Relativity, 4-momentum and 4-velocity

1. Feb 28, 2009

Mathmos6

1. The problem statement, all variables and given/known data

A particle with 4-momentum $\textbf{P}$ is detected by an observer with four-velocity U. Show that the speed, v, of the detected particle in the observers rest frame is given by $\sqrt{1-\frac{(\textbf{P}.\textbf{P})c^2}{(\textbf{P}.\textbf{U})^2}}$

2. Relevant equations
I have a feeling $E=mc^2=\sqrt{p^2c^2+(m_0c^2)^2}$ will be of use...

3. The attempt at a solution I really don't know where to start on this one, relativity confuses me! :( Any help would be hugely appreciated - thanks!

2. Feb 28, 2009

tiny-tim

Hi Mathmos6!
Nooo … the question and answer use 4-vectors, not their components (E and p).

Hint: 4-momentum = rest-mass times 4-velocity, so P.P = … ?

Then use the Lorentz transformation.

3. Mar 1, 2009

LongLiveYorke

Could the OP double check that he posted the question properly. There are some thing about it that don't seem to make sense to me. Can anyone confirm if they've solved it or not.

4. Mar 2, 2009

Mathmos6

Yahuh it's posted correctly!

5. Mar 2, 2009

LongLiveYorke

Correct me where I went wrong:

P.P = $m^{2}$ (or maybe with a minus sign depending on the metric, but let's assume it's positive)

U.U = $c^{2}$

P = mU, so P.U = m

Hence, $\frac{P.P} { (P.U)^{2}} = \frac{m^{2}} { (m)^{2}} = 1$

6. Mar 2, 2009

Mathmos6

I'm sorry but I still can't see how to do it :( I played around with the vectors for a while but didn't really get anywhere - a little further guidance maybe?

7. Mar 2, 2009

gabbagabbahey

Since when does 4-momentum have units of mass?

What makes you think that the observer is traveling at the speed of light?

8. Mar 2, 2009

gabbagabbahey

Start by answering tiny tim's question....what is P.P in terms of rest mass and the 4-velocity of the particle?

9. Mar 3, 2009

LongLiveYorke

Maybe I'm missing something, then. When I write U.U, I mean the 4-dot product between 4 vectors. Is that what the problem means as stated when it writes U.P and P.P?

ie

U.U = $g_{\mu \nu} U^{\mu} U^{\nu}$

10. Mar 4, 2009

Mmmm

I think your problem is that you are taking U to be the 4-velocity of the particle whereas it is actually the 4-velocity of the observer relative to some other frame. P is also relative to this other frame.

Last edited: Mar 4, 2009
11. Mar 4, 2009

LongLiveYorke

Ah, okay, this problem makes 1000% more sense now. Thanks for the help.

12. Mar 4, 2009

LongLiveYorke

To the OP:

There's a really nice trick to this problem. Remember that a quantity like U.P (where they are both 4 vectors) is an invariant, meaning that it will be the same when measured in any frame. So, the key is to pick the right frame to calculate this dot product.

13. Mar 8, 2009

Mathmos6

Ahh brilliant! Thankyou so much for your help everyone, got it sorted now :)