Let's consider a generic two-body decay $A \rightarrow B + C$. In the center-of-mass frame (i.e, the rest frame of particle $A$), we know that the four-momentum of particle $A$ can be written as $p^{\mu}_{A} = \left(m_{A} , \vec{0} \right)$.

Now, we can also easy calculate the energy of particles $B$ and $C$ as following:

$$E_{B} = \frac{m_{A}^2 + m_{B}^2}{2m_A}, \qquad E_{C} = \frac{m_{A}^2 + m_{C}^2}{2m_{A}}.$$

Adding $E_{B}$ +and $E_C$, I obtain $$E_{B} + E_{C} = m_{A} + \frac{m_{B}^2 + m_{C}^2}{2m_{A}} \ne E_{A} = m_{A}.$$

Clearly, energy conservation cannot be violated. But where is the error then?


Okay, I actually found my own mistake: In fact, the formula $E_{B} = \frac{m_{A}^2 + m_{B}^2}{2m_{A}}$ is only valid if particle $C$ would be massless, i.e. if we had a(n) (anti-)neutrino for example.

The correct formulas read: $$E_{B} = \frac{m_{A}^2 + m_{B}^2 - m_{C}^2}{2m_{A}}, \qquad E_{C} = \frac{m_{A}^2 + m_{C}^2 - m_{B}^2}{2m_{A}}.$$

Adding these two formulas, we get $E_{A} = m_{A}$, showing energy conservation, as $E_{B}$ and $E_{C}$ were derived in the centre-of-mass frame of particle $A$.

  • $\begingroup$ These formulas are essentially the Law of Cosines in Minkowski spacetime. $E_B=m_B\cosh\theta_B$ where $\theta_B$ Is the rapidity of B wrt A. $\endgroup$ – robphy Feb 9 at 16:20
  • $\begingroup$ I am not that familiar with this. But we get the formulas by applying four-momentum conservation and by going into the restframe of particle $A$. $\endgroup$ – user248824 Feb 9 at 16:26
  • $\begingroup$ Draw a vector diagram in energy-momentum space where the particle 4-momenta satisfy $\tilde A=\tilde B+\tilde C$ (conservation of 4-momenta). Assume we are given all of the 4-vector magnitudes (squares of each rest mass). In the process of finding the [Minkowski-]angles (called rapidities) between 4-vectors, one can use Law of Cosines (by taking dot-products of a vector with itself). For instance, write $\tilde B=\tilde A-\tilde C$, then proceed. Example: physics.stackexchange.com/a/469191/148184 $\endgroup$ – robphy Feb 9 at 20:58

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