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	<title>Comments on: Mathematics of the Environment (Part 5)</title>
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		<title>By: Mathematics of the Environment (Part 10) &#171; Azimuth</title>
		<link>http://johncarlosbaez.wordpress.com/2012/11/06/mathematics-and-the-environment-part-5/#comment-22847</link>
		<dc:creator><![CDATA[Mathematics of the Environment (Part 10) &#171; Azimuth]]></dc:creator>
		<pubDate>Fri, 07 Dec 2012 20:44:31 +0000</pubDate>
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		<description><![CDATA[[...] Part 5 &#8211; A model showing bistability of the Earth&#8217;s climate due to the ice albedo effect: statics. [...]]]></description>
		<content:encoded><![CDATA[<p>[...] Part 5 &#8211; A model showing bistability of the Earth&#8217;s climate due to the ice albedo effect: statics. [...]</p>
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		<title>By: John Roe</title>
		<link>http://johncarlosbaez.wordpress.com/2012/11/06/mathematics-and-the-environment-part-5/#comment-22628</link>
		<dc:creator><![CDATA[John Roe]]></dc:creator>
		<pubDate>Sun, 02 Dec 2012 14:37:55 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=12855#comment-22628</guid>
		<description><![CDATA[I set up a version of this model on InsightMaker, you can find it &lt;a href=&quot;http://insightmaker.com/insight/3545&quot; rel=&quot;nofollow&quot;&gt; here.&lt;/a&gt; If you want to play around, note that A, B are as above; C is rescaled to measure time in days; G is twice gamma. Please, let me know if I messed something up.

The time scale is quicker than I had expected, this is because the heat capacity C is quite low.  In particular, the latent heat of fusion of a 1cm layer of ice appears to be of the same order of magnitude as C... one might try to incorporate this in some way, e.g. with an &quot;ice quantity&quot; variable driving c, and the &quot;ice quantity&quot; evolving according to temperature.]]></description>
		<content:encoded><![CDATA[<p>I set up a version of this model on InsightMaker, you can find it <a href="http://insightmaker.com/insight/3545" rel="nofollow"> here.</a> If you want to play around, note that A, B are as above; C is rescaled to measure time in days; G is twice gamma. Please, let me know if I messed something up.</p>
<p>The time scale is quicker than I had expected, this is because the heat capacity C is quite low.  In particular, the latent heat of fusion of a 1cm layer of ice appears to be of the same order of magnitude as C&#8230; one might try to incorporate this in some way, e.g. with an &#8220;ice quantity&#8221; variable driving c, and the &#8220;ice quantity&#8221; evolving according to temperature.</p>
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	<item>
		<title>By: gymnosperm</title>
		<link>http://johncarlosbaez.wordpress.com/2012/11/06/mathematics-and-the-environment-part-5/#comment-22583</link>
		<dc:creator><![CDATA[gymnosperm]]></dc:creator>
		<pubDate>Sat, 01 Dec 2012 03:49:05 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=12855#comment-22583</guid>
		<description><![CDATA[Very nice. When one extends the timeline back to our horizons, basically the Proterozoic for semi reliable temperature, it is clear that bistability applies not only to glacial and interglacial periods within a snowball earth, but to the alternation between snowball and greenhouse climates. At the larger scale the warm state appears more stable.]]></description>
		<content:encoded><![CDATA[<p>Very nice. When one extends the timeline back to our horizons, basically the Proterozoic for semi reliable temperature, it is clear that bistability applies not only to glacial and interglacial periods within a snowball earth, but to the alternation between snowball and greenhouse climates. At the larger scale the warm state appears more stable.</p>
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		<title>By: Mathematics of the Environment (Part 8) « Azimuth</title>
		<link>http://johncarlosbaez.wordpress.com/2012/11/06/mathematics-and-the-environment-part-5/#comment-22175</link>
		<dc:creator><![CDATA[Mathematics of the Environment (Part 8) « Azimuth]]></dc:creator>
		<pubDate>Tue, 20 Nov 2012 17:12:50 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=12855#comment-22175</guid>
		<description><![CDATA[In Part 5 and Part 6, we studied a model where the Earth can be bistable.  [...]]]></description>
		<content:encoded><![CDATA[<p>In Part 5 and Part 6, we studied a model where the Earth can be bistable.  [...]</p>
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		<title>By: Renato Iturriaga</title>
		<link>http://johncarlosbaez.wordpress.com/2012/11/06/mathematics-and-the-environment-part-5/#comment-21936</link>
		<dc:creator><![CDATA[Renato Iturriaga]]></dc:creator>
		<pubDate>Tue, 13 Nov 2012 19:28:48 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=12855#comment-21936</guid>
		<description><![CDATA[Hi John

In the homepage of Kerry Emanuel (http://eaps4.mit.edu/faculty/Emanuel/) there are a lot of things. In particular there are presentations of a  lot of courses. In one of them  (Tropical meteorology)  there is a model for the equilibrium radiation. The simplest model has two layers, the atmosphere and the surface.  The atmosphere is completely transparent to  visible light and completely opaque to infrared.  Let $latex T_e$ be the temperature of equilibrium (250 K), then the equilibrium is achieved at  at $latex T_e$ in the atmosphere and $latex 2^{\frac{1}{4}} T_e$ in the surface.  

A 3 layer model predicts $latex T_e$ in the outer part and $3^{\frac{1}{4}} T_e$ in the surface.  This is way to hot and the explanation is partially because it is not completely opaque and because of convection. Something else should be wrong because if we do the same thing with n layers, we would get a surface temperature of  $latex n^{\frac{1}{4}} T_e.$  But it is funny that the outer part is always the equilibrium temperature.  Finally there is a complete calculation, that unfortunately is not clear how it  is done, that gives a vertical profile of the temperature if only radiation is taken into account. The profile is very hot in the surface, again the explanation is the lack of the convection in the model.]]></description>
		<content:encoded><![CDATA[<p>Hi John</p>
<p>In the homepage of Kerry Emanuel (<a href="http://eaps4.mit.edu/faculty/Emanuel/" rel="nofollow">http://eaps4.mit.edu/faculty/Emanuel/</a>) there are a lot of things. In particular there are presentations of a  lot of courses. In one of them  (Tropical meteorology)  there is a model for the equilibrium radiation. The simplest model has two layers, the atmosphere and the surface.  The atmosphere is completely transparent to  visible light and completely opaque to infrared.  Let <img src='http://s0.wp.com/latex.php?latex=T_e&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='T_e' title='T_e' class='latex' /> be the temperature of equilibrium (250 K), then the equilibrium is achieved at  at <img src='http://s0.wp.com/latex.php?latex=T_e&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='T_e' title='T_e' class='latex' /> in the atmosphere and <img src='http://s0.wp.com/latex.php?latex=2%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D+T_e&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='2^{&#92;frac{1}{4}} T_e' title='2^{&#92;frac{1}{4}} T_e' class='latex' /> in the surface.  </p>
<p>A 3 layer model predicts <img src='http://s0.wp.com/latex.php?latex=T_e&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='T_e' title='T_e' class='latex' /> in the outer part and $3^{\frac{1}{4}} T_e$ in the surface.  This is way to hot and the explanation is partially because it is not completely opaque and because of convection. Something else should be wrong because if we do the same thing with n layers, we would get a surface temperature of  <img src='http://s0.wp.com/latex.php?latex=n%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D+T_e.&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n^{&#92;frac{1}{4}} T_e.' title='n^{&#92;frac{1}{4}} T_e.' class='latex' />  But it is funny that the outer part is always the equilibrium temperature.  Finally there is a complete calculation, that unfortunately is not clear how it  is done, that gives a vertical profile of the temperature if only radiation is taken into account. The profile is very hot in the surface, again the explanation is the lack of the convection in the model.</p>
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		<title>By: Mathematics and the Environment (Part 6) « Azimuth</title>
		<link>http://johncarlosbaez.wordpress.com/2012/11/06/mathematics-and-the-environment-part-5/#comment-21792</link>
		<dc:creator><![CDATA[Mathematics and the Environment (Part 6) « Azimuth]]></dc:creator>
		<pubDate>Sat, 10 Nov 2012 02:48:32 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=12855#comment-21792</guid>
		<description><![CDATA[Last time we saw a &#8216;bistable&#8217; climate model, where the temperatures compatible with a given amount of sunshine can form an S-shaped curve like this: [...]]]></description>
		<content:encoded><![CDATA[<p>Last time we saw a &#8216;bistable&#8217; climate model, where the temperatures compatible with a given amount of sunshine can form an S-shaped curve like this: [...]</p>
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		<title>By: John Baez</title>
		<link>http://johncarlosbaez.wordpress.com/2012/11/06/mathematics-and-the-environment-part-5/#comment-21772</link>
		<dc:creator><![CDATA[John Baez]]></dc:creator>
		<pubDate>Fri, 09 Nov 2012 16:38:25 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=12855#comment-21772</guid>
		<description><![CDATA[Renato wrote:

&lt;blockquote&gt;
To see the stability or instability of a particular equilibrium is convenient to look at the graph of $latex \frac {dT}{dt}$.
&lt;/blockquote&gt;

Thanks for explaining this!  I hope all the students in my class read your explanation, which is better than mine.]]></description>
		<content:encoded><![CDATA[<p>Renato wrote:</p>
<blockquote><p>
To see the stability or instability of a particular equilibrium is convenient to look at the graph of <img src='http://s0.wp.com/latex.php?latex=%5Cfrac+%7BdT%7D%7Bdt%7D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;frac {dT}{dt}' title='&#92;frac {dT}{dt}' class='latex' />.
</p></blockquote>
<p>Thanks for explaining this!  I hope all the students in my class read your explanation, which is better than mine.</p>
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		<title>By: John Baez</title>
		<link>http://johncarlosbaez.wordpress.com/2012/11/06/mathematics-and-the-environment-part-5/#comment-21771</link>
		<dc:creator><![CDATA[John Baez]]></dc:creator>
		<pubDate>Fri, 09 Nov 2012 16:34:48 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=12855#comment-21771</guid>
		<description><![CDATA[Renato wrote:

&lt;blockquote&gt;I have a couple of questions and comments. About the values of $latex A= 218 \frac{W}{m^2} $  and $latex B=1.9 \frac{W}{m^2 K} $, the value of $latex A$ is the emission of a body at -23, very close to the radiate equilibrium of -18, I guess this is not a coincidence. Or is it? The derivative of  $latex \sigma T^4$ at this value is 3.5,  B is in this order of magnitude, but not as close one could imagine.  Is there some reason for that?&lt;/blockquote&gt;

Good question!  It would be nice to understand this in detail.

The main reason the observed infrared power emission of the Earth doesn&#039;t match the power emission of a blackbody is the greenhouse effect.  Quite a bit of infrared doesn&#039;t make it through the atmosphere.   So, the simplest thing we can say is that we should have

$latex A + B T &lt; \sigma T^4 $

for temperatures in the range we see on Earth.  But it would be nice to predict the (linearized) observed average power emission per square meter, $latex A + B T,$ starting from a simple model. 

Could Simpson&#039;s model in &lt;a href=&quot;http://johncarlosbaez.wordpress.com/2012/10/13/mathematics-of-the-environment-part-3/&quot; rel=&quot;nofollow&quot;&gt;Part 3&lt;/a&gt; be good enough to approximately calculate $latex A$ and $latex B$?  Or is it too simple?

In its simplest form, this model amounts to saying that the fraction of infrared radiation that escapes to space is a function 

$latex 0 \le  F(\lambda) \le 1$ 

of the wavelength $latex \lambda$, given as follows.  To a crude approximation, the atmosphere is:

&#8226; completely opaque from 5.5 to 7 micrometers (due to water vapor),

&#8226; partly transparent from 7 to 8.5 micrometers (interpolating between opaque and transparent),

&#8226; completely transparent from 8.5 to 11 micrometers,

&#8226; partly transparent from 11 to 14 micrometers (interpolating between transparent and opaque),

&#8226; completely opaque above 14 micrometers (due to carbon dioxide and water vapor).

Simpson&#039;s function $latex F$ is 0 where the atmosphere is &#039;completely opaque&#039;, 1 where it&#039;s &#039;completely transparent&#039;, and it linearly interpolates between these values for wavelengths where the atmosphere is &#039;partly transparent&#039;.

One could multiply the Planck distribution at temperature $latex T$ by this function of wavelength, then integrate it over wavelengths, to estimate the power emitted per square meter as a function of $latex T$.

This neglects many effects like &#039;water vapor feedback&#039; (there&#039;s more water vapor when it&#039;s hotter), &#039;lapse rate feedback&#039; and &#039;cloud feedback&#039;, as mentioned in this post.  But it would still be worth doing!]]></description>
		<content:encoded><![CDATA[<p>Renato wrote:</p>
<blockquote><p>I have a couple of questions and comments. About the values of <img src='http://s0.wp.com/latex.php?latex=A%3D+218+%5Cfrac%7BW%7D%7Bm%5E2%7D+&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='A= 218 &#92;frac{W}{m^2} ' title='A= 218 &#92;frac{W}{m^2} ' class='latex' />  and <img src='http://s0.wp.com/latex.php?latex=B%3D1.9+%5Cfrac%7BW%7D%7Bm%5E2+K%7D+&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='B=1.9 &#92;frac{W}{m^2 K} ' title='B=1.9 &#92;frac{W}{m^2 K} ' class='latex' />, the value of <img src='http://s0.wp.com/latex.php?latex=A&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='A' title='A' class='latex' /> is the emission of a body at -23, very close to the radiate equilibrium of -18, I guess this is not a coincidence. Or is it? The derivative of  <img src='http://s0.wp.com/latex.php?latex=%5Csigma+T%5E4&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;sigma T^4' title='&#92;sigma T^4' class='latex' /> at this value is 3.5,  B is in this order of magnitude, but not as close one could imagine.  Is there some reason for that?</p></blockquote>
<p>Good question!  It would be nice to understand this in detail.</p>
<p>The main reason the observed infrared power emission of the Earth doesn&#8217;t match the power emission of a blackbody is the greenhouse effect.  Quite a bit of infrared doesn&#8217;t make it through the atmosphere.   So, the simplest thing we can say is that we should have</p>
<p><img src='http://s0.wp.com/latex.php?latex=A+%2B+B+T+%3C+%5Csigma+T%5E4+&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='A + B T &lt; &#92;sigma T^4 ' title='A + B T &lt; &#92;sigma T^4 ' class='latex' /></p>
<p>for temperatures in the range we see on Earth.  But it would be nice to predict the (linearized) observed average power emission per square meter, <img src='http://s0.wp.com/latex.php?latex=A+%2B+B+T%2C&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='A + B T,' title='A + B T,' class='latex' /> starting from a simple model. </p>
<p>Could Simpson&#8217;s model in <a href="http://johncarlosbaez.wordpress.com/2012/10/13/mathematics-of-the-environment-part-3/" rel="nofollow">Part 3</a> be good enough to approximately calculate <img src='http://s0.wp.com/latex.php?latex=A&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='A' title='A' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=B&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='B' title='B' class='latex' />?  Or is it too simple?</p>
<p>In its simplest form, this model amounts to saying that the fraction of infrared radiation that escapes to space is a function </p>
<p><img src='http://s0.wp.com/latex.php?latex=0+%5Cle++F%28%5Clambda%29+%5Cle+1&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='0 &#92;le  F(&#92;lambda) &#92;le 1' title='0 &#92;le  F(&#92;lambda) &#92;le 1' class='latex' /> </p>
<p>of the wavelength <img src='http://s0.wp.com/latex.php?latex=%5Clambda&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;lambda' title='&#92;lambda' class='latex' />, given as follows.  To a crude approximation, the atmosphere is:</p>
<p>&bull; completely opaque from 5.5 to 7 micrometers (due to water vapor),</p>
<p>&bull; partly transparent from 7 to 8.5 micrometers (interpolating between opaque and transparent),</p>
<p>&bull; completely transparent from 8.5 to 11 micrometers,</p>
<p>&bull; partly transparent from 11 to 14 micrometers (interpolating between transparent and opaque),</p>
<p>&bull; completely opaque above 14 micrometers (due to carbon dioxide and water vapor).</p>
<p>Simpson&#8217;s function <img src='http://s0.wp.com/latex.php?latex=F&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='F' title='F' class='latex' /> is 0 where the atmosphere is &#8216;completely opaque&#8217;, 1 where it&#8217;s &#8216;completely transparent&#8217;, and it linearly interpolates between these values for wavelengths where the atmosphere is &#8216;partly transparent&#8217;.</p>
<p>One could multiply the Planck distribution at temperature <img src='http://s0.wp.com/latex.php?latex=T&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='T' title='T' class='latex' /> by this function of wavelength, then integrate it over wavelengths, to estimate the power emitted per square meter as a function of <img src='http://s0.wp.com/latex.php?latex=T&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='T' title='T' class='latex' />.</p>
<p>This neglects many effects like &#8216;water vapor feedback&#8217; (there&#8217;s more water vapor when it&#8217;s hotter), &#8216;lapse rate feedback&#8217; and &#8216;cloud feedback&#8217;, as mentioned in this post.  But it would still be worth doing!</p>
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		<title>By: Renato Iturriaga</title>
		<link>http://johncarlosbaez.wordpress.com/2012/11/06/mathematics-and-the-environment-part-5/#comment-21687</link>
		<dc:creator><![CDATA[Renato Iturriaga]]></dc:creator>
		<pubDate>Thu, 08 Nov 2012 03:08:46 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=12855#comment-21687</guid>
		<description><![CDATA[Hi John 

I have a couple of questions and comments. About the values of $latex A= 218 \frac{W}{m^2} $  and $latex B=1.9 \frac{W}{m^2 K} $, the value of $latex A$ is the emission of a body at -23, very close to the radiate equilibrium of -18, I guess this is not a coincidence. Or is it? The derivative of  $latex \sigma T^4$ at this value is 3.5,  B is in this order of magnitude,  but not as close one could imagine.  Is there some reason for that?

To see the stability or instability of a particular equilibrium is convenient to look at the graph of $latex \frac {dT}{dt}$. In this case $latex -A-BT+Qc(T)$. In particular the zeros give the equilibrium points.  For large negative values  of  $latex T $, the function is positive, and for large positive temperatures the function is negative so there is a zero somewhere, an equilibrium. Almost surely there is an odd number of them: The first time you cross the zero (the coldest equilibrium) you cross it downwards, and if you go up  and  reach zero you will cross it upwards,  afterwards you will have to come down again.   So we have either one or three,  the only exception is if the zero is double, meaning that the derivative is also zero. The  equilibrium points where the derivative is negative, there is  local negative feedback effect: if your are slightly above the equilibrium you tend to go down if you are slightly down you tend to go up, so the point is stable, in the points where you have positive derivative -when you cross  upwards- there is a positive feedback: if you are a little bit up you tend to go upwards and if you are a little bit down  you tend to downwards, hence unstable. So if you have three points, the coldest and the hottest are stable and the middle one is unstable.]]></description>
		<content:encoded><![CDATA[<p>Hi John </p>
<p>I have a couple of questions and comments. About the values of <img src='http://s0.wp.com/latex.php?latex=A%3D+218+%5Cfrac%7BW%7D%7Bm%5E2%7D+&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='A= 218 &#92;frac{W}{m^2} ' title='A= 218 &#92;frac{W}{m^2} ' class='latex' />  and <img src='http://s0.wp.com/latex.php?latex=B%3D1.9+%5Cfrac%7BW%7D%7Bm%5E2+K%7D+&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='B=1.9 &#92;frac{W}{m^2 K} ' title='B=1.9 &#92;frac{W}{m^2 K} ' class='latex' />, the value of <img src='http://s0.wp.com/latex.php?latex=A&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='A' title='A' class='latex' /> is the emission of a body at -23, very close to the radiate equilibrium of -18, I guess this is not a coincidence. Or is it? The derivative of  <img src='http://s0.wp.com/latex.php?latex=%5Csigma+T%5E4&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;sigma T^4' title='&#92;sigma T^4' class='latex' /> at this value is 3.5,  B is in this order of magnitude,  but not as close one could imagine.  Is there some reason for that?</p>
<p>To see the stability or instability of a particular equilibrium is convenient to look at the graph of <img src='http://s0.wp.com/latex.php?latex=%5Cfrac+%7BdT%7D%7Bdt%7D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;frac {dT}{dt}' title='&#92;frac {dT}{dt}' class='latex' />. In this case <img src='http://s0.wp.com/latex.php?latex=-A-BT%2BQc%28T%29&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='-A-BT+Qc(T)' title='-A-BT+Qc(T)' class='latex' />. In particular the zeros give the equilibrium points.  For large negative values  of  <img src='http://s0.wp.com/latex.php?latex=T+&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='T ' title='T ' class='latex' />, the function is positive, and for large positive temperatures the function is negative so there is a zero somewhere, an equilibrium. Almost surely there is an odd number of them: The first time you cross the zero (the coldest equilibrium) you cross it downwards, and if you go up  and  reach zero you will cross it upwards,  afterwards you will have to come down again.   So we have either one or three,  the only exception is if the zero is double, meaning that the derivative is also zero. The  equilibrium points where the derivative is negative, there is  local negative feedback effect: if your are slightly above the equilibrium you tend to go down if you are slightly down you tend to go up, so the point is stable, in the points where you have positive derivative -when you cross  upwards- there is a positive feedback: if you are a little bit up you tend to go upwards and if you are a little bit down  you tend to downwards, hence unstable. So if you have three points, the coldest and the hottest are stable and the middle one is unstable.</p>
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		<title>By: amarashiki</title>
		<link>http://johncarlosbaez.wordpress.com/2012/11/06/mathematics-and-the-environment-part-5/#comment-21667</link>
		<dc:creator><![CDATA[amarashiki]]></dc:creator>
		<pubDate>Wed, 07 Nov 2012 20:29:08 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=12855#comment-21667</guid>
		<description><![CDATA[By the way... Zeta functions are everywhere, even they arise in graph theory, hypergraph theory and it can also appear, I presume, in network theory. How? I am not sure, since I began to learn about zeta seriously only some time ago...And it is hard to keep you updated with the zeta theory...]]></description>
		<content:encoded><![CDATA[<p>By the way&#8230; Zeta functions are everywhere, even they arise in graph theory, hypergraph theory and it can also appear, I presume, in network theory. How? I am not sure, since I began to learn about zeta seriously only some time ago&#8230;And it is hard to keep you updated with the zeta theory&#8230;</p>
]]></content:encoded>
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