# Extra factor of u in the expression for relativistic momentum?

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1. Oct 27, 2014

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

$E = \gamma m u$

It then uses the 4-momentum transformations for the second particle (the one that is moving in both frames), noting that

$(p^0)' = \gamma (p^0 - \frac{v}{c} (p^1)')$

But then it goes on to define $p^1$ as

$p^1 = \gamma m u (-u)$

This is where I made the mistake in my working, I used

$p^1 = \gamma m (-u)$

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.

#### Attached Files:

• ###### q4a.png
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Last edited: Oct 27, 2014
2. Oct 27, 2014