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Relativity, 4-momentum and 4-velocity |
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| Feb28-09, 12:29 PM | #1 |
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Relativity, 4-momentum and 4-velocity
1. The problem statement, all variables and given/known data
A particle with 4-momentum [latex]\textbf{P}[/latex] 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 [latex]\sqrt{1-\frac{(\textbf{P}.\textbf{P})c^2}{(\textbf{P}.\textbf{U})^2}}[/latex] 2. Relevant equations I have a feeling [latex]E=mc^2=\sqrt{p^2c^2+(m_0c^2)^2}[/latex] 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! |
| Feb28-09, 06:29 PM | #2 |
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Hi Mathmos6!
![]() Hint: 4-momentum = rest-mass times 4-velocity, so P.P = … ? Then use the Lorentz transformation.
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| Mar1-09, 12:01 PM | #3 |
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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.
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| Mar2-09, 07:08 PM | #4 |
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Relativity, 4-momentum and 4-velocity |
| Mar2-09, 08:57 PM | #5 |
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P.P = [latex]m^{2}[/latex] (or maybe with a minus sign depending on the metric, but let's assume it's positive) U.U = [latex]c^{2}[/latex] P = mU, so P.U = m Hence, [latex] \frac{P.P} { (P.U)^{2}} = \frac{m^{2}} { (m)^{2}} = 1[/latex] |
| Mar2-09, 09:08 PM | #6 |
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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?
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| Mar2-09, 09:13 PM | #7 |
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Recognitions:
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| Mar2-09, 09:21 PM | #8 |
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Recognitions:
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| Mar3-09, 04:20 PM | #9 |
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ie U.U = [latex] g_{\mu \nu} U^{\mu} U^{\nu} [/latex] |
| Mar4-09, 08:09 AM | #10 |
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| Mar4-09, 09:56 AM | #11 |
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Ah, okay, this problem makes 1000% more sense now. Thanks for the help. |
| Mar4-09, 10:03 AM | #12 |
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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. |
| Mar8-09, 02:33 PM | #13 |
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Ahh brilliant! Thankyou so much for your help everyone, got it sorted now :)
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