Hi I'm trying to derive the Compton scattering formula and I don't understand where I'm going wrong. I keep getting that the momentum of the outgoing photon is equal to the momentum of the incoming photon (and thus their wavelengths are the same) which I know (intuitively and from the actual formula given) is wrong. If anyone could point out where I went wrong that would be great. Thanks a lot

$E_e$ is the energy of the still electron.

my workings so far where$E_e$ is the energy of the still electron.


2 Answers 2


We can derive the Compton scattering formula using the 4-momentum. The 4-momentum before the collision is equal to the 4-momentum after the collision: $$\tag{1}P_{b}=P_{a}\label{1}$$ The 4-momentum before the collision is given by $P_{\gamma}+P_{e}$ and the 4-momentum after the collision is given by $P'_{\gamma}+P'_{e}$. In general case, the 4-momentum components are: $$\tag{2}p_{\mu}=\left(\frac{E}{c},\vec{p}\right)\label{2}$$ The dot product of 4-momentum with itself is: $$\tag{3}p_{\mu}\cdot p^{\mu}=\frac{E^2}{c^2}-p^2=m^2c^2\label{3}$$ Because photons are massless, the dot product of their 4-momentum is zero. We can rewrite equation (\ref{1}) in terms of photon and electron 4-momenta: $$\tag{4}P_{\gamma}+P_{e}=P'_{\gamma}+P'_{e}\label{4}$$ Rearranging equation (\ref{4}) and squaring: $$\tag{5}\left(P_{\gamma}+P_{e}-P'_{\gamma}\right)^2=\left(P'_{e}\right)^2\label{5}$$ Considering that the incoming photon is traveling along the $x$ axis, we can write its 4-momentum as $P_{\gamma}=\left(\frac{E_{\gamma}}{c},p_{\gamma},0,0\right)$. After the collision, the photon is scattered and will have a spatial momentum component in both $x$ and $y$ direction, so its 4-momentum can be written as $P'_{\gamma}=\left(\frac{E'_{\gamma}}{c},p'_{\gamma}\cdot \cos(\theta),p'_{\gamma}\cdot\sin(\theta),0\right)$. Also, considering the electron at rest before the collision, its 4-momentum is $P_{e}=\left(\frac{E_{e}}{c},0,0,0\right)$. Going back to the equation (\ref{5}) and using the 4-momentum expressions we obtain: $$\tag{6}0+m^2c^2+0+2\cdot\frac{E_{\gamma}E_{e}}{c^2}-2\left(\frac{E_{\gamma}E'_{\gamma}}{c^2}-p_{\gamma}\cdot p_{\gamma}'\cdot\cos(\theta)\right)-2\cdot\frac{E'_{\gamma}E_{e}}{c^2}=m^2c^2\label{6}$$ $$\tag{7}2\cdot\frac{E_{\gamma}E_{e}}{c^2}-2\cdot\frac{E'_{\gamma}E_{e}}{c^2}=2\left(\frac{E'_{\gamma}E_{\gamma}}{c^2}-p_{\gamma}\cdot p_{\gamma}'\cdot\cos(\theta)\right)\label{7}$$ $$\tag{8}E_{e}\left(E_{\gamma}-E'_{\gamma}\right)=E_{\gamma}E'_{\gamma}-p_{\gamma}c\cdot p_{\gamma}'c\cdot\cos(\theta)\label{8}$$ $$\tag{9}mc^2\left(\frac{hc}{\lambda}-\frac{hc}{\lambda'}\right)=\frac{hc}{\lambda}\frac{hc}{\lambda'}\left(1-\cos(\theta)\right)\label{9}$$ $$\tag{10}\left(\lambda'-\lambda\right)\frac{hc}{\lambda\lambda'}mc^2=\frac{hc}{\lambda}\frac{hc}{\lambda'}\left(1-\cos(\theta)\right)\label{10}$$ So we get the Compton scattering formula: $$\tag{11}\lambda'-\lambda=\frac{h}{mc}\left(1-\cos(\theta)\right)\label{11}$$


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This is my derivation of it; hopefully you’ll find it useful. You’re work seemed right until you equated the two expressions for $P_e^2c^2$

  • $\begingroup$ Hi sorry how did I equate them wrong? $\endgroup$
    – Thomas
    Commented May 27, 2017 at 15:45
  • $\begingroup$ It appears you equated them right but the left hand side of that equation (what you got from energy) is carried out in your work as if it were a vector; it should be carried out as a scalar $\endgroup$ Commented May 27, 2017 at 16:05
  • $\begingroup$ Ok thanks I'll carry that through and see if it works. $\endgroup$
    – Thomas
    Commented May 27, 2017 at 17:29

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