# Equivalent tensor order parameters of nematic liquid crystals?

I found in the literatures two different definitions of the tensor order parameter of nematic liquid crystals.

One is $$Q_{ij}=\frac{S}{2}(3n_{i}n_{j}-\delta_{ij}),$$ where $S$ is the scalar order parameter and $\mathbf{n}$ the director, $i=x, y, z$ . The other definition is $$Q_{ij}=\langle u_{i}u_{j}-\frac{1}{3}\delta_{ij}\rangle,$$ where $\mathbf{u}$ is a unit vector along the axis of one molecule which describes its orientation, $i=x, y, z$, and $\langle\cdot\rangle$ represents a thermal average.

In some special cases, the two definitions are the same (e.g., in a perfectly aligned nematic). Are the two definitions identical to each other in general? If yes how to prove it?

• Are you sure about the thermal averages? My book of choice (Chaikin-Lubensky p.168) has $\langle Q_{ij}\rangle=S(n_in_j-\frac{1}{3}\delta_{ij})$, where $S = \frac{1}{2}\langle3\cos^2\theta-1\rangle$. With your latter model (without $S$), I don't see how you can get the isotropic/nematic transition, but I think you might use it to figure out how the nematics are aligned (Frank free energy; e.g. Freedericksz transition). – alarge Nov 24 '14 at 21:25
• @alarge The second definition can be found in Denis Andrienko's lecture notes. (eqn (1), p.8) – sash Nov 24 '14 at 22:28
• Ok, so the definition in the lecture notes (your def'n 2) is more general (I misread your original question and did not notice that you were talking about single molecules). The usual case is that of the uniaxial nematic, which is eq. (11) of the notes (same as your def'n 1). In general you can find a matrix representation $\langle Q\rangle = \begin{pmatrix}2S/3&0& 0\\0&-S/3+\eta&0\\0&0&-S/3-\eta\end{pmatrix}$. With $\eta=0$ you get the uniaxial system (see end of p.9/beginning of p.10 of the notes). – alarge Nov 24 '14 at 22:50
• @alarge thank you. but I am still unable to derive the first equation directly from the second one (with the uniaxial condition). – sash Nov 24 '14 at 23:30
• Please don't cross-post the question (as you have, to chemistry SE), as this might result in the deletion of the questions: meta.stackexchange.com/questions/64068/…. – alarge Nov 25 '14 at 13:56

## 1 Answer

This is a matrix algebra heavy solution. I think there might be a better way to do this, but here goes.

Take from Denis Andrienko's notes: $$Q_{ij} = \frac{1}{N}\sum_\alpha(u_i^\alpha u_j^\alpha - \frac{1}{3}\delta_{ij})$$ where $\mathbf{u}^\alpha$ is the unit vector pointing along the long axis of molecule $\alpha$ located at $\mathbf{x}^\alpha$, and $N$ the number of molecules.

Or if you prefer matrix notation, $$Q = \frac{1}{N}\sum_\alpha((\mathbf{u}^\alpha)(\mathbf{u}^\alpha)^T - \frac{1}{3}I)$$

We write out $\langle Q \rangle$ in an orthonormal basis $(\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3)$, where one of the axes is set by the director $\mathbf{e}_1 = \langle\sum_\alpha \mathbf{u}^\alpha\rangle \big/ \|\langle\sum_\alpha \mathbf{u}^\alpha\rangle\|$: $$\langle Q\rangle = U^T \frac{1}{N}\sum_\alpha \langle (\mathbf{u}^\alpha)(\mathbf{u}^\alpha)^T \rangle U - \frac{1}{3}I$$ where the matrix $U = [\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3]$. This can be written as $$\langle Q\rangle = \frac{1}{N}\sum_\alpha \langle ((\mathbf{u}^\alpha)^TU)^T((\mathbf{u}^\alpha)^TU) \rangle - \frac{1}{3}I$$

Writing this out in its full glory: $$\langle Q\rangle = \frac{1}{N}\sum_\alpha \begin{pmatrix} \langle((\mathbf{u}^\alpha)^T\mathbf{e}_1)^2\rangle & \langle(\mathbf{u}^\alpha)^T\mathbf{e}_1(\mathbf{u}^\alpha)^T\mathbf{e}_2\rangle & \langle(\mathbf{u}^\alpha)^T\mathbf{e}_1(\mathbf{u}^\alpha)^T\mathbf{e}_3\rangle \\ \langle(\mathbf{u}^\alpha)^T\mathbf{e}_1(\mathbf{u}^\alpha)^T\mathbf{e}_2\rangle & \langle((\mathbf{u}^\alpha)^T\mathbf{e}_2)^2\rangle & \langle(\mathbf{u}^\alpha)^T\mathbf{e}_2(\mathbf{u}^\alpha)^T\mathbf{e}_3\rangle \\ \langle(\mathbf{u}^\alpha)^T\mathbf{e}_1(\mathbf{u}^\alpha)^T\mathbf{e}_3\rangle & \langle(\mathbf{u}^\alpha)^T\mathbf{e}_2(\mathbf{u}^\alpha)^T\mathbf{e}_3\rangle & \langle((\mathbf{u}^\alpha)^T\mathbf{e}_3)^2\rangle \end{pmatrix} - \frac{1}{3}I$$

Writing $\mathbf{e}_1$ out, we find that $\langle((\mathbf{u}^\alpha)^T\mathbf{e}_1)^2\rangle$ = $\langle\cos^2\theta^\alpha\rangle$. If the molecule is uniaxial, we have $\langle((\mathbf{u}^\alpha)^T\mathbf{e}_2)^2\rangle = \langle((\mathbf{u}^\alpha)^T\mathbf{e}_3)^2\rangle = \langle\frac{1}{2}\cos^2(\frac{\pi}{2}-\theta^\alpha)\rangle$ (if its not uniaxial, one gets more weight while the other the same amount less). The offdiagonals vanish and the sum is just an average over all molecules.

Thus we have $$\langle Q \rangle = \begin{pmatrix} \frac{2}{3}S & 0 & 0 \\ 0 & -\frac{1}{3}S & 0 \\ 0 & 0 & -\frac{1}{3}S \end{pmatrix}$$ where $S = \frac{1}{2}\langle 3\cos^2\theta^\alpha - 1 \rangle$.

That is to say that we can write $$\langle Q_{ij}\rangle = S(n_i n_j - \frac{1}{3}\delta_{ij})$$ where the unit vector $\mathbf{n}$ specifies the direction of the principal axis of $\langle Q_{ij}\rangle$ in the transformed coordinates (e.g. $\textbf{n} = (1, 0, 0)$ points towards $\textbf{e}_1$ etc.).

• This is a nice solution. I found another: set $x$ axis to be parallel to the director, then $u_x=\cos\theta,u_y=\sin\theta\cos\psi,u_z=\sin\theta\sin\psi$. Under the uniaxial condition, the distribution function of the molecules $f(\theta,\psi)$ is independent of $\psi$. the thermal average can be calculated: $<u_iu_j>=\int u_iu_j f(\theta,\psi)\sin\theta d\theta d\psi$. At last we get the same result. – sash Nov 25 '14 at 18:00
• oh, they're in fact the same solution – sash Nov 25 '14 at 18:15