3 Macroscopic / Phenomenological Theories of Superconductivity

After the discovery of the Meissner effect, numerous theories of superconductivity began to be developed. Some of these theories agreed closely with some experimental data, others less so. ``The assumptions made in some models were difficult, if not impossible, to justify physically'' (Parks: 1969, p.27). The 1930's to the 1950's was a period of experimentation for theories of superconductivity. While many approaches had modest success, none of the theories could explain what was happening inside a superconductor at the microscopic level, nor could any one theory explain both the thermodynamic and electrodynamic properties of a superconductor. Below is a quick look at some of the more successful theories.

3.1 Thermodynamic Theories⁷

After the discovery of the Meissner effect, superconductivity could be viewed as an equilibrium thermodynamic problem. This approach was used to attempt to explain numerous experimental results ranging from the $T^{3}$ dependence of the specific heat to the parabolic curve of the critical field versus temperature (equation 1). Below are a few of these derivations to show how sometimes relatively simple thermodynamic arguments can yield successful, measurable predictions.

3.1.1 Gibbs Free Energy

The difference in Gibbs free energy determines whether a given reaction will occur spontaneously forwards or backwards; states in equilibrium have the same Gibbs free energy because both directions are equally favored. In order to calculate observable thermodynamic properties, we must first derive the change in Gibbs free energy when a metal becomes superconducting with no external magnetic field. Treating superconductivity as a change of state, then equation (1) is the boundary between the two states and points on that curve (when magnetic field, $H=H_{c}$) are in equilibrium and have the same Gibbs free energy, therefore $G_{n}(T)=G_{s}(T,H_{c})$. Solving for the difference in Gibbs free energy between the normal state and the superconducting state with no magnetic field yields

$$\begin{array}{ll} \Delta G & \equiv G_{n}(T)-G_{s}(T,0)\\ & \\ & =G_{s}(T,H_{c})-G_{s}(T,0).\\ \end{array}$$ (2)

To calculate the Gibbs free energy in the superconducting state, we need to calculate the work done by changing the magnetic field - magnetic work. Consider a cylinder with length L, current I, cross-sectional area A, and n turns per unit length. The magnetic intensity, $\mathcal{H}$, of a cylinder is

$$\mathcal{H}=nI.$$ (3)

By Faraday's law, the induced Emf is

$$\mathcal{E}=nLA\, \frac{dH}{dt}.$$ (4)

Because power equals current times voltage,

$$dW = InLA\frac{dH}{dt}\, dt = InV\, dH$$

$$= \mathcal{H}V\, dH,$$ (5)

where V is the volume of the cylinder. The above equation is not specific to a cylinder, but is in general magnetic work. When the work done is from the expansion or contraction of a gas, where $dW=PdV$, the Gibbs free energy is

$$dG=Vdp-SdT.$$ (6)

From equation (5), substitute $\mathcal{H}\, V$ for $P$ and $H$ for $V$ in the above equation to get the Gibbs free energy in terms of magnetic work,

$$dG=H\, d\mathcal{H}V-SdT.$$ (7)

Since temperature is constant in equation (2), volume (V) is constant, and the magnetic field $H$ equals $\mathcal{H}\mu _{0},$

$$dG=\frac{HV}{\mu _{0}}\, dH.$$ (8)

The change in Gibbs free energy from the normal state to the superconducting state at constant temperature is,

$$\Delta G=\frac{V}{\mu _{0}}\int _{0}^{H_{c}}HdH$$ (9)

$$\Delta G=\frac{VH_{c}^{2}}{2\mu _{0}}.$$ (10)

3.1.2 Entropy and Specific Heat

Once $\Delta G$ is known, calculating the change in entropy, specific heat and other thermodynamic properties becomes relatively easy. Since $G=E+PV-TS$, $\Delta S=-\frac{\partial }{\partial T}\Delta G$. Remembering that $H_{c}$ is temperature dependent, we find

$$\Delta S=-\frac{V}{\mu _{0}}H_{c}\frac{\partial H_{C}}{\partial T}.$$ (11)

The latent heat associated with a phase transition is $T\Delta S$, so

$$\textrm{Latent Heat }=\frac{V}{\mu _{0}}TH_{c}\frac{\partial H_{c}}{\partial T}.$$ (12)

Since the specific heat (C), equals $\frac{\partial Q}{\partial T}=\frac{T\partial S}{\partial T}$, by the product rule,

$$\Delta C(T) = T\frac{\partial }{\partial T}\Delta S$$

$$= -\mu _{0}VT\left[\left(\frac{\partial H_{C}}{\partial T}\right)^{2}+H_{c}\left(\frac{\partial ^{2}H_{C}}{\partial T^{2}}\right)\right].$$ (13)

Assuming from experimental evidence (or derivations not covered here) that $\Delta C\propto T^{3}$, then both terms inside the brackets in the above equation must be proportional to $T^{2}$, forcing $H_{c}(T)$ to be parabolic, which agrees with the experimentally determined equation (1).

Remembering that $H_{c}=0$ at $T=T_{c}$, we see that the discontinuous jump in the specific heat at $T=T_{c}$ is

$$\Delta C(T_{c})=-\mu _{0}VT_{c}\left(\frac{\partial H_{C}}{\partial T}\right)_{T=T_{c}}^{2}$$ (14)

From relatively basic thermodynamic arguments we have predicted the discontinuous jump in the specific heat and the parabolic shape of $H_{c}(T)$ and calculated the latent heat and change in entropy associated with the superconducting transition with no external magnetic field. In general these and other thermodynamic relationships fit the basic shapes of experimental data, but more precise data often shows consistent deviations between theory and experiment, particularly with certain superconductors.

3.2 The London-London-Pippard Theory⁸

The London brothers - Fritz and Heinz - proposed in 1935 two electrodynamic equations to account for a superconductor's perfect diamagnetism and conductivity. The London equations gave predictions generally corresponding with experimental data, but detailed analysis showed serious discrepancies. In 1953, Pippard discovered that dissolving 3% indium in tin doubled the penetration depth while leaving electrodynamic and thermodynamic quantities basically unchanged, demonstrating an inadequacy with thermodynamic theories and with the Londons' original formulation. Pippard modified the London equations to include a `coherence length' parameter, which represents the length of some long range interaction within the metal lattice. Solving the London equations or detailing Pippard's modifications is beyond the scope of this paper, but below is a simplified derivation of the two London equations and the penetration depth.

Starting with Newton's famous equation $\mathbf{F}=m\mathbf{a}$, we can say that in a superconductor the only force is from the electric field since there is no energy dissipation, so $\mathbf{F}=q\mathbf{E}=-e\mathbf{E}.$

$$\mathbf{F} = m\mathbf{a}$$

$$-e\mathbf{E} = m\frac{d\mathbf{v}}{dt}$$ (15)

The current density $(\mathbf{j}_{s})$ is equal to the charge per electron times the density of electrons $(n_{s})$ times the velocity of the electrons $(\mathbf{v})$:

$$\mathbf{j}_{s}=-en_{s}\mathbf{v}.$$ (16)

Taking the time derivative of the above equation and substituting in equation (15) yields

$$\frac{\partial \mathbf{j}_{s}}{\partial t}=\frac{e^{2}n_{s}}{m}\mathbf{E}=\frac{1}{\Lambda _{L}}\mathbf{E},$$ (17)

where we define $\Lambda _{L}\equiv m/e^{2}n_{s}$. Equation (17) is the first London equation. From Maxwell's equations we know that $\nabla \times \mathbf{E}=-\dot{\mathbf{B}}$, therefore

$$\frac{\partial }{\partial t}\left(\nabla \times \Lambda _{L}\mathbf{j}_{s}\right)=-\dot{\mathbf{B}.}$$ (18)

Integrating both sides and setting the integration constant to zero, so that there is no magnetic field produced when the current is zero, yields

$$\nabla \times \mathbf{j}_{s}=-\frac{1}{\Lambda _{L}}\mathbf{B}.$$ (19)

Equation (19) is the second London equation. Setting the integration constant to zero accounts for the Meissner effect since if there is no bulk current in the superconductor, there is no magnetic field inside the metal, regardless of the external magnetic field. Taking the curl of the Maxwell equation⁹ $\nabla \times \mathbf{B}=\mu _{0}\mathbf{j}_{s}$ and substituting in the above equation yields

$$\nabla \times \nabla \times \mathbf{B}=\mu _{0}\nabla \times \mathbf{j}_{s}=-\frac{\mu _{0}}{\Lambda _{L}}\mathbf{B}.$$ (20)

As experiments showed, a superconductor is not a perfect diamagnetic at the surface - there is some distance (the penetration depth) over which the magnetic field penetrates into the metal. If we have a large square superconductor with an external magnetic field in the $\hat{z}$ direction and the $\hat{y}$ direction is into the metal (see figure 8),

Figure 8: An external magnetic field decays as it penetrates into the superconductor.

then the magnetic field is a function of penetration distance (y) and in the $\hat{z}$ direction: $\mathbf{B}=(0,0,B_{z}(y))$. Remembering the vector identity $\nabla \times \nabla \times \mathbf{A}=\nabla (\nabla \cdot \mathbf{A})-\nabla ^{2}\mathbf{A}$ and noting that $\nabla \cdot \mathbf{B}=0$ by Maxwell's equations,

$$\nabla ^{2}\mathbf{B} = \frac{\mu _{0}}{\Lambda _{L}}\mathbf{B}$$ (21)

$$\frac{\partial ^{2}B_{z}}{\partial y^{2}} = \frac{\mu _{0}}{\Lambda _{L}}B_{z}.$$ (22)

Defining $\lambda _{L}^{2}\equiv \mu _{0}/\Lambda _{L}$ the differential equation has a solution of the form,

$$B_{z}(y)=B_{z_{0}}e^{-y/\lambda _{L}}.$$ (23)

A field perpendicular to the surface of a superconductor decays exponentially with a characteristic length of $\lambda _{L}$, which ranges from 500Å to 10,000Å depending on the superconductor. The exponential shape of the decay is in agreement with experimental observations and the length of the decay is of the right order of magnitude, but with varying degrees of accuracy for different materials.

3.3 The Two-Fluid Model¹⁰

The two-fluid model assumes that every electron is either in the normal state or the superconducting state. All thermodynamic quantities are linear combinations of the contributions from the normal and the superconducting electrons. For example,

$$F(T)=x(T)F_{s}(T)+\left(1-x(T)\right)F_{n}(T),$$ (24)

where $F(T)$ is the Helmholtz free energy and x(T) is the fraction of electrons in the superconducting state. This equation does not permit x(T) to be determined from the equilibrium condition $\left(\frac{\partial F}{\partial x}\right)_{T}=0$, so, in an ad hoc method common in many phenomenological theories, the equation was modified to

$$F(T)=x(T)F_{s}(T)+\left(\sqrt{1-x(T)}\right)F_{n}(T),$$ (25)

with the only justification that then $x(T)$ could be solved for and more importantly it led to the relationship $C_{s}(T)\propto T^{3}$, which was then believed to be correct. For a correct theory that can explain superconductivity without ad hoc assumptions, we now turn to the development of BCS theory.

Footnotes

...sec:Thermodynamic-Theories⁷

The derivations covered here are all based on the first chapter in Parks (1969, p.19-26), except for the derivation of magnetic work which comes from Mandl (1988, p.21-28).

...Londonsec⁸

The derivation of the London equations and the penetration depth is based on Tinkham (1996); Ketterson and Song (1999); Parks (1969); Ibach and Lüth (1996). Pippard's modifications to the London equations comes from the first chapter in Parks (1969, p.27-42).

... equation⁹

The Maxwell equation is actually $\nabla \times \mathbf{B}=\mu _{0}\mathbf{j}_{s}+\frac{\partial }{\partial t}\mathbf{E}$, but we can ignore the second term because the electric field is nearly constant with respect to time.

... Model¹⁰

Source: Parks (1969, p.24-26).