Carsten Timm: Theory of superconductivity

Note that in two dimensions the curl is a scalar.
Now it is always possible to decompose a vector field into a rotation-free and a divergence-free component,
with
v = v ph + v v (7.18)
∇ × v ph = 0, (7.19)
∇ · v v = 0. (7.20)
Since the vortex concentation associated with v ph vanishes, the component v ph does not contain any vortices.
Alternatively, note that the first equation implies that there exists a single-valued scalar field Ω(r) so that
v ph = ∇Ω. (7.21)
Ω is a single-valued component of the phase, which cannot be due to vortices. This is the contribution from small
phase fluctuations, which we have already discussed. Conversely, v v is the vortex part, for which
∇ × v v = 2π n v , (7.22)
∇ · v v = 0. (7.23)
This already suggests an electrodynamical analogy, but to formulate this analogy it is advantageous to rescale
and rotate the field v v :
√
E(r) := − −2γ α ẑ × v v (r). (7.24)
β
} {{ }
> 0
Then the energy density far from vortex cores, where |ψ| ∼ = √ −α/β, is
w = −γ α β (∇ϕ v) · ∇ϕ v = −γ α β v v · v v = −γ α β
(
−2γ α ) −1
E · E = 1 E · E. (7.25)
β 2
Also, we find
√
∇ · E = − −2γ α √
β (−∇ × v v) = 2π −2γ α β n v (7.26)
and
∇ × E = 0. (7.27)
These equations reproduce the fundamental equations of electrostatics if we identify the “charge density” with
√
ρ v = −2γ α β n v. (7.28)
(The factor in Gauss’ law is 2π instead of 4π since we are considering a two-dimensional system.) We can now
derive the pseudo-electric field E(r) for a single vortex,
∮
√
da · E = 2πr E = 2π −2γ α (7.29)
β
√
−2γ α β
⇒ E(r) =
(7.30)
r
and thus
E(r) =
√
−2γ α β
55
r
ˆr. (7.31)