Elastic Energy $E_{el}$

The elastic energy $E_{el}$ is bilinear in $a$

	$$E_{el}$$	$=$	$$\frac{1}{2}\sum_{ij} \frac{c_L(ij)-c_T(ij)}{2\vert\mathbf R_{ij}\vert^2} (\mathbf R_{ij}^T\bar a \mathbf R_{ij})^2$$	(80)
			$$+ \frac{c_T(ij)}{2} \mathbf R_{ij}^T\bar a^T\bar a \mathbf R_{ij}$$
		$=$	$$\frac{1}{2} \sum_{ij,\alpha\beta\gamma\delta} \frac{c_L(ij)-c_T(i... ...ha}R_{ij}^{\beta}R_{ij}^{\gamma}R_{ij}^{\delta} a_{\alpha\beta}a_{\gamma\delta}$$
			$$+ \frac{c_T(ij)}{2} R_{ij}^{\alpha} R_{ij}^{\delta} a_{\alpha\beta} \delta_{\beta\gamma} a_{\gamma\delta}$$

We calculate it's derivative with respect to $a_{\alpha\beta}$:

	$$\frac{\partial E_{el}}{\partial a_{\alpha\beta}}$$	$=$	$$\frac{1}{2}\sum_{ij,\gamma\delta} \frac{c_L(ij)-c_T(ij)}{\vert\ma... ...2} R_{ij}^{\alpha}R_{ij}^{\beta}R_{ij}^{\gamma}R_{ij}^{\delta} a_{\gamma\delta}$$
			$$+ \frac{c_T(ij)}{2} R_{ij}^{\alpha}R_{ij}^{\delta} \delta_{\beta\gamma} a_{\gamma\delta} +$$
			$$\frac{c_T(ij)}{2} R_{ij}^{\gamma} R_{ij}^{\beta} a_{\gamma\delta} \delta_{\delta\alpha}$$	(81)

An we make also use of the definition of elastic constants:

	$$c^{\alpha\beta\gamma\delta}$$	$=$	$$\frac{\partial^2E_{el}}{\partial a_{\alpha\beta} \partial a_{\gamma\delta}}$$	(82)
		$=$	$$\frac{1}{2}\sum_{ij}\frac{c_L(ij)-c_T(ij)}{\vert\mathbf R_{ij}\vert^2} R_{ij}^{\alpha}R_{ij}^{\beta}R_{ij}^{\gamma}R_{ij}^{\delta}$$
			$$+ \frac{c_T(ij)}{2} R_{ij}^{\alpha}R_{ij}^{\delta} \delta_{\beta\gamma} +$$
			$$\frac{c_T(ij)}{2} R_{ij}^{\gamma} R_{ij}^{\beta} \delta_{\delta\alpha}$$	(83)

we make use of the fact that the strain $\bar \epsilon$ is a symmetric tensor ( $\epsilon_{\alpha\beta}=\epsilon_{\beta\alpha}$) and a rotation is antisymmetric ( $\omega_{\alpha\beta}=-\omega_{\beta\alpha}$) and the linear transformation $\bar a$ can be written as $\bar a=\bar \epsilon + \bar \omega$. Thus it is possible to rewrite the elastic energy in the well known fashion

$$E_{el}$$

$=$

$$\frac{1}{2}\sum_{\alpha\beta\gamma\delta} c^{\alpha\beta\gamma\delta} \epsilon_{\alpha\beta}\epsilon_{\gamma\delta}$$

(84)

Note that we have neglected the fact, that nonzero transversal springs will result in a dependence of the elastic energy on the rotation tensor $\bar \omega$ as can been seen by inserting a rotation into the second part of (84). ¹² Therefore transversal springs have to be used with caution in the description of a phonon spectrum.

Note that the transversal spring part in the elastic constants as defined in (86) is not symmetric upon exchange of indices. It is convenient to use a symmetrized version of the elastic constants, which will yield the same elastic energy, because the strain tensor is symmetric:

	$$c^{\alpha\beta\gamma\delta}$$	$=$	$$\frac{1}{2}\sum_{ij}\frac{c_L(ij)-c_T(ij)}{\vert\mathbf R_{ij}\vert^2} R_{ij}^{\alpha}R_{ij}^{\beta}R_{ij}^{\gamma}R_{ij}^{\delta}$$
			$$+ \frac{c_T(ij)}{4} ( R_{ij}^{\alpha}R_{ij}^{\delta} \delta_{\beta\gamma} +$$
			$$R_{ij}^{\gamma} R_{ij}^{\beta} \delta_{\delta\alpha}$$
			$$+ R_{ij}^{\beta}R_{ij}^{\delta} \delta_{\alpha\gamma} + R_{ij}^{\gamma}R_{ij}^{\alpha} \delta_{\delta\beta} )$$

With the help of elastic constants we can rewrite the derivative of the elastic energy (88), again ignoring the rotation dependence:

$$\frac{\partial E_{el}}{\partial \epsilon_{\alpha\beta}}$$

$=$

$$\sum_{\gamma\delta=1,2,3} c^{\alpha\beta\gamma\delta} \epsilon_{\gamma\delta}$$

(85)

To easy the indexing we apply the notation of Voigt

	$$\epsilon_1$$	$=$	$$\epsilon_{11}$$	(86)
	$$\epsilon_2$$	$=$	$$\epsilon_{22}$$
	$$\epsilon_3$$	$=$	$$\epsilon_{33}$$
	$$\epsilon_4$$	$=$	$$2\epsilon_{23}=2\epsilon_{32}$$
	$$\epsilon_5$$	$=$	$$2\epsilon_{31}=2\epsilon_{13}$$
	$$\epsilon_6$$	$=$	$$2\epsilon_{12}=2\epsilon_{21}$$

Note the elastic constants do not contain any prefactor in Voigt notation, i.e.

	$$c^{11}$$	$=$	$$c^{1111}$$	(87)
	$$c^{44}$$	$=$	$$c^{2323}$$

There are 21 independent elastic constants, $c^{\alpha\beta}$ with $\beta=1,...,6$ and $\alpha \le \beta$ (or less, see http://koski.ucdavis.edu/BRILLOUIN/CRYSTALS/Elasticities.html). The other 60 elastic constants can be obtained from the symmetry relations $c^{\alpha\beta}=c^{\beta\alpha}$, $c^{\alpha\beta\gamma\delta}=c^{\beta\alpha\gamma\delta}=c^{\alpha\beta\delta\gamma}$.

The elastic energy in Voigt notation is given by

$$E_{el}$$

$=$

$$\frac{1}{2}\sum_{\alpha\gamma=1,..,6} c^{\alpha\gamma} \epsilon_{\alpha}\epsilon_{\gamma}$$

(88)