Plane waves in conductors, dielectric-conductor boundaries

Problem:

A "tenuous" plasma consists of free electric charges of mass m and charge -q_e (where q_e is positive).  There are n charges per unit volume.  Assume that the density is uniform and that the interactions between the charges may be neglected.  Also assume that the charges can be treated classically.
A linearly-polarized electromagnetic wave of frequency ω is incident on the plasma.
Let the electric field component of the plane wave be E = E₀exp(i(kx - ωt)).
(a)  Solve the equation of motion for a single charge and find the current density j and the conductivity σ of the plasma as a function of ω.
(b)  Assume a plane wave of the form E = E₀exp(i(kx - ωt)) propagate in the plasma with conductivity σ.  Find the dispersion relation (the relation between k and ω) for the electromagnetic wave in the plasma and the index of refraction as a function of ω.

Solution:

• Concepts:
  Newton's 2nd law, charge and current density, plane waves in a conducting medium
• Reasoning:
  The "tenuous" plasma is a conducting medium.  We can find its conductivity by solving for the motion of the free charges using Newton's second law.
• Details of the calculation:
  (a)  Equation of motion for a single charge: md²r/dt² = -q_eE = -q_eE₀exp(i(kx - ωt)).
  r(t) = (q_eE₀/(mω²))exp(i(kx - ωt)).  v(t) = (q_eE₀/(imω))exp(i(kx - ωt)) = (q_e/(imω))E.
  j(t) = -nq_ev(t) = -(nq_e²/(imω))E(t).  j = σE.  σ = (inq_e²/(mω)).

  (b)  ∇×B = μ₀σE + μ₀ε₀∂E/∂t.
  ∇×(∇×B) = ∇(∇∙B) - ∇²B = μ₀σ(∇×E) + μ₀ε₀∂(∇×E)/∂t.
  ∇²B - μ₀σ∂B/∂t - μ₀ε₀∂²B/∂t² = 0.
  Similarly:  ∇²E - μ₀σ∂E/∂t - μ₀ε₀∂²E/∂t² = 0.
  Plane wave solutions of the form E = E₀exp(i(kx - ωt)) have to satisfy
  -k² + iωμ₀σ + ω²/c² = 0,  k² = ω²/c² - (μ₀nq_e²/m ) = (1/c²)(ω² - ω_p²),
  k = (1/c)(ω² - ω_p²)^½,
  where ω_p = μ₀nq_e²c²/m = nq_e²/(ε₀m).
  c/n = ω/k.  n = ck/ω = (1 - ω_p²/ω²)^½.  For ω_p > ω the index of refraction n is imaginary, the field decreases exponentially.

Problem:

In a medium with conductivity σ but no net charge, write down Maxwell's equations, and derive the wave equation for the electric field, E, in this medium.

Solution:

• Concepts:
  Maxwell's equations
• Reasoning:
  In regions with  ρ_f = 0  and  j_f = σ_cE  Maxwell's equations can be used to show that both E and B satisfy the damped wave equation.
• Details of the calculation:
  Assume ε = ε₀, μ = μ₀, ρ_f = 0 and j_f = σ_cE in the conductor.  Then Maxwell's equations become
  ∇∙E = 0,  ∇×E = -∂B/∂t,  ∇∙B = 0,  ∇×B = μ₀σE + μ₀ε₀∂E/∂t.
  ∇×(∇×E) = ∇(∇∙E) - ∇²E = -∂(∇×B)/∂t = -μ₀∂j/∂t - μ₀ε₀∂²E/∂t².
  ∇²E = μ₀σ_c∂E/∂t + μ₀ε₀∂²E/∂t², or
  ∇²E - μ₀σ_c∂E/∂t - μ₀ε₀∂²E/∂t² = 0.
  This is the damped wave equation for E.
  Similarly,
  ∇×(∇×B) = ∇(∇∙B) - ∇²B = μ₀σ(∇×E) + μ₀ε₀∂(∇×E)/∂t.
  ∇²B - μ₀σ∂B/∂t - μ₀ε₀∂²B/∂t² = 0.
  This is the damped wave equation for B.

Problem:

Starting from Maxwell's equations, prove that a plane EM wave propagates in a good conductor such that the electric and magnetic fields are out of phase by 45^o.

Solution:

• Concepts:
  Maxwell's equations
• Reasoning:
  In regions with  ρ_f = 0  and  j_f = σ_cE  Maxwell's equations can be used to show that both E and B satisfy the damped wave equation.
• Details of the calculation:
  Maxwell's equations:
  ∇∙E = ρ/ε₀,  ∇×E = -∂B/∂t,  ∇∙B = 0,  ∇×B = μ₀j + (1/c²)∂E/∂t.
  Assume ε = ε₀, μ = μ₀, ρ_f = 0 and j_f = σ_cE in the conductor.  Then
  ∇×B = μ₀σ_cE + μ₀ε₀∂E/∂t.
  ∇×(∇×B) = ∇(∇∙B) - ∇²B = μ₀σ_c(∇×E) + μ₀ε₀∂(∇×E)/∂t.
  ∇²B - μ₀σ_c∂B/∂t - μ₀ε₀∂²B/∂t² = 0.
  Similarly:  ∇²E - μ₀σ_c∂E/∂t - μ₀ε₀∂²E/∂t² = 0.
  Both E and B satisfy the damped wave equation.
  Try solutions of the form E(r,t) = E₀ exp(i(k∙r - ωt)).
  Then k² = iμ₀σ_cω + μ₀ε₀ω² = μ₀ε₀ω²(1 + iσ_c/(ε₀ω)).
  k² = μ₀ε₀ω²(1 + σ_c²/(ε₀ω)²)^½ e^iφ.  tanφ = σ_c/(ε₀ω).
  k = (μ₀ε₀)^½ω(1 + σ_c²/(ε₀ω)²)^¼ e^iφ/2 = |k|e^iφ/2.
  For a plane wave propagating into the z-direction we have
  E(r,t) = E₀ exp(ikz)exp(-iωt).
  ∇×E = ik×E = -∂B/∂t.
  B(r,t) = B(r) exp(-iωt).  Therefore  ik×E  = iωB.  ik(z/z)×E  = iωB.
  Choose the coordinate system so that E = E(x/x), then B(r) = (y/y)(k/ω)E₀exp(ikz).
  B(r) = (y/y)(|k|/ω)E₀e^ikze^iφ/2 = (y/y)B₀e^ikze^iφ/2,
  where  B₀ = (|k|/ω)E₀.  B₀ = (y/y)B₀.
  E(r,t) = E₀ exp(i(k∙r - ωt)).  B(r,t) = B₀ exp(i(k∙r - ωt + φ/2)).
  For a good conductor σ_c/(ε₀ω) >> 1.  tanφ --> ∞.  φ -- > 90^o.  φ/2 --> 45^o.
  [Example:  copper
  σ_c = 5.98*10⁷ (Ωm)^-1, ε₀ = 8.85*10^-12 C²/(Nm²), σ_c/(ε₀ω) = 6.75*10¹⁸ s^-1/ω.]

Problem:

An electromagnetic wave with frequency of 10⁶ Hz "travels" along the z-axis in an aluminum medium located at z ≥ 0.  The conductivity of aluminum is 38.2∙ 10⁶ (Ωm)^-1 and its relative permeability is k_m = 1.  Just inside the conductor at z = +0, the electric field amplitude is E₀ i. 

(a)  Write down an expression for the electric field inside the conductor.
(b)  Find the skin depth, wave velocity, and wavelength of the wave in aluminum.
(c)  Determine the corresponding magnetic field.
(d)  Find the phase difference between the electric and magnetic fields at each fixed location in aluminum.

Solution:

• Concepts:
  Maxwell's equations, plane waves in conductors.
• Reasoning:
  In regions with  ρ_f = 0  and  j_f = σ_cE  Maxwell's equations can be used to show that both E and B satisfy the damped wave equation.
• Details of the calculation:
  (a)  Maxwell's equations:
  ∇∙E = ρ/ε₀,  ∇×E = -∂B/∂t,  ∇∙B = 0,  ∇×B = μ₀j + (1/c²)∂E/∂t.
  Assume ε = ε₀, μ = μ₀, ρ_f = 0 and j_f = σ_cE in the conductor.  Then
  ∇×B = μ₀σ_cE + μ₀ε₀∂E/∂t.
  ∇×(∇×B) = ∇(∇∙B) - ∇²B = μ₀σ_c(∇×E) + μ₀ε₀∂(∇×E)/∂t.
  ∇²B - μ₀σ_c∂B/∂t - μ₀ε₀∂²B/∂t² = 0.
  Similarly:  ∇²E - μ₀σ_c∂E/∂t - μ₀ε₀∂²E/∂t² = 0.
  Both E and B satisfy the damped wave equation.
  Plane wave solutions are of the form E(r,t) = E₀ exp(i(k∙r - ωt)).
  k² = iμ₀σ_cω + μ₀ε₀ω² = μ₀ε₀ω²(1 + iσ_c/(ε₀ω)) = μ₀ε₀ω²(1 + σ_c²/(ε₀ω)²)^½ e^iφ.  tanφ = σ_c/(ε₀ω).
  k is a complex number, k = β + iα/2.
  k = (μ₀ε₀)^½ω(1 + σ_c²/(ε₀ω)²)^¼ e^iφ/2 = |k|e^iφ/2.
  To find the real and imaginary parts we have to find cos(φ/2) and sin(φ/2) in terms of tanφ.
  cos(φ/2) = ((1 + cosφ)/2)^½.  sin(φ/2) = ((1 - cosφ)/2)^½.  cosφ = (1 + tan²φ)^-½ in the first quadrant.
  Therefore cos(φ/2) = 2^-½(1 + (1 + (σ_c/(ε₀ω))²)^-½)^½. 
  sin(φ/2) = 2^-½(1 - (1 + (σ_c/(ε₀ω))²)^-½)^½.
  k = (μ₀ε₀/2)^½ω[(1 + (σ_c/(ε₀ω))²)^½ + 1]^½ + i(μ₀ε₀/2)^½ω[(1 + (σ_c/(ε₀ω))²)^-½ - 1]^½.
  β = (μ₀ε₀/2)^½ω[(1 + (σ_c/(ε₀ω))²)^½ + 1]^½.  α/2 = (μ₀ε₀/2)^½ω[(1 + (σ_c/(ε₀ω))²)^½ - 1]^½.
  (b)  The skin depth δ is the distance it takes to reduce the amplitude by a factor of 1/e.
  δ = 2/α.  σ_c/(ε₀ω) =  (38.2*10⁶ /Ωm)/(8.85*10^-12 C²/(Nm²))/(2π*10⁶/s) = 6.87*10¹¹.
  σ_c/(ε₀ω) >> 1, α/2 ≈ (μ₀ε₀/2)^½ω(σ_c/(ε₀ω))^½ = (μ₀σ_cω/2)^½,
  δ = (2/(μ₀σ_cω))^½ = (2/(4π*10^-7)/(38.2*10⁶)/(2π*10⁶))^½m = 8.14*10^-5m.
  β ≈ α/2 ≈ (μ₀σ_cω/2)^½ = 1/δ.
  Wave velocity = ω/β = ω*δ = (2π*10⁶)(8.14*10^-5) m/s = 5.1*10² m/s.
  Wavelength = 2π/β = 2π*δ = 5.1*10^-4 m.
  All numbers are for aluminum.
  (c)  Assume the plane wave is propagating into the z-direction.
  E(r,t) = E₀ exp(ikz)exp(-iωt).
  ∇×E = ik×E = -∂B/∂t.
  B(r,t) = B(r) exp(-iωt).  Therefore  ik×E  = iωB.  ik(z/z)×E  = iωB.
  Choose the coordinate system so that E = E(x/x), then B(r) = (y/y)(k/ω)ε₀exp(ikz).
  B(r) = (y/y)(|k|/ω)E₀e^ikze^iφ/2 = (y/y)B₀e^ikze^iφ/2,
  where  B₀ = (|k|/ω)E₀.  |k| = (μ₀ε₀)^½ω(1 + σ_c²/(ε₀ω)²)^¼ ≈ (μ₀σ_cω)^½ = 2^½/δ.
  B₀ = (y/y)B₀.
  E(r,t) = E₀ exp(i(kz - ωt)) = E₀ e^-αz/2exp(i(βz - ωt)).
  B(r,t) = B₀ exp(i(kz - ωt + φ/2)) = B₀ e^-αz/2exp(i(βz - ωt + φ/2)).
  (d)  For aluminum σ_c/(ε₀ω) >> 1.  tanφ --> ∞.  φ -- > 90^o.  φ/2 --> 45^o.
  B₀ e^-αz/2exp(i(βz - (ωt - π/4))).
  "B lags behind E" by φ/2 = π/4.

Problem:

An electric field E = i E₀exp(-iωt)is applied at the interface of a vacuum (z < 0) and a conductor (z > 0) of real average conductivity σ_c.  Assume ε = ε₀, μ = μ₀, in the conductor.
(a)  For σ_c >> ε₀ω, calculate how deeply the electric field penetrates into the conductor, i.e. calculate the depth at which the electric field amplitude has decreased to 1/e of its value at the surface.
(b)  Calculate dW/dt =  j∙EdV, the rate at which work is done by the field on the charges in a volume dV, as a function of z.

Solution:

• Concepts:
  Maxwell's equations
• Reasoning:
  In regions with ρ_f = 0 and j_f = σ_cE  Maxwell's equations can be used to show that both E and B satisfy the damped wave equation.
• Details of the calculation:
  Assume j_f = σ_cE in the conductor.  Then
  ∇×B = μ₀σ_cE + μ₀ε₀∂E/∂t.
  ∇×(∇×E) = ∇(∇∙E) - ∇²E = -∂(∇×B)/∂t. = -μ₀σ_c∂E/∂t - μ₀ε₀∂²E/∂t.
  Therefore  ∇²E - μ₀σ_c∂E/∂t - μ₀ε₀∂²E/∂t² = 0.
  The tangential component of E is continuous across the boundary, therefore E(z = 0) = i E₀exp(-iωt) just inside the conductor.
  ∂²E_x/∂z² + iμ₀ωσ_cE_x + μ₀ε₀ω²E_x = 0 inside the conductor.
  Since σ_c >> ε₀ω, we can write ∂²E_x/∂z² + iμ₀ω∂E_x = 0.
  Try a solutions of the form E_x(z,t) = E₀ exp(i(kz - ωt)).
  Then k² = iμ₀ωσ_c, k = (μ₀ωσ_c)^½exp(iπ/4) = 2^-½(1 + i)(μ₀ωσ_c)^½.
  Im(k) = Re(k) = (μ₀ωσ_c/2)^½.
  E_x(z,t) = E₀ exp(-Im(k)z) exp(i(Re(k)z - ωt)).
  Skin depth:  E₀ exp(-Im(k)d) = 1/e.  d = (2/(μ₀ωσ_c))^½.
  (b)  dW/dt =  j∙EdV = rate at which work is done by the field on the charges in dV.
  When evaluating products, use the real parts of the complex expressions.
  j∙E = σ_cE₀² exp(-2Im(k)z) cos²(Re(k)z - ωt).  <j∙E> = σ_c½E₀² exp(-2Im(k)z).
  dW/dt = σ_cE₀² exp(-2Im(k)z) cos²(Re(k)z - ωt)dV.
  <dW/dt> = σ_c½E₀² exp(-2Im(k)z)dV.