Electron-positron annihilation in two inertial frames

Question

I have a question about the electron-positron annihilation example worked out in Greiner's Classical Mechanics, vol. 1, p. 468 (English edition).

If we consider an electron-positron collision in the center-of-mass inertial frame, he deduces from momentum conservation that at least two photons must be produced. Then he uses energy conservation to find the photon's wavelength (which turns out to be equal to Compton's wavelength).

And that's where he stops. It seems to me, however, that if we consider any other inertial frame that is moving with nonzero velocity with respect to the c.o.m. frame, then we do not end up with the restriction on the number of photons. Thus, we have managed to construct a preferential frame in which there may not be a single photon while there are infinitely many other frames where there may be a single photon.

How does this agree with the principle of relativity? [I guess there must be some additional restrictions on what may be produced in such reactions (from QFT...?). Is that the case?]

The following picture shall clarify the question. On top, the frame is the c.o.m. frame. On the bottom, frame $O'$ is moving to the left with constant velocity. In order to conserve momentum in the c.o.m. frame, at least two photons must be produced, with oppositely equal momentum. In frame $O'$, one photon is enough to conserve momentum.

Answer 1

How do you get 1 photon in a moving frame? The invariant mass is as follows:

The 4-momenta in the COM are:

$$ p^{\mu}_+ = (E, \vec p)$$ $$ p^{\mu}_- = (E, -\vec p)$$

so

$$ p^{\mu} = p^{\mu}_+ + p^{\mu}_- = (2E, 0) $$

$$ W^2 = p^2 = 4E^2 $$

That is invariant in all frames. For any single photon:

$$ k^{\mu} = (k, k) $$

so

$$ W_{\gamma}^2 = k^2 = 0 \ne 4E^2$$

It just can't happen.

Answer 2

The energy of a photon is completely determined by its momentum, $E=c|\vec{p}|$. In no frame do the total energy and momentum of the system satisfy that relation, so they cannot be carried by a single photon. This is most easily seen in the center-of-mass frame, and since $E^{2}-c^{2}\vec{p}\,^{2}$ is a Lorentz-invariant scalar, if $E^{2}-c^{2}\vec{p}\,^{2}\neq 0$ in one frame, it holds in any frame, meaning the energy cannot be carried by a single photon.

However, it is also easy to seen an arbitrary frame. Whatever the three-momenta of the electron and positron are (call them $\vec{p}_{-}$ and $\vec{p}_{+}$), the total energy is $$E=\sqrt{mc^{2}+\vec{p}\,^{2}_{-}c^{2}}+\sqrt{mc^{2}+\vec{p}\,^{2}_{+}c^{2}},$$ which is always greater than $c|\vec{p}_{-}|+c|\vec{p}_{+}|$ because of the mass energy terms. However, the total momentum is $\vec{p}_{-}+\vec{p}_{+}$, whose maximum magnitude is $|\vec{p}_{-}|+|\vec{p}_{+}|$ (achieved only if $\vec{p}_{-}$ and $\vec{p}_{+}$ are parallel). So the total energy and total momentum never satisfy $E=c|\vec{p}|$ and thus cannot be carried by a single outgoing photon.

Answer 3

For two massive particles producing a single photon,
conservation of total 4-momentum would read $$\tilde m_1 + \tilde m_2 = \tilde k,$$ where $\tilde k$ is lightlike ($\tilde k \cdot \tilde k=0$).

However, the sum of two future-directed timelike vectors is future-directed timelike.
So, $(\tilde m_1 + \tilde m_2)$ is not lightlike.

[If two future-directed timelike vectors could be lightlike, then one could send a sequence of timelike signals instead of a single light-signal. But that doesn't happen in special relativity...

(You can't reach an event-Z on the future light cone of event-A with a sequence of future-timelike-displacements starting at A.) ]