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	<title>4by12</title>
	
	<link>http://4by12.com/blog</link>
	<description>by Guy Gur-Ari</description>
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		<title>Gmail-Like Thread View in the Mac Mail App</title>
		<link>http://feedproxy.google.com/~r/4by12/~3/oxrotWy3JvQ/307</link>
		<comments>http://4by12.com/blog/archives/307#comments</comments>
		<pubDate>Sat, 21 Aug 2010 00:01:49 +0000</pubDate>
		<dc:creator>Guy Gur Ari</dc:creator>
				<category><![CDATA[Apple]]></category>

		<guid isPermaLink="false">http://4by12.com/blog/?p=307</guid>
		<description><![CDATA[While Mail supports threaded conversation view, it shows only the received messages of a given inbox thread. So if you have a back-and-forth with someone, you only see his side of the conversation. Gmail handles this better by including the relevant sent messages in the thread. Others have found this annoying, and none of the [...]]]></description>
			<content:encoded><![CDATA[<p>While Mail supports threaded conversation view, it shows only the received messages of a given inbox thread. So if you have a back-and-forth with someone, you only see his side of the conversation. Gmail handles this better by including the relevant sent messages in the thread. Others have <a href="http://forums.macrumors.com/showthread.php?t=913013">found</a> <a href="http://forums.macrumors.com/showthread.php?t=498578">this</a> <a href="http://smokingapples.com/software/tips/leopard-tips-coversations-mail/">annoying</a>, and none of the suggested workarounds I found worked well.</p>
<p>Here&#8217;s how to get Gmail-like threads in your Mail inbox that include both sides of the conversation. The instructions refer to Mail version 4.3.</p>
<p>1. Enable threaded view: Go to your inbox and click View, Organize by Thread. If you like, you can change the color of threads under Preferences, Viewing, Message threading.</p>
<p>2. Go to Preferences, Composing, and check &#8220;Automatically Bcc: myself&#8221;. When you reply to someone, you will get a copy of your reply in your inbox, and it will be added to the conversation thread.</p>
<p><a href="http://4by12.com/blog/wp-content/uploads/2010/08/Mail-Bcc-Myself.png"><img src="http://4by12.com/blog/wp-content/uploads/2010/08/Mail-Bcc-Myself.png" alt="" title="Mail - Bcc Myself" width="630" height="678" class="alignnone size-full wp-image-309" /></a></p>
<p>3. The problem now is that, when you send a new message (rather than a reply) you will also get a copy of that message in your inbox. To get rid of those messages, add a rule that deletes a message coming from you if it doesn&#8217;t refer to an existing message (replace steve@apple.com with your email address),</p>
<p><a href="http://4by12.com/blog/wp-content/uploads/2010/08/Mail-Delete-non-re-or-fwd-copies.png"><img src="http://4by12.com/blog/wp-content/uploads/2010/08/Mail-Delete-non-re-or-fwd-copies.png" alt="" title="Mail - Delete non re or fwd copies" width="570" height="260" class="alignnone size-full wp-image-339" /></a></p>
<p>To get that &#8216;References&#8217; field in the second row, choose &#8216;Edit Header List&#8230;&#8217; in the drop-down menu and add &#8216;References&#8217; to the list.</p>
<p><a href="http://4by12.com/blog/wp-content/uploads/2010/08/Mail-Edit-Header-List.png"><img src="http://4by12.com/blog/wp-content/uploads/2010/08/Mail-Edit-Header-List-150x150.png" alt="" title="Mail - Edit Header List" width="150" height="150" class="alignnone size-thumbnail wp-image-340" /></a></p>
<p>4. Finally, add a rule that marks as read all messages coming from your email address. This will keep the reply/forward Bcc copies from showing up as unread emails.</p>
<p><a href="http://4by12.com/blog/wp-content/uploads/2010/08/Mail-Mark-self-mail-as-read.png"><img src="http://4by12.com/blog/wp-content/uploads/2010/08/Mail-Mark-self-mail-as-read.png" alt="" title="Mail - Mark self mail as read" width="570" height="232" class="alignnone size-full wp-image-310" /></a></p>
<p>That&#8217;s it.</p>
<p>If you&#8217;re syncing Mail with a Gmail account, don&#8217;t worry about the extra Bcc copies cluttering your Gmail threads. Gmail somehow knows they&#8217;re dups and handles them correctly.</p>
<p>Enjoy your threads!</p>
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		<title>The Windows Experience – An Anecdote</title>
		<link>http://feedproxy.google.com/~r/4by12/~3/xvEKNg0aUuw/279</link>
		<comments>http://4by12.com/blog/archives/279#comments</comments>
		<pubDate>Sat, 14 Aug 2010 21:53:54 +0000</pubDate>
		<dc:creator>Guy Gur Ari</dc:creator>
				<category><![CDATA[Microsoft]]></category>

		<guid isPermaLink="false">http://4by12.com/blog/?p=279</guid>
		<description><![CDATA[My niece came over for a visit, toting an HP Pavilion. Her laptop couldn&#8217;t connect to the wifi at my house, or at my parents&#8217; house. The dialog box cited an &#8216;unknown reason&#8217; as the reason for this problem. She was about to spend a week at my parents&#8217;, so this was kind of annoying. [...]]]></description>
			<content:encoded><![CDATA[<p>My niece came over for a visit, toting an HP Pavilion. Her laptop couldn&#8217;t connect to the wifi at my house, or at my parents&#8217; house. The dialog box cited an &#8216;unknown reason&#8217; as the reason for this problem. She was about to spend a week at my parents&#8217;, so this was kind of annoying.</p>
<p>The network in question generally works well, accommodating other laptops with no problem. I vaguely recalled that Windows has trouble handling WPA-PSK, the authentication method I&#8217;m using. Now, WPA-PSK is not a particularly new technology, and the laptop in question isn&#8217;t very old (it runs Vista), but I figured it was worth a shot.</p>
<p>I went to HP&#8217;s site (on a wired connection &#8211; thank God for those) to look for updated drivers. I was greeted by a large banner, splattered over half the screen, asking me to participate in an HP survey. No, thanks. (Having just entered the site, do I really have anything of value to say?)</p>
<p>Searching for &#8216;HP Pavilion Wifi Drivers&#8217; was no help. It wasn&#8217;t `specific enough&#8217;. I had to enter the full model number from the back of the laptop, something like &#8216;d5v fgh123&#8242;, to get anywhere. I doubt most users would be able to find the correct number to copy off the laptop, but then again, most users wouldn&#8217;t get this far anyway. So I guess it&#8217;s reasonable in some backward sort of way. </p>
<p>Or at least it would be reasonable if this query led to the correct drivers. It did not. I was presented with three options for network drivers: One for the wired card (despite having searched for &#8216;wifi&#8217;), and two for wireless cards: Intel/PRO and Broadcom. Wait, what? Even given the exact model number, HP <i>still</i> doesn&#8217;t know which driver I need? Ugh.</p>
<p>Oh well, right-clicking on My Computer -> Properties -> Device Manager -> Network Adapters (is this <i>still</i> the best way to get this info? seriously?) showed a Broadcom adapter. Another survey request later, and the Broadcom driver was downloading. Double-click, next, next, next, and&#8230; nothing is installed. Again&#8230; double-click, next, next&#8230; oh I see, it just extracts the files to some arcanely-named directory. Open arcanely-named directory, double-click Setup.exe, next, next, next, &#8216;drivers successfully installed&#8217;. A restart for good measure, connect to wireless network, problem solved. Yey.</p>
<p>I didn&#8217;t even mention Norton Antivirus bugging me every time the computer started; the LCD screen being configured to a lower-than-native resolution (by default no doubt), leading to blurry text; the horrendously tiny trackpad; the awful build quality of the whole thing, etc. In other words, more or less the same experience I&#8217;m familiar with from every Windows laptop I ever used.</p>
<p>In this particular case, most of the blame goes to HP. But that&#8217;s just a coincidence. This level of incompetence is run-of-the-mill in the PC world. Like the build quality and the trackpad size, it&#8217;s the sort of thing that doesn&#8217;t come up when you compare PCs to Macs using just the hardware specs. As always, God is in the details.</p>
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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 [...]]]></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>
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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>
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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 this seems [...]]]></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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		<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 [...]]]></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>
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		<pubDate>Sat, 10 Nov 2007 14:20:51 +0000</pubDate>
		<dc:creator>Guy Gur Ari</dc:creator>
				<category><![CDATA[Physics]]></category>

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		<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 [...]]]></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>&#8216;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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