Undergrad How is the derivative of an inexact differential defined?

[spin_100]

This is from Callen's thermodynamics. What does the differentiation with respect to T means for an inexact differential like dQ. Also why is T treated as a constant if we start by replacing dQ by TdS? Any references to the relevant mathematics will be much appreciated.

 

[Demystifier]

https://en.wikipedia.org/wiki/Inexact_differential

 

[gleem]

This is from Callen's thermodynamics. What does the differentiation with respect to T means for an inexact differential like dQ.

I am a little rusty but I think that this should not be interpreted as a differentiation but as a ratio of two incremental changes in values. In this case the change in the heat of a system d'Q ^* that occurs when the temperature increases by dT at a temperature of T.
\frac{d'Q}{dt }\neq \frac{d}{dt}\left ( Q \right )


^* Thermodynamics by Sears always distinguishes inexact differentials with a prime superscript on d indicating a small change in the value of the quantity..

 

[Demystifier]

this should not be interpreted as a differentiation but as a ratio of two incremental changes in values.

Exactly.

 

[Demystifier]

Also why is T treated as a constant if we start by replacing dQ by TdS?

The formula $d'Q=TdS$ does not imply that $T$ is constant.

For analogy, consider classical mechanics of a particle moving in one dimension. The infinitesimal path $dx$ during the time $dt$ is $dx=vdt$, but it does not imply that the velocity $v(t)$ is constant. Instead, it means that $v(t)$ is defined as
$$v(t)=\frac{dx(t)}{dt}$$
which physicists write in the infinitesimal form $dx=vdt$.

Indeed, similarly to $v$, the $T$ can also be thought of as defined by a derivative formula. But it is not $T=d'Q/dS$ or $T=dQ/dS$, because such things are not defined as derivatives. Instead, starting from the 1st law of thermodynamics
$$dU=TdS-PdV$$
we see that $U$ must be a function of $S$ and $V$, because then we have the mathematical identity
$$dU(S,V)=\left( \frac{\partial U(S,V)}{\partial S} \right)_V dS +
\left( \frac{\partial U(S,V)}{\partial V} \right)_S dV$$
which is compatible with the 1st law above. The compatibility implies that $T$ can be defined as
$$T(S,V)=\left( \frac{\partial U(S,V)}{\partial S} \right)_V$$
which is a function of $S$ and $V$, not a constant.