# Phase change boundary near Lifshitz Point

I am following the text of Chaikin and Lubensky.

They breifly discuss the results of the following equations but I find myself confused by some of the details:

$$F = \frac12 \int \text{d} \vec{x} (r\phi^2 + 2u\phi^4 + c_\parallel (\nabla_\parallel\phi)^2 + c_\perp (\nabla_\perp\phi)^2 + D (\nabla^2\phi))^2.$$

They state that $$c_\parallel,D, u$$ are all positive values and that $$c_\perp$$ can be positive or negative.

From here they state that the phase diagramss can be shown to be:

They state and I agree, that in the $$c_\perp>0$$ case the solution should be uniform and so we can ignore the gradient terms. This makes it very easy to find that the phase transition occurs when $$r$$ changes sign giving us the line to the left of the lifshitz point.

However, there is little detail for the $$c_\perp <0$$ case which I will now outline my thinking.

We will use the Euler-Poisson equation to find the minimization of the free-energy. We assume that the perpendicular direction is along $$z$$ and that the solution is still uniform in the parallel direction. I find that this minimization gives us:

$$2r\phi + 8u\phi^3 - 2 c_\perp\partial_z^2\phi + 2D\partial_z^4\phi=0,$$

I further assume that the $$\phi$$ must have the form of a modulation (based on the fact that the region is called the modulation region). If we assume that $$\phi = a_0\cos(b_0 z)$$ the above solution reduces and removes the pesky derivatives to give:

$$2r\phi + 8u\phi^3 + 2b_0^2 c_\perp\phi + 2b_0^4 D\phi=0.$$

From which I clearly see 2 solutions:

$$\phi=0,$$

and

$$\phi = \pm\sqrt{-\frac{r+c_\perp b_0^2 + D b_0^4}{D}}.$$

From this I can predict that there is phase change when the numerator in the square-root becomes a negative value as this is the first time the two solutions become real valued. So I would get that one of the lines occurs as:

$$r = -c_\perp b_0^2 - D b_0^4.$$

At this point I can not spot another possible phase change. However, I am already a little dubious of the above answer as firstly it is only linearly with $$c_\perp$$ but in the textbook diagram it appears to be quadratic. Secondly, it does not connect to the the Lifshitz point unless $$b_0=0$$ which is useless.

Any advice would be greatly appreciated. I'm not sure what I'm doing wrong at this point.

Here is a partial answer that seems to give the right phase boundary in the $$r>0, c_\perp<0$$ part of the phase diagram.

Following the argument in the post, I will consider the following free energy

$$F = \frac12\int dz\, [r\phi^2+2u\phi^4+c_\perp(\partial_z\phi)^2+D(\partial_z^2\phi)^2]$$

Assuming $$\phi=a_0\cos b_0z$$ and plug the ansatz into the expression for $$F$$, the free energy density is given by

$$f= \frac{1}{4}[ (r+c_\perp b_0^2+Db_0^4)a_0^2+3ua_0^4]$$

So now the task is to minimize $$f$$ with respect to both $$b_0$$ and $$a_0$$. We simply solve $$\frac{\partial f}{\partial b_0}=0, \frac{\partial f}{\partial a_0}=0$$:

$$b_0(c_\perp + 2Db_0^2)=0$$ $$a_0[R(b_0)+6ua_0^2]=0$$

If $$c_\perp>0$$, then the only solution for $$b_0$$ is $$b_0=0$$, and we are back to the standard Landau case.

If $$c_\perp<0$$, but $$r>0$$, the $$b_0=0$$ solution leads to a paramagnetic state with $$a_0=0$$. The $$b_0^2=-\frac{c_\perp}{2D}$$ solution gives $$a_0^2=\frac{1}{6u}(c_\perp^2/4D-r)$$. If $$r>\frac{c_\perp^2}{4D}$$, again the only minimum is the paramagnetic one. If $$0, then the minimum is the modulated solution. So we obtain the quadratic portion of the phase boundary in the $$r>0$$ half (i.e. $$r=c_\perp^2/4D$$).

Next we consider $$c_\perp<0, r<0$$. The $$b_0=0$$ solution leads to $$a_0^2=\frac{-r}{6u}$$, and $$b_0^2=-\frac{c_\perp}{2D}$$ again gives $$a_0^2=\frac{1}{6u}(\frac{c_\perp^2}{4D}-r)$$. The latter has lower free energy, regardless of the values of $$c_\perp$$ or $$r$$ as long as both are negative. As $$c_\perp$$ goes to 0, $$b_0$$ also vanishes, continuously connected to the uniform case. So still does not agree with the lower portion of the phase diagram...

• Hi Meng, thank you for you answer :). In terms of the second phase boundary, this is just a notation the textbook has been using to denote a first-order phase transition instead of a second order. A couple questions, how did you find the free energy density? When I try to calculate I end up with cos^3 terms. I did as you suggested of inserting the ansatz directly into the lagrangian (and presumably taking a taylor series) but do not get your $f$. Could you expand on how you are interpreting the second solution?... Commented Apr 5, 2022 at 0:18
• ... it appears to me that the solution $6a_0^2 u + \frac{c_\perp^2}{ 4 D} = r$ also suffers from the problem of not crossing meeting at the point $(c_\perp,r)=(0,0)$. Commented Apr 5, 2022 at 0:19
• Please ignore my first point, after typing that I went back and found my error in my free energy density :). Commented Apr 5, 2022 at 0:23
• @akozi The phase boundary should not contain $b_0$ or $a_0$, since $b_0$ and $a_0$ are variables that need to be minimized over. $a_0^2=\frac{\frac{c_\perp^2}{4D}-r}{6u}$ is just the solution in that parameter regime. Commented Apr 5, 2022 at 0:24
• I see. Sorry, I may be getting to tired to continue studying for the night :D. So at this point, we would just want to insert these definitions for $a_0$ and $b_0$ back into the definition of $\phi$ and reanalyze the minimization of $F$ with these definitions? Commented Apr 5, 2022 at 0:36

I recently encountered the same problem in the same book as your question. Although an answer has already been accepted, I think I can solve some of the issues in Meng Cheng's solution.

I will follow the same procedure of evaluating the free energy using the ansatz $$\phi(\vec{x})=a \cos(bz)$$ for a scalar field in (2+1)-dimensions, where $$x,y$$ correspond to the parallel ($$\parallel$$) components and $$z$$ corresponds to the perpendicular ($$\perp$$) components.

Then, $$F=\frac{1}{2}\int d^3x \left[ ra^2 \cos^2(bz) +2ua^4 \cos^4(bz)+ c_{\perp}(-ab\sin(bz))^2 +D(-ab^2\cos(bz))^2\right]$$

Integrals of the form $$\int d^3x \cos^2(bz)$$ don't have a well-defined value, so instead of studying the total free energy, we switch to the free energy density $$f=F/V$$. Thus we replace $$\int d^3x \cos^2(bz)$$ by its average value over space.

In other words, we use $$\frac{\int d^3x \cos^2(bz)}{V} \implies \lim_{L\to \infty}\frac{\int_{-L}^{L} dz \cos^2(bz)}{2L} = \frac{1}{2}$$ Similarly $$\frac{\int d^3x \sin^2(bz)}{V} \implies \lim_{L\to \infty}\frac{\int_{-L}^{L} dz \sin^2(bz)}{2L} = \frac{1}{2}$$ $$\frac{\int d^3x \cos^4(bz)}{V} \implies \lim_{L\to \infty}\frac{\int_{-L}^{L} dz \cos^4(bz)}{2L} = \frac{3}{8}$$

There is, however, a hidden assumption here and it is at the crux of Meng Cheng's problems. These results are only valid if $$b\neq 0$$. If $$b$$ equals zero, then the averages must be 1!!! (or zero) This follows from:

$$\frac{\int d^3x \cos^2(bz)}{V} \implies_{b \to 0} \frac{\int d^3x 1}{V} = 1$$

Thus, we have two free energy densities: $$f(b \neq 0) = \frac{1}{2} \left[ \frac{1}{2} r a^2 + \frac{3}{4} u a^4 + \frac{1}{2}c_{\perp}a^2b^2 + \frac{1}{2}Da^2b^4\right] = \frac{1}{4} \left[ (r + c_{\perp}b^2 + Db^4)a^2 + \frac{3}{2} u a^4 \right]$$

$$f(b=0)= \frac{1}{2} \left[ r a^2 + 2 u a^4\right]$$

The next steps are just finding the (a,b) parameters that minimize the densities.

Case 1: $$b \neq 0$$

We find the extrema: $$\frac{\partial f}{\partial a}=0 \iff a(r+4ua^2) = 0 \iff a=0 \quad \lor \quad a^2 = -\frac{r}{4u}$$

We study their stability by the second derivative criteria. I will skip those steps (a simple calculus exercise). The result is that the solution $$a=0$$ (Paramagentic) is only stable when $$r>0$$. The solution $$a\neq 0$$ (Ferromagnetic) is a local minimum when $$r<0$$ otherwise, the solution does not exist because of the condition $$a^2 = -\frac{r}{4u} \geq 0$$.

I illustrate those results in the following figures:

Case 2: $$b \neq 0$$

Here we have to maximize both $$a$$ and $$b$$. We can find the extrema as follows:

$$\frac{\partial f}{\partial b} = 0 \iff (c_\perp+2Db^2)ba = 0 \iff a=0 \quad \lor \quad b^2 = - \frac{c_\perp}{2D}$$ Notice that here we are using the fact that this free energy is valid only when $$b \neq 0$$

$$\frac{\partial f}{\partial a} = 0 \iff a(r + c_{\perp}b^2 + Db^4 + 3 u a^2) = 0 \iff a=0 \quad \lor \quad a^2 = -\frac{r + c_{\perp}b^2 + Db^4}{3u}$$

The solution $$a=0$$ corresponds to the paramagnetic solution studied before. Thus, we are left with the equations: $$b^2 = - \frac{c_\perp}{2D}$$ $$a^2 = -\frac{r + c_{\perp}b^2 + Db^4}{3u}$$

By noticing that $$b^2 \geq 0$$ we conclude that the modulated solution can only exist when $$-c_\perp \geq 0$$. With that requirement:

$$a^2 = -\frac{r + c_{\perp}(- \frac{c_\perp}{2D}) + D(- \frac{c_\perp}{2D})^2}{3u} = - \frac{r-\frac{c_\perp^2}{8D}}{3u}$$

By noticing that $$a^2 \geq 0$$ we conclude that $$r \leq \frac{c^2_\perp}{8D}$$. This explains the upper parabola just as in Meng Cheng's answer.

I illustrate this region in the following figure:

Now, we could study the second derivative to check the limit of metastability of the modulated phase, but I will skip that to go straight into the phase transition boundary.

First order transition:

As you can see from the images, there is a region in which both the ferromagnetic and modulated phases can exist. The region is given by $$r<0$$ and $$-c_\perp > 0$$. To determine which phase will exist in equillibrium, we need to determine which of the two phases has lower free energy.

By substituting the values of $$a$$ and $$b$$ into their respective free energy densities we get:

$$f_{\text{Ferro}}= \frac{1}{2} \left[ r \left(-\frac{r}{4u}\right) + 2 u \left(-\frac{r}{4u}\right)^2\right] = - \frac{r^2}{16u}$$

$$f_{\text{Mod}} = \frac{1}{4} \left[ \left(r + c_{\perp}\left(- \frac{c_\perp}{2D}\right) + D\left(- \frac{c_\perp}{2D}\right)^2\right) \left(- \frac{r-\frac{c_\perp^2}{8D}}{3u}\right) + \frac{3}{2} u \left(- \frac{r-\frac{c_\perp^2}{8D}}{3u}\right)^2 \right] = -\frac{\left( r- \frac{c_\perp^2}{8D}\right)^2}{24 u}$$

The ferromagnetic phase will be more stable whenever:

$$- \frac{r^2}{16u} < -\frac{\left( r- \frac{c_\perp^2}{8D}\right)^2}{24 u}$$

$$\iff r^2 > \frac{2}{3} \left(r- \frac{c_\perp^2}{8D}\right)^2 \iff r^2 + 2r \left(\frac{c_\perp^2}{4D}\right) - \frac{1}{2}\left(\frac{c_\perp^2}{4D}\right)^2$$

By doing some algegraic manipulation keeping in mind that $$c_\perp <0$$ and $$r<0$$, it is possible to reduce that inequality to the condition:

$$-r > \left(\sqrt{\frac{3}{2}}-1\right) \left(\frac{c_\perp^2}{4D}\right)$$

which yields the lower parabola as shown in the following image:

We can note that in general, both parabolas have different curvatures, just like the one shown by Chaikin and Lubensky, and the values for $$a^2$$ change discontinuously. Thus a first-order transition.

Final remarks:

The solution and the phase diagram given by Meng Cheng correctly describe the phase diagram for a vector order parameter. I encourage you to try and corroborate that yourself!