While preparing for my exams, I found the following question on a past paper for which I am getting a different answer to what the question says I should be getting, I can't see where I am going wrong or whether what I have is equivalent to the answer but I suspect it isn't. (Note: This is not for an assignment, it is just some extra revision I am doing to be better prepared).

Question: In the laboratory frame an electron travelling with velocity $u$ collides with a positron at rest. They annihilate, producing two photons of frequencies $ν_1$ and $ν_2$ that move off at angles $θ_1$ and $θ_2$ to $u$, in the directions of the unit vectors $e_1$ and $e_2$ respectively. By considering 4-momenta in the laboratory frame, or otherwise, show that $\dfrac{1+\cos(\theta_1+\theta_2)}{\cos(\theta_1)+\cos(\theta_2)} = \sqrt{\dfrac{\gamma-1}{\gamma+1}}$ where $\gamma$ is the Lorentz factor for the electron.

Here is what I have achieved so far:

Let $m$ be the mass of the positron/election. Consider the frame in which the election is initially travelling in the x-axis. By conservation of 4-momentum: $$ m\gamma\pmatrix{c \\ u \\ 0 \\ 0} + m\pmatrix{c \\ 0 \\ 0 \\ 0} = \frac{h\nu_1}{c}\pmatrix{1 \\ \cos(\theta_1) \\ \sin(\theta_1) \\ 0} + \frac{h\nu_2}{c}\pmatrix{1 \\ \cos(\theta_2) \\ \sin(\theta_2) \\ 0}$$

The ratio of the second and first lines gives us: $$ \frac{\gamma u}{(1+\gamma)c} = \frac{\nu_1\cos(\theta_1)+\nu_2\cos(\theta_2)}{\nu_1+\nu_2}$$

But $ (\frac{\gamma u}{(1+\gamma)c})^2 = \frac{1}{(1+\gamma)^2}(\frac{\gamma u}{c})^2$ and $(\frac{\gamma u}{c})^2 = \gamma^2 u^2/c^2$ = $\frac{u^2/c^2}{1-u^2/c^2} = -1 + \frac{1}{1-u^2/c^2} = \gamma^2-1$. So $(\frac{\gamma u}{(1+\gamma)c})^2 = \frac{\gamma^2-1}{(\gamma+1)^2} = \frac{\gamma-1}{\gamma+1}$. So $\frac{\gamma u}{(1+\gamma)c} = \sqrt{\frac{\gamma-1}{\gamma+1}}$. So we have:

$$ \sqrt{\frac{\gamma-1}{\gamma+1}} = \frac{\nu_1\cos(\theta_1)+\nu_2\cos(\theta_2)}{\nu_1+\nu_2}$$

Now the third line in the Conservation of 4-momentum equation tells us: $\frac{h}{c}(\nu_1\sin(\theta_1)+\nu_2\sin(\theta_2)) = 0$ So $\nu_2 = -\nu_1\frac{\sin(\theta_1)}{\sin(\theta_2)}$ (division by $0$ not a problem here as that would imply the frequency of one of the photons was $0$).

Putting this in to our result from before (and cancelling $\nu_1$ and simplifying):

$$ \sqrt{\frac{\gamma-1}{\gamma+1}} = \frac{\nu_1\cos(\theta_1)-\nu_1\frac{\sin(\theta_1)}{\sin(\theta_2)}\cos(\theta_2)}{\nu_1-\nu_1\frac{\sin(\theta_1)}{\sin(\theta_2)}} = \frac{\cos(\theta_1)-\frac{\sin(\theta_1)}{\sin(\theta_2)}\cos(\theta_2)}{1-\frac{\sin(\theta_1)}{\sin(\theta_2)}} = \frac{\cos(\theta_1)\sin(\theta_2)-\sin(\theta_1)\cos(\theta_2)}{\sin(\theta_2)-\sin(\theta_1)} = \frac{\sin(\theta_2-\theta_1)}{\sin(\theta_2)-\sin(\theta_1)}$$

However this is not the result we are supposed to prove. Can anyone see where I went wrong with this?

Thanks in advance.

  • $\begingroup$ Not sure where you went wrong, but if you change the minus signs in your result to plus signs, you get the result you are supposed to show. $\endgroup$
    – Jens
    Commented May 29, 2017 at 19:32
  • $\begingroup$ @Jens Hmm, yes you are right, however the only place I introduce a - sign is when I said that $\nu_2 = -\nu_1\dfrac{\sin(\theta_1)}{\sin(\theta_2)}$ and I can't see the problem with that bit (unless I am being exceptionally slow...) $\endgroup$
    – Hadi Khan
    Commented May 29, 2017 at 19:45
  • $\begingroup$ Also changing the - to a + causes other problems as the L.H.S is always < 1 and the $\theta_i$ have different parities (needed for 4-momentum conservation) but if the - is a + we can make R.H.S > 1 $\endgroup$
    – Hadi Khan
    Commented May 29, 2017 at 19:49
  • $\begingroup$ The formula you should prove cannot hold: for $u=0$ r.h.s. vanishes, whereas l.h.s. diverges ($\theta_2=\pi+\theta_1$). $\endgroup$ Commented May 29, 2017 at 21:03
  • $\begingroup$ Yes, you are right, but as this is from a past exam paper, I am pretty sure that lots of thought was put into the question and that someone would managed to solve the question or there would have been a lot of complaining (which I can't recall). The paper is here (last question): maths.cam.ac.uk/sites/www.maths.cam.ac.uk/files/pre2014/… $\endgroup$
    – Hadi Khan
    Commented May 29, 2017 at 22:03

1 Answer 1


If you need a minus sign, put it.

$$m\gamma\pmatrix{c \\ u \\ 0 \\ 0} + m\pmatrix{c \\ 0 \\ 0 \\ 0} = \frac{h\nu_1}{c}\pmatrix{1 \\ \cos(\theta_1) \\ \sin(\theta_1) \\ 0} + \frac{h\nu_2}{c}\pmatrix{1 \\ \cos(\theta_2) \\ -\sin(\theta_2) \\ 0}$$

You suppose the angles are defined both the same way, anticlockwise from the $x$ axis, but usually in collision exercises, one is measured clockwise and the other anticlockwise (e.g.)

The $u=0$ thing is not a problem as you take limits ($u\to0$), the sum can be finite and the numerator is clearly, with the "redefinition", zero. And the point about the possibility the formula gives for the rhs to be greater than $1$, you almost said it: some angles are forbidden. We need more information for a complete/determinate description or this is impossible, at most this relation we are commenting between them (not sure about possible/impossible).

I finally succeed proving the hint Jens gave, but it took me a too long while. I think there is a simpler way to prove it.


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