(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

A^{μ} = (-Ex^{2}/c,
-(B/2)x^{2}, (B/2)x^{1}, 0)

along the x^{1} direction with v = cβ
= E/B, (x^{0}, x^{1}, x^{2}, x^{3}) are the usual coordinates. Note E and B are constants and B
≠
0. Find the electric and magnetic fields
before and after the Lorentz transformation.

Solution:

- Concepts:

Lorentz transformation of the 4-vector potential,

E**∇**Φ - ∂**A**/∂t,**B**=**∇**×**A**(SI units) - Reasoning:

We are asked to Lorentz transform a given 4-vector potential, and from the transformed potential find the transformed fields. - Details of the calculation:

(a) Maxwell's equations in differential form are:

(SI units) **∇∙E**= ρ/ε_{0},**∇**×**E**= -∂**B**/∂t**∇∙B**= 0,**∇**×**B**= μ_{0}**j**+ (1/c^{2})∂**E**/∂t(b) Maxwell's equations in integral form in SI units are:

(1) ∮ _{A}**E**∙d**A**= Q_{inside}/ε_{0}(2) ∮ _{Γ}**E**∙d**s**= -∂Φ_{B }**/**∂t(3) ∮ _{A}**B**∙d**A**= 0(4) ∮ _{Γ}**B**∙d**s**= μ_{0}I_{through Γ}+ (1/c^{2})∂Φ_{E }**/**∂tΦ

_{B }= ∫**B∙**d**A**is the flux of B through the area enclosed by the curve Γ.

Φ_{E }= ∫**E∙**d**A**is the flux of E through the area enclosed by the curve Γ.(c) A

^{μ}= (Φ/c, A_{x}, A_{y}, A_{z}) = (-Ey/c, -By/2, Bx/2, 0)

**E**= -**∇**Φ – ∂**A**/∂t = E**j**=

B**∇**×**A**= B**k**These are crossed electric and magnetic fields, they act as a velocity filter. A charged particle moving with velocity

**v**=**i**E/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.

Lorentz transformation of the 4-vector potential:

Here γ = (1 - β^{2})^{–½}, β = (E/B)**i**.

The coordinates transform as

or

.

We therefore have after the Lorentz transformation:

A'^{μ}= (Φ'/c, A'_{x}, A'_{y}, A'_{z}) = (-γEy'/2c, γE^{2}y'/Bc^{2}- γBy'/2, γEt'/2 + γBx'/2, 0)**E**' = -**∇**'Φ' – ∂**A**'/∂t' = (γE/2 - γE/2)**j**= 0**B**' =**∇**×**A**' =**k**(γB/2 - γE^{2}/B + γB/2) =**k**(-γE^{2}/Bc^{2}+ γB)

A point magnetic moment **m** is at rest in frame K' and in that
frame produces a vector potential **A**' = (μ_{0}**/**4π)**m**' x** r**'/r'^{3}
(SI units)
and no scalar potential (Φ = 0). Frame K'
moves with constant velocity v << c along the x-axis of frame K, so
that an observer in K sees the moment moving with velocity **v **= βc**i**. Show that to first order in
β the
observer in K detects an electric dipole moment **p **=** β
**x** m**/c
as well as an undiminished (to first order) magnetic moment **m**.

Solution: needs work

- Concepts:

Lorentz transformation of 4-vector potential A'^{μ} - Reasoning:

We transform the 4-vector potential from K' to K and check if it transforms into a vector potential due to a magnetic dipole**m**and a scalar potential due to an electric dipole**p**to first order. - Details of the calculation:

In the rest frame K' of the magnetic dipole we have A'^{μ}= (0,A'_{x},A'_{y},A'_{z}).

The dipole is at rest at the origin in K'.

K moves with velocity -βc**i**= - v**i**with respect to K'.

In K we have

A^{μ}= (γβA'_{x}, γA'_{x}, A'_{y}, A'_{z}).

Since v << c we have γ ~ 1. We therefore write

A^{μ}= (βA'_{x}, A'_{x}, A'_{y}, A'_{z}).

x^{μ}= (ct', x' + βct', y', z'), i.e. t = t' x = x' + βct', y = y', z = z'.

**A**(**r**') =**A**'(**r**').**A**(**r**) =**A'**(**r**- βct**i**), the observer in K detects an undiminished (to first order) magnetic moment**m**=**m**' moving with velocity βc**i**.

Φ(**r**') = cβA'_{x}(**r**') =**v**∙**A**' = (μ_{0}**/**4π)**v**∙(**m**'×**r**'/r'^{3})

= (μ_{0}**/**4π)(**r**'/r'^{3})∙(**v**×**m**') = (1**/(**4πε_{0}))(**r**'/r'^{3})∙**p**,

with**p**=**v**×**m**'/c^{2}=**v**×**m**/c^{2}=**β**x**m**/c**.**

Φ(**r**) = (1**/(**4πε_{0}))((**r**- βct**i**)/|**r**- βct**i**|^{3})∙**p**.

Φ(**r**) is the scalar potential of a dipole**p**moving with velocity βc**i**.

In reference frame K a long,
straight, neutral wire with a circular cross sectional area A =
πr^{2} lies centered on the z-axis and carries a current with uniform current
density j**k**.

(a) Find the Φ, **A**, **E**, and **B** at a point P on the x-axis a distance
x > r from the wire.

(b) In a frame K' moving with velocity v**k** with respect to K, find
the ρ, **j**, Φ, **A**, **E**, and
**B** at the point P.

(c) In a frame K'' moving with velocity v**i** with respect to K,
find the ρ, **j**, Φ, **A**, **E**, and
**B** at the point
P.

Solution:

- Concepts:

Lorentz transformation of 4-vector current, 4-vector potential and electric and magnetic field - Reasoning:

We are asked to find ρ,**j**Φ,**A**,**E**, and**B**in 3 different reference frames. - Details of the calculation:

(a) In K we have**j**= j**k**inside the wire,**j**= 0 outside the wire, ρ = 0 everywhere.

Outside a neutral wire carrying a current I = jA in the z-direction we have= 0, Φ = 0,

E**B**= μ_{0}I/(2πr)(**r/**r),**A**= -[μ_{0}I/(2π)] ln(r/r_{0})**k**.

Here r is a cylindrical coordinate.

At point P on the x-axis**B**= μ_{0}I/(2πx)**j**,**A**= C - [μ_{0}I/(2π)] ln(x)**k**.

C is an arbitrary constant. Chose C = 0.

(b) In a frame K' moving with velocity v**k**with respect to K we have

cρ' = γcρ - γβj = -γβj. j_{z}' = γj inside the wire, ρ' =**j**' = 0 outside the wire.

The wire is not neutral in this frame.

Given ρ' and**j**' we find Φ' = [γ(v/c^{2})I/(2πε_{0})]ln(x)

and**A**' = -[μ_{0}γI/(2π)]ln(x)**k**at point P.

Note x' = x.

From the potentials we obtain**E**' = (-**i**) γ(v/c^{2})I/(2πε_{0}x) = (-**i**) μ_{0}γvI/(2πεx)

and**B**' = μ_{0}γI/(2πx)**j**at point P.

We can also transform the 4-vector potential directly.

Φ'/c = - γβA_{z}= [γβμ_{0}I/(2π)]ln(x), A_{z}' = γA_{z},**A**' = -[μ_{0}γI/(2π])ln(x)**k**at point P.Transforming the fields directly we obtain E_{z}' = B_{z}' = E_{y}' = B_{x}' = 0,

B_{y}' = μ_{0}γI/(2πx), E_{x}' = -μ_{0}γvI/(2πε_{0}x) at point P.

All the transformations yield consistent results.

(c) In a frame K'' moving with velocity v**i**with respect to K we have

ρ' = 0, j_{z}' = j inside the wire, ρ' =**j**' = 0 outside the wire.

The wire is neutral in this frame.

Transforming the 4-vector potential we find Φ' = 0,

**A**' = -[μ_{0}I/(2π)]ln(γ(x'_{ }+ vt'))**k**at point P.

At point P we have**B**' = B_{y}'**j**, B_{y'}= -∂A_{z}'/∂x' = μ_{0}γI/(2π(γ(x'_{ }+ vt')),' = -∂

E**A**'/∂t' = μ_{0}γvI/(2πε_{0}γ(x'_{ }+ vt'))**k**.

Transforming the fields directly we obtain

E_{x}' = B_{x}' = 0, B_{y}' = μ_{0}γI/(2πγ(x'_{ }+ vt')), E_{z}' = μ_{0}γvI/(2πε_{0}γ(x'_{ }+ vt')) at point P.

Even thought we have no charge density, we have an electric field. Why?

In reference frame K
a wire with a circular cross sectional area A =
πr^{2} forms a
square of side length L. The square lies in the xz-plane. A current
with uniform current density j flows clockwise through the wire, I = jπr^{2}.

(a)
In a frame K' moving with velocity v**i** with respect to K, find
the charge density ρ' and the current density **j**'.

(b) Find the current flowing in the wire in K'.

Solution:

- Concepts:

Lorentz transformation of 4-vector current - Reasoning:

(cρ,**j)**is a 4-vector. - Details of the calculation:

In K we have**j**= j**k**inside the wire for side 1,**j**= -j**k**inside the wire for side 3,**j**= j**i**inside the wire for side 2,**j**= -j**i**inside the wire for side 4,**j**= 0 outside the wire, ρ = 0 everywhere.

In a frame K' moving with velocity v**i**with respect to K we have**j**= 0, ρ = 0 outside the wire.

for side 1: cρ' = γcρ = 0. j_{z}' = j_{z}= j at a position inside the wire

for side 3: cρ' = γcρ = 0. j_{z}' = j_{z}= -j at a position inside the wire.

Wires 1 and 3 are neutral in K'.

for side 2: cρ' = γcρ - γβj_{x}= -γβj, j_{x}' = γj_{x}= γj at a position inside the wire,

for side 4: cρ' = γcρ - γβj_{x}= γβj, j_{x}' = γj_{x}= -γj at a position inside the wire,

Wires 2 and 4 is not neutral in this frame.

(b) The cross sectional area of wires 1 and 3 is an ellipse of area πr^{2}/γ. The current flowing in those wires in frame K' is I' = jπr^{2}/γ = I/γ. The same current must flow in all wires.

The cross sectional area of wires 2 and 4 in K' is πr^{2}, the same as in K, But the current density at a point that at a time t' lies inside the moving wire is due to both the current in the wire and the moving charged wire.

Consider side 2: γj = -ρ'v + j'' = γβ^{2}j + j'', where j'' is the current density due to the current flowing in the wire.

j'' = γj(1 - β^{2}) = j/γ. I'' = j''πr^{2}= I/γ.

Consider a point charge q at rest at the origin in reference frame K.
Reference frame K' moves with velocity **v** = cβ**i** with respect to K.
At t = 0 the origins of the two frames coincide.

(a) Write down expressions for the 4-vector potential A^{μ} and the
4-vector current j^{μ} in reference frame K.

(b) Find A'^{μ} and j'^{μ} in reference frame K'.

Solution:

- Concepts:

Lorentz transformation of the 4-vector potential and 4-vector current - Reasoning:

We are asked to find and Lorentz transform the 4-vector potential and 4-vector current. - Details of the calculation:

(a) A^{μ}= (Φ/c, A_{x}, A_{y}, A_{z}) = (Φ/c,**A**), j^{μ}= (cρ, j_{x}, j_{y}, j_{z}) = (cρ,**j**).

In K we have**j**= 0 and we choose the gauge so**A**= 0.

ρ(r) = qδ(**r**),

Φ(r) = kq/r, with k = 1/(4πε_{0}).

A^{μ}= (kq/((x^{2 }+ y^{2}+ z^{2})^{½}c), 0, 0, 0),

j^{μ}= (cqδ(x**i**+ y**j**+ z**k**), 0 0, 0) = (cqδ(x) δ(y) δ(z), 0 0, 0)(b) j'

^{0}= γcρ, j^{1}= -γβcρ

A'^{0}= γΦ(r)/c, A'^{1}= -γβΦ(r)/c

x = γβct' + γx', y = y', z = z'.

A'^{μ}= γkq/[((γβct' + γx')^{2 }+ y'^{2}+ z'^{2})^{½}c], -γβkq/[((γβct' + γx')^{2 }+ y'^{2}+ z'^{2})^{½}c], 0, 0),

j'^{μ}= (γcqδ(γ(βct' + x')) δ(y) δ(y), -γβcqδ(γ(βct' + x')) δ(y) δ(y), 0, 0)

= (cqδ(βct' + x') δ(y) δ(y), -βcqδ(βct' + x') δ(y) δ(y), 0, 0),

since δ(ax) = δ(x)/|a|.