ph1bA_08_hw4s

ph1bA_08_hw4s - Ph1b (analytic track): Solution to Homework...

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Ph1b (analytic track): Solution to Homework 4 Chefung Chan (cchan@caltech.edu), Winter 2008 Problem 8-14 (a) (1 point) In the center-of-mass frame, the system has zero total momentum. Since producing one photon would give a non-zero momentum, there must be at least two photons produced. (See page 238 and Fig. 8-11 in particular). (b) (2 points) From the conservation of energy, E A + E B = E C + E D ( m + K )+ m = E C + E D ( ) From conservation of momentum, p A + p B = p C + p D p ( m + K ) 2 m 2 = E C E D , ( ∗∗ ) where in the last step we have used the fact that, for photons, E = | p | .So lv ing ( )and( ∗∗ ) for the unknowns E C and E D gives: E C = m + K 2 + 1 2 p K (2 m + K ) E D = m + K 2 1 2 p K (2 m + K ) . (c) (2 points) (1) When K ± m (or equivalently, K/m ± 1) we have E C = E D m . (2) When K/m ² 1, we write the square root term as: 1 2 p K (2 m + K )= K 2 r 2 m K +1 K 2 ± 1+ 1 2 2 m K ² = K 2 + m 2 . Therefore, E C K , and E D m 2 . Problem 8-17 (a) (2 points) We are told that the protons and antiprotons are accelerated up to 0.9 TeV and collide head on. Since this is the center-of-mass frame, all the energy is available for the interaction, that is: E T = E p + + E p =1 . 8TeV . 1
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(b) (3 points) Now we consider the the lab frame, in which an (even faster) an- tiproton collides with a proton at rest. lab: p + →← p rest: p + ←← p Since this frame moves at the speed of the proton (corresponding to 0.9 TeV) with respect to the center-of-mass frame, we Fnd that E p + = m p + γ p + γ p + = 900 GeV 938 MeV 959 v p + = s 1 1 γ 2 p + 0 . 999 999 5 . Comparing the proton’s energy and momentum E = and p = mγv we Fnd that these are almost identical if v is very close to one. The proton is highly relativistic.
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ph1bA_08_hw4s - Ph1b (analytic track): Solution to Homework...

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