2
$\begingroup$

In Nakahara section 4.9, "Defects in nematic liquid crystals", it is discussed that the order parameter for a nematic should be the real projective plane $\mathbb{R}P^2$, which has fundamental group $\pi_1(\mathbb{R}P^2) \cong \mathbb{Z}_2 = \{0,1\}$. The real projective plane's fundamental group is generated by two homotopy classes, one being a trivial loop and the other a non-trivial loop which winds around the space. As discussed earlier in the text, the fact that the fundamental group is isomorphic to $\mathbb{Z}_2$ is equivalent to the fact that winding around the non-trivial loop twice creates a trivial loop, as demonstrated in this represetation, where antipodal points of the disk's boundary are identified:

enter image description here

Correspondingly, it is then stated that there are two types of line defects in a nematic, one of which can be deformed to a uniform configuration and one which cannot. These are shown in the following two figures,

enter image description here

enter image description here

However, I am having trouble visualizing how the configuration in 4.25 can be deformed to a uniform configuration. I was trying to visualize it in the same way as in Figure 4.16: take a circle $S^1$ surrounding the line defect (like the curve $\alpha$ in 4.24) and think of how it maps into a curve on $\mathbb{R}P^2$ ala Figure 4.16, but have thus far I am still very confused as to how to visualize the deformation in 4.16 in terms of the configuration shown in 4.25.

Can anyone explain how the deformation should be done and how it relates to the deformation in 4.16? And for that matter, how the configuration in 4.24 maps into $\mathbb{R}P^2$?

$\endgroup$
1

1 Answer 1

1
$\begingroup$

I think your confusion stems from the fact that the drawings are only 2D spatial projections of a three-dimensional nematic liquid crystal. In 2D, the order parameter space would be $\mathbb R P^1$ which is isomorphic to the circle $S^1$. In this case, the defects would be point defects classified by $\pi_1(S^1) / \{ n \sim -n \}$ and both defects would be stable under continuous deformation, with winding number $1$ and $2$ for Fig. 4.24 and Fig. 4.25, respectively.

However, for a 3D nematic liquid crystal the defects are line defects. If we consider a sideways view of the line defect of Fig. 4.25 it becomes obvious how to deform the configuration without affecting the boundary:

Continuous deformation of line defect to uniform configuration.

What does this correspond with in order parameter space? I will take the disk with antipodal points on the boundary identified $D^2 / \{ \mathbf r \sim - \mathbf r, |\mathbf r|=1\}$ to represent the real projective plane $\mathbb R P^2$. If we assume the director lies in the $xy$ plane then the loop $\alpha$ in Fig. 4.24 runs along the boundary of the disk:

Loop in configuration space corresponding to Fig. 4.24.

Note that this is a loop since the points $(1,0)$ and $(-1,0)$ are identified. Similarly, we have for Fig. 4.25:

Loop in configuration space corresponding to Fig. 4.25.

where the total loops consists of first going around the red loop and then going around the blue loop. Now, keeping the base point $(1,0) \sim (-1,0)$ fixed, we can deform the loops as follows:

Continuous deformation of the loop corresponding to Fig. 4.25.

Now one can contract the total loop as in Fig. 4.16. Note that this is not possible for the liquid crystal in two spatial dimensions, where there is no interior. Also note that points in the interior correspond to directors that point out of the $xy$ plane. Hence, we are performing the deformation shown in the figure above.

$\endgroup$
3
  • $\begingroup$ thank you! I understand what you wrote but I still can't visualize the continuous deformation of the liquid crystal configuration from the one with the (trivial) line defect to a uniform one. $\endgroup$
    – Kai
    Jul 19, 2021 at 15:02
  • $\begingroup$ @Kai Think of the line defect (Fig. 4.25) in three spatial dimensions. You have a column of directors that all lie in parallel planes normal to the $z$ axis and they all point to the $z$ axis. Now imagine moving all directors out of the plane so that all of them point along the $z$ axis. This is what is showed in the first figure of my answer (it only shows a sideways cut) and it results in a uniform configuration with director $n_0 = (0,0,1)$. $\endgroup$
    – Praan
    Jul 19, 2021 at 15:18
  • $\begingroup$ thanks, I can see that, I'll have to think more about how to connect the 3d deformation of the director field with the deformation of the loop on the projective plane to a uniform one. Great answer none the less! $\endgroup$
    – Kai
    Jul 19, 2021 at 19:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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