9.5: Optical Theorem

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The differential scattering cross-section $ f(\theta)$ . The total cross-section is given by

$ = \oint d{\mit\Omega}\,\vert f(\theta)\vert^{\,2}$ $= \frac{1}{k^2} \oint d\varphi \int_{-1}^{1} d\mu \sum_l \sum_{l'... ...elta_l-\delta_{l'})]\, \sin\delta_l \,\sin\delta_{l'}\, P_l(\mu)\, P_{l'}(\mu),$ \ref{980}

where $ \sigma_{\rm total} = \frac{4\pi}{k^2} \sum_{l=0,\infty} (2\,l+1)\,\sin^2\delta_l,$ \ref{981}

where use has been made of Equation \ref{967}. A comparison of this result with Equation \ref{979} yields

$ P_l\ref{1} = 1$ . This result is known as the optical theorem. It is a reflection of the fact that the very existence of scattering requires scattering in the forward ($ \sigma_{\rm total} = \sum_{l=0,\infty} \sigma_l,$ \ref{983}
where
$ l$ th partial cross-section: i.e., the contribution to the total cross-section from the $ l$ th partial cross-section occurs when the phase-shift $ \pi/2$ .

Contributors

Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)

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