# Conservation of 4-momentum in special relativity

I understand that the inner product of two 4-vectors is conserved under the Lorentz transformations, so that the absolute value of the four momentum is the same in any reference frame. This is what I (most likely mistakenly) thought was meant by the conservation of momentum. I don't understand why equations such as

$P_1=P_2+P_3$

($P_i$ are 4-momentum vectors for different particles in a collision for example)

should hold, within a reference frame. I've been told that you can't just add four velocities together on collision of particles, so why should you be able to do this with the momentum vectors?

• I just want to point out that you are confusing "conserved" with "invariant".
– Y2H
Oct 21, 2019 at 15:46

## 3 Answers

I understand that the inner product of two 4-vectors is conserved under the Lorentz transformations

Yes, $p_1.p_2$ is a Lorentz invariant

So that the absolute value of the four momentum is the same in any reference frame.

It is not correct to speak about the "absolute value" of a (quadri)vector. Which is conserved in a Lorentz transformation is $p^2 = (p^o)^2 - \vec p^2$

This is what I (most likely mistakenly) thought was meant by the conservation of momentum.

No, conservation of momentum is a completely different thing. Ultimately, you have some theory describing fields and interactions, describing by an action which is invariant by some symmetries. If the action is invariant by space and time translations, then there is a conserved quantity which is momentum/energy.

I don't understand why equations such as P 1 =P 2 +P 3 (P i are 4-momentum vectors for different particles in a collision for example) should hold, within a reference frame. I've been told that you can't just add four velocities together on collision of particles, so why should you be able to do this with the momentum vectors?

If the theory action is invariant by space/time translations, then the momentum/energy is conserved, so the total momentum/energy of the initial particles is the same as the total momentum/energy of the final particles :

$$(p_\textrm{tot})_\textrm{in}^\mu = (p_\textrm{tot})_\textrm{out}^\mu\tag{1}$$

If there are several initial particles, they are considered as independent (the global state is the tensor product of the states of the initial particles). The independence means that you have :

$$(p_\textrm{tot})_\textrm{in}^\mu = \sum_i p_i^\mu\tag{2}$$ where the sum is about all the initial particles. A similar equation holds for the final particles.

In special relativity, if you add two velocities, you have to use the formula

$$v = (v_1+v_2)\left(1+\frac{v_1v_2}{c^2}\right)^{-1} \text{ .}$$

So you cannot simply add two velocities together. Usually, velocity is not a good variable to work with in special relativity. It's much easier to use four-momentum conservation, which is simply given by

$$p = p_1 + p_2 \text{ ,}$$

for a particle collision where two particles with $p_1$ and $p_2$ collide and then stick together and have the momentum $p$. Since the four-momentum is given by

$$p = \begin{pmatrix}E/c \\ \vec{p}\end{pmatrix} \text{ ,}$$

the conservation of four-momentum is nothing else than the conservation of Energy $E$ and the conservation of three-momentum $\vec{p}$.

To answer your questions:

Why can we add four-momentums in a particle collision? Because energy and momentum conservation also holds in relativity.

Why can't we add four-velocities in a particle collision? Because there is no such thing as "velocity conservation", neither classically, nor in relativity.

• This answer was great. I have a clarifying question - will $(P_1 + P_2)^2$ be invariant, thus $(P_1 + P_2)^2= -(m_1+m_2)^2c^2$?
– inya
Oct 24, 2017 at 22:38

You can just verify each component and they are just momentum conservation in 3-momentum. There is no velocity conservation so you cannot add them together.