ass2: Process feedback from Roos on draft

superconductivity_assignment2_kvkempen.tex

@@ -57,21 +57,19 @@ This point we call $F_0(T)$ with order parameter $\psi_0(T)$.
 For this, we will take the derivative of $F$ with respect to $\psi$ and equate it to zero.
 
 \[
-	0 = \pfrac{\mathcal{F}}{\psi} = \pfrac{}{\psi} \left[ a(T-T_c)\psi^2 \right] = 2a(T-T_c)\psi + 2\beta\psi^3
+	0 = \pfrac{\mathcal{F}}{\psi} = \pfrac{}{\psi} \left[ a(T-T_c)\psi^2 + \frac{\beta}{2}\psi^4 \right] = 2a(T-T_c)\psi + 2\beta\psi^3
 \]
 
 Extreme points are found at $\psi = 0$ and $\psi = \pm\sqrt{\frac{-a}{\beta}(T-T_c)}$.
 
 For $T \geq T_c$, $\psi_0(T \geq T_c) = 0$ gives the minimum, i.e. $\mathcal{F}_0(T \geq T_c) = 0$.
 
-For $T \leq T_c$, $\psi_0(T \leq T_c) = \sqrt{\frac{-a}{\beta}(T-T_c)}$ is the minimum,
+For $T \leq T_c$, $\psi_0(T \leq T_c) = \pm\sqrt{\frac{-a}{\beta}(T-T_c)}$ is the minimum,
 giving free energy
 \[
-	\mathcal{F}_0(T \leq T_c) = \frac{-a^2}{\beta}(T-T_c)^2 + \frac{a^2}{2\beta}(T-T_c)^2 = \frac{-a^2}{2\beta}(T-T_c)^2 \leq \mathcal{F}_0(T \geq T_c)
+	\mathcal{F}_0(T \leq T_c) = \frac{-a^2}{\beta}(T-T_c)^2 + \frac{a^2}{2\beta}(T-T_c)^2 = \frac{-a^2}{2\beta}(T-T_c)^2 \leq \mathcal{F}_0(T \geq T_c).
 \]
-where we chose the positive of the $\pm$ as the order parameter is understood to increase from finite at the phase transition.
-\todo{Is this a reasonable statement? It actually does not really matter that much as mostly $\psi^2$ is used, but the physical meaning is totally different. It implies some kind of symmetry, too. It seems that also \cite{abrikosov} mentions this.}
-	
+
 For the specific heat, we find
 \[
 	C(T) = -T\pfrac{^2\mathcal{F}}{T^2} =
@@ -81,13 +79,20 @@ For the specific heat, we find
 	\end{cases}.
 \]
 There is thus a discontinuity in $C(T)$ at $T = T_c$ with size $\Delta C(T) = \frac{a^2}{\beta}T_c$.
+See the following sketches of the $T$-dependence of the derived quantities.
+
+\begin{figure}[H]
+	\centering
+	\includegraphics[width=.9\textwidth]{ass2-landau-theory-T.pdf}
+	\caption{For the Landau theory, we find the drawn temperature dependences for equilibrium values of $\mathcal{F}_0$, $\psi_0$ and $C$. Note the minus sign for the free energy $\mathcal{F}_0$. At $T = 0$, there are  $y$ axis intersections for all three quantities, namely a minimum $\mathcal{F}_0(0) = \frac{-a^2}{2\beta}T_c^2$, $\psi_0(0) = \pm\sqrt{\frac{-a}{b}T_c}$, and $C(0) = 0$, which I forgot to indicate in the sketches. Do also note that there thus is an intersection in the $\mathcal{F}_0(T)$ curve at $T = 0$, although the drawing may look asymptotic.}
+\end{figure}
 
 \section{Type-I superconducting foil}
 \begin{enumerate}
 \item
 The screening equation is given as
 \[
-	\nabla^2\vec{B} = \frac{\vec{B}}{\lambda}.
+	\nabla^2\vec{B} = \frac{\vec{B}}{\lambda^2}.
 \]
 For easy of calculation, we will use cartesian coordinates,
 and put the external magnetic field $B_E$ along the $x$ axis:
@@ -98,17 +103,15 @@ So we define the magnitude of the field $|\vec{B}| = B(z)$.
 
 Using the screening equation, we look for a solution.
 \[
-	\nabla^2\vec{B} = \nabla(\nabla\cdot\vec{B}) - \nabla\times(\nabla\times\vec{B})
+	\nabla^2\vec{B} = \left[ \pfrac{^2}{x^2} + \pfrac{^2}{y^2} + \pfrac{^2}{z^2} \right] \vec{B}
 \]
-$\vec{B}$ is divergenceless, so we are left with the latter term.
-Next, we take the curls writing $B_i$ for the $i$th component of $\vec{B}$,
-and realize that we only have $z$ dependence, and $B_y = 0 = B_z$.
+we realize that we only have $z$ dependence, and $B_y = 0 = B_z$.
 \[
-	-\nabla\times(\nabla\times\vec{B}) = -\nabla\times(\pfrac{B_x}{z} \hat{y}) = -(-\pfrac{^2B_x}{z^2} \hat{x})
+	\nabla^2\vec{B} = \pfrac{^2B_x}{z^2} \hat{x}
 \]
 Rewriting yields
 \[
-	\vec{B} = \lambda \pfrac{^2B_x}{z^2}\hat{x}.
+	\vec{B} = \lambda^2 \pfrac{^2B_x}{z^2}\hat{x}.
 \]
 For this we know the general solution:
 \[
@@ -158,12 +161,16 @@ Next, we can equate the previously found supercurrent for our foil to the Ginzbu
 \[
 	\vec{A} = \frac{-B_E m \sinh{\frac{z}{\lambda}}}{4\lambda\mu_0 e^2 n_s \cosh{\frac{a}{2\lambda}}} \hat{y}
 \]
+This we can rewrite using $\lambda = \sqrt{\frac{m}{\mu_0 n_s e^2}}$ for the London penetration depth as
+\[
+	\vec{A} = \frac{-B_E \lambda \sinh{\frac{z}{\lambda}}}{4 \cosh{\frac{a}{2\lambda}}} \hat{y}.
+\]
 
 \item
 \[
 	\nabla \cdot \vec{A} = \pfrac{A_x}{x} + \pfrac{A_y}{y} + \pfrac{A_z}{z} = \pfrac{0}{x} + \pfrac{A_y}{y} + \pfrac{0}{z} = 0,
 \]
-as $A_y \perp \hat{y}$, giving zero partial derivative.
+as $A_y$ is independent of $y$.
 
 In our case, indeed the rigid gauge choice gives the criterium for the London gauge ($\nabla \cdot \vec{A} = 0$).
 
@@ -181,11 +188,12 @@ and take the divergence,
 	\frac{2e}{\hbar}\nabla \cdot \vec{A} = -\frac{m}{2e\hbar n_s} \nabla \cdot \vec{J_s} - \Delta\theta = 0
 	\implies \Delta\theta = -\frac{m}{2e\hbar n_s} \nabla \cdot \vec{J_s}.
 \]
-This is what only the London gauge implies.
-But when is then the rigid gauge applied by this?
-This is the case for $\nabla \cdot \vec{J_s}$, or, in words, when there is no conservation of supercurrent.
-If this is not the case (if the divergence is non-zero), there is conversion between normal current and supercurrent.
-This result seems to Waldram's conclusion in \cite[p. 24--26]{waldram}.
+This is what the London gauge implies.
+Now the question is under what circumstances the rigid gauge follows from the London gauge.
+This is the case for $\nabla \cdot \vec{J_s} = 0$, or in words, when there is conservation of superelectrons.
+If this is not the case (if the divergence is non-zero), there is conversion between normal electrons and superelectrons.
+This would take place if the temperature is lowered, as more superelectrons allow for a larger supercurrent, thus a larger critical magnetic field.
+This result seems to agree with Waldram's conclusion in \cite[p. 24--26]{waldram}.
 
 \item
 We apply a gauge transformation as follows.
@@ -195,7 +203,6 @@ We apply a gauge transformation as follows.
 	\phi &\to \phi - \pfrac{\chi}{t} = \phi + \frac{\hbar}{2e} \omega
 \end{align}
 
-\todo{Do I really need to put in the previously found $\vec{A}$?}
 \end{enumerate}
 
 \section{Type II superconductors and the vortex lattice}
@@ -209,10 +216,10 @@ Superconductors are characterized by perfect diamagnetism and zero resistance.
 Perfect diamagnetism is the ability by superconductors to have a net zero magnetic field inside.
 If you apply an external magnetic field, this thus means that a superconductor will let a current flow on its inside to generate a field to counteract this external field $\vec{H}$.
 This generated current is called a supercurrent.
-This is, however, a phase of the material.
+Superconductivity is, however, a phase of the material.
 Superconductors only have these properties below a certain temperature, its critical temperature $T_c$, and can only expel a maximum external magnetic field, its critical magnetic field $B_c(T)$, which is a function of the temperature.
 
-The class of superconductors we have a model for, is the class of conventional superconductors, which are explained by a theory called BCS (and some extensions).
+The class of superconductors we have a model for, is the class of conventional superconductors.
 In this class, there are two types, called type-I and type-II superconductors.
 % End copy from philosophy
 
@@ -225,7 +232,9 @@ This state is reached for $T < T_c$ and $B_E < B_{c1}(T)$.
 The other state is a mixed state that allows some flux to pass through the material.
 This passing through is done by creating normally conducting channels throughout the material where a fixed amount of flux can pass through.
 This fixed amount is a multiple of the flux quantum $\Phi_0$.
-The material generates current around these channels cancelling the field on the inside of the superconducting part of the material.
+The material generates current around these channels in accordance to the Maxwell-Amp\`ere law, conforming to the let through magnetic field inside the vortex and cancelling the field on the outside the vortex.
+
+
 
 ---
 
@@ -244,7 +253,11 @@ In the mixed or vortex state, superconductors let through a number of finite flu
 Some small regions of the material are not superconducting, but in the normal state.
 Flux passes through these regions in multiples of $\Phi_0$, but usually just one $\Phi_0$ per region,
 and a supercurrent is generated to expel the field from the rest of the material.
-This supercurrent moves around these region in a vortex-like shape.
+These flux allowing regions are called vortices, due to their shape and direction of current flow.
+Vortices look like channels (or tubes), and supercurrents move around these channels in a spiraling fashion.
+One can visualize this as current through a coil such that on the inside of the coil, the field is in one direction perpendicular to it, and on the outside it is the opposite direction.
+The current direction is governed by the Maxwell-Amp\`ere equation.
+In this case, the current is such that the field inside the cylinder but outside these channels is counteracted.
 
 Please see the figure below for a beautiful drawing.
 It was not specified what the direction of $\vec{B_E}$ was with respect to the cylinder orientation, so I chose what I thought was most reasonable as an example.
@@ -252,7 +265,7 @@ It was not specified what the direction of $\vec{B_E}$ was with respect to the c
 \begin{figure}
 	\centering
 	\includegraphics[width=.6\textwidth]{SchermafbeeldingKees-vortex-by-fleur-ahlers.png}
-	\caption{The direction of $\vec{J_s}$ is such that a magnetic field is generated to counteract and even expel the external field. Around the vortices, that menas that the supercurrents run anti-clockwise. Around the outside border of the cylinder, however, $\vec{J_s}$ runs clockwise to cancel $\vec{B_E}$.}
+	\caption{The direction of $\vec{J_s}$ is such that a magnetic field is generated to counteract and even expel the external field outside the vortices inside the material. Around the vortices, that means that the supercurrents run anti-clockwise. The field is then along $\vec{B_E}$ inside the vortices, but along $-\vec{B_E}$ outside the vortices but inside the material. Around the outside border of the cylinder, however, $\vec{J_s}$ runs clockwise and again cancels $\vec{B_E}$ on the inside of the material.}
 \end{figure}
 
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