Page 402 - Electrical Properties of Materials

384 Superconductivity

According to experiment, H c varies linearly with temperature in the
neighbourhood of the critical temperature. Thus, for this temperature range
we may make eqn (14.46) agree with the experimental results by choosing
c 1 and c 2 are independent of tem-
a 1 = c 1 (T – T c ) and a 2 = c 2 . (14.47)
perature.
If you now believe that eqn (14.41) was a reasonable choice for the G s (0),
we may substitute it into eqn (14.40) to get our final form for Gibbs free
energy,

2
G s (B)= G n (0) + a 1 |ψ| + a 2 |ψ| 4
1 2 1 2
+ (∇ × A – B a ) + |i ∇ψ –2eAψ| , (14.48)
2μ 0 2m
where the relationship

B = ∇ × A (14.49)

has been used.
The arguments used above are rather difficult. They come from various
sources (thermodynamics, quantum mechanics, electrodynamics, and ac-
tual measured results on superconductors) and must be carefully combined
to give an expression for the Gibbs free energy.
6. The Gibbs free energy for the entire superconductor may be obtained by
integrating eqn (14.48) over the volume

G s (B)dV.
V
The integrand contains two undetermined functions, ψ(x, y, z) and A(x, y, z)
which, according to Ginzburg and Landau, may be obtained from the condition
that the integral should be a minimum.
The problem belongs to the realm of variational calculus. Be careful; it is
not the minimum of a function we wish to find. We want to know how A and ψ
vary as functions of the coordinates x, y, and z in order to minimize the above
definite integral.
We shall not solve the general problem here but shall restrict the solution to
the case of a half-infinite superconductor that fills the space to the right of the
x = 0 plane. We shall also assume that the applied magnetic field is in the z-
direction and is independent of the y- and z-coordinates, reducing the problem
to a one-dimensional one, where x is the only independent variable.
In view of the above assumptions,

A y is the only component of A, and dA y
B z = . (14.50)
will be simply denoted by A. dx
Since ∇ψ is a vector in the x-direction, it is perpendicular to A, so that

∗ Taking ψ real reduces the mathem- A ·∇ψ = 0. (14.51)
atical labour and, fortunately, does not
restrict the generality of the solution. Under these simplifications the integrand takes the form, ∗