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