Please check the image provided. If somebody can explain why the last two terms were considered and how it happened to be so. According to some text books, its because of expulsion of applied field by superconductor. But how to figure it out.
Note: - CGS units has been used
Source: V. V. Schmidt, The physics of superconductors.
$\begingroup$
$\endgroup$
1
-
$\begingroup$ Consider to spell out acronyms in title. $\endgroup$– Qmechanic ♦Commented Aug 10, 2018 at 12:59
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
The last two terms are simply the energy stored in the microscopic magnetic field. In CGS units a magnetic field $B$ has energy density $B^2/8\pi$. If $H$ is the total magnetic field at the point of interest and $H_0$ an applied external field, then the magnetic field due to the superconductor is $H-H_0$. Therefore the magnetic energy density is $(H-H_0)^2/8\pi$, which reproduces the last two terms upon neglecting the constant term proportional to $H_0^2$.
-
$\begingroup$ Thank you sir. But as you said the term proportional to Ho^2 should be neglected. But on the LHS of the equation, we have free energy in superconducting state due to applied field H. How can we neglect the constant term. $\endgroup$ Commented Aug 14, 2018 at 6:07
-
1$\begingroup$ We're only interested in the thermodynamics of the superconductor. The applied field $H_0$ is static and totally independent of any thermodynamic variables, so it can be treated as a constant. Adding a constant term to your free energy doesn't change any observable quantities in the same way that adding a constant term to a Lagrangian or Hamiltonian doesn't affect the equations of motion. $\endgroup$ Commented Aug 14, 2018 at 14:26