Problem 1:

The 21 cm line (1420.4 MHz) line of neutral hydrogen atoms from a gas cloud is monitored with a radio telescope on Earth.
(a)  If the distance galaxy is moving with speed 0.2c perpendicular to the line of sight, what will be the observed frequency?
(b)  If the distant galaxy is receding with speed 0.2c along the line of sight, what will be the observed frequency?

Solution:

• Concepts:
  The relativistic Doppler shift
• Reasoning:
  The frequency f of a sinusoidal electromagnetic wave with wave vector k (k = 2π/λ = 2πf/c) in a reference frame K is measured as f' in a reference frame K' moving with uniform velocity v with respect to K.
  f' = γf(1 - (v/c)cosθ), where θ is the angle between the directions of k and v.
• Details of the calculation:
  (a)  f' = γf since cosθ = 0.  γ = (1 - v²/c²)^-½ = 1.02, f' = 1449.7 MHz.
  (b)  cosθ = 1 if the galaxy is receding or cosθ = -1 if the galaxy is approaching.
  cosθ = 1:  f' = γf(1 - (v/c)) = f([1 - (v/c))/(1 + v/c)]^½ = 946.9 MHz.
  cosθ = -1:  f' = γf(1 + (v/c)) = f([1 + (v/c))/(1 - v/c)]^½ = 2130.6 MHz.

Problem 2:

Two spaceships, each measuring 100 m in length in its own rest frame, pass by each other traveling in opposite directions.  Instruments on spaceship 1 determine that the front end of spaceship 1 requires (5/3)*10^-7 s to traverse the full length of spaceship 2.  What is the relative speed of the two ships?

Solution:

• Concepts:
  Lorentz contraction
• Reasoning:
  To an observer in motion relative to an object, the dimensions of the object are contracted by a factor of 1/γ in the direction of motion.
• Details of the calculation:
  The length of ship 2 in the rest frame of ship 1 is 100 m/γ, with  γ = (1 - v²/c²)^-½, and v the relative speed of the two ships.
  In SI units:  100/(γv) = t.  ct = 100*(1 - v²/c²)^½/(v/c),  0.5 = (1 - v²/c²)^½/(v/c),  (v/c)² = 1/1.25 = 4/5.  v = 0.89 c.

Problem 3:

A beam of antiprotons is incident on a hydrogen target at rest.  In a proton-antiproton reaction, a lambda particle and an anti-sigma particle are produced.
p(bar) + p --> Λ + Σ⁰(bar).
(a)  What is the minimum kinetic energy of the antiproton to make this reaction possible.
(b)  What is the velocity of the reaction products right after the collision?
The rest masses are m_p(bar) = m_p = 938.27 MeV/c², m_Λ = 1115.68 MeV/c²,  m_Σ0(bar) = 1192.64 MeV/c²

Solution:

• Concepts:
  Relativistic collisions, energy and momentum conservation, frame transformations
  For each component p_μ of the 4-vector (p₀,p₁,p₂,p₃) we have
  ∑_{particles_in} p_μ = ∑_{particles_out} p_μ, or ∑_i (p_i)_μ = ∑_j (p_j)_μ,
  where i denotes the particles going into the collision and j denotes the particles emerging from the collision.
  For transformations between reference frames we have
  (P₀,P)∙(P₀,P) = (P₀',P')∙(P₀',P').
  Here  P₀ = ∑_particles p₀  and  P = ∑_particles p.
• Reasoning:
  In relativistic collisions between free particles energy and momentum are always conserved.  The antiproton has minimum energy when the reaction products are at rest in the CM frame.
• Details of the calculation:
  (a)  After the collision we have in the CM frame
  P₀² - P² = P₀² = (∑p₀)² = (m_Λc + m_Σ0c)².
  The square of the magnitude of the momentum 4-vector is (m_Λc + m_Σ0c)² before and after the collision in any frame.
  Before the collision we have in the lab frame
  P₀² - P² = (p_0p(bar) + p_0p)² - (p_pbar + p_p)² = p_0p(bar)² + p_0p² - p_0p(bar)² + 2p_0p(bar)p_0p.
  For any free particle we have p₀²- p² = m²c².  We therefore have:
  P₀² - P² = (m_Λc + m_Σ0c)² = 2m_p²c² + 2p_0p(bar)p_0p.
  p_0p(bar)p_0p  =  (m_Λc + m_Σ0c)²/2 - m_p²c².  p_0p(bar) = (m_Λc + m_Σ0c)²/(2m_pc) - m_pc.
  Inserting the numbers: p_0p(bar) = 1901 MeV/c.
  The minimum total energy of the antiproton to make this reaction possible is E = 1901 MeV, the minimum kinetic energy of the antiproton to make this reaction possible is T = E - m_pc² = 961.91 MeV.
  
  We can also use straight energy and momentum conservation.
  Momentum conservation:  p_before = p_after = p.
  Energy conservation:  (m_p²c⁴ + p²c²)^½ + m_pc² = ((m_Λ+ m_Σ0)²c⁴ + p²c²)^½.
  m_p²c⁴ + p²c² + m_p²c⁴ + 2E_p(bar)m_pc² = (m_Λ+ m_Σ0)²c⁴ + p²c².
  2E_p(bar)m_pc² = (m_Λ+ m_Σ0)²c⁴ - 2m_p²c⁴.
  E_{p(bar) =} (m_Λ+ m_Σ0)²c²/2m_p - m_pc².
  
  (b) The total energy before and after the collision is 1901 MeV + 938.27 MeV.
  The momentum before and after the collision is found from
  p²c² = E_p(bar)² - m_p²c⁴ = (1901 MeV)² - (938.27 MeV)².  pc = 1553.52 MeV
  The speed of the reaction products is v/c = pc/E = 1553.52/2839.45 = 0.58.

Problem 4:

In inertial frame O a rod of length L is oriented along the x-axis and moving with velocity u in the positive y direction.  This rod is then viewed from an inertial reference frame O' moving with velocity v in the positive x direction.
(a)  What is the velocity of the rod in O'?
(b)  What is the length of the rod in O'?
(c)  What angle does the rod make with respect to the x' axis?

Solution:

• Concepts:
  Relativistic kinematics
• Reasoning:
  We are asked to relate observations in an inertial frame O to observations in an inertial frame O'.
• Details of the calculation:
  (a)  The stick moves with velocity u j in O.  It is oriented parallel to the x-axis, so its length in O is L.
  O' moves with velocity v i with respect to O.  The velocity u' of the stick in O' can be found from the velocity addition formulas.
  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.

  Here v = v i,  u_|| = 0, u_⊥ = u j.
  Therefore v_x' = -v, v_y' = u/γ = (1 - v²/c²)^½u.

  We can also proceed by transforming the coordinates of events.  Let us look at the positions of the left and the right end of the stick at t = 0 and at t = t₁ in O and O'.
  To transform the coordinates of an event from O to O' we use the matrix equation below.

   

  	x0'
  	x'
  	y'
  	z'

    =  

   

  	γ	-γβ	0	0
  	-γβ	γ	0	0
  	0	0	1	0
  	0	0	0	1

    

  	x0
  	x
  	y
  	z

   .

  
  

   

   

  Event:	coordinates in O	coordinates in O'
  (1)  left edge at t = 0	(0, 0, 0, 0)	(0, 0, 0, 0)
  (2)  right edge at t = 0	(0, L, 0, 0)	(-γβL, γL, 0, 0)
  (3)  left edge at t = t	(ct, 0, ut, 0)	(γct, -γβct, ut, 0)
  (4)  right edge at t = t	(ct, L, ut, 0)	(γct - γβL, -γβct + γL, ut, 0)

  From events 1 and 3 we find u_x' = ∆x'/∆t' = -γβct/γt = -βc = -v.
  From events 1 and 3 we find u_y' = ∆y'/∆t' = u/γ.
  (b)  In O' at t' = 0 the left edge of the stick is at x' = 0, y' = 0.
  At t' = -γβL/c the right edge of the stick is at x' = γL, y' = 0.
  The stick travels with velocity u' = -v i + (u/γ) j.  Therefore  at t' = 0 the right edge of the stick is at
  x' = γL - vγβL/c,  y' = (u/γ)γβL/c, or x' = γL(1 - β²) = L/γ,  y' = (u/c)βL.
  The length of the stick in O' is (x'² + y'²)^½ = L(1 - β² + ((u/c)β)²)^½.
  (c)  tanα' = y'/x' = (u/c)βγ.

Problem 5:

An observer moves horizontally away from a flashlight with a speed 0.6c. 
(a)  The flashlight is pointed in the direction of the observer and emits light.  Prove that the speed of the light determined by the observer is exactly c. 
(b)  The flashlight is now turned perpendicular to the direction of motion for the observer and light is emitted.  Demonstrate that the observer again will deduce that the speed of the light as measured in her frame is c.
(c)  What angle will the light velocity vector make with the horizontal axis in the observer's frame for case b?

Solution:

• Concepts:
  Velocity addition
• Reasoning:
  A particle moves in K with velocity u = dr/dt.  K' moves with respect to K with velocity v.  The particle'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.
• Details of the calculation:
  (a)  Let the observer move towards the right, in the positive x-direction.
  K is the frame of the flashlight, K' the frame of the observer, u and v are parallel to each other.
  u = ci, is the velocity of the light in K, v = vi is the velocity of K' with respect to K.
  u'i = i(c - v)/(1- cv/c²) = ci is the velocity of the light in K'.
  (b)  Now v∙u = 0.  u'_|| = -v,  u_⊥ =  c/γ,  u'² = v² + c²(1 - v²/c²) = c²,  u' = c.
  (c)  |tanθ| = |u_⊥/ u'_|||  = c/(vγ) = 1.33.  θ = (180 - 53)^o.