PS1-solutions

PS1-solutions - Physics 253a 1 Problem Set 1 Solutions...

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Physics 253a 1 Problem Set 1 Solutions September 21, 2010 1. (a) In the COM frame, one of the protons has energy E cm 2 = γm = m (1 - v 2 ) - 1 2 (where c = 1). Solving for v gives v 1 - ± with ± = 2 m 2 E 2 cm . The mass of the proton is about 1GeV, so ± 2 14000 2 10 - 8 , or about 6 mph. (b) The protons have velocities v 1 = v,v 2 = - v in the center of mass frame, so we can use the velocity addition formula to boost to the frame of one of the protons: v 0 = v 1 - v 2 1 - v 1 v 2 = 2 v 1 + v 2 = 2 - 2 ± 2 - 2 ± + ± 2 1 - ± 2 2 Alternatively, note that in a proton’s rest frame, the energy-momentum 4-vectors are p 1 = m (1 , 0 , 0 , 0) and p 2 = m ( γ 0 ,v 0 γ 0 , 0 , 0) where γ 0 = (1 - v 0 2 ) - 1 2 . Then E 2 cm = ( p 1 + p 2 ) 2 = 2 m 2 + 2 p 1 · p 2 = 2 m 2 + 2 m 2 γ 0 2 m 2 γ 0 (since γ 0 ± 1). Solving for v 0 gives v 0 = 1 - ± 0 with ± 0 = 2 m 4 E 4 cm = ± 2 2 10 - 8 mph. 2. (a) Starting from the blackbody distribution n ω = 1 π 2 ω 2 e ω kT - 1 ( c = ~ = 1!), we can evaluate the average energy h ω i = R 0 ω 3 e ω/kT - 1 R 0 ω 2 e ω/kT - 1 = π 4 30 ζ (3) kT 10 - 3 eV We’ll call this E γ . (b) Naively, we only need m π - E γ m π of extra energy to turn a pair into a pair. We might try just giving this all to the proton, so that it has initial energy m p + m π . But this won’t be enough in a process because of conservation of momentum. To see this, assume the proton and photon approach each other from opposite directions along the z axis. The incoming 4-momenta in the t - z plane are p γ = ( E γ , - E γ ) and p p = ( E,p ). At threshold, the result of the collision is a proton and a pion, both at rest in the center of mass frame.
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Physics 253a 2 Conservation of energy gives
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This note was uploaded on 02/04/2012 for the course PHYS 253A at Harvard.

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PS1-solutions - Physics 253a 1 Problem Set 1 Solutions...

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