Gibbs free energy for superconductor in intermediate state

[patric44]

Homework Statement

derive Gibbs free energy for superconductor in intermediate state

Relevant Equations

G(Ha) = G(0)-u0 ∫MdHa

hi guys
I am trying to derive the Gibbs free energy for a superconductor in the intermediate state , the book(Introduction to Superconductivity by A.C. Rose-Innes) just stated the equation as its :
$$
G(Ha) = Vgs(0)+\frac{V\mu_{o}H_{c}}{2n}[H_{a}(2-\frac{H_{a}}{H_{c}})-H_{c}(1-n))]
$$
I am not sure how he got there, i tried to evaluate that from the integral :
$$G(Ha) = G(0)-\int_{0}^{Ha}\mu_{o}MdHa$$
where M is the magnetic moment, i multiplied end divided the right hand side by V and got
$$G(Ha) = Vgs(0)-\mu_{o}V\int_{0}^{Ha}IdHa$$
where I is the magnetization, and since in the intermediate state $I = (\eta-1)H_{c}$, and the limits for the integral would be , (1-n)Hc to Hc
$$G(Ha) = Vgs(0)-\mu_{o}V\int_{Hc}^{(1-n)Hc}(\eta-1)H_{c}dHa$$
I am not sure that this is the correct approach , beside how does he still has Ha inside that expression after evaluating the integral ?
i will appreciate any hint , thanks in advance

 

[Nate810]

A:In the intermediate state, you have two types of regions in the superconductor: normal and superconducting. The normal regions are demagnetized and contain no supercurrents. In the superconducting regions, the magnetic field is completely expelled due to the Meissner effect. This excludes the type of integration you suggested, since magnetization in the superconducting regions is zero.The solution of the problem lies in considering that the London equations are satisfied in both regions. The London equations hold a relation between the parallel and perpendicular components of the magnetic field:$$H_\perp = \lambda_L^{-2}J_\parallel,$$where $\lambda_L$ is the London penetration depth and $J_\parallel$ is the parallel component of total supercurrent density.The total supercurrent density is the sum of the two contributions coming from the normal ($J_{\parallel,n}$) and superconducting ($J_{\parallel,s}$) regions and it is required to be constant throughout the sample. Therefore, we can write:$$J_{\parallel} = J_{\parallel,n} + J_{\parallel,s} = const$$The parallel component of the supercurrent in the normal region is determined by the normal conductivity of the material and the applied field, while in the superconducting region is determined by the applied field and the critical field:$$J_{\parallel,n} = \frac{\mu_0 \sigma H_a}{2}, \qquad J_{\parallel,s} = \frac{\mu_0 H_c^2}{2 H_a} \left ( n - \frac{H_a}{H_c}\right )$$The Gibbs free energy is proportional to the volume of the sample. The contribution from the normal region is given by the product of the volume fraction and the free energy difference between the normal and superconducting states:$$G(H_a) = V \left ( g_s - g_n \right )$$where $V$ is the volume of the sample, $g_s$ and $g_n$ the Gibbs free energy per unit volume in the superconducting and normal states, respectively. The difference between $g_s