1
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

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

$\endgroup$
2
  • $\begingroup$ @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. $\endgroup$ – Chris2807 Feb 20 '15 at 22:28
  • $\begingroup$ i made an answer with a faster way to read de answer :), hope it helps. $\endgroup$ – Héctor Feb 21 '15 at 0:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.