Electromagnetism
Physics 15b
Lecture #13
Special Relativity
Purcell 5.3–5.9, 6.7
What We Did Last Time
Biot-Savart Law
IdL
dA =
cr
IdL × r̂
dB =
cr 2
dA
r12
dL
I
Special relativity recap


Lorentz transformation, time dilation, length contraction
Ladder and shed puzzle
I expect that you’ve reviewed your 15a/16 notes on
uy
ux − v
uy′ =
  Velocity addition u ′ =
x
2
1− vux c
γ 1− vux c 2

Momentum, energy
(
)
1
Today’s Goals
Study how Relativity and
Electromagnetism work together

How do E and B fields transform
between two inertial frames?
Electricity
Magnetism
Case study: a charged particle
near a line of current

Connection between electric and
magnetic forces
I
q
v
Transformation of Forces
In frame O, a force F = (Fx, Fy, 0)
acts on an object of mass m,
initially at rest (p = 0 at t = 0)

Observe this for a very short time Δt in O
y
y´
F
O
O´
x´
x
Force is defined by F = dp/dt
z´
  Momentum grows as Δp = FΔt
z
1
1F
  Position changes as Δx =
a(Δt)2 =
(Δt)2
2
2m
2
1 (Δp)
1 (FΔt)2
  Energy increases as ΔE =
=
2 m
2 m
Both Δx and ΔE are proportional to (Δt)2

That means dx/dt
= 0, dE/dt = 0
2
Transformation of Forces
Observe the same motion in O′

For the x′ component ⎡ ⎛

For the y′ component
=0
v ⎞⎤
d ⎢γ ⎜ px − 2 E ⎟ ⎥ dp − v dE F − v dE
x
x
⎝
c ⎠⎦
dp′
c2
c 2 dt = F
Fx′ ≡ x = ⎣
=
=
x
v
v dx
dt ′
⎡ ⎛
v ⎞⎤
dt − 2 dx
1− 2
d ⎢γ ⎜ t − 2 x ⎟ ⎥
c
c dt = 0
⎣ ⎝ c ⎠⎦
Fy′ ≡
dpy′
dt ′
=
dpy
⎡ ⎛
v ⎞⎤
d ⎢γ ⎜ t − 2 x ⎟ ⎥
⎣ ⎝ c ⎠⎦
  Same for the z′ component
=
dpy
⎛
⎞
v
γ ⎜ dt − 2 dx ⎟
⎝
⎠
c
Forces transform as F′ = F + F⊥
γ
=
Fy
⎛
v dx ⎞
γ ⎜ 1− 2 ⎟
⎝ c dt ⎠
=
Fy
γ
=0
NB: O is the rest frame
of the object
Relativity of Electromagnetism
Ready to tackle E&M in moving frames?
We assume relativity:
The laws of electromagnetism (Maxwell and others)
are the same in any inertial frame
We further assume:
Values of electric charges are the same in any inertial frame


“Electric charge is a Lorentz scalar”
Reasonable, but not obviously true
We will build our theory, and see where it leads us
Is the theory self-consistent?
  Does the theory match experimental facts?

3
Current and Moving Charge
A charged particle is moving near an infinitely-long current
q

I
v
x
In the laboratory frame O, the wire contains:
  Positive ions: line density λ0 esu/cm, velocity 0
  Electrons:
line density –λ0 esu/cm, velocity v0
+
+
–
v0
+
–
+
–
+
–
+
–
+
+
–
+
–
–
–
Net current on wire: I = λ0v 0
  Net charge on wire = 0
  Therefore no electric field around the wire

Current and Moving Charge
Observe in the charge’s rest frame O’
–v
+
+
–
v0’
+
–
+
–
+
–
+
–
+
–
+
–
q
Positive ions are moving with velocity –v
  Length contraction increases line density to γ λ0

Electrons are moving with velocity v0’ given by

v0 − v
vv
1− 02
c
+
–
–
x’

v 0′ =
+
or
β0′ =
β0 − β
1− β0 β
The line density is?
4
Line Density of Electrons
Length contraction is relative to “natural” length at rest

Electrons are not at rest in either O or O′
Line density –λ0 esu/cm was given in the lab frame O

The rest-frame density must be λrest =

Moving to the O′ frame, the density becomes

λ0
λ
1
=− 0
2
γ 0 1− β ′
γ0
0
−
1
− λ0
where γ 0 ≡
γ0
1− v 02 c 2
1− ββ0
(1− β 2 )(1− β02 )
= − λ0γ (1− ββ0 )
Total line charge of the wire in the O′ frame is
γλ0 − λ0γ (1− ββ0 ) = γββ0 λ0

The wire appears electrically charged from the moving particle
Force on the Charge
+
–v
–
+
v0’
+
–
+
+
–
+
–
r = r’
q
F′
In O’
+
–
–
Electric field due to wire charge is E ′ =
  Repels if v and v0 are parallel
+
+
–
–
F′ = q
2γββ0 λ0
r′
x’
E′

+
–
2γββ0 λ0
r′
Going back to the O (laboratory) frame
F ′ 2qββ0 λ0 2qvI Note that O′ is the particle’s
F=
=
= 2
γ
r
c r rest frame

Particle receives a velocity-dependent force from the current

Can rewrite as
F=
⎛ 2I
q
v × ⎜−
c
⎝ cr
⎞
φ̂ ⎟ Lorentz force and B field
⎠
5
What Happened?
Starting from
Special Relativity,
Lorentz invariance of the electric charge, and
  Coulomb’s Law,


we have found a velocity-dependent force on a moving
charge near a current

This is the force we used earlier to define magnetic field
Magnetism is a natural consequence of
electricity and special relativity
More generally
Electricity and magnetism are two manifestations
of the same physics = Electromagnetism
Transformation of E, B Fields
We have just seen that a B field in one frame appears to be
an E field in another frame
There must be a set of transformation rules that can
connect (E, B) in O to (E′, B′) in O′
Expect the rules to be linear, i.e., each component of E′, B′ is a
linear function of the components of E, B
  So that the superposition principle works in both frames
  The rules must agree with the transformation of forces

I’ll show how E transforms into (E′, B′) first

B transforming into (E′, B′) is similar, and is discussed in the
textbook, so I may skip if time is short
6
y´
y
E Field in Motion
v
O
Create uniform E with parallel plates
Plates are stationary in O, and parallel to
the x-z plane
  Charge density +σ (and −σ) gives
E = 4πσ ŷ
Look at this in O′
Plates are moving with velocity −v
  Charge is unchanged
  But the plate length L shrinks to L ′
  Charge density increases σ ′ = γσ
z´
z

x´
x
O´
L
–
–
–
–
–
–
–
–
+
+
+
+
+
+
+
+

L/γ
=L γ
––––––––
E′⊥ = γ E ⊥
++++++++
Electric field perp. to motion is increased by a factor γ
E Field  B Field
In O′ the charges are moving in −x′
Two current sheets create a uniform B field
in between
  Current densities are J ′ = σ ′v = γσ v
  Careful with the directions of J !
L/γ

––––––––
J′
J′
++++++++
This produces uniform B field between
the current sheets in the −z′ direction
B′⊥ = −
4π J ′
4πγσ v
ẑ = −
ẑ = −γ β × E⊥
c
c
Electric field perp. to motion creates magnetic field
7
+
+
+
+
+
+
–
L
–

–
J is parallel to x-axis and uniform in y-z
A must be parallel to x−axis and uniform in y-z
  ∇ × A = 0

–
E field parallel to motion does not create B
+
–
Electric field parallel to motion is unchanged
+
–
E′ = E
+
–
–
L
–
Distance between the plates shrink
  But the charge density remains same

–
Now orient the plates in the y-z plane
Boost again from O to O′
+
E Field in Motion
Consistency
Combining what we found so far, we have
E′ = E + γ E⊥ , B′ = −γ β × E⊥

Is this consistent with the force transformation?
Consider a charged particle at rest in the lab (O) frame


In O, the force on the particle is F = qE
In O′, the Lorentz force on the same particle is
F′ = qE′ − qβ × B′
Velocity of the particle is –v in O′
= q(E + γ E⊥ ) + qβ × (γβ × E⊥ )
= qE + qγ E⊥ + qγ ⎡⎣β(β ⋅E⊥ ) − E⊥ (β ⋅ β) ⎤⎦
= qE + qγ (1− β 2 )E⊥
= qE + q E⊥ γ = F + F⊥ γ
Consistent with the force
transformation
8
y´
y
Boosting B Fields
v
O
Create uniform B field by a solenoid in O
and observe it in O′

z
x´
x
O´
z´
Solenoid shrinks, so
the winding density
increases
n′ = γ n

How about the current?
I=

L/γ
L
ΔQ′ ΔQ I
I′ =
=
=
Δt ′ γΔt γ
ΔQ
Δt
The two effects cancel each other in B field
B=
B′ = B
4π nI
c
Magnetic field parallel to motion is unchanged
Boosting B Fields
It’s harder to boost B field perpendicular to motion
We’ve already done boosting current on a straight wire

In the lab frame O
+
+
–
v0
+
–
+
–
+
–
+
–
+
–
+
+
–
–
–
Positive ions: line density λ0 esu/cm, velocity 0
  Electrons:
line density –λ0 esu/cm, velocity v0
  In a moving frame O’

+
+
–
v0’
+
–
+
–
+
–
+
–
+
–
+
–
+
+
–
–
Positive ions: line density γλ0 esu/cm, velocity –v
v −v
  Electrons:
line density − λ0γ (1− ββ0 ) esu/cm, velocity 0
1− ββ0

9
Boosting B Fields
Net charge density λ and net current I on the wire are
  In the lab frame O
λ = λ0 − λ0 = 0
I = λ0v 0

In the moving frame O’
λ ′ = γλ0 − λ0γ (1− ββ0 ) = γββ0 λ0 =
I ′ = γλ0 ⋅v + λ0γ (1− ββ0 ) ⋅
γβ I
c
v0 − v
= γλ0v 0 = γ I
1− ββ0
E and B fields at distance r from the wire are
B⊥ =
2I
cr
2I ′
=
cr
2λ ′
E ⊥′ =
=
r
B⊥′ =
2γ I
= γ B⊥
cr
2γβ I
= γβ B⊥
cr
Magnetic field
increases
Electric field
is created
Transformation of E, B Fields
Complete set of field transformations:
E′ = E
E′⊥ = γ (E⊥ + β × B ⊥ )
B′ = B B′⊥ = γ (B ⊥ − β × E⊥ )
Striking symmetry between E and B
  Taking v in the +x direction as usual

⎧ E′ = E
x
⎪ x
⎪ E = γ (E − β B )
⎨ y′
y
z
⎪
⎪⎩ E z′ = γ (E z + β By )
⎧ B′ = B
x
⎪ x
⎪ B = γ (B + β E )
⎨ y′
y
z
⎪
⎪⎩ Bz′ = γ (Bz − β E y )
Linear transformation,
as expected
Note: E and B are not Lorentz four-vectors

(φ, A/c) is — but that will be another lecture
10
Summary
Moving charge near current receives force proportional to
its velocity
Identified as Lorentz force due to magnetic field
  Electricity and magnetism are connected through relativity

Relativistic transformation of E and B fields
E′ = E
E′⊥ = γ (E⊥ + β × B ⊥ )
B′ = B B′⊥ = γ (B ⊥ − β × E⊥ )
Linear transformation
  Consistent with the force transformation

F′ = F + F⊥ γ
11