Working with Maxwell's equations

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    Maxwell's equations
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The discussion revolves around a mathematical problem in thermodynamics, specifically demonstrating the relationship (δT/δV)s = - (δP/δS)v. The initial confusion stems from understanding the notation and the substitution process in the derivation. The user references the fundamental equation dU = TdS - PdV and compares it to the differential form involving partial derivatives. The key point is that by recognizing the relationships between temperature, pressure, and internal energy, the user is guided to the conclusion that the notation can be clarified through proper manipulation of partial derivatives. Ultimately, the solution confirms the validity of the initial equation, resolving the user's confusion.
AndrewBworth
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Hello all - I've been trying to work out an example from a book, and I don't quite understand the math.

show that (δT/δV)s = - (δP/δS)v

solution (δ/δV (δU(S,V)/δS)v)s = (δ/δS(δU(S,V)/δV)s)v
(δ/δV (δ(TdS - PdV)/δS)v)s = (δ/δS(δ(TdS-PdV)/δV)s)v
(δT/δV)s = -(δP/δS)v

I don't understand the substitution or the last step
 
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I don't understand the notation. The way I learned it is as follows:

##dU=TdS-PdV##

We must also have that:

$$dU = \left(\frac{\partial U}{\partial S}\right)_VdS+\left(\frac{\partial U}{\partial T}\right)_SdV$$

Therefore, comparing both equations, we have:

$$T=\left(\frac{\partial U}{\partial S}\right)_V$$
$$-P=\left(\frac{\partial U}{\partial T}\right)_S$$
 
Right I don't understand it either they've inserted the differential expression into a partial fraction I don't know how to work with it.
 
AndrewBworth said:
Right I don't understand it either they've inserted the differential expression into a partial fraction I don't know how to work with it.
So their notation in hinky. Can you get to the final result from my last two equations?

Chet
 
Sure no sweat.
 
Ah I figured it out
 
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