Invariance of U_harm of crystal to rotation

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Homework Statement



Show that because a pure rotational displacement field u(r) has no effect, that the energy of a crystal only depends on the symmetric strain tensor epsilon_ij.

Homework Equations



As in Ashcroft & Mermin (22.72), the energy of a crystal as a function of displacement field u_i(r) and dynamical matrix D_ij(R) (R = direct lattice vector) is:
U_harm = 1/2 Integral[dr Sum[ijkl, (du_j/dx_i)(du_l/dx_k)E_ijkl ]]
where
E_ijkl = -1/2 Sum[R, R_i D_jl(R) R_k] (ijkl indices go over x,y,z)

We need to show that this only depends on the strain in the symmetric combination
epsilon_ij = 1/2( du_i/dx_j + du_j/dx_i ) (Ashcroft & Mermin 22.77)

The Attempt at a Solution



For simplicity, consider the 2D case. Then u(x,y) = (-ay, ax) for small a is an infinitesimal rotation of the plane. Writing (22.72) for the crystal energy, we get

0 = U_harm = 1/2 Integral[dr (du_y/dx)(du_y/dx) E_xyxy + (du_y/dx)(du_x/dy)E_xyyx + (du_x/dy)(du_y/dx)E_yxxy + (du_x/dy)(du_x/dy)E_yxyx]
= 1/2 Integral[dr a^2(E_xyxy - E_xyyx - Eyxxy + E_yxyx) ] = 0

Or, E_xyxy - E_xyyx - Eyxxy + E_yxyx = 0.

Now, we rewrite du_y/dx = eps_xy + ant_xy = 1/2(du_y/dx+du_x/dy) + 1/2(du_y/x-du_x/dy), and similarly du_x/dy = eps_xy - ant_xy as symmetric and antisymmetric parts. Then we can substitute in (22.72):

U_harm = 1/2 Integral[dr (du_y/dx)(du_y/dx) E_xyxy + (du_y/dx)(du_x/dy)E_xyyx + (du_x/dy)(du_y/dx)E_yxxy + (du_x/dy)(du_x/dy)E_yxyx + (remaining terms)]
= 1/2 Integral[ dr, (eps_xy + ant_xy)(eps_xy + ant_xy)(E_xyxy) + (eps_xy + ant_xy)(eps_xy - ant_xy)(E_xyyx + E_yxxy) + (eps_xy - ant_xy)(eps_xy - ant_xy)(E_yxyx) + (remaining terms)]
= 1/2 Integral[ dr, eps_xy^2 (E_xyxy + E_xyyx + Eyxxy + E_yxyx) + (eps_xy ant_xy)(2 E_xyxy - 2 E_yxyx) + ant_xy^2(E_xyxy - E_xyyx - Eyxxy + E_yxyx) + (remaining terms)]

Now the coefficient of ant_xy^2 vanishes, but I can't see why the coefficient for eps_xy*ant_xy would vanish. The goal is to remove all dependence on antisymmetric terms. I see no reason from what I read in Ashcroft & Mermin that E_xyxy = E_yxyx generally.
 

Answers and Replies

  • #2
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I figured it out
 

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