2
$\begingroup$

I can't get one of the coefficients in the equation for $T < T_c$ in the bottom, specifically the equation with the factor of two. any help appreciated.

Consider an ising type expansion of the free energy density in the order parameter $\psi$. (the mean values of $\psi$ corresponding to a uniform configuration.)

\begin{equation} f(T,\psi) = f_0 + G(\nabla\psi)^2+ A\psi^2 + B\psi^4 \end{equation} \begin{equation} \bar{\psi} = \begin{cases} \pm\sqrt{\frac{-A}{2B}} & \qquad T < T_c \\ 0 & \qquad T > T_c \end{cases} \end{equation}

these solutions come from minimizing the free energy density. Expanding the coefficients near the critical temperature, we have \begin{equation} A = at + ... \end{equation} \begin{equation} B = B_0 + ... \end{equation} \begin{equation} G= G_0 + ... \end{equation} Where $t=T-T_c$. Then evaluating the free energy density at $\psi = \bar\psi$ we get \begin{equation} \left. f(T,\psi)\right|_{\psi=\bar{\psi}} = \begin{cases} f_0 - \frac{a^2t^2}{4B_0} & \qquad T < T_c \\ f_0 & \qquad T>T_C \end{cases} \end{equation} To study fluctuations in the order paramater define, $\delta\psi = \psi -\bar{\psi}$.

\begin{equation} \delta\psi = \begin{cases} \psi & \text{in symmetrical phase} \\ \psi - \sqrt{\frac{-at}{2B_0}} & \text{in disordered phase} \end{cases} \end{equation}

We now calculate the change in free energy though I am not sure how exactly to calculate this, for some reason I am missing something simple I'm sure (!).

\begin{equation} \Delta F(T,\delta\psi) = \begin{cases} \int d^{d}r\,\left( G(\nabla\delta \psi)^2 + at\delta \psi^2\right) & \qquad T > T_c \\ \int d^{d}r\,\left( G(\nabla\delta \psi)^2 -2at\delta \psi^2\right) & \qquad T < T_c \end{cases} \end{equation}

I have spent some time trying to calculate the coefficient above for $\delta\psi^2$ for $T < T_c$. I am clearly not understanding something here, since when I calculate $f(\psi) - f(\bar{\psi})$ I do not get the above expression. In fact I am not sure exactly how to go about calculating this. Landau and Lifshitz gives the above coefficient in section 146 "Fluctuations of the Order Parameter". I am also confused because it seems like there are two values of the mean $\bar{\psi} = \pm (\frac{-at}{2B_0})^\frac{1}{2} $ however in the book I only see a reference to the positive square root. This topic is a bit obscure so I'm not sure anybody really is gonna be able to help, but I'd appreciate any insight.

$\endgroup$
1
$\begingroup$

To get the coefficient, you just have to expand the free energy up to $\delta\psi^2$. The linear term gives zero (because you're expanding around the minimum) whereas the quadratic term is $A+6B\bar\psi^2$, which gives the expected result. You can see that this result is independent of the minimum choose (independent of the sign).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.