Problems

Problem 1:

A neutral conducting sphere is at rest with its center at the origin in reference frame K.  A uniform magnetic field B = B0k is present.  Reference frame K’ moves with uniform velocity v = vi with respect to K.  Find E an B inside the sphere as observed in K’.

Solution:

Concepts, principles, relations that apply to the problem:
Lorentz transformation of the electromagnetic fields
Why do they apply?
We are asked to transform the electromagnetic fields from the laboratory frame K to a frame moving with uniform velocity v = vi with respect to K.
How do they apply?
In SI units he transformation of the electromagnetic fields from the laboratory frame to a frame moving with velocity v with respect to the laboratory frame is given by:
E'|| = E||,  B'|| = B||,  E'^ = g(E + v´B)^,  B'^ = g(B - (v/c2)´E)^.
In K we have E = 0, B = B0k.
In K' we have Ex = 0,  Bx = 0, E'^ = gv´B = -vgB0j,  B'^ = gB = gB0k.
Details of the calculation:
Is there a force on the electrons on the conductor?
F' = -qe(E' + v'´B'), where v' = -vi and qe = 1.6*10-19C.
F' = -qe(-vgB0j - (vi)´(gB0k)) = 0.
The total force on the electrons inside the conductor is zero.

Problem 2:

In a certain reference frame a static uniform electric field E0 is at an angle q0 ¹ p/2 to a static, uniform, magnetic induction B0, (B0 = E0/c), in the SI system of units.
(a)  Determine the magnitude of E and B in a reference frame in which the angle between E and B is q.
(b)  Is there a reference frame for which E and B are perpendicular to each other?  If so, what is it?

Solution:

Concepts, principles, relations that apply to the problem:
The Lorentz invariance of the electromagnetic field-strength tensor
Why do they apply?
In SI units E2 - c2B2 and (E×B)2 are invariant under a Lorentz transformation.
How do they apply?
(a)  E2 - c2B2 and (E×B)2 are the same in every reference frame.
In K we have E0×B0 = E0B0cosq0 = (E02/c)cosq0.
In K' we have E'×B' = E'B'cosq.
Since E02 - c2B02 = 0 in K, we have  E'2 - c2B'2 = 0 in K', therefore E' = cB'.
We then have E'×B' = (E'2/c)cosq.
E0×B0 = E'×B' -->  E'2 = E02cosq0/cosq.
Details of the calculation:
(b)  If E and B are perpendicular to each other, then E×B = 0.  But E×B is invariant and not zero in K.  There is no such frame.

Problem 3:

A proton of mass m0 moving with speed bz in the z-direction, encounters a quadrupole magnetic field of the form B = B0xj + B0yi over a length L along the z-direction.  Both the magnetic field B and the length L are as observed in the laboratory frame.
(a)    Find the electric field experienced by the particle as a function of x and y in its rest frame.
(b)   If the magnet is short enough, such that the acceleration of the particle going through the field can be considered an impulse, (i.e. the acceleration changes the direction, but not the position), find the impulse observed in the lab and the entering particle’s rest frame, for a particle that enters the magnet at x = x0, y = 0.  Explain the difference.

Solution:

Concepts, principles, relations that apply to the problem:
Lorentz transformation of the electromagnetic fields
Why do they apply?
We are asked to transform the electromagnetic fields from the laboratory frame to the rest frame of the proton.
How do they apply?
(a)  In SI units he transformation of the electromagnetic fields from the laboratory frame to a frame moving with velocity v with respect to the laboratory frame is given by:
E'|| = E||,  B'|| = B||,  E'^ = g(E + v´B)^,  B'^ = g(B - (v/c2)´E)^.
In the laboratory frame we have B = B0xj + B0yi.
With v = vk we have v ´ B = -vB0xi + vB0yj.
In a frame moving with velocity vk (i.e. in the original rest frame of the proton) we have:
E'z = 0,  B'z = 0,  E'x = -gvB0x',  B'x = gB0y’,  E'y = gvB0y',   B'y = gB0x'
(We have used x = x’, y = y’.)
Details of the calculation:
(b)  In the lab frame the force on the proton is F = qv ´ B = -qvB0xi + qvB0yj.
This force acts for a short time Dt = L/v. 
For a proton entering at x = x0, y = 0 the impulse is I = Dp = -qLB0x0i.
In its original rest frame the force on the proton at x’ = x0, y’ = 0 is F = qE = -gqvB0x0i . 
This force acts for a short time Dt = L’/v.  The impulse is I’ = Dp’ = -gqL’B0x0i .
The momentum is a 4-vector.  The components perpendicular to the relative velocity are the same in two frames moving with respect to each other.  We need Dp = Dp’, L’ = L/g, we need the Lorentz contraction.

Problem 4:

Plot E and B as a function of time at a point P one cm away from the path of a 10 MeV proton.  Set P at (0, 0.01, 0) with the charge at (vt, 0, 0).

Solution:

Concepts, principles, relations that apply to the problem:
The electromagnetic fields of a moving point charge, the Lorentz transformation of the electromagnetic fields
Why do they apply?
In the rest frame of the proton 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 vi.
How do they apply?
Let K be the rest frame of the proton.  In K,  E = (qe/(4pe0r2))(r/r),  B = 0.
Let K' be the frame in which the proton is moving with velocity vi.  K' is moving with velocity -vi with respect to K.
In SI units he 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'^ = g(E + v´B)^,  B'^ = g(B - (v/c2)´E)^.
Therefore
E'x = Ex = (qe/(4pe0))(x/(x2 + y2 + z2)3/2
= (gqe/(4pe0))(x' - vt')/(g2(x' - vt')2 + y'2 + z'2)3/2,
E'^ = gE^, E'y = (gqe/(4pe0))y'/(g2(x' - vt')2 + y'2 + z'2)3/2,  
E'z = (gqe/(4pe0))z'/(g2(x' - vt')2 + y'2 + z'2)3/2.
B'x = 0,  B'^ = g(vi/c2)´E,  B'y = -g(v/c2)Ez = -(v/c2)E'z,  B'z = g(v/c2)Ey = (v/c2)E'y.
Details of the calculation:
In K' at x' = 0, y' = y0' = 0.01, z' = 0 we have
E'x = -(gqe/(4pe0))vt'/(g2(vt')2 + y0'2)3/2, E'y = (gqe/(4pe0))y0'/(g2(vt')2 + y0'2)3/2,  E'z = 0.
 B'x = 0,  B'y = 0, B'z = (v/c2)E'y.

10 MeV proton:  gmc2 = (938.28 + 10)MeV,  g = 1.01,  b = 0.145, v = 4.35*107 m/s.
E'x = -1.01*9*109*1.6*10-19*4.35*107 t(s)/[(1.01*4.35*107)2t2 + 10-4]N/C
= 6.3*10-2t/(1.9*1015t2 +10-4)N/C.
E'y = -1.01*9*109*1.6*10-19*10-4/[(1.01*4.35*107)2t2 + 10-4]N/C
= 1.5*10-11/(1.9*1015t2 +10-4)N/C.

Here g = 1.01 is nearly equal to 1.  The electric field in K' differs very little from the instantaneous Coulomb field.  But using the transformation of the fields we can calculate the magnetic field in K'.

Problem 5:

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, principles, relations that apply to the problem:
The Lorentz transformation of the electromagnetic fields, the Lorentz force
Why do they apply?
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.
How do they apply?
Assume a charge is at rest in K.  Assume K is moving with velocity v with respect to K'.  So in K' the charge moves with velocity v.
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^ = g(E' + v´B')^,  B^ = g(B' - (v/c2)´E')^.
Details of the calculation:
In K the force on the charge is F = dp/dt = dp/dt, where t is the proper time, a Lorentz invariant.
In K' the force on the charge is F' = dp'/dt' = (1/g)dp'/dt, gF' = dp'/dt.
Here   g = (1 - b2)-1/2, b = v/c.
From the transformation properties of the momentum 4-vector
pm = (gmc,gmv) = (E/c,p) = under a Lorentz transformation,
p'|| = g(p|| + bE/c), p'^ = p^, we have
 gF'^ = dp'^/dt  = dp^/dt  = dp^/dt = qE^ = gq(E' + v´B')^.
 F'^ = q(E'^ + v´B').
gF'|| = dp'||/dt = gdp||/dt = gdp||/dt = gqE|| = gqE'||.
 F'|| = qE'||.
We have used that in K d(E/c)/dt  = (1/c)dE/dt = 0 since in K the charge is at rest.
Therefore F' = q(E' + v´B').  This is the expression for the Lorentz force.

Problem 6:

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, principles, relations that apply to the problem:
The Lorentz transformation
Why do they apply?
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).
How do they apply?
Assume the electrons move with velocity vi in the x-direction in the laboratory frame K.  Electron 1 has coordinates y = z = 0 and electron 2 has coordinates y = y0 = 1 mm, z = 0.
In the rest frame K' of the electrons the force electron 1 exerts on electron 2 is
F' = j kqe2/y'02 = j kqe2/y02.
F' = j 9*109*(1.6*10-19)2/10-6 N = j 2.3*10-22 N
F' = dp'/dt' = dp'/dt, where t is the proper time, a Lorentz invariant.
In K the force on the charge is F = dp/dt = (1/g)dp/dt, gF = dp/dt.
Here   g = (1 - b2)-1/2, b = vi/c.
From the transformation properties of the momentum 4-vector
pm = (gmc,gmv) = (E/c,p) = under a Lorentz transformation,
p||= g(p'|| + bE'/c), p^ = p'^, we have
 gF^ = dp^/dt  = dp'^/dt  = dp'^/dt = j kqe2/y02, F^ = j (1/g)kqe2/y02.
 gF|| = dp||/dt  = gdp'||/dt  = gdp'||/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/g)kqe2/y02.
Details of the calculation:
1 eV = (1/2)mv2, g = 1, F = j  2.3*10-22 N.
1 MeV = (g - 1)mc2, (g - 1) = 1.96, g = 2.96, F = j  7.8*10-23 N.

Problem 7:

(a)  State Maxwell’s equations in differential form.
(b)  State Maxwell’s equations in integral form.
(c)  Carry out the Lorentz transformation for the vector potential

Am = (-Ex2, -(B/2)x2, (B/2)x1, 0)

along the x1 direction with b = E/B, (x0, x1, x2, x3) are the usual coordinates.  Note E and B are constants and B ¹ 0.  Find the electric and magnetic fields before and after the Lorentz transformation.

Solution:

Concepts, principles, relations that apply to the problem:
Lorentz transformation of the 4-vector potential,
E = -Ñf - (1/c)A/t, B = Ñ´A (Gaussian units)
Why do they apply?
We are asked to Lorentz transform a given 4-vector potential, and from the transformed potential find the transformed fields.
How do they apply?
(a)  Maxwell’s equations in differential form are:
(SI units)   (Gaussian units)
 
 

(b)  Maxwell's equations in integral form in SI units are:

(1)

(2)

(3)

(4)

FB = òB×dA is the flux of B through the area enclosed by the curve G.
FE = òE×dA is the flux of E through the area enclosed by the curve G.

(c) I use Gaussian units because Am is given in Gaussian units.
In the original frame, before the Lorentz transformation:

Am = (f, Ax, Ay, Az) = (-Ey, -By/2, Bx/2, 0) is
E = -Ñf – (1/c)A/t = Ej
B = Ñ ´ A = Bk

These are crossed electric and magnetic fields, they act as a velocity filter.  A charged particle moving with velocity v = icE/B through these fields will experience no force.  The Lorentz transformation must therefore produce a vector potential that yields E' = 0 in the rest frame of that particle.

Details of the calculation:
Lorentz transformation of the 4-vector potential:

Here g = (1 - b2)–½ ,  b = (E/B)i.

The coordinates transform as

or 

.

We therefore have after the Lorentz transformation:
A'm = (f', A'x, A'y, A'z) = (-gEy'/2, gE2y'/B - gBy'/2, gEct'/2 + gBx'/2, 0)
E' = - Ñ'f' – (1/c)A'/t' = (gE/2 - gE/2)j = 0
B' = Ñ ´ A' = k(gB/2 - gE2/B + gB/2) = k(-gE2/B+ gB)

Problem 8:

An aluminum disk of radius R, thickness d, conductivity s and mass density r 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 W0, how many revolutions will it make before it comes to rest?

Solution:

Concepts, principles, relations that apply to the problem:
The Lorentz transformation of the electromagnetic fields, the Lorentz force
Why do they apply?
Even at non-relativistic speeds, problems involving Eddy currents are often easier analyzed after a frame transformation.
How do they apply?

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 = WRi.  In this observer's rest frame K' there exists an electric field E'^ = g(E + v´B)^ = -gvBj.
All velocities are non-relativistic, so  g @ 1 and E'^ @ -vBj = -WRBj.
There also exist a magnetic field B'^ = gB^ @ Bk.
In K' the current density is  jc' = sE' = -sWRBj and the Lorentz force on the small volume DV = Ad is
F' = jc'´B' DV = -sWRB2Adi.  Because g @ 1the force F observed in the laboratory frame is equal to F'.
In the laboratory frame F produces a torque about the axis of rotation.
t = -Rj´F = -sWR2B2Adk.
The moment of inertia of the disk is I = (1/2)MR2 = (1/2)rdpR4.
The equation of motion is tz = IdW/dt or -sWR2B2Ad = (1/2)rdpR4dW/dt.
dW/dt = -W2sB2A/(rpR2).  W(t) = W0exp(-2sB2At/(rpR2)).
The number of revolutions the disk makes before it comes to rest is
N = qtot/2p = (1/2p)ò0¥W(t)dt = (W0/2p)ò0¥exp(-2sB2At/(rpR2))dt = (W0rR2/4sB2A).
Details of the calculation:
None

Problem 9:

A light beam of intensity I0 and frequency n0 propagating along the positive z-axis is reflected normally from a perfect mirror moving along the z-axis with speed v.  What is the intensity and the frequency of the reflected light?

Solution:

Concepts, principles, relations that apply to the problem:
The Doppler effect, frame transformations.
Why do they apply?
In the rest frame of the mirror, the incident and the reflected beam have the same frequency and intensity.  We use the Doppler formula and transform from the laboratory frame to the mirror frame and back to the laboratory frame.
To find the ratio of the intensities of the incident and reflected beam in the laboratory frame we transform the electric field amplitude from the laboratory frame to the mirror frame and back to the laboratory frame.
How do they apply?

In the rest frame of the mirror the incident beam has frequency n' = n0[(c-v)/c+v)]1/2.
The reflected beam has the same frequency.
In the lab frame the reflected beam has frequency
n = n'[(c-v)/c+v)]1/2 = n0[(c-v)/c+v)].
Details of the calculation:
The intensity of a light beam is proportional to E^2.  Denote the electric field amplitude of the incident beam in the laboratory by E0^2.

In the rest frame of the mirror we have E'^ = g(E0 + v´B)^ = gE0^(1 - v/c),
since for a plane wave B = (1/c)(k/k)´E, where k/k is the direction of propagation of the wave.
Here k/k denotes the z-direction and v is in the z-direction.
The reflected beam in the lab frame has amplitude
E''^ = g(E' - v´B')^ = gE'^(1 - v/c) = g2E0^(1 - v/c)2 = E0^(c-v)/c+v).
Note that for the reflected wave k/k denotes the -z-direction.
The intensity is proportional to the square of the amplitude.
I/I0 = [(c-v)/c+v)]2.