Physics 3401, Homework Set #4, Rohlf Ch 4#3,6,8,9,13,14,17,18,21,22,29,36

#3  Given time shift of aircraft, calculate speed.
 Time difference =   4.00E-09 seconds,    c= 3.00E+08
 Distance 3.00E+06 meters in earth frame
 a. Earth time interval is longer, stationary observer.
 b. Pilot sees 3000 km as length contracted.
 Both observers measure same value for v
 t(earth) = 3000/v
 t(pilot) = (3000)sqrt(1-v^2/c^2)/v
 3000/v(1 - sqrt(1-v^2/c^2)) = 4ns
 Use first two terms of binomial expansion:
  sqrt(1-a) approx= 1 - a/2

 3000000(v/2c^2) = 4 ns
 v = 239.7 m/s

6. Proper lifetime of particle
 Proper lifetime means lifetime in rest frame of particle
 v =.99c with respect to lab frame
 Lab frame distance =  0.001 m
 Particle frame distance =   0.000141
 Two frames agree on the 0.99c velocity
 Particle frame time = distance/velocity =   4.75 E-13 sec

8. Muon and pion
 a. Muon proper lifetime =  2.20E-06 seconds
 speed = .99c
 Distance before decay =   652.96 meters in muon frame
 The relativity factor is =  7.0888
 In the lab frame, this would be  4629 meters in lab frame.

 b. Charged pion lifetime =  2.60E-08 seconds
 speed = .99c
 relativity factor =  7.0888
 Lab frame distance =  54.7 meters
9. Twin paradox
 a. Earthbound twin determines time to be shorter in travelers frame.
 b. In the absence of acceleration, the traveler would measure the earthbound twin's time to be shorter.
 c. The acceleration is in the travelers frame, so the earth frame is considered to be the inertial frame.
 The traveling twin would be younger.

13. Momentum and speed of a 1 GeV proton.
 From the relativistic kinetic energy expression, the speed must be     0.875 c From the relativistic momentum expression, its momentum is then = 9.06E-19 kg m/s
At this speed, the non-relativistic momentum expression would give 4.38E-19 kg m/s
which would be in error by 51%

Another approach is to use pc =sqrt(E^2 - (m0c^2)^2) and express p in MeV/c
Must note that the E in this expression includes both KE and the rest mass energy.
pc=sqrt((1GeV +.938GeV)^2-(.938 GeV)^2)) = 1.696 GeV
The momentum can then be expressed as p=1.70 GeV/c

14.  Radius of curvature of 1 MeV electron in 1 Tesla magnetic field
 Force equation mv^2/r = evB, so r = mv/eB, but relativistic momentum must be used
 From the relativistic kinetic energy expression, the velocity must be=    0.941 c
 for the kinetic energy to be 1 MeV.
 From the relativistic momentum relationship, p =   7.59E-22 kg m/s
 e= 1.60E-19 C
 Resulting radius =  4.74E-03 m

17. Calculated mass of alpha particle
 mass of proton= 938.3 MeV in energy equivalents
 mass of neutron= 939.6 MeV
 Binding energy = 28.4 MeV
 mass of alpha= 3727.4 MeV
 mass of alpha= 6.64E-27 kg

18. Carbon-14 decay energy
 From Appendix K of Rohlf
 mass of carbon 13040.976 MeV
 mass of nitrogen 13040.309 MeV
 mass of electron 0.511 MeV
 energy release 0.156 MeV

21. Electron KE=mass energy, find photon energy for photon with same momentum
 Electron mass energy =  0.511 MeV
 KE of electron = 0.511 MeV requires speed of    0.866 c
 The momentum of this electron is then =    4.73E-22 kg m/s
 The momentum of a photon = E/c, so E=pc =   1.42E-13 joules
    8.85E+05 eV

22. Speed to red-shift to 1% of frequency
 To shift frequency to 1% of its source frequency requires a recession
 velocity of  0.9998 c

29. Z-zero decay into electron and positron
 Mass energy of Z-zero =  91.2 GeV
 Particle is at rest, so electron and positron must have equal but
 oppositely directed momenta.
 Energy of each is 45.6 GeV p=45.6 GeV/c
 For this extreme relativistic case, the mass energy can be neglected
 in calculating the momentum, and the photon expression p=E/c used.
 With this approximation, p=  152.1014009 eV s/m
 The required speed is so close to c that you can't take the square root
 to calculate c.
 The relativistic gamma factor is = 45,600MeV/.511MeV =
 gamma = 89236.8
 (use b and g for beta and gamma since Greek letters can't be depended upon.)
 We can assess the departure from c with the fractional departure from c, 1 -v/c
 The trick is to find an expression for 1-v/c = 1-b which we can evaluate.
 Since g^2 = 1/(1-b^2) = 1/((1-b)(1+b)), then we can write
 1-b = 1/((1+b)g^2) or approximately: 1-b = 1/2g^2
 This gives 1-b= 6.28E-11

36. Back-scattered 10GeV photon
 a) High energy collisions like this need to be viewed in the center-of-mass
 frame. The photon energy does not change in the collision as viewed in the
 COM frame. Transforming back to the laboratory frame, the electron has
 essentially all the energy. Take electron energy = 10 GeV, and the photon
 energy can be approximated by:

 E' = mc^2 E/(mc^2 + 2E) approx= mc^2/2 = .511 MeV/2 = .26 MeV

 b) Again the collision is examined in the COM frame. This time the photon has
 essentially all the energy. The photon energy is about 10 GeV; the electron
 energy is about mc^2

Relativistic projectile problem.
 Enterprise traveling at 0.8c as seen by external observer
 Projectile fired at 0.6c as seen by Enterprise
 Klingon vessel approaches at 0.7c in opposite direction as seen by external observer

 Projectile speed as seen by external observer =   0.9459 c
 Approach speed of Klingons as seen by Enterprise =   -0.9615 c
 Projectile speed as seen from Klingon ship =    0.9902 c