2
$\begingroup$

The Maxwell relations are often given as for example

$$\left(\frac{\partial T}{\partial V}\right)_S = -\left(\frac{\partial P}{\partial S}\right)_V.$$

What does the $S$ and the $V$ in the index of the parantheses mean? I guess that $S$ and $V$ should stay constant for the derivation, but is this not already in the definition of the partial derivative?

$\endgroup$
4
  • $\begingroup$ en.wikipedia.org/wiki/Maxwell_relations Best place to solve this doubt. $\endgroup$ Mar 12, 2018 at 9:55
  • $\begingroup$ It does not explain the notation, does it? $\endgroup$
    – Alduno
    Mar 12, 2018 at 9:59
  • $\begingroup$ The derivation part makes it pretty clear as to what the variables mean and what the notations are $\endgroup$ Mar 12, 2018 at 10:01
  • 2
    $\begingroup$ There are many more than two parameters used in thermodynamics. This just makes clear exactly which ones are being held constant. $\endgroup$ Mar 12, 2018 at 12:53

3 Answers 3

5
$\begingroup$

Basically in Thermodynamics different functions get the same name if they refer to the same quantity. So, for example, the inner energy $U(p,V,N)$ and $U(T,V,N)$ both are called $U$ although they are not the same function.

To my understanding this is why you write the constant variables next to the brackets. It's used to further differentiate between the different functions.

$\endgroup$
0
4
$\begingroup$

Your system has two degrees of freedom. So any of your quantities $V$, $E$, $P$, $T$, $S$ can be viewed as a function of any two of the others. The expression $$\left(\frac{\partial T}{\partial V}\right)_S$$ means "the derivative of $T$ with respect to $V$ when viewing it as a function of $V$ and $S$ (i.e. $T(V,S)$)". Likewise $$\left(\frac{\partial P}{\partial S}\right)_V$$ is the derivative of $P(S,V)$ with respect to $S$.

$\endgroup$
0
$\begingroup$

This is just a common notation for partial derivatives. The index means that you take the partial derivative while holding the index variable (of the parenthesis) constant.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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