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	<title>4by12</title>
	
	<link>http://4by12.com/blog</link>
	<description>by Guy Gur-Ari</description>
	<lastBuildDate>Tue, 01 Sep 2009 11:28:28 +0000</lastBuildDate>
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		<title>The Missing Comment</title>
		<link>http://feedproxy.google.com/~r/4by12/~3/vvNvs7XEqxA/274</link>
		<comments>http://4by12.com/blog/archives/274#comments</comments>
		<pubDate>Tue, 01 Sep 2009 11:28:28 +0000</pubDate>
		<dc:creator>Guy Gur Ari</dc:creator>
				<category><![CDATA[Uncategorized]]></category>

		<guid isPermaLink="false">http://4by12.com/blog/?p=274</guid>
		<description><![CDATA[Over on The Gauge Connection I was engaged in a lively discussion about the arrow of time with the blog owner, Marco Frasca. For some reason he decided to censor my latest comment, and notify me he will not publish anything else by me on his blog. Touchy, touchy.
Seeing as I have an acute allergy [...]]]></description>
			<content:encoded><![CDATA[<p>Over on <a href="http://marcofrasca.wordpress.com/2009/08/31/the-question-of-the-arrow-of-time/">The Gauge Connection</a> I was engaged in a lively discussion about the arrow of time with the blog owner, Marco Frasca. For some reason he decided to censor my latest comment, and notify me he will not publish anything else by me on his blog. Touchy, touchy.</p>
<p>Seeing as I have an acute allergy to censorship, I will publish my last comment here, as long with anything else Marco chooses to censor. If you are interested in the nature of the arrow of time, you may want to head over to <a href="http://marcofrasca.wordpress.com/2009/08/31/the-question-of-the-arrow-of-time/">The Gauge Connection</a> for the discussion, and then come back to read my missing comment (below). You can decide for yourself who is right.</p>
<p>For more information, <a href="http://motls.blogspot.com/2009/08/arrow-of-time-understood-for-100-years.html">The Reference Frame has a good post</a> on the subject. I am also working on a series of posts that I hope will demystify the second law of thermodynamics, entropy, and the arrow of time.</p>
<p>Here is my missing comment:</p>
<blockquote><p>>> Of course, even if we are physicists (are you?),</p>
<p>Yes.</p>
<p>>> definitions are important. For one reason, when one goes to do measurements, vagueness can be of no help and our main tool is mathematics. Sloppiness should be rejected on any ground.</p>
<p>Was I vague? Sloppy? I defined exactly a test for whether or not there is an arrow of time. You can do this test in a laboratory, and you can do it in a simulation.</p>
<p>The test is experimental, not theoretical, so the only mathematics you need is the ability to count. Before we try to build a theory, we should have some measurements that we are trying to explain, no? We are talking about some physical phenomenon here, right? Or do you want to have a philosophical discussion?</p>
<p>It’s kind of funny that earlier you lectured me about what is science, and now you make this strange claim that seems to suggest mathematics comes before experiment…</p>
<p>You keep talking about the arrow of time, yet you refuse to discuss what is or is not an example of an arrow of time. You say we are drowning in definitions, so I suggest a simple test. You don’t even say whether you accept this test; you wave it away claiming it’s `vague’, while clearly the opposite is true. And yet, you refuse to provide a definition, an example, or a test of your own for what is the arrow of time… Why are you trying to keep this discussion on a philosophical level, instead of actually drilling down to the heart of the matter?</p>
<p>>> I do not need to do your simulation.</p>
<p>Again, you are ignoring my simple question. Here it is again: What is the result of the test I suggested? Will you not even grant me the answer to this simple question?</p>
<p>I am not yet trying to draw any conclusions, I’m just asking — what is the result of the experiment? What is the result of the simulation?</p>
<p>>> I take two liquids both in equilibrium</p>
<p>Well no, not really. I am taking two liquids in a specific microstate. This is not an equilibrium. I am repeating the experiment many times, but each time I am starting in a specific microstate. Not in equilibrium.</p>
<p>Talking about equilibrium is already trying to model the system using statistical mechanics. But the confusion lies in the transition from deterministic mechanics to statistical mechanics, so I am not using statistical mechanics — I’m sticking to the most basic things.</p>
<p>>> and I make them mix.</p>
<p>No. I let them evolve according to Newton’s laws. We want to see whether they mix or not, and this is why we do the experiment.</p>
<p>>> You can change these distributions as you like, making them unphysical if you want, but the problem will remain.</p>
<p>Again, what is this `problem’? All I did was suggest an experiment and a simulation. I haven’t drawn any conclusions. How can there already be a problem?</p>
<p>>> The question should be: Who puts such deterministic systems with such initial probability distributions?</p>
<p>What do you mean by `who puts’? I am trying to learn how a deterministic system evolves in time. I say: If the laws of nature are deterministic, and I do this experiment, what will be the result? I am not claiming that this describes nature.</p>
<p>>> In conclusion, you introduce an arrow of time since the start and you are a step below Boltzmann.</p>
<p>How did I introduce an arrow of time? Does determinism introduce an arrow of time? Does the distribution for the initial conditions introduce it? Please explain.</p></blockquote>
<img src="http://feeds.feedburner.com/~r/4by12/~4/vvNvs7XEqxA" height="1" width="1"/>]]></content:encoded>
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		<title>How to push GMail, Twitter to your iPhone with Prowl</title>
		<link>http://feedproxy.google.com/~r/4by12/~3/mGuVeA_KAUc/159</link>
		<comments>http://4by12.com/blog/archives/159#comments</comments>
		<pubDate>Fri, 24 Jul 2009 23:26:42 +0000</pubDate>
		<dc:creator>Guy Gur Ari</dc:creator>
				<category><![CDATA[Software]]></category>

		<guid isPermaLink="false">http://4by12.com/blog/?p=159</guid>
		<description><![CDATA[Officially, Prowl is described as a &#8216;Growl client for the iPhone&#8217;, which doesn&#8217;t mean much to most non-Mac users. Here&#8217;s an alternative description: Prowl lets you send push notifications to your iPhone. Unlike most programs that use push, Prowl doesn&#8217;t use it for a specific purpose; You can send any notification you like using their [...]]]></description>
			<content:encoded><![CDATA[<p>Officially, <a href="http://prowl.weks.net/">Prowl</a> is described as a &#8216;Growl client for the iPhone&#8217;, which doesn&#8217;t mean much to most non-Mac users. Here&#8217;s an alternative description: <em>Prowl lets you send push notifications to your iPhone</em>. Unlike most programs that use push, Prowl doesn&#8217;t use it for a specific purpose; You can send any notification you like using their <a href="http://prowl.weks.net/api.php">API</a>. I&#8217;m currently using Prowl and some Perl scripts to push new GMail messages and Twitter mentions. In this post I&#8217;ll explain how to create such a setup, but you&#8217;ll probably need some Perl skills to make it work with my scripts. It&#8217;s also easy to add other types of notifications.</p>
<p>I should mention first that there are simpler alternatives. You can use <a href="http://growlforwindows.com/gfw/">Growl</a> to push GMail, Twitter, and lots of other stuff (see their <a href="http://growlforwindows.com/gfw/apps.aspx">apps</a> page). The latest version supports <a href="http://lifehacker.com/5322057/gmail-growl-15-improves-push-notifications-for-iphone-with-prowl">true GMail push</a> using IMAP IDLE. YMMV but when I tried it Growl was unreliable and constantly lost notifications. There are also iPhone apps that push specific types of notifications; For instance <a href="http://boxcar.io/">Boxcar</a> pushes Twitter mentions and DMs. Those seem to me like a waste of money (with one small caveat &#8212; below) since Prowl can handle that and more. Also, when using Prowl it is your own computer that handles the notifications, so you don&#8217;t need to give anyone your GMail or Twitter password.</p>
<p>The only disadvantage to using Prowl is that the &#8216;Slide to View&#8217; function on the iPhone takes you to Prowl instead of to the relevant app. For example, with Boxcar, &#8216;Slide to View&#8217; takes you to your Twitter client. In my experience this is a non-issue.</p>
<p>Finally, the instructions and scripts below are provided AS IS and without any warranty. The scripts are licensed under the <a href="http://www.perl.com/pub/a/language/misc/Artistic.html">Perl license</a>.</p>
<p><H2>The Setup</H2><br />
Here is what you&#8217;ll need:</p>
<ol>
<li>An iPhone running 3.0 or higher with push notifications enabled</li>
<li>A computer that is connected to the internet 24/7, which will run the notification scripts</li>
<li>Perl installed on said computer (this is needed if you want to use my scripts)</li>
</ol>
<p>Install Prowl on your iPhone (it costs $3) and register an account. Login on <a href="http://prowl.weks.net/">their site</a> and go to the <a href="https://prowl.weks.net/settings.php">settings</a> page. Click &#8216;Generate API key&#8217; and save the API key to a file. This key is like a password you use for sending yourself notifications.</p>
<p>If you want push GMail, <a href="http://mail.google.com/support/bin/answer.py?hl=en&#038;answer=77695">enable IMAP access</a> to your GMail account.</p>
<p>Next, to run my scripts you&#8217;ll need some Perl modules that you can get from CPAN. They are:</p>
<ul>
<li><code>FindBin</code></li>
<li><code>List::Util</code></li>
<li><code>Data::Dumper</code></li>
<li><code>WebService::Prowl</code></li>
<li><code>Net::IMAP</code> (for GMail notifications)</li>
<li><code>Net::Twitter</code> (for Twitter notifications)</li>
</ul>
<p><a href='http://4by12.com/blog/wp-content/uploads/2009/07/prowl_scripts.zip'>Download the scripts</a> and insert your details:</p>
<ul>
<li>Place your Prowl API key that you created above in <code>prowl_apikey</code></li>
<li>Enter your GMail username and password in <code>push_gmail.pl</code></li>
<li>Enter your Twitter username and password in <code>push_twitter.pl</code></li>
</ul>
<p>Run <code>push_gmail.pl</code> to get GMail notifications. Run <code>push_twitter.pl</code> to get Twitter notifications. </p>
<p><H2>How It Works</H2><br />
Push GMail works using the IMAP IDLE command, which lets a client leave a connection open to the server, and when a new message comes in the server notifies the client. I used the <code>Net::IMAP</code> package, which sadly didn&#8217;t support IDLE. So I added this support in a very inelegant way &#8212; you can find it in <code>Client.pm</code>, which is a modified version of the original <code>Net/IMAP/Client.pm</code> the comes with <code>Net::IMAP</code>.</p>
<p>Push Twitter is more straightforward and works by simply polling Twitter every minute, using their API.</p>
<img src="http://feeds.feedburner.com/~r/4by12/~4/mGuVeA_KAUc" height="1" width="1"/>]]></content:encoded>
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		<title>A Short Note on the Poincaré Algebra</title>
		<link>http://feedproxy.google.com/~r/4by12/~3/Ndi_qbCFJrQ/131</link>
		<comments>http://4by12.com/blog/archives/131#comments</comments>
		<pubDate>Fri, 22 May 2009 17:03:02 +0000</pubDate>
		<dc:creator>Guy Gur Ari</dc:creator>
				<category><![CDATA[Mathematics]]></category>
		<category><![CDATA[Physics]]></category>

		<guid isPermaLink="false">http://4by12.com/blog/?p=131</guid>
		<description><![CDATA[As physicists, we learn that the Poincaré algebra has two Casimirs,  and , together describing the mass and spin of a particle. A standard question is then, &#8220;why does the algebra have two Casimirs?&#8221; and the standard answer is, &#8220;because it is a rank 2 algebra, for instance taking  as the Cartan&#8221;. Well [...]]]></description>
			<content:encoded><![CDATA[<p>As physicists, we learn that the Poincaré algebra has two Casimirs, <img src='/latexrender/pictures/ab38acaab90d464bdf88fa9c6fbfe333.png' title='p_\mu p^\mu' alt='p_\mu p^\mu' align=absmiddle> and <img src='/latexrender/pictures/59fc9404b7bfe27a4079ac851a4bdbfc.png' title='W_\mu W^\mu' alt='W_\mu W^\mu' align=absmiddle>, together describing the mass and spin of a particle. A standard question is then, &#8220;why does the algebra have two Casimirs?&#8221; and the standard answer is, &#8220;because it is a rank 2 algebra, for instance taking <img src='/latexrender/pictures/b06726df018922e6758f3214ead7da1e.png' title='\left\{ p_0,J_3 \right\}' alt='\left\{ p_0,J_3 \right\}' align=absmiddle> as the Cartan&#8221;. Well this seems wrong, since we can also take <img src='/latexrender/pictures/3e282892fd59f396c2f72049253e93a6.png' title='\left\{p_\mu\right\}_\mu' alt='\left\{p_\mu\right\}_\mu' align=absmiddle> as the Cartan, which is of dimension 4. </p>
<p>It is a standard result that any Cartan subalgebra of a (complex) semisimple Lie algebra has the same size, so what&#8217;s going on?</p>
<p>The answer is simple: Poincaré isn&#8217;t a <em>semisimple</em> Lie algebra. Therefore we have to be careful about how to define the rank. First, let&#8217;s see why Poincaré isn&#8217;t semisimple.</p>
<p><b>Definition.</b> If <img src='/latexrender/pictures/b2f5ff47436671b6e533d8dc3614845d.png' title='g' alt='g' align=absmiddle> is a complex Lie algebra, then an <em>ideal</em> in <img src='/latexrender/pictures/b2f5ff47436671b6e533d8dc3614845d.png' title='g' alt='g' align=absmiddle> is a complex subalgebra <img src='/latexrender/pictures/2510c39011c5be704182423e3a695e91.png' title='h' alt='h' align=absmiddle> of <img src='/latexrender/pictures/b2f5ff47436671b6e533d8dc3614845d.png' title='g' alt='g' align=absmiddle> such that, for all <img src='/latexrender/pictures/0f0aec368cf183b239ab385863abd4c1.png' title='X \in g' alt='X \in g' align=absmiddle> and <img src='/latexrender/pictures/d42f7caac220102a2e6a1724081e407e.png' title='H \in h' alt='H \in h' align=absmiddle>, <img src='/latexrender/pictures/ac35b9c3c2753689beabe9a005c524b4.png' title='[X,H] \in h' alt='[X,H] \in h' align=absmiddle>.</p>
<p>The brackets of a Lie algebra can be thought of as a product of two elements in that algebra. Then, an ideal (as always), is a sort of &#8216;zero&#8217;, making anything it multiplies a member of itself (just like <img src='/latexrender/pictures/7d114bf80931553e0acb29417fd29fbb.png' title='x \cdot 0 = 0' alt='x \cdot 0 = 0' align=absmiddle> for any <img src='/latexrender/pictures/9dd4e461268c8034f5c8564e155c67a6.png' title='x' alt='x' align=absmiddle>).</p>
<p><b>Definition.</b> A complex Lie algebra <img src='/latexrender/pictures/b2f5ff47436671b6e533d8dc3614845d.png' title='g' alt='g' align=absmiddle> is called <em>simple</em> if <img src='/latexrender/pictures/3c95306078da63d736c61a0819a7acbd.png' title='dim g \ge 2' alt='dim g \ge 2' align=absmiddle>, and the only ideals in <img src='/latexrender/pictures/b2f5ff47436671b6e533d8dc3614845d.png' title='g' alt='g' align=absmiddle> are <img src='/latexrender/pictures/b2f5ff47436671b6e533d8dc3614845d.png' title='g' alt='g' align=absmiddle> and <img src='/latexrender/pictures/49565e389414292f8fcf95678b9d3ab6.png' title='\left\{0\right\}' alt='\left\{0\right\}' align=absmiddle>.</p>
<p><b>Definition.</b> A complex Lie algebra is called <em>semisimple</em> if it&#8217;s (isomorphic to) a direct sum of simple Lie algebras.</p>
<p>We can now see why the Poincaré algebra isn&#8217;t simple. Translations, namely <img src='/latexrender/pictures/3e282892fd59f396c2f72049253e93a6.png' title='\left\{p_\mu\right\}_\mu' alt='\left\{p_\mu\right\}_\mu' align=absmiddle>, form a basis for an ideal: Translations commute, and the commutator of a translation with a rotation <img src='/latexrender/pictures/5fa13b39f73a0dbe5285a81e9addde23.png' title='M_{\mu \nu}' alt='M_{\mu \nu}' align=absmiddle> is a sum of translations. It is also not semisimple, which I guess can be seen by considering the <img src='/latexrender/pictures/7b7281d259c4febeb99ab5f28cc43c74.png' title='SU(2) \times SU(2)' alt='SU(2) \times SU(2)' align=absmiddle> decomposition of the Lorentz subalgebra, then adding translations which will &#8216;link&#8217; the two components.</p>
<p>The Cartan can be defined for non-semisimple algebras, and it turns out it is the Cartan of the largest semisimple subalgebra. In the case of Poincaré, the largest semisimple subalgebra is the Lorentz subalgebra. So we can still define the rank to be the size of the Cartan, with this more general definition. I don&#8217;t know if the relation between the number of Casimirs and the rank still holds in this case, but at least for the Poincaré algebra it does turn out correctly, since the rank of the Lorentz algebra is 2.</p>
<img src="http://feeds.feedburner.com/~r/4by12/~4/Ndi_qbCFJrQ" height="1" width="1"/>]]></content:encoded>
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		<title>Correct arXiv authors in Google Reader</title>
		<link>http://feedproxy.google.com/~r/4by12/~3/EcX67yF3eRo/128</link>
		<comments>http://4by12.com/blog/archives/128#comments</comments>
		<pubDate>Sun, 18 Jan 2009 20:24:37 +0000</pubDate>
		<dc:creator>Guy Gur Ari</dc:creator>
				<category><![CDATA[Physics]]></category>

		<guid isPermaLink="false">http://4by12.com/blog/?p=128</guid>
		<description><![CDATA[Google Reader has an annoying problem when viewing arXiv.org feeds: The author names get mixed up with some HTML code and become unreadable. 
I tried using Yahoo! Pipes to solve the problem, but Pipes somehow messed up the whole feed, even when it wasn&#8217;t supposed to do anything. 
Anyway, to fix this I wrote a [...]]]></description>
			<content:encoded><![CDATA[<p>Google Reader has an annoying problem when viewing <a href="http://arxiv.org/">arXiv.org</a> feeds: The author names get mixed up with some HTML code and become unreadable. </p>
<p>I tried using Yahoo! Pipes to solve the problem, but Pipes somehow messed up the whole feed, even when it wasn&#8217;t supposed to do anything. </p>
<p>Anyway, to fix this I wrote a small &#8216;feed proxy&#8217; that removes the offending HTML code. If you subscribe to it instead of the official feed, you&#8217;ll get good looking author names.</p>
<p>The feed proxy address for hep-th is:<br />
 <a href="http://4by12.com/cgi-bin/arxiv_feed.pl?feed=hep-th">http://4by12.com/cgi-bin/arxiv_feed.pl?feed=hep-th</a></p>
<p>For other feeds, replace &#8216;hep-th&#8217; by any other arXiv topic. For example, try <a href="http://4by12.com/cgi-bin/arxiv_feed.pl?feed=gr-qc">gr-qc</a> or <a href="http://4by12.com/cgi-bin/arxiv_feed.pl?feed=math-ph">math-ph</a>.</p>
<p>Disclaimer: This feed proxy is provided as-is. I can&#8217;t guarantee uptime, data integrity, or anything else regarding this service.</p>
<p>Enjoy!</p>
<img src="http://feeds.feedburner.com/~r/4by12/~4/EcX67yF3eRo" height="1" width="1"/>]]></content:encoded>
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		<item>
		<title>Meditation: A Short Tutorial</title>
		<link>http://feedproxy.google.com/~r/4by12/~3/dysouQY8CTs/122</link>
		<comments>http://4by12.com/blog/archives/122#comments</comments>
		<pubDate>Wed, 08 Oct 2008 20:00:28 +0000</pubDate>
		<dc:creator>Guy Gur Ari</dc:creator>
				<category><![CDATA[Personal]]></category>
		<category><![CDATA[Random Thoughts]]></category>

		<guid isPermaLink="false">http://4by12.com/blog/archives/122</guid>
		<description><![CDATA[Meditation is exercise for the mind. In my experience, practicing meditation for several months has the following effects:

Much improved ability to concentrate
The mind feels &#8216;lucid&#8217; and calm rather than &#8216;murky&#8217; and agitated
Improved memory (things become easier to recall)
Improved creativity
A general feeling of being one step ahead of things (chores and such) instead of one step [...]]]></description>
			<content:encoded><![CDATA[<p>Meditation is exercise for the mind. In my experience, practicing meditation for several months has the following effects:</p>
<ul>
<li>Much improved ability to concentrate</li>
<li>The mind feels &#8216;lucid&#8217; and calm rather than &#8216;murky&#8217; and agitated</li>
<li>Improved memory (things become easier to recall)</li>
<li>Improved creativity</li>
<li>A general feeling of being one step ahead of things (chores and such) instead of one step behind</li>
</ul>
<p>Meditation is tightly linked with several far-eastern religions and philosophies such as Buddhism. As can be expected, these belief systems attribute many mystical properties to meditation. Nonetheless, meditation can be successfully detached from these beliefs and used as a tool to sharpen the mind. It is just an example of a practical discovery made by religiously-minded people, and I will stick to its practical aspects.</p>
<p>There are many forms of meditation. The most commonly practiced form, which is the one I practice, involves concentrating on one&#8217;s breathing. It is called <a href="http://en.wikipedia.org/wiki/Anapanasati">Anapanasati</a>. The technique is very simple: Sit on a chair, feet straight on the floor, back straight, hands resting on lap. Set a timer to ring in say fifteen minutes. Close your eyes and mouth, and breath normally through your nose. Concentrate on the air coming in and out of your nostrils, counting each breath as you exhale. Count the breaths silently: One, two, three, four; one, two, three, four; and so on. Continue until the timer goes off.</p>
<p>Meditating is difficult because as you try to concentrate on your breaths, your mind wanders and you start thinking about other things. When this happens, you should gently yet firmly divert your attention back to the breathing, back to the air flowing in and out of your nostrils. At first this will happen a lot. After several meditations the intruding thoughts will appear less often, and there will be stretches of pure concentration without disturbances.</p>
<p>After some practice, you will come to a stage where a thought occasionally creeps into your consciousness, you note it and let it fade back to where it came from. If you think of your mind as a bucket filled with water and sand, the everyday mind is constantly stirred so the sand muddies the water. Meditation pauses the stirring, letting the sand sink down, making the water lucid.</p>
<p>Once your mind is calm enough so that you can count your breaths for fifteen or twenty minutes without losing count, you can go on to the next level and concentrate more deeply on your breathing and on other areas of your body. For example, you can gradually let your awareness cover greater areas surrounding your nose: First your nose and your mouth, then your whole face, then on to your entire body. You should do this slowly, all the while keeping your concentration and not letting your mind wander. When you reach this level you should seek further instruction, for instance by reading a book or taking a course on the subject.</p>
<p>Some general guidelines:</p>
<ul>
<li>If at any point you feel meditation is somehow bad or wrong for you &#8212; stop doing it. It&#8217;s not everybody&#8217;s cup of tea.</li>
<li>Try not to move while meditating. Don&#8217;t open your eyes or mouth. After several minutes your hands or legs may feel numb &#8212; this is normal. On the other hand, if you feel pain you should stop and change your position.</li>
<li>Meditation should be practiced at least once a day, otherwise it is not effective. At first, sit down for ten minutes at a time. When this becomes easy, go up to fifteen minutes. Then to twenty or more.</li>
<li>Do not meditate when you are tired, for example before going to sleep. You will quickly lose concentration and fall asleep. By the way, if you are having trouble sleeping, meditation can sometimes solve the problem.</li>
</ul>
<p>Here are some tips I picked up along the way:</p>
<ul>
<li>When the mind wanders, you may feel bad about failing to keep your concentration. The fact is that you should actually feel pleased, because diverting your thought back to the breathing is exactly what improves your concentration in your everyday life. It took me a long time to appreciate this.</li>
<li>Itches make for excellent practice. An itch is something concrete that tries to grab your attention. Try to resist for as long as you can and keep thinking about your breathing. Often, the itching sensation goes away after a while.</li>
<li>Pain makes for terrible practice. As explained above, if you feel pain you should act to stop it.</li>
<li>In time, this form of meditation may become boring, and you may be tempted to switch to another form or try variations of breath-counting. There&#8217;s no problem with trying out other forms <em>in addition</em> to the one you&#8217;re practicing, but it&#8217;s a huge mistake to just switch because of boredom. For a beginner, the value of meditation lies in the fact that it is hard and boring. Concentrating on something boring is exactly how you improve your concentration. So shuffling things around when you&#8217;re bored defeats the whole purpose of what you&#8217;re doing.</li>
</ul>
<p>If you want to read further, there&#8217;s a wealth of information available on the <a href="http://www.google.com/search?q=meditation">web</a> and in <a href="http://www.amazon.com/s/qid=1223495682/ref=sr_pg_3?ie=UTF8&#038;rs=&#038;keywords=meditation&#038;rh=i%3Aaps%2Ck%3Ameditation&#038;page=1">books</a>. Amazon doesn&#8217;t carry the books I learned from, but it looks like there are many other good ones.</p>
<img src="http://feeds.feedburner.com/~r/4by12/~4/dysouQY8CTs" height="1" width="1"/>]]></content:encoded>
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		<title>Complex Square Root</title>
		<link>http://feedproxy.google.com/~r/4by12/~3/gswdsiCrCH0/119</link>
		<comments>http://4by12.com/blog/archives/119#comments</comments>
		<pubDate>Wed, 13 Feb 2008 23:45:47 +0000</pubDate>
		<dc:creator>Guy Gur Ari</dc:creator>
				<category><![CDATA[Mathematics]]></category>

		<guid isPermaLink="false">http://4by12.com/blog/archives/119</guid>
		<description><![CDATA[Proof that 1 = -1:


This function should be taken outside and shot.
]]></description>
			<content:encoded><![CDATA[<p>Proof that 1 = -1:</p>
<p><center><img src='/latexrender/pictures/2413efa13bef96fa4ca860c81ee2db5c.png' title=' \frac{1}{\sqrt{i}} = \sqrt{\frac{1}{i}} = \sqrt{-i} = \sqrt{-1} \sqrt{i} = i \sqrt{i}' alt=' \frac{1}{\sqrt{i}} = \sqrt{\frac{1}{i}} = \sqrt{-i} = \sqrt{-1} \sqrt{i} = i \sqrt{i}' align=absmiddle></center><br />
<center><img src='/latexrender/pictures/6eef5215009310a2a934be514e73ccbf.png' title=' \Rightarrow 1 = i \sqrt{i} \sqrt{i} = i^2 = -1 ' alt=' \Rightarrow 1 = i \sqrt{i} \sqrt{i} = i^2 = -1 ' align=absmiddle></center></p>
<p>This function should be taken outside and shot.</p>
<img src="http://feeds.feedburner.com/~r/4by12/~4/gswdsiCrCH0" height="1" width="1"/>]]></content:encoded>
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		<title>Legendre Transform</title>
		<link>http://feedproxy.google.com/~r/4by12/~3/RYxmoX4EnIg/117</link>
		<comments>http://4by12.com/blog/archives/117#comments</comments>
		<pubDate>Sat, 05 Jan 2008 02:19:04 +0000</pubDate>
		<dc:creator>Guy Gur Ari</dc:creator>
				<category><![CDATA[Mathematics]]></category>
		<category><![CDATA[Physics]]></category>

		<guid isPermaLink="false">http://4by12.com/blog/archives/117</guid>
		<description><![CDATA[The Legendre transform is a simple and useful tool in some branches of physics such as Thermodynamics and Mechanics. In my experience though, this transform is usually explained in a very confusing way. 
I want to give my own derivation, which I hope is clear and straightforward. If you&#8217;re already familiar with the transform, you [...]]]></description>
			<content:encoded><![CDATA[<p>The Legendre transform is a simple and useful tool in some branches of physics such as Thermodynamics and Mechanics. In my experience though, this transform is usually explained in a very confusing way. </p>
<p>I want to give my own derivation, which I hope is clear and straightforward. If you&#8217;re already familiar with the transform, you can skip straight to <i>Derivation of the Legendre Transform</i>.</p>
<h2>Some Background</h2>
<p>Let&#8217;s start with the definition. If you have a function f(x), its Legendre transform is a function g(p), where p is thought of as f(x)&#8217;s derivative. g(p) is defined as:</p>
<p><center><img src='/latexrender/pictures/b658c0d439db87fed08de704df38bd59.png' title='g(p) = xp-f(x)' alt='g(p) = xp-f(x)' align=absmiddle></center></p>
<p>This definition is already confusing, because x is considered a function of p. This function is found by solving the following equation for x:</p>
<p><center><img src='/latexrender/pictures/7a338881435ffcc84748515586d1ef4f.png' title='p = \frac{df}{dx}(x)' alt='p = \frac{df}{dx}(x)' align=absmiddle></center></p>
<p>To clarify matters, let&#8217;s consider an example. Take <img src='/latexrender/pictures/d271cedde6675e55152d3c7a4236f775.png' title='f(x)=x^2' alt='f(x)=x^2' align=absmiddle>, then:</p>
<p><center><img src='/latexrender/pictures/b43bbe8703dac6d17a59bdcb2d2141b8.png' title='p=\frac{df}{dx}(x)=2x' alt='p=\frac{df}{dx}(x)=2x' align=absmiddle></center><br />
<center><img src='/latexrender/pictures/e0bd34610fdc577ac311571354bffce0.png' title='x=p/2' alt='x=p/2' align=absmiddle></center></p>
<p>The Legendre transform is then:<br />
<center><img src='/latexrender/pictures/a8bc3ada7dcf28cdb00c87ef4d4ef1c0.png' title='g(p) = x(p)p-f(x(p))=\frac{p}{2}p-(\frac{p}{2})^2=\frac{p^2}{4}' alt='g(p) = x(p)p-f(x(p))=\frac{p}{2}p-(\frac{p}{2})^2=\frac{p^2}{4}' align=absmiddle></center></p>
<p>Okay, what is it good for? The usual explanation starts with some geometric construction like the one you see on <a href="http://en.wikipedia.org/wiki/Legendre_transformation">Wikipedia</a>(*). Then, to explain why the transform works, you are shown the differential:</p>
<p><center><img src='/latexrender/pictures/e22c87959ef4568019722ad190dab438.png' title='df=\frac{df}{dx}dx=pdx' alt='df=\frac{df}{dx}dx=pdx' align=absmiddle></center><br />
<center><img src='/latexrender/pictures/5d5c6d67173005c7fa3e7b2c33fd2033.png' title='dg=d(px)-df=xdp+pdx-pdx=xdp' alt='dg=d(px)-df=xdp+pdx-pdx=xdp' align=absmiddle></center></p>
<p>And the conclusion is that g is indeed a function of p, which is f&#8217;s derivative. Mathematically speaking, this argument is quite unconvincing, because saying that g is a function of f&#8217;(x) has no meaning. g is a function of some real number, and this number has no intrinsic connection to any derivatives. </p>
<p>One possible relation could be that g and f agree if you give the corresponding arguments:</p>
<p><center><img src='/latexrender/pictures/08fc33f328166c9edf9133731fb4ef50.png' title='g(\frac{df}{dx}(x))=f(x)' alt='g(\frac{df}{dx}(x))=f(x)' align=absmiddle></center></p>
<p>but this is not the case! Some better explanation is obviously needed.</p>
<h2>Derivation of the Legendre Transform</h2>
<p>We define as before:<br />
<center><img src='/latexrender/pictures/75c78038c497b73beec28d5fb7f4bfc8.png' title='p(x)=\frac{df}{dx}(x)' alt='p(x)=\frac{df}{dx}(x)' align=absmiddle></center></p>
<p>And we are looking for a function g(p). The crucial point is this: We require that g&#8217;s derivative will correspond to x:<br />
<center><img src='/latexrender/pictures/c75ab3b8e7cc1fdc00dda6965f7313d9.png' title='x=\frac{dg}{dp}(p)' alt='x=\frac{dg}{dp}(p)' align=absmiddle></center></p>
<p>Rigorously, this means:</p>
<p><center><img src='/latexrender/pictures/eaf1fc9c5cb5dc1b2e411783c781fe1c.png' title='x=\frac{dg}{dp}(p(x))' alt='x=\frac{dg}{dp}(p(x))' align=absmiddle></center></p>
<p>Where p(x) is the function defined by the first requirement.</p>
<p>Now we have something we can use. We look at the function g(p(x)) and we calculate:</p>
<p><center><img src='/latexrender/pictures/1451fc7f5eb233435ab1a450122e8f0e.png' title='\frac{d}{dx}g(p(x))=\frac{dg}{dp}(p(x)) \cdot \frac{dp}{dx}(x) = x \cdot \frac{d^2f}{dx^2}(x) ' alt='\frac{d}{dx}g(p(x))=\frac{dg}{dp}(p(x)) \cdot \frac{dp}{dx}(x) = x \cdot \frac{d^2f}{dx^2}(x) ' align=absmiddle></center></p>
<p>Both requirements were used in the last transition. We now integrate this equation by dx:</p>
<p><center><img src='/latexrender/pictures/3a6a8467d6cba077f9d1c7c65807e1d8.png' title='\int \frac{d}{dx}g(p(x)) dx=\int x \cdot \frac{d^2f}{dx^2}(x) dx ' alt='\int \frac{d}{dx}g(p(x)) dx=\int x \cdot \frac{d^2f}{dx^2}(x) dx ' align=absmiddle></center></p>
<p>Integrating by parts on the right-hand side: </p>
<p><center><img src='/latexrender/pictures/453e22468691df320229e9ee1f251ac4.png' title='g(p(x))=x \cdot \frac{df}{dx}(x)-\int \frac{df}{dx}(x) dx = xp(x)-f(x) + C' alt='g(p(x))=x \cdot \frac{df}{dx}(x)-\int \frac{df}{dx}(x) dx = xp(x)-f(x) + C' align=absmiddle></center></p>
<p>Thus g(p) is defined up to a constant, and choosing C=0 we get the familiar Legendre transform:</p>
<p><center><img src='/latexrender/pictures/b81cb17580cf7054289d25293caaf606.png' title='g(p)=xp-f(x)' alt='g(p)=xp-f(x)' align=absmiddle></center></p>
<h2>Motivation</h2>
<p>Finally, I want to comment on this second requirement that x=g&#8217;(p), which is how I came up with this derivation. In Thermodynamics, the important parameters of a problem always come in pairs: Pressure and Volume, Temperature and Entropy, Number of particles and the Chemical constant, and so on. They are paired by the First Law, which is conservation of energy:</p>
<p><center><img src='/latexrender/pictures/c0e068a2c0d723e3fecb7e8ab4b156a8.png' title='dU = TdS-PdV + \mu dN + ~ \cdots' alt='dU = TdS-PdV + \mu dN + ~ \cdots' align=absmiddle></center></p>
<p>So here, the internal energy U is a natural function of S,V,N. Its partial derivatives are T,-P, and mu. Of course it can depend on other variables and there are other derivatives, but these are the ones that are easiest to calculate.</p>
<p>The Legendre transform is used to get new forms of energy that are naturally dependent on other parameters. This is done by switching between pairs of variables. For instance, if you know the temperature T instead of the entropy S, you can transform like this:</p>
<p><center><img src='/latexrender/pictures/4011c60e02d111b524ddbd6dd1c2c14e.png' title='F = U-TS' alt='F = U-TS' align=absmiddle></center></p>
<p>F is called the &#8216;Helmholtz Free Energy&#8217;, and its natural parameters are T,V,N: </p>
<p><center><img src='/latexrender/pictures/8227eac1256b84954a5c271ce4f05591.png' title='dF = -SdT-PdV + \mu dN + ~ \cdots' alt='dF = -SdT-PdV + \mu dN + ~ \cdots' align=absmiddle></center></p>
<p>So T, which was a derivative before, is now a parameter. But equally as important, S is now a derivative. This is very important because it allows us to calculate S very easily. If we had a new function F=F(T,V,N) that wouldn&#8217;t allow simple calculation of S, it would be useless. This is what makes the Legendre transform so useful.</p>
<p>Having said that, we now see that we can define other transforms by changing this requirement. We still get functions that can be thought of as &#8216;functions of the derivative of f&#8217;. For example, require:</p>
<p><center><img src='/latexrender/pictures/183fb4e6bba69cbe5c3bbcb15e24cad3.png' title='ax+b=\frac{dg}{dp}(p)' alt='ax+b=\frac{dg}{dp}(p)' align=absmiddle></center></p>
<p>Where a,b are some arbitrary parameters. Then we get a new transform:</p>
<p><center><img src='/latexrender/pictures/6d1a23c978b7e257c3ccbe9781febdcf.png' title='g(p) = a(xp-f(x)) + bp' alt='g(p) = a(xp-f(x)) + bp' align=absmiddle></center></p>
<p>Perhaps this method can generate some other useful transforms.</p>
<p><small>(*) I just saw that Wikipedia has something that is related to my explanation under &#8216;Another definition&#8217;, although it goes through a different route and still uses the geometric requirements.</small></p>
<img src="http://feeds.feedburner.com/~r/4by12/~4/RYxmoX4EnIg" height="1" width="1"/>]]></content:encoded>
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		<title>Relativistic Bohr Model</title>
		<link>http://feedproxy.google.com/~r/4by12/~3/LsLGDEMa1PI/112</link>
		<comments>http://4by12.com/blog/archives/112#comments</comments>
		<pubDate>Sat, 10 Nov 2007 14:20:51 +0000</pubDate>
		<dc:creator>Guy Gur Ari</dc:creator>
				<category><![CDATA[Physics]]></category>

		<guid isPermaLink="false">http://4by12.com/blog/archives/112</guid>
		<description><![CDATA[I recently needed to find the energy levels of hydrogen-like atoms (i.e. atoms with a single electron), taking into account relativistic effects. These effects become important for atoms with a large number of protons.
I only had a couple of days to do this, and I figured the easiest thing would be to create a relativistic [...]]]></description>
			<content:encoded><![CDATA[<p>I recently needed to find the energy levels of hydrogen-like atoms (i.e. atoms with a single electron), taking into account relativistic effects. These effects become important for atoms with a large number of protons.</p>
<p>I only had a couple of days to do this, and I figured the easiest thing would be to create a relativistic version of the <a href="http://en.wikipedia.org/wiki/Bohr_model">Bohr model</a>. This is a toy model that predicts the energy levels of a non-relativistic hydrogen atom.</p>
<p>The first thing I guessed was that the basic assumption from the Bohr model regarding the angular momentum still holds. That is:</p>
<p><center><img src='/latexrender/pictures/723159a5f0d5d415ac907996f9c0df56.png' title='L=n\hbar' alt='L=n\hbar' align=absmiddle></center></p>
<p>where <img src='/latexrender/pictures/7b8b965ad4bca0e41ab51de7b31363a1.png' title='n' alt='n' align=absmiddle> is some integer and <img src='/latexrender/pictures/e395749c6a6a497d729be52525d5d71d.png' title='\hbar' alt='\hbar' align=absmiddle> is <a href="http://en.wikipedia.org/wiki/Hbar">hbar</a>.</p>
<p>Also, the electron should still travel in a circular orbit, so that:</p>
<p><center><img src='/latexrender/pictures/11b59afa6ff68e70757c8892d7d5c47c.png' title='L=rp' alt='L=rp' align=absmiddle></center></p>
<p>where <img src='/latexrender/pictures/4b43b0aee35624cd95b910189b3dc231.png' title='r' alt='r' align=absmiddle> is the orbit radius and <img src='/latexrender/pictures/83878c91171338902e0fe0fb97a8c47a.png' title='p' alt='p' align=absmiddle> is the linear momentum. It is easy to show that this is still correct relativistically (*).</p>
<p>Next we have the energy, which for a free particle now obeys the dispersion relation:</p>
<p><center><img src='/latexrender/pictures/007d76885a175ed1f86c0af692380ec9.png' title='E^2 = p^2 c^2 + m^2 c^4' alt='E^2 = p^2 c^2 + m^2 c^4' align=absmiddle></center></p>
<p>Our electron is in the electric field created by the nucleus, so this should be (**) :</p>
<p><center><img src='/latexrender/pictures/18ebe7ec402ae5cd35264592e006b01d.png' title='E = \sqrt{p^2 c^2 + m^2 c^4}-\frac{Z e^2}{r}' alt='E = \sqrt{p^2 c^2 + m^2 c^4}-\frac{Z e^2}{r}' align=absmiddle></center></p>
<p>Where <img src='/latexrender/pictures/e1671797c52e15f763380b45e841ec32.png' title='e' alt='e' align=absmiddle> is the electron charge and <img src='/latexrender/pictures/6f8f57715090da2632453988d9a1501b.png' title='m' alt='m' align=absmiddle> is its rest mass.</p>
<p>Let&#8217;s find the possible orbit radii. In the classical Bohr model this is usually done by calculating the force acting on the particle. But here we already wrote down this nice expression for the energy so we might as well use it&#8230; To do that, we will venture one last guess.</p>
<p>We assume that, given angular momentum <img src='/latexrender/pictures/723159a5f0d5d415ac907996f9c0df56.png' title='L=n\hbar' alt='L=n\hbar' align=absmiddle>, the system chooses the radius where it has minimal energy.</p>
<p>You can verify for example that when you make this assumption in the non-relativistic version (instead of using forces and such), you still get the same result.</p>
<p>To get the orbit radii we then need to solve:</p>
<p><center><img src='/latexrender/pictures/ca896ec866d5132aba58a9b1562945a9.png' title='\frac{dE}{dr} = 0' alt='\frac{dE}{dr} = 0' align=absmiddle></center></p>
<p>which comes to:</p>
<p><center><img src='/latexrender/pictures/e34099c84424f1276f81fed418175c22.png' title='r = \frac{L}{mc} \sqrt{ \left( \frac{Lc}{Ze^2} \right)^2-1 }' alt='r = \frac{L}{mc} \sqrt{ \left( \frac{Lc}{Ze^2} \right)^2-1 }' align=absmiddle></center></p>
<p>Plugging this into the expression for the energy we get the atom&#8217;s energy levels:</p>
<p><center><img src='/latexrender/pictures/c7de67eba3c05dbee03dfdf2f6d21e24.png' title='E_n = mc^2 \cdot \sqrt{1-\left(\frac{Ze^2}{n \hbar c}\right)^2}' alt='E_n = mc^2 \cdot \sqrt{1-\left(\frac{Ze^2}{n \hbar c}\right)^2}' align=absmiddle></center></p>
<p>And there you have it. I have reason to believe this model works, at least partially, because I used it on measurements of large-Z atoms and it fit the data splendidly (while the non-relativistic model failed miserably). In addition, taking <img src='/latexrender/pictures/943aed88e89e4e57d4c8de06570d7f07.png' title='Z \rightarrow 0' alt='Z \rightarrow 0' align=absmiddle> recovers the non-relativistic energies, as expected.</p>
<p>Having said that, you may have noticed a problem in the last expression. For large enough Z, it is imaginary, and energy isn&#8217;t imaginary. How large a Z? Well if you&#8217;re a physicist may have also noticed that the coefficient is the <a href="http://en.wikipedia.org/wiki/Fine_structure_constant">fine structure constant</a>:</p>
<p><center><img src='/latexrender/pictures/12c8e2037c60bb06fb37eda4f6f851d8.png' title='\alpha = \frac{e^2}{\hbar c} \approx \frac{1}{137}' alt='\alpha = \frac{e^2}{\hbar c} \approx \frac{1}{137}' align=absmiddle></center></p>
<p>So for the ground state <img src='/latexrender/pictures/6d24e2bc97c5e4283dd8e34674afe7ea.png' title='n=1' alt='n=1' align=absmiddle>, the largest possible Z is about 137. Above that, according to the model, the system is unstable. </p>
<p>You might be thinking that a more plausible explanation is that I made some mistake in my series of, uhm, guesses. That could be, but a quick look in the periodic table shows that it goes all the way up to&#8230; 118.</p>
<p>Well! Could this little toy model tell us something deep about the stability of atoms? Apparently it can. I googled for <i>&#8217;stability of large z atoms&#8217;</i> and came up with <a href="http://arxiv.org/pdf/math-ph/0401004">this review</a>, where the instability is discussed. In page 10 it says: </p>
<p><small></p>
<blockquote><p>The moral to be drawn from this is that relativistic kinematics plus quan-<br />
tum mechanics is a â€˜criticalâ€™ theory (in the mathematical sense). This fact<br />
will plague any relativistic theory of electrons and the electromagnetic field<br />
â€“ primitive or sophisticated.
</p></blockquote>
<p></small></p>
<p>&#8230; we are on the &#8216;primitive&#8217; side. <img src='http://4by12.com/blog/wp-includes/images/smilies/icon_wink.gif' alt=';)' class='wp-smiley' /> </p>
<p><small><br />
(*) This can be verified by taking <img src='/latexrender/pictures/f3f16efcb37bf491633ba5e3db2df6a7.png' title='L=p_\theta' alt='L=p_\theta' align=absmiddle> and making a change of coordinates:</p>
<p><center><img src='/latexrender/pictures/ef46052982f3676cb940b70738c2b4c2.png' title='L = p_\theta = \frac{\partial x_i}{\partial \theta} p_{x_i}' alt='L = p_\theta = \frac{\partial x_i}{\partial \theta} p_{x_i}' align=absmiddle></center></p>
<p>Place the electron on the intersection of the X axis and the circular orbit to obtain the answer.</p>
<p>(**) This guess can be justified by writing down the Lagrangian for a relativistic charged particle (c=1):</p>
<p><center><img src='/latexrender/pictures/3f802745f7f97470500e3260178c8901.png' title='L =-m u_\mu u^\mu-A_\mu J^\mu' alt='L =-m u_\mu u^\mu-A_\mu J^\mu' align=absmiddle></center></p>
<p>And converting to a Hamiltonian. Taking the Hamiltonian to be the energy, one arrives at this formula.<br />
</small></p>
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		<title>Twin Paradox in a Closed Universe</title>
		<link>http://feedproxy.google.com/~r/4by12/~3/4qtB5q0FXn4/115</link>
		<comments>http://4by12.com/blog/archives/115#comments</comments>
		<pubDate>Fri, 21 Sep 2007 01:14:50 +0000</pubDate>
		<dc:creator>Guy Gur Ari</dc:creator>
				<category><![CDATA[Physics]]></category>

		<guid isPermaLink="false">http://4by12.com/blog/archives/115</guid>
		<description><![CDATA[If you&#8217;re familiar with the twin paradox, you can skip straight to the Closed Universe section.
Suppose you&#8217;re sitting in a car heading east at 100 km/h, while another car passes you by heading west at 100 km/h. According to special relativity, if you compare your clock with the clock of the passenger in the other [...]]]></description>
			<content:encoded><![CDATA[<p><i>If you&#8217;re familiar with the twin paradox, you can skip straight to the <b>Closed Universe</b> section.</i></p>
<p>Suppose you&#8217;re sitting in a car heading east at 100 km/h, while another car passes you by heading west at 100 km/h. According to special relativity, if you compare your clock with the clock of the passenger in the other car (let&#8217;s call her Alice), you will notice that her clock runs slower than yours.</p>
<p>The closer your relative speed is to the speed of light, the more noticeable the difference will be. For example, if your relative speed is 90% the speed of light, you will see Alice&#8217;s clock running about 2.3 times slower than yours. Of course, it&#8217;s not just the clock that will run slower &#8212; <i>everything</i> will run slower, including, for example, biological processes. Alice will age 2.3 times slower than you.</p>
<p>But what makes you so special? Nothing, really. Because according to Alice, it&#8217;s <i>you</i> whose aging slowly. So whose right? Well, the short answer is that you&#8217;re both right, because time isn&#8217;t an absolute thing &#8212; it depends on who&#8217;s measuring it.</p>
<p>This line of reasoning leads to the famous &#8216;twin paradox&#8217;: Two twins, Alice and Bob, are born on earth (if you must know then yes, it&#8217;s the same Alice). At birth, Alice is put aboard a spaceship that can reach extremely high speeds and sent off into space for a long trip. She reaches the edge of the galaxy, makes a U-turn, and returns to earth. This round-trip takes her 40 years in earth-time.</p>
<p>Meanwhile, Bob remains on earth. When Alice gets back, Bob is 40 years old. Because Alice traveled at high speed, Bob saw her aging very slowly, so when she gets back she is much younger than Bob, say 20 years old. But according to Alice, it&#8217;s Bob who aged slowly, so actually Alice should be older than Bob. </p>
<p>That&#8217;s the twin paradox, but it isn&#8217;t really a paradox. The problem is that in order to make this round-trip, Alice has to accelerate &#8212; she can&#8217;t go away and back again with constant speed. Therefore, the problem is no longer symmetrical: Alice is accelerating, Bob isn&#8217;t. To calculate exactly what happens in this case you need to use general relativity, but the answer is that Bob is right and Alice is wrong: Alice will have aged more slowly than Bob.</p>
<h2>Closed Universe</h2>
<p>I&#8217;ve been wondering, what happens to the twin paradox if the universe is closed? What I mean by &#8216;closed&#8217; is that if you travel in a straight line, you end up back where you started. So the universe looks like a closed loop, only in three dimensions. (*)</p>
<p>In such a universe, the Alice twin can leave earth, travel with constant speed, and return to earth by simply going straight (**). Meanwhile, the Bob twin remains on earth like before. No one is accelerating, so the usual explanation of &#8216;Alice is accelerating and that&#8217;s cheating&#8217; doesn&#8217;t apply here. Apparently, the situation is symmetrical: Bob thinks Alice is moving and aging slowly, Alice thinks Bob is. Whose right?</p>
<p>Well the answer is rather interesting. In special relativity we are used to saying that there are no &#8216;preferred frames of reference&#8217; &#8212; that all inertial observers are the same, and that there are no absolute speeds, only relative ones. It turns out that in a closed universe this is no longer the case.</p>
<p>Consider the following experiment. Bob, who is on non-moving earth, simultaneously fires two photons at opposite directions. The photons travel around the universe and return to Bob, both at exactly the same instant. </p>
<p>Now let Alice perform this experiment on her spaceship. She sends out two photons, traveling at the speed of light. Let&#8217;s look at this experiment from Bob&#8217;s frame of reference, where the photons are still traveling at the speed of light (***). Bob sees the photons going around the universe, but meanwhile Alice is moving. So one photon will reach Alice before the other. Alice must observe the same thing (although they will not agree on the timing).</p>
<p>The same experiment, conducted by Bob or by Alice, produces different results. This is because Bob&#8217;s frame is special: It has zero speed, an <i>absolute</i> speed. Thus the symmetry between Bob and Alice is broken. Further calculations are required, but it turns out that Bob is correct: Alice ages less than Bob.</p>
<p>If you want more details, I found an <a href="http://arxiv.org/abs/gr-qc/0503070">interesting paper</a> that discusses this paradox, and also develops the Lorentz transformations for such a universe.</p>
<p><small><br />
(*) The mathematical term for this is actually &#8216;compact&#8217;, but that&#8217;s because mathematicians enjoy complicating things.</p>
<p>(**) You may be wondering how it&#8217;s possible to go in a circle without accelerating. You can try looking at it this way: In circular motion with constant speed, the vector of acceleration points inward toward the center of the circle. If you live in a 2-dimensional world, you can measure this vector because it points in a direction you can move in. For instance, you can hang a ball from a piece of string and watch the string stretch.</p>
<p>But if you live in a closed, 1-dimensional world, and you&#8217;re going in a circle, where is the acceleration vector pointing? It can&#8217;t point inward, because for you there&#8217;s no &#8216;inward&#8217;. In your 1D world there&#8217;s only front and back. As a consequence, you can&#8217;t measure this acceleration experimentally. And what you can&#8217;t measure experimentally doesn&#8217;t exist.</p>
<p>By the way, in terms of differential geometry, this is what <a href="http://en.wikipedia.org/wiki/Geodesic_curvature">geodesic curvature</a> is all about: It&#8217;s the part of the acceleration that&#8217;s due to the shape of the manifold.</p>
<p>(***) This is one of the premises of special relativity &#8212; that light travels at the same speed for all observers. If you&#8217;re unconvinced it&#8217;s still true in a closed universe, consider the following. The metric of a cylinder is the same as that of a plane &#8212; a cylinder isn&#8217;t curved. Similarly, the metric of a closed, 1-dimensional universe is the same as that of an open one:</p>
<p><center><img src='/latexrender/pictures/1222c3301137e3ebf3f69fdf44ab3957.png' title='ds^2 =-dt^2 + dx^2' alt='ds^2 =-dt^2 + dx^2' align=absmiddle></center></p>
<p>The constancy of light speed follows from the metric, and from assuming that the correct transformations (boosts) preserve the metric.<br />
</small></p>
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		<title>It’s like deja vu all over again</title>
		<link>http://feedproxy.google.com/~r/4by12/~3/OguXqU1Y2Ew/114</link>
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		<pubDate>Thu, 20 Sep 2007 18:21:06 +0000</pubDate>
		<dc:creator>Guy Gur Ari</dc:creator>
				<category><![CDATA[Random Thoughts]]></category>
		<category><![CDATA[Software]]></category>

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		<description><![CDATA[6th grader raises $6.5 million in round A funding. The bubble&#8217;s back, folks.
]]></description>
			<content:encoded><![CDATA[<p>6th grader <a href="http://www.techcrunch.com/2007/09/19/playspan-takes-65m-series-a-founder-in-grade-6/">raises $6.5 million</a> in round A funding. The bubble&#8217;s back, folks.</p>
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