diff --git a/SchermafbeeldingKees-vortex-by-fleur-ahlers.png b/SchermafbeeldingKees-vortex-by-fleur-ahlers.png index edbbe46..d6b46f1 100755 Binary files a/SchermafbeeldingKees-vortex-by-fleur-ahlers.png and b/SchermafbeeldingKees-vortex-by-fleur-ahlers.png differ diff --git a/ass2-landau-theory-T.pdf b/ass2-landau-theory-T.pdf new file mode 100755 index 0000000..2ebe4b1 Binary files /dev/null and b/ass2-landau-theory-T.pdf differ diff --git a/superconductivity_assignment2_kvkempen.tex b/superconductivity_assignment2_kvkempen.tex index 8cbf2a1..5df7dc5 100755 --- a/superconductivity_assignment2_kvkempen.tex +++ b/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} \bibliographystyle{vancouver}