A reference frame K' is moving with uniform velocity **v **= v**i**
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 - v^{2}/c^{2})^{-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)).

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) = (x_{0}, 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**= v**k**or**v**= v**i**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**/c^{2})×**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.

j_{x}' = j_{y}' = 0, j_{z}' = -γβσcδ(x').

In K:**E**_{||}=**B**= 0,**E**_{⊥}= σ/(2ε_{0})**i**for x > 0,**E**_{⊥}= -σ/(2ε_{0})**i**for x < 0.

In K':**E**'_{||}=**B**'_{||}= 0,**E**'_{⊥}= γ**E**_{⊥},**E**'_{⊥}= γσ/(2ε_{0})**i**for x' > 0,**E**'_{⊥}= -γσ/(2ε_{0})**i**for x' < 0.

**B**'_{⊥}= γ(**β**/c)×**E**).**B**'_{⊥}= -γβσ/(2cε_{0})**j**for x' > 0,**B**'_{⊥}= γβσ/(2cε_{0})**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|.

j_{y}' = j_{z}' = 0, j_{x}' = -βσcδ(x' + vt').

In K:**E**_{⊥}=**B**= 0,**E**_{||}= σ/(2ε_{0})**i**for x > 0,**E**_{||}= -σ/(2ε_{0})**i**for x < 0.

In K':**E**_{||}= σ/(2ε_{0})**i**for x' > -vt',**E**_{||}= -σ/(2ε_{0})**i**for x' < -vt',**E**'_{⊥}=**B**' = 0.

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**= q**E**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 v**i**in the x-direction in the laboratory frame K. Electron 1 has coordinates y = z = 0 and electron 2 has coordinates y = y_{0}= 1 mm, z = 0.

In the rest frame K' of the electrons the force electron 1 exerts on electron 2 is' =

F**j**kq_{e}^{2}/y'_{0}^{2}=**j**kq_{e}^{2}/y_{0}^{2}.

**F**' =**j**9*10^{9}*(1.6*10^{-19})^{2}/10^{-6}N =**j**2.3*10^{-22}N.

**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**/dτ, γ**F**= d**p**/dτ.

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

From the transformation properties of the momentum 4-vector

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

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

γ**F**_{⊥}= d**p**_{⊥}/dτ = d**p**'_{⊥}/dτ = d**p**'_{⊥}/dt' =**j**kqe^{2}/y_{0}^{2},**F**_{⊥}=**j**(1/γ)kq_{e}^{2}/y_{0}^{2}.

γ**F**_{||}= d**p**_{||}/dτ = γd**p**'_{||}/dτ = γd**p**'_{||}/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)kq_{e}^{2}/y_{0}^{2}.

1 eV = ½mv^{2}, γ = 1,**F**=**j**2.3*10^{-22}N.

1 MeV = (γ - 1)mc^{2}, (γ - 1) = 1.96, γ = 2.96,**F**=**j**7.8*10^{-23}N.

(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** = v_{0}**j**,
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**= d**p**/dt,**p**= γm**v**. - 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) = p_{0}+ 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 << mc^{2}= 0.511 MeV * 1.6*10^{-13}J/MeV.

v_{x}= v_{0}- qEt/m = 0 --> t = v_{0}m/(qE).

x = ½at^{2}= ½mv_{0}^{2}/(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) = p_{0}.

Energy E_{t}= (m^{2}c^{4}+ p^{2}c^{2})^{½}= (E_{t0}^{2}+ (qcEt)^{2})^{½}.

Notation: E_{t}denotes the energy, E denotes the field strength.

E_{t}= γmc^{2},**p**= γm**v**,**v**/c =**p**c/E_{t}.

v_{x}/c = cqEt/(E_{t0}^{2}+ (qcEt)^{2})^{½}, v_{y}/c = cp_{0}/(E_{t0}^{2}+ (qcEt)^{2})^{½}.

dx/dt = c^{2}qEt/(E_{t0}^{2}+ (qcEt)^{2})^{½}, x(t) = (qE)^{-1}∫_{0}^{qcEt}t'dt'/(E_{t0}^{2}+ t'^{2})^{½}.

x(t) = (qE)^{-1}(E_{t0}^{2}+ (qcEt)^{2})^{½}- E_{t0}/(qE).

dy/dt = c^{2}p_{0}/(E_{t0}^{2}+ (qcEt)^{2})^{½}= c^{2}p_{0}E_{t0}/(1 + (qcEt/E_{t0})^{2})^{½}.

y(t) = (c^{2}p_{0}/E_{t0})_{ }∫_{0}^{u}dt'/(1 + t'^{2})^{½}, with u = qcEt/E_{t0}.

y(t) = (p_{0}c/(qE)) sinh^{-1}(qcEt/E_{t0}).

The work done on the electron by the field is

W(t) = (E_{t0}^{2}+ (qcEt)^{2})^{½}- E_{t0}.

(c) Express t in terms of y(t): sinh(qEy/(p_{0}c)) = qcEt/E_{t0}.

x(t) = E_{t0}(qE)^{-1}(1 + (qcEt/E_{t0})^{2})^{½}- E_{t0}/(qE).

x(y) = [E_{t0}/(qE)][(1 + sinh^{2}(qEy/(p_{0}c)))^{½}- 1] = [E_{t0}/(qE)] [(cosh(qEy/(p_{0}c)) - 1].

Starting with the transformation of the electromagnetic fields under a
Lorentz transformation show that

(a) if **E** is normal to **B** in an inertial frame, it must be true in
all other inertial frames, and

(b) if |E| > c|B| in an inertial frame, it must be true in all other inertial
frames.

Solution:

- Concepts:

Vector products - Reasoning:

Lorentz transformation of the electromagnetic fields:

**E**'_{||}=**E**_{||},**B**'_{||}=**B**_{||},

**E**'_{⊥}= γ(**E**+**v**×**B**)_{⊥},

**B**'_{⊥}= γ(**B**- (**v**/c^{2})×**E**)_{⊥}.

We can show that (**E**∙**B**)^{2}and**E**^{2 }- c^{2}**B**^{2}are invariant under a Lorentz transformation. - Details of the calculation:

(a) We can choose a Cartesian coordinate system and evaluate the products using Cartesian coordinate.

Let the direction of the relative velocity of K' with respect to K be the z-direction.

Then**E**_{||}= E_{z}**k**and**E**_{⊥}= E_{y}**i**+ E_{y}**j**,**B**_{||}= B_{z}**k**and**B**_{⊥}= B_{y}**i**+ B_{y}**j**,

**E**∙**B**= E_{x}B_{x}+ E_{y}B_{y}+ E_{z}B_{z}.

**E**'∙**B**' =**E**'_{||}∙**B**'_{||}+**E**'_{⊥}∙**B**'_{⊥}.**E**'_{||}∙**B**'_{||}= E'_{z}B'_{z}= E_{z}B_{z}.

**E**'_{⊥}= E'_{y}**i**+ E'_{y}**j**.**B**'_{⊥}= B'_{y}**i**+ B'_{y}**j**.

E'_{x}= γ(E_{x}- vB_{y}), E'_{y}= γ(E_{y}+ vB_{x}), B'_{x}= γ(B_{x}+ (v/c^{2})E_{y}), B'_{y}= γ(B_{y}- (v/c^{2})E_{x}).

E'_{x}B'_{x}= γ^{2}(E_{x}B_{x}+ (v/c^{2})E_{x}E_{y}- vB_{y}B_{x}- (v^{2}/c^{2})B_{y}E_{y}).

E'_{y}B'_{y}= γ^{2}(E_{y}B_{y}- (v/c^{2})E_{y}E_{x}+ vB_{x}B_{y}- (v^{2}/c^{2})B_{x}E_{x}).

E'_{x}B'_{x}+ E'_{y}B'_{y}= γ^{2}(1 - v^{2}/c^{2})( E_{x}B_{x}+ E_{y}B_{y}) = E_{x}B_{x}+ E_{y}B_{y}.

(**E**∙**B**)^{2}is invariant under a Lorentz transformation.

If**E**∙**B**is zero in one frame, it is zero in every frame.

(b)**E**^{2 }- c^{2}**B**^{2}= E_{x}E_{x}+ E_{y}E_{y}+ E_{z}E_{z}- c^{2}(B_{x}B_{x}+ B_{y}B_{y}+ B_{z}B_{z}).

E'_{z}E'_{z}- c^{2}B'_{z}B'_{z}= E_{z}E_{z}- c^{2}B_{z}B_{z}.

E'_{x}E'_{x}= γ^{2}(E_{x}^{2}- 2vE_{x}B_{y}+ v^{2}B_{y}^{2}).

E'_{y}E'_{y}= γ^{2}(E_{y}^{2}+ 2vE_{y}B_{x}+ v^{2}B_{x}^{2}).

B'_{x}B'_{x}= γ^{2}(B_{x}^{2}+ 2(v/c^{2})E_{y}B_{x}+ (v^{2}/c^{4})E_{y}^{2}).

B'_{y}B'_{y}= γ^{2}(B_{y}^{2}- 2(v/c^{2})E_{x}B_{y}+ (v^{2}/c^{4})E_{x}^{2}).

E'_{x}E'_{x}+ E_{y}'E'_{y}- c^{2}(B'_{x}B'_{x}+ B'_{y}B'_{y})

= γ^{2}(E_{x}^{2}+ v^{2}B_{y}^{2}+ E_{y}^{2}+ v^{2}B_{x}^{2}- c^{2}B_{x}^{2}- (v^{2}/c^{2})E_{y}^{2}- c^{2}B_{y}^{2}- (v^{2}/c^{2})E_{x}^{2})

= γ^{2}(1 - v^{2}/c^{2})(E_{x}^{2}+ E_{y}^{2}) - γ^{2}(c^{2}- v^{2})(B_{x}^{2}+ B_{y}^{2}) = E_{x}E_{x}+ E_{y}E_{y}- c^{2}(B_{x}B_{x}+ B_{y}B_{y}).

**E**^{2 }- c^{2}**B**^{2}is invariant under a Lorentz transformation.

If |E| > c|B| in one inertial frame then |E| > c|B in every inertial frame.