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enter image description here

I feel I am missing something about deriving Maxwell relations. I have read http://ocw.mit.edu/courses/physics/8-044-statistical-physics-i-spring-2013/readings-notes-slides/MIT8_044S13_notes.Max.pdf and as far as I can see if we have $dE=TdS+FdL$, the only maxwell relation will be

$$\frac{\partial F}{\partial S}=\frac{\partial T}{\partial L} $$

So how is the given maxwell relation in the picture found?

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As you said, the relation you have is $$ dE = TdS + FdL $$ So one maxwell relation is $$ \left(\frac{\partial T}{\partial L}\right)_S = \left(\frac{\partial F}{\partial S}\right)_L $$ Which is the one you have obtained. For get the other do the following transformation $$ dE = TdS + FdL = d(ST) - SdT + FdL $$ $$ \implies d(E-ST) = -SdT + FdL $$ Since $-SdT + FdL$ is an exact differential, the cross derivatives coincides, so you can read the other maxwell relation $$ -\left(\frac{\partial S}{\partial L} \right)_T = \left(\frac{\partial F}{\partial T} \right)_L $$ Which is the one on your textbook.

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Specifically the result you have is $$\left(\frac{\partial F}{\partial S}\right)_L=\left(\frac{\partial T}{\partial L}\right)_S$$

So using the cyclic rule for partial derivatives we can write that $$ \left(\frac{\partial T}{\partial L}\right)_S=-\left(\frac{\partial T}{\partial S}\right)_L \left(\frac{\partial S}{\partial L}\right)_T$$

So if we sub this in and multiply both sides by $\left(\frac{\partial S}{\partial T}\right)_L$ we have $$\left(\frac{\partial F}{\partial S}\right)_L \left(\frac{\partial S}{\partial T}\right)_L=-\left(\frac{\partial T}{\partial S}\right)_L \left(\frac{\partial S}{\partial T}\right)_L \left(\frac{\partial S}{\partial L}\right)_T$$

and since $$\left(\frac{\partial F}{\partial S}\right)_L \left(\frac{\partial S}{\partial T}\right)_L=\left(\frac{\partial F}{\partial T}\right)_L$$ and $$\left(\frac{\partial T}{\partial S}\right)_L \left(\frac{\partial S}{\partial T}\right)_L=1$$

you get the required result

$$\left(\frac{\partial F}{\partial T}\right)_L=-\left(\frac{\partial S}{\partial L}\right)_T$$

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  • $\begingroup$ @1234 I think there is probably a faster way to do this if the question says we can "read off" the answer but it isn't obvious to me. $\endgroup$
    – ChrisM
    Feb 20, 2015 at 22:28
  • $\begingroup$ i made an answer with a faster way to read de answer :), hope it helps. $\endgroup$
    – Héctor
    Feb 21, 2015 at 0:01

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