Scattering Differential Cross Section Laboratory Frame

Question

Let the differential cross section of a scattering experiment given by $\frac{\text{d}\sigma_{c}}{\text{d}\Omega_{c}}(\vartheta_{c})$, where $\vartheta_{c}$ describes the scattering angle in the center of mass frame. The relation between $\vartheta_{c}$ and the scattering angle in the laboratory frame is given by $$ \tan\vartheta_{L} = \frac{\sin\vartheta_{c}}{\frac{m_{1}}{m_{2}}+\cos\vartheta_{c}} \qquad (1). $$ Now the differential cross section in the laboratory reference frame can be calculated via $$ \frac{\text{d}\sigma_{L}}{\text{d}\Omega_{L}}(\vartheta_{L}) = \frac{\text{d}\sigma_{c}}{\text{d}\Omega_{c}}(\vartheta_{c}(\vartheta_{L}))\cdot \frac{\sin(\vartheta_{c}(\vartheta_{L}))}{\sin(\vartheta_{L})}\cdot \frac{\text{d}\vartheta_{c}}{\text{d}\vartheta_{L}}(\vartheta_{L}) \qquad (2) $$ For Rutherford scattering, the differential cross section in the center of mass frame is given by $$ \frac{\text{d}\sigma_{c}}{\text{d}\Omega_{c}}(\vartheta_{c}) = \left( \frac{1}{4\pi \epsilon_{0}} \frac{Z_{1}Z_{2}e^{2}}{4E_{c}} \right)^{2} \frac{1}{\sin^4\left( \frac{\vartheta_{c}}{2} \right)} \qquad (3) $$ In the laboratory frame it is apparently given by $$ \frac{\text{d}\sigma_{L}}{\text{d}\Omega_{L}}(\vartheta_{L}) = \left( \frac{1}{4\pi \epsilon_{0}} \frac{Z_{1}Z_{2}e^{2}}{4E_{L}} \right)^{2} \frac{4}{\sin^4\vartheta_{L}} \frac{\left( \cos\vartheta_{L}+\sqrt{1-\left( \frac{m_{1}}{m_{2}}\sin\vartheta_{L} \right)^{2}} \right)^{2}}{\sqrt{1-\left( \frac{m_{1}}{m_{2}}\sin\vartheta_{L} \right)^{2}}} \qquad (4) $$ I tried for hours to derive formula (4) using (1),(2), and (3). I didnt get anywhere. Could you help me? Thanks in advance

Answer 1

After an Odyssey up the reference tree i found the answer in

W. K. Chu, J. W. Mayer, and M. -A. Nicolet, Backscattering Spectrometry (Academic Press, New York, 1978)

So you don't have to search it here it is:

Starting with the formula: $$ \frac{\text{d}\sigma}{\text{d}\Omega_{L}} = \frac{\sin\vartheta_{c}}{\sin\vartheta_{L}}\frac{\text{d}\sigma}{\text{d}\Omega_{c}}\frac{\text{d}\vartheta_{c}}{\text{d}\vartheta_{L}} \qquad (1) $$ we first want to rewrite $\frac{\text{d}\vartheta_{c}}{\text{d}\vartheta_{L}}$. For that we will use the fact that $$ \begin{align} \sin(\vartheta_{c}-\vartheta_{L}) &= \sin\vartheta_{c}\cos\vartheta_{L}-\cos\vartheta_{c}\sin\vartheta_{L} \\ &= K\sin\vartheta_{L}\left(\frac{1}{K}\sin\vartheta_{c}\cot\vartheta_{L}-\frac{1}{K}\cos\vartheta_{c}\right) \\ &= K\sin\vartheta_{L} \end{align} $$ Now it follows that $$ \begin{align} \frac{\text{d}\vartheta_{c}}{\text{d}\vartheta_{L}} &= \frac{\text{d}}{\text{d}\vartheta_{L}}[\arcsin(K\sin\vartheta_{L})+\vartheta_{L}] \\ &= \frac{K\cos\vartheta_{L}}{\sqrt{1-K^2\sin^2\vartheta_{L}}}+1 = \frac{K\cos\vartheta_{L}}{\cos(\vartheta_{c}-\vartheta_{L})}+1 \\ &= \frac{\sin\vartheta_{L}\cos(\vartheta_{c}-\vartheta_{L})+K\sin\vartheta_{L}\cos\vartheta_{L}}{\sin\vartheta_{L}\cos(\vartheta_{c}-\vartheta_{L})} \\ &= \frac{\sin(\vartheta_{L}+\vartheta_{c}-\vartheta_{L})}{\sin\vartheta_{L}\cos(\vartheta_{c}-\vartheta_{L})} = \frac{\sin\vartheta_{c}}{\sin\vartheta_{L}\cos(\vartheta_{c}-\vartheta_{L})} \end{align} $$ Plug this in (1): $$ \begin{align} \frac{\text{d}\sigma}{\text{d}\Omega_{L}} = \left(\frac{1}{4\pi\varepsilon_0}\frac{Z_{1}Z_{2}e^{2}}{2E_{L}}\right)^{2}\left(\frac{(1+K)\sin\vartheta_{c}}{2\sin\vartheta_{L}\sin^2\left(\frac{\vartheta_{c}}{2}\right)}\right)^2\frac{1}{\cos(\vartheta_{c}-\vartheta_{L})} \end{align} $$ Where $E_{L} = E_{c}(1+K)$. Now one only needs to express every $\vartheta_{c}$ via $\vartheta_{L}$. Starting with the middle term in the brackets: $$ \frac{(1+K)\sin\vartheta_{c}}{2\sin^2\left({\frac{\vartheta_c}{2}}\right)} = (K+1)\cot\left(\frac{\vartheta_{c}}{2}\right) $$ Where the half angle identities $\sin^2(\frac{\vartheta}{2}) = \frac{1-\cos\vartheta}{2}$ and $\tan\left(\frac{\vartheta}{2}\right) = \frac{1-\cos\vartheta}{\sin\vartheta}$ were used. Now use $1+K = \frac{\sin(\vartheta_{c}-\vartheta_{L})+\sin\vartheta_{L}}{\sin\vartheta_{L}}$ to get $$ (K+1)\cot\left(\frac{\vartheta_{c}}{2}\right) = \frac{(\sin(\vartheta_{c}-\vartheta_{L})+\sin\vartheta_{L})(\cos(\vartheta_{c}-\vartheta_{L})+\cos\vartheta_{L})}{(\cos(\vartheta_{c}-\vartheta_{L})+\cos\vartheta_{L})\sin\vartheta_{L}\tan\left(\frac{\vartheta_{c}}{2}\right)} $$ Now use another half angle identity $\tan\left(\frac{\alpha+\beta}{2}\right) = \frac{\sin\alpha+\sin\beta}{\cos\alpha+\cos\beta}$ to get $$ \frac{(1+K)\sin\vartheta_{c}}{2\sin^2\left({\frac{\vartheta_c}{2}}\right)} = \frac{\cos(\vartheta_{c}-\vartheta_{L})+\cos\vartheta_{L}}{\sin\vartheta_{L}} $$ These simplifications leave us with $$ \frac{\text{d}\sigma}{\text{d}\Omega_{L}} = \left(\frac{Z_{1}Z_{2}\xi^{2}}{2E_{L}}\right)^{2}\frac{(\cos(\vartheta_{c}-\vartheta_{L})+\cos\vartheta_{L})^2}{\sin^4\vartheta_{L}\cos(\vartheta_{c}-\vartheta_{L})} $$ Lastly, with $\cos(\vartheta_{c}-\vartheta_{L}) = \sqrt{1-K^{2}\sin^{2}\vartheta_{L}}$ we get $$ \frac{\text{d}\sigma}{\text{d}\Omega_{L}} = \left(\frac{Z_{1}Z_{2}\xi^{2}}{2E_{L}}\right)^{2} \frac{\left(\cos\vartheta_{L}+\sqrt{1-K^{2}\sin^2\vartheta_{L}}\right)^{2}}{\sin^4\vartheta_{L}\sqrt{1-K^{2}\sin^2\vartheta_{L}}} $$