A. S. KOMPANEETS

DIFFUSION FROM AN INSTANTANEOUS SOURCE IN A GRAVITATIONAL FIELD

(Presented by Academician V. N. Kondrat’ev, 1 XI 1962)

Assuming that the gas in which diffusion takes place is distributed according to the barometric law, we write for the diffusion coefficient:

\[ D=\delta e^{\lambda z}, \qquad \lambda=\frac{mg}{kT}. \tag{1} \]

We seek the density of the diffusing gas in the form

\[ n=N(r,z,t)e^{-\lambda_1 z}, \qquad \lambda_1=\frac{m_1 g}{kT}. \tag{2} \]

Putting also \(\tau=\delta t\), we obtain the equation for \(N\):

\[ e^{-\lambda z}\frac{\partial N}{\partial \tau} = \frac{1}{r}\frac{\partial}{\partial r}r\frac{\partial N}{\partial r} + \frac{\partial^2 N}{\partial z^2} + (\lambda-\lambda_1)\frac{\partial N}{\partial z}. \tag{3} \]

We have a system of partial solutions orthonormal in \(r\) and \(z\):

\[ N_{p,q} = e^{-p^2\tau}J_0(qr) \sqrt{\frac{\lambda}{2}}\, e^{(\lambda_1-\lambda)z/2} J_\nu\!\left(\frac{2p}{\lambda}e^{-\lambda z/2}\right). \tag{4} \]

Here

\[ \nu= \sqrt{\frac{4q^2}{\lambda^2} + \left(\frac{\lambda_1-\lambda}{\lambda}\right)^2}. \tag{5} \]

A point unit instantaneous source at \(r=0,\ z=0\) corresponds to the expression:

\[ n= \frac{2}{\pi\lambda}e^{-(\lambda+\lambda_1)z/2} \int_0^\infty\!\!\int_0^\infty q\,dq\,p\,dp\,e^{-p^2\tau} J_0(qr) J_\nu\!\left(\frac{2p}{\lambda}e^{-\lambda z/2}\right) J_\nu\!\left(\frac{2p}{\lambda}\right). \tag{6} \]

Integration with respect to \(p\) is easily performed by standard formulas, which gives

\[ n= \frac{1}{2\pi\tau\lambda} e^{-(\lambda_1+\lambda)z/2} \exp\!\left[ -\frac{1+e^{-\lambda z}}{\tau\lambda^2} \right] \int_0^\infty q\,dq\,J_0(qr)\, I_\nu\!\left(\frac{2e^{-\lambda z/2}}{\tau\lambda^2}\right). \tag{7} \]

The last integral cannot be evaluated in general form. Therefore let us consider the following special cases:

1) \(2e^{-\lambda z}/\tau\lambda^2 \gg 1\). This applies either to small times or to large depths downward, \(e^{-\lambda z/2}\gg 1\). Taking (5) into account, we obtain for \(I_\nu\) the asymptotic expression

\[ I_\nu = \sqrt{\frac{\tau\lambda^2}{4\pi}} \exp\!\left\{ \frac{\lambda z}{4} + \frac{2e^{-\lambda z/2}}{\tau\lambda^2} - \tau e^{\lambda z} \left[ q^2+\frac{1}{4}(\lambda_1-\lambda)^2 \right] \right\}. \tag{8} \]

After this the integration is carried out, and for \(n\) we find:

\[ n= \frac{1}{(4\pi\tau)^{3/2}} \exp\!\left[ -\left(\frac{5\lambda}{4}+\frac{\lambda_1}{2}\right)z -\frac{r^2}{4\tau}e^{-\lambda z} -\frac{(1-e^{-\lambda z/2})^2}{\tau\lambda^2} -\frac{\tau}{4}(\lambda_1-\lambda)^2e^{\lambda z} \right]. \tag{9} \]

For small \(z\) this becomes the standard solution for a point instantaneous source.

2) \(2e^{-\lambda z}/\tau\lambda^2 \ll 1\), which corresponds to large times or large heights, \(e^{-\lambda z/2}\ll 1\). In this case \(I_\nu\) may be replaced by the first term of the expansion
\(I_\nu(x)\sim \dfrac{1}{\Gamma(1+\nu)}\left(\dfrac{x}{2}\right)^\nu\). It is convenient to represent the gamma function by a contour integral in order to express it through an exponential. We obtain the approximate equality

\[ I_\nu\left(\frac{2e^{-\lambda z/2}}{\tau\lambda^2}\right) = \frac{1}{2\pi i}\int_{-\infty}^{(0+)} \frac{e^s}{s}e^{-\lambda\nu\alpha/2}\,ds, \tag{10} \]

where

\[ \alpha=\frac{2}{\lambda}\ln(s\tau\lambda^2 e^{\lambda z/2}). \tag{11} \]

After this, the integration with respect to \(q\) is performed:

\[ \int_0^\infty q\,dq\,J_0(qr)e^{-\alpha\nu\lambda/2} = \frac{\alpha}{r^2+\alpha^2} \left( \frac{|\lambda_1-\lambda|}{2} + \frac{1}{\sqrt{r^2+\alpha^2}} \right) e^{-|\lambda_1-\lambda|\sqrt{r^2+\alpha^2}/2}. \tag{12} \]

The integration with respect to \(s\) can be carried out in two cases:

2a) The point directly above the source, \(r=0\). Assuming that \(s\) under the logarithm is of order unity and taking \(\alpha\) out from under the integral, we obtain

\[ n= \frac{|\lambda_1-\lambda|}{4\pi\tau} \frac{(\tau\lambda^2)^{-|\lambda_1-\lambda|/\lambda}}{\ln(\tau\lambda^2 e^{\lambda z/2})} \exp\left(-\frac{1+e^{-\lambda z/2}}{\tau\lambda^2}\right) \begin{cases} \dfrac{e^{-\lambda z}}{\Gamma(\lambda_1/\lambda)}, & \lambda<\lambda_1, \tag{13'}\\[1.2em] \dfrac{e^{-\lambda z}}{\Gamma(2-\lambda_1/\lambda)}, & \lambda>\lambda_1. \tag{13''} \end{cases} \]

Formula (13′) shows that a light gas in a heavy one does not by any means immediately pass over to the equilibrium Boltzmann distribution. At first it spreads in height almost according to the same law as a heavy gas. This distribution also satisfies the diffusion equation, since

\[ e^{\lambda z}\left(\frac{\partial}{\partial z}+\lambda_1\right)e^{-\lambda z}=\mathrm{const}, \]

which corresponds to a constant flux. The density vanishes earlier than it becomes equilibrium. For \(\lambda_1<\lambda\), equilibrium is reached approximately earlier.

2b) A point at the level of the source, \(z=0\), and large times \(\tau\lambda^2\gg 1\). Then, if \(r\gg\alpha\), we have

\[ n= \frac{|\lambda_1-\lambda|}{2\pi\tau\lambda^2} \frac{e^{-2/(\tau\lambda)^2}}{r^2} e^{-|\lambda_1-\lambda|r/2} \left[\ln(\tau\lambda^2)-0.577\right]. \tag{14} \]

In the case when \(\lambda_1=\lambda\):

\[ n= \frac{e^{-2/(\tau\lambda)^2}}{\pi\tau\lambda^2 r^3} \left[\ln(\tau\lambda^2)-0.577\right]. \tag{15} \]

The inequality \(r>\alpha\) actually means \(r>2\ln\tau\lambda^2/\lambda\). As equations (14) and (16) show, the density at large distances from the source decreases rapidly. Thus, in contrast to diffusion in a homogeneous medium, lateral spreading in a gravitational field is noticeably suppressed.

Let us also calculate the amount of substance that has gone upward. For \(\tau\lambda^2\gg 1\) it is easy to integrate the general expression (7):

\[ 2\pi\int_0^\infty r\,dr\int_0^\infty dz\,n = \begin{cases} \dfrac{1}{\Gamma(1+\lambda_1/\lambda)(\tau\lambda^2)^{\lambda_1/\lambda}}, & \lambda_1>\lambda, \tag{16'}\\[1.2em] \dfrac{1}{\Gamma(2-\lambda_1/\lambda)(\tau\lambda^2)^{2-\lambda_1/\lambda}}, & \lambda_1<\lambda. \tag{16''} \end{cases} \]

In both cases the main mass of the substance goes downward. This is connected with the distributions (13′) and (13″).

The problem is greatly simplified if diffusion occurs only with respect to height from a flat horizontal layer. Then, instead of formula (7), we have the closed expression

\[ n(z,t)=\frac{1}{\tau\lambda}\exp\left[-\frac{(\lambda+\lambda_1)z}{2}-\frac{1+e^{-\lambda z}}{\tau\lambda^2}\right] I_{|\lambda_1-\lambda|/\lambda}\left(\frac{2e^{-\lambda z/2}}{\tau\lambda^2}\right). \tag{17} \]

For the point where the density is maximal, from this we obtain the equation

\[ e^{-\lambda z_m}=\tau\lambda^2 \begin{cases} \times\,\dfrac{\lambda_1}{\lambda}, & \lambda_1>\lambda, \\[6pt] \times\,1, & \lambda_1<\lambda. \end{cases} \tag{18′, 18″} \]

Assuming that \(\tau\lambda^2\gg 1\), one could replace \(I_{|\lambda_1-\lambda|/\lambda}(x)\) by the zero term of its expansion.

The width of the distribution is estimated by the usual formulas of fluctuation theory. This gives

\[ \sqrt{\overline{(z-z_m)^2}}= \begin{cases} \dfrac{1}{\sqrt{\lambda\lambda_1}}, & \lambda_1>\lambda, \\[6pt] \dfrac{1}{\lambda}, & \lambda_1<\lambda. \end{cases} \tag{19′, 19″} \]

Thus, in contrast to free diffusion, the width of the distribution tends to a finite limiting value of order \(\lambda\).

In conclusion I express my gratitude to V. L. Tal’roze for posing the problem.

Institute of Chemical Physics
Academy of Sciences of the USSR

Received
25 X 1962