**Problem 1:**

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)). |

**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) = (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 = 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/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'_{⊥}
= γE_{⊥} 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. |

**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 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'
= 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 - β ^{2})^{-½}, β
= vi/c.From the transformation properties of the momentum 4-vector p ^{μ}
= (γmc,γmv)
= (E/c,p) = under a Lorentz transformation,
p _{||}= γ(p'_{||} + βE'/c),
p_{⊥}
= p'_{⊥}, we haveγ F_{⊥}
= dp_{⊥}/dτ
= dp'_{⊥}/dτ
= dp'_{⊥}/dt'
= j kqe^{2}/y_{0}^{2},
F_{⊥}
= j (1/γ)kq_{e}^{2}/y_{0}^{2}.γ F_{||}
= 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)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. |

**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** = 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 = 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.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^{-19} J/MeV.v _{x} = v_{0} - qEt/m = 0 --> t = v_{0}m/(qE).x = ½ at ^{2} = ½ mv_{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}.E _{t} = γmc^{2}, p = γmv, v/c
= pc/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_{0t})^{2})^{½}.y(t) = (c ^{2}p_{0}/E_{t0})_{ }∫_{0}^{u}
dt'/(1 + t'^{2})^{½}, with u = qcEt/E_{0t}.y(t) = (p _{0}c/(qE)) sinh^{-1}(qcEt/E_{0t}).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_{0t}.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]. |

**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** = p_{0} **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/(4πε_{0})) (1/r^{3})[3(p∙r)r/r^{2} - p],
B(r,t) = 0.p = p_{0} 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/r^{2} - p = 3p_{0}cosθsinθ
(ρ/ρ) + 3p_{0}cos^{2}θ k - p_{0} k =
3p_{0}cosθsinθ (ρ/ρ) + p_{0}(3os^{2}θ -1) k= 3p_{0}(ρ/ρ) z/(ρ^{2} + z^{2}) (ρ/ρ) + p_{0}(2z^{2}
- ρ^{2)}/(ρ^{2} + z^{2}) k.E(r,t) = (1/(4πε_{0})) (1/(ρ^{2}
+ z^{2})^{5/2})[3p_{0}ρz (ρ/ρ) + p_{0}(2z^{2}
- ρ^{2}) 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/c^{2})×E)_{⊥}.E'_{||} = (1/(4πε_{0})) (1/(ρ^{2} + z^{2})^{5/2})p_{0}(2z^{2}
- ρ^{2)} k. B'_{||} = 0.z = γ(z' - vt'), ρ = ρ'. E'_{||} = (1/(4πε_{0})) (1/(γ^{2}(z' - vt')^{2}
+ ρ'^{2})^{5/2})p_{0}(2γ^{2}(z' - vt')^{2}
- ρ'^{2)} k.E'_{⊥} = γE_{⊥} = (γ/(4πε_{0})) (1/(ρ^{2}
+ ρ'^{2})^{5/2})3p_{0}ρz (ρ/ρ)= (γ ^{2}/(4πε_{0})) (1/(γ^{2}(z' - vt')^{2}
+ ρ'^{2})^{5/2})3p_{0}ρ'(z' - vt') (ρ/ρ).B'_{⊥} = γ(v/c^{2})×E = (v/c^{2})(γ^{2}/(4πε_{0}))(1/(γ^{2}(z'
- vt')^{2} + ρ'^{2})^{5/2})3p_{0}ρ'(z' - vt')
(φ/φ). |