Relativity, 4-momentum and 4-velocity

[Mathmos6]

Homework Statement

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 $\sqrt{1-\frac{(\textbf{P}.\textbf{P})c^2}{(\textbf{P}.\textbf{U})^2}}$

Homework Equations

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

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!

 

[tiny-tim]

Hi Mathmos6!

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

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.

 

[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.

 

[Mathmos6]

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.

Yahuh it's posted correctly!

 

[LongLiveYorke]

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.

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

 

[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?

 

[gabbagabbahey]

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)

Since when does 4-momentum have units of mass?

U.U = $c^{2}$

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

 

[gabbagabbahey]

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?

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

 

[LongLiveYorke]

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

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

 

[Mmmm]

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

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.

 

[LongLiveYorke]

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.

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

 

[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.

 

[Mathmos6]

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