# How they got this Identity

• pivoxa15
In summary, the conversation discusses the identity (dS/dV)T=(dP/dT)V and how it was derived from thermodynamics using exact differentials. The Helmholtz function and Clairaut's theorem are mentioned as important tools in the derivation process. The conversation also touches on the use of differential forms in this context.

## Homework Statement

If dU=TdS-PdV then
(dS/dV)T=(dP/dT)V

the T and V at the end means that T and V are constant
How did they get this identity? It came from a thermodynamics hence for their notations.

I have tried ways like rearranging but it dosen't seem to work. I think it has something to do with exact differentials.

Are you familiar with the Helmholtz function?

by Clairaut's theorem,
let [itex]z=z(x,y)[/tex]
$$\left ( \frac{\partial^2 z}{\partial x\partial y}\right )=\left ( \frac{\partial^2 z}{\partial y\partial x}\right )$$

or more explicitly, using "thermodynamic notations":
$$\left [\frac{\partial}{\partial y}\left ( \frac{\partial z}{\partial x}\right )_y\right ]_x=\left [\frac{\partial}{\partial x}\left ( \frac{\partial z}{\partial y}\right )_x\right ]_y$$

then look at:
$$F=U-TS$$

$$dF=-SdT-PdV$$
so F is a function of T and V, ie. [itex]F=F(T,V)[/tex] (when N is treated as a constant)

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I see. The Clairaut's theorem is a formal way of stating exact differentials isn't it?

Helmholtz function is F=U-TS and sub dU=TdS-PdV and apply Clairaut's theorem.

Using F=U-TS was the important bit. The sheet didn't mention Helmoholtz anywhere so it would have been hard to know what to use in order to derive the relation. Wouldn't you say?

not really.. hehe, looking at the derivatives you posted. One easily notice that the variables involved are V and T.

so you should know the one thermodynamic potential that is a function of V and T. (which is F)

in general, you have
U(S,V)
H(S,P)
F(T,V)
G(T,P)

H=U+PV
F=U-TS
G=U-TS+PV

so when you see an identity, check what variables it involves and use the appropriate thermodynamic functions.

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There's a neat calculation involving differential forms.
$$0=d(dU)=d(TdS-PdV)=dT \wedge dS+Td(dS) - dP\wedge dV-Pd(dV)=dT\wedge dS-dP\wedge dV$$
Then write
$$dS=\left(\frac{\partial S}{\partial T}\right)_V dT + \left(\frac{\partial S}{\partial V}\right)_T dV$$
and
$$dP=\left(\frac{\partial P}{\partial T}\right)_V dT + \left(\frac{\partial P}{\partial V}\right)_T dV$$.
Note that $$dT\wedge dT=0$$ and $$dT\wedge dV= - dV\wedge dT$$.

tim_lou said:
not really.. hehe, looking at the derivatives you posted. One easily notice that the variables involved are V and T.

so you should know the one thermodynamic potential that is a function of V and T. (which is F)

in general, you have
U(S,V)
H(S,P)
F(T,V)
G(T,P)

H=U+PV
F=U-TS
G=U-TS+PV

so when you see an identity, check what variables it involves and use the appropriate thermodynamic functions.

OK. But you could interchange these functions couldn't you to get them into different variables intermingled together.

robphy said:
There's a neat calculation involving differential forms.
$$0=d(dU)=d(TdS-PdV)=dT \wedge dS+Td(dS) - dP\wedge dV-Pd(dU)=dT\wedge dS-dP\wedge dV$$
Then write
$$dS=\left(\frac{\partial S}{\partial T}\right)_V dT + \left(\frac{\partial S}{\partial V}\right)_T dV$$
and
$$dP=\left(\frac{\partial P}{\partial T}\right)_V dT + \left(\frac{\partial P}{\partial V}\right)_T dV$$.
Note that $$dT\wedge dT=0$$ and $$dT\wedge dV= - dV\wedge dT$$.

What is the upside down V in the equations?

the notation looks insane.

just one question, why does d(dU)=0??

and I am very confused by the info from mathworld...

so
$$dT\wedge dS$$

equals what in terms of partial derivatives?

is it like a vector cross product or something like that?
suppose a=a(x,y), b=b(x,y)

what would
$$da\wedge db$$
be?

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tim_lou said:
the notation looks insane.

just one question, why does d(dU)=0??
This comes from the definition of the exterior derivative
and I am very confused by the info from mathworld...

so
$$dT\wedge dS$$

equals what in terms of partial derivatives?
Plug dS and dP into the first equation in robphy's post.
is it like a vector cross product or something like that?
suppose a=a(x,y), b=b(x,y)

what would
$$da\wedge db$$
be?

Look up the exterior derivative above.

wow.. thanks for the info. differential forms are awesome! time for me to pick up a book on such topic.