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How can I use $T=\left(\frac{\partial E}{\partial S}\right)_V$ and $P=-\left(\frac{\partial E}{\partial V}\right)_S$ to derive

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

The book (Statistical Physics by F. Mandl) suggests that $\frac{\partial^2 E}{\partial V \partial S} = \frac{\partial^2 E}{\partial S \partial V}$, but I can't make the connection as to how this helps. He calls this "cross-differentiation", i.e. to form the two derivatives given.

However, how can his suggestion be true? If we take the 2nd derivative of $E$ using the equation for $T$ and $P$ given above, then

$$\partial T=\left(\frac{\partial^2 E}{\partial S^2}\right)$$

$$\partial P=-\left(\frac{\partial^2 E}{\partial V^2}\right)$$

Qmechanic
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whatwhatwhat
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1 Answers1

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Assuming the functions are well-behaved (continuous and differentiable), you can change the order of differentiation.

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

hsnee
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