Past exam of the theoretical physics I course of the University of Cambridge / 2012 / Paper 1 / 6 / b / Solution

The spatial transform of $\cos (px)$ is $\pi [\delta (k-p)+\delta (k+p)]$. Convolution with the impulsive source and the inverse transform therefore give simply $T(x,t)=\cos (px)G_p(t-t_0)$. Take $p\ge 0$ without loss of generality and set $\tau =t-t_0$.

For $\alpha >p/c$, let $d_p=\sqrt{{\alpha }^2{c}^4-p^2{c}^2}$. ThenFor $\alpha <p/c$, let ${\Omega }_p=\sqrt{p^2{c}^2-{\alpha }^2{c}^4}$. ThenBoth satisfy $T(t_0^{+})=0$ and ${\partial }_{t}T(t_0^{+})=c^2\cos (px)$, so the impulse changes the time derivative rather than discontinuously changing temperature. The first case rises and then relaxes without sign oscillations for a nonzero mode; the second oscillates as its amplitude decays. The limiting equality gives $T=\Theta (\tau )c^2\tau e^{-\alpha c^2\tau }\cos (px)$, and $p=0$ gives the nondecaying uniform increment discussed above.

A cosine source has both positive and negative regions, so this linear solution can be understood as a temperature perturbation about a background; its negative values are not assertions of negative absolute temperature. The temporal oscillations arise from the second-time-derivative term. Dropping that term changes the initial-jump structure as well as removing the underdamped regime.