# 5.9: More on the Equation of Transfer

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Refer to equation 5.5.1. We see from what had been subsequently discussed that \([\alpha(\nu) + \sigma(\nu)]dx = d\tau(\nu)\) and that \(j_\nu dx = d\tau(\nu)\). Therefore

\[\frac{dI_v}{d\tau(\nu)} = S_\nu - I_\nu , \label{5.9.1}\]

and this is another form of the *equation of transfer*.

Now consider a spherical star with a shallow atmosphere ("plane parallel atmosphere"). In figure \(\text{V.6}\), radial distance \(r\) is measured radially outwards from the centre of the star. Optical depth is measured from outside towards the centre of the star. The thickness of the layer is \(dr\). The coordinate \(z\) is measured from the centre of the star towards the observer, and the path length through the atmosphere in that direction at angle \(\theta\) is \(dz = dr \sec \theta\). The equation of transfer can be written

\[dI_\nu (\theta) = - [\kappa (\nu) I_\nu (\theta) - j_\nu] dz. \label{5.9.2}\]

Now \(\kappa (\nu) dz = -\sec \theta d\tau(\nu)\) and \(j_\nu = \kappa (\nu) S_\nu\). Therefore

\[\cos \theta \frac{dI_{\nu}(\theta)}{d\tau (\nu)} = I_{\nu} (\theta) - S_\nu \label{5.9.3}\]

This is yet another form of the equation of transfer. The quantity \(\cos\theta\) is often written \(\mu\), so that equation \(\ref{5.9.3}\) is often written

\[\mu\frac{dI_v(\theta)}{d\tau(\nu)} = I_\nu(\theta) - S_\nu \label{5.9.4}\]

\(\text{FIGURE V.6}\)

Let us do \(\frac{1}{4\pi} \oint d\omega\) to each term in equation \(\ref{5.9.4}\). By \(\oint\) I mean integrate over \(4\pi\) steradians. In spherical coordinates \(d \omega = \sin \theta d \theta d \phi\). We obtain

\[\frac{1}{4\pi} \oint \frac{dI_\nu (\theta)}{d \tau(\nu)} \cos \theta d \omega = \frac{1}{4 \pi} \oint I_\nu d \omega - \frac{1}{4 \pi} \oint S_\nu d \omega \label{5.9.5}\]

The left hand side is \(dH_\nu/d\tau(\nu)\) and the first term on the right hand side is \(J_\nu\). (See the definitions - equations 4.5.2 and 4.7.1.) In the case of isotropic scattering, the source function is isotropic so that, in this case

\[\frac{dH_\nu}{d\tau(\nu)} = J_\nu - S_\nu, \label{5.9.6}\]

and this is another form of the equation of transfer.

On the other hand, if we do \(\frac{1}{4\pi} \oint \cos \theta d \omega\) to each term in equation 5.15, we obtain

\[\frac{1}{4\pi} \oint \frac{dI_\nu(\theta)}{d\tau(\nu)} \cos^2 \theta d \omega = \frac{1}{4\pi} \oint I_\nu \cos \theta d \omega - \frac{1}{4\pi} \oint S_\nu \cos \theta d \omega \label{5.9.7}\]

In the case of isotropic scattering the last integral is zero, so that

\[\frac{dK_\nu}{d\tau(\nu)} = H_\nu , \label{5.9.8}\]

and this is yet another form of the equation of transfer.

Now \(H_\nu\) is independent of optical depth (why? - in a plane parallel atmosphere, this just expresses the fact that the flux (watts per square metre) is conserved), so we can integrate equation \(\ref{5.9.8}\) to obtain

\[K_\nu = H_\nu \tau(\nu) + \text{constant} \label{5.9.9}\]

Note also that \(H_\nu = F_\nu /(4\pi)\), and, if the radiation is isotropic, \(K_\nu = J_\nu/ 3\) so that,

\[J_\nu = \frac{3 F_\nu \tau(\nu)}{4\pi} + J_\nu(0) \label{5.9.10}\]

where \(J_\nu(0)\) is the mean specific intensity (radiance) at the surface, which is half the specific intensity at the surface (since the radiance of the sky above the surface is zero). Thus

\[J_\nu (0) = \frac{1}{2} I_\nu (0) = F_\nu / (2\pi) \label{5.9.11}\]

Therefore

\[J_\nu = \frac{F_\nu}{2\pi} \left( 1 + \frac{3}{2} \tau ( \nu ) \right) \label{5.9.12}\]

This shows, to this degree of approximation (which includes the approximation that the radiation in the atmosphere is isotropic - which can be the case exactly only at the centre of the star) how the mean specific intensity increases with optical depth.

Let \(T\) be the temperature at optical depth \(\tau\).

Let \(T_0\) be the surface temperature.

Let \(T_{\text{eff}}\) be the effective temperature, defined by \(F(0) = \sigma T^4_{\text{eff}}\),

We also have \(\pi J = \sigma T^4\) and \( \pi J (0) = \sigma T_0^4 = \frac{1}{2} F\).

From these we find the following relations between these temperatures:

\[T^4 = \left( 1 + \frac{3}{2} \tau \right) T^4_0 = \frac{1}{2} \left( 1 + \frac{3}{2} \tau \right) T^4_{\text{eff}} \label{5.9.13}\]

\[T_0^4 = \frac{2}{2+3\pi}T^4 = \frac{1}{2} T^4_{\text{eff}} \label{5.9.14}\]

\[T^4_{\text{eff}} = \frac{4}{2+3\tau}T^4 = 2 T^4_0 \label{5.9.15}\]

Note also that \(T = T_{\text{eff}}\) at \(\tau = 2/3\), and \(T = T_0\) at \(\tau = 0\).