$\sqrt{\epsilon }$ Law: Centennial of the First Exact Result of Classical Radiative Transfer Theory

Abstract

Nowadays probably nobody knows when and where the term “$\sqrt{\epsilon }$ law” was coined. But as early as in 1975 in the title of one of the papers we find “$\sqrt{\epsilon }$ revisited” Frisch and Frisch (1975).

Addendum

In this Appendix the Wiener–Hopf integral equation for the source function $S(\tau )$ with the free term $Q=Q(\tau )$ and parameter $\lambda \equiv 1-\epsilon$, i.e., the equation

$$\begin{aligned} S(\tau )=\lambda \int \limits _0^\infty K(\tau -\tau ')\,S(\tau ')\,d\tau '+Q(\tau ) \end{aligned}$$

(A.1)

will be written in short as

$$\begin{aligned} S=\lambda \varLambda S+Q. \end{aligned}$$

(A.2)

The Green function $G(\tau ,\tau _1)$ of Eq. (A.1)

$$\begin{aligned} G=\lambda \varLambda G+\delta (\tau -\tau _1) \end{aligned}$$

(A.3)

provides the solution of Eq. (A.2) for an arbitrary $Q(\tau )$:

$$\begin{aligned} S(\tau )=\int \limits _0^\infty G(\tau ,\tau _1)\,Q(\tau _1)\,d\tau _1. \end{aligned}$$

(A.4)

Now, let $G_\infty (\tau )$ be the Green function for an infinite medium, i.e., the solution of

$$\begin{aligned} G_\infty =\lambda \varLambda _\infty G_\infty +\delta (\tau ), \end{aligned}$$

(A.5)

where $\varLambda _\infty$ is the operator of the same form as $\varLambda$ is, but with integration with respect to $\tau '$ from $-\infty$ to $\infty$, and not $(0, \infty )$.

The aim of this Addendum is to show, following Ivanov (1994), how the explicit expression for the H–function given by Eq. (65) can be derived by elementary means. First of all, we note that H(z) is essentially the Laplace transform of the surface Green function $\overline{G}_0$, or, to be more accurate,

$$\begin{aligned} H(1/s)=\overline{G}_0(s)\equiv \int \limits _0^\infty G_0(\tau )\,\mathrm{e}^{-s\tau }\,d\tau , \end{aligned}$$

(A.6)

so that we want to show that

$$\begin{aligned} \ln {\overline{G}_0(s)}=-\,\frac{s}{2\pi }\,\int \limits _{-\infty }^{\infty } \ln {\bigl (1-\lambda \widetilde{K}(\omega )\bigr )}\, \frac{d\omega }{s^2+\omega ^2}\,. \end{aligned}$$

(A.7)

Here $\widetilde{K}(\omega )$ is the Fourier transform of the kernel function $K(\tau )$ (Eq. (47).

To get a hint on how we can get the desirable result, let us compare (A.7) with the expression for the Green function $G_\infty$ for the whole space

$$\begin{aligned} G_\infty (\tau )= \varPhi _\infty (\tau )+\delta (\tau ) = \frac{1}{2\pi }\int \limits _{-\infty }^{\infty } \frac{\lambda \widetilde{K}(\omega )}{1-\lambda \widetilde{K}(\omega )}\; \mathrm{e}^{-i\tau \omega }\,d\omega +\delta (\tau ), \end{aligned}$$

(A.8)

which is easily obtained by applying the Fourier transform to the convolution equation (A.5) for $G_\infty$ for the whole space. The comparison enables one to reveal that

$$\begin{aligned} \lambda \frac{\partial \ln {\overline{G}_0(s)}}{\partial \lambda }=\overline{\varPhi }_{\infty }(s), \end{aligned}$$

(A.9)

where $\overline{\varPhi }_{\infty }(s)$ is the Laplace transform of the non-singular part of $G_\infty (\tau ).$

Hence to get (A.7), it suffices to prove (A.9). And this is a non-linear relation between the Laplace transforms of the Green functions for full space and for half-space. Somewhat unexpectedly, it involves the derivative with respect to the parameter $\lambda$. This derivative with respect to $\lambda$ is the key in obtaining Eq. (A.7) by elementary means.

Now it is easily seen how we have to proceed. Using the equation that defines the half-space surface Green function

$$\begin{aligned} G_0=\lambda \varLambda G_0+\delta (\tau ) \end{aligned}$$

(A.10)

one gets

$$\begin{aligned} \frac{\partial G_0}{\partial \lambda }=\lambda \varLambda \,\frac{\partial G_0}{\partial \lambda }+ \varLambda \, G_0. \end{aligned}$$

(A.11)

By (A.10), the term $\varLambda \, G_0$ in the right hand side (RHS) of (A.11) equals $\lambda ^{-1}\bigl (G_0(\tau )-\delta (\tau )\bigr )$. Thus, Eq. (A.11) may be looked at as the equation of our usual form (A.2) for the function ${\partial G_0}/{\partial \lambda }$ with the free term

$$\begin{aligned} Q(\tau )=\lambda ^{-1}\bigl (G_0(\tau )-\delta (\tau )\bigr ). \end{aligned}$$

(A.12)

Substituting this expression for Q into (A.4) and using the symmetry of $G(\tau ,\tau _1)$, we obtain

$$\begin{aligned} \lambda \frac{\partial {{G}_0(\tau )}}{\partial \lambda }=\int \limits _0^\infty G(\tau _1,\tau )\bigl [G_0(\tau _1)-\delta (\tau _1)\bigr ]\,d\tau _1. \end{aligned}$$

(A.13)

In the RHS of this equation there is a product of three functions $G_0$ (see Eq. (9) for $G(\tau ,\tau _1)$). Now, we make the Laplace transformation in $\tau$ (with the parameter of the transformation s) and substitute for $\overline{G}(\tau _1, s)$ in the RHS the expression

$$\begin{aligned} \overline{G}(\tau _1, s)=\overline{G}_0(s)\int \limits _0^{\tau _1} G_0(\tau _1-t)\,\mathrm{e}^{-st}\,dt. \end{aligned}$$

(A.14)

This expression for $\overline{G}(\tau _1, s)$ is obtained from Eq. (9) by applying the Laplace transformation in $\tau$. As a result, we get the double integral. By first changing the order of integration and then putting $\tau _1=\tau +t$ we obtain

$$\begin{aligned}&\lambda \frac{\partial {\overline{G}_0(s)}}{\partial \lambda }=\overline{G}_0(s)\int \limits _0^\infty \int \limits _0^{\tau _1} G_0(\tau _1-t)\,\Bigl [G_0(\tau _1)-\delta (\tau _1)\Bigr ]\,\mathrm{e}^{-st}\,dt\,d\tau _1 \\&\qquad \qquad =\overline{G}_0(s)\int \limits _0^\infty \int \limits _t^{\infty }\Biggl ( G_0(\tau _1-t)\,\Bigl [G_0(\tau _1)-\delta (\tau _1)\Bigr ]\,d\tau _1\Biggr )\,\mathrm{e}^{-st}\,dt \nonumber \end{aligned}$$

(A.15)

$$\begin{aligned}\qquad \quad =\overline{G}_0(s)\int \limits _0^\infty \Biggl (\int \limits _0^{\infty } G_0(\tau )\,G_0(\tau +t)\,d\tau -\delta (t)\Biggr )\mathrm{e}^{-st}\,dt. \end{aligned}$$

(A.16)

Using Eq. (13) (p. 5) and taking into account that $G_\infty (t)=\varPhi _\infty (t)+\delta (t)$, we find

$$ \lambda \frac{\partial {\overline{G}_0(s)}}{\partial \lambda }=\overline{G}_0(s)\int \limits _0^\infty \varPhi _\infty (t)\,\mathrm{e}^{-st}\,dt, $$

and this is identical to (A.9). Equation (A.7) is thus proved.