# Energy conservation in two-body decays?

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 $$A$$.
• 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. Feb 9 '21 at 16:20
• 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 $A$.
• 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 Feb 9 '21 at 20:58