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PROBLEM SOLVED - the worked example I was referring too was wrong :/
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Hello, I've been stuck on a question in one of my SR problem sets for some time now, and managed to find a worked solution to a similar problem online. I've attached an image of the problem (the missing text just says that two particles of equal mass are accelerated to a speed u).
It begins by showing that in frame S, both particles have the same energy, which is
[itex]E = \gamma m u[/itex]
It then uses the 4-momentum transformations for the second particle (the one that is moving in both frames), noting that
[itex](p^0)' = \gamma (p^0 - \frac{v}{c} (p^1)')[/itex]
But then it goes on to define [itex]p^1[/itex] as
[itex]p^1 = \gamma m u (-u)[/itex]
This is where I made the mistake in my working, I used
[itex]p^1 = \gamma m (-u)[/itex]
I don't understand why the proper momentum of the second particle has an additional factor of u associated with it. Any explanation would be greatly appreciated.
Cheers.
----------------------------
Hello, I've been stuck on a question in one of my SR problem sets for some time now, and managed to find a worked solution to a similar problem online. I've attached an image of the problem (the missing text just says that two particles of equal mass are accelerated to a speed u).
It begins by showing that in frame S, both particles have the same energy, which is
[itex]E = \gamma m u[/itex]
It then uses the 4-momentum transformations for the second particle (the one that is moving in both frames), noting that
[itex](p^0)' = \gamma (p^0 - \frac{v}{c} (p^1)')[/itex]
But then it goes on to define [itex]p^1[/itex] as
[itex]p^1 = \gamma m u (-u)[/itex]
This is where I made the mistake in my working, I used
[itex]p^1 = \gamma m (-u)[/itex]
I don't understand why the proper momentum of the second particle has an additional factor of u associated with it. Any explanation would be greatly appreciated.
Cheers.
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