- #1

Grasshopper

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- TL;DR Summary
- I'm looking to figure out why the gamma factor is needed

Now, deriving relativistic momentum isn't terribly difficult, but that's not the same as understanding it. I'm trying to figure out why conservation of momentum in special relativity

When I looked at conservation of momentum in elementary physics, we basically just took two masses and two speeds, worked out a collision, and showed that the total momentum was the same before and after the collision. For example, suppose there are two equal masses moving at speeds equal in magnitude but opposite in direction along the same axis, and they collide in a perfectly inelastic collision. For example, ##\frac{1}{2}mu + \frac{1}{2}m(-u)= m(0) = 0##. Both before and after collision total momentum equals 0, so momentum is conserved.

What I would like to know is, how would I go about showing that in special relativity, momentum is not conserved unless I include the gamma factor coefficient for each object? I'm thinking that a simple collision in which I can make the objects approaching along an axis isn't going to do the job. Is it more related to something like an observer passing by wouldn't agree that the momentum is conserved in the collision? I can see why this might be the case, since the transformation equation includes both energy and momentum, and since velocities add using relativistic velocity addition. Or can it be shown in a single reference frame?

.

I'm just looking for some general direction about how I can prove to myself that the gamma factor in the momentum equation is necessary,

Anything helps. Thanks.

*requires*the gamma factor.When I looked at conservation of momentum in elementary physics, we basically just took two masses and two speeds, worked out a collision, and showed that the total momentum was the same before and after the collision. For example, suppose there are two equal masses moving at speeds equal in magnitude but opposite in direction along the same axis, and they collide in a perfectly inelastic collision. For example, ##\frac{1}{2}mu + \frac{1}{2}m(-u)= m(0) = 0##. Both before and after collision total momentum equals 0, so momentum is conserved.

What I would like to know is, how would I go about showing that in special relativity, momentum is not conserved unless I include the gamma factor coefficient for each object? I'm thinking that a simple collision in which I can make the objects approaching along an axis isn't going to do the job. Is it more related to something like an observer passing by wouldn't agree that the momentum is conserved in the collision? I can see why this might be the case, since the transformation equation includes both energy and momentum, and since velocities add using relativistic velocity addition. Or can it be shown in a single reference frame?

.

I'm just looking for some general direction about how I can prove to myself that the gamma factor in the momentum equation is necessary,

*without*just someone telling me the answer, if that's possible.Anything helps. Thanks.