PS1_253A_solutions.pdf - Problem Set 1 Solutions Physics...

This preview shows page 1 - 4 out of 8 pages.

Problem Set 1 Solutions Physics 253A Hofie Hannesdottir September 17, 2018 Problem 1. a) In the center-of-mass frame, each proton has E p = 6 . 5 TeV energy. The speed of each proton is then (in natural units) v = | ~ p | E p = q E 2 p - m 2 p E p 1 - 1 2 m 2 p E 2 p , (1) where the Taylor approximation in the last equation is justified since the proton mass m p = 1 GeV is much less than E p = 6 . 5 TeV. The difference between the speed of light and the proton velocity is then Δ v c - v = 1 2 m 2 p E 2 p c = 1 2 938 MeV 6 . 5 TeV 2 3 . 00 · 10 8 m s = 11 km h . (2) Problem 1. b) The 4-vectors of the protons can be written, in the rest frame of one of them, as p 1 = ( m p , 0 , 0 , 0) , p 2 = ( E 2 , 0 , 0 , | ~ p 2 | ) , (3) with E 2 2 = | ~ p 2 | 2 + m 2 p . The quantity ( p 1 + p 2 ) 2 is frame independent, so labeling the total center-of-mass energy with Q , we get Q 2 = ( p 1 + p 2 ) 2 = ( m p + E 2 ) 2 - p 2 2 = 2 m p p 2 , (4) where we have dropped subleading terms in m 2 p E 2 2 . The speed of one proton relative to the other is then v 2 = p 2 E 2 = p 2 q p 2 2 + m 2 p = 1 - 1 2 m 2 p p 2 2 = 1 - 2 m 4 p Q 4 , (5) and the difference between this speed and the speed of light is Δ v = c - v 2 = 2 m 4 p Q 4 = 5 . 9 · 10 - 8 km h . (6) Problem 2. a) Note that from dimensional analysis, we expect an answer on the order of k B T 2 · 10 - 4 eV. We assume that the universe is a hot box of light with size L in thermal equilibrium. According to equation (1.5) in Schwartz, the expectation value of energy in each mode with angular frequency ω n is h E n i = ~ ω n e ~ ω n β - 1 , (7) with β = 1 k B T , and T = 2 . 73 K. The total energy of the universe is then (the factor of 2 is from the two 1
possible polarizations of photons): E tot = 2 Z all ~n d 3 ~n ~ ω n e ~ ω n β - 1 = 2 1 Z - 1 d cos θ 2 π Z 0 Z 0 d | ~n | | ~n | 2 ~ ω n e ~ ω n β - 1 = 2 · 4 π ~ L 3 8 π 3 c 3 Z 0 0 ω 0 3 e ~ ω 0 β - 1 = L 3 ~ π 2 c 3 1 ( ~ β ) 4 Z 0 u 3 e u - 1 = π 2 15 L 3 c 3 ~ 3 β 4 , (8) where we have used that ω n = 2 π L | ~n | c , and Z 0 du u 3 e u - 1 = π 4 15 . (9) The average number of photons in such box is N tot = 2 Z all ~n d 3 ~n 1 e ~ ω n β - 1 = 2 1 Z - 1 d cos θ 2 π Z 0 Z 0 d | ~n | | ~n | 2 e ~ ω n β - 1 = 2 · 4 π L 3 8 π 3 c 3 Z 0 0 ω 0 2 e ~ ω 0 β - 1 = L 3 π 2 c 3 1 ( ~ β ) 3 Z 0 u 2 e u - 1 = 2 ζ (3) π 2 L 3 c 3 ~ 3 β 3 , (10) where we have used that Z 0 du u 2 e u - 1 = 2 ζ (3) . (11) The average energy of photons in outer space is therefore h E i = E tot N tot = π 4 30 ζ (3) k B T = 6 . 4 · 10 - 4 eV . (12) Problem 2. b) Assuming that the Earth is at rest in a frame with isotropic distribution of photons of energy E γ = 6 . 4 eV, the minimum amount of proton energy required to create a pion is obtained when the proton and photon collide head on. Furthermore, the minimum energy required to create a pion is obtained when the outgoing pion and proton are at rest in the center-of-mass frame. We label the incoming proton and photon momenta with p p and p γ respectively, and the outgoing proton and pion momenta with q p and q π respectively. Then, in Earth’s frame, p p = ( E p , 0 , 0 , | ~ p p | ) p γ = ( E γ , 0 , 0 , - E γ ) , (13) 2
with E 2 p = p 2 p + m 2 p

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture