Blundell Thermal Physics · Problem 10.6

The thermal diffusion equation

Physics — Statistical Mechanics · modified-thermal-diffusion-equation-group-velocity

Problem

For thermal waves, calculate the magnitude of the group velocity. This shows that the thermal diffusion equation cannot hold exactly since the velocity of propagation can become larger than that of any particles that could carry heat through the material. We now consider a modification of the thermal diffusion equation which fixes this problem. Consider the number density nn of thermal carriers in a material. In equilibrium, n=n0n = n_0, so that (∂n∂t)=−v⋅∇n+n−n0τ,(10.79)\left(\frac{\partial n}{\partial t}\right) = -\boldsymbol{v} \cdot \nabla n + \frac{n - n_0}{\tau}, \tag{10.79} where τ\tau is a relaxation time and v\boldsymbol{v} is the carrier velocity. Multiply this equation by ωτv\omega \tau \boldsymbol{v}, where ω\omega is the energy of a carrier, and sum over all k\boldsymbol{k} states. Using the fact that ∑kn0v=0\sum_{\boldsymbol{k}} n_0 \boldsymbol{v} = 0 and J=∑kωnv\boldsymbol{J} = \sum_{\boldsymbol{k}} \omega n \boldsymbol{v}, and that ∣n−n0∣≪n0|n - n_0| \ll n_0 show that J+τdJdt=−κ∇T,(10.80)\boldsymbol{J} + \tau \frac{\mathrm{d}\boldsymbol{J}}{\mathrm{d}t} = -\kappa \nabla T, \tag{10.80} and hence the modified thermal diffusion equation becomes ∂T∂t+τ∂2T∂t2=D∇2T.(10.81)\frac{\partial T}{\partial t} + \tau \frac{\partial^2 T}{\partial t^2} = D\nabla^2 T. \tag{10.81} Show that this modified equation gives a group velocity whose magnitude remains finite. Is this modification ever necessary?

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