SimoBartz

@Feynman_00 thankyou, it is useful instead

this in case the variables are T,V,B

$\delta W= BdM=B \frac {\partial M} {\partial T}|_{v,B} dT$

why?

@Feynman_00 are you saying that $a=\delta Q/dT=\left(\frac{\partial U}{\partial T}\right)_{V, B}$?

today i spent hours on this and the more i think the more i'm confused. I think i'm misunderstanding something obvious

sorry for bothering, i'm honestly struggling in understanding the concept. Can you write 2 expression to give me an example?

@Feynman_00 B magnetic field is fixed and M=M(T,V) so there aren't additional degree of freedom

in the last expression i wrote i assumed v=constant

@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 $\delta W= BdM=B \frac {\partial M} {\partial T}|_v dT$

thanks

ok, i wait

how do you know that the first two terms are equivalent to $dU$ without assuming $\delta W=PdV$

you wrote $$\delta Q=\underbrace{\left(\frac{\partial U}{\partial V}\right)_TdV+\left(\frac{\partial U}{\partial T}\right)_VdT}_{dU}+p(T,V)dV$$

can i ask?

hi

@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?

@GiorgioP the problem is that the two definitions of specific heat that I wrote are different in this case

@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 $M=M(V,T)$

so yes

at the beginning is just potential energy

i think yes, but it is asking you to calculate in the moment a touch the floor

i mean, why should it stop

why should it?

hi

I'm getting crazy with this apparently stupid thing

Hi everyone, could someone give me an example of use of the Helmholtz Potential Minimum Principle to find the equilibrium state of a pure gas system?

answered

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bonjour :D

i'm from italy

it's 2 am also in my zone

yes i understand what you mean now. It should be like considering the em field as a system with its own degree of freedom coupled with the dipole

this conversation is really helpful for me, i'm an autodidact in physics

thankyou!

sense

it makes sene

ah ok

and conjugate momentum

?

you mean the space of the degrees of freedom=

configuration space?

while i should consider it something external?

if the B field rotates with the system is part of the system

if i rotate the frame of reference is like the system rotates (opposite direction)

wait maybe i get what you mean