# Momentum in spec relativity

1. Oct 17, 2006

### thenewbosco

here is my question and my solution, id like to see where i have gone wrong and how to rectify it:

A particle of mass M, at rest, decays into two smaller particles, masses m1 and m2. What are their energies and 4-momenta (given: $$\sqrt{p^2 + m_1^2c^2}+\sqrt{p^2 + m_2^2c^2}=Mc$$ $$E_1 = \sqrt {m_1^1c^4 + p^2c^2}$$ $$E_2=\sqrt{m_2^2c^4+p^2c^2}$$? then the part i am working on: solve the problem again for m2=0. solve the equations for p and e1, and take the limit m1 -> 0.

i find it kind of ambiguous, but to clarify, is the correct answer obtained by using the equation for Mc and subbing m2=0 in there, then using the two equations you then have to solve for E2?

Last edited: Oct 17, 2006
2. Oct 17, 2006

### OlderDan

Relativistic momentum has to be conserved. The decaying particle has no momentum, so the two decay products must have equal and opposite momenta. Since the masses are assumed known, you just need to find their respective momenta subject to conservation of energy (your first equation).

For a massless particle, the energy equation reduces to E = pc

Last edited: Oct 17, 2006
3. Oct 18, 2006

### thenewbosco

i can just put into the first equation that m2=0 correct?

4. Oct 18, 2006

### OlderDan

Yes. I can't post just yes, so p is the same for both, which I think you already know.