The definition of specific heat at constant volume is $C_v=\frac {\delta Q} {dT}|_v$ but sometimes i find this expression instead: $C_v=\frac {\partial E} {\partial T}|_v$ . I guess that the reason why these two expression are both used is that they are equivalent, however i don't know how to de...
It is not clear where you get the last relation. If both $V$ and $M$ are present, you should specify what else is taken constant when partial derivatives are evaluated.
@GiorgioP thank you for the observation. I tried to edit the question to make it clear. About the $\delta W=-PdV+BdM$: i think the magnetic field is taken constant and also the number of particles while $M=M(V,T)$
Then, in our last case, you should be working with a $C_{v,B}= \left.\frac{\partial{E}}{\partial{T}}\right|_{v,B}$. The independent variables are $V$, $T$, and $B$.
However, I do not see why the introduction of an additional variable should solve your starting problem.
@GiorgioP I don't think i0m understanding your point, $C_v=\frac {\delta Q} {dT}|_v$ should be the formal definition of $C_v$. I'm trying to show that this is equivalent to $\frac {\partial E} {\partial T}|_v$. If the equivalence is true then when i change the variables both $\delta Q$ and $dE$ should change. Have you seen the edit to the question?
I think the point is quite simple. In a simple system, i.e. described by N,V,T variables or an equivalent set, if Delta V =0 there is no work exchanged between system and environment. Therefore Q = Delta U and that's all.
@SimoBartz Regarding your discussion above with Giorgio, specific heat is a thermodynamic coefficient that establishes "how much" heat is exchanged under given conditions. In the case of a $pVT$ system, thanks to the equation of state only two of them are independent. If you fix one of them (volume in our case), you get a coefficient. If you add an additional degree of freedom, you can't fix only volume
@Giorgio this is interesting and it would solve any doubts except i don't see how work exchange is zero if $dV=0$. Take for example the case $M=M(T,V)$, then $\delta W= BdM=B \frac {\partial M} {\partial T}|_v dT$
in the last expression i wrote i assumed v=constant
@Feynman_00 B magnetic field is fixed and M=M(T,V) so there aren't additional degree of freedom
@SimoBartz As I wrote, that is the simple system case (no magnetic variable). If you need to add magnetic or other variables, it will be necessary to introduce additional specifications for the the heat capacity. For example "at constant field" or "at constant magnetization".
@SimoBartz When you say $B$ magnetic field is fixed what you are really doing is considering a 3 variable system along paths where $dB=0$ ($B$ is the additional variable)
$\delta Q=a(T,V,B)dT+b(T,V,B)dV+c(T,V,B)dB$
This is the differential form you are considering right now. By definition specific heat is the coefficient $a$
What is $a$? The component of this differential form along $dT$. How do we get $a$ (handwavy way)? Consider $dB=0$ and $dV=0$ paths. Then $a=\delta Q/dT$ as you wish
What are we really doing? The coefficient $dT$ will have no additional contribution adding magnetic work, there is just the $\left(\frac{\partial U}{\partial T}\right)_{V, B}dT$ coming from $dU$.
You're right. The one I wrote down would be the constant magnetization (and volume) specific heat
So what I wrote above but with variables (T,V,M) and $M$ is held constant in taking the derivatives
If we were to find the constant magnetic field specific heat, I would say we also consider what you wrote
So to sum it up
$c_{V,M}=\left(\frac{\partial U}{\partial T}\right)_{V,M}$ and $c_{V,B}\left(\frac{\partial U}{\partial T}\right)_{V,B}+B\left(\frac{\partial M}{\partial T}\right)_{V,B}$
I don't know if this helps but I don't think I can help any further with this problem. I'm not comfortable with thermodynamics of magnetic systems to be honest
Discussion between SimoBartz and Gior…
Discussion between SimoBartz and Gior…