## Assignment 5, solutions

Problem 1:

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 45o 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)).

Problem 2:

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. (a)  In K:  ρ = σδ(x), j = 0.  In K':  cρ' = γcρ,  ρ' = γσδ(x'), since x' = x. jx' = jy' = 0, jz' = -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)  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|. jy' = jz' = 0, jx' = -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 3:

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. In the rest frame K' of the electrons the force electron 1 exerts on electron 2 is F' = j kqe2/y'02 = j kqe2/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 pμ = (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)kqe2/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)kqe2/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.

Problem 4:

(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. dpx/dt = qE,  dpy/dt = dpz/dt = 0. px(t) = p0 + qEt, py(t) = pz(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*10-19 J/MeV. vx = v0 - qEt/m = 0  -->  t = v0m/(qE). x = ½ at2 = ½ mv02/(qE) = (5*10-17 J)/(1.6*10-19 C * 1000 N/C) = 0.3125 m. (b)  dpx/dt = qE,  dpy/dt = dpz/dt = 0. pz(t) = 0,  px(t) = qEt,  py(t) = p0. Energy Et = (m2c4 + p2c2)½ = (Et02 + (qcEt)2)½. Et = γmc2,  p = γmv,  v/c = pc/Et. vx/c = cqEt/(Et02 + (qcEt)2)½,  vy/c = cp0/(Et02 + (qcEt)2)½. dx/dt =  c2qEt/(Et02 + (qcEt)2)½,  x(t) = (qE)-1∫0qcEt t'dt'/(Et02 + t'2)½. x(t) = (qE)-1(Et02 + (qcEt)2)½ - Et0/(qE). dy/dt = c2p0/(Et02 + (qcEt)2)½ = c2p0Et0/(1 + (qcEt/E0t)2)½. y(t) = (c2p0/Et0) ∫0u dt'/(1 + t'2)½, with u = qcEt/E0t. y(t)  = (p0c/(qE)) sinh-1(qcEt/E0t). The work done on the electron by the field is W(t) =  (Et02 + (qcEt)2)½ - Et0. (c)  Express t in terms of y(t):  sinh(qEy/(p0c)) = qcEt/E0t. x(t) = Et0(qE)-1(1 + (qcEt/Et0)2)½ - Et0/(qE). x(y) = [Et0/(qE)][(1 + sinh2(qEy/(p0c)))½ - 1] = [Et0/(qE)] [(cosh(qEy/(p0c)) - 1].

Problem 5:

Determine the E and B fields of an electric dipole moving along the z-axis with a velocity v = v k relative to the laboratory using the Lorentz transformation of the fields.  The dipole moment in its rest frame is p = p0 k.

Solution:

 Concepts: The dipole field, the transformation of E and B from one Lorentz frame to another Reasoning: In the rest frame of the dipole, with the dipole at the origin, E(r,t) = (1/(4πε0)) (1/r3)[3(p∙r)r/r2 - p],  B(r,t) = 0.  p = p0 k. We have axial symmetry about the z-axis. Let is find the components of the electric field parallel and perpendicular to the z-direction. 3(p∙r)r/r2 - p = 3p0cosθsinθ (ρ/ρ) + 3p0cos2θ k - p0 k =  3p0cosθsinθ (ρ/ρ) + p0(3os2θ -1) k = 3p0(ρ/ρ) z/(ρ2 + z2) (ρ/ρ) + p0(2z2 - ρ2)/(ρ2 + z2) k. E(r,t) = (1/(4πε0)) (1/(ρ2 + z2)5/2)[3p0ρz (ρ/ρ) + p0(2z2 - ρ2) k]. Details of the calculation: The laboratory frame is moving with speed v = -v k with respect to the rest frame of the dipole. We find the fields in the laboratory frame using E'|| = E||,  B'|| = B||,  E'⊥ = γ(E + v×B)⊥,  B'⊥ = γ(B - (v/c2)×E)⊥. E'|| = (1/(4πε0)) (1/(ρ2 + z2)5/2)p0(2z2 - ρ2) k.  B'|| = 0. z = γ(z' - vt'),  ρ = ρ'. E'|| = (1/(4πε0)) (1/(γ2(z' - vt')2 + ρ'2)5/2)p0(2γ2(z' - vt')2 - ρ'2) k. E'⊥ = γE⊥ = (γ/(4πε0)) (1/(ρ2 + ρ'2)5/2)3p0ρz (ρ/ρ) = (γ2/(4πε0)) (1/(γ2(z' - vt')2 + ρ'2)5/2)3p0ρ'(z' - vt') (ρ/ρ). B'⊥ = γ(v/c2)×E = (v/c2)(γ2/(4πε0))(1/(γ2(z' - vt')2 + ρ'2)5/2)3p0ρ'(z' - vt') (φ/φ).