# Four-Momentum Invariant and Conservation Laws Yielding Contradictory Results

1. Dec 1, 2011

### n+1

Hello PF community!

I'm having trouble with what strikes me as an inconsistency within conservation of energy, conservation of momentum, and the four-momentum invariant equation (E2-p2c2 = m2c4). For the sake of this question, I'll be using non-relativistic mass--i.e. mass is the same in all reference frames and p=γmv and E = γmc2.

Here is the situation I'm imagining: a particle of mass M moves in the +x direction with some velocity when suddenly it decays into two particles of mass (2/5)M moving in any ol' direction. By the conservation of momentum and energy, we know that some of our mass before the decay has been converted to kinetic energy. That's all okay.

The issue: since energy and momentum are conserved, combinations of them are conserved. For problem solving, a convenient combination is Ei2-pi2c2 = Ef2-pf2c2. But if we use the four-momentum invariant equation, we can substitute in mi2c4 and mf2c4 respectively. This leaves us with a problem; in the example, the Mf is less than the Mi, but we just derived that the final mass energy squared is equal to the initial mass energy squared. These statements are contradictory!

So where does the physics/math go awry?

Any and all help is appreciated!

2. Dec 1, 2011

### Staff: Mentor

Hi n+1, welcome to PF, and excellent first post.

The key to recognize is that the conservation laws apply to the whole system. So the invariant mass of the system is indeed conserved before and after the decay. However, the invariant mass of the system is not equal to the sum of the invariant masses of the particles. The best way to see that is simply to work out the problem you have outlined with some concrete numbers.

3. Dec 1, 2011

### Staff: Mentor

The equation you wrote relating energy, momentum, and mass applies to a single particle. In the final state, you have two particles, not one. You have to write a separate equation relating E, p, and m for each final particle; these equations then become additional constraints on the final energy and momentum of each final particle, in addition to the conservation laws (sum of final energies equals initial energy, and vector sum of final momenta equals initial momentum).

A tip: try working the problem in a frame in which the initial particle is at rest; that way the initial momentum is zero and the equations get quite a bit simpler.

4. Dec 1, 2011

### Staff: Mentor

I wrote my previous post before seeing DaleSpam's post; having seen it, I need to amend the above. Actually, you *can* write the equation E2 - p2 = m2 (in units where c = 1) for the entire system in the "final" state, where there are two particles; and when you do, as the OP discovered, you get that the total invariant mass m of the system is the *same* as it was before! This is a feature, not a bug. As DaleSpam says, the invariant mass m of the system as a whole is not the sum of the invariant masses of the individual particles, for multi-particle systems. You determine the invariant mass of the entire system, in fact, by the procedure just described: you know the total energy and momentum of the system from the conservation laws, so just subtract their squares (and take the square root of the result).

All that doesn't help in trying to figure out how the total energy and momentum of the system is divided up among its parts; that's where the stuff I posted previously comes in.

5. Dec 1, 2011

### n+1

Ah, I see. I was implicitly assuming that (E1 + E2)2-(p1+p2)2 = (E12 - E22) + (p12 - p22) by saying that total invariant mass equals the sum of the individual particles' invariant masses. So naturally my invariant mass was off by (2E1E2 - 2p1p2).

Makes sense!

Thanks for your help Dale and Peter!