Solution:

• Concepts:
  Lorentz transformation of 4-vector current, 4-vector potential and electric and magnetic field
• Reasoning:
  We are asked to Lorentz transform the 4-vector current density and the 4-vector potential..
• Details of the calculation:
  In K we have j = ji inside the wire, j = 0 outside the wire,  ρ = 0 everywhere.
  At point P on the z-axis A = -[μ₀I/(2π)] ln(z) i. 
  In a frame K' moving with velocity vi with respect to K we have
  cρ' = γcρ - γβj = -γβj.  j_x' = γj inside the wire,
  ρ' = j' = 0 outside the wire.
  The wire is not neutral in this frame.
  To find Φ' and A' at point P, we can transform the 4-vector potential
  Φ'/c  = -γβA_x = [γ(v/c) μ₀I/(2π)]ln(z'), A_x' = γA_x, A' = -[μ₀γI/(2π])ln(z') i at point P.

Problem 2:

Calculate the force, as observed in the laboratory, between two electrons moving side by side along parallel paths 1 mm apart, if they have a kinetic energy of 1 eV and 1 MeV.

Solution:

• Concepts:
  The Lorentz transformation
• Reasoning:
  We can calculate the force between the electrons in the rest frame of the electrons from F = qE and transform this force to the laboratory frame, or we can calculate the electric field of one of the electrons in its rest frame, transform it into and electric and magnetic field in the laboratory frame, and calculate the force between the electrons from F = q(E + v×B).
• Details of the calculation:
  Assume the electrons move with velocity vi in the x-direction in the laboratory frame K.  Electron 1 has coordinates y = z = 0 and electron 2 has coordinates y = y₀ = 1 mm, z = 0.
  
  (i)  Transform the force.
  In the rest frame K' of the electrons the force electron 1 exerts on electron 2 is
  F' = j kq_e²/y'₀² = j kq_e²/y₀².
  F' = j 9*10⁹*(1.6*10^-19)²/10^-6 N = j 2.3*10^-22 N.
  F' = dp'/dt' = dp'/dτ, where τ is the proper time, a Lorentz invariant.
  In K the force on the charge is F = dp/dt = (1/γ)dp/dτ, γF = dp/dτ.
  Here   γ = (1 - β²)^-½, β = vi/c.
  From the transformation properties of the momentum 4-vector
  p^μ = (γmc,γmv) = (E/c,p) = under a Lorentz transformation,
  p_||= γ(p'_|| + βE'/c), p_⊥ = p'_⊥, we have
  γF_⊥ = dp_⊥/dτ  = dp'_⊥/dτ  = dp'_⊥/dt' = j kqe²/y₀², F_⊥ = j (1/γ)kq_e²/y₀².
  γF_|| = dp_||/dτ  = γdp'_||/dτ  = γdp'_||/dt'  = 0.
  We have used that in K' d(E'/c)/dt'  = (1/c)dE'/dt' = 0 since K' is the rest frame of the charges.
  The force between the electrons in the lab frame is F_⊥ = j (1/γ)kq_e²/y₀².
  1 eV = ½mv², γ = 1, F = j  2.3*10^-22 N.
  1 MeV = (γ - 1)mc², (γ - 1) = 1.96, γ = 2.96, F = j  7.8*10^-23 N.
  
  (ii)  Transform the fields.
  In K' the electric field produced by electron 1 at the position of electron 2 is E'_⊥ = -j kq_e/y'₀²,  E'_||= 0.
  In K' the magnetic field produced by electron 1 is B = 0.
  Frame K moves with velocity v'= -v i with respect to K'.
  Therefore at the position of electron 2 E_||= B_||= 0.
  E_⊥ = γ(E' + v'×B')_⊥ = γE'_⊥= -j γkq_e/y'₀².
  B_⊥ = γ(B' - (v'/c²)×E')_⊥ = -k (v/c²)γkq_e/y'₀².
  F = -q_e(E + v'×B) = j (γkq_e²/y'₀² - (v/c²)γkq_e²/y'₀²)
  = j γ(1 - v/c²)kq_e²/y'₀² = j (1/γ)kq_e²/y₀².