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I was looking at this statement (found in a problem set online):

The decay $\gamma \longrightarrow \pi^+ + \pi^-$ is impossible according to special relativity

So I considered the following method of proof:

Let us denote $p_\gamma, p_+, p_-$ the 4-momentum for the photon, the pion + and the pion -. We shall have conservation of the total 4-momenta; thus: $$ p_\gamma = p_+ + p_- $$

Yet, taking the Minkowskian norm (where $p^2$ denotes $p^2 = p^\alpha p_\alpha = - (p^0)^2 + \vec p.\vec p $): $$ 0 = p_+^2 + p_-^2 + 2p_+\cdot p_- = -2m^2 + 2p_+\cdot p_- $$

Let set up in the rest frame of $\pi^+$. We thus have $p_+ = (m,0,0,0)~;~ p_- = (E,\vec p_-)$ in this frame. Then: $$ p_+\cdot p_- = -mE $$ Here's my problem: if we showed that $E\neq m$ it would be OK. However, I don't know how to show this. I considered $m^2 = E^2 - \vec p.\vec p$ which shows that $E^2 \geq m^2$.

  • If $\vec p \neq \vec 0$ we showed what we wanted. However, I'm afraid there is nothing that prevents the rest frame of $\pi^+$ being the one of $\pi^-$.
  • If $E\geq 0$ we thus get $2m^2 = -2mE$ and it is a contradiction. But why would it be positive?

EDIT: looking at https://physics.stackexchange.com/a/334990/146705 , I realized my misunderstanding came from the fact that in any reference frame, the energy $p^0$ seems to be always positive. Why is that so?

Qmechanic
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MiKiDe
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    See https://physics.stackexchange.com/q/22916 for the analogous question for an electron-positron pair – Nihar Karve Dec 22 '20 at 11:59
  • The proper orthochronous Lorentz group always preserves the sign of the zeroth component of any four vector. – RedGiant Dec 24 '20 at 00:31

3 Answers3

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The photon is an elementary particle of mass zero. If it could decay into two pions, it would violate Lorenz invariance, because four vector invariant mass does not change, that is why it is called invariant. The photon should have had at least the summed mass of the two pions.

The length of the energy-momentum 4-vector is given by

fourv

The length of this 4-vector is the rest energy of the particle. The invariance is associated with the fact that the rest mass is the same in any inertial frame of reference.

anna v
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As with photon non-decay to $e^+e^-$, the trouble is that the massive system always has a rest frame, but the photon does not.

In a comment, you write

I believe here's a problem in your statements. The quantity that is conserved is the 4-momentum, not the 3-momentum. Thus, there is no contradiction in having no 3-momentum conservation.

Here you are mixing up two different ideas. Four-momentum is conserved under Lorentz transformations: if you and I are in different reference frames, I would see that some photon has energy and momentum $(E, \vec p)$, while you would see some different $(E', \vec p')$. Lorentz symmetry guarantees that we'll get the same $E^2 - p^2$ for the two cases. But once you've chosen a reference frame to analyze your interaction, energy and three-momentum are conserved separately. There's no reference frame where a finite-energy photon has zero three-momentum.

rob
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If $\gamma \rightarrow\pi^+ + \pi^-$ is valid process, then the conservation of total 4-momentum equation $$\tilde \gamma = \tilde \pi_+ + \tilde \pi_-$$ expresses a lightlike vector as the sum of two future-timelike vectors.

UPDATE:

  • To address your last question

in any reference frame, the energy $^0$ seems to be always positive. Why is that so?

  • The answer is: Since $m>0$ and $E=\gamma m$ (where, here, $\gamma$ is the time-dilation factor), real massive particles have future-directed timelike 4-momenta.
    And as @RedGiant commented, this future-directed property is preserved by transformations in the proper orthochronous Lorentz group.

If instead of 4-momenta, these were displacements in spacetime,
then this would imply that a sequence of two future-timelike displacements from A to B to C
would equal a lightlike displacement A to C. But that can't happen!
(The sum of two future timelike-vectors is a future-timelike vector.)

As a calculation, using your $(-,+,+,+)$ signature, I get $$\begin{align} 0 &=\tilde \gamma\cdot \tilde \gamma \\ &= \tilde \pi_+ \cdot \tilde \pi_+ + \tilde \pi_- \cdot \tilde \pi_- + 2\tilde \pi_+ \cdot \tilde \pi_- \\ &= (-m_{\pi}^2)\ +\ (-m_{\pi}^2) \ +\ 2\tilde \pi_+ \cdot \tilde \pi_-\\ &= \quad\ \ -2m_{\pi}^2 \quad\ \ + 2\tilde \pi_+ \cdot \tilde \pi_- \end{align} $$

So, for the total 4-momenta to be lightlike, we must have $$ \tilde \pi_+ \cdot \tilde \pi_- = m_{\pi}^2 $$ But this can't happen for a pair of two future-timelike vectors
since, in this $(-,+,+,+)$ signature,
the dot-product of two future-timelike [or two past-timelike] vectors is negative
(just like the dot-product of a timelike vector with itself is negative).

Thus, $ \tilde \pi_+ \cdot \tilde \pi_- = m_{\pi}^2 $ implies that one of these timelike 4-momenta must be past-timelike
and the other future-timelike ... in fact, $\tilde \pi_+ = -\tilde \pi_-$.
So, $$\tilde\gamma =\quad \tilde \pi_+ + \tilde \pi_- =\quad (- \tilde \pi_-) + \tilde \pi_- =\quad \tilde 0$$

In other words, the polygon (here, a triangle) of 4-momenta that describes this process is degenerate, where the photon leg is the zero 4-momentum vector.

So, I conclude that $\gamma \rightarrow\pi^+ + \pi^-$ is not a valid process.

UPDATE:

  • Here is a Desmos visualization: https://www.desmos.com/calculator/fqmryiui20
  • robphy-Desmos-relativity-decay
  • Try to get violet point (the tip of the sum of two timelike 4-momenta ) to lie on the light-cone (the black dashed figure, where $E^2=p^2$).
    If both timelike 4-momenta are future-timelike, you can't do it.
  • If they have opposite time-orientation, then you can do it only when the timelike 4-momenta are negatives of each other (resulting in a zero photon 4-momentum).
  • If you allow unequal-mass particles, then their 4-momenta must still have opposite time-orientation. One can get a nonzero photon 4-momentum. The resulting triangle can be re-interpreted as a Doppler effect.
robphy
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