Question

Derive an expression for the velocity of the transverse wave in the [100] direction in a
cubic crystal.

Answer 1

Answer on Question #51238-Physics-Solid State Physics

Derive an expression for the velocity of the transverse wave in the [100] direction in a cubic crystal.

Solution

The equation of motion in the $x$ direction is

$$\rho \frac {\partial^ {2} u}{\partial t ^ {2}} = C _ {1 1} \frac {\partial^ {2} u}{\partial x ^ {2}} + C _ {4 4} \left(\frac {\partial^ {2} u}{\partial y ^ {2}} + \frac {\partial^ {2} u}{\partial z ^ {2}}\right) + \left(C _ {1 2} + C _ {4 4}\right) \left(\frac {\partial^ {2} v}{\partial x \partial y} + \frac {\partial^ {2} w}{\partial x \partial z}\right). \quad (a)$$

The equation of motion in the $y$ direction is

$$\rho \frac {\partial^ {2} v}{\partial t ^ {2}} = C _ {1 1} \frac {\partial^ {2} v}{\partial x ^ {2}} + C _ {4 4} \left(\frac {\partial^ {2} v}{\partial y ^ {2}} + \frac {\partial^ {2} v}{\partial z ^ {2}}\right) + \left(C _ {1 2} + C _ {4 4}\right) \left(\frac {\partial^ {2} u}{\partial x \partial y} + \frac {\partial^ {2} w}{\partial y \partial z}\right). \quad (b)$$

The equation of motion in the $z$ direction is

$$\rho \frac {\partial^ {2} w}{\partial t ^ {2}} = C _ {1 1} \frac {\partial^ {2} w}{\partial x ^ {2}} + C _ {4 4} \left(\frac {\partial^ {2} w}{\partial y ^ {2}} + \frac {\partial^ {2} w}{\partial z ^ {2}}\right) + \left(C _ {1 2} + C _ {4 4}\right) \left(\frac {\partial^ {2} u}{\partial x \partial z} + \frac {\partial^ {2} v}{\partial y \partial z}\right). \quad (c)$$

here $\rho$ is the density and $u, v, w$ are the components of the displacement, $C_{ab}$ are the elastic stiffness constants.

Consider a transverse or shear wave with the wavevector along the $x$ cube edge and with the particle displacement $v$ in the $y$ direction:

$$v = v _ {0} e ^ {i (K x - \omega t)}.$$

On substitution in (b) this gives the dispersion relation

$$\omega^ {2} \rho = C _ {4 4} K ^ {2};$$

thus the velocity $\frac{\omega}{K}$ of a transverse wave in the [100] direction is

$$v _ {s} = \left(\frac {C _ {4 4}}{\rho}\right) ^ {2}.$$

The identical velocity is obtained if the particle displacement is in the $z$ direction.

Thus for $K$ parallel to [100] the two independent shear waves have equal velocities. This is not true for $K$ in a general direction in the crystal.

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