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GR0177 #99
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Problem
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This problem is still being typed. |
Special Relativity }Conservation of Energy
The tricky part of this problem is to note that the momentum of the photon is shared equally between all three final particles. (This insight was supplied by Felipe Birk). Thus,  , where p is the momentum of the photon and  is the momentum of the final electron or photon.
The initial energy before the photon strikes the electron is  , which is just the energy of the photon plus the rest energy of the electron.
The final energy after the collision is ^2+(m_e c^2)}) , which is the sum total energy of all three final particles, i.e., the positron and two electrons. (A positron is the electron's antiparticle, and thus they have the same mass.) Note that the momentum split relation mentioned above is used to equate the final particle momentum with the initial photon momentum.
Conjure up the good conservation of energy idea. Equating  , one gets ^2+(m_ec^2)^2 + 2pcm_ec^2 = 9((pc/3)^2+(m_ec^2)^2)) . Canceling the ^2) terms on both side, then solving, one arrives at  , which is choice (D). |
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Alternate Solutions |
dryu
2008-10-17 22:32:47 | It's simplest to use four vectors. In natural units:
The initial four-momentum is ) and the final four-momentum is ) . The invariant mass, e.g. these four vectors squared, is conserved:
^2-E^2=9(\gamma^2m^2-\gamma^2m^2v^2))
To convert this into SI units, just let  . |  |
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Comments |
zmburell
2009-09-30 13:05:08 | The electron is inititially at rest, so in the expression for the initial 4-momentum squared, why isnt (pc)=0,
That is, if the electron is initially at rest then all of its energy is in its rest mass
zmburell
2009-10-03 10:36:56 |
Nevermind, Im retarded, I figured it out
disregard my last post
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|  | matonski
2009-04-01 00:35:29 | In a reference frame moving along with the three particles at the end, the magnitude of the momentum 4 vector is simply  . Since this is an invariant, it must also be the magnitude of the 4 vector at the beginning. Letting  be the initial photon energy and setting  , we have ^2) and  . Therefore, its magnitude is ^2 = (m+x)^2 - x^2) Solve for x and you are done. | | Poop Loops
2008-11-02 14:41:44 | I'm confused... if they don't say how fast the 3 particles are moving, how can we deduce the energy of the photon? It was at least  since it created a positron and an electron and then moved them. Okay, that makes sense.
But if the collision sent all  going at  then the photon would have had more energy than if they were all going at say  afterwards. What gives? | | dryu
2008-10-17 22:32:47 | It's simplest to use four vectors. In natural units:
The initial four-momentum is and the final four-momentum is . The invariant mass, e.g. these four vectors squared, is conserved:
To convert this into SI units, just let . | | ssp
2008-10-17 12:35:10 | Why does not common sense work?rnrnYou create  pair, that is an energy of rnYou put energy into the resting  that gets some energy  rnThen you need energy to make the pair move, so you need  for the total energy of those particles.rnrnAdding up all energy you need to get something above 3 at least... you are down to 2 answers and 5 seems a bit large for  so you pick 4 | | Ning Bao
2008-02-01 08:26:25 | If we set c and m=1 and let p=a*m=a The math becomes much simpler. | | Tommy Koulax
2007-11-01 22:08:02 | I don't understand why p is divided by three. If the photon is no longer, then shouldn't the p be divided by 2?
alpha
2008-02-07 11:02:47 |
i think it is because there are 3 final particles?
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2007-09-13 18:14:56 | Although the overall solution seems okay, I think there is a small typo. ^2+(m_e c^2)^2}) . Note the second squared on the second term.
sravani
2008-10-09 19:05:48 |
TigerTed8 is right. Solution has to be corrected by squaring the second term. Yosun, you got to correct it...
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zylstra
2009-04-01 18:35:24 |
Please excuse my ignorance. How did you get this formula for  ?
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| | angiep
2005-11-11 22:17:50 | im confused, the problem says the photon is destroyed. you say all three share the momtentum, but there are only 2...
angiep
2005-11-11 22:18:29 |
nevermind. two electrons, duh
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| | Void
2005-11-09 23:08:19 | Hi Yosun. Great site. I have a few solutions to GR9677 that aren't on your PDF yet. Do you take submissions? How should I send them, if so?
yosun
2005-11-09 23:36:54 |
Hi, please post alternate solutions on the website in the comments section corresponding to the particular problem. (The pdf I posted in the yahoo physicsgre group is incomplete and contains a number of errors that I have corrected here on this site. Of course, users like you might have more elegant solutions than what I have posted---so, feel free to post away. This site aims to be the most convenient one-stop source for all solutions to released ETS GRE Physics exams. You can help by contributing.)
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| | yosun
2005-11-09 22:08:33 | This solution has been corrected. |  |
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