Elastic Energy $E_{el}$

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

$$E_{el} = \frac{1}{2}\sum_{ij} \frac{c_L(ij)-c_T(ij)}{2\vert\vec R_{ij}\vert^2} (\vec R_{ij}^T\bar a \vec R_{ij})^2$$ (79)

$$+ \frac{c_T(ij)}{2} \vec R_{ij}^T\bar a^T\bar a \vec R_{ij}$$

$$= \frac{1}{2} \sum_{ij,\alpha\beta\gamma\delta} \frac{c_L(ij)-c_T(i... ...ha}R_{ij}^{\beta}R_{ij}^{\gamma}R_{ij}^{\delta} a_{\alpha\beta}a_{\gamma\delta}$$

$$+ \frac{c_T(ij)}{2} R_{ij}^{\alpha} R_{ij}^{\delta} a_{\alpha\beta} \delta_{\beta\gamma} a_{\gamma\delta}$$

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

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

$$+ \frac{c_T(ij)}{2} R_{ij}^{\alpha}R_{ij}^{\delta} \delta_{\beta\gamma} a_{\gamma\delta} +$$

$$\frac{c_T(ij)}{2} R_{ij}^{\gamma} R_{ij}^{\beta} a_{\gamma\delta} \delta_{\delta\alpha}$$ (80)

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 derivatives with respect to the strain component $\epsilon_{\alpha\beta}$ with $\alpha,\beta=1,2,3,\alpha \le \beta$ can be written as

$$\frac{\partial}{\partial \epsilon_{\alpha\beta}} = \sum_{\gamma\delta=1,2,3}\frac{\partial a_{\gamma\delta}}{\partial \epsilon_{\alpha\beta}} \frac{\partial}{\partial a_{\gamma\delta}}$$ (81)

$$= \frac{\partial}{\partial a_{\alpha\beta}} + (1-\delta_{\alpha\beta}) \frac{\partial}{\partial a_{\beta\alpha}}$$ (82)

We calculate the derivative of the elastic energy with respect to the strain:

$$\frac{\partial E_{el}}{\partial \epsilon_{\alpha\beta}} = \frac{\partial E_{el}}{\partial a_{\alpha\beta}} + \frac{\partial... ...\alpha}}-\delta_{\alpha\beta} \frac{\partial E_{el}}{\partial a_{\alpha\alpha}}$$ (83)

$$= \frac{1}{2}\sum_{ij,\gamma\delta} (2-\delta_{\alpha\beta})\frac{c... ...2} R_{ij}^{\alpha}R_{ij}^{\beta}R_{ij}^{\gamma}R_{ij}^{\delta} a_{\gamma\delta}$$

$$+ \frac{c_T(ij)}{2} a_{\gamma\delta} ( R_{ij}^{\alpha}R_{ij}^{\delta} \delta_{\beta\gamma} + R_{ij}^{\gamma} R_{ij}^{\beta} \delta_{\delta\alpha}$$

$$+ R_{ij}^{\beta} R_{ij}^{\delta} \delta_{\alpha\gamma} + R_{ij}^{\gamma}R_{ij}^{\alpha} \delta_{\delta\beta}$$

$$- R_{ij}^{\alpha}R_{ij}^{\delta} \delta_{\alpha\beta}\delta_{\bet... ...} - R_{ij}^{\gamma} R_{ij}^{\beta} \delta_{\delta\alpha} \delta_{\alpha\beta} )$$

An we make also use of the definition of elastic constants to find

$$c^{\alpha\beta\gamma\delta} = \frac{\partial^2E_{el}}{\partial \epsilon_{\alpha\beta} \partial \epsilon_{\gamma\delta}}$$ (84)

$$= \frac{1}{2}\sum_{ij}(2-\delta_{\alpha\beta})(2-\delta_{\gamma\del... ...\vec R_{ij}\vert^2} R_{ij}^{\alpha}R_{ij}^{\beta}R_{ij}^{\gamma}R_{ij}^{\delta}$$

$$+ c_T(ij) ( R_{ij}^{\alpha}R_{ij}^{\delta} \delta_{\beta\gamma} + R_{ij}^{\gamma}R_{ij}^{\beta} \delta_{\delta\alpha}$$

$$+ R_{ij}^{\beta} R_{ij}^{\delta} \delta_{\alpha\gamma} + R_{ij}^{\gamma}R_{ij}^{\alpha} \delta_{\delta\beta}$$

$$- R_{ij}^{\alpha}R_{ij}^{\delta} \delta_{\alpha\beta}\delta_{\bet... ...} - R_{ij}^{\gamma}R_{ij}^{\alpha} \delta_{\delta\alpha} \delta_{\alpha\beta} +$$

$$- R_{ij}^{\beta}R_{ij}^{\alpha} \delta_{\alpha\gamma} \delta_{\ga... ...lta} - R_{ij}^{\gamma}R_{ij}^{\alpha} \delta_{\delta\beta}\delta_{\gamma\delta}$$

$$+ R_{ij}^{\alpha}R_{ij}^{\alpha} \delta_{\alpha\beta}\delta_{\beta\gamma} \delta_{\gamma\delta} )$$

and rewrite the elastic energy in the well known fashion

$$E_{el} = \frac{1}{2}\sum_{\alpha\beta\gamma\delta} c^{\alpha\beta\gamma\delta} \epsilon_{\alpha\beta}\epsilon_{\gamma\delta}$$ (85)

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 (83). ¹⁴ Therefore transversal springs have to be used with caution in the description of a phonon spectrum.

With the help of elastic constants we can rewrite the derivative of the elastic energy (87), again ignoring the rotation dependence:

$$\frac{\partial E_{el}}{\partial \epsilon_{\alpha\beta}} = \sum_{\gamma\delta=1,2,3,\gamma \le \delta} c^{\alpha\beta\gamma\delta} \epsilon_{\gamma\delta}$$ (86)

To easy the indexing we apply the notation of Voigt

$$\epsilon_1 = \epsilon_{11}$$ (87)

$$\epsilon_2 = \epsilon_{22}$$

$$\epsilon_3 = \epsilon_{33}$$

$$\epsilon_4 = 2\epsilon_{23}=2\epsilon_{32}$$

$$\epsilon_5 = 2\epsilon_{31}=2\epsilon_{13}$$

$$\epsilon_6 = 2\epsilon_{12}=2\epsilon_{21}$$

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

$$c^{11} = c^{1111}$$ (88)

$$c^{44} = c^{2323}$$

There are 21 independent elastic constants, $c^{\alpha\beta}$ with $\beta=1,...,6$ and $\alpha \le \beta$. 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

$$E_{el} = \frac{1}{2}\sum_{\alpha\gamma=1,..,6} c^{\alpha\gamma} \epsilon_{\alpha}\epsilon_{\gamma}$$ (89)