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This question is about physical theory, but my question is pure nathematical, so I post it here. I don't think you have to know physics in order to answer it.

I am studying liquid crystal theory with the book Kleman, Lavrentovich, Soft Matter Physics.

In the Ericksen-Leslie theory, Frank-OSeen energy is:

$$ f=0.5*(K_1*div^2 (n)+K_2 *(n*curl(n))^2+K_3*(n \times curl(n) )^2) $$

Here $\overrightarrow{n}=\overrightarrow{n}(x,y,z,t)$ is director (almost like a vector, for this problem you can treat it is a vector field), and summation over repeated indexes is used. $K_1,K_2,K_3$ are constants. "$*$" denotes scalar product (or product of two scalar quantities), "$ \times$" denotes vector product.

Later they take a full derivative with respect to time:

$$ {{df} \over {dt}}= {\partial f \over \partial n_i} {dn_i \over dt} + {\partial f \over \partial (\partial n_i / \partial x_j)} {d \over dt} {\partial n_i \over \partial x_j} $$

Summation over repeated indexes is used. Variables $x,y,z = x_1, x_2,x_3$ respectively are spatial variables, as usual in tensor algebra.

The above formula for $df/dt$ looks like the famous formula for the total derivative, but if the independent variables were $t$ and spatial derivatives of the director components.

1) Am I write that expression for $df/dt$ contains a sum of 10 terms?

2) How can I make sure (proove), that I indeed can use above 10 variables as independent, instead of standard 4 variables $t,x,y,z$.

Note: I think, the precise formula for $f(\overrightarrow{n})$ is not really important. I posted it just in case, so you can better understand what is going on.

Hope, that my question is clear. Thanks

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  • $\begingroup$ What is n? If it is a director then how can * be a scalar product? What does curl mean if n has 4 components? $\endgroup$
    – Paul
    Mar 25, 2015 at 19:13
  • $\begingroup$ $n$ has three components. It depends on 4 variables. Directors are almost like vectors, except of the presence of reversal symmetry. All vector operations are well-defined for directors. You can here consider $n$ as a vector $\endgroup$ Mar 25, 2015 at 19:15
  • $\begingroup$ Just treat $n$ as usual vector field. $\endgroup$ Mar 25, 2015 at 19:17
  • $\begingroup$ Crossposted from physics.stackexchange.com/q/172342/2451 $\endgroup$
    – Qmechanic
    Mar 26, 2015 at 0:11
  • $\begingroup$ Yes, it reffred both to physics and math $\endgroup$ Mar 26, 2015 at 21:05

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