A moving electron with energy $E$ hits a stationary electron. Question is to find the scattering angle (the angle between the paths of the two electrons after collision) in terms of the energy $E$ and the electron mass $m$.

What can we say about the angle between the electrons after collision?

Are both making the same angle to the horizontal axis?

Is the angle between them $\pi/2$ ?

EDIT 3: *By squaring and adding equations (2) and (3) written below, we can show that $$\cos\phi = \frac{\gamma_1\gamma_2 - \gamma}{\sqrt{\gamma_1^2-1}\sqrt{\gamma_2^2-1}} $$

EDIT 2: In Newtonian mechanics, this information is sufficient to show that the angle between them after collision is $\pi/2$. I can show the proof also. I just can't be sure if this information is not enough in relativistic mechanics.

What extra information, if necessary, can be added to find the angle between them after collision?


In the center of mass frame, all I can do is find the total energy, and from there, find the equal and opposite momenta of each particle in vertical axis. $E_{com}$ is $m\sqrt{2(\gamma + 1)}$. Equal mass and equal momentum gives equal energy for both ($E^2 = p^2 + m^2$). $E = E_{com}/2 = m\sqrt{(\gamma + 1)/2}$. So $p$ along $y$-axis for each is $m(\sqrt{(\gamma + 1)/2}-1) = m\sqrt{(\gamma - 1)/2}$. I am not sure how I can proceed further to get my angle. For that I would require individual momenta along horizontal axis.

enter image description here

My approach:

So here the red solid line represents the path between the two electrons before collision, the blue and green lines show the paths of the two moving electrons after collision. Total angle between them, which we need to determine, let's call it $\phi$. And the angle that the blue electron makes with the horizontal axis be $\alpha$.

I am working in natural units, so my $c=1$. I assume my energy $E$ is $\gamma m$ and the energies of the electrons after collision are $\gamma_1 m$ for blue and $\gamma_2 m$ for green electron. I have following constraint equations.

For momentum conservation equations, I used $\gamma v = \sqrt{\gamma^2 -1}$

  • Energy conservation: $$\gamma + 1 = \gamma_1 + \gamma_2 \tag1$$

  • Horizontal momentum conservation: $$\sqrt{\gamma^2-1} = \sqrt{\gamma^2_1 - 1}\cos(\phi-\alpha) + \sqrt{\gamma^2_2 -1} \cos\alpha \tag2$$

  • Vertical momentum conservation: $$0 = \sqrt{\gamma^2_1-1}\,\sin(\phi - \alpha) - \sqrt{\gamma^2_2 -1}\,\sin\alpha.\tag3$$

And for my four unknowns $\gamma_1$, $\gamma_2$, $\alpha$ and $\phi$, I need a fourth constraint equation. I believe that should be:

  • $P_\mu P^\mu$ before and after collision: $${(\gamma+1)^2}m^2 - \gamma^2 m^2 v^2 = 2m^2 + 2\gamma_1 \gamma_2 m^2 - 2\gamma_1\gamma_2 m^2 v_1 v_2 \cos\phi$$

Simplifying $${(\gamma+1)^2}- (\gamma^2 -1) = 2 + 2\gamma_1 \gamma_2 - 2\sqrt{\gamma^2_1 -1}\sqrt{\gamma^2_2 -1} \cos\phi$$ $$\gamma = \gamma_1 \gamma_2 - \sqrt{\gamma^2_1 -1}\,\sqrt{\gamma^2_2 -1}\,\cos\phi.\tag4$$

I am ready with 4 constraint equations needed to solve for 4 unknowns. But, this looks quite complicated and maybe I missed some simpler trick to do this the easy way. Can anyone help me out here?

  • $\begingroup$ Normally one would do the collision in the center of mass frame, and then translate to whatever frame you prefer. $\endgroup$ – Jon Custer Mar 8 '19 at 16:31
  • $\begingroup$ @JonCuster I added an edit. That basically says that I can get my energies in the center of mass frame. And also the momenta in vertical axis. $\endgroup$ – MycrofD Mar 8 '19 at 17:11
  • $\begingroup$ It looks mathematical to me rather than conceptual. You are asking for help solving a problem, doing the calculation. You are not asking for an explanation of a concept. $\endgroup$ – sammy gerbil Mar 9 '19 at 1:32
  • 3
    $\begingroup$ @sammygerbil a) You really don't have to look at the math shown after the figure, that says "my approach".. please just answer my question which comes much before.. I think that's pretty conceptual, eh.. the solid physics question being "Are both making the same angle to the horizontal axis?"... b) Being mathematical doesn't take away its conceptual nature, does it? c)Not showing the math would demand the question, "where's your math that shows how much you have tried this?".. I just need answers to my physics question. I guess, I will put that in bold. $\endgroup$ – MycrofD Mar 9 '19 at 2:04
  • $\begingroup$ I hope people don't close this question. I am struggling to show if enough info is there for relativistic collisions, as it is enough for newtonian mechanics. What new information can be added to make it sufficient to work out the angle? $\endgroup$ – MycrofD Mar 10 '19 at 19:33

I've edited your equations to improve readability. I also added tags I'll make use of in the following.

Your main error is in believing that your data may determine final state. It's not so, not even in newtonian mechanics. The reason is that the assumption of an exactly central collision isn't tenable (otherwise you'd have all momenta aligned). So there's an unsaid unknown: the disalignment between balls centres.

Energy and momentum conservation still do hold, but are unable to completely determine the outcome: there is an equation lacking.

You've tried to obtain a fourth equation by conservation of $P_\mu P^\mu$ but it's an illusion. Since you already used up conservation of $P_\mu$ your last eq. (4) must be a consequence of the preceding three. Try it, by computing $(1)^2 - (2)^2 - (3)^2$.

The really interesting thing is to prove $\phi<\pi/2$. Would you like to engage in the proof?


$\let\g=\gamma$ Using (1) you get $$\cos\phi = {(\g_1-1)\,(\g_2-1) \over \sqrt{\g_1^2-1}\,\sqrt{\g_2^2-1}} > 0.$$

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  • $\begingroup$ I think I was looking for this kind of answer, thanks. I suspected the 4th equation to be a combination of first 3. But I couldn't prove so concretely. $\endgroup$ – MycrofD Mar 9 '19 at 17:00
  • $\begingroup$ However, in Newtonian mechanics, it can be proved that the angle $\phi$ between them is $\pi/2$ with the given information. The data is sufficient for that, I believe. And yes, I would like to engage in the proof for $\phi$ to be $< \pi/2$ in relativistic mechanics. $\endgroup$ – MycrofD Mar 9 '19 at 17:02
  • $\begingroup$ Oh, I found that in relativistic mechanics, $\cos\phi$ was indeed non-zero, and therefore, $\phi$ was $<$ $\pi/2$. :) $\endgroup$ – MycrofD Mar 17 '19 at 13:11
  • $\begingroup$ I made some edits to the $\cos\phi$ equation, because my energy conservation was missing the energy of the stationary particle. Now I am not sure how to show that $\phi < \pi/2$. Could you please help? $\endgroup$ – MycrofD Mar 20 '19 at 17:46
  • $\begingroup$ @MicrofD Read the edit to my answer. $\endgroup$ – Elio Fabri Mar 20 '19 at 20:48

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