Ångström method with linear heat loss ($\alpha =\omega /(2ab)$)

Side losses need not destroy the method. Let $\theta$ be the small temperature disturbance about a steady background in surroundings at fixed temperature. For a rod of cross-sectional area $S$ and perimeter $\mathcal{P}$, a uniform real surface heat transfer coefficient $h_s$ produces a loss per unit volume of $H\theta$, with $H=h_s\mathcal{P}/S$. Assume constant $k,\rho ,c_p,H$, a cross-section nearly uniform in temperature, and no distributed heat source. The balance becomesHere $H$ has units $\mathrm{W}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$, while $\lambda$ has units ${\mathrm{s}}^{-1}$. This is a distributed version of Newton's law of cooling; a constant linearized contribution from thermal radiation may be included over a small temperature range. It does not represent arbitrary time-dependent coupling to a surrounding material with its own thermal storage.

For the decaying exponential branch, the phasor equation givesso thatReal positive side loss increases $a$ and decreases $b$ at fixed angular frequency and thermal diffusivity. Their product eliminates $\lambda$, yielding the combined inferenceThis result requires that attenuation and lag belong to the same exponential complex wavenumber at the same frequency, and that the loss term is real. It does not generally survive a superposition caused by a nearby finite end, spatially varying coefficients, transverse temperature gradients, or unequal sensor delays. The inferred $\lambda =\alpha (a^2-b^2)$ supplies an additional model check; a negative value signals inconsistency with this passive-loss model rather than negative physical cooling.