An observer in the laboratory observes a beam of electrons
with charge density ρ moving at velocity **v** = v**k**. 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/(2ε_{0}), the direction of**E**is radial inward.

B = μ_{0}ρrv/2, the direction of**B**is the -φ direction.

The force on the electron has magnitude F = [q_{e}ρr/(2ε_{0})](1 - v^{2}/c^{2}), pointing radially outward.**In the rest frame of the electron**F' = q_{e}ρ'r'/(2ε_{0}), 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 - v^{2}/c^{2}) = ρ(1 - v^{2}/c^{2})^{½}.

F' = [q_{e}ρr/(2ε_{0})](1 - v^{2}/c^{2})^{½}, since r' = r.

or

**F**' = d**p**'_{⊥}/dt' = d**p**'_{⊥}/dτ in the rest frame of the electron

**F**' = d**p**_{⊥}/dτ = γd**p**_{⊥}/dt = γ**F**, since**p**is a 4-vector and**p**'_{⊥}=**p**_{⊥}.

An electric field **E** exerts a force
**F **= q**E
**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**= q**E**.

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**/c^{2})×**E**')_{⊥}.

In K the force on the charge is**F**= d**p**/dt = d**p**/dτ, where τ is the proper time, a Lorentz invariant.

In K' the force on the charge is**F**' = d**p**'/dt' = (1/γ)d**p**'/dt, γ**F**' = d**p**'/dτ.

Here γ = (1 - β^{2})^{-½},**β**=**v**/c.

From the transformation properties of the momentum 4-vector

p^{μ}= (γmc,γm**v**) = (E_{k}/c,**p**) = under a Lorentz transformation,

**p**'_{|| }= γ(**p**_{||}+ βE_{k}/c),**p**'_{⊥}=**p**_{⊥}, we have

γ**F**'_{⊥}= d**p**'_{⊥}/dτ = d**p**_{⊥}/dτ = d**p**_{⊥}/dt = q**E**_{⊥}= γq(**E**' +**v**×**B**')_{⊥}.

**F**'_{⊥}= q(**E**'_{⊥}+**v**×**B**').

(Note: we use E_{k}to distinguish energy from the magnitude of the electric field.)

γ**F**'_{||}= d**p**'_{||}/dτ = γd**p**_{||}/dτ = γd**p**_{||}/dt = γq**E**_{||}= γq**E**'_{||}._{}**F**'_{||}= q**E**'_{||}.

We have used that in K d(E_{k}/c)/dt = (1/c)dE_{k}/dt = 0 since in K the charge is at rest.

[E_{k}= mc^{2}/(1 - u^{2}/c^{2})^{½}, dE_{k}/dt = [m/(1 - u^{2}/c^{2})^{3/2}] udu/dt = 0 if u = 0.]

Therefore**F**' = q(**E**' +**v**×**B**'). This is the expression for the Lorentz force.

At t = 0 a particle with mass m and negative charge -q
leaves the origin with a relativistic velocity **v** = v_{0}**k**.
It moves in a region with **E** = E_{0}**k**,
**B** = 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**= d**p**/dt. Energy = (m^{2}c^{4}+ p^{2}c^{2})^{½}= γmc^{2},**p**= γm**v**. - Details of the calculation:

d**p**/dt = -q**E**, p_{z}(t) = p_{0}- qE_{0}t, p_{x}(t) = p_{y}(t) = 0.

At the highest point p_{z}(t) = 0, t = p_{0}/qE_{0}. The particle returns to the origin at t = 2p_{0}/qE_{0}.

The particle returns at t = 2γ_{0}mv_{0}/(qE_{0}) = 2(1 - v_{0}^{2}/c^{2})^{-½}mv_{0}/qE_{0}.

The maximum distance from the origin is z_{max}.

z_{max}= ∫_{0}^{p0/qE0}v dt. v/c = pc/E = (p_{0}- qE_{0}t)/(m^{2}c^{2}+ (p_{0}- qE_{0}t)^{2})^{½}.

z_{max}= (c/qE_{0})∫_{0}^{p0}y dy/(m^{2}c^{2}+ y^{2})^{½}

= (1/qE_{0})[(m^{2}c^{4}+ p_{0}^{2}c^{2})^{½}- mc^{2}]

= (γ_{0}- 1)mc^{2}/(qE_{0}) = mc^{2}/(qE_{0}) [(1 - v_{0}^{2}/c^{2})^{-½}- 1].

or,

U(z_{max}) = qE_{0}z_{max}. E = mc^{2}+ qE_{0}z_{max}= γ_{0}mc^{2}. (energy conservation)

z_{max}= (γ_{0}- 1)mc^{2}/(qE_{0}) = mc^{2}/(qE_{0}) [(1 - v_{0}^{2}/c^{2})^{-½}- 1].

Two electric charges, q_{1} and q_{2}, are moving with the
same velocity **V**. Let **R** be the vector pointing from q_{2} to q_{1},
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 q_{2} exerts on q_{1} 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**= q**E**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 R^{2}= (Δx)^{2}+ (Δy)^{2}, Δx = Rcosθ, Δy = Rsinθ.

The rest frame of the two charges moves with respect to the lab frame with velocity V**i**.

The lab frame moves with respect to the rest frame with velocity -V**i**.

In the rest frame of the charges R'^{2}= (Δx')^{2}+ (Δy')^{2}, Δx' = γRcosθ, Δy' = Rsinθ.

Here γ = (1 - β^{2})^{-½}, β = V/c.

R'^{2}= γ^{2}((Δx)^{2}+ (Δy)^{2}/γ^{2}) = γ^{2}R^{2}(cos^{2}θ + sin^{2}θ/γ^{2})

= γ^{2}R^{2}(1 - sin^{2}θ + (1 - β^{2})sin^{2}θ) = γ^{2}R^{2}(1 - β^{2}sin^{2}θ).

Method 1:

In the rest frame of the charges let**F**' denote the force q_{2}exerts on q_{1}.

F'_{x}= (kq_{1}q_{2}/R'^{2})(Δx'/R'), F'_{y}= (kq_{1}q_{2}/R'^{2})(Δy'/R').

**F**' = d**p**'/dt' = d**p**'/dτ, where τ is the proper time, a Lorentz invariant.

In the lab frame the force on q_{1}is**F**= d**p**/dt = (1/γ)d**p**/dτ, γ**F**= d**p**/dτ.

From the transformation properties of the momentum 4-vector p^{μ}= (γmc,γmv) = (E/c,p) under a Lorentz transformation,

**p**_{||}= γ(**p**'_{||}+**β**E'/c),**p**_{⊥}=**p**'_{⊥}, we have

γF_{y}= dp_{y}/dτ = dp'_{y}/dτ = dp'_{y}/dt' = (kq_{1}q_{2}/R'^{2})(Δy'/R'), F_{y}= (1/γ)(kq_{1}q_{2}/R'^{2})(Δy'/R').

F_{y}= (1/γ)(kq_{1}q_{2}Rsinθ/(γ^{2}R^{2}(1 - β^{2}sin^{2}θ))^{3/2}= (kq_{1}q_{2}sinθ/R^{2})(1 - β^{2})^{2}/(1 - β^{2}sin^{2}θ)^{3/2}.

γF_{x}= dp_{x}/dτ = γdp'_{x}/dτ = γdp'_{x}/dt' = γ(kq_{1}q_{2}/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.

F_{x}= (kq_{1}q_{2}/R'^{2})(Δx'/R') = (kq_{1}q_{2}cosθ/R^{2})(1 - β^{2})/(1 - β^{2}sin^{2}θ)^{3/2}.

Summary:

F_{y}= (kq_{1}q_{2}sinθ/R^{2})(1 - β^{2})^{2}/(1 - β^{2}sin^{2}θ)^{3/2},

F_{x}= (kq_{1}q_{2}cosθ/R^{2})(1 - β^{2})/(1 - β^{2}sin^{2}θ)^{3/2}.

Method 2:

In the rest frame of the charges let**E**' and**B**' denote the fields produced by q_{2}at the location of q_{1}.

E'_{x}= (kq_{2}/R'^{2})(Δx'/R'), E'_{y}= (kq_{2}/R'^{2})(Δy'/R').**B**' = 0.

Transformation properties of the fields:

E_{||}= E'_{||}, B_{||}= B'_{||}, E_{⊥}= γ(E' + v×B')_{⊥}, B_{⊥}= γ(B' - (v/c^{2})×E')_{⊥}.

Here E_{x}= E_{x'}= (kq_{2}/R'^{2})(Δx'/R') = (kq_{2}cosθ/R^{2})(1 - β^{2})/(1 - β^{2}sin^{2}θ)^{3/2}. B_{x}= 0.

E_{y}= γE'_{y}= γ(kq_{2}/R'^{2})(Δy'/R') = (kq_{2}sinθ/R^{2})(1 - β^{2})/(1 - β^{2}sin^{2}θ)^{3/2}.

B_{y}= 0, B_{z}= γ(β/c)(kq_{2}/R'^{2})(Δy'/R') = (β/c)(kq_{2}sinθ/R^{2})(1 - β^{2})/(1 - β^{2}sin^{2}θ)^{3/2 }F_{x}= q_{1}E_{x }= (kq_{1}q_{2}cosθ/R^{2})(1 - β^{2})/(1 - β^{2}sin^{2}θ)^{3/2}.

F_{y}= q_{1}E_{y }- VB_{z}= (kq_{1}q_{2}sinθ/R^{2})(1 - β^{2})/(1 - β^{2}sin^{2}θ)^{3/2}- β^{2}(kq_{2}sinθ/R^{2})(1 - β^{2})/(1 - β^{2}sin^{2}θ)^{3/2}.

F_{y}= (kq_{1}q_{2}sinθ/R^{2})(1 - β^{2})^{2}/(1 - β^{2}sin^{2}θ)^{3/2}.

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 Ω_{0}, 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**= B**k**is perpendicular to the disk over a small area A. Consider an observer passing through the region of the magnetic field with speed**v**= ΩR**i**. In this observer's rest frame K' there exists an electric field**E**'_{⊥}= γ(**E**+**v**×**B**)_{⊥}= -γvB**j**.

All velocities are non-relativistic, so γ ~ 1 and**E**'_{⊥}~ -vB**j**= -ΩRB**j**.

There also exist a magnetic field**B**'_{⊥}= γ**B**_{⊥ }~ B**k**.

In K' the current density is**j**_{c}' = σ**E**' = -σΩRB**j**and the Lorentz force on the small volume ΔV = Ad is' =

F**j**_{c}'×**B**' ΔV = -σΩRB^{2}Ad**i**. 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.

**τ**= -R**j**×**F**= -σΩR^{2}B^{2}Ad**k**.

The moment of inertia of the disk is I = ½MR^{2}= ½ρdπR^{4}.

The equation of motion is τ_{z}= IdΩ/dt or -σΩR^{2}B^{2}Ad = ½ρdπR^{4}dΩ/dt.

dΩ/dt = -Ω2σB^{2}A/(ρπR^{2}). Ω(t) = Ω_{0}exp(-2σB^{2}At/(ρπR^{2})).

The number of revolutions the disk makes before it comes to rest is

N = θ_{tot}/2π = (½π)∫_{0}^{∞}Ω(t)dt = (Ω_{0}/2π)∫_{0}^{∞}exp(-2σB^{2}At/(ρπR^{2}))dt = (Ω_{0}ρR^{2}/4σB^{2}A).

A line of charge with charge density
λ C/m is fixed at rest along the x' axis of a reference frame S'. A test charge q is at rest in S' at (0, 0, z' =
d). S' is in constant motion with velocity **v** = v**i** with respect to
a reference frame S.

(a) Calculate the electric field of the line of charge in the rest frame S' and
the force on q.

(b) Calculate the electric and magnetic fields of the line of charge measured
by an observer at rest in S.

(c) Calculate the force measured by the observer in S on the test charge q.