Chapter 9, Problem 9.17P

To determine

The Eddington approximation leads to expression for the mean intensity, radiative flux and radiation pressure.

Answer to Problem 9.17P

The Eddington approximation leads to expression for the mean intensity is $\frac{1}{2}\left(I_{\text{out}}+I_{\text{in}}\right)$, radiative flux is $\pi \left(I_{\text{out}}-I_{\text{in}}\right)$ and radiation pressure is. $\frac{4\pi }{3c}\langle I\rangle$.

Explanation of Solution

Write the expression for average value of intensity.

    $\langle I\rangle =\frac{1}{4\pi }{\displaystyle \int Id\Omega }$        (1)

Here, $\Omega$ is the solid angle and $I$ is the intensity.

Write the expression for specific radiative flux.

    $F_{\text{rad}}={\displaystyle \int I\cos \theta }d\Omega$        (2)

Write the expression for radiation pressure.

    $P_{\text{rad}}=\frac{1}{c}{\displaystyle \int I{\cos }^2\theta }d\Omega$        (3)

Conclusion:

Substitute $I_{\text{out}}+I_{\text{in}}$ for $I$ in equation (1)

    $\begin{array}{c}\langle I\rangle =\frac{1}{4\pi }{\displaystyle \int \left(I_{\text{out}}+I_{\text{in}}\right)}d\Omega \\ =\frac{1}{4\pi }{\displaystyle \int I_{\text{out}}}d\Omega +\frac{1}{4\pi }{\displaystyle \int I_{\text{in}}}d\Omega \end{array}$        (4)

Substitute $\sin \theta d\theta d\varphi$ for $d\Omega$ in equation (4)

    $\langle I\rangle =\frac{1}{4\pi }{\displaystyle \int I_{\text{out}}}\sin \theta d\theta d\varphi +\frac{1}{4\pi }{\displaystyle \int I_{\text{in}}}\sin \theta d\theta d\varphi$        (5)

Consider the limits $0$ to $\pi$ for $\theta$ and $0$ to $2\pi$ for $\varphi$ in equation (5)

    $\begin{array}{c}\langle I\rangle =\frac{1}{4\pi }{\displaystyle \underset{0}{\overset{\pi /2}{\int }}{I}_{\text{out}}\sin \theta }d\theta {\displaystyle \underset{0}{\overset{2\pi }{\int }}d\varphi }+\frac{1}{4\pi }{\displaystyle \underset{\pi /2}{\overset{\pi }{\int }}{I}_{\text{in}}\sin \theta d\theta }{\displaystyle \underset{0}{\overset{2\pi }{\int }}d\varphi }\\ =\frac{1}{4\pi }{I}_{\text{out}}{\left[-\cos \theta \right]}_0^{\pi /2}{\left[\varphi \right]}_0^{2\pi }+\frac{1}{4\pi }{I}_{\text{in}}{\left[-\cos \theta \right]}_{\pi /2}^{\pi }{\left[\varphi \right]}_0^{2\pi }\\ =\frac{1}{4\pi }{I}_{\text{out}}\left[-\cos \left(\frac{\pi }{2}\right)+\cos 0°\right]\left[2\pi -0\right]+\frac{1}{4\pi }{I}_{\text{in}}\left[-\cos \pi +\cos \left(\frac{\pi }{2}\right)\right]\left[2\pi -0\right]\\ =\frac{I_{\text{out}}}{2}+\frac{I_{\text{in}}}{4\pi }\end{array}$

Solve further as,

    $\langle I\rangle =\frac{1}{2}\left(I_{\text{out}}+I_{\text{in}}\right)$

Substitute $\sin \theta d\theta d\varphi$ for $d\Omega$ and $I_{\text{out}}+I_{\text{in}}$ for $I$ in equation (2)

    $F_{\text{rad}}={\displaystyle \int \left(I_{\text{out}}+I_{\text{in}}\right)}\cos \theta \sin \theta d\theta d\varphi$        (6)

Consider the limits $0$ to $\pi$ for $\theta$ and $0$ to $2\pi$ for $\varphi$ in equation (6)

    $\begin{array}{c}{F}_{\text{rad}}={\displaystyle \underset{0}{\overset{\pi /2}{\int }}{I}_{\text{out}}\cos \theta \sin \theta d\theta }{\displaystyle \underset{0}{\overset{2\pi }{\int }}d\varphi }+{\displaystyle \underset{\pi /2}{\overset{\pi }{\int }}{I}_{\text{in}}\cos \theta \sin \theta d\theta }{\displaystyle \underset{0}{\overset{2\pi }{\int }}d\varphi }\\ =\frac{I_{\text{out}}}{4}\left[-\cos \pi +\cos 0\right]\left[2\pi -0\right]+\frac{I_{\text{in}}}{4}\left[-\cos 2\pi +\cos \pi \right]\left[2\pi -0\right]\\ =\frac{I_{\text{out}}}{4}\left[1+1\right]\left(2\pi \right)+\frac{I_{\text{in}}}{4}\left[-1-1\right]\left(2\pi \right)\\ =\frac{I_{\text{out}}}{4}\left(4\pi \right)+\frac{I_{\text{in}}}{4}\left(-4\pi \right)\end{array}$

Solve further as,

    $F_{\text{rad}}=\pi \left(I_{\text{out}}-I_{\text{in}}\right)$

Substitute $\sin \theta d\theta d\varphi$ for $d\Omega$ and $I_{\text{out}}+I_{\text{in}}$ for $I$ in equation (3)

    $P_{\text{rad}}=\frac{1}{c}{\displaystyle \int \left(I_{\text{out}}+I_{\text{in}}\right){\cos }^2\theta }\sin \theta d\theta d\varphi$        (7)

Consider the limits $0$ to $\pi$ for $\theta$ and $0$ to $2\pi$ for $\varphi$ in equation (7)

    $\begin{array}{c}{P}_{\text{rad}}=\frac{1}{c}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}{I}_{\text{out}}{\cos }^2\theta \sin \theta d\theta }{\displaystyle \underset{0}{\overset{2\pi }{\int }}d\varphi }+\frac{1}{c}{\displaystyle \underset{\pi /2}{\overset{\pi }{\int }}{I}_{\text{in}}{\cos }^2\theta \sin \theta d\theta }{\displaystyle \underset{0}{\overset{2\pi }{\int }}d\varphi }\\ =\frac{1}{c}{I}_{\text{out}}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}{\cos }^2\theta \sin \theta d\theta }{\displaystyle \underset{0}{\overset{2\pi }{\int }}d\varphi }+\frac{1}{c}{I}_{\text{in}}{\displaystyle \underset{\pi /2}{\overset{\pi }{\int }}{\cos }^2\theta \sin \theta d\theta }{\displaystyle \underset{0}{\overset{2\pi }{\int }}d\varphi }\\ =\frac{1}{c}\left(2\pi I_{\text{out}}\right){\displaystyle \underset{0}{\overset{\pi /2}{\int }}{\cos }^2\theta \sin \theta d\theta }+\frac{1}{c}\left(2\pi I_{\text{in}}\right){\displaystyle \underset{\pi /2}{\overset{\pi }{\int }}{\cos }^2\theta \sin \theta d\theta }\\ =\frac{2\pi }{3c}\left(I_{\text{out}}+I_{\text{in}}\right)\end{array}$        (8)

Substitute $I_{\text{out}}+I_{\text{in}}=I$ in equation (8)

    $\begin{array}{c}{p}_{\text{rad}}=\frac{2\pi }{3c}\left(2I\right)\\ =\frac{4\pi }{3c}\end{array}$

Thus, the Eddington approximation leads to expression for the mean intensity is $\frac{1}{2}\left(I_{\text{out}}+I_{\text{in}}\right)$, radiative flux is $\pi \left(I_{\text{out}}-I_{\text{in}}\right)$ and radiation pressure is. $\frac{4\pi }{3c}\langle I\rangle$.