Electron Scattering

Now we take a look at an electron with the four-momentum $p_1$ that is scattered at a much heavier particle such as a proton or nucleus with a four-moment $p_2$. Momentum conservation leads to the following equation: $$p_1 + p_2 = p_1' + p_2'$$ Squaring both leads results in $$p_{1}^2 + 2 p_1 p_2 + p_{2}^{2} = p_{1}^{'2} + 2p_{1}'p_{2}' + p_{2}^{'2}$$ In the case of elastic scattering, the masses $m_e$ of the electron and $M$ of the collision partner remain constant. Because of the invariant masses $$p_1^2 = p_1^{'2}=m_e^2c^4$$ and $$p_2^2 = p_2^{'2}=M^2c^4$$ it follows after cancelling out the terms on boths sides: $$p_1p_2 = p_1^{'2}p_2^{'2}$$ Using the law of momentum conservation again, the left side can be rewritten as $$p_1p_2 = p_1'(p_1+p_2-p_1')$$ Multiplying the left side out results in $$p_1'p_1 + p_1'p_2 - m_e^2c^2$$ Now we assume that the collision partner is in rest. Then it follows: $$EMc^2 = E'E - \vec{p}_1\vec{p}_2c^2 + E' M c^2 - m_e^2c^4$$ If the energy of the electron is large enough, the term $m_ec^2$ can be neglected and $E=pc$ be assumed: $$EMc^2 = E'E(1-\cos\theta) + E'Mc^2$$ Finally, this equation can be solved for E' to obtain the energy of the scattered electron in the lab system. $$\boxed{E' = \frac{E}{1+\frac{E}{Mc^2}(1-\cos\theta)}}$$ The angle $\theta$ is called the scattering angle. The energy difference $E - E'$ is fully transferred to the target. The larger the relativistic mass of the electron compared to the target mass, the larger the energy that is transferred to the target. We can now have a look at the following 2 special cases: 1. $\theta = 0$: The electron is not scattered, and the cosine of $\theta$ is equal to $1$. This leads to $E=E'$. Therefore, no energy is transferred to the target. 2. $\theta = 180^\circ$: This means that the electron is scattered back. In this case, the cosine gets $0$ and the energy transfer to the target reaches a maximum.