= Thermal diffusion rate of a Fourier mode
{title2=$\beta=\xi k^2$}
With <thermal diffusivity> $\xi$, a <Fourier mode> of angular <wavenumber> $k$ decays at rate $\beta=\xi k^2$. To avoid confusing this $k$ with <thermal conductivity>, use $\alpha=\xi$ and angular <wavenumber> $K$ below: the same rate is $\beta=\alpha K^2$, with units $\mathrm{s^{-1}}$.
Consider an insulated bar $0\leq x\leq L$ with constant <thermal diffusivity>, no sources and no <heat flux> through either end or the sides. An original initial <temperature> profile and its exact evolution are
$$
T(x,0)=\overline T+A_0\cos\frac{n\pi x}{L},\qquad
T(x,t)=\overline T+A_0e^{-\alpha K_n^2t}\cos\frac{n\pi x}{L},\qquad
K_n=\frac{n\pi}{L},\quad n=1,2,\ldots.
$$
The <cosine> has zero <temperature gradient> at both ends, satisfying the insulated <Neumann boundary conditions>. Its spatial mean is zero, so <conservation of energy> leaves the mean <temperature> $\overline T$ unchanged while the nonuniform part disappears. A short derivation substitutes $T=\overline T+A(t)\cos(K_nx)$ into the <heat equation>: the second spatial <derivative> supplies $-K_n^2$, hence $A'=-\alpha K_n^2A$. The <exponential decay> time is $\tau_n=1/(\alpha K_n^2)$; doubling angular <wavenumber> quarters this time. If cycles-per-<length> <spatial frequency> $f_x$ is used instead, $K=2\pi f_x$ and the rate is $4\pi^2\alpha f_x^2$.
\Image[/thermal-diffusivity-insulated-fourier-mode-decay.png]
{title=Insulated-bar <Fourier mode> decay: equal <thermal conductivity>, different <heat capacities>}
{description=Original constant-property model with $L=0.010\,\mathrm m$ and $n=1$. The normalized initial <temperature> profile is identical for both stipulated materials in <thermal diffusivity>. The larger <volumetric heat capacity> gives a smaller <thermal diffusivity> and four times the decay time. Profiles at $t=10\,\mathrm s$ and amplitude histories are analytical predictions, not measured records.}
For the two model materials in <thermal diffusivity> and $L=0.010\,\mathrm m$, $\tau_{1,A}\approx5.07\,\mathrm s$ and $\tau_{1,B}\approx20.3\,\mathrm s$. After $10\,\mathrm s$, the remaining amplitude fractions are approximately $0.139$ and $0.610$. A narrow hot-and-cold pattern disappears faster than a broad one because its <temperature> curvature, and hence the imbalance in conductive <heat flux>, is larger.
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