Problem 1:

A rebellion ship moving at v = 0.999c flies past a stationary star destroyer.  The star destroyer fires a laser canon with a very short energy pulse of 105.3 μJ, such that it strikes the rebellion ship at the point of closest approach.   What is the energy of the laser pulse according to the rebellion ship?

Solution:

• Concepts:
  Doppler shift
• Reasoning:
  Choose the coordinate system so that the ship (frame K' is moving with uniform velocity v = vi with respect to the star destroyer (frame K) in the xy plane.
  In K, a laser pulse with frequency ω is traveling in the j direction. 
  The frequency ω' of the pulse in K' is ω' = γ(ω - v∙k) = γω.
• Details of the calculation:
  ω'/ω = γ.  The energy of each photon in the pulse is increased by a factor of (1 - v²/c²)^-½ = 22.37.
  The energy of the pulse in the ship's frame is 2355 μJ.

Problem 2:

At noon a rocket ship passes the Earth at speed β = 0.8.  Observers on the ship and on Earth agree that it is noon.  Answer the following questions, and draw complete space-time diagrams in both the Earth and rocket ship frames, showing all events and worldlines.

(a)  At 12:30 pm, as read by a rocket ship clock, the ship passes an interplanetary navigational station that is fixed relative to the Earth and whose clocks reads Earth time.  What time is it at the station?
(b)  How far from Earth, in Earth coordinates, is the station?
(c)  At 12:30 pm rocket time, the ship reports by radio back to Earth.  When does Earth receive this signal (in Earth time)?
(d)  The earth replies immediately. When does the rocket receive the response (in rocket time)?

Use a grid for the space-time diagrams. Let the spacing of the gridlines be (c*hour)/3 for the Earth frame diagram and (c*hour)/5 for the rocket ship diagram.

Solution:

• Concepts:
  The Lorentz transformation, space-time diagrams
• Reasoning:
  Define the following events:
  Event A:  Ship passes Earth.
  Event B:  Ship passes station and sends signal.
  Event C:  Earth receives signal and replies.
  Event D:  Ship receives reply.
• Details of the calculation:
  (a)  The two events A and B happen at the same location in the ship's frame, so the time between them (½ h) is the proper time interval.  We can find the interval in the Earth frame via the time dilation formula.
  γ = (1 - β²)^-½ = 5/3.
  τ = γτ₀ = (5/3)½ = 5/6 h or 50 minutes. Thus, when the ship passes the station, the station's clock reads 12:50.
  
  (b)  In the Earth frame, it takes 5/6 h for the ship, moving at speed v = 4c/5, to get to the station. Thus, the distance to the station is d = (4c/5)(5/6 h) = 2/3 c*h.
  
  (c)  The signal must travel a distance of 2/3 c*h in the Earth frame.  This takes the light 2/3 h or 40 minutes.  So the signal is received on Earth at 12:50 + 40 minutes or 13:30.
  
  (d)  In the 1.5 hours it takes for the earth to receive the signal, the rocket travels a distance (4c/5)(1.5 h) = 6/5 c*h (in the Earth frame). This is the separation between the Earth and rocket when the signal is relayed.  The relative velocity between the signal and the ship is c − v = c − 4c/5 = c/5, so the time it takes (in the Earth frame) for the light signal to catch up with the ship is ∆t = (6/5 c*h)/(c/5) = 6 hours.  If we add this to 13:30 (the time the Earth relays the signal), we see that the ship receives the signal at 19:30 Earth time.  The two events A and D happen at the same spatial location in the rocket ship's frame (at the ship itself) and thus the time interval between them is the proper interval.  Thus, we can use the time-dilation formula to convert the time interval in the Earth frame to the proper time interval in the ship's frame.  In the Earth frame, the time interval between events A and D is τ = 7.5 hours, so in the ship's frame, it is τ₀ = τ/γ = 7.5/(5/3) = 4.5 hours.  Adding this to noon, we see that the light signal is received at 16:30 in ship time.

The space-time diagrams in the Earth and rocket ship frames are shown above. Each gridline corresponds to (c*h)/3 in the Earth frame and (c*h)/5 in the rocket ship frame. One c*h is a unit of distance equal to the distance traveled by light in 1 hour. The four relevant events have been marked.

Problem 3:

An object moves in K with velocity u = dr/dt.  K' moves with respect to K with velocity v. 
Show that the object's velocity in K', u' = dr'/dt', is given by
u'_|| = (u_|| - v)/(1 - v∙u/c²),
u'_⊥ =  u_⊥/(γ(1 - v∙u/c²)),
where parallel and perpendicular refer to the direction of the relative velocity v.

Solution:

• Concepts:
  The Lorentz transformation
• Reasoning Choose your coordinate system so that v = vi, and u lies in the xy-plane.
  Using the Lorentz transformation we transform dx, dy, and dt from K to K'.
  dx' = -γβcdt + γdx,   dy' = dy,  cdt' = γcdt - γβdx.
  u'_|| = (dx'/dt)(dt/dt')=  (-γβc + γu_||)(dt/dt').
  u'_⊥ = (dy'/dt)(dt/dt') = u_⊥(dt/dt').
  dt'/dt = γ - γ(β/c)u_||= γ(1 - vu_||/c²) = γ(1 - v∙u/c²).
  u'_|| = (-γv + γdu_||)/(γ(1 - v∙u/c²)) = (u_|| - v)/(1 - v∙u/c²).
  u'_⊥ =  u_⊥/(γ(1 - v∙u/c²)).