- #1
stallm
- 7
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Hi, I've been struggling with this question for a while, can't help but think I'm missing something obvious
Let's say I have a particle of rest mass 2m. In the frame S, it moves with velocity v. S' is the rest frame of the particle. In S, the momentum of the particle (c=1)
[itex]p=\frac{2mv}{\sqrt{1-v^2}}[/itex]
Now, in S', the particle splits into two particles of mass m with velocities v and -v. The particles with -v will be stationary in S, the particle with v will have velocity vnew
[itex]v_{new}=\frac{v+v}{1+v^2}=\frac{2v}{1+v^2}[/itex]
by the velocity addition formula
So the momentum in S
[itex]p=\frac{mv_{new}}{\sqrt{1-v_{new}^2}}[/itex]
This doesn't appear to be equal to the p before (just by trying v=0.5), so something's clearly gone wrong. Can someone help?
Thanks
Let's say I have a particle of rest mass 2m. In the frame S, it moves with velocity v. S' is the rest frame of the particle. In S, the momentum of the particle (c=1)
[itex]p=\frac{2mv}{\sqrt{1-v^2}}[/itex]
Now, in S', the particle splits into two particles of mass m with velocities v and -v. The particles with -v will be stationary in S, the particle with v will have velocity vnew
[itex]v_{new}=\frac{v+v}{1+v^2}=\frac{2v}{1+v^2}[/itex]
by the velocity addition formula
So the momentum in S
[itex]p=\frac{mv_{new}}{\sqrt{1-v_{new}^2}}[/itex]
This doesn't appear to be equal to the p before (just by trying v=0.5), so something's clearly gone wrong. Can someone help?
Thanks