= Ångström method with linear heat loss
{c}
{title2=$\alpha=\omega/(2ab)$}
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 becomes
$$
\rho c_p\partial_t\theta=k\partial_x^2\theta-H\theta,\qquad
\partial_t\theta=\alpha\partial_x^2\theta-\lambda\theta,\qquad
\lambda=\frac{H}{\rho c_p}\geq0.
$$
Here $H$ has units $\mathrm{W\,m^{-3}\,K^{-1}}$, 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 gives
$$
\gamma^2=\frac{\lambda+i\omega}{\alpha},\qquad
a^2-b^2=\frac{\lambda}{\alpha},\qquad
2ab=\frac{\omega}{\alpha},
$$
so that
$$
a=\sqrt{\frac{\sqrt{\lambda^2+\omega^2}+\lambda}{2\alpha}},\qquad
b=\sqrt{\frac{\sqrt{\lambda^2+\omega^2}-\lambda}{2\alpha}}.
$$
Real positive side loss increases $a$ and decreases $b$ at fixed <angular frequency> and <thermal diffusivity>. Their product eliminates $\lambda$, yielding the combined inference
$$
\boxed{\alpha=\frac{\omega}{2ab}
=\frac{\omega\Delta x^2}{2\log(A_1/A_2)\,\Delta\phi}
=\frac{\Delta x^2}{2\log(A_1/A_2)\,\Delta t}},\qquad
\Delta t=\frac{\Delta\phi}{\omega}.
$$
This 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.
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