Physics 171.201

Midterm Exam 1

 

October 9^th, 2000

 

Answer all three problems. Be sure that you pace yourself so that you have enough time to work on each question. Partial credit will be given for partially correct answers. Be sure to show your working.

 

 

List of potentially useful formulae

 

 

 

        E/c

4-momentum =                  p_x

     p_y

     p_z

 

For a single photon, E²/c² = p² = p_x² + p_y² + p_z²

 

g = (1 – v²/c²)^–1/2

 

n’ = n (1 + v/c)^1/2 (1 – v/c)^–1/2

 

t = g (vx’/c2 + t’)       t’ = g (–vx/c2 + t)

x = g (x’ + vt’)              x’ = g (x – vt)

 

 

 

1. (30 points) Two relativistic rockets are approaching Earth from opposite directions at speed v = 0.6 c. One rocket emits a pulse of radio waves of frequency 100 MHz (in its own frame). What is the frequency of the radio waves as measured

 

a.    from the Earth?

 

n’ = n (1 + v/c)^1/2 (1 – v/c)^–1/2 = 100 MHz (1.6/0.4) ^1/2 = 200 MHz

 

b. from the other rocket?

 

         Apply transformation again

 

n’ = n (1 + v/c)^1/2 (1 – v/c)^–1/2 = 200 MHz (1.6/0.4) ^1/2 = 400 MHz

 

or use velocity addition law:

 

rel. velocity between rockets = (0.6 + 0.6) c / (1 + 0.6 x 0.6) = 0.8823 c and then apply transformation just once

 

n’ = n (1 + v/c)^1/2 (1 – v/c)^–1/2 = 100 MHz (1.8823/0.1176)^1/2 = 400 MHz

 

2. (30 points) An astronaut travels in a relativistic rocket to a planet 20 light years distant from Earth. (Assume the planet to be stationary with respect to Earth).

 

a. At what speed does the rocket need to travel if the astronaut’s trip is to take only 10 years as measured on her wristwatch?

 

Time in Earth frame, Dt = (20/b) yr, where b = v/c

Proper time = Dt/g = (20/bg) yr = 10 yr

è bg = 2 è b² = 4/g² = 4(1 – b²) è b = (4/5)^1/2 è v = (4/5)^1/2 c

 

b. In the astronaut’s reference frame, how far has she traveled in reaching the planet (answer in light years)?

 

The correct answer is zero, since the astronaut does not move relative to her own reference frame.

However, I had really intended to ask you the distance between Earth and the planet in the astronaut’s frame

 

Distance in astronaut’s frame = 10 (v/c) l.y. = 0.8944 l.y.

 

Or alternatively, distance = proper length / g = 20 l.y. / (1 – [4/5])^1/2

= 20 / (5^1/2 ) l.y. = 8.944 l.y.

 

Either answer gets full marks.

 

c. If the astronaut returns immediately to Earth at the same speed at which she came, how much will her friends and family on Earth have aged by the time she returns home?

 

Dt = 2 x (20/b) yr = 40/(4/5)^1/2 = 44.72 yr

 

or

 

Dt = 2 x 10g yr = 44.72 yr

 

3. (40 points) In the lab frame, a photon of energy E₁ travels along the x-axis and collides with a photon of energy E₂ that is traveling along the y-axis

(i.e in a perpendicular direction)

 

a. Write down the total momenergy 4-vector for the pair of photons as measured in the lab-frame?

 

(E/c,p_x,p_y,p_z) = (E₁/c, E₁/c, 0, 0) for photon 1

and (E₂/c, 0, E₂/c, 0) for photon 2

 

è total 4-momentum = ([E₁ + E₂] /c, E₁/c, E₂/c, 0)

 

 

b. Show that the total energy in the center-of-mass frame (i.e. the frame in which the total momentum is zero) is equal to (2E₁E₂)^1/2

 

In any frame, E² – p²c² = (E₁ + E₂)² – E₁² – E₁² = 2 E₁ E₂

è In the center of mass frame, p=0 è E_cm² = 2 E₁ E₂

è E_cm = (2E₁E₂)^1/2

 

 

 

c. If E₁ is 1 keV, what is the minimum energy E₂ needed to create an electron positron pair via the process g + g à e^– + e⁺

 

Required condition is E_cm =(2E₁E₂)^1/2 > 2m_ec²

 

è minimum E₂ = [2m_ec²] ²/ 2E₁ = 2 [511 keV]² / 1keV

= 522 MeV

 

(NB Rest-mass energy of the electron (or positron) = m_ec² = 511 keV)