Momentum Conservation Law of Ultra-Relativistic Particles (Is it true?)
- Thread starter TonyYuan
- Start date
Elastic or inelastic collision?
Elastic collision
OK. Then it's pretty straightforward. They exchange momentums.
Can you give your answer directly?
What are the velocities of A and B?
Please post your answer first, Tony. If you have an answer, that is.
Remember the problem you had with your previous relativistic scenario? You couldn't solve it yourself.
Are you able to solve your own problem here? Or are you asking for help?
Is this your homework, Tony?
Also, is this a straight-line collision, or a glancing collision? Remember how you underspecified your scenario in a previous thread, Tony? Yet here you are again, failing to include important information in the scenario description.
If you leave out necessary information, how do you expect us to help you with your homework?
Remember the problem you had with your previous relativistic scenario? You couldn't solve it yourself.
Are you able to solve your own problem here? Or are you asking for help?
Is this your homework, Tony?
Also, is this a straight-line collision, or a glancing collision? Remember how you underspecified your scenario in a previous thread, Tony? Yet here you are again, failing to include important information in the scenario description.
If you leave out necessary information, how do you expect us to help you with your homework?
Last edited:
James R, my question has been clearly stated (aligned with the center of the ball).
I'm here to ask a question, because some SR/GR experts are gathered here, so I think I can find the answer here.
James R, do you know the answer? I will listen carefully to your analysis.
James R, can you give an answer? I don't really know the answer, it looks like you know what the answer is, can you just share your answer here?
exchemist
Valued Senior Member
If this were simple Newtonian mechanics, i.e. at relative speeds <<c, what would be the answer, Tony?
OK, I can give the answer: Suppose the speeds of A and B after the collision are v1 and v2 respectively.
According to the law of conservation of momentum:
m*0+m*v = m*v1 + m*v2..........(1)
According to the law of conservation of kinetic energy:
1/2m*0*0 + 1/2m*v*v = 1/2mv1*v1+1/2mv2*v2..........(2)
Organized to get:
v1 = v
v2 = 0
If A had velocity V and B had velocity 0, then after the collision A has velocity 0 and B has velocity V.
Have you ever thought about such a problem, the mass of A is measured when A is stationary, now A has a velocity of 0.8660254c, according to SR, the mass of A is no longer m, and the momentum of A is no longer mv , the kinetic energy of A is no longer 1/2mvv. What do you think? Is there really no need to consider SR here?
exchemist
Valued Senior Member
Why not do the algebra again, using γ, then?
That's why I'm here, because I really don't know how to use SR to calculate this problem. If you can give an answer, you can share it here directly, you are always full of wisdom.
If you are actually curious, Wiki has a good explanation, just look up "Elastic Collision".
exchemist
Valued Senior Member
There is actually a derivation at the site Origin refers to, using the SR formulae for energy and momentum. The trick seems to be to use the centre of momentum frame of reference. That gets you away from the rather hideous calculations you would get if you tried to expand γ. Here is the link: https://en.wikipedia.org/wiki/Elastic_collision.
“The trick seems to be to use the centre of momentum frame of reference”, this method seems to greatly simplify the calculation, but when the reference frame changes, does the mass m of A and B also need to change? When A and B are stationary relative to the earth, we measured their mass as m, and now the reference system is no longer the earth, so is their mass still m?
But from the analysis on the wiki, m1 and m2 have not changed due to the change of the reference system, which makes us wonder, is the earth the absolute reference system in the universe?
Of course not. Nothing from that analysis suggests anything of the sort.