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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?

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  • $\begingroup$ 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). $\endgroup$
    – alarge
    Nov 24, 2014 at 21:25
  • $\begingroup$ 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). $\endgroup$
    – alarge
    Nov 24, 2014 at 22:50
  • $\begingroup$ 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/…. $\endgroup$
    – alarge
    Nov 25, 2014 at 13:56
  • $\begingroup$ related: physics.stackexchange.com/q/604604/226902 $\endgroup$
    – Quillo
    Jul 4, 2022 at 10:27

1 Answer 1

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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.).

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