More Problems

Problem 1:

A spaceship is approaching its home planet with velocity v = 0.8c i, when radio waves arriving from two stations on the planet signal that the stations are under attack at the same time (t = 0) recorded on the planet.  The captain of the ship knows that station 1 has planet coordinate x = 0, and station 2 has planet coordinate x = L = 10⁴ km.
What is the time difference between the two attacks in the ship's reference frame, and which station was attacked first.

Solution:

• Concepts:
  The Lorentz transformation, the simultaneity problem
• Reasoning:
  We are given the space-time coordinates of two events in one reference frame.  We are asked to transform to another reference frame.
• Details of the calculation:
  Let K be the planet's fame and K' be the ship's frame.
  To transform the coordinates of an event from K to K' 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

   .

  
  
  

  ct₁' = γct₁ - γβx₁ = 0.
  ct₂' = γct₂ - γβx₂ = -γβL.
  t₂' - t₁' = -44.4 ms.
  Station 2 is attacked first.

Problem 2:

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.

Problem 3:

Find the threshold energy for the process γ + p --> π + p in which a single π-meson, or pion, is produced when an energetic photon (or gamma ray) strikes a proton at rest.  The threshold energy is the minimum energy of the photon.  The rest energy of a π⁰ is 135 MeV, and that of the proton is 938 MeV.

Solution:

• Concepts:
  Relativistic collisions, energy and momentum conservation, frame transformations
• Reasoning:
  In relativistic collisions between free particles energy and momentum are always conserved.  Often the physics is best visualized in the center of momentum frame.
• Details of the calculation:
  The photon has minimum energy when the reaction products are at rest in the CM frame.  In the lab frame the proton and pion move together as one particle with mass M = m_π0 + m_p.
  In the CM frame after the collision we have for the "length"² of the momentum 4-vector
  P₀² - P² = P₀²  = (∑p₀)² = (Mc)².
  The "length"² of the total momentum 4-vector is (Mc)² before and after the collision in the lab frame.
  Before the collision we have
  P₀² - P² = (hf/c + m_pc)² - (hf/c)² = m_p²c² + 2hfm_p = (Mc)².
  hf = m_π0(m_π0 + 2m_p)c²/(2m_p)  = 144.7 MeV is the minimum photon energy.
  
  or, working in the lab frame:
  
  momentum conservation:  p_before = hf/c = p_after.
  energy conservation:  E_before = m_pc² + hf =  E_after = ((m_π0 + 2m_p)²c⁴ + h²f²)^½.
  --> hf = m_π0(m_π0 + 2m_p)c²/(2m_p).