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Phys 801 - Homework 1 Special and General relativity, Relativistic kinematics 1. Show that Newtonʼs laws of motion are not invariant under a transformation to a frame that is uniformly accelerated with respect to an inertial frame of Newton mechanics. What is the equation of motion in the accelerated frame? [Hint: consider transformations between 2 frames with aligned x-axes and acceleration along x]. 2. A resonance of significance for experiments on ultra-high energy neutrinos in astrophysics is the Glashow resonance: ν e + e − →W − . Assuming that the target electrons in the cosmos are at rest, show that this resonance would be excited for antineutrino energies of about 6,400 TeV. Some expressions for the cross sections are given below. Chose at least one of them € and plot it (you should get something similar to the figure below). Then derive the value of the cross section around the resonance in an interval equal to the boson W width. Use any software tool you like for numerical integrations and plots and the values of the constants in the table. 3. Calculate the minimum photon energy in the rest frame of the proton for direct + pion production in the reaction p-gamma→π + X, given that the mass of the charged pion is 139.57 MeV. Infer what can be X from the fact that baryon number and charge must be conserved in the reaction. 4. Demonstrate that the neutrino takes ~1/4 of the pion energy (rest mass) in the + + charged pion decay: π → µ + ν µ , working in the rest frame of the pion. The muon mass is 105.66 MeV. 5. Demonstrate that an isolated electron cannot absorb a photon. € 6. [Optional for undergraduate students] Find the total time to the Big crunch in the case of a closed universe with a curvature term with k = +1 and mass 1023 MSun, where MSun = 2 x 1030 kg. Hints: we found in class the value of the maximum radius . You need to calculate the time to reach this radius using the change of variable: and then calculate the time to the Big Crunch where the universe collapses to Rr = 0. 7. [Optional] Calculate the minimum projectile energy in the rest frame of the target proton for a pp reaction that produces a single π0 (remember Baryon conservation!). The rest mass of the π0 is 134.97 MeV, and the mass of the proton is mp = 938.27 MeV. Suggested solutions 1. We consider the uniform acceleration along the x axis so that transformations are: ⎧ 1 2 ⎪ x' = x − 2 at ⎪ y' = y ⎨ ⎪ z' = z ⎪ t' = t ⎩ € Calculating the second derivatives in time we do not get zero from the first one so (xʼyʼzʼ) frame is not inertial. d 2 x' d 2 x We get = 2 − at ⇒ F' = ma − mat dt 2 dt € 2. Glashow.C script available on request. I get 2.2e-31 cm^2 GeV for formula 6 + 3. p+gamma→π + X X can be a neutron for baryon number and charge conservation and because it is lighter than other baryons. In the rest frame of the proton (the target) we calculate the minimum photon energy of the photon projectile: (mn + mπ ) 2 − m 2p (939.57 + 139.57) 2 − 938.27 2 Eγ = = = 151.45MeV 2m p 2 × 938.27 → µ + + ν µ In the rest frame of the decaying pion: Energy conservation (and considering the neutrino as a massless particle so that E =| p |: 4. π € + ν ν 2 p µ + mµ2 + Eν mπ = and for momentum conservation: p = -‐p ν µ 2 (mπ − Eν ) 2 = p µ + mµ2 = Eν2 + mµ2 ⇒ mπ2 + Eν2 − 2 Eν mπ = Eν2 + mµ2 ⇒ Eν = mπ2 − mµ2 2mπ = 29.8MeV So the fraction of pion energy is 2 2 Eν mπ − mµ = = 0.21 mπ 2mπ2 This is approximately 1/4 of pion energy. 5. 4-momentum conservation: ( pγ )µ + ( pe )µ = ( p'e )µ ⇒ ( pγ )µ ( pe )µ = 0 € € E r in the frame of the electron initially at rest: ( pe ) µ = (me c,0), ( pγ ) µ = ( , pγ ) c r me E − 0⋅ pγ = 0 so energy-momentum conservation requires a photon with 0 energy, hence the process is impossible. € 6. From the Friedmann equation: For k = 1 � �2 Ṙ R Substituting: Differentiating: And the extreme of integration are: for = 8πGρ 3 − c2 r 2 R2 7. The reaction is (conservation of baryon number and charge): pp→π0pp In the rest frame of the proton: (2m p + mπ ) 2 − (2m p ) 2 (2 × 938.27 + 134.97) 2 − 4 × 938.27 2 Ep = = = 279.65MeV 2m p 2 × 938.27