A confined rod: mean field theory for hard rod-like particles

We first show that, independent of the shape of the cell, the Cauchy stress tensor $\bm{\sigma}$, associated at equilibrium with the free energy $F$, is isotropic, and so its effect on the walls of the cell reduces to uniform pressure. Here we consider the minimiser $\mathcal{C}$ over a family of cells described as a linear, volume preserving transformation of a reference cell $\mathcal{C}_{0}$. We assume that the cell $\mathcal{C}_{0}$ is of some arbitrary shape (say, a parallelepiped, for definiteness), and let $\mathbf{F}$ denote a corresponding deformation gradient. Thus $F$ can be regarded as a function of $\mathbf{F}$, whose explicit form need not be specified at this stage, provided that it is sufficiently smooth. The Cauchy stress $\bm{\sigma}$ exerted on the walls of $\mathcal{C}$ is [11]

$$\bm{\sigma}=\frac{1}{\det{\mathbf{F}}}\frac{\partial F}{\partial{\mathbf{F}}}\cdot{\mathbf{F}}^{T}.$$

(12)

For $\mathcal{C}$ and $\mathcal{C}_{0}$ to have the same volume $V_{cell}$, $\mathbf{F}$ must obey the constraint

$$\det\mathbf{F}\equiv 1.$$

(13)

Stationarity of $F$ over this constraint requires that

$$\frac{\partial F}{\partial\mathbf{F}}=\lambda\frac{\partial}{\partial\mathbf{F}}\det\mathbf{F},$$

(14)

where $\lambda$ is a Lagrange multiplier. Since, for an invertible tensor $\mathbf{F}$,

$$\frac{\partial}{\partial\mathbf{F}}\det{\mathbf{F}}=(\det\mathbf{F})\mathbf{F}^{-T},$$

(15)

it follows from Eqs. (12), (14), and (15) that

$$\bm{\sigma}=\frac{1}{\det{\mathbf{F}}}\frac{\partial F}{\partial{\mathbf{F}}}\cdot{\mathbf{F}}^{T}=\lambda\mathbb{I}.$$

(16)

Eq. (16) indicates that the pressure is isotropic in equilibrium, and the Lagrange multiplier reduces to the (isotropic) pressure.

We remark that a stress also acts on the particle inside the cell, and if the pressure is isotropic on the cell walls, it will not, in general, be isotropic on the particle - and vice versa.

Here we consider the case when the cell is a parallelepiped and the particle is a uniaxial ellipsoid. We conjecture, as suggested by numerics, that the orientational averaged free volume of an ellipsoid in a parallelepiped cannot attain its maximum and free energy its minimum when the cell faces are not orthogonal. Furthermore, if the particle has uniaxial symmetry, the maximum free volume occurs when the cell is uniaxial. Therefore here we consider uniaxial rectangular cells, and apply the mean field theory of Section 2 to the case of a rigid uniaxial ellipsoid in a uniaxial rectangular cell.

Let $a$ and $b$ be the lengths of the semiaxes of the uniaxial ellipsoid in the directions along and perpendicular to the symmetry axis. The equation of the ellipsoid is

$$\mathbf{rAr}=1,$$

(17)

where

$$\mathbf{A}=\frac{1}{b^{2}}\mathbb{I}+(\frac{1}{a^{2}}-\frac{1}{b^{2}})\mathbf{\hat{m}\hat{m},}$$

(18)

and $\mathbf{\hat{m}}$ is along the symmetry axis. The aspect ratio, the ratio of the length of the distinguished semi-axis to that of one of the others, is $\ \eta=a/b$. Alternately, we can consider the uniaxial affine deformation of a sphere of radius $R$ to form this ellipsoid; the principal stretches are $\lambda_{1}=a/R$, $\lambda_{2}=\lambda_{3}=b/R$. If the volume is conserved, $\lambda_{2}=\lambda_{3}=1/\sqrt{\lambda_{1}}$, and the aspect ratio

$$\eta=\lambda^{3/2},$$

(19)

where $\lambda$ indicates stretch along the distinguished principal direction. The aspect ratio $\eta$ and stretch $\lambda$ will be used interchangeably to describe particle and cell shapes.

The particle volume is $v_{p}=4\pi ab^{2}/3$ and the occupied volume fraction is $\phi=v_{p}/V_{cell}=v_{p}\rho_{0}$.

The extent of the ellipsoid with orientation $\mathbf{\hat{m}}$ in direction $\mathbf{\hat{N}}$ is

$$h(\mathbf{\hat{N})}=\sqrt{a^{2}(\mathbf{\hat{m}\cdot\hat{N}})^{2}+b^{2}(1-(\mathbf{\hat{m}\cdot\hat{N}})^{2})},$$

(20)

and in the principal directions of the cell this is

$$h_{i}=\sqrt{a^{2}m_{i}^{2}+b^{2}(1-m_{i}^{2})},i=1,2,3.$$

(21)

The free volume available to the center of the ellipsoid is then simply

$$V_{f}(\mathbf{\hat{m}},\mathcal{C})=\prod_{i=1}^{3}\max\{d_{i}-2h_{i},0\},$$

(22)

where $d_{i}$’s are lengths of the cell along its principal directions. A 2D illustration of the free volume of an ellipse in a rectangular cell is given in Fig. 1.

The free energy of one cell, from Eq. (7), is

$$F_{c}=-kT\ln\int_{S^{2}}\prod_{i=1}^{3}\max(d_{i}-2h_{i}(\mathbf{\hat{m}}),0)d\mathbf{\hat{m}},$$

(23)

where we have omitted $-3\ln\Lambda_{dB}$. For simplicity, we will drop the subscript, and with the understanding that the free energy $F$ is for a single cell. The free energy $F$ can therefore be calculated by straightforward numerical integration of Eq. (23) using, say, the quasi-Monte Carlo method. Local minima/maxima can be identified via standard gradient descent/ascent methods.

Let $d_{2}=d_{3}$ be the lengths of the cell along two principal directions, and $d_{1}$ be the length along the third, the distinguished principal direction. The aspect ratio of the cell is $\eta_{cell}=d_{1}/d_{2}$; the stretch is $\lambda_{cell}=\eta_{cell}^{2/3}$. The cell is considered prolate if $\eta_{cell}>1$ and oblate if $\eta_{cell}<1$. In calculating the equilibrium free energy, at each volume fraction, we use $\eta_{cell}$, which minimizes the free energy, to satisfy self-consistency.

We note that if the maximum (linear) dimension of the ellipsoid exceeds the minimum dimension of the cell, then the cell will exclude some orientations of the ellipsoid. In the case of a cubic domain, with $\eta_{cell}=1$, this occurs at a critical volume fraction; the corresponding critical volume fractions are

$$\phi_{cp}=\frac{\frac{4\pi}{3}ab^{2}}{(2a)^{3}}=\frac{\pi}{6}\left(\frac{b}{a}\right)^{2}$$

(24)

for prolate and

$$\phi_{co}=\frac{\frac{4\pi}{3}ab^{2}}{(2b)^{3}}=\frac{\pi}{6}\left(\frac{a}{b}\right)$$

(25)

for oblate ellipsoids.

Figure 2 shows the results for two representative ellipsoids with aspect ratios $2$ and $1/2$, respectively, as function of occupied volume fraction. Volume fraction was varied by changing particle volumes at fixed particle aspect ratio. Anisotropy at large volume fractions is apparent both in the cell shape, and in the orientational distribution function. Four scenarios have been observed: prolate or oblate particles in prolate or oblate cells. If the particle and cell shape are similar, then $S_{2}>0$. If the particle and cell shapes are dissimilar, then $S_{2}<0$.

For the prolate ellipsoid with $\eta_{particle}=2$, the system is dilute with $\phi<\phi_{cp}$, the whole orientation state space is accessible, the cubic cell with $\eta_{cell}=1$ is the only minimiser of $F$, and the orientational order parameter $S_{2}=0$, the system is isotropic. At $\phi=\phi_{cp}$, the isotropic state loses its stability and two orientationally ordered solutions bifurcate from the isotropic branch. The prolate branch with $S_{2}>1,\eta_{cell}>1$ has lower energy and pressure than those of the oblate branch with $S_{2}<0,\eta_{cell}>1$ for $\phi>\phi_{cp}$.

For the oblate ellipsoid with $\eta_{particle}=0.5$, although the bifurcation is at $\phi_{c}$, orientational order can be observed both below and above $\phi_{c}$. In fact, there are two first order transitions from isotropic to both prolate nematic phase and oblate nematic phase at two different volume fractions slightly below $\phi_{c}$. The prolate branch with $S_{2}>0,\eta_{cell}<1$ possesses lower energy and pressure than those of the oblate branch with $S_{2}>0,\eta_{cell}>1$.

For sufficiently large volume fractions, certain orientations are disallowed by the shape of the cell, and the orientational distribution function has nontrivial compact support. This is indicated in Fig. 3(b-e), where representative equilibrium orientational probability density functions for a prolate ellipsoid with $\eta_{particle}=2$ in equilibrium prolate $\eta_{cell}>1$ (Figs. 3(b) and (c)) and oblate $\eta_{cell}<1$ (Figs. 3(d) and (e)) cells are shown. One can readily see that the distribution function lacks axial symmetry, even in the very dilute limit where $S_{2}=0$, as shown in Fig. 3(a). This is because an uniaxial ellipsoid has more free volume when it aligns along the diagonals of a rectangular cell compared to other orientations.

As the volume fraction is increased, a topological transition takes place: the connected accessible orientational space becomes disconnected forming a number of disconnected regions, where the orientation of the ellipsoid is locked in a small region, as can be observed in going from Figs. 3(b) to (c) and from (d) to (e). These topological changes also manifest themselves as kinks on the bifurcation diagrams as in Fig. 2 and Fig. 4, although the topological change on the prolate branch is not as apparent as the one on the oblate branch.

Since the orientational distribution function is known, in addition to $S_{2}$, higher moments $S_{n}=\langle\mathcal{P}_{n}(\mathbf{\hat{m}\cdot\hat{e}})\rangle$ can be readily calculated. Here $\mathcal{P}_{n}$ is the Legendre polynomial of degree $n$. Although in the isotropic phase $S_{2}=0$, interestingly, higher moments may exist as nonzero functions of the occupied volume fraction $\phi$, as shown in Fig. 4(b). Given the edges and corners of the box, it is not surprising that the distribution function has higher moments which are nonzero at all volume fractions.

The free volume of an arbitrarily shaped particle in a rectangular cell can be calculated using a support function based approach, where the support function plays the role of the function $h$ in Eq. (22). We also remark that as long as the cell is convex, the free volume for a concave particle will be identical to that of its convex hull. Applying the support function approach to extend our model to particles with more complex shapes has already been initiated by one of us (JMT).

We realize that cell shapes other than rectangular boxes may allow more free volume to ellipsoidal particles; we expect that to every particle shape there corresponds an optimal cavity shape with the lowest free energy in equilibrium. The rectangular box is just our first attempt; an ellipsoidal cell is considered in Section 3.3.

Molecular dynamics (MD) is a useful tool for studying properties of systems of interacting particles. MD simulations keep track of the motion of individual particles relying on classical mechanics. The results of MD simulations can be used to test the validity of statistical mechanical models. Here, we perform MD simulations of a hard rigid ellipsoidal particle moving in a rectangular cell, undergoing collisions with the cell walls. We imagine that this approximates the real physical situation where thermally excited molecules collide with nearest neighbors. As we show in this subsection, MD gives the same predictions for the equilibrium shape of the cell, for the average orientation of the ellipsoid, and for the pressure as our statistical mechanical model.

The dynamics consists of free flights and collisions with cell walls. When the particle does not undergo a collision, the position of its center of mass is updated as

$$\mathbf{r(}t_{i+1})=\mathbf{r}(t_{i})+\mathbf{v}_{cm}\Delta t_{i}.$$

(26)

However, the angular velocity of the particle changes in time even without external torque. The angular acceleration in an inertial reference frame is given by Euler’s equations for rigid body dynamics,

$$\dot{\bm{\omega}}(t_{i})=\mathbf{-I}^{-1}\cdot(\bm{\omega}\times(\mathbf{I}\cdot\bm{\omega})),$$

(27)

where $\mathbf{I}$ is the tensor of inertia of the ellipsoid, given by

$$\mathbf{I}=(\frac{2}{5}mb^{2}-\frac{1}{5}m(a^{2}+b^{2}))\mathbf{\hat{m}\hat{m}}+\frac{1}{5}m(a^{2}+b^{2})\mathbb{I}\mathbf{,}$$

(28)

where $m$ is its uniformly distributed mass. We then update the angular velocity and the orientation of the particle as follows,

	$\displaystyle\bm{\omega}(t_{i})$	$\displaystyle=$	$\displaystyle\bm{\omega}(t_{i-1})+\bm{\dot{\omega}}\Delta t_{i},$		(29)
	$\displaystyle\mathbf{\hat{m}}(t_{i+1})$	$\displaystyle=$	$\displaystyle\frac{\mathbf{\hat{m}}(t_{i})+\bm{\omega}(t_{i})\times\mathbf{\hat{m}}(t_{i})\Delta t_{i}}{|\mathbf{\hat{m}}(t_{i})+\bm{\ \omega}(t_{i})\times\mathbf{\hat{m}}(t_{i})\Delta t_{i}|},$		(30)

where we use the default value of $\Delta t_{i}=10^{-3}$. If the updated configuration causes any part of the particle to leave the cell, we return to the previous step, and halve $\Delta t_{i}$. The process is repeated until the new configuration of the particle is completely within the cell. When the time step size reaches a lower threshold (e.g., $\Delta t_{i}=10^{-14}$), collision with the wall occurs.

We assume that both the particle and cell walls are frictionless, thus the impulses are along the normals to the surfaces in contact. In a collision, the conservation of linear momentum, angular momentum and kinetic energy, gives the postcollision center of mass velocity $\mathbf{v}_{f}$ and angular velocity $\bm{\omega}_{f}$ by ([12])

$$\mathbf{v}_{f}=\mathbf{v}_{i}-\frac{J}{m}\mathbf{\hat{N}},$$

(31)

$$\bm{\omega}_{f}=\bm{\omega}_{i}-J\mathbf{I}^{-1}\cdot(\mathbf{p\times\hat{N}}),$$

(32)

where $\mathbf{v}_{i}$ and $\bm{\omega}_{i}$ are precollision velocity and angular velocity, $\mathbf{\hat{N}}$ is the outward normal of the ellipsoid pointing towards the cell wall. The vector $\mathbf{p}$ is a body-fixed vector joining the center of mass of the ellipsoid to the point of contact with the cell wall at the instant of collision,

$$\mathbf{p}=\frac{\mathbf{A}^{-1}\cdot\mathbf{\hat{N}}}{\sqrt{\mathbf{\hat{N}}\cdot\mathbf{A}^{-1}\cdot\mathbf{\hat{N}}}}$$

(33)

and the positive scalar $J$ is the magnitude of the impulse transferred to the ellipsoid at the collision,

$$J=2\frac{(\mathbf{\mathbf{v}}_{i}+\mathbf{\mathbf{\omega}}_{i}\times\mathbf{p)\cdot\hat{N}}}{1/m-\mathbf{\hat{N}}\cdot\mathbf{p}\times\mathbf{I}^{-1}\times\mathbf{p}\cdot\mathbf{\hat{N}}}.$$

(34)

The pressure on each wall is the average impulse per area on the wall. If the pressure happens not to be isotropic, we then update the relevant length of the cell, denoted $d_{l}$, according to

$$d_{l}\rightarrow d_{l}+\gamma\frac{P_{l}-P_{t}}{P_{l}},$$

(35)

where $P_{l}$ and $P_{t}$ represent the averaged pressures on two square walls and four rectangular walls, respectively, and $\gamma=0.01$ plays the role of compressibility. The updates for the other two dimensions are determined by maintaining constant volume and a uniaxial shape.

A very great deal of work has been done on simulations of systems of hard ellipsoids[13],[14],[15],[16]. Our system is different, however, due to the mean field aspect of our model, where only a single particle is present, enclosed by a cell representing the effects of the other particles. Confinement induced ordering of hard ellipsoids has also been studied between two hard walls [17] however our confinement is in 3D; furthermore the shape of the confining shape is adaptive.

The scalar order parameter is calculated according to $S_{n}=\langle\mathcal{P}_{n}(\mathbf{\hat{m}\cdot\hat{e}})\rangle$ where the average is over all instantaneous orientations of the symmetry axis $\mathbf{\hat{m}}$ of the ellipsoid between collisions. All simulation results were averaged over $10^{3}$ random initial conditions.

Figure 4 shows the agreement of the results of free volume calculation in Sect. 3.1 and MD simulations for an ellipsoid with aspect ratio $4$. The transition from the isotropic phase to the prolate phase is continuous, whereas the transition from the isotropic phase to the oblate phase is discontinuous. This is different from the case of $\eta_{particle}=2$ where both transitions are continuous.

In this subsection, we consider a needle within an ellipsoidal cell. This geometry allows for an analytic free volume expression, and reveals a topology common to the case of very slender particles.

Consider $\mathcal{C}_{0}$ to be a spherical cavity $\mathscr{B}_{R}$ of radius $R$ enclosing a rigid rod of length $2a$ (see Fig. 5(a)) and subject the sphere to a homogeneous, isochoric deformation $\bm{f}$ with gradient $\mathbf{F}$, which, modulo a rotation, changes the sphere into the ellipsoid $\mathcal{C}$ depicted in Fig. 5(b).

Now, the unit vector $\mathbf{\hat{m}}$ denotes the orientation of the rod, such that if the center of the rod is at the point $\bm{y}$, its ends occupy the points $\bm{y}_{\pm}=\bm{y}\pm a\mathbf{\hat{m}}$, respectively. We want to determine the volume of the free region $\mathscr{R}_{\mathrm{f}}(\mathbf{\hat{m}})$ accessible to $\bm{y}$ within the ellipsoid $\mathcal{C}$, for a given orientation $\mathbf{\hat{m}}$ of the rod. Writing $\bm{y}_{\pm}=\mathbf{F}\bm{x}_{\pm}$ and $\bm{y}=\mathbf{F}\bm{x}$, so that $\bm{x}_{\pm}$ are the preimages of $\bm{y}_{\pm}$, and $\bm{x}$ the preimage of $\bm{y}$, the region $\bm{f}^{-1}(\mathscr{R}_{\mathrm{f}})$ comprises all points $\bm{x}$ such that

$$\mathbf{F}\bm{x}\pm a\mathbf{\hat{m}}=\mathbf{F}\bm{x}_{\pm}\quad\text{for both}\quad\bm{x}_{\pm}\in\mathscr{B}_{R},$$

(36)

from which, since $\mathbf{F}$ is invertible, we have that

$$\bm{x}\pm a\mathbf{F}^{-1}\mathbf{\hat{m}}=\bm{x}_{\pm},\quad\bm{x}_{\pm}\in\mathscr{B}_{R},$$

(37)

and so

$$\bm{f}^{-1}(\mathscr{R}_{\mathrm{f}})=\{\bm{x}:|\bm{x}+a\mathbf{F}^{-1}\mathbf{\hat{m}}|\leq R\}\cap\{\bm{x}:|\bm{x}-a\mathbf{F}^{-1}\mathbf{\hat{m}}|\leq R\}.$$

(38)

Geometrically, this set is the region comprised of two spheres of radius $R$ whose centers are $2b$ apart, where

$$b:=a|\mathbf{F}^{-1}\mathbf{\hat{m}}|.$$

(39)

Since $\det\mathbf{F}=1$, the free volume $V_{\mathrm{f}}$ is the same as the volume of $\bm{f}^{-1}(\mathscr{R}_{\mathrm{f}})$, that is, twice the volume of a spherical frustum of radius $r:=\sqrt{R^{2}-b^{2}}$ and height $h:=R-b$,

$$V_{\mathrm{f}}=\frac{\pi h}{3}(3r^{2}+h^{2})=V_{cell}\left(1-\frac{3}{2}\alpha\beta+\frac{1}{2}\alpha^{3}\beta^{3}\right),$$

(40)

where $V_{cell}$ is the volume of $\mathscr{B}_{R}$ and we have set

$$\alpha=\frac{a}{R}\leq 1\quad\text{and}\quad\frac{b}{R}=\alpha\beta\leq 1.$$

(41)

It readily follows from (38) and (40) that

$$\beta=\sqrt{\mathbf{\hat{m}}\cdot\mathbf{B}^{-1}\mathbf{\hat{m}}},$$

(42)

where $\mathbf{B}=\mathbf{F}\mathbf{F}^{\mathsf{T}}$ is the left Cauchy-Green tensor, which in the frame $(\mathbf{\hat{e}}_{1},\mathbf{\hat{e}}_{2},\mathbf{\hat{e}}_{3})$ of the principal directions of stretching is represented as

$$\mathbf{B}=\lambda_{1}^{2}\mathbf{\hat{e}}_{1}\mathbf{\hat{e}}_{1}+\lambda_{2}^{2}\mathbf{\hat{e}}_{2}\mathbf{\hat{e}}_{2}+\lambda_{3}^{2}\mathbf{\hat{e}}_{3}\mathbf{\hat{e}}_{3},$$

(43)

where $\lambda_{1}$, $\lambda_{2}$, and $\lambda_{3}$ are the principal stretches subject to $\lambda_{1}\lambda_{2}\lambda_{3}=1$. Letting $m_{i}$ be the components of $\mathbf{\hat{m}}$ in the frame $(\mathbf{\hat{e}}_{1},\mathbf{\hat{e}}_{2},\mathbf{\hat{e}}_{3})$, they can be written as

$$m_{1}=\sin\theta\cos\varphi,\quad m_{2}=\sin\theta\sin\varphi,\quad m_{3}=\cos\theta,$$

(44)

for $\theta\in[0,\pi]$ and $\varphi\in[0,2\pi]$. By use of (44) and (43) in (42) , we arrive at

$$\beta=\sqrt{\frac{1}{\lambda_{1}^{2}}\sin^{2}\theta\cos^{2}\varphi+\frac{1}{\lambda_{2}^{2}}\sin^{2}\theta\sin^{2}\varphi+\frac{1}{\lambda_{3}^{2}}\cos^{2}\theta}.$$

(45)

For simplicity, we shall assume that $\mathbf{B}$ has the uniaxial symmetry around $\hat{\mathbf{e}}_{3}$, so that $\lambda_{1}=\lambda_{2}=1/\sqrt{\lambda_{3}}$ and $\beta$ reduces to

$$\beta=\sqrt{{\lambda}\sin^{2}\theta+\frac{1}{\lambda^{2}}\cos^{2}\theta},\text{ with }\lambda=\lambda_{3}.$$

(46)

With this choice, the ellipsoid is prolate along $\hat{\mathbf{e}}_{3}$ if $\lambda>1$ and oblate if $\lambda<1$.

For given $\lambda$, the domain $I_{\lambda}$ of admissible values of $\theta$ is restricted by the second inequality in (41). A direct inspection shows that where

$$I_{\lambda}=\left\{\begin{array}[]{cc}[\chi,\pi-\chi],&0\leq\lambda\leq\alpha\\ [0,\pi],&\alpha\leq\lambda\leq\frac{1}{\alpha^{2}}\\ [0,\chi]\cup[\pi-\chi,\pi],&\lambda\geq\frac{1}{\alpha^{2}}\end{array}\right.$$

(47)

where

$$\chi=\arccos\left(\frac{\lambda}{\alpha}\sqrt{\frac{1-\lambda\alpha^{2}}{1-\lambda^{3}}}\right).$$

(48)

The free energy of the rod (in units $kT$) is then given by

$$F(\alpha,\lambda)=-\ln\left(\pi\int_{I_{\lambda}}(2-3\alpha\beta+\alpha^{3}\beta^{3})\sin\theta d\theta\right)+\ln[2\pi(\alpha-1)^{2}(\alpha+2)],$$

(49)

where the added constant has been gauged so as to ensure that $F(\alpha,1)=0$. The integral in (49) can be expressed in terms of elementary functions, not given explicitly here in the interest of space. Plots of $F$ as function of $\lambda$ are shown in Fig. 6 for several values of $\alpha$.

For $\alpha<\alpha_{\mathrm{c}}^{(1)}\doteq 0.737$, $F$ has a single critical point at $\lambda=1$, its absolute minimum representing the isotropic state where the spherical globule remains undeformed. For $\alpha>\alpha_{\mathrm{c}}^{(1)}$, a local minimum and a local maximum develop in the oblate branch (where $\lambda<1$), while the isotropic state remains the absolute minimum, until $\alpha$ reaches the transition value $\alpha_{\mathrm{t}}\doteq 0.757$, where both the isotropic state and the oblate minimum have the same energy. A first-order transition takes place at $\alpha=\alpha_{\mathrm{t}}$, with the oblate minimum at $\lambda\doteq 0.518$. At $\alpha=\alpha_{c}^{(2)}\dot{=}0.805$, an inflection point develops on the prolate branch, from which a local maximum and a local minimum emanate, the former approaching the equilibrium isotropic state at $\lambda=1$. At $\alpha=\sqrt{\frac{2}{3}}\doteq 0.816$, the isotropic state becomes unstable and two branches emanate from it, a local prolate minimum and a local oblate maximum: the former connects with the prolate maximum at $\alpha=\alpha_{c}^{(3)}\dot{=}0.826$, while the latter connects with the oblate minimum at $\alpha=\alpha_{c}^{(1)}$. The locally stable prolate branch converges to $\lambda\doteq 2.627$ as $\alpha\rightarrow 1$, whereas the globally stable oblate branch converges to $\lambda\dot{=}0.254$ as $\alpha\to 1$.

Figure 7 summarizes the above results, which are also compared with the bifurcation diagrams for a needle in a rectangular cell. We note that the topologies are mildly distinct. Whilst the oblate branch has a similar topology in each case, two unstable branches, one prolate and one oblate, bifurcate from the isotropic state at the critical value $\alpha=1$ in the cuboidal case, with the former being a very small region that is difficult to see in the figure.