The coefficient of linear expansion of crystal in one direction is $\alpha_1$ and that in every direction perpendicular to it is $\alpha_2$ . The coefficient of cubical expansion is
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$\mathbf{v} = \mathbf{v}_0 (1 + \gamma \Delta \theta)$
$\mathrm{L}^3 = \mathrm{L}_0 (1 + \alpha_1 \Delta \theta) \mathrm{L}_0^2 (1 + \alpha_2 \Delta \theta)^2 = \mathrm{L}_0^3 (1 + \alpha_1 \Delta \theta) (1 + \alpha_2 \Delta \theta)^2$
Since $\mathrm{L_0^3 = V_0}$ and $L^{3} = V$
Hence $1 + \gamma \Delta \theta = (1 + \alpha_1 \Delta \theta)(1 + \alpha_2 \Delta \theta)^2$
$\cong (1 + \alpha_1 \Delta \theta)(1 + 2 \alpha_2 \Delta \theta)$ $\simeq (1 + \alpha_1 \Delta \theta + 2 \alpha_2 \Delta \theta)$
⇒ ⇒ γ γ = α α 1 + 2 α α 2
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