/2kz^2$

for the equations of motion it is

$ \displaystyle \dot{p}_\theta=-\frac{\partial H}{\partial \theta}=0 $

$ \displaystyle \dot{p}_z=-\frac{\partial H}{\partial z}=-kz $

$ \displaystyle \dot{\theta}=\frac{\partial H}{\partial p_\theta}=\frac{p_\theta}{mR^2} $

$ \displaystyle \dot{z}=\frac{\partial H}{\partial p_z}=\frac{p_z}{m} $

From the previous relationships one sees that the angular momentum about the $ {z}$ axis is constant $ {p_\theta=mR^2\dot{\theta}}$. Which is equivalent to saying that the $ {z}$ axis is a axis of symmetry of the system.

We also have $ {m\ddot{z}=-kz\Rightarrow m \ddot{z}+kz=0\Rightarrow \ddot{z}+k/mz=0\Rightarrow\ddot{z}+\omega_0^2}$ with $ {\omega_0^2=k/m}$. Which means that the particle describes an harmonic motion along the $ {z}$ axis.

To finalize our treatment of classical dynamics let us just make a quick summary of Lagrangian and Hamiltonian dynamics.

The generalized coordinates and the respective generalized momenta are said to be canonical coordinates.
Coordinates that don't appear explicitly on the expressions ok $ {K}$ and $ {U}$ are called cyclical coordinates.
A coordinate that is cyclical in $ {H}$ also is cyclical in $ {L}$.
A generalized coordinate and and it's corresponding generalized momentum are said to be canonical coordinates.

Newtonian Mechanics 05

2014-02-27T23:43:00.000-08:00

— 4. Hamilton's Principle — Minimum principles have a long history in the history of Physics:

Heron explained the law of light reflection using a principle of minimum time.
Fermat corrected Heron's Principle by stating that light travels between two points in the shortest time available.
Maupertuis stated his minimum action principle that postulated that a particle's dynamics always minimized the action.
Gauss postulated his principle of least constraint.
Hertz postulaed his principle of minimum curvature.

In modern Physics one uses a more general extremum principle and the focus of this section will be to state this principle and flesh out its consequences.

Definition 2 The lagrangian (sometimes called the lagrangian function), $ {L}$, of a particle is the difference between its kinetic and potential energies.

$ \displaystyle L=K-U \ \ \ \ \ (1)$

Definition 3 The action, $ {S}$, of a particle's movement (be it a real or virtual one) is:

$ \displaystyle \int_{t_1}^{t_2}(K-U)dt \ \ \ \ \ (2)$

Axiom 1 Given a collection of paths that a particle can take between points $ {x_1}$ and $ {x_2}$ in the the time interval $ {\Delta t= t_2-t_1}$ the actual path that the particle takes is the one that makes the action stationary

$ \displaystyle \delta S=\delta \int_{t_1}^{t_2}(K-U)dt=0 \ \ \ \ \ (3)$

For rectangular coordinates it is $ {T=T(x_i)}$, $ {U=U(x_i)}$, so $ {L=T-U=L(x_i,\dot{x}_i)}$ (where $ {\dot{x}_i=\dfrac{dx_i}{dt}}$ is called Newton's notation).

The function $ {L}$ can be identified with the function $ {f}$ that we saw on Newtonian Mechanics 04 if one makes the obvious analogies

$ {x \rightarrow t}$
$ {y_i(x) \rightarrow x_i(t)}$
$ {y\prime_i(x) \rightarrow x\prime_i(t)}$
$ {f(y_i(x),y\prime_i (x),x) \rightarrow L(x_i,\dot{x}_i,t)}$

In this case the Euler equations are called the Euler-Lagrange equations and it is

$ \displaystyle \frac{\partial L}{\partial x_i}-\frac{d}{dt}\frac{\partial L}{\partial \dot{x}_i}=0 $

Example 3 Let us study the harmonic oscillator under Langrangian formalism

$ \displaystyle L=T-U=1/2m\dot{x}^2-1/2kx^2 \ \ \ \ \ (4)$

First it is $ {\dfrac{\partial L}{\partial x_i}=-kx}$.
Then we have $ {\dfrac{d}{dt}\dfrac{\partial L}{\partial \dot{x}_i}=\dfrac{d}{dt}m\dot{x}=m\ddot{x}}$.
Hence it is $ {\dfrac{\partial L}{\partial x_i}-\dfrac{d}{dt}\dfrac{\partial L}{\partial \dot{x}_i}=0 \Rightarrow m\ddot{x}+kx=0 \Rightarrow m\ddot{x}=-kx}$ which is just the harmonic oscillator dynamic equation that we already know.

Example 4 Consider a planar pendulumplanar pendulum write its Lagrangian and derive its equation of motion.
The Lagrangian for the planar pendulum is

$ \displaystyle L=1/2ml^2\dot{\theta}^2-mgl(1-\cos \theta) \ \ \ \ \ (5)$

If we consider $ {\theta}$ to be a rectangular coordinate (which it isn't!) it follows that the equation of motion is:

$ \displaystyle \ddot{\theta}+g/l\sin \theta=0 $

This is precisely the equation of motion of a planar pendulum and this result is apparently unexpected since we only analyzed the Lagrangian for rectangular coordinates.

— 5. Generalized coordinates —

Consider a mechanical system constituted by $ {n}$ particles. In this case one would need $ {3n}$ quantities to describe the position of all particles (since we have 3 degrees of freedom). In the case of having any kind of restraints on the motion of the particles the number of quantities needed to describe the motion of particle is less than $ {3n}$. Suppose that one has $ {m}$ restrictions than the degrees of freedom are $ {3n-m}$.

Let $ {s=3n-m}$. These $ {s}$ coordinates don't need to be rectangular, polar, cylindrical nor spherical.

These coordinates can be of any kind provided that they completely specify the mechanical state of the system.

Definition 4 The set of $ {s}$ coordinates that totally specify the mechanical state of $ {n}$ particles is defined to be the set of generalized coordinates.
The generalized coordinates are represented by

$ \displaystyle q_1,q_2,\cdots,q_s$

Since we defined the generalized coordinates of a system of particles one can also define its set of generalized velocities.

Definition 5 The set of $ {s}$ velocities of a system of $ {n}$ particles described by a set of generalized coordiinattes is defined to be the set of generalized velocities.
The generalized velocities are represented by

$ \displaystyle \dot{q_1},\dot{q_2},\cdots,\dot{q_s}$

Let $ {\alpha}$ denote the particle, $ {\alpha=1,2,\cdots,n}$, $ {i}$ represent the degrees of freedom $ {i}$, $ {i=1,2,3}$ and $ {j}$ the number generalized coordinates $ {j=1,2,\cdots,s}$.

$ \displaystyle x_{\alpha,i}=x_{\alpha,i}(q_1,q_2,\cdots,q_s,t)=x_{\alpha,i}(q_j,t) \ \ \ \ \ (6)$

For the generalized velocities it is

$ \displaystyle \dot{x}_{\alpha,i}=\dot{x}_{\alpha,i}(q_j,t) \ \ \ \ \ (7)$

The inverse transformations are

$ \displaystyle q_j=q_j(x_{\alpha,i},t) \ \ \ \ \ (8)$

and

$ \displaystyle \dot{q_j}=\dot{q}_j(x_{\alpha,i},t) \ \ \ \ \ (9)$

Finally let us note that we also need $ {m=3n-s}$ equations of constraint

$ \displaystyle f_k=f_k(x_{\alpha,i},t) \ \ \ \ \ (10)$

with $ {k=1,2,\cdots,m}$.

Example 5 Consider a point particle that moves along the surface of a semi-sphere of radius $ {R}$ whose center is the origin of the coordinate system.

The relevant equations are $ {x^2+y^2+z^2-R^2\geq 0}$ and $ {z\geq 0}$.

Let $ {q_1=x/R}$, $ {q_2=y/R}$ and $ {q_3=z/R}$ be our generalized coordinates.
Furthermore we have the condition $ {q_1^2+q_2^2+q_3^2=1}$ as a constraint equation. Hence $ {q_3=\sqrt{1-(q_1^2+q_2^2)}}$

The time evolution of a mechanical system can be represented as a curve in the configuration space.

— 6. Euler-Lagrange Equations in generalized coordinates —

Since $ {K}$ and $ {U}$ are scalar functions $ {L}$ is also a scalar function. Therefore $ {L}$ is an invariant for coordinate transformations.

Hence it is

$ \displaystyle L=K(\dot{x}_{\alpha,i})- U(x_{\alpha,i})=T(q_j,\dot{q}_j,t)-U(q_j,t) \ \ \ \ \ (11)$

and $ {L=L(q_j,\dot{q}_j,t)}$.

Hence we can write Hamilton's Principle (Section 4) in the form

$ \displaystyle \delta \int_{t_1}^{t_2} L(q_j,\dot{q}_j,t) dt=0 \ \ \ \ \ (12)$

That is

$ {x \rightarrow t}$
$ {y_i(x) \rightarrow q_j(t)}$
$ {y\prime_i(x) \rightarrow q\prime_j(t)}$
$ {f(y_i(x),y\prime_i (x),x) \rightarrow L(q_j,\dot{q}_j,t)}$

are the analogies to be made now.

Finally the Euler-Lagrange Equations are

$ \displaystyle \frac{\partial L}{\partial q_j}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_j}=0 \ \ \ \ \ (13)$

for $ {j=1,2,\cdots,s}$

To finalize this section let us note the conditions of validity for the Euler-Lagrange equations:

The system is conservative.
The equations of constraint have to be functions between the coordinates of the particles and can also be a function of time.

Definition 6 Configuration space is the vector space defined by the generalized coordinates

Example 6 Consider the motion of a particle of mass $ {m}$ along the surface of a half-angle cone under the action of the force of gravity

.

The equations are $ {z=r\cot\alpha}$ and $ {v^2=\dot{r}^2\csc^2\alpha+r^2\dot{\theta}^2}$

Now for the potential energy it is $ {U=mgz=mgr\cot\alpha}$. Thus the lagrangian is

$ \displaystyle L=1/2m(\dot{r}^2\csc^2\alpha+r^2\dot{\theta}^2)-mgr\cot\alpha$

Since $ {\dfrac{\partial L}{\partial \theta}=0}$ it is $ {\dfrac{d}{dt}\dfrac{\partial L}{\partial \dot{\theta}}=0}$.

Hence it is $ {\dfrac{\partial L}{\partial \dot{\theta}}=mr^2\dot{\theta}=\mathrm{const}}$.

The angular momentum about the $ {z}$ axis is $ {mr^2\dot{\theta}=mr^2\omega}$.

Thus $ {mr^2\omega=\mathrm{const}}$ expresses the conservation of angular momentum about the axis of symmetry of the system.

It is left as an exercise for the reader to find the Euler-Lagrange equation for $ {r}$.

Newtonian Mechanics 04

2014-02-27T23:30:00.000-08:00

— 1. Variational Calculus —

Definition 1 A functional is a mapping from vector spaces to into real numbers.

Let $ {\displaystyle J = \int _{x_1}^{x^2} f\{y(x),y\prime (x),x\}dx}$. Suppose that $ {x_1}$ and $ {x_2}$ are constants, the functional form of $ {f}$ is known.

According to definition 1 $ {J}$ is a functional and the goal of the Calculus of Variations is to determine $ {y(x)}$ such that the value of $ {J}$ is an extremum.

Let $ {y=y(\alpha, x)}$ be a parametric representation of $ {y}$ such that $ {y(0,x)=y(x)}$ is the function that makes $ {J}$ an extremum.

We can write $ {y(\alpha, x)=y(0,x)+ \alpha\eta(x}$, where $ {\eta (x)}$ is a function of $ {x}$ of the class $ {C^1}$ (that means that $ {\eta}$ is a continuous function whose derivative is also continuous) with $ {\eta (x_1)=\eta (x_2)=0}$.

Now $ {J}$ is of the form $ {\displaystyle J(\alpha) = \int _{x_1}^{x^2} f\{y(\alpha, x),y\prime (\alpha, x),x\}dx}$. Therefore the condition for $ {J}$ to be an extremum is

$ \displaystyle \frac{dJ}{d\alpha}(\alpha=0)=0$

Example 1 Let $ {y(x)=x}$. Take $ {y(\alpha, x)= x+ \alpha\sin x}$ as a parametric representation of $ {y}$. Let $ {f=\left(dy/dx\right)^2}$, $ {x_1=0}$ and $ {x_2=2\pi}$. Given the previous parametric equation find $ {\alpha}$ such that $ {J}$ is a minimum. Now $ {\eta (0)=\eta (2\pi)=0}$ and $ {dy/dx=1+\alpha\cos x}$. Hence $ {\displaystyle J(\alpha)= \int_0^{2\pi}(1+2\alpha\cos x +\alpha^2\cos ^2x)dx=2\pi+\alpha^2\pi}$. By the previous expression $ {J(\alpha)}$ it is trivial to see that the minimum value is reached when $ {\alpha=0}$

Exercise 1 Given the points $ {(x_1,y_1)=(0,0)}$ and $ {(x_2,y_2)=(1,0)}$, calculate the equation of the curve that minimizes the distance between the points. Now $ {y(\alpha, x)=y(0,x)+\alpha \eta (x) = 0+\alpha(x^2-x)}$. It is $ {\eta (x) = x^2-x}$, $ {ds=\displaystyle \sqrt{dx^2+dy^2}=\sqrt{1+(dy/dx)^2}dx}$ And it is $ {s= \displaystyle \int _0^1 \sqrt{1+(dy/dx)^2}dx}$ with $ {dy/dx=\alpha (2x-1)}$. The rest is left as an exercise for the reader.

— 2. Euler Equations —

In the following section we'll analyze the condition for $ {J}$ to be an extremum:

$ {\begin{aligned} \frac{\partial J}{\partial \alpha} &= \frac{\partial}{\partial \alpha} \int _{x_1}^{x_2}f(y,y\prime,x)dx \\ &= \int _{x_1}^{x_2}\left(\frac{\partial f}{\partial y}\frac{\partial y}{\partial \alpha}+ \frac{\partial f}{\partial y\prime}\frac{\partial y\prime}{\partial \alpha}\right) dx \end{aligned}}$

Since it is $ {\partial y /\partial \alpha = \eta (x)}$ and $ {\partial y\prime /\partial \alpha = d\eta/dx}$ it follows

$ \displaystyle \frac{\partial J}{\partial \alpha}= \int _{x_1}^{x_2}\left(\frac{\partial f}{\partial y}\eta (x)+ \frac{\partial f}{\partial y\prime}\frac{d \eta}{dx}\right) dx$

Now $ {\displaystyle \int _{x_1}^{x_2}\frac{\partial f}{\partial y\prime}\frac{d \eta}{dx}dx=\frac{\partial f}{\partial y\prime}\eta (x)|_{x_1}^{x_2}- \int _{x_1}^{x_2}\frac{d}{dx}\left( \frac{\partial f}{\partial y\prime} \right)\eta (x) dx}$.

For the first term it is $ {\frac{\partial f}{\partial y\prime}\eta (x)|_{x_1}^{x_2}=0}$ since $ {\eta (x_1)=\eta (x_2)=0}$ by hypothesis.

Hence

$ {\begin{aligned} \frac{\partial J}{\partial \alpha} &= \int _{x_1}^{x_2}\left(\frac{\partial f}{\partial y}\frac{\partial y}{\partial \alpha}- \frac{d}{dx}\left( \frac{\partial f}{\partial y\prime} \right) \frac{\partial y}{\partial \alpha}\right)dx \\ &= \int _{x_1}^{x_2}\left( \frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y\prime} \right)\eta (x) dx \end{aligned}}$

Remembering that $ {\partial J / \partial\alpha(\alpha=0)=0}$ and taking into account the fact that $ {\eta (x)}$ is an arbitrary function one can conclude that

$ \displaystyle \frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y\prime}=0 $

The previous equation is known as the Euler's Equation

Example 2 As an example let us try to derive the equations of motion of a particle that moves in a constant force field starting its movement from the state of rest. The particles starts from point $ {x_1, y_1}$ and goes to point $ {x_2, y_2}$.From the enunciate it follows $ {K+U=c}$. Let us take our original point as being our reference point for the potential. Then it is $ {k+U=0}$. As always it is $ {k=1/2mv^2}$. For the potential it is $ {U=-Fx=-mgx}$. From the previous equations it follows that $ {v=\sqrt{2gx}}$. From the definition of velocity it follows that

$ \displaystyle t=\int _{x_1,y_1}^{x_2,y_2} \frac{ds}{v}=\int _{x_1,y_1}^{x_2,y_2}\frac{\sqrt{dx^2+dy^2}}{\sqrt{2gx}}=\int _{x_1,y_1}^{x_2,y_2}\frac{\sqrt{1+y\prime^2}}{\sqrt{2gx}}dx$

Let $ {f=\sqrt{\frac{1+y\prime^2}{x}}}$ since $ {(2g)^{-1/2}}$ is only a constant factor and can be omitted from our analysis. Given the functional form of $ {f}$ it is $ {df/dy=0}$ and Euler's Equation just is:

$ \displaystyle \frac{d}{dx}\frac{\partial f}{\partial y\prime}=0 $

From the previous relationship it is

$ \displaystyle \frac{\partial f}{\partial y\prime}=(2a)^{-1/2}=\mathrm{const} $

Hence it is $ {\begin{aligned} \frac{y\prime^2}{x(1+y\prime^2)} &= \frac{1}{2a} \Rightarrow\\ y &= \int \frac{x}{\sqrt{2ax-x^2}}dx \end{aligned}}$ Making the change of variables $ {x=a(1-\cos \theta)}$ it follows $ {dx=a\sin \theta d\theta}$. Hence the expression for $ {y}$ is $ {y=\int a(1-\cos \theta)d\theta\Rightarrow y=a(\theta-\sin \theta)+A}$. Since our particle starts from the origin it is $ {A=0}$. Thus the solution to our initial problem is $ {\begin{aligned} x &= a(1-\cos \theta) \\ y &= a(\theta-\sin \theta) \end{aligned}}$ Which are the parametric equations of a cycloid. Cycloid[/caption]

To close our thoughts on the Euler equation let us say that there also is a second form for the Euler equation. The second form is

$ \displaystyle f-y\prime\frac{\partial f}{\partial y\prime}= \mathrm{const} $

and is used in the cases where $ {f}$ doesn't depend explicitly on $ {x}$.

— 3. Euler Equation for $ {n}$ variables —

Let $ {f}$ be of the form $ {f=f\{ y_1(x),y\prime _1(x),y_2(x),y\prime _2(x),\cdots,y_n(x),y\prime _n(x), x \}}$.

Now we have $ {y_i(\alpha, x)= y_i(0,x)+\alpha \eta (x)}$ and $ {\displaystyle \int _{x_1}^{x_2}\left( \frac{\partial f}{\partial y_i}-\frac{d}{dx}\frac{\partial f}{\partial y _i\prime} \right)\eta _i (x) dx}$ for each of the values of $ {i}$. Since $ {\eta _i(x)}$ are independent functions it follows that for $ {\alpha=0}$

$ \displaystyle \frac{\partial f}{\partial y_i}-\frac{d}{dx}\frac{\partial f}{\partial y _i\prime}=0 $

That is to say we have $ {n}$ independent Euler equations.

Newtonian Mechanics 03

2014-02-27T23:21:00.004-08:00

— 1. One Particle Conservation Theorems —

— 1.1. Linear considerations —

Let $ {\vec{s}}$ be a constant vector such as $ {\vec{F}\cdot\vec{s}=0}$. Then $ {\dfrac{d\vec{p}}{dt}\cdot\vec{s}=\vec{F}\cdot\vec{s}=0}$. Hence $ {\vec{p}\cdot\vec{s}}$ is constant. The previous derivation shows that if $ {\vec{F}}$ is null along a given direction ($ {\vec{s}}$), then the momentum component along that direction is a constant quantity.

— 1.2. Rotational considerations —

Definition 1 Given a reference point one can define the angular momentum of a particle relative to that point.

$ \displaystyle \vec{L}=\vec{r}\times\vec{p} \ \ \ \ \ (1)$

The angular momentum is a measure of the amount a rotation that a particle has relative to a given point. For example if a particle moves in a straight line relative to point its angular momentum is $ {\vec{L}=\vec{r}\times\vec{p}=0}$ since $ {\vec{r}}$ is parallel to $ {\vec{p}}$ and the vector product of two parallel vectors is $ {0}$ by definition (Do you see why? If not go to this post). see the definition of vector product and prove the previous statement.

Just like we had forces in rectilinear motion to account for the variations of momentum one has the torque in curvilinear motion to account for the variation of angular momentum-

Definition 2 Given a reference point one can define the torque of a particle relative to that point.

$ \displaystyle \vec{\tau}=\vec{r}\times\vec{F} \ \ \ \ \ (2)$

Given the definitions of angular momentum it follows that

$ \displaystyle \frac{d}{dt}\left(\vec{r}\times\vec{p}\right)= \frac{d\vec{r}}{dt}\times\vec{p}+\vec{r}\times\frac{d\vec{p}}{dt} $

It is $ {\dfrac{d\vec{r}}{dt}\times\vec{p}=\dfrac{d\vec{r}}{dt}\times m\vec{v}=m\dfrac{d\vec{r}}{dt}\times\dfrac{d\vec{r}}{dt}=0}$ by definition.

Hence $ {\vec{\tau}=\vec{r}\times \dfrac{d\vec{p}}{dt}=\dfrac{d\vec{L}}{dt}}$.

Thus if $ {\vec{\tau}=0}$, $ {\dfrac{d\vec{L}}{dt}=0}$ and $ {\vec{L}}$ is constant in time.

— 1.3. Energetic considerations —

Let's consider a particle moves under the action of a force and evolves from mechanical state $ {1}$ to mechanical state $ {2}$.

Definition 3 The amount work done by a force against the inertial mass along the trajectory that leads from mechanical state $ {1}$ to mechanical state $ {2}$ is

$ \displaystyle W_{12}=\int_1^2\vec{F}\cdot d\vec{r} \ \ \ \ \ (3)$

It is

$ {\begin{aligned} \vec{F}\cdot d\vec{r} &= m\frac{d\vec{v}}{dt}\cdot\frac{d\vec{r}}{dt}dt \\ &= m\frac{d\vec{v}}{dt}\cdot\vec{v}dt \\ &= \frac{m}{2}\frac{d}{dt}(\vec{v}\cdot\vec{v})dt \\ &= \frac{m}{2}\frac{d}{dt}(v^2)dt \\ &= d\left(\frac{1}{2}mv^2\right) \end{aligned}}$

Hence, the integrand function for $ {W_{12}}$ is a total differential (note that we assumed that $ {\vec{F}}$ doesn't depend explicitly on time nor on the velocity). Hence

$ \displaystyle W_{12}=\frac{1}{2}m(v_2^2-v_1^2)=K_2-K_1 $

If $ {W_{12}}$ depends uniquely on the mechanical states $ {1}$ and $ {2}$ and not on the trajectory that connects them one says that $ {\vec{F}}$ derives from a potential. In this case the force can be written as the negative gradient of a function that is said to be the potential energy function: $ {\vec{F}=-\nabla U}$

It follows $ {\begin{aligned} \int_1^2 \vec{F}\cdot d\vec{r} &= -\int_1^2 \nabla U\cdot d\vec{r} \\ &= \int_1^2 \sum_i \frac{\partial U}{\partial x_i}dx_i \\ &= \int_1^2 dU \\ &= U_1-U_2 \end{aligned}}$ Let us define

Definition 4 The mechanical, $ {E}$, energy of a mechanic system is the sum of the kinetic energy and the potential energy

$ \displaystyle E=K+U \ \ \ \ \ (4)$

Now

$ {\displaystyle E = T+U \Rightarrow \frac{dE}{dt} = \frac{dT}{dt} + \frac{dU}{dt}}$

and

$ {\displaystyle \vec{F}\cdot d \vec{r}=d\left(1/2mv^2\right)=dT\Rightarrow \frac{dT}{dt}=\vec{F}\cdot \frac{d \vec{r}}{dt} }$

For $ {dU/dt}$ it is

$ {\begin{aligned} \frac{dU}{dt} &= \sum_i\frac{\partial U}{\partial x_i}\frac{\partial x_i}{\partial t}+ \frac{\partial U}{\partial t} \\ &= \nabla U \cdot \frac{d \vec{r}}{dt}+ \frac{\partial U}{\partial t} \end{aligned}}$

Finally

$ {\begin{aligned} \frac{dE}{dt} &= \vec{F}\cdot \frac{d \vec{r}}{dt}+\nabla U \cdot \frac{d \vec{r}}{dt}+ \frac{\partial U}{\partial t} \\ &= \left( \vec{F}+\nabla U \right)\cdot \frac{d \vec{r}}{dt}+ \frac{\partial U}{\partial t} \\ &= \frac{\partial U}{\partial t} \end{aligned}}$

If $ {U}$ isn't an explicit function of time it follows that $ {dE/dt=0}$ and the mechanical system is said to be conservative.

Newtonian Mechanics 02

2014-02-27T23:14:00.000-08:00

— 3.4. Gravitational Field —

Stating the Law of Universal Gravitation in usual terms one usually says that all particles in the Universe attract eachother with a force that is directly proportional to the masses of the particles and inversely proportional to the square of the distance between them. Stated in these terms it seems like the force of gravitation is an instantaneous one. Hence one has to resort to field theoretical language in order to describe the action of a gravitational field in terms that are acceptable to us.

Definition 16 Gravitational field: vectorial field, $ {\vec{g}}$, created by a body of mass $ {m_1}$ in all points of space (except on the point where the particle is situated) which is responsible for the gravitational interaction.

$ \displaystyle \vec{g}=G\frac{m_1}{r^2}\hat{r} \ \ \ \ \ (13)$

If a particle of mass $ {m_2}$ interacts with a gravitic field $ {\vec{g}}$ the particle experiences the force $ {\vec{F}_g}$:

$ \displaystyle \vec{F}_g=\vec{g}m_2=G \frac{m_1 m_2}{r^2}\hat{r} \ \ \ \ \ (14)$

$ {\hat{r}}$ is a unit vector whose direction points from the position of $ {m_2}$ to the position of $ {m_1}$. For the particular case of a body of mass $ {m}$ which is suspended from height $ {h}$ and interacts with the gravitational field of the Earth the gravitational force is

$ \displaystyle F_g=G \frac{M_T m}{(R_T+h)^2}$

Since $ {\vec{F}=m\vec{a}}$ holds for body of constant mass, one can write for the intensity of gravitational acceleration:

$ \displaystyle g=G\frac{M_T}{(R_T+h)^2}$

Definition 17 When two bodies of mass $ {m_1}$ and mass $ {m_2}$ interact via gravity an energy which derives from the gravitational field is established between them. This energy has the name of gravitational potential energy:

$ \displaystyle U=-G \frac{m_1 m_2}{r} \ \ \ \ \ (15)$

— 4. Waves and Oscillations —

Definition 18 Period is the minimum time interval necessary for two points in the same oscillatory phenomenon to be i the same mechanical state. It is represented by $ {T}$.

Definition 19 Frequency is the number of cycles of an oscillatory phenomenon that occur in a second. It is represented by $ {f}$ and can be calculated by $ {f=1/T}$.

Definition 20 Angular frequency is $ { \omega = 2\pi/T=2\pi f }$

— 4.1. Oscillations —

In this subsection one will study the harmonic motion. This is an important kind of movement since that in first approximation one can always use this model to study oscillatory m otions. Let us suppose that a particle moves along a straight line under the effect of a force $ {F}$.

Definition 21 A motion is said to be harmonic when in an oscillatory movement the force is proportional to the displacement (initial, also called equilibrium, position is taken as the origin) and as the opposite direction of the movement.

$ \displaystyle F=-k x $

Using Newton's Axiom 2 and introducing $ {k/m=\omega^2}$ one can write:

$ \displaystyle \frac{\partial ^2 x}{\partial t^2}=-\omega ^2 x \ \ \ \ \ (16)$

Solutions to this equation can be of the form $ {x(t)=A\cos (\omega t + \theta)}$ where $ {A}$ is the maximum displacement relative to the equilibrium position and $ {\theta}$ is the specific phase which identifies the particle's initial position. In the case of harmonic motion the definitions 18 and e 19 Can be written in the form $ {T=2\pi \sqrt{m/k} }$ e $ {f=1/(2\pi) \sqrt{k/m} }$. For an harmonic motion the kinetic and the potential energy are:

$ {K=\dfrac{1}{2} m \omega^2 A^2 \sin^2( \omega t + \theta ) }$
$ {U=\dfrac{1}{2} k A^2 \cos^2( \omega t + \theta ) }$

Thus the total energy of the system is $ {E=\dfrac{1}{2}kA^2}$

— 4.2. Waves —

Definition 22 A wave is a propagation in a medium that propagates itself transporting energy.

Definition 23 Wavelength, $ {\lambda}$ is the minimum distance between two points in a wave that are in the same mechanical state.

Definition 24 The speed of a wave of wavelength $ {\lambda}$ and period $ {T}$ is $ {c=\lambda/T=\lambda f}$

Definition 25 The wavenumber of a wave is é $ {k=2\pi/\lambda}$

It is possible to demonstrate mathematically that the equation that governs the propagation of a perturbation that moves with constant speed $ {c}$:

$ \displaystyle \frac{\partial ^2 \phi}{\partial x^2}=\frac{1}{c^2}\frac{\partial ^2 \phi}{\partial t^2} \ \ \ \ \ (17)$

With the previous definitions it is trivial to see that equations of the form $ {f_1=A\sin(kx \pm \omega t)}$ and $ {f_2=A\cos(kx \pm \omega t)}$ are solutions of equation 17. These functions are called trignometric functions, $ {A}$ is the amplitude and it represents the maximum displacement of the entity that is vibrating. In general one can say that a progressive wave that propagates to the right is always of the form $ {f=f(x-ct)}$, while a wave that moves to the left is always of the form $ {g=g(x+ct)}$. Since the wave equation is a linear partial derivative equation any linear combination of solutions of equation 17 is still a solution of the wave equation. In order for our solutions to have physical sense one has to impose certain conditions that the equations must follow in certain regions of space (and time).

These conditions are called boundary conditions and its effect is to restrict the set of values that the solutions might take.

The solutions of the wave equation that follow from boundary conditions are said to be normal modes of vibration. When is propagating and it encounters the boundary between two different media two things can happen:

Transmission: some of the energy of the wave propagates into the second medium.
Reflection: all of the wave's energy propagates in the first medium, but with opposite direction.

When two trigonometric waves of the same amplitude and frequency propagate in the same medium with opposite directions interact thet create a resulting wave whose mathematical expression is given by $ {f=2A\sin kx \cos \omega t}$. This is the mathematical expression of a stationary wave.

— 4.3. Interference —

When two waves of equal wavelength and constant phase difference interact they are said to interfere. If the two waves are in the same region of space and are of equal phase the interference is said to be constructive and the amplitude of the resulting wave equals the sum of the individual amplitudes of each original wave.

If the two waves are in the same region of space in phase opposition the interference is said to be destructive and the amplitude of the resulting wave is equal to the subtraction of the amplitudes of the two original waves.

The following diagram is a schematic representation of an experimental realization of an interference pattern:

— 4.4. Diffraction —

When light of a well defined wavelength passes through a barrier with a slit of width $ {d}$ the phenomenom that occurs is called diffraction. Each portion of the slit acts as if it is an independent source of a propagation and as a consequence waves coming from different portions of the slit have different phases. From this interaction one can observe destructive or constructive interference. The following diagram shows a schematic representation of an experimental realization of the phenomenon of diffraction:

Newtonian Mechanics 01

2014-02-27T23:05:00.001-08:00

— 2. Initial Considerations —
Without being too far away from the truth one can say that modern Fundamental Physics rests on these three conceptions

The concept of field
The theory of relativity
Quantum Physics

The concept of field is essential to all of our future discussion hence we'll define it right away:

Definition 1 Field is an mathematical structure with a defined value in a given set of points.

Definition 2 A field is said to be a vectorial field when it assumes vector values.

Definition 3 A field is said to be a scalar field when it assumes scalar values.

The field equations that we are going to use always represent linear interactions. Hence one can consider each interaction resulting from a field as being independent of all other interactions being analyzed and the resulting interaction is just the sum of all interactions.

Associated to the concept of field we have the concept of potential energy. This energy arises as a result of the result of the interaction of the particle with the field $ {\vec{A}}$ and in general it is proportional to $ {\displaystyle\int_a^b\vec{A}\cdot d\vec{s}}$ where $ {d\vec{s}}$ represents an infinitesimal displacement.

— 3. Mechanics —
This section will be a very brief and shallow introduction to the results and triumphs of the first modern physical theory. Nevertheless one hopes that its internal elegance and depth can be glimpsed through.

— 3.1. Basic concepts and preliminary definitions —
All mechanical quantities can be expressed in units that derive from the units of the following quantities:

Length $ {L}$.
Time $ {T}$.
Mass $ {M}$. In classical mechanics the mass of a body its a measure of its resistance to alter its state of movement. This characteristic has the name of inertia.

The units that one uses to express the previous quantities are totally conventional. In the following we'll use the International System of Units:

$ {\left[ L \right] =m}$
$ {\left[ T \right] =s}$
$ {\left[ M \right] = \mathrm{Kg}}$

Definition 4
A frame is a set of axes that represent the degrees of freedom of the system that is being studied and an arbitrary point that serves has its origin.

Definition 5 A frame is said to be inertial when it possesses the following properties:

Space is homogeneous (all points are equivalent) and isotropic (there is no special orientation)
Time is homogeneous (all time instants are equivalent)

Definition 6 Position is the geometric localization of a particle in a frame.

Definition 7 Trajectory is the geometric place of the successive positions of a particle in a frame in a given time interval.

Definition 8 Displacement it's the difference between the final position and the initial position of a particle. The displacement is represented by the symbol $ {\Delta \vec{x}}$.

We know by everyday experience that bodies move by different displacements during different time intervals. The concept that expresses how a particle position changes in a given time interval it's called velocity.

Definition 9 Average velocity: vectorial quantity that expresses that rate of change in a particles position for a given time interval:

$ \displaystyle \vec{v}_m=\dfrac{\Delta \vec{x}}{\Delta t} \ \ \ \ \ (4)$

Definition 10 Velocity: Vectorial quantity that expresses a particles velocity in a given time instant:

$ \displaystyle \vec{v}=\lim_{\Delta t\rightarrow 0}\dfrac{\Delta \vec{x}}{\Delta t}=\dfrac{d\vec{x}}{dt} \ \ \ \ \ (5)$

Since a particle's velocity varies in time one can introduce the concept of acceleration.

Definition 11 Average acceleration: Vectorial quantity that expresses the rate of change of velocity for a given time interval.

$ \displaystyle \vec{a}_m=\dfrac{\Delta \vec{v}}{\Delta t} \ \ \ \ \ (6)$

Definition 12 Acceleration: vectorial quantity that expresses the change in velocity in a given time instant.

$ \displaystyle \vec{a}=\lim_{\Delta t\rightarrow 0}\dfrac{\Delta \vec{v}}{\Delta t}=\dfrac{d\vec{v}}{dt} \ \ \ \ \ (7)$

Definition 13 Kinetic energy is the energy associated with the movement of a particle and is defined by:

$ \displaystyle K=\dfrac{1}{2}m\vec{v}\cdot\vec{v}=\dfrac{1}{2}mv^2=\dfrac{1}{2}m\left( \dfrac{d\vec{x}}{dt}\right)^2 \ \ \ \ \ (8)$

Definition 14 Linear momentum : vectorial quantity that is associated with the movement of a particle

$ \displaystyle \vec{p}=m \vec{v}=m \dfrac{d\vec{x}}{dt} \ \ \ \ \ (9)$

Definition 15
The mechanical state of a particle is specified by the simultaneous determination (and with infinite precision) of its coordinates and linear momentum.

— 3.2. Newton's Axioms —

Thus far we have defined the actors in our stage play but we haven't defined by which rules do they interact. These rules are given by Newton's Axioms.

Axiom 1 There is an inertial frame where the linear momentum of a free particle (a particle that experiences no interactions) is constant

Our definition 5 of an inertial frame is, strictly speaking, a mathematical definition. As such nothing in it implies that such a frame exists in our world. As such the function of Newton's first Axiom is to guarantee that in inertial frame exists in our world.

One important detail is that Axiom 5 only posits the existence of one inertial frame, but we can conclude that there exists an infinite number of inertial frames.

Since an inertial frame space is homogeneous and isotropic the point one picks as the origin is arbitrary. Hence one can make a translation of the first origin point and consider the final origin point as being the origin of this new frame, which has to be a new inertial frame.

Moreover one can apply a rotation to our firs inertial frame and obtain a new frame (the time the axes have different orientations). Since space is isotropic this orientation can't change the nature of the frame and we can conclude that the final frame is also an inertial one.

Additionally one can consider and inertial that moves with constant velocity relatively to an inertial frame. Since space is homogeneous this operation can't change the nature of the the frame in question and once again it still has to be an inertial frame.

Likewise, since time time is homogeneous one can consider a frame that differs of an inertial frame only in the instant of time that was taken as its origin. Since time is homogeneous this second frame also has to be an inertial one.

Finally let us just notice that any finite or infinite composition of these transformations also produces an inertial frame.

Axiom 2
An in inertial frame the change of linear momentum of a particle is caused by the action of a force $ {\vec{F}}$.
$ {\vec{F}= \dfrac{d\vec{p}}{dt}}$.

When the mass of the particle being considered is constant this axiom becomes $ {\vec{F}=m\vec{a}}$.
In what follows the mass of a particle is to be considered constant unless stated otherwise.

Axiom 3
When two particles interact the force $ {\vec{F}_{12}}$ (force that object $ {1}$ exerts on object $ {2}$) has an equal intensity as $ {\vec{F}_{21}}$ but opposite direction$ {\vec{F}_{12}=-\vec{F}_{21}}$

— 3.3. Kinematics and Dynamics —

On this section we'll introduce very briefly the concepts that describe (kinematics) and explain (dynamics) the movement of a particle.

— 3.3.1. Equations of movement —

From the definitions of velocity and acceleration that we introduced in section 3 it follows that

$ \displaystyle d\vec{v}= \vec{a}dt \Rightarrow \int_{t_0}^t d\vec{v}= \int_{t_0}^t \vec{a}dt \Rightarrow \vec{v}(t)-\vec{v}(t_0)=\int_{t_0}^t \vec{a}dt \ \ \ \ \ (10)$

Since the functional relationship between the acceleration and time is unknown the right hand side of the equality can not be calculated.

It also is

$ \displaystyle d\vec{x}= \vec{v}dt \Rightarrow \int_{t_0}^t d\vec{x}= \int_{t_0}^t \vec{v}dt \Rightarrow \vec{x}(t)-\vec{x}(t_0)=\int_{t_0}^t \vec{v}dt \ \ \ \ \ (11)$

Since $ {\vec{v}(t)}$ is also an unknown function the calculation has to stop.

If we assume that $ {\vec{a}}$ is constant in time (uniformly accelerated motion) we can solve equation 10, $ { \vec{v}=\vec{v}_0+\vec{a}(t-t_0)}$. After substituting in equation 11 it follows

$ \displaystyle \vec{x}(t)=\vec{x}_0+\vec{v}_0(t-t_0)+\frac{1}{2}\vec{a}(t-t_0)^2 \ \ \ \ \ (12)$

For the special case $ {\vec{a}=\vec{0}}$ the motion is said to be uniform.

— 3.3.2. Galileu Transformations —

After axiom 1 we concluded that there is an infinite number of inertial frames. Hence it makes sense to ask how can calculate the coordinates (velocity) of particle in an inertial frame when knowing its coordinates (velocity) in a first inertial frame.

Let $ {S}$ and $ {S'}$ be two inertial frames whose origins coincide at $ {t=0}$. Moreover $ {S'}$ moves with velocity $ {\vec{v}_0}$ relative to $ {S}$.

By simple vector addition it is $ {\vec{v}_0 t+\vec{r}'=\vec{r}}$ which we can write in component form:

$ {\begin{aligned} x' &= x-v_{0x}t\\ y'&= y-v_{0y}t\\ z' &= z-v_{0z}t \end{aligned}}$

Deriving in order to $ {t}$ it is

$ {\begin{aligned} v'_x &= v_x-v_{0x}\\ v'_y &= v_y-v_{0y}\\ v'_z &= v_z-v_{0z} \end{aligned}}$

Galileu transformations are equivalent to the affirmation that the form of the equations of mechanics don't depend on the inertial frame that one chooses to study the motion of a particle.

Matrices, Scalars, Vectors and Vector Calculus 03

2014-02-27T08:47:00.000-08:00

After introducing some mathematical machinery with our first and second posts it is now time for us to look into some Newtonian Physics, after a brief look into vector integration.

— 1. Vector Integration —
When dealing with vectors and the mathematical operation we have three basic options:

Volume integration
Surface integration
Line (contour) integration

The result of integrating a vector, $ {\vec{A}=\vec{A}(x_i)}$, over a volume is also a vector and the result is given by the following expression:

$ \displaystyle \int_V \vec{A}dv = \left( \int_V A_1 dv, \int_V A_2 dv, \int_V A_3 dv \right) \ \ \ \ \ (1)$

Hence, the result of vector integration is just three separate integration operations (one of each spatial dimension).

The result of integrating the projection of a vector $ {\vec{A}=\vec{A}(x_1)}$ over an area is what is called surface integration.

Surface integration is always done with the normal component of $ {\vec{A}}$ over the surface $ {S}$ in question. Thus what we need to define first is the normal of a surface at a given point. $ {d\vec{a}=\vec{n}da}$ will be this normal. We still have the ambiguity of having two possible directions for the normal at any given point, but this is taken care of by defining the normal to be on the outward direction of a closed surface.

Hence the quantity of interest is $ {\vec{A}\cdot d\vec{a}=\vec{A}\cdot \vec{n} da}$ ($ {da_1=dx_2dx_3}$ for example) with

$ \displaystyle \int_S \vec{A}\cdot d\vec{a} = \int_S \vec{A}\cdot \vec{n}da \ \ \ \ \ (2)$

As for the line integral it is define along the path between two points $ {B}$ and $ {C}$. Again we have to consider the normal of $ {\vec{A}=\vec{A}(x_i)}$, but this time the quantity of interest is:

$ \displaystyle \int_{BC} \vec{A}\cdot d\vec{s} \ \ \ \ \ (3)$

The quantity $ {d\vec{s}}$ is an element of length along $ {BC}$ and is taken to be positive along the direction in which the path is defined.

Matrices, Scalars, Vectors and Vector Calculus 02

2014-02-24T13:44:00.000-08:00

In the last post we took our first step in the mathematical introduction to classical Mechanics. In this post we'll introduce a few physical concepts and some more mathematical tools.

— 1. Velocity and Acceleration —

The first thing to notice is that in Physics velocity is a vector concept. That means that in order for us to specify the velocity of an entity we must not only indicate its magnitude but also its direction.

First of we need to define a frame.

Definition 1
A frame is a set of axis and a special point that is taken as the origin that allows one to describe the motion of a particle.

When to get to real Physics we'll talk a little bit more of what is a particle in Physics but for now we'll just leave it to intuition and/or background knowledge.

If we trace the position of a particle for a given period of time what we get is curve whose parameter is time and this curve has the name of trajectory.

If we want to keep track of the variations of the position of a particle when it is describing a given trajectory we can do that using the concept of displacement.

Definition 2 Displacement is the difference between the final and initial positions of a particle. It is represented by the symbol $ {\Delta\vec{r}} $ and is calculated in the following way:

$ \displaystyle \Delta\vec{r}=\vec{r}_f - \vec{r}_i \ \ \ \ \ (1) $

Now we all know that when a body describes a trajectory some parts of it may take a longer time and others a shorter time. To rigorously describe the notion of a particle we have to take into account its variability in the motion and we do this by introducing the concept of velocity.

Definition 3
The average velocity of a particle is the rate of change in position of a particle for a given time interval.

$ \displaystyle \vec{v}=\frac{\Delta\vec{r}}{\Delta t}=\frac{\vec{r}_f - \vec{r}_i}{t_f-t_i} \ \ \ \ \ (2) $

Definition 4
Sometimes we aren't interested in the average velocity and we want to know what is the velocity of a particle in a given instant. Intuitively we know that if we measure the average velocity in a sufficiently small interval we won't be making a very big error in determining the value of the velocity.
If we let the time interval go to to zero what the have is the velocity of the particle for the instant considered. We can be rest assured that such a limit always exist since we are dealing with real bodies and their movements are always smooth.

$ \displaystyle \vec{v}=\lim_{t_f \rightarrow t_i}\frac{\Delta\vec{r}}{\Delta t}=\lim_{t_f \rightarrow t_i}\frac{\vec{r}_f - \vec{r}_i}{t_f-t_i}=\frac{d\vec{r}}{dt} \ \ \ \ \ (3) $

Another way to mathematically define the average velocity and velocity is to make a change of variable and define $ {h=t_f-t_i} $ and calculate $ {\vec{r}(t)} $, $ {\vec{r}(t+h)} $ and $ {\lim_{h \rightarrow 0}} $.

The other quantity of physical interest is average acceleration and it is the the rate of change in velocity of a particle in a given time interval.

The definitions are analogous to the ones of average velocity and we just have to change $ {\vec{r}} $ to $ {\vec{v}} $.
We'll just state the equation for acceleration here:

$ \displaystyle \vec{a}=\dfrac{d\vec{v}}{dt}=\frac{d^2\vec{r}}{dt^2}=\ddot{\vec{r}} \ \ \ \ \ (4) $

Where $ {\ddot{\vec{r}}} $ is the so called Newton notation and indicates that we are differentiating $ {\vec{r}} $ two times with respect to $ {t} $. Obviously that one dot indicates that we are differentiating one time, and etc.

If we use Cartesian coordinates we know that it is $ {\vec{r}=\sum_i x_i\vec{e}_i} $. Since our basis vectors are fixed it follows that $ {\vec{v}=\dot{\vec{r}}=\sum_i \dot{x_i}\vec{e}_i} $ and that $ {\vec{a}=\dot{\vec{v}}=\sum_i \ddot{x_i}\vec{e}_i} $.

— 2. Curvilinear Coordinates —

Sometimes to use Cartesian Coordinates in order to describe the motion of a particle is very awkward and one instead uses some other kind of curvilinear coordinates. The ones that are most used are, polar coordinates, cylindrical coordinates and spherical coordinates.

— 2.1. Polar Coordinates —

If we are dealing with two dimensional problems that have circular symmetry then the best way to deal with them is to use polar coordinates.

In this system of coordinates instead of the $ {_1x} $ and $ {x_2} $ coordinates, one uses a radial coordinate $ {r} $ that expresses the distance that a point is from the origin and an angular coordinate $ {\theta} $. This angular coordinate is measured from the $ {x_1} $ axis to the position vector in counter clockwise direction.

The range of values for these two variables are $ {0<r<\infty} $ and $ {0\leq \theta \leq 2\pi} $. Notice that in $ {r=0} $ $ {\theta} $ isn't defined and thus the origin can be thought to be a kind of singularity, but this is only so because we chose a particular set of coordinate and it isn't intrinsic to the point itself.

As indicated in the picture the transformation of coordinates is:

$ {x_1=r \cos \theta} $
$ {x_2=r \sin \theta} $
$ {\theta=\arctan \left(\dfrac{x_2}{x_1}\right)} $

— 2.2. Cylindrical Coordinates —

Sometimes when dealing with three dimensional problems we have to deal with circular symmetry around a certain axis. If this happens the most apt way to deal with the said problem is to use cylindrical coordinates.

Again we'll use a mix of radial distances and angular distances. First we'll only think about the plane formed by the $ {x_1} $ and $ {x_2} $ axes. In this plane we'll project the position vector of a given point. The angle between $ {x_1} $ and this projection is $ {\theta} $. The other two coordinates are $ {r} $ which is the distance between the point in question and the origin, and $ {x_3} $ (or $ {z} $).

The range of values for these three coordinates are $ {0<r<\infty} $, $ {0\leq \theta \leq 2\pi} $ and $ {-\infty<z<\infty} $,

As indicated in the pictures the transformation of coordinates is:

$ {x_1=r \cos \theta} $
$ {x_2=r \sin \theta} $
$ {x_3=z} $

— 2.3. Spherical Coordinates —

The last kind of curvilinear coordinates that we'll look at are spherical coordinates. This kind of coordinates are extremely useful when a problem has spherical symmetry.

This time we'll have one radial coordinate and two angular coordinate. Like we did for cylindrical coordinates first we'll project the position vector into the $ {x_1} $, $ {x_2} $ plane and define $ {\phi} $ to be the angle between $ {x_1} $ and the projection of the position vector. $ {r} $ is the distance between the origin and the point in question. The second angular coordinate is $ {\theta} $ and it is measured between $ {x_3} $ and the position vector.

As can be seen from the picture the change of variables from spherical coordinates to Cartesian coordinates is:

$ {x_1=r \sin \theta \cos \phi} $
$ {x_2=r \sin \theta \sin \phi} $
$ {x_3=r \cos \theta} $

The range of each coordinate is defined to be $ {0<r<\infty} $, $ {0 \leq \phi \leq 2\phi} $ and $ {0 \leq \theta \leq \pi} $.

— 3. Velocity and Acceleration in Curvilinear Coordinates —

Having defined the previous three systems of coordinates it is time for us to know how to write the velocity and acceleration of a particle in these systems of coordinates.
As we already saw for Cartesian coordinates one has the following relationships:

$ {ds^2=dx_1^2+dx_2^2+dx_3^2} $
$ {v^2=\dot{x}_1^2+\dot{x}_2^2+\dot{x}_3^2} $
$ {v=\dot{x}_1\vec{e}_1+\dot{x}_2\vec{e}_2+\dot{x}_3\vec{e}_3} $

Our task will be to see how these equations translate when we are dealing with polar, cylindrical and spherical coordinates.

— 3.1. Polar Coordinates —

Has stated previously when we are using curvilinear coordinates the basis vector change from point to point. Hence one has to use Leibniz rule when calculating derivatives.

If we have two points that are infinitely close to each other, $ {P^1} $ and $ {P^2} $ (the reason for using superscripts will be evident right away) the infinitesimal variation in $ {\vec{e}_r} $ is $ {d\vec{r}_r=\vec{e}_r^2-\vec{e}_r^1} $ and for the angular infinitesimal variation it is $ {d\vec{e}_\theta=\vec{e}_\theta^2-\vec{e}_\theta^1} $

The previous relationships allows one to see that the infinitesimal variation for $ {\vec{e}_r} $ is perpendicular to $ {\vec{e}_r} $ and that the infinitesimal variation for $ {\vec{e}_\theta} $ is perpendicular to $ {\vec{e}_\theta} $. One can show this rigorously but I'll only state the result that it is

$ {d\vec{e}_r=d\theta\vec{e}_\theta} $
$ {d\vec{e}_\theta=-d\theta\vec{e}_r} $

Hence
$ {\begin{aligned} \vec{v} &=\dfrac{d\vec{r}}{dt}\\ &=\dfrac{d}{dt}\left( r\vec{e}_r \right)\\ &= \dot{r}\vec{e}_r+r\dot{\vec{e}}_r\\ &= \dot{r}\vec{e}_r+r\dot{\theta}\vec{e}_\theta \end{aligned}} $

Thus the velocity can be thought has having a radial component and an angular component. The radial component is $ {\dot{r}} $ and is a measure for the variation of velocity in magnitude. The radial component is $ {r\dot{\theta}} $ and it measures the rate of change in direction for the position vector.

The $ {\dot{\theta}} $ term is so important and appears so many times that it deserves its own special name. It called the angular velocity, it is represented by the symbol $ {\omega} $ and if one is interested in an vector representation it is $ {\vec{v}=\vec{\omega}\times \vec{r}} $.

For the acceleration it is
$ {\begin{aligned} \vec{a} &= \dfrac{d}{dt}\left( \dot{r}\vec{e}_r+r\dot{\theta}\vec{e}_\theta \right)\\ &= (\ddot{r}-r\dot{\theta}^2)\vec{e}_r+(r\ddot{\theta}+2\dot{r}\dot{\theta})\vec{e}_\theta \end{aligned}} $

Hence the radial component of the acceleration is $ {\ddot{r}-r\dot{\theta}^2} $ and the angular component is $ {r\ddot{\theta}+2\dot{r}\dot{\theta}} $.

— 3.2. Cylindrical Coordinates —

For cylindrical coordinates the infinitesimal line element is $ {ds^2=dr^2+r^2d\theta^2+dz^2} $. Hence one can show that it is

$ {v^2=\dot{r}^2+r^2\dot{\theta}^2+\dot{z}^2} $
$ {\vec{v}=\dot{r}\vec{e}_r+r\dot{\theta}\vec{e}_\theta+\dot{z}\vec{e}_z} $

— 3.3. Spherical Coordinates —

For spherical coordinates the infinitesimal line element is $ {ds^2=dr^2+r^2d\theta^2+r^2\sin ^2 \theta d\phi ^2} $. And for the velocity it is

$ {v^2=\dot{r}^2+r^2\dot{\theta}^2+r^2 \sin ^2 \theta \dot{\phi}^2} $
$ {\vec{v}=\dot{r}\vec{e}_r+r\dot{\theta}\vec{e}_\theta+r\sin \theta \dot{\phi}\vec{e}_\phi} $

— 4. Vector Calculus —

In this part of our post we'll take a very brief look at some concepts of vector calculus that are useful to Physics.

— 4.1. Gradient, Divergence and Curl —

If one has $ {\varphi=\varphi(x_1,x_2,x_3)} $ a scalar function, continuous and differentiable, it is by definition $ {\varphi'(x_1',x_2',x_3')=\varphi(x_1,x_2,x_3)} $. From this it follows

$ \displaystyle \frac{\partial \varphi'}{\partial x_i'}=\sum_j \frac{\partial \varphi}{\partial x_j }\frac{x_j}{x_i'} $

The inverse coordinate transformation is $ {x_j= \displaystyle \sum _k\lambda_{kj}x'_k} $. If we calculate the derivative of $ {x_j} $ with respect to $ {x_i'} $ it is
$ {\begin{aligned} \dfrac{\partial x_j}{\partial x_i'} &= \dfrac{\partial}{\partial x'_i}\left( \displaystyle \sum _k\lambda_{kj}x'_k \right)\\ &= \displaystyle \sum _k\lambda_{kj}\dfrac{\partial x'_k}{\partial x_i'}\\ &= \displaystyle \sum _k\lambda_{kj}\delta_{ik}\\ &=\lambda_{ij} \end{aligned}} $

Hence it is $ {\displaystyle \frac{\partial \varphi'}{\partial x_i'}=\sum_j \lambda_{ij}\frac{\partial \varphi}{\partial x_j }} $. From what we've seen in the last post this means that the function $ {\dfrac{\partial \varphi}{\partial x_j}} $ is the j-th component of a vector.

If we introduce the operator $ {\nabla} $ (sometimes called del) whose definition is

$ \displaystyle \nabla= \left( \frac{\partial}{\partial x_1},\frac{\partial}{\partial x_2}, \frac{\partial}{\partial x_3} \right)= \sum_i \vec{e}_i \frac{\partial}{\partial x_i} $

we can define the gradient of a scalar function, $ {\varphi} $ to be $ {\nabla\varphi=\left( \dfrac{\partial \varphi}{\partial x_1},\dfrac{\partial \varphi}{\partial x_2}, \dfrac{\partial \varphi}{\partial x_3} \right)} $.
The divergence of a vector field, $ {\vec{A}} $, is a scalar and is defined to be:

$ \displaystyle \nabla\cdot\vec{A} $

If the divergence of vector field is positive at a given point that means that in that point a source of the field is present. If the divergence of the field is negative that means that a sink of the field is present.
Another vector entity that one can define with the del operator is the curl of a vector field. The curl is a vector entity too and its definition is:

$ \displaystyle \nabla \times \vec{A}=\sum_{i,j,k}\varepsilon _{ijk}\vec{e}_i\nabla _j A_k $

The last operator that we'll define in this section is the Laplacian. This operator is the result of applying the del operator two times and it is

$ \displaystyle \nabla\cdot\nabla=\sum_i \frac{\partial}{\partial x_i}\frac{\partial}{\partial x_i}=\sum_i \frac{\partial ^2}{\partial x^2_i}=\nabla ^2 $

A quick note on Relativity

2014-02-24T13:33:00.001-08:00

This post is intended to be a very concise rebuttal to one of the most popular misconceptions about the theory of Special Relativity.

— 1. Brief Historical Review —

For the ones that really want to have a better grasp of this complicated issue you can go to:

But what I really want to say is that we can basically say that the catalyst of Relativity was the Electromagnetic Theory of Maxwell. After deriving the full set Maxwell's equations Maxwell was able to show that the Electromagnetic Field propagated itself like a wave whose speed was $ {c=1/\sqrt{\epsilon_0 \mu_0}} $.

This simple fact had at least two problems to it:

According to Galileo's transformations any speed was relative to the frame were it was being measured. But this value of $ {c} $ was an invariant.
According to classical Physical ideas if some phenomenon was described by a wave equation than something had to be waving.

In the case of the electromagnetic field that question of what exactly was waving was a very though nut to crack. Just remember the simple fact that we can see the Sun even though the space between us and the Sun is essentially void.

To solve these two problems with one go the Physicists proposed that there was a substance that existed all over the space that waved whenever the electromagnetic radiation was being carried. And the speed of light was indeed an invariant, but it was an invariant relatively to the aether...

But that of course begged the question of why didn't we find any variation in the speed of light since the Earth's movement relative to the aether was changing all the time?

The answer was that the methods weren't good enough to detect such a little variation. And the hunt was on to devise a method with enough sensibility to detect the more than obvious changes.

A lot of things were tried, all of them failed, a lot of partial answers were proposed to explain this failure but a coherent picture was missing!

— 2. Enter Einstein —

This of course changed in 1905 when a young man named Einstein publish some of the best articles in the history of our business.

One of these articles was a major rethinking of the concepts of space and time and really is a beauty to be read for any self-respecting physicist. But I digress...

With just two simple (and seemingly contradictory axioms) Einstein was able to show why no variation in the value of the speed of light could be detected and he also predicted many more phenomena that initiated a whole new cultural Zeitgeist for the years to come.

His two principles were:

The same laws of electrodynamics and optics will be valid for all frames of reference for which the equations of mechanics hold good.
Light is always propagated in empty space with a definite velocity c which is independent of the state of motion of the emitting body

— 3. The misconception —

After this somewhat lengthy introduction is time for me to state what I consider to be the popular misconception: "The theory of relativity says that nothing can travel faster than light."

Well, we surely don't see that being said by Einstein in the previous quote! What he says is that the speed of light is an invariant (even though even in University textbooks we can read the first postulate as saying that the speed of light is the limit of all speeds in the Universe and that it is the same for all observers).

— 3.1. Moving Frames —

Well if it isn't a postulate (if it were a postulate it would always be up for experimental scrutiny) maybe it is something that we can derive in Special Relativity?

Well, sort of...

Imagine that you have two inertial frames $ {S} $ and $ {S'} $ and you consider $ {S} $ to be a stationary frame and that $ {S'} $ has speed $ {v} $ relative to $ {S} $. If you know that the speed of a given body to be $ {u'_x} $ in $ {S'} $ than you can show that the velocity of the same body in the inertial frame $ {S} $, $ {u_x} $, to be:

$ \displaystyle u_x=\frac{u'_x+v}{1+\frac{u'_xv}{c^2}} $

Now if you admit that $ {u'_x < c} $ and that $ {v < c} $ you can show that $ {u_x < c} $. But if you don't admit the initial hypothesis than the value of $ {u_x} $ isn't bounded above by $ {c} $.

And now for a proof of the previous assertion:

If $ {u'x < c} $ and $ {v < c} $, that for some $ {\alpha, \beta > 0} $ it is $ {u'_x=c-\alpha} $ and $ {v =c-\beta} $.

Now $ {u_x} $ is

$ {\begin{aligned} u_x &= \dfrac{c-\alpha+c-\beta}{1+(c-\alpha)(c-\beta)/c^2} \\ &= \dfrac{2c-(\alpha+\beta)}{1+(c^2-(\alpha+\beta)+\alpha\beta)/c^2} \\ &= \dfrac{2c-(\alpha+\beta)}{2c^2-(\alpha+\beta)c+\alpha\beta}c^2 \\ &< \dfrac{2c-(\alpha+\beta)}{2c^2-(\alpha+\beta)c}c^2 \\ &= \dfrac{2c-(\alpha+\beta)}{2c-(\alpha+\beta)}c \\ &= c \end{aligned}} $

Which is $ {u_x < c} $ just like we said it would be.

Again all of that this shows is that one can't boost a body to have a speed bigger than $ {c} $ by using an inertial frame whose speed is smaller than $ {c} $ if the speed of the body in $ {S'} $ is smaller than $ {c} $ too.

— 3.2. Accelerating bodies —

But we still have another chance, though. What if we don't use boosts? What if we accelerate a particle during a sufficient long time interval? Doesn't a speed bigger than $ {c} $ happens eventually?

Let's see.

First we have to define the linear momentum of a particle in Special Relativity. If we want the linear momentum to be conserved in relativistic collisions the definition that makes sense is

$ \displaystyle \vec{p}=\gamma m \vec{u} $

Where $ {\gamma} $ has its usual meaning and $ {\vec{u}} $ is the speed of the body.

The definition of force still is $ {\vec{F}=\dfrac{d\vec{p}}{dt}} $. Thus $ {\vec{F}=\dfrac{d}{dt}(\gamma m \vec{u})} $.

It is possible to show that for a body of constant mass (which is always the case if something doesn't get aggregated to the body or if the body doesn't get broken up - summing up: we aren't falling for the "relativistic mass" erroneous idea) one has

$ \displaystyle a=\frac{F}{m}(1-u^2/c^2)^{3/2} $

Now imagine that you are pushing on a body and its speed is getting bigger and bigger. Than what we have is $ {u \rightarrow c} $ but in this case it is $ {a \rightarrow 0} $. Thus the closer the speed of the body gets to $ {c} $ the less it is accelerating until that finally its acceleration is $ {0} $.

The conclusion is that if a body starts with $ {u < c} $ than it would take an infinite amount of time to get to a speed of $ {c} $ and from then on its speed would be $ {c} $ for ever.

— 4. Conclusion —

Contrary to what is normally said the fact that the speed of light is limit of all speeds on the real world isn't a postulate.

It does hold with the caveat that the bodies have to have speeds less than $ {c} $ to begin with, but nothing in Special Relativity forces the bodies to have speeds less than $ {c} $ when they start moving.

— 5. Appendix —

I think that in order to finish this post I just have to mention that it is possible to derive the theory of Special Relativity with just the first postulate and one doesn't need to make any mention to the speed of light.

The idea is the following one: imagine that you have three inertial frames $ {S_1} $, $ {S_2} $ and $ {S_3} $ whose axes are parallel to each other.

Then you boost $ {S_2} $ relatively to $ {S_1} $ with a velocity $ {v_{21}} $.

You also boost $ {S_3} $ relatively to $ {S_2} $ with a velocity $ {v_{32}} $.

The question now is: Are the axes of $ {S_1} $ and $ {S_3} $ still parallel?

If you answered the previous question with a resounding yes, than what you're doing is saying that we always live an Euclidean timespace and that the dynamics that rules us are the classic ones.

If on the other hand you just said: Gee, I don't really know... than you're saying that is none of our business to tell the spacetime continuum how it ought to behave and you are willing to study it and only then conclude what its true nature is.

This is just what Feigenbaum did in this great article: The Theory of Relativity - Galileo's Child whose reading I highly reccomend and whose abstract I just have to quote (no bold in the original):

We determine the Lorentz transformations and the kinematic content and dynamical framework of special relativity as purely an extension of Galileo's thoughts. No reference to light is ever required: The theories of relativity are logically independent of any properties of light. The thoughts of Galileo are fully realized in a system of Lorentz transformations with a parameter $ {1/c^2} $, some undetermined, universal constant of nature; and are realizable in no other. Isotropy of space plays a deep and pivotal role in all of this, since here three-dimensional space appears at first blush, and persists until the conclusion: Relativity can never correctly be fully developed in just one spatial dimension.

Matrices, Scalars, Vectors and Vector Calculus 01

2014-02-24T13:24:00.001-08:00

Let us imagine that we have a system of coordinates $ {S} $ and a system of coordinates $ {S'} $ that is rotated relatively to $ {S} $. Let us consider a point $ {P} $ that has coordinates $ {(x_1,x_2,x_3)} $ on $ {S} $ and coordinates $ {(x'_1,x'_2,x'_3)} $ on $ {S'} $.
In general it is obvious that $ {x'_1=x'_1(x_1,x_2,x_3)} $, $ {x'_2=x'_2(x_1,x_2,x_3)} $ and that $ {x'_3=x'_3(x_1,x_2,x_3)} $.
Since the transformation from $ {S} $ to $ {S'} $ is just a rotation we can assume that the transformation is linear. Hence we can write explicitly
$ {\begin{aligned} x'_1 &= \lambda _{11}x_1+ \lambda _{12}x_2 +\lambda _{13}x_3 \\ x'_2 &= \lambda _{21}x_1+ \lambda _{22}x_2 +\lambda _{23}x_3 \\ x'_3 &= \lambda _{31}x_1+ \lambda _{32}x_2 +\lambda _{33}x_3 \end{aligned}} $
Another way to write the three previous equations in a more compact way is:

$ \displaystyle x'_i=\sum_{j=1}^3 \lambda_{ij}x_j $

In case you don't see how the previous equation is a more compact way of writing the first equations I'll just lay out the $ {i=1} $ case.

$ \displaystyle x'_1=\sum_{j=1}^3 \lambda_{1j}x_j $

Now all that we have to do is to sum from $ {j=1} $ to $ {j=3} $ and we get the first equation. For the other two a similar reasoning applies.
If we want to make a transformation from $ {S'} $ to $ {S} $ the inverse transformation is

$ \displaystyle x_i=\sum_{j=1}^3 \lambda_{ji}x'_j $

The previous notation suggests that the $ {\lambda} $ indexes can be arranged in a form of a matrix:

$ \displaystyle \lambda= \left(\begin{array}{ccc} \lambda_{11} & \lambda_{12} & \lambda_{13} \\ \lambda_{21} & \lambda_{22} & \lambda_{23} \\ \lambda_{31} & \lambda_{32} & \lambda_{33} \end{array} \right) $

In the literature the previous matrix has the name of rotation matrix or transformation matrix.

— 1. Properties of the rotation matrix —

For the transformation $ {x'_i=x'_i(x_i)} $

$ \displaystyle \sum_j \lambda_{ij}\lambda_{kj}=\delta_{ik} $

Where $ {\delta_{ik}} $ is a matrix known as Kronecker delta and its definition is

$ \displaystyle \delta_{ik}=\begin{cases} 0 \quad i\neq k\\ 1 \quad i=k \end{cases} $

For the inverse transformation $ {x_i=x_i(x'_i)} $ it is

$ \displaystyle \sum_i \lambda_{ij}\lambda_{ik}=\delta_{jk} $

The previous relationships are called orthogonality relationships.

— 2. Matrix operations, definitions and properties —

Let us represent the coordinates of a point $ {P} $ by a column vector

$ \displaystyle x = \left(\begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array}\right) $

Using the usual notation of linear algebra we can write the transformation equations as $ {x'=\mathbf{\lambda} x} $
Where we define the matrix product, $ {\mathbf{AB}=\mathbf{C}} $, to be possible only when the number of columns of $ {\mathbf{A}} $ is equal to the number of rows of $ {\mathbf{B}} $
The way to calculate a specific element of the matrix $ {\mathbf{C}} $, we will denote this element by the symbol $ {\mathbf{C}_{ij}} $ is,

$ \displaystyle \mathbf{C}_{ij}=[\mathbf{AB}]_{ij}=\sum_k A_{ik}B_{kj} $

Given the definition of a matrix product it should be clear that in general one has $ {\mathbf{AB} \neq \mathbf{BA}} $
As an example let us look into;

$ \displaystyle \mathbf{A}=\left( \begin{array}{cc} 2 & 1\\ -1 & 3 \end{array}\right) ;\quad \mathbf{B}=\left( \begin{array}{cc} -1 & 2\\ 4 & -2 \end{array}\right) $

With

$ \displaystyle \mathbf{AB}=\left( \begin{array}{cc} 2\times (-1)+1\times 4 & 2\times 2+1\times (-2)\\ -1\times (-1)+3\times 4 & -1\times 2+3\times (-2) \end{array}\right)=\left( \begin{array}{cc} 2 & 2\\ 13 & -8 \end{array}\right) $

and

$ \displaystyle \mathbf{BA}=\left( \begin{array}{cc} -4 & 5\\ 10 & -2 \end{array}\right) $

We'll say that $ {\lambda^T} $ is the transposed of $ {\lambda} $ and calculate the matrix elements of the transposed matrix by $ {\lambda_{ij}^T=\lambda_{ji}} $. In a more pedestrian way one can say that in order to obtain the transpose of a given matrix one needs only to exchange its rows and columns.
For a given matrix $ {\mathbf{A}} $ it exists another matrix $ {\mathbf{U}} $ such as $ {\mathbf{AU}=\mathbf{UA}=\mathbf{A}} $. The matrix $ {\mathbf{U}} $ is said to be the unit matrix and usually one can represent it by $ {\mathbf{U}=\mathbf{1}} $.
If $ {\mathbf{AB}=\mathbf{BA}=\mathbf{1}} $, then $ {\mathbf{A}} $ and $ {\mathbf{B}} $ are said to be the inverse of each other and $ {\mathbf{B}=\mathbf{A}^{-1}} $, $ {\mathbf{A}=\mathbf{B}^{-1}} $.
Now for the rotation matrices it is
$ {\begin{aligned} \lambda \lambda ^T &= \left( \begin{array}{cc} \lambda_{11} & \lambda_{12}\\ \lambda_{21} & \lambda_{22} \end{array}\right)\left( \begin{array}{cc} \lambda_{11} & \lambda_{21}\\ \lambda_{12} & \lambda_{22} \end{array}\right) \\ &= \left( \begin{array}{cc} \lambda_{11}^2+\lambda_{22}^2 & \lambda_{11}\lambda_{21}+\lambda_{12}\lambda_{22}\\ \lambda_{21}\lambda_{11}+\lambda_{22}\lambda_{12} & \lambda_{21}^2+\lambda_{22}^2 \end{array}\right)\\ &=\left( \begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right)\\ &= \mathbf{1} \end{aligned}} $
Where the second last equality follows from what we've seen in section 1.
Thus $ {\lambda ^T=\lambda ^{-1}} $.
Just to finish up this section let me just mention that even though, in general, matrix multiplication isn't commutative it still is associative. Thus $ {(\mathbf{AB})\mathbf{C}=\mathbf{A}(\mathbf{BC})} $. Also matrix addition has just the definition one would expect. Namely $ {C_{ij}=A_{ij}+B_{ij}} $.
If one inverts all three axes at the same time the matrix that we get is the so called inversion matrix and it is

$ \displaystyle \left( \begin{array}{ccc} -1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & -1 \end{array}\right) $

Since it can be shown that rotation matrices always have their determinant equal to $ {1} $ and that the inversion matrix has a $ {-1} $ determinant we know that there isn't any continuous transformation that maps a rotation into an inversion.

— 3. Vectors and Scalars —

In Physics quantities are either scalars or vectors (they can also be tensors but since they aren't needed right away I'll just pretend that they don't exist for the time being). These two entities are defined according to their transformation properties.
Let $ {\lambda} $ be a coordinate transformation, $ {\displaystyle\sum_j\lambda_{ij}\lambda_{kj}=\delta_{ij}} $, if it is:

$ {\displaystyle\sum_j\lambda_{ij}\varphi=\varphi} $ then $ {\varphi} $ is said to be a scalar.
$ {\displaystyle\sum_j\lambda_{ij}A_j=A'_i} $ for $ {A_1} $, $ {A_2} $ and $ {A_3} $ then $ {(A_1,A_2,A_3)} $ is said to be a vector.

— 3.1. Operations between scalars and vectors —

I think that most people in here already know this but in the interest of a modicum of self containment I'll just enumerate some properties of scalars and vectors.

$ {\vec{A}+\vec{B}=\vec{B}+\vec{A}} $
$ {\vec{A}+(\vec{B}+\vec{C})=(\vec{A}+\vec{B})+\vec{C}} $
$ {\varphi+\psi=\psi+\varphi} $
$ {\varphi+(\psi+\xi)=(\varphi+\psi)+\xi} $
$ {\xi \vec{A}= \vec{B}} $ is a vector.
$ {\xi \varphi=\psi} $ is a scalar.

As an example we will show the second proposition 5 and the reader has to show the veracity of the last proposition.
In order to show that $ {\xi \vec{A}= \vec{B}} $ is a vector we have to show that it transforms like a vector.
$ {\begin{aligned} B'_i &= \displaystyle\sum_j \lambda_{ij}B_j\\ &= \displaystyle\sum_j \lambda_{ij}\xi A_j\\ &= \xi\displaystyle\sum_j \lambda_{ij} A_j\\ &= \xi A'_i \end{aligned}} $
Hence $ {\xi A} $ transforms like a vector.

— 4. Vector "products" —

The operations between scalars are pretty much well know by everybody, hence we won't take a look at them, but maybe it is best for us to take a look at two operations between vectors that are crucial for our future development.

— 4.1. Scalar product —

We can construct a scalar by using two vectors. This scalar is a measure of the projection of one vector into the other. Its definition is

$ \displaystyle \vec{A}.\cdot\vec{B}=\sum_i A_i B_i = AB\cos (A.B) $

For this operation deserve its name, one still has to prove that the result indeed is a scalar.
First one writes $ {A'_i=\displaystyle \sum_j\lambda_{ij}A_j} $ and $ {B'_i=\displaystyle \sum_k\lambda_{ik}B_k} $, where one changes the index of the second summation because we'll have to multiply the two quantities and that way the final result can be achieved much more easily.
Now it is
$ {\begin{aligned} \vec{A}'\cdot \vec{B}' &= \displaystyle\sum_i A'_i B'_i \\ &= \displaystyle \sum_i \left(\sum_j\lambda_{ij}A_j\right)\left( \sum_k\lambda_{ik}B_k \right)\\ &= \displaystyle \sum_j \sum_k \left( \sum_i \lambda_{ij}\lambda_{ik} \right)A_j B_k\\ &= \displaystyle \sum_j \left(\sum_k \delta_{jk}A_jB_k \right)\\ &= \displaystyle \sum_j A_j B_j \\ &= \vec{A}\cdot \vec{B} \end{aligned}} $
Hence $ {\vec{A}\cdot \vec{B}} $ is a scalar.

— 4.2. Vector product —

First we have to introduce the permutation symbol $ {\varepsilon_{ijk}} $. Its definition is $ {\varepsilon_{ijk}=0} $ if two or three of its indices are equal; $ {\varepsilon_{ijk}=1} $ if $ {i\,j\,k} $ is an even permutation of $ {123} $ (the even permutations are $ {123} $, $ {231} $ and $ {312} $); $ {\varepsilon_{ijk}=-1} $ if $ {i\,j\,k} $ is an odd permutation of $ {123} $ (the odd permutations $ {132} $, $ {321} $ and $ {213} $).
The vector product, $ {\vec{C}} $, of two vectors $ {\vec{A}} $ and $ {\vec{B}} $ is denoted by $ {\vec{C}=\vec{A}\times \vec{B}} $.
To calculate the components the components of the vector $ {\vec{C}} $ the following equation is to be used:

$ \displaystyle C_i=\sum_{j,k}\varepsilon_{ijk}A_j B_k $

Where $ {\displaystyle\sum_{j,k}} $ is shorthand notation for $ {\displaystyle\sum_j\sum_k} $.
As an example let us look into $ {C_1} $
$ {\begin{aligned} C_1 &= \sum_{j,k}\varepsilon_{1jk}A_j B_k\\ &= \varepsilon_{123}A_2 B_3+\varepsilon_{132}A_3 B_2\\ &= A_2B_3-A_3B_2 \end{aligned}} $
where we have used the definition of $ {\epsilon_{ijk}} $ throughout the reasoning.
One can also see that (this another exercise for the reader) $ {C_2=A_3B_1-A_1B_3} $ and that $ {C_3=A_1B_2-A_2B_1} $.
If one only wants to know the magnitude of $ {\vec{C}} $ the following equation should be used $ {C=AB\sin (A,B)} $.
After choosing the three axes that define our frame of reference one can choose as the basis of this space a set of three linearly independent vectors that have unit norm. These vectors are called unit vectors.
If we denote these vectors by $ {\vec{e}_i} $ any vector $ {\vec{A}} $ can be written as $ {\vec{A}=\displaystyle \sum _i \vec{e}_i A_i} $. We also have that $ {\vec{e}_i\cdot \vec{e}_j=\delta_{ij}} $ and $ {\vec{e}_i\times \vec{e}_j=\vec{e}_k} $. Another way to write the last equation is $ {\vec{e}_i\times \vec{e}_j=\vec{e}_k\varepsilon_{ijk}} $.

— 5. Vector differentiation with respect to a scalar —

Let $ {\varphi} $ be a scalar function of $ {s} $: $ {\varphi=\varphi(s)} $. Since both $ {\varphi} $ and $ {s} $ are scalars we know that their transformation equations are $ {\varphi=\varphi '} $ and $ {s=s'} $. Hence it also is $ {d\varphi=d\varphi '} $ and $ {ds=ds'} $
Thus it follows that for differentiation it is $ {d\varphi/ds=d\varphi'/ds'=(d\varphi/ds)'} $.
In order to define the derivative of a vector with respect to a scalar we will follow an analogous road.
We already know that it is $ {A'_i=\displaystyle \sum_j \lambda _{ij}A_j} $ hence
$ {\begin{aligned} \dfrac{dA'_i}{ds'} &= \dfrac{d}{ds'}\left( \displaystyle \sum_j \lambda _{ij}A_j \right)\\ &= \displaystyle \lambda _{ij}\dfrac{d A_j}{ds'}\\ &= \displaystyle \lambda _{ij}\dfrac{d A_j}{ds}\ \end{aligned}} $
where the last equality follows from the fact that $ {s} $ is a scalar.
From what we saw we can write

$ \displaystyle \frac{d A'_i}{ds'}= \left( \frac{d A_i}{ds} \right)'=\sum_j \lambda _{ij}\frac{d A_j}{ds} $

Hence $ {dA_j/ds} $ transforms like the coordinates of a vector which is the same as saying that $ {d\vec{A}/ds} $ is a vector.
The rules for differentiating vectors are:

$ {\dfrac{d}{ds}(\vec{A}+\vec{B})= \dfrac{d\vec{A}}{ds}+\dfrac{d\vec{B}}{ds}} $
$ {\dfrac{d}{ds}(\vec{A}\cdot\vec{B})= \vec{A}\cdot\dfrac{d\vec{B}}{ds}+\dfrac{d\vec{A}}{ds}\cdot \vec{B}} $
$ {\dfrac{d}{ds}(\vec{A}\times\vec{B})= \vec{A}\times\dfrac{d\vec{B}}{ds}+\dfrac{d\vec{A}}{ds}\times \vec{B}} $
$ {\dfrac{d}{ds}(\varphi\vec{A})= \varphi\dfrac{d\vec{A}}{ds}+\dfrac{d\varphi}{ds}\vec{A}} $

The proof of these rules isn't needed in order for us to develop any kind of special skills but if the reader isn't very used to this, then it is better for him to do them just to see how things happen.

Watchmath script

2011-05-05T14:02:00.000-07:00

I don't know what's wrong with it but the previous JavaScript I used is down. Thus I had to find another way to display the equations you are so fond of.

The solution was provided by this post in the blog Dysfunctional. It uses the MathJax JavaScript and the equations do look mighty fine.

Mathematical trick in Statistical Physics

2011-04-13T11:43:00.000-07:00

The other day I was looking into some Statistical Physics and ended up doing an exercise that was kinda cute. The exercise in question is on Mandl's Statistical Physics, Second Edition page 66.

The first time I did the exercise I did it the normal way, but then I noticed something that ended up simplifying my calculations and here I am posting it.

A system has three energy levels $ {E_1= \epsilon } $, $ {E_2=2 \epsilon } $ and $ {E_3=3 \epsilon } $ with degeneracies $ {g(E_1)=g(E_3)=1} $, $ {g(E_2)=2} $. Find the heat capacity of the system.

Don't worry if you don't understand all the terms in here since the gist of this post is not Statistical Physics but the mathematical trick I ended up using.

To solve this problem we need to calculate the partition function $ {Z} $.

$ \displaystyle \begin{array}{rcl} Z &=& \displaystyle \sum_{E_r}g(E_r)e^{-\beta E_r} \\ &=& 1+2e^{-\beta \epsilon}+e^{-\beta 2\epsilon} \\ &=& e^{\beta \epsilon}+2+2e^{-\beta \epsilon} \\ &=& 2(1+ \cosh (\beta \epsilon)) \end{array} $

After having calculated the partition function we have to calculate the average energy, $ {\bar{E}} $, of this system. By definition it is:

$ \displaystyle \begin{array}{rcl} \bar{E} &=& -\dfrac{\partial}{\partial \beta} \log Z \\ &=& -\dfrac{\partial}{\partial \beta} \log 2(1+ \cosh (\beta \epsilon)) \\ &=& -\dfrac{2 \epsilon \sinh (\beta \epsilon)}{2(1+\cosh (\beta \epsilon))} \\ &=& -\dfrac{\epsilon \sinh (\beta \epsilon)}{1+\cosh (\beta \epsilon)} \\ \end{array} $

Keeping in mind that in Statistical Physics it is $ {\beta = 1/(kT)} $ the heat capacity, $ {c} $, is (this is where I started using the trick):

$ \displaystyle \begin{array}{rcl} c &=& \dfrac{\partial \bar{E}}{\partial T} \\ &=& k \dfrac{\partial \bar{E}}{\partial (kT)} \\ &=& k \dfrac{\partial \bar{E}}{\partial (1/\beta)} \\ &=& -\beta ^2 k \dfrac{\partial \bar{E}}{\partial \beta}\\ &=& \beta ^2 k \dfrac{\partial}{\partial \beta}\left( \dfrac{\epsilon \sinh (\beta \epsilon)}{1+\cosh (\beta \epsilon)}\right) \end{array} $

Notice that up until now we haven't calculated a thing at all. All that we have done is just to change variables in order to ease the difficulty in the derivative we'll have to calculate.

Taking the derivative with respect to $ {T} $ in the last expression isn't that hard, but it sure is boring and if one isn't careful errors are expected to creep in.

On the other hand the last expression is much easier to differentiate, but since we are caught up in the moment we'll just do one more change of variable.

$ \displaystyle \begin{array}{rcl} c &=& \beta ^2 k \dfrac{\partial}{\partial \beta}\left( \dfrac{\epsilon \sinh (\beta \epsilon)}{1+\cosh (\beta \epsilon)}\right) \\ &=& \beta ^2 \epsilon ^2 k \dfrac{\partial}{\partial (\beta \epsilon)}\left( \dfrac{ \sinh (\beta \epsilon)}{1+\cosh (\beta \epsilon)}\right) \\ &=& x ^2 k \dfrac{\partial}{\partial x}\left( \dfrac{ \sinh x}{1+\cosh x}\right) \\ &=& x ^2 k \dfrac{\cosh x + \cosh ^2 x - \sinh ^2 x}{(1+ \cosh x)^2} \\ &=& x ^2 k \dfrac{\cosh x + 1}{(1+ \cosh x)^2} \\ &=& \dfrac{x ^2 k}{1+ \cosh x} \end{array} $

Yes, in this case the simplification wasn't that great but I think that one shouldn't lose sight of the fact that this type of reasoning can greatly simplify some other physical derivations.

As an afterthought just let me point out the fact that we cold get a much simpler life if we just remembered a very easy algebraic identity:

$ \displaystyle \begin{array}{rcl} Z &=& 1+2e^{-\beta \epsilon}+e^{-\beta 2\epsilon} \\ &=& (1+e^{-\beta \epsilon})^2 \\ \end{array} $

With this expression the calculation of $ {\bar{E}} $ and $ {c} $ is a lot easier and the reader is urged to try it out.

More talk about numbers

2010-12-01T07:24:00.000-08:00

At the end of this post we had arrived at the rational numbers and were able to derive four mathematical operations. At first sight this might look like an impressive achieving, taking into account the tools we set ourselves to use, but due to some shortcomings that plague our construction we think that things can get better.

— 1. Problems —

Our operations are binary
By that we mean that our mathematical operations only make sense when applied to two numbers. So for instance $ {2+3+4} $ has no answer in our current state of affairs. But worry not dear reader because things can be formalized in a fast and straightforward way. And there's nothing too fancy, or technical in this formalization: it is nothing more than common sense put to action.
How do we add $ {2+3+4} $? We just do $ {2+3=5} $ and $ {5+4=9} $. So the correct formalization of $ {a_1 + a_2 + a_3 + \ldots + a_n} $ just is:

$ \displaystyle ((((a_1 + a_2) + a_3) + \ldots )+ a_n) $

That is to say that we define addition of $ {n} $ numbers to be done two by two (this means that addition is still binary but can be done repeatedly).
Of course that all of this is valid for the three remaining operations even though we haven't explicitly said so. Thus this is one possible problem that was taken care of.
Another problem is that we have two inverse operations and due to the existence of them we could expand the set of the available numbers. But we never said nothing about what happens to the properties of the operations while we continuously do that. And for that matter we never said anything at all about the properties of the operations in the first place!
Another problem that we have is that our number system is rather incomplete.

— 2. Properties of the mathematical operations —

To solve our second problem it's time for us to talk about the properties of addition, multiplication, subtraction, and division.

Associativity

$ \displaystyle (m+n)+p = m+(n+p) $

$ \displaystyle (m \times n) \times p = m\times (n \times p) $
Commutativity

$ \displaystyle m+n=n+m $

$ \displaystyle m \times n = n \times m $
Distributivity

$ \displaystyle m \times (n+p)= m \times n + m \times p $
Neutral Element
For every natural number it is:

$ \displaystyle m+0 = m $

and

$ \displaystyle m \times 1 = m $

If we think about addition and multiplication in the simple terms of the first post in this series it is easy to see why these properties hold. But remember that those simple terms apply to the natural numbers. As we went along in enlarging the set of numbers that we were working with we never cared about what happened to those properties.

Do they still hold in $ { \mathbb{Z} } $ and $ { \mathbb{Q} } $?

We can take the lazy out and just define those sets as being sets in which the properties hold or we can check how things really are. Even though I'll take the lazy way out I want readers of this blog to know that things can be done in an intellectually satisfying way.

— 3. Powers —

To solve our third problem we'll just continue to use our ambition and see where it leads us.

Just like we introduced multiplication as a device of allowing us to write more succinctly a sum of $ {n} $ equal numbers we will now introduce the power operation as a more succinct way of writing a product of equal factors.

So for instance we'll write $ {2^5} $ for $ {2 \times 2 \times 2 \times 2 \times 2} $. In general if we have a number $ {a} $, this number is called the base, multiplying with itself $ {n} $ times, this number is called the exponent, we'll write $ {a^n} $ for the end result.

Now that we have introduced powers as a shorthand notation for multiplication of equal factors we have to know how this new entity behaves with the previous four operations.

When we are summing (subtracting) powers we have to first calculate the result of the powers and after this we sum (subtract) those same results. For example:

$ \displaystyle 2^3+ 3^2 = 8+9=17 $

$ \displaystyle 4^2 + 3^2 = 16+9=25 $

$ \displaystyle 5^3-10^2=125-100=25 $

But when we are multiplying or dividing there are some rules that can be used in order to get the right result in a faster way.

— 3.1. Multiplying powers —

Unlike addition (subtraction) of powers when one is multiplying powers we can do it in a more cleaner way in some number of cases:

Same base
For instance:

$ \displaystyle 2^3 \times 2^4=2\times 2 \times 2 \times 2 \times 2 \times 2 \times 2=2^7 $
Thus the final power has the same base as the initial ones and its exponent is just the sum of the exponents of the initial power.
Of course that the numbers $ {2} $, $ {3} $ and $ {4} $ that we choose initially aren't in any way special and as a general rule it always is:

$ \displaystyle a^n\times a^m = a^{n+m} $
Same exponent
For instance:

$ \displaystyle 3^3 \times 4^3= 3 \times 3 \times 3 \times 4 \times 4 \times 4= (3\times 4) \times (3\times 4) \times (3\times 4)= 12^3 $

Once again the numbers we initially choose have nothing special about them so we can state in full generality that

$ \displaystyle a^n\times b^n = (a\times b)^n $

Yes we have stated these two properties for binary multiplication but they can be extended to any number of factors without any problems whatsoever.

— 3.2. Dividing powers —

In the case of division we'll just state the theorems and won't bother to present any example since this can be done very easily by an interested reader once he grasps the idea behind the examples used for multiplication.

Same base

$ \displaystyle a^n/a^m=a^{n-m} $
Same exponent

$ \displaystyle a^n/b^n=(a/b)^n $

— 4. Here comes trouble! —

Since we haven't abandoned our principle of ambition thus far the questions that are being begged to be answered are if we can introduce the inverse operation of a power? and where such course of action might take us?

The answer to the first question is an emphatic yes. The answer to second question is that the inverse operation of a power takes us to wonderful places and open the floodgates of some wonderful mathematics.

The answer of these two questions will be the subject of our next, and possibly final, installment and in these series of posts.

A first and fast look into Kuhn

2010-11-17T17:39:00.001-08:00

Thomas Kuhn, born in 1922 and passed away in 1996, got interested about the history of Physics during his Ph.D. studies

He was in charge of a course, for humanities students, that was about the episodes in the history of science . To prepare for his lectures he decided to read the original texts as much has he could. Once he started to read Physics by Aristotle he was totally surprised with the new Universe that unfolded before him: the ideas expressed in that book seemed to come from outer space!

But they just had come from another time and space. Kuhn realized that taking into account the historical context and the remarkable internal consistency of Aristotle's Physics, it wasn't bad Physics: it was just another kind of Physics. From that moment on Kuhn was more and more drawn to reading the original sources while always stressing that original sources need to be read and analyzed in context.

Essential to Kuhn's analysis of science is the notion of paradigm. A paradigm is a highly ample concept in Kuhn's usage: it is simultaneously all of the laws, all of the theories, all methodological rules, all applications and extensions, all the models, metaphysical suppositions, conceptual frameworks and even the vague and imprecise scheme of the way Nature works.

Of course that such a broad concept had to face its fair share of scrutiny. To respond to his critics Kuhn dropped the notion of paradigm and focused on the much less vague notion of "disciplinary matrix". A disciplinary matrix has four subdivisions:

Symbolic generalizations: laws and definitions of symbols contained in the laws.
Metaphysical assumptions: shared beliefs of a scientific community during a certain time interval.
Values: the criteria of simplicity, internal coherence, accuracy, elegance, plausibility shared by the scientific community.
Exemplars: the designated body of knowledge contained in lab activities, exercises, textbooks, etc, that will help the student acquire his/her scientific formation.

After knowing what Kuhn's notion of a disciplinary matrix is we can summarize his view of how scientific development occurs.

At first we have the period of pre-paradigmatic science, a period of competition between multiple schools of thought where the explanation of certain natural phenomena cause divergences on what is to be studied and how it is to be studied (this occurrence of competing views of Nature is what Kuhn called incommensurability - a rather controversial, bum rapped and poorly understood concept).

After this period a school of thought has won preponderance and one can speak of an existing scientific community. This community has now a set of procedures, problems and a world view that are more or less broadly shared and is engaged in what mostly is puzzle solving activities. This is the period of normal science and as Kuhn rightly pointed out scientists in this stage have a kind of aversion to novelties on their fields of expertise.

Anomalies in the reigning paradigm are seen as minor nuisances that have to be solved in the context of the scientific zeitgeist. If the anomaly resists numerous attempts of a solution for a long enough period of time it will cause an erosion of the rules that are present in normal science and a period of renewed uncertainty about the disciplinary matrix is sure to follow (this convoluted period that has a lot of similarities with the pre-paradigmatic stage of science is sometimes called multi-paradigmatic science).

Following this either a new theory emerges where the discrepancies are mostly taken care of, or the old theory, with some little modifications, is able to accommodate the results that seemed to be out of its range. The first possibility is what is called a scientific revolution and after it a new period of normal science will follow in which the breadth and depth of the new disciplinary matrix will increase.

Two things are worth noticing in order to finish this very short introduction:

Even though the period of normal science is adverse to scientific novelties it, nevertheless, is a catalyst for them given its constant poking at the limits of its own domains of application.
On Kuhn's approach to the development of science, progress is achieved not in a cumulative way: science consists of long stretches of time where scientists engage in what Kuhn called normal science that are interrupted whenever a sufficient amount of anomalous results are too big to be ignored. In Kuhn's terminology this sets off a period that is called a crisis. After the resolution of this crisis by means of a scientific revolution a new world view emerges from which the old world view results can be derived (for instance Newtonian Dynamics results can be derived from Special Relativity by formally taking the limit $ {c \rightarrow \infty} $), but the underlying metaphysical assumptions of the two theories are profoundly at odds with each other.