Problem 1:

Muons decay with a half-life of t_½ = 1.56 μs.  They are produced with very high speeds in the upper atmosphere when cosmic rays collide with air molecules.  Assume the height of the atmosphere of Earth is 200 km in the reference frame of the earth.
(a)  If the muons are traveling at 0.99998c, how long does it take for the muons to pass through Earth's atmosphere as viewed from the Earth?
(b)  How many half-lives is this?
(c)  How long does it take as viewed from the rest frame of the muons?
(d)  What is the height of the atmosphere in the rest frame of the muons?

Solution:

• Concepts:
  Relativistic kinematics, the proper length, the proper time
• Reasoning:
  Proper time τ = t/γ,  proper length L₀ = γL,  γ = (1 - v²/c²)^-½.
• Details of the calculation:
  (a)  t_earth = (2*10⁵ m)/ 0.99998c = 6.67*10^-4 s.
  (b)  t_earth/t_½ = 427.
  (c)  t_muon = τ = t_earth/γ, γ = (1 - v²/c²)^-½ = 158.  t_muon = 4.22*10^-6 s.
  (d)  L = (200/158) km = 1.26 km.

Problem 2: 

Consider two particles with mass m.  The first particle, with energy E, collides elastically with the second particle, which is at rest.
(a)  If both scatter at an angle θ with respect to the incoming particle's momentum, write down the value of θ in terms of E and m.
(b)  What value does θ approach in the relativistic limit?
(c)  What value does θ approach in the non-relativistic limit?

Solution:

• Concepts:
  Energy and momentum conservation in relativistic collisions
• Reasoning:
  In relativistic collisions between free particles energy and momentum are always conserved.
• Details of the calculation:
  (a)  Since both particles have the same mass m and the component of the total momentum perpendicular to the incoming particle's momentum is zero, both particles must have the same energy E' = (E + mc²)/2 after the collision.
  For each particle we have tanθ = p'_⊥ / p'_||.
  Energy conservation:
  E'² = (E² + m²c⁴ + 2Emc²)/4 = p'_⊥²c²  +  p'_||²c² + m²c⁴.
  Momentum conservation:
  p'_||²c² = (E² - m²c⁴)/4 .
  Combining:
  p'_⊥²c² = E²/4 + m²c⁴/4 + Emc²/2 - E²/4 + m²c⁴/4 - m²c⁴ = Emc²/2 - m²c⁴/2 = (E - mc²)mc²/2.
  tan²θ = 2(E - mc²)mc²/(E² - m²c⁴) = 2mc²/(E + mc²).
  
  (b)  In the relativistic limit, mc² becomes negligible compared to E. 
  Then tan²θ ~ 2mc²/ E --> 0.
  θ approaches 0^o in the relativistic limit.
  
  (c)  E = T + mc². 
  tan²θ = 2mc²/(T + 2mc²) = 1/(1 + T/2mc²) ~ 1
  if T << mc² (nonrelativistic limit).
  θ approaches 45^o in the non-relativistic limit.

Problem 3:

A horizontal one-meter-long stick is moving with velocity u j along the y−direction in the lab frame S.  Reference frame S' is moving with velocity v = v i along the x-direction relative to S.

(a)  Using the Lorentz transformation, derive the velocity u' of the stick measured by an observer at rest in S' in terms of u, v, and c.
(b)  In the S frame the stick is oriented horizontally.  Assume that the stick is crossing the x-axis at t = 0, with its left and right ends at x_L = −0.5 m, y_L = 0 and x_R = +0.5 m, y_R = 0 respectively.
The rod is not horizontal for observers in the S' frame.  Find the vertical distance that the rod travels while crossing the x'-axis according to an observer at rest in S', in terms of u, v and c. 
(c)  Find the length of the stick and the angle θ that the stick makes with the x'-axis in terms of u, v, and c according to the observer in the S' frame.

Solution:

• Concepts:
  Relativistic kinematics
• Reasoning:
  We are asked to relate observations in an inertial frame S to observations in an inertial frame S'.
• Details of the calculation:
  (a)  The stick moves with velocity u j in S.  The velocity of the stick in S' can be found from the velocity addition formulas.
  S' is moving with velocity v i with respect to S.   A particle moves in S with velocity u = dr/dt. The particle's velocity in S', 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.
  Here v = v i,  u_|| = 0, u_⊥ = uj.
  Therefore u_x' = -v,  u_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 S and S'.
  To transform the coordinates of an event from S to S' b 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 S	coordinates in S'
  (1)  left edge at t = 0	(0, -0.5, 0, 0)	(0.5γβ, -0.5γ , 0, 0)
  (2)  right edge at t = 0	(0, 0.5, 0, 0)	(-0.5γβ, 0.5γ, 0, 0)
  (3)  left edge at t = t	(ct, -0.5, ut, 0)	(γct+0.5γβ, -γβct-0.5γ, ut, 0)
  (4)  right edge at t = t	(ct, 0.5, ut, 0)	(γct-0.5γβ, -γβct+0.5γ, ut, 0)

  (All quantities are in SI units.)

  From events 1 and 3 we find u_x' = Δx'/Δt' = -γβct/γt = -βc = -v.
  From events 1 and 3 we find u_y' = Δy'/Δt' = ut/γt = u/γ.
  
  (b)  In S' at t' = 0.5γβ/c the left edge of the stick is at x' = -0.5γ, y' = 0.
  At t' = -0.5γβ/c the right edge of the stick is at x' = 0.5γ, y' = 0.
  The stick travels with velocity u' = -vi + (u/γ)j. 
  It takes the stick ∆t' = γβ/c to cross the x-axis.  In this time interval it travels a distance u'∆t' = uβ/c meter in the y'-direction.
  
  (c)  At t' = 0.5γβ/c the right edge of the stick is at x' = 0.5γ - vγβ/c, y' = uβ/c.
  For the length of the stick in x' we have
  L_x' = x_R' - x_L' = γ(1 - β²) m = 1/γ m, L_y' = y_R' - y_L' = uβ/c m.
  (Note:  we have to measure the position of both ends at the same time in S'.)
  The length of the stick in S' is (x'² + y'²)^½ = (1 - β² + (uβ/c)²)^½ m.
  The angle θ that the stick makes with the x'-axis is found from tanθ' = L_y'/L_x' = γuβ/c.

Problem 4:

A relativistic particle is launched at the origin (0,0) with initial momentum p(0) = (p_x(0), p_y(0)), with p_x(0) > 0 and p_y(0) > 0, and is subject to a constant force pointing in the negative y direction.
(a)  Solve the equations of motion for x(t) and y(t).
(b)  Determine the time T at which the particle reaches the x-axis again (i.e. y(T) = 0).
(c)  Find the trajectory of the particle, i.e. y = y(x).

NOTE: Give all answers for the laboratory frame.

Problem 5:

Consider an electron propagating in empty space with speed v, and hence with energy E_e = γmc².  Show that the electron cannot emit a photon.

Solution:

• Concepts:
  Energy and momentum conservation
• Reasoning:
  In the inertial frame that is the initial rest frame of the electron, conservation of momentum requires that the electron recoils when emitting a photon.  Conservation of energy requires mc² = γmc² + hf =  mc² + T + hf, which is impossible because all quantities are positive.