- #1
Botttom
- 15
- 0
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 [itex]G_n(H) = G_n(H=0)[/itex] and at the transition point [itex]H_c[/itex] the superconducting phase energy equals to the normal state [itex]G_s(H_c) = G_n (H_c)[/itex] 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 [itex]H_c(T)=H_c(0)(1-(\frac{T}{T_c})^2)[/itex].
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
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 [itex]G_n(H) = G_n(H=0)[/itex] and at the transition point [itex]H_c[/itex] the superconducting phase energy equals to the normal state [itex]G_s(H_c) = G_n (H_c)[/itex] 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 [itex]H_c(T)=H_c(0)(1-(\frac{T}{T_c})^2)[/itex].
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