### The transformation of the force

#### Problem:

An observer in the laboratory observes a beam of electrons with charge density ρ moving at velocity v = vk.  If the beam has circular radius R find the repulsive force acting on an electron inside the beam (r < R) both with respect to the rest frame of the electrons and with respect to the observer in the laboratory frame, respectively.

Solution:

• Concepts:
The Lorentz force, the Lorentz transformation
• Reasoning:
Because of the cylindrical symmetry we can find the electric and magnetic fields from Gauss' law and Ampere's law respectively.  We can then find the force F = q(E + v×B) on an individual particle with charge q.
• Details of the calculation:
In the laboratory frame at r < R:  E = ρr/(2ε0), the direction of E is radial inward.
B = μ0ρrv/2, the direction of B is the -φ direction.
The force on the electron has magnitude F = [qeρr/(2ε0)](1 - v2/c2), pointing radially outward.

In the rest frame of the electron F' = qeρ'r'/(2ε0), radially outward.
The Lorentz transformation of the 4 vector current to a frame moving with β = (v/c)k with respect to the lab frame yields
cρ' = γ(cρ - βρv),  ρ' = ργ(1 - v2/c2)  = ρ(1 - v2/c2)½
F' = [qeρr/(2ε0)](1 - v2/c2)½, since r' = r.
or
F' = dp'/dt' = dp'/dτ in the rest frame of the electron
F' = dp/dτ = γdp/dt = γF,  since p is a 4-vector and p' = p.

#### Problem:

An electric field E exerts a force F = qE on a charge q at rest.  Show that the expression for the Lorentz force follows directly from the Lorentz transformation properties of the fields.

Solution:

• Concepts:
The Lorentz transformation of the electromagnetic fields, the Lorentz force
• Reasoning:
We are asked to derive the expression for the Lorentz force from the transformation properties of the fields.  The Lorentz force is the force exerted by electromagnetic fields on a charge moving with velocity v.
• Details of the calculation:
let K be the frame in which the charge is at rest.
In K we have external fields E and B and the force on the charge is F = qE.
In K' we observe an electric field E' and a magnetic field B'.
We have from the Lorentz transformation of the fields
E|| = E'||B|| = B'||,  E = γ(E' + v×B')B = γ(B' - (v/c2E').
In K the force on the charge is 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'/dt, γF' = dp'/dτ.
Here   γ = (1 - β2), β = v/c.
From the transformation properties of the momentum 4-vector
pμ = (γmc,γmv) = (Ek/c,p) = under a Lorentz transformation,
p'|| = γ(p|| + βEk/c), p' = p, we have
γF' = dp'/dτ  = dp/dτ  = dp/dt = qE = γq(E' + v×B').
F' = q(E' + v×B').
(Note: we use Ek to distinguish energy from the magnitude of the electric field.)
γF'|| = dp'||/dτ = γdp||/dτ = γdp||/dt = γqE|| = γqE'||.
F'|| = qE'||.
We have used that in K d(Ek/c)/dt  = (1/c)dEk/dt = 0 since in K the charge is at rest.
[Ek = mc2/(1 - u2/c2)½,  dEk/dt = [m/(1 - u2/c2)3/2] udu/dt = 0 if u = 0.]
Therefore F' = q(E' + v×B').  This is the expression for the Lorentz force.

#### Problem:

At t = 0 a particle with mass m and negative charge -q leaves the origin with a relativistic velocity v = v0k.  It moves in a region with  E = E0kB = 0.   When does it return to the origin?  What is its maximum distance from the origin?  Neglect radiation.

Solution:

• Concepts:
Relativistic dynamics
• Reasoning:
The particle may move with relativistic speed.
F = dp/dt.  Energy = (m2c4 + p2c2)½ = γmc2p = γmv.
• Details of the calculation:
dp/dt = -qE,  pz(t) = p0 - qE0t,  px(t) = py(t) = 0.
At the highest point pz(t) = 0, t = p0/qE0.  The particle returns to the origin at t = 2p0/qE0
The particle returns at t = 2γ0mv0/(qE0) = 2(1 - v02/c2)mv0/qE0.
The maximum distance from the origin is zmax.
zmax = ∫0p0/qE0 v dt.  v/c = pc/E = (p0 - qE0t)/(m2c2 + (p0 - qE0t)2)½ .
zmax = (c/qE0)∫0p0 y dy/(m2c2 + y2)½
= (1/qE0)[(m2c4 + p02c2)½ - mc2]
= (γ0 - 1)mc2/(qE0) = mc2/(qE0) [(1 - v02/c2) - 1].
or,
U(zmax) = qE0zmax.  E = mc2 + qE0zmax = γ0mc2.  (energy conservation)
zmax =  (γ0 - 1)mc2/(qE0) = mc2/(qE0) [(1 - v02/c2) - 1].

#### Problem:

Two electric charges, q1 and q2, are moving with the same velocity V.  Let R be the vector pointing from q2 to q1, and let θ be the angle between R and V.  The speed V is not small compared to the speed of light.  Find the force q2 exerts on q1 in this coordinate system (the lab frame).  Give all the components of the force vector.  Assume that the charges are moving in the xy-plane, and that the positive x-direction is the direction of motion.

Solution:

• Concepts:
The Lorentz transformation
• Reasoning:
We can calculate the force between the charges in the rest frame of the charges from F = qE and transform this force to the laboratory frame, or we can calculate the electric field of one of the charges in its rest frame, transform it into and electric and magnetic field in the laboratory frame, and calculate the force between the charges from F = q(E + v×B).
• Details of the calculation:
In the lab frame R2 = (Δx)2 + (Δy)2, Δx = Rcosθ, Δy = Rsinθ.

The rest frame of the two charges moves with respect to the lab frame with velocity Vi
The lab frame moves with respect to the rest frame with velocity -Vi
In the rest frame of the charges R'2 = (Δx')2 + (Δy')2, Δx' = γRcosθ, Δy' = Rsinθ.
Here   γ = (1 - β2), β = V/c.
R'2 = γ2((Δx)2 + (Δy)22) = γ2R2(cos2θ + sin2θ/γ2)
= γ2R2(1 - sin2θ + (1 - β2)sin2θ) = γ2R2(1 - β2sin2θ).
Method 1:
In the rest frame of the charges let F' denote the force q2 exerts on q1.
F'x = (kq1q2/R'2)(Δx'/R'),  F'y = (kq1q2/R'2)(Δy'/R').
F' = dp'/dt' = dp'/dτ, where τ is the proper time, a Lorentz invariant.
In the lab frame the force on q1 is F = dp/dt = (1/γ)dp/dτ, γF = dp/dτ.
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
γFy = dpy/dτ  = dp'y/dτ  = dp'y/dt' = (kq1q2/R'2)(Δy'/R'), Fy = (1/γ)(kq1q2/R'2)(Δy'/R').
Fy = (1/γ)(kq1q2Rsinθ/(γ2R2(1 - β2sin2θ))3/2 =  (kq1q2sinθ/R2)(1 - β2)2/(1 - β2sin2θ)3/2.
γFx = dpx/dτ  = γdp'x/dτ  = γdp'x/dt'  =  γ(kq1q2/R'2)(Δx'/R')
We have used that in K' d(E'/c)/dt'  = (1/c)dE'/dt' = 0 since K' is the rest frame of the charges.
Fx = (kq1q2/R'2)(Δx'/R') = (kq1q2cosθ/R2)(1 - β2)/(1 - β2sin2θ)3/2.
Summary:
Fy = (kq1q2sinθ/R2)(1 - β2)2/(1 - β2sin2θ)3/2,
Fx = (kq1q2cosθ/R2)(1 - β2)/(1 - β2sin2θ)3/2.

Method 2:
In the rest frame of the charges let E' and B' denote the fields produced by q2 at the location of q1.
E'x = (kq2/R'2)(Δx'/R'),  E'y = (kq2/R'2)(Δy'/R').  B' = 0.
Transformation properties of the fields:
E|| = E'||,  B|| = B'||,  E = γ(E' + v×B'),  B = γ(B' - (v/c2)×E').
Here Ex = Ex' = (kq2/R'2)(Δx'/R') = (kq2cosθ/R2)(1 - β2)/(1 - β2sin2θ)3/2.  Bx = 0.
Ey = γE'y = γ(kq2/R'2)(Δy'/R') =  (kq2sinθ/R2)(1 - β2)/(1 - β2sin2θ)3/2.
By = 0,  Bz = γ(β/c)(kq2/R'2)(Δy'/R') = (β/c)(kq2sinθ/R2)(1 - β2)/(1 - β2sin2θ)3/2
Fx = q1Ex = (kq1q2cosθ/R2)(1 - β2)/(1 - β2sin2θ)3/2.
Fy = q1Ey - VBz = (kq1q2sinθ/R2)(1 - β2)/(1 - β2sin2θ)3/2 - β2(kq2sinθ/R2)(1 - β2)/(1 - β2sin2θ)3/2.
Fy = (kq1q2sinθ/R2)(1 - β2)2/(1 - β2sin2θ)3/2.

#### Problem:

An aluminum disk of radius R, thickness d, conductivity σ and mass density ρ is mounted on a frictionless vertical axis.  It passes between the poles of a magnet near its rim, which produces a B field perpendicular to the plane of the disk over a small area A of the disk.  If the initial angular speed of the disk is Ω0, how many revolutions will it make before it comes to rest?

Solution:

• Concepts:
The Lorentz transformation of the electromagnetic fields, the Lorentz force
• Reasoning:
Even at non-relativistic speeds, problems involving Eddy currents are often easier analyzed after a frame transformation.
• Details of the calculation:

In the laboratory frame K, the magnetic field B = Bk is perpendicular to the disk over a small area A.  Consider an observer passing through the region of the magnetic field with speed v = ΩRi.  In this observer's rest frame K' there exists an electric field E' = γ(E + v×B) = -γvBj.
All velocities are non-relativistic, so  γ ~ 1 and E' ~ -vBj = -ΩRBj.
There also exist a magnetic field B' = γB ~ Bk.
In K' the current density is  jc' = σE' = -σΩRBj and the Lorentz force on the small volume ΔV = Ad is
F
' = jcB' ΔV = -σΩRB2Adi.  Because γ ~ 1, the force F observed in the laboratory frame is equal to F'.
In the laboratory frame F produces a torque about the axis of rotation.