Problem 1:
In reference frame K a long,
straight, neutral wire with a circular cross sectional area A =
πr02 lies centered on the x-axis and carries a current with uniform current
density j i. For r > r0, choose the scalar potential to
be Φ = 0 and the vector potential to be
A = -[μ0I/(2π)] ln(r)
i.
In a frame K' moving with velocity vi with respect to K,
find the ρ', j', Φ' and A' at the point
P on the z-axis a distance z = z' > r0 from the wire.
Solution:
- Concepts:
Lorentz transformation of 4-vector current, 4-vector potential and electric and
magnetic field
- Reasoning:
We are asked to Lorentz transform the 4-vector current density and the
4-vector potential..
- Details of the calculation:
In K we have j = ji inside the wire,
j = 0 outside the
wire, ρ = 0 everywhere.
At point P on the z-axis A = -[μ0I/(2π)] ln(z)
i.
In a frame K' moving with velocity vi with respect
to K we have
cρ' = γcρ - γβj = -γβj. jx' = γj inside the wire,
ρ' =
j'
= 0 outside the wire.
The wire is not neutral in this frame.
To find Φ'
and A' at point P, we can transform the 4-vector potential
Φ'/c = -γβAx =
[γ(v/c) μ0I/(2π)]ln(z'),
Ax' = γAx,
A' = -[μ0γI/(2π])ln(z') i at point P.
Problem 2:
(a) Write the relativistic equation of motion for a
particle of charge q and mass m in an electromagnetic field. Consider these
equations for the special case of motion in the x-direction only, in a Lorentz
frame that has a constant electric field E pointing in the positive
x-direction.
(b) Show that a particular solution of the equations of motion is given by
x = (mc2/qE) cosh(qEτ/(mc), t = (mc/qE) sinh(qEτ/(mc),
(c) Show explicitly that the
parameter τ used to describe the world-line of
the charge q is the proper time along this world-line by showing that c2dτ2
= c2dt2 - dx2.
Solution:
- Concepts:
Relativistic dynamics, F = dp/dt,
p = γmv,
the Lorentz force, proper time
- Reasoning:
A particular solution to F = dp/dt is given. Using the
appropriate expressions for p and τ, we can verify this solution.
- Details of the calculation:
(a) F = dp/dt = q(E +
v×B).
Here E = Ei, B = 0.
For the special case of motion in the x-direction only we have
dpx/dt = (dpx/dτ)(dτ/dt) = qE is the equation of motion.
(b) Let us verify that the given expressions satisfy the equation of
motion.
px = γm dx/dt = γm(dx/dτ)(dτ/dt) = m(dx/dτ), since dt = γdτ.
x = (mc2/qE) cosh(qEτ/(mc), dx/dt = (dx/dτ)/(dt/dτ).
dx/dτ = csinh(qEτ/(mc). px = mdx/dτ = mcsinh(qEτ/(mc).
dpx/dτ = qEcosh(qEτ/(mc).
t = (mc/qE) sinh(qEτ/(mc), dt/dτ = cosh(qEτ/(mc).
dpx/dt = qE.
(c)
dx = csinh(qEτ/(mc) dτ, dt = cosh(qEτ/(mc) dτ,
c2(dt)2 - dx2 = c2cosh2(qEτ/(mc)
dτ2 - c2sinh2(qEτ/(mc) dτ2
= c2dτ2.
Problem 3:
A positive point charge q is moving with constant
velocity v = 103 m/s k along the z axis.
(a) When the charge crosses the origin, find the magnetic field produce by
the point charge at any point (x, y, z) away from the origin as a function of the
Cartesian coordinates x, y, and z.
(b) Express this field on terms of the spherical coordinates (r, θ, φ).
Solution:
- Concepts:
The electromagnetic fields of a moving point charge, the
Lorentz transformation of the electromagnetic fields
- Reasoning:
In the rest frame of the point the electric field is the Coulomb field and
the magnetic field is zero. We can transform those field to a frame in
which the proton is moving with velocity vk.
- Details of the calculation:
(a)
Let K be the rest frame of the point charge. Let the origin of
the two frames coincide at t = 0.
In K,
E = (qe/(4πε0r2))(r/r),
B = 0.
Let K' be the frame in which the point charge is moving with velocity vk.
K' is moving with velocity -vk with respect to K.
In SI units the transformation of the electromagnetic fields to a frame K' moving with velocity
v with respect to the
frame K is given by:
E'|| = E||,
B'|| =
B||, E'⊥
=
γ(E + v×B)⊥,
B'⊥
=
γ(B - (v/c2)×E)⊥.
Here v << c, therefore we can set γ = 1.
Therefore we have
B'|| =
Bz = 0,
B'
=
-(v/c2)×E.
k×E = -Ey i + Ex j.
Ey = (qe/(4πε0))(y/(z2
+ x2 + y2)3/2.
Ex = (qe/(4πε0))(x/(z2
+ x2 + y2)3/2.
B'
=
-(v/c2)Ey i + (v/c2)Ex
j
= (v/c2)(qe/(4πε0))[-(y/(z2
+ x2 + y2)3/2) i + (x/(z2
+ x2 + y2)3/2) j]
Since v << c, (x' y', z') = (x, y, z).
(b) Transforming to spherical coordinates:
z2
+ x2 + y2 = r2.
B' = (v/c2)(qe/(4πε0r3))(-y
i + x j)
= (v/c2)(qesinθ/(4πε0r2))
eφ
= (μ0/(4πr2))qev sinθ eφ
= (μ0/(4πr2))qev×er.