Problem 1:

A dipole p = pk is fixed at the origin in reference frame K.  Reference fame K' moves with velocity -vk with respect to K.  At t = 0, the coordinate axes of the two frames are parallel and the origins coincide.
What are the electric and magnetic fields at a point (x', y' = z' = 0) in frame K'?

Solution:

• Concepts:
  Lorentz transformation of the electromagnetic fields
• Reasoning:
  In K the dipole is at rest.  E(r,t) = (1/(4πε₀)) (1/r³)[3(p∙r)r/r² - p],  B = 0 everywhere.
  3(p∙r)r/r² - p = 3pcosθsinθ (ρ/ρ) + 3pcos²θ k - p k.
  In the y = 0 plane E(r,t) = (1/(4πε₀))(x² + z²)^-3/2[3pz(xi + zk)/(x² + z²) - pk].
  E_|| = (1/(4πε₀))(x² + z²)^-3/2[3pz²)/(x² + z²) - p]
  = 1/(4πε₀))(x² + z²)^5/2[2pz² - px²].
  E_⊥ = (1/(4πε₀))(x² + z²)^-3/23pzxi /(x² + z²)
  = (1/(4πε₀))(x² + z²)^-5/23pzxi.
  In K' we have E'_|| = E_||,  B'_|| = B_||,  E'_⊥ = γ(E + v×B)_⊥,  B'_⊥ = γ(B - (v/c²)×E)_⊥.
• Details of the calculation:
  E'_|| = E_||,  B'_|| = 0,
  E'_⊥ = γE_⊥, B'_⊥ = γ(v/c²)×E.
  
  Transformation of the coordinates:
  We want to express x and z in terms of x' and z'.
  K moves with velocity vk with respect to K'
  x = x',  z = -γ(v/c)ct' + γz'.  ct= γct' - γ(v/c)z,
  At z' = 0 --> z = -γvt'.
  
  E'_|| = (1/(4πε₀))(x'² + (γvt')²)^-5/2[2p(γvt')² - px'²]k.
  E'_⊥ = -(γ²/(4πε₀))(x'² + (γvt')²)^-5/23pvt'x'i.
  E = (1/(4πε₀))(x'² + (γvt')²)^-5/2[2p(γvt')² - x'²)k - 3γ²pvt'x'i ].
  B = γ(v/c²)×E = γ(v/c²)k×E.
  = -(v/c²)γ³/(4πε₀))(x'² + (γvt')²)^-5/23pvt'x'j.

Problem 2:

(a)  Write the relativistic equation of motion for a particle of charge q and mass m in an electromagnetic field.  Consider these equations for the special case of motion in the x-direction only, in a Lorentz frame that has a constant electric field E pointing in the positive x-direction.
(b)  Show that a particular solution of the equations of motion is given by
x = (mc²/qE) cosh(qEτ/(mc),  t = (mc/qE) sinh(qEτ/(mc),
(c)  Show explicitly that the parameter τ used to describe the world-line of the charge q is the proper time along this world-line by showing that c²dτ² =  c²dt² - dx².

Solution:

• Concepts:
  Relativistic dynamics, F = dp/dt,  p = γmv, the Lorentz force, proper time
• Reasoning:
  A particular solution to F = dp/dt is given.  Using the appropriate expressions for p and τ, we can verify this solution. 
• Details of the calculation:
  (a)  F = dp/dt =  q(E + v×B).  Here E = Ei, B = 0.
  For the special case of motion in the x-direction only we have
  dp_x/dt = (dp_x/dτ)(dτ/dt) = qE is the equation of motion.
  
  (b)  Let us verify that the given expressions satisfy the equation of motion.
  p_x = γm dx/dt = γm(dx/dτ)(dτ/dt) = m(dx/dτ), since dt = γdτ.
  x = (mc²/qE) cosh(qEτ/(mc),  dx/dt = (dx/dτ)/(dt/dτ).
  dx/dτ = csinh(qEτ/(mc). p_x = mdx/dτ = mcsinh(qEτ/(mc).
  dp_x/dτ = qEcosh(qEτ/(mc). 
  t = (mc/qE) sinh(qEτ/(mc),  dt/dτ = cosh(qEτ/(mc).
  dp_x/dt  = qE.
  
  (c)  dx = csinh(qEτ/(mc) dτ,  dt = cosh(qEτ/(mc) dτ,
  c²(dt)² - dx² = c²cosh²(qEτ/(mc) dτ² -  c²sinh²(qEτ/(mc) dτ² = c²dτ².

Problem 3:

Consider an infinite sheet of charge with uniform charge density ρ = σδ(x) in the y-z plane. 
(a)  An observer moves on a trajectory r(t) = (x₀, 0, vt).  Calculate the 4-vector current density and electromagnetic fields E and B in the rest frame of this observer.
(b)  Calculate the 4-vector current density and electromagnetic fields E and B in the rest frame of an observer moving along the x-axis in the positive x-direction with speed v with respect to the sheet of charge.

Solution:

• Concepts:
  Lorentz transformation of the electromagnetic fields and the 4-vector current  (cρ,j)
• Reasoning:
  We are asked to transform the electromagnetic fields from the laboratory frame K to a frame moving with uniform velocity v = vk or v = vi with respect to K.
• Details of the calculation:
  The 4-vectors (cρ,j) transform as cρ' = γ(cρ - β∙j), j'_|| =  γ(j_|| - βcρ), j'_⊥ = j_⊥.
  For the electromagnetic fields we have in SI units
  E'_|| = E_||,  B'_|| = B_||,
  E'_⊥ = γ(E + v×B)_⊥, B'_⊥ = γ(B - (v/c²)×E)_⊥.
  Here || and ⊥ refer to the direction of the relative velocity.
  
  Let K be the lab frame and K' be the rest frame of the observer.  K' moves with velocity vk with respect to K.
  (a)  In K:  ρ = σδ(x), j = 0. 
  In K':  cρ' = γcρ,  ρ' = γσδ(x'), since x' = x.
  j_x' = j_y' = 0, j_z' = -γβσcδ(x').
  
  In K: E_|| = B = 0, E_⊥ = σ/(2ε₀) i for x > 0,  E_⊥ = -σ/(2ε₀) i for x < 0.
  In K':  E'_|| = B'_|| = 0, E'_⊥ = γE_⊥,  E'_⊥ = γσ/(2ε₀) i for x' > 0,  E'_⊥ = -γσ/(2ε₀) i for x' < 0.
  B'_⊥ = γ(β/c)×E).  B'_⊥ = -γβσ/(2cε₀) j for x' > 0,  B'_⊥ = γβσ/(2cε₀) j for x' < 0.
  
  (b)  Now K' moves with velocity vi with respect to K.
  In K:  ρ = σδ(x), j = 0. 
  In K':  cρ' = γcρ,  ρ' = σδ(x' + vt'), since x  = γ(x '+ vt') and δ(ax) = δ(x)/|a|.
  j_y' = j_z' = 0, j_x' = -βσcδ(x' + vt').
  
  In K:  E_⊥ = B = 0, E_|| = σ/(2ε₀) i for x > 0,  E_|| = -σ/(2ε₀) i for x < 0.
  In K':  E_|| = σ/(2ε₀) i for x' > -vt',  E_|| = -σ/(2ε₀) i for x' < -vt',  E'_⊥ = B' = 0.