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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?

Qmechanic
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1 Answers1

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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 $A$.