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Disclaimer: this is a homework question, so I am happy with just a hint or the expressions needed to proceed with my understanding.

I am working on the momentum conservation of a particle/anti-particle annihilation process, and I have been asked to show that the annihilation of a particle with a finite mass and its anti-particle cannot lead to the emission of only one photon.

I understand why this happens: the conservation of momentum. However, I would like show this in a more sophisticated 4-momentum proof...how would I go about showing that momentum is conserved for two photons but it is not conserved when the annihilation process creates just one photon...?

This may be a duplicate of: Proving the conservation of 4-momentum for a particle collision $A+B\to C+D$

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    $\begingroup$ Go to the center-of-momentum frame. $\endgroup$
    – Danu
    Commented Oct 17, 2015 at 12:24
  • $\begingroup$ I know that a photon has a non-zero momentum, so is it as easy as saying that for momentum to be conserved it has to have a photon moving in the opposite direction? (With the same momentum value, obviously resulting in a negative momentum relative to the CoM frame?) $\endgroup$
    – DarthPlagueis
    Commented Oct 17, 2015 at 12:30

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There are many possible proofs. Here is one that involves some practice with four-vectors. I write with mostly-minus metric st $p^2=m^2$.

You can write four-momentum conservation as $$ p_1 + p_2 = a $$ Now Minkowski-square, finding $$ 2m^2 + 2 p_1 \cdot p_2 = 0 \Rightarrow p_1 \cdot p_2 < 0$$ Try to show that the latter inequality is impossible. Hint: evaluate the Minkowski product in the rest frame of $p_1$ or $p_2$.

Alternatively, as @Danu suggests, think about the centre of momentum frame, in which $\vec p_1 + \vec p_2 =0$. Can a photon have zero momentum but non-zero energy?

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  • $\begingroup$ The answer to the last question is no, as it has zero rest mass. Would it be as simple as that...? $\endgroup$
    – DarthPlagueis
    Commented Oct 17, 2015 at 13:16
  • $\begingroup$ I advise you to do both proofs, and make sure you understand them. I don't want to just tell you "that is a proof" $\endgroup$
    – innisfree
    Commented Oct 17, 2015 at 13:25
  • $\begingroup$ No worries, I guess I need the confidence to approach a proof. $\endgroup$
    – DarthPlagueis
    Commented Oct 17, 2015 at 13:48

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