Liquid Crystals. Iam-Choon Khoo

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odd function of Q ensures that states with some nonvanishing value of Q (e.g. due to some alignment of molecules) will have different free‐energy values depending on the direction of the alignment. For example, the free energy for a state with an order parameter Q of the form

      (2.24a)upper Q 1 equals Start 3 By 3 Matrix 1st Row 1st Column negative xi 2nd Column 0 3rd Column 0 2nd Row 1st Column 0 2nd Column negative xi 3rd Column 0 3rd Row 1st Column 0 2nd Column 0 3rd Column 2 xi EndMatrix

Schematic illustration of free energies F(Q) for different temperatures T.

      (2.24b)upper Q 2 equals Start 3 By 3 Matrix 1st Row 1st Column xi 2nd Column 0 3rd Column 0 2nd Row 1st Column 0 2nd Column xi 3rd Column 0 3rd Row 1st Column 0 2nd Column 0 3rd Column minus 2 xi EndMatrix equals minus upper Q 1

      (which signifies some alignment of the molecules in the xy plane).

      2.4.2. Free Energy in the Presence of an Applied Field

      In the presence of an externally applied field (e.g. dc or low‐frequency electric, magnetic, or optical electric field), a corresponding interaction term should be added to the free energy.

      For an applied magnetic field H, the energy associated with it is

      (2.25)upper F Subscript i n t Baseline equals minus integral Subscript 0 Superscript upper H Baseline bold upper M dot d bold upper H comma

      (2.26)upper M Subscript alpha Baseline equals sigma-summation Underscript beta Endscripts chi Subscript italic alpha beta Baseline upper H Subscript beta Baseline period

      Thus

      (2.27)upper F Subscript i n t Baseline equals minus one half sigma-summation Underscript alpha Endscripts chi Subscript italic alpha beta Superscript m Baseline upper H Subscript beta Baseline upper H Subscript alpha Baseline period

      Using Eq. (2.9), we can rewrite Fint as

      (2.29)upper F Subscript i n t Superscript upper H Baseline equals minus one half sigma-summation Underscript alpha comma beta Endscripts upper Q Subscript italic alpha beta Baseline upper H Subscript alpha Baseline upper H Subscript beta Baseline period

      Therefore, the total free energy of a liquid crystal in the isotropic phase, under the action of an externally applied magnetic field, is given by

      (2.30)StartLayout 1st Row upper F equals upper F prime 0 plus one half upper A left-parenthesis upper T right-parenthesis sigma-summation Underscript alpha comma beta Endscripts upper Q Subscript italic alpha beta Baseline upper Q Subscript italic beta alpha Baseline plus one third upper B left-parenthesis upper T right-parenthesis sigma-summation Underscript alpha comma beta comma gamma Endscripts upper Q Subscript italic alpha beta Baseline upper Q Subscript italic beta gamma Baseline upper Q Subscript italic gamma alpha Baseline 2nd Row minus one half sigma-summation Underscript alpha comma beta Endscripts upper Q Subscript italic alpha beta Baseline upper H Subscript alpha Baseline upper H Subscript beta Baseline period EndLayout

      Without solving the problem explicitly, we can infer from the magnetic interaction term that a lower energy state corresponds to some alignment of the molecules in the direction of the magnetic field (for Δχ m > 0).

      Using a similar approach, we can also deduce that the electric interaction contribution to the free energy is given by (in inks units)

      (2.31)upper F Subscript i n t Superscript upper E Baseline equals minus one half integral Subscript 0 Superscript upper E Baseline bold upper D dot bold upper E equals minus one half epsilon Subscript up-tack Baseline upper E squared minus one half normal upper Delta epsilon left-parenthesis ModifyingAbove n With ampersand c period circ semicolon dot bold upper E right-parenthesis squared period

      (2.32)upper F Superscript upper E Baseline equals minus one half normal upper Delta epsilon left-parenthesis ModifyingAbove n With ampersand c period circ semicolon dot bold upper E right-parenthesis squared equals minus one half sigma-summation 
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