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I just had a glimpse of Feynman's statistical mechanics. On page 8, I saw a 'third' definition of pressure:

$$ P = - \frac{\partial U }{\partial V} (V, S) . $$

I am curious why it is $S$ but not $T$. Could anyone give an argument? At least personally, I tend to treat energy $U$ as a function of $V$ and $T$ instead of $V$ and $S$.

The question is, why is it not

$$ P = - \frac{\partial U }{\partial V} (V, T) $$

poisson
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2 Answers2

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While you can choose any two independent variables as your basis function to express any quantity, you need to be careful in the choice of your variables when defining new quantities as derivatives. Depending which variable you fix, you will get another result.

In your case, you can define $P$ as: $$ P = -\left(\frac{\partial U}{\partial V}\right)_S $$ but then the other quantity is: $$ \begin{align} \tilde P &:= -\left(\frac{\partial U}{\partial V}\right)_T\\ &= P-\left(\frac{\partial U}{\partial S}\right)_V\left(\frac{\partial S}{\partial V}\right)_T \\ &= P-T\left(\frac{\partial P}{\partial T}\right)_V \end{align} $$ As you can see the two quantities are different. The issue now is to know why the first definition is the correct one for pressure. This comes from the definition of pressure work: $$ \delta W = -PdV $$ which is related to energy increments for adiabatic processes: $$ dU = \delta W $$ An adiabatic reversible process is isentropic, so $S$ is the quantity to keep constant, not $T$.

Hope this helps.

LPZ
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After some thought, I think why $S$ should be kept constant can be understood as follows.

Microscopically, the pressure exerted by an eigenstate of the system on the wall is

$$ P_n = - \frac{\partial E_n }{\partial V }, $$

where $E_n$ is the eigenenergy, which generally depends on the volumn $V$.

Now the macroscopic pressure should be an average of the individial pressures, that is,

$$ P = \sum_n p_n P_n = -\sum_n p_n \frac{\partial E_n}{\partial V } \neq -\frac{\partial U}{\partial V}= -\frac{\partial }{\partial V }(\sum_n p_n E_n ),$$

where $p_n (V, T)$ is the probabality of the system being in the $n$th state. The point is that, we do not have the term $\partial p_n/\partial V $.

Now we know the entropy is a function of the distribution of $p_n$. To keep $p_n$ constant means to keep $S$ constant.

poisson
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