Direct observation of coupling between orientation and flow fluctuations in a nematic liquid crystal at equilibrium

To demonstrate coupling between orientation and flow fluctuations in a nematic liquid crystal at equilibrium, we simultaneously observe the intensity change due to director fluctuations under a polarizing microscope and the Brownian motion of a fluorescent particle trapped weakly by optical tweezers. The calculated cross-correlation function of the particle position and the spatial gradient of the intensity is nonzero, clearly indicating the existence of coupling.

Figure 1

(a) Coupling between bend deformation of $\delta n_x$ and shear flow of $v_x$, which both fluctuate sinusoidally with a wave number along the $y$ axis. At the point $\mathrm{P}, v_x$ becomes maximum, whereas the spatial gradient of $\delta n_x$ becomes minimum. (b) Twist deformation along the $z$ axis. The amplitude of the bend deformation in the $x-y$ plane depends on $z$: it is maximum at $z=d/4\mathrm{and}3d/4$, whereas it vanishes at $z=d/2$. It should be noted that the phases at $z=d/4\mathrm{and}3d/4$ are opposite, indicating that the corresponding flow velocities are also opposite.

Figure 2

Typical image obtained by combining a polarizing microscope and a fluorescence microscope. The angle between the polarizer and analyzer, θ, is set to be ${72}^{\circ }$. The spatial fluctuations of $\delta n_x$ are represented by the intensity changes, and the temporal change can be seen in a video [17]. A particle is trapped at the center.

Figure 3

(a) Time correlation functions of image intensity at $q_x=0$ and several values of $q_y$. (b) Inverse of $\langle {|\delta I\left(q_x,q_y\right)|}^2\rangle$, obtained from Fig. 3 by least-squares fitting, as a function of $q_x^2\mathrm{at}{q}_y=0$ and as a function of $q_y^2\mathrm{at}{q}_x=0$.

Figure 4

Mean-squared displacements (MSDs) obtained from Fig. 2 in the $x$ and $y$ directions.

Figure 5

Cross-correlation functions $\langle \left(x\left(t\right)-x\left(0\right)\right)\left(g_y\left(t\right)-g_y\left(0\right)\right)\rangle$ obtained at $z=d/4,d/2$, and $3d/4$. Solid lines are calculated from Eq. (9).