Problem 1:

Submit this problem on Canvas as Assignment 9.  If you used an AI as a Socratic tutor, submit a copy of your session leading to your solution and reflect on your session.  If you did not need any help or worked with another student, explain your reasoning, do not just write down formulas. 

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.

Problem 2:

(a)  A fast electron (kinetic energy = 5*10^-17 Joule) enters a region of space containing a uniform electric field of magnitude E = -i 1000 V/m.  The field is parallel to the electron's motion and in a direction such as to decelerate it.  How far does the electron travel before it is brought to rest?  Neglect radiation losses.
(b)  Now assume that the initial velocity of the electron is v = v₀j, perpendicular to the direction of the electric field.  Assume that at t = 0 the electron moves through the origin.  Find the speed of the electron, its position, and the work done by the field on the electron as a function of time.
(c)  Find the trajectory of the electron, x(y).

Solution:

• Concepts:
  F = dp/dt, p = γmv.
• Reasoning:
  The electron is acted on by a constant force.  We neglect radiation losses.
• Details of the calculation:
  (a)  E = E i, q = -1.6*10^-19 C.
  dp_x/dt = qE,  dp_y/dt = dp_z/dt = 0.
  p_x(t) = p₀ + qEt, p_y(t) = p_z(t) = 0.
  Note:  both q and E are negative, so qE is positive.
  This part of the problem can be treated non-relativistically,
  since 5*10^-17 J << mc² = 0.511 MeV * 1.6*10^-13 J/MeV.
  v_x = v₀ - qEt/m = 0  -->  t = v₀m/(qE).
  x = v₀t - ½at² = ½mv₀²/(qE) = (5*10^-17 J)/(1.6*10^-19 C * 1000 N/C) = 0.3125 m.
  
  (b)  dp_x/dt = qE,  dp_y/dt = dp_z/dt = 0.
  p_z(t) = 0,  p_x(t) = qEt,  p_y(t) = p₀.
  Energy E_t(t) = (m²c⁴ + p²c²)^½ = (E_t0² + (qcEt)²)^½.
  E_t0² = m²c⁴ + p₀²c².
  Notation:  E_t denotes the total energy, E denotes the field strength.
  E_t = γmc²,  p = γmv,  v/c = pc/E_t.
  v_x(t)/c = cqEt/(E_t0² + (qcEt)²)^½,  v_y(t)/c = cp₀/(E_t0² + (qcEt)²)^½.
  dx/dt =  c²qEt/(E_t0² + (qcEt)²)^½,  x(t) = (qE)^-1∫₀^qcEt t'dt'/(E_t0² + t'²)^½.
  x(t) = (qE)^-1(E_t0² + (qcEt)²)^½ - E_t0/(qE).
  dy/dt = c²p₀/(E_t0² + (qcEt)²)^½ = c²p₀E_t0/(1 + (qcEt/E_t0)²)^½.
  y(t) = (c²p₀/E_t0) ∫₀^u dt'/(1 + t'²)^½, with u = qcEt/E_t0.
  y(t)  = (p₀c/(qE)) sinh^-1(qcEt/E_t0).
  The work done on the electron by the field is
  W(t) =  (E_t0² + (qcEt)²)^½ - E_t0.
  (c)  Express t in terms of y(t):  sinh(qEy/(p₀c)) = qcEt/E_t0.
  x(t) = E_t0(qE)^-1(1 + (qcEt/E_t0)²)^½ - E_t0/(qE).
  x(y) = [E_t0/(qE)][(1 + sinh²(qEy/(p₀c)))^½ - 1] = [E_t0/(qE)] [(cosh(qEy/(p₀c)) - 1].
  E_t0 =  mc² + T =  8.18*10^-14 J,  qE = 1.6*10^-16 N,  E_t0/(qE) = 511 m,
  p₀c = 2.86*10^-15 J, qE/(p₀c) = 0.056/m.
  x(y) = 511 m (cosh(y*0.056/m) - 1).
  
    

Problem 3:

In the lab frame K, in some volume of space, a constant electric field E and a constant magnetic field B are present. 
Can you find an inertial reference frame K' moving relative to frame K with velocity β = v/c , such that the field E and B are parallel to each other?  If yes, what is the velocity β of K' relative to K?

Solution:

• Concepts:
  Lorentz transformation of the electromagnetic fields:
• Reasoning:
  E'_|| = E_||,  B'_|| = B_||,
  E'_⊥ = γ(E + v×B)_⊥,
  B'_⊥ = γ(B - (v/c²)×E)_⊥.
  (E∙B)² and E² - c²B² are invariant under a Lorentz transformation.
• Details of the calculation:
  (E∙B)² is invariant under a Lorentz transformation.  If it is zero in K, then it must be zero in K'.  If E and B are perpendicular to each other in K,  then no reference frame K' can be found in which E' and B' are parallel to each other.
  
  Assume E∙B is nonzero.   Orient your coordinate system so that E points in the positive x-direction and B lies in the xy plane.
  Let K' move along the z-axis with velocity vk.
  E and B are perpendicular to v, so E'_|| = B'_|| = 0.
  E' = γ(E + v×B), B' = γ(B - (v/c²)×E).
  We want E'×B' = 0.
  There are various ways to evaluate the cross product.
  In terms of the Cartesian components we write
  (Ei + vB_xj - vB_yi )×(B_xi + B_yj - (v/c²)Ej) = 0.
  EB_y - (v/c²)E² - vB_x² - vB_y² + (v²/c²)EB_y = 0.
  (v²/c²) - (v/c)(cB² + E²/c) /(EB_y) + 1 = 0.
  v/c = β = (cB² + E²/c)/(2EB_y) - [(cB² + E²/c)²/(4E²B_y²) - 1]^½.
  The square root is real.  We need to choose the negative square root so that v/c < 1.
  
  [For any a, b we have (a - b)² > 0, or  a² + b² > 2ab.
  Therefore (cB² + E²/c)/(2EB) > 1.
  (cB² + E²/c)/(2EB_y) > (cB² + E²/c)/(2EB) > 1.}
  
  So in a reference frame moving with velocity
  v = ((c²B² + E²)/(2EB_y) - [(c²B² + E²)²/(4E²B_y²) - 1]^½) k
  with respect to K the fields E' and B' are parallel to each other.
  Note: EB_y =|E×B|.  This lets you write the v independent of the chosen coordinate system.
  Are there other frames K''?
  In any frame K'' that move with respect to K' parallel to E' and B', E'' and B'' are also parallel to each other.

Problem 4:

A reference frame K' is moving with uniform velocity v = vi with respect to reference frame K.
(a)  In K, a plane wave with angular frequency ω is traveling in the i direction.  What is its frequency in K'?
(b)  In K, a plane wave with angular frequency ω is traveling in the j direction.  What is its frequency in K'?
(c)  In K, a plane wave with angular frequency ω is traveling in a direction making an angle of 45^o with respect to the i direction and the j direction.  What is its frequency in K'?

Solution:

• Concepts:
  The relativistic Doppler shift
• Reasoning:
  Given the angular  frequency ω of a plane wave in reference frame K we are asked to find the angular frequency ω' in reference frame K' moving with constant velocity with respect to K.
• Details of the calculation:
  (a)  ω' = ω[(1 - v/c)/(1 + v/c)]^½ if k and v are parallel to each other.
  (b)  ω' = γω if k and v are perpendicular to each other.  γ = (1 - v²/c²)^-½.
  There is a transverse Doppler shift, even if θ = π/2.
  (c)  ω' = γ(ω - v·k) = γω(1 - vcosθ/c) = γω(1 - v/(√2c)).