# General formula for relativistic Energy/momentum conservation?

Say we have a particle of mass $M$, which is at rest in a laboratory. This particle then decays into a particle of mass $m$ and speed $v$ and a photon of frequency $f$. How would one describe the conservation of Energy/Momentum of such a decay? I know that their conservation can be combined into a single equation of conservation of 4-momentum but I don't really know how to use 4-momentum properly yet. Is there any general Equation/technique one should use here?

## 1 Answer

The equation is just $$\mathbf{p}_f = \mathbf{p}_i \;,$$ which decomposes into \begin{align} E_f &= E_i\\ p_{x,f} &= p_{x,i} \\ p_{y,f} &= p_{y,i} \\ p_{z,f} &= p_{z,i} \;, \end{align} which are exactly the expressions you know and (hopefully) love from classical physics. As in classical physics to deal with a system you add the four-momentum of all the parts to get the total for the system and it is that total that is conserved.

The things that change are the expressions you can use for the various components. For free particles we have things like \begin{align} E &= \gamma m c^2 \tag{total energy} \\ K &= (\gamma -1) m c^2 \tag{kinetic energy} \\ p &= \gamma m v \tag{magnitude of momentum} \;, \end{align} and perhaps the most important relation of all the definition of mass1 $$\left( mc^2 \right)^2 \equiv E^2 - \left(\vec{p}c\right)^2 \;. \tag{*}$$ Note that unlike classical physics you can not add the mass of the parts of the system to get the mass of the system. You must find the mass of the system from the total four-momentum of the system.

I'm using the particle physics metric (+---), here. Some people reverse the signs for no good reason.

The reason I like to emphasize the importance of (*) is two-fold:

1. The mass defined this way is a Lorentz invariant. You can compute it in one frame and use it in another without fear. Exploit this without mercy.
2. It does not involve the Lorentz factor $\gamma$. Anytime you introduce an expression involving an inverse square-root of a difference of a squared thing you are just asking for an algebraic headache. Avoid it when you can.

It is worth keeping in mind that sometime you may be able to treat part of the problem with the Newtonian expressions (non-relativistic) or in the ultra-relativistic limit (in effect ignoring the mass).

1 The invariant mass, previously known as the "rest mass", but since we don't the term "relativistic mass" in the modern parlance, we rarely bother to append any adjective at all.