Gibbs and Helmholtz energies of a superconductor

[Botttom]

Hello,
I consider an ideal superconductor with the gibbs-energy $$ d G=-SdT + VdP - \mu_0 M V dH$$
and helmholtz energy $$ dF = -SdT -P dV + \mu_0 V H dM$$
Assuming, that in the normal state the magnetization is too small, so that $G_n(H) = G_n(H=0)$ and at the transition point $H_c$ the superconducting phase energy equals to the normal state $G_s(H_c) = G_n (H_c)$ I get a continuous gibbs energy function with the superconducting-normal- states energy differency $$G_n (T)-G_s(T) =\frac{1}{2} H^2_c(T) V$$, when $H_c(T)=H_c(0)(1-(\frac{T}{T_c})^2)$.

Why one cannot use the helmholtz energy for the same calculation of the energy differences and would the function of the helmholtz energy would be continuous as well?

Thanks

 

[DrDu]

I suppose you could use Helmholtz energy, too, but which experimental setup would correspond to this situation? You assumed that the magnetisation in the normal state is very small, so if you want to keep M constant, it would have to be so in the superconducting phase, too. So basically you want to discuss the field free case.

 

[Botttom]

No, i just assume that the magnetization is small enough in the normal state, because the sample is not diamagnetic in normal state. The diamagnetism is only seen in the superconducting state with $$M=H,$$ and hence can not be assumed to be small enough to be neglected

 

[DrDu]

I understand this. But the Gibbs potential is so useful in this case because H is not changing on the transition, in contrast to M.

 

[Botttom]

But the helmholtz energy should still be continuous at $T_c$ like the gibbs energy, right?

 

[DrDu]

I don't think so. The two differ by a term MH. But below Tc, M=H, while below M=0. So if one is continuous, the other one jumps by $H^2$.

 

[DrDu]

Just realsed that at Tc H=0, so both are continuous.

 

[Botttom]

Ok, thanks