Problem 1:
A reference frame K' is moving with uniform velocity v = vi
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 45o
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 - v2/c2)-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:
In reference frame K a long, straight, neutral wire with a circular cross
sectional area A = πR2 lies along the x-axis and carries a steady current
I in the +x direction.
The current density is j(x, y, z) = I/A i for r = (y2 +
z2)1/2 < R and zero everywhere else;
We write j = j- i = ρ-<v->i,
since the current is carried by the electrons. Both ρ- and <v->
are negative numbers, j- is a positive number.
In general j = (j- + j+). ρ = ρ- + ρ+.
In reference frame K j+ = 0, ρ- = -ρ+.
(a) Consider the negative charge and current density and the positive
charge and current density seperately.
In a frame K' moving with velocity vi with respect to K, find the ρ-',
j-' and ρ+', j+' and ρ', j' by transforming the
appropriate 4-vector current densities.
Write your answers in terms of ρ- and <v->. Is
the wire neutral in K'? If not, is it positively or negatively
charged? What is the direction of the current flow in K'?
Hint: The y and z components of all current densities are zero in K and K'.
The cross sectional area of the wire is the same in K and K'. Inside the
wire the current densities are constant in K and K'. There is no need to
transform the coordinates.
(b) Now assume that v = <v->, i.e. K' moves in the
negative x-direction with a velocity equal to the drift velocity of the
electrons.
Write down the expressions for ρ-', j-' and ρ+',
j+' and ρ', j' for this special case. Is the wire neutral
in K'? If not, is it positively or negatively charged? Can you
argue why this should be the case by considering length contraction?
Solution:
- Concepts:
Lorentz transformation of 4-vector current
- Reasoning:
We are asked to transform the 4-vector (cρ, j) from one to another
inertial frame.
- Details of the calculation:
(a) For the negative 4-vector current density we have in K'
cρ-' = γcρ- - γβj-x = γcρ-(1
- (v/c2)<v->),
ρ-' = γρ-(1 - (v/c2)<v->),
j-x' = -γβcρ- + γj-x = γρ-(-v
+ <v->).
If v > 0 then
j-x' is bigger than
j-x.
For the positive 4-vector current density we have in K'
cρ+' = γcρ+ = -γcρ-,
ρ+' = -γρ-,
j+x' = -γβcρ+ = +γρ-v.
If v > 0, then
j+' point into the negative x-direction.
For the total4-vector current density we have in K'
ρ' = ρ+' + ρ-' = -γρ-(v/c2)<v->,
j' = j+x' + j-x' = γρ-<v->).
For v ≠ 0 the wire is not neutral. The sign of the charge density
depends on the direction of v. If v > 0. then the wire has negative charge density in K'.
The direction of the current flow is the positive x-direction in K'.
(b) If v = <v->, the v points in the
negative -x-direction, and we have γ = (1 - (<v->2/c2))-½
and
ρ-' = γρ-(1 - (<v->2/c2))
= ρ-/γ
j-x' = 0.
ρ+' = -γρ-.
j+x' = γρ-<v->.
ρ' = ρ+' + ρ-' = -γρ-(<v->2/c2).
j' = j+x' + j-x' = γρ-<v->.
The wire will be positively charged in K'.
The direction of the current flow is the positive x-direction in K'.
Consider the wire to be a superposition of two cylinders one with negative
and one with positive charge density.
In K the cylinder with positive charge density is at rest and the cylinder
with negative charge density is moving.
The length of each section of the negative cylinder is contracted by a
factor of 1/γ, the volume of each section is shrunk by a factor of 1/γ.
Since the magnitude of the negative and positive charge in the same cylinder
length in K is the same, the charge density is smaller in the rest frame of
the negative cylinder, since in that frame the volume of the cylinder is
larger by a factor of γ. The positive cylinder's
charge density is a factor of γ greater in its rest frame than the negative cylinder's charge
density in its rest frame.
In K' the cylinder with negative charge density is at rest and the
cylinder with positive charge density is moving. The charge density of
the negative cylinder is its rest-frame charge density, the charge density
of the positive cylinder increases by a factor of γ over its rest-frame
charge density. In K' the wire is therefore not neutral, but
positively charged, ρ+' = -γ2ρ-'.
Problem 3:
A point charge of magnitude q moves on the trajectory
z(t) = vt,
y(t) = x(t) = 0.
(a) Find the electric and
magnetic fields at an arbitrary point (x,y,z) as a function of the time
t.
(b) Find the energy current density (Poynting vector) at each point as a function of
time.
(c) What is the total amount of electromagnetic energy, which is radiated out of the
surface of a very large sphere of radius R centered at the origin?
Solution:
- Concepts:
The Coulomb field, the transformation of E and B from one
Lorentz frame to another
- Reasoning:
Let the laboratory frame be K, and the rest frame of the point charge be K'.
In
K', with the point charge at the origin,
E'(r',t) = (1/(4πε0))
(q/r'3)r',
B'(r',t) = 0.
We have axial symmetry about the z-axis.
Let is find the components of the electric field parallel and perpendicular
to the z-direction.
E'(r',t) = (1/(4πε0)) (q/(ρ'2
+ z'2)3/2)[ρ' (ρ'/ρ') + z'
k].
- Details of the calculation:
The laboratory frame K is moving with speed v = -v k with
respect to the rest frame of the point charge.
We find the fields in the laboratory frame using
E|| = E'||,
B|| = B'||,
E⊥ = γ(E' + v×B')⊥,
B⊥ = γ(B' - (v/c2)×E')⊥.
Therefore
E|| = (q/(4πε0)) (1/(ρ'2 + z'2)3/2)
z' k. B|| = 0.
E⊥ = γE'⊥ = (γq/(4πε0)) (1/(ρ'2
+ z'2)3/2) ρ' (ρ'/ρ'),
B⊥ = -γ(v/c2)×E'⊥.
The Lorentz transformation of the 4-vector (ct, z, x, y) in K to K' moving with
velocity v k with respect to K yields
z' = γ(z - vt), ρ' = ρ.
E|| = (q/(4πε0)) (1/(γ2(z - vt)2
+ ρ2)3/2)γ(z - vt) k.
E⊥ = (γq/(4πε0)) (1/(γ2(z - vt)2
+ ρ2)3/2) ρ (ρ/ρ).
B⊥ = (v/c2)(γq/(4πε0)) (1/(γ2(z - vt)2
+ ρ2)3/2)ρ
(φ/φ).
So E = (γq/(4πε0)) (1/(γ2(z - vt)2
+ ρ2)3/2) ((z - vt) k + ρ (ρ/ρ)).
B = -(v/c2)×E = (v/c2)(γq/(4πε0)) (1/(γ2(z - vt)2
+ ρ2)3/2)ρ
(φ/φ).
(b) S = (1/μ0)E×B.
E||×B⊥ = (γq/(4πε0))2(v/c2)(1/(γ2(z - vt)2
+ ρ2)3)(z - vt)ρ (ρ/ρ).
E⊥×B⊥
= (γq/(4πε0))2(v/c2)(1/(γ2(z - vt)2
+ ρ2)3) ρ2
k.
S = (γ2q2v/(16π2ε0))(1/(γ2(z - vt)2
+ ρ2)3)[(vt - z)ρ (ρ/ρ) + ρ2
k].
(c) No acceleration --> no radiation field proportional to 1/r.
Problem 4:
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 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.
(i) Transform the force.
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'/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 - β2)-½, β
= vi/c.
From the transformation properties of the momentum 4-vector
pμ
= (γmc,γmv)
= (E/c,p) = under a Lorentz transformation,
p||= γ(p'|| + βE'/c),
p⊥
= p'⊥, we have
γF⊥
= dp⊥/dτ
= dp'⊥/dτ
= dp'⊥/dt'
= j kqe2/y02,
F⊥
= j (1/γ)kqe2/y02.
γF||
= 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)kqe2/y02.
1 eV = ½mv2,
γ = 1, F = j 2.3*10-22 N.
1 MeV = (γ - 1)mc2, (γ
- 1) = 1.96,
γ = 2.96, F =
j 7.8*10-23 N.
(ii) Transform the fields.
In K' the electric field produced by electron 1 at the position of electron 2 is
E'⊥ = -j
kqe/y'02, E'||=
0.
In K' the magnetic field produced by electron 1 is B = 0.
Frame K moves with velocity v'=
-v i with respect to K'.
Therefore at the position of electron 2 E||=
B||= 0.
E⊥ = γ(E'
+ v'×B')⊥
= γE'⊥= -j γkqe/y'02.
B⊥
= γ(B' - (v'/c2)×E')⊥
= -k (v/c2)γkqe/y'02.
F = -qe(E + v'×B) =
j (γkqe2/y'02 - (v/c2)γkqe2/y'02)
= j γ(1 - v/c2)kqe2/y'02
= j (1/γ)kqe2/y02.
Problem 5:
A particle accelerator accelerates
electrons to a speed of 0.999 999 999 7 c, which is very nearly equal to the
speed of light.
(a) What is the rest energy (in MeV) of the electrons?
(b) Find the magnitude of the relativistic momentum of the electrons.
Comparing it with the non-relativistic value, is it bigger, smaller? By what
factor?
(c) Compute the electron's total energy (in GeV).
(d) Assume the electrons collide with the nuclei of "stationary" hydrogen atoms
in a gas cell. Assume the collision products are a proton, an electron, and a
particle-antiparticle pair. What is the rest energy of the most massive
particle (or antiparticle) that can be created in such a collision?
Solution:
Relativistic expression for energy and momentum
p = γmv, E = γmc2 = (m2c4 + p2c2)½.
(a) Erest= mc2 = 8.2*10-14 J = 0.511 MeV.
(b) γ = (1 - v2/c2)-½ = 4.08*104.
p = γmv = 1.1*10-17 kgm/s = 2.09*104 MeV/c.
The non-relativistic value would be mv = 2.73*10-22 kgm/s.
The relativistic value is bigger by a factor of γ = 4.08*104.
(c) E = γmc2 = 2.09*104 MeV = 20.9 GeV.
(d) To create the most massive particles, the velocities of all collision
products with respect to each other have to be zero with respect to each other.
Let M be the rest mass of all the collision products,
M = mp + me + 2mparticle.
Energy conservation: mpc2 + γmec2 =
(M2c4 + P2c2)½.
Momentum conservation: pe = P.
0.938 GeV + 20.9 GeV = (M2c4 + (20.9 GeV)2)½.
Mc2 = 6.33 GeV.
mparticlec2 = 2.7 GeV.