# Scattering Differential Cross Section Laboratory Frame

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

• The derivation is apparently really horrendous; there are a lot of complications coming from Equation (2). Even when a textbook is kind enough to cover this is more detail, they would be skipping quite a lot of steps. I've been shelving this for a while and if any one of us manages it, we should help each other. Commented Sep 14, 2023 at 5:43
• Yes if i get anywhere ill post it here :) Commented Sep 14, 2023 at 9:36

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 $$\frac{\text{d}\sigma}{\text{d}\Omega_{L}} = \left(\frac{Z_{1}Z_{2}\xi^{2}}{2E_{L}}\right)^{2} \frac{\left(\cos\vartheta_{L}+\sqrt{1-K^{2}\sin^2\vartheta_{L}}\right)^{2}}{\sin^4\vartheta_{L}\sqrt{1-K^{2}\sin^2\vartheta_{L}}}$$