6. Maxwell Equations and Conservation Laws
6A. Maxwell’s Equations
Inconsistency of Ampere’s Law
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Ampere’s Law says  H  J
Take the divergence:
0     H    J

But conservation of charge says:
J    0
t
So Ampere’s Law (as written) only applies to static situations
 

Substitute Coulomb’s Law   D  
  J  D  0
We now have
t 

So it is not J, but this combination that has no divergence
And this combination is therefore writeable as a curl
This suggests Ampere’s Law becomes
D
H  J 
t
Maxwell’s Equations
• We now have all four equations known as Maxwell’s Equations
• We also need constitutive equations
D  D E
relating D to E and B to H
B  B H
D  E
• Most commonly, we’ll
B  H
just assume linear relations
• In general, these will also be
functions of frequency, so
D       E  
B       H  
• In vacuum these relations are simply
• In vacuum, generally rewrite
equations in terms of only B and E
• The constant 00 has units of
an inverse velocity squared
• c is, of course, the speed of light
D   0E
B  0 H
1
 0 0  2
c
D  

E  B  0
t

H  J  D
t
B  0
  E   0

E  B  0
t

  B   0 J   0 0 E
t
B  0
c  2.99792458 108 m/s
Vector and Scalar Potentials
• The two homogenous Maxwell equations:

  E  B  0,   B  0
t
• Anything with a vanishing divergence is a curl
• Substitute this into Faraday’s law
 


0  E   A  E  A
t 
t

• Anything with a vanishing curl is a divergence
• In summary, we have
B   A
E

A  
t
B   A

E  A  
t
Gauge Choice and Gauge Transformations
B   A
• B and E are physical; they can be directly measured
• A and  cannot be measured, so they may be non-unique

E   A  
• Suppose (A,) and (A',') both produce the same electric
t
and magnetic fields, so that



B    A    A and E   A     A  
t
t
• This implies that   A  A   0
• Anything with a vanishing curl is a divergence, so A  A  
–  is an arbitrary function of space and time
• Substitute into our other equation




0   A  A                 
t
t

• This implies
   
t
• These are the equations for a gauge transformation
 

 
    
 t

A  A  

   
t
Coulomb and Lorentz Gauges
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We previously defined Coulomb gauge by the demand that Coulomb:   A  0
Suppose this is not satisfied, so  A  f  x, t 
A  A  
We want to find a gauge transformation such that  A  0

We therefore want: 0    A     f  x, t   2
   
2
Can we solve:     f  x, t 
t
We already know the
f  x, t  d 3x
  x, t   
solution to this equation:
4 x  x
1 
• Another useful gauge is Lorentz gauge, defined by Lorentz:   A  2
0
c t
• Suppose this is
1 
A  2
 f  x, t 
not satisfied, so
c t
1 
  A  2
0
• We want to find a gauge transformation such that
c t
2
• We therefore want
1 
 2   2 2   f  x, t 
c t
• We will be solving this equation soon, and it has a solution (actually several)
6B. Solving Maxwell in Vacuum
Non-Uniqueness of Solutions
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We want to solve Maxwell’s Equation in vacuum
Given  and J, can we find E and B?
There are solutions that have no sources
We can always add these homogenous solutions to nonhomogenous solutions
We will first try to get solutions assuming there are no
fields without the sources
We assume the fields can only appear after their sources
We will try to find the vector and scalar potential
This form automatically satisfies homogenous equations
Because we have gauge freedom, A and  are not unique
We will have to pick a gauge
– Coulomb or Lorentz gauge
  E   0

B0
t
1 
  B  0 J  2 E
c t
B  0
E 
B   A

E   A  
t
Equations for the Vector and Scalar Potentials
B   A
  E   0

1 
A  
 B  0 J  2 E
t
c t
• We still have two Maxwell equations to satisfy
• Substitute into each of them


2

 


2




A



     A        A   
0
t
0
t
 t

E

1 
1  2

0 J    B  2 E     A   2  2 A   
c t
c  t
t

2
1

1 
0 J      A    2 A  2 2 A  2 
c t
c t
• These equations are hard because they are coupled
• An appropriate choice of gauge can help us solve them
Solving in Coulomb Gauge


    A   2
0
t
2
1

1 
    A    2 A  2 2 A  2   0 J
c t
c t
• Let’s try solving these in Coulomb gauge, A = 0
2
1

1 
2
2
• Then these formulas     

A

A

  0 J
0
2
2
2
simplify to
c t
c t
• The first of these
  x, t  d 3x
we know how to solve:   x, t    4 x  x
0
• Treating this as completed, we
rewrite the second one as
• This is the second time we’ve
encountered the equation
• We still need to solve this
• We will do so shortly
2
1

1 
2
 A  2 2 A   0 J  2 
c t
c t
2
1

 2   2 2    f  x, t 
c t
Solving in Lorentz Gauge


    A   2
0
t
2
1

1 
    A    2 A  2 2 A  2   0 J
c t
c t
1 
• Let’s try solving these in Lorentz gauge, in which
A  2
0
c t
• Get rid of A in each equation
2
2

1

1 
1 
1 
2
2





 2
  A  2 2 A  2   0 J
2
2
 0 c t
c
t
c t
c t
• Cancel and rewrite slightly:
1 2
 A  2 2 A   0 J
c t
1  2

  2 2 
c t
0
2
1

• This is the third and fourth time
 2   2 2    f  x, t 
c t
we encountered this equation
• If we solve this, then we are done with both cases
• So, it’s time to solve it
2
2
Green’s Function for d’Alambert Operator (1)
2
• We need to solve this equation
1

2

  2 2    f  x, t 
Strategy:
c t
• Solve it for a point source
2
1

2
– I’m not including 4 like the book
   2 2    3  x  x    t  t  
c t
• We will call the solution
G  x, t; x, t 
the Green’s function
• Pretty clearly, the Green’s function will depend only on the difference between
x and x', and between t and t' , so we write
G  x, t; x, t   G  x  x, t  t  
• This simpler
1 2
2
Green’s function is
 G  x, t   2 2 G  x, t    3  x    t 
c t
a solution of
• It is certainly going to be spherically symmetric, so G  x, t   G  r, t 
• Write it in spherical coordinates:
1 2
1 2
3
rG

G





 x   t 
2
2
2
r r
c t
Green’s Function for d’Alambert Operator (2)
• Recall, we are interested in solutions
1 2
1 2
3
rG

G





 x   t 
2
2
2
where the source causes the fields
r r
c t
• This implies that for t < 0, G  r, t  0  0
  
 
we must have

  c   c   rG   0
• For t > 0, right side is zero, multiply by
r  t
r 
 t
– c2r and “factor” the derivatives
• One of the two derivatives must vanish
• The solution to this equation is of the form rG  r, t  0  f  t  r c   h t  r c 
• For t > 0, right side must be well-behaved at r = 0, so
0  f t   h t 
rG  r, t  0  f  t  r c   f t  r c 
• Function and time derivative must also be continuous at t = 0 if r > 0, so
0  f   r c   f  r c  and 0  f    r c   f   r c  for r  0
• The only solution to these equations is f(r) = constant, which then cancels out
• So effectively, f = 0 except at zero
rG  r , t  0   f 0   t  r c     t  r c  
• It must be a delta-function
Green’s Function for d’Alambert Operator (3)
rG  r , t  0   f 0   t  r c     t  r c  
1 2
3
 G  x, t   2 2 G  x, t     x    t 
c t
G  r, t  0  0
• We note that since t > 0 and r > 0, the
1
second term actually never contributes
G  r , t   f 0  t  r c 
• And for t < 0 and r > 0, first term vanishes everywhere, so
r
• To find f0, integrate the equation over time from –  to + 
2

 2  f0  r 
3

t

dt



 t 2  r  c 
  x   t  dt
t 
1 f0 
First term and right side can be
2 1
f 0  2
  t  r c   t    3  x 
done via delta-function
r c r t
Second term can be done by fundamental theorem of calculus
Second term vanishes at both limits f  2 1   3  x 
0
r
But we already know
1
1
G  r, t  
 t  r c 
So f0 = 1/4
 2  4 3  x 
4 r
r

1
2 f0


t

r
c
dt


 r 
c2
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
Green’s Function for d’Alambert Operator (4)
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2
1

1
2
3
 G  x, t   2 2 G  x, t     x    t 
G  r, t  
 t  r c 
c t
4 r
2
1

3
Now change the problem back 2G 
G



 x  x   t  t 
2
2
c t
to a different point source:
We can translate the solution in space and time:
 t  t   x  x c 
G  x, t; x, t   
1 2
To solve the more 2
4 x  x
   2 2   f  x, t 
general problem
c t
We now use superposition
 t  t   x  x c 
3
  x, t    G  x, t ; x, t   f  x, t   d xdt   
f  x, t   d 3xdt 
4 x  x
• Do the t' integral
using the delta function
  x, t   
f  x, t  x  x c 
4 x  x
d 3x
Application to Coulomb and Lorentz Gauge
2
f  x, t  x  x c  3
1


2
   2 2   f  x, t 
  x, t   
d x
c t
4 x  x
• In Coulomb
, t  d 3x
2

x

1


gauge, we have
2
  x, t   
 A  2 2 A   0 J  
4 0 x  x
• We now have the
c t
t
solution to the
1



3





A
x
,
t


J
x
,
t



x
,
t
d
x
second equation:
  



 0 



4 x  x 
 t 
 t t  x x c
2
1

• In Lorentz gauge,
 2 A  2 2 A   0 J
we have
c t
• We now have the solution to these equations
2



2  2  
t
0
J  x, t  x  x c  3
  x, t  x  x c  3
d x
  x, t   
d x A  x, t   0 
4 x  x
4 0 x  x
Sample Problem 6.1 (1)
A point charge q moves along the z-axis at velocity v. Find the fields everywhere.
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z  vt , x  y  0.
The equation of motion for the particle is
  x   q  x    y    z  vt 
The charge density is
The current density is
J  x   qvzˆ   x    y    z  vt   vzˆ   x, t 
Work in Lorentz gauge
0 J  x, t  x  x c  3
  x, t  x  x c  3
A  x, t   
d x
  x, t   
d x
4 x  x
4 0 x  x
• Scalar and vector
A  x, t   0 0vzˆ   x, t 
potential are related
• It remains to find the scalar potential
vzˆ
A  x, t   2   x , t 
c
q  x    y    z  v  t  x  x c  3
q  z  vt  v x  z zˆ c 
dz
  x, t   
d x  
4 0 x  zzˆ
4 0 x  x
Sample Problem 6.1 (2)
A point charge q moves along the z-axis at velocity v. Find the fields everywhere.
2
2
2




q

z

vt

v
x

y

z

z
c


q  z  vt  v x  z zˆ c 


dz 
  x  
dz  
2
4 0 x  zzˆ
4 0 x 2  y 2   z  z  
• Recall the rule for doing
f  z0 
integrals with delta functions:  f  z    g  z  dz  g   z 
0
• We therefore have:

v  z  z 
q

  x 
1
2
2
2
2
2
2



4 0 x  y   z  z   c x  y   z  z 
• Let’s switch to
cylindrical coordinates
 2  x2  y 2
q  2
2
  x 
   z  z  v  z  z 
4 0 
c

1
where g  z0   0.




1
Sample Problem 6.1 (3)
A point charge q moves along the z-axis at velocity v. Find the fields everywhere.
q  2
2

  x 
   z  z   v  z  z 
4 0 
c

1
2
2
• In this equation, z' is the root of


z  vt  v    z  z  c  0
the argument of the -function
• Do a LOT of algebra:
q
vzˆ


A 2
• Recall our expression
2
2
2
2
4 0  z  vt    1  v c 
c
relating vector and
scalar potentials:

vzˆ 


1 
• Now we find the
E   A     2   zˆ   ρˆ
  φˆ

electric field:
t
c t
z

 
E
q   v 2  z  vt  zˆ c 2   z  vt  zˆ  1  v 2 c 2   ρˆ 
4 0  z  vt    1  v c  


2
2
2
2
3/2
q 1  v 2 c 2   z  vt  zˆ   ρˆ 

3/2
2
2
2
2
4 0  z  vt    1  v c  


Sample Problem 6.1 (4)
A point charge q moves along the z-axis at velocity v. Find the fields everywhere.
2
2
q
1

v
c
 z  vt  zˆ   ρˆ 


ˆ
vz

A 2  E
2
2
2
2
3/2
2
4 0  z  vt    1  v c 
2
2
2
c
4 0  z  vt    1  v c  


2
2
vq

1

v
c
φˆ


• Now for
Az
1 Az

 φˆ
B   A  ρˆ
3/2
2
the magnetic
2 
2
2
2 
 

4 0 c  z  vt    1  v c 


fields:
q
B
0vq  1  v 2 c 2  φˆ
4  z  vt    1  v c  


2
2
2
2
• Interestingly, electric fields
always point directly away
from the charge’s current position
3/2
Comments on Fields from a Point Charge
•
•
•
•
It is interesting that E always points away from the current position of the charge
According to relativity, information must travel no faster than light
The formulas we used do not contradict this:
  x, t  x  x c  3
  x, t   
d x
Suppose at t = 0, the charge
4 0 x  x
suddenly stopped at the origin
• At this point the electric fields all
point back towards the origin
• For a while, they must continue to
point as if the charge kept moving
• Until the information reaches them
• There is a smarter/easier way to do this
whole computation
– Using relativity
Relativity and Gauge Choice
• According to relativity, information cannot travel faster than light
• In Lorentz gauge, the scalar and vector potentials respond to the charge
distribution and current at the retarded time
  x, t  x  x c  3
J  x, t  x  x c  3
  x, t   
d x
A  x, t   0 
d x
4 0 x  x
4 x  x
• Lorentz gauge is, in fact, Lorentz invariant
• In contrast, in Coulomb gauge, the potential
  x, t  d 3x
  x, t   
reacts instantaneously to the charge distribution
4 0 x  x
– Apparent causality violation!
• Nonetheless, E and B are invariant, and hence the same in all gauges
• Causality violation in  cancels causality violation in A in Coulomb gauge
• Coulomb gauge is manifestly not Lorentz invariant
6C. Cons. of Energy & the Poynting Vector
Energy Density
• For linear media, we know the energy density:
• Even when this is not true, we were able to
demonstrate that the rate of change of the total
electromagnetic energy is
• So we would guess that the local
energy density satisfies
• For electric charge conservation
we had a local version:
• Can we find a similar formula for
conservation of electromagnetic energy?
u  12  E  D  B  H 
dU
B  3
 D
  E 
 H d x
dt
t 
 t
u
D
B
 E
 H
t
t
t

J 
0
t
The Poynting Vector
u
D
B
 E
 H
t
t
t

E  B  0
t

and   H  J  D
t
• Faraday’s Law and Ampere’s Law:
• Rewrite using these
u
 E     H  J   H   E
t
• Pull out one of those fancy product rules:   E  H   H  E  E  H
• We therefore have
u
 E  J     E  H 
t
• We define the Poynting vector:
• We therefore have:
S  E H
u
 S 
 E  J
t
Where Does the Energy Go?
S  E H
u
 S 
 E  J
t
• This looks like conservation of energy, except for the right-hand side
• For example, if we
d
U field    S  nˆ da   E  J d 3x
integrate it over a
S
V
dt
volume V, it yields
• Second term on the right is work being done on d
U mech   E  J d 3 x
V
particles being pushed by electric fields
dt
d
U field  U mech    S S  nˆ da
dt
• Mechanical energy may appear as macroscopic motion, heating, etc.
• Right hand side is now flow of energy out of the region
– Poynting vector shows the direction energy is flowing
Sample Problem 6.2 (1)
A point charge q moves at velocity v. Find the energy flow out of a cylinder of radius a
centered on the path of the charge (a) ahead of the charge, and (b) behind the charge
• We can arbitrarily have the charge move along z-axis, then we know the fields:
q 1  v 2 c 2   z  vt  zˆ   ρˆ 
E
3/2
2
2
2
2
4 0  z  vt    1  v c  


0vq  1  v 2 c 2  φˆ
B
4  z  vt    1  v c  


2
2
2
2
3/2
1
• The Poynting vector is then S  E  H  0 E  B

  z  vt  zˆ  ρˆ   φˆ  q v 1  v c    zˆ   z  vt  ρˆ 
 z  vt    1  v c  
16   z  vt    1  v c  




q v  1  v c
2
2
16  0
2
2 2
• The flow out of
the cylinder is
2
2
2
2
2
3
2
2 2
2
2
2
2
3
2
0

S
q v  1  v c
2
S  nˆ da  S  ρˆ da
2
16 2 0

2 2
2
 
0
  z  vt   d dz
z  vt 

2
  1  v c  

2
2
2
3
Sample Problem 6.2 (2)
A point charge q moves at velocity v. Find the energy flow out of a cylinder of radius a
centered on the path of the charge (a) ahead of the charge, and (b) behind the charge
q v  1  v c
2
 S  nˆ da 
2

2 2
16 2 0
S
  z  vt   d dz
2
 
0
z  vt 

2
  1  v c  

2
2
2
3
• Let radius be a, and change variables to w  z  vt
q v 1  v c
2

S
S  nˆ da 
2

2 2
a
8 0
• We now substitute limits
– From 0 to  ahead of
– From – to 0 behind
2

 w dw

 w  a 1  v c  
2
2
q v 1  v c
2
2
2
3

2

2 2
a2
32 0  w  a 1  v c  
q 2 v
ahead S  nˆ da  8 0a 2 ,
2
2
2
2
q 2v
behind S  nˆ da  8 0a 2
2
6D. Conservation of Momentum in E and M
Rate of Change of Mechanical Momentum
Is there a similar formula for conservation of momentum?
• We expect three formulas, one for each component of p
• The forces on particles is given by
F   qi E  xi   v i  B  xi  
• If we have a continuum of
i
different types of particles with
F    ni  x  qi E  x   v i  x   B  x   d 3x
i
number density ni, this becomes
• This expression can be
  x    ni  x  qi , J  x    ni  x  qi vi  x 
simplified using
i
i
• So we have
F    E  J  B  d 3x

• Force is the time derivative of momentum, so
d
p mech     E  J  B  d 3x
dt
Go Nuts with Maxwell’s Equations
•
•
•
•
d
p mech     E  J  B  d 3x
dt

  D   and   H  J  D
Use Ampere’s Law and Coulomb’s Law
t
So we have:
d

   3

p mech   E    D      H  D   B  d x
V
dt
t  



D
B
Use the product rule
D B   B  D
t
t
t
So we have d

B


3
p mech  
D  B  d x   E    D   D 
    H   B  d 3x

V t
V
dt
t


B
• Use Faraday’s Law, and
E 
 0 and   B  0
t
Gauss’s Law for magnetism
d
d
p mech    D  B  d 3x   E    D   D     E   H    B   B     H   d 3x
V
dt
dt V
When the Going Gets Tough:
d
d
p mech    D  B  d 3x   E    D   D     E   H    B   B     H   d 3x
V
dt
dt V
• We’d like to interpret second term on the left as momentum
• Then we would have momentum density: g  D  B
• To make this identification, we need to show right side is a total divergence
• It is tricky because there may be momentum shared with the medium
• Let’s assume we are in vacuum:
d
d
p mech   gd 3x    0 E    E   E     E    0  H    H   H     H   d 3x
V
dt
dt V

• There is clearly a similarity between the E and the H terms

Can We Write It as a Divergence?


d
d
p mech   g d 3x    0 E    E   E     E    0  H    H   H     H   d 3x
dt
dt
• Consider just the z-component of the E-terms
E y
 Ex E y Ez 
Ex
Ez
Ez


 Ex
 Ey
 Ey
E    E   E    E  z  Ez 
  Ex
y
z 
x
z
y
z
 x


1 
  Ez Ex    Ez E y  
Ez2  Ex2  E y2      Ez E   1  E2     E E  1 zˆ E2 

z
2
x
y
2 z
2 z
• Very similar formula works for the H-terms
• Define the z vector components Tz   0 Ez E  0 H z H  12 zˆ   0E2  0 H 2 
of the stress tensor
d
d
• Then we have
pmech,z   g z d 3 x     Tz d 3x
dt
dt
The Maxwell Stress Tensor
2
2
d
d
1
3
3
ˆ
T


E
E


H
H

z

E


H


z
0 z
0
z
0
0
2
 p mech  z   g z d x     Tz d x
dt
dt
g  D B
d
d
3
p

g
d
x   nˆ  Tz da




mech z
• Use the divergence theorem:

z
S
dt
dt V
Interpretation:
• gz is the electromagnetic momentum density in the z-direction
• Tz is (minus) the flow of z-momentum
• Generalize this to get
Tij   0 Ei E j  0 H i H j  12  ij  0 E2  0 H 2 
the Maxwell stress tensor:
d
d

• Then we find:
3
pmech,i   gi d x   
Tij d 3x
V
• So g is momentum
dt
dt V
j x j
density
• And Tij is (minus) the flow of i-momentum in the j-direction
Comments on Momentum Density
g  D B
• Recall, we have really only worked this out for vacuum
– Jackson says it gets complicated otherwise
• So we have:
1
g   0E  0 H  2 E  H  1 S
c
c2
g
1
S
2
c
6E. Symmetries of Fields
Rotations
• It will be helpful, for the moment, to
x   x, y, z    x1 , x2 , x3 
denote components of a vector by subscripts:
• We believe the universe is invariant under translations and rotations
– Laws of physics do not depend on choice of origin or choice of directions
for axes
• What makes rotations and translations the correct transformations?
• Consider the distance formula
2
2
2
2
s 2   x  y    x1  y1    x2  y2    x3  y3 
between two points x and y:
• If you translate both x and y by the same amount, x  x  x  a
the distance between them won’t change
• If you rotate them with a rotation
xi  xi   Rij x j
matrix R, they will also not change
j
– But only if we pick the matrix R judiciously!
Proper and Improper Rotations
xi  xi   Rij x j
• We want to keep, for example, distance to origin fixed x  x
• We therefore have:
2


2
2
Rij x j  Rik xk   x j xk  Rij Rik
i xi  i xi  i  j Rij x j   
j
k
k
i
j
k


• This will work if (and only if)
 Rij Rik   jk
2
2
j
k
• Written as a 33 matrix, this implies RT R  1
• Take the determinant of this equation, using the fact that det(RT) = det(R)
1  det  RT R   det  RT  det  R   det  R  
det  R   1
Those with det(R) = +1 are called proper rotations
– They can be achieved by a combination of small rotations
 1 0 0 


Those with det(R) = –1 are called improper rotations
P   0 1 0 
Simple example of improper is the parity rotation
 0 0 1


All other improper are combinations of parity plus proper
2
•
•
•
•
Scalars, Vectors, and Tensors
xi  xi   Rij x j
j
• So, we say that (for example) energy U is a scalar and momentum p is a vector
– What does this mean?
U U   U
• A scalar quantity is a quantity that
is unchanged when you rotate it
pi  pi   Rij p j
• A vector quantity is a quantity that
j
rotates like the coordinates
Tij  Tij   Rik R jlTkl
• We can also have (rank 2) tensors with two
k
l
indices, both of which must be rotated
Tij  xi p j
• You can make them out of lower rank tensors, for example
• This can be generalized to tensors of arbitrary order
– A scalar is a rank-0 tensor
– A vector is a rank-1 tensor
– Rank-2 tensors are sometimes simply called tensors
Scalars, Vector, and Tensor Fields
xi  xi   Rij x j
j
• We also talk about  as a scalar function and A as a vector function
– What does this mean?
• A scalar function, under a rotation,   x     x    x    RT x


has its coordinates change
• A vector function also has its
components get mixed up
Ai  x   Ai  x    Rij Aj  x 
• You can make tensor functions as well
j
Tij  x   Tij  x    Rik R jlTkl  x 
k
l
Invariant Tensors
There are a some tensors that are invariant under rotations
• For example, if we rotate the Kronecker delta-function, we get
 ij   ij   Rik R jl kl   Rik R jk   RRT 
ij
k
l
k
  ij
1 even perm. of 123

 1 odd perm. of 123
0
otherwise

• Consider the Levi-Civita tensor, defined by
 ijk
• It is completely anti-symmetric, that is
if you change any pair of indices you get
same thing times –1
 ijk   ijk   Rii R jj  Rkk  ij k 
• Under a rotation, this becomes
i
j k 
• Easy to show that ' is completely antisymmetric
– Which means it is proportional to 
 ijk  det  R   ijk
• The proportionality constant is the determinant
– This is definition of determinant
• So it is invariant under proper rotations but not improper rotations
Dot and Cross Products, Pseudovectors
Suppose we wanted to multiply two vectors a and b
• We could simply make a rank-2 tensor out of them:
• Or we could combine them using Kronecker delta:
Tij  ai b j
S    ij ai b j  a  b
i
j
• Or we could combine them using Levi-Civita: v 
 ijk ai b j v  a  b

k
• That’s why these are the particular ways we
i
j
normally combine them
• Not hard to show that S transforms as a scalar under all rotations S  S   S
• In contrast, if a and b are two vectors,
vi  vi  det  R   Rij v j
because Levi-Civita has a minus sign
j
• A vector that behaves like this is called a pseudovector
– Those that don’t behave this way are called true vectors or vectors
• We can similarly make scalar quantities with odd behavior
– For example, consider the triple scalar product of three true vectors
P  a  b  c
• Under rotation, pretty easy to see that
P  P  det  R  P
Examples
• True vectors and true scalars behave the way
you expect when you perform an improper rotation
• Consider, for example, a rotating moving object
• Reflect it in a mirror
• The velocity will reflect like you expect
v
• But the angular momentum is reversed!
• This implies angular momentum is a pseudovector
• Not surprising, since
L  rp
L
vR
LR
Symmetry of Gradient and Divergence
xi   Rij x j
• We can take derivatives to change fields of one type into another
• We note that
xi
xi
 Rij and
 RijT  R ji
x j
xj
j
xi   RT ij xj
j
• Consider first the gradient
  x     x    x 
• Suppose  is a true scalar field
x j 

• Then  transforms 



x


  Rij
  x
  x 
  x   x x
as a true vector:
j
x j
i
j
xi
xi
j
  x i   Rij   x j
• Now consider the divergence
j
• The divergence converts a true vector into a true scalar

 
xi 

Rij Aj  x   
Aj  x  
 A  x     A  x  


 x Aj  x 
i xi  j
i
j x j xi

j
 A  x     A  x   A  x 
j
The Curl

• The curl is a bit more complicated
Ak  x 
  A  x   k    ijk
xi
• Let it act on a true vector field,
i, j
then when we perform a rotation, we have
x

  A  x   k    A  x   k    ijk  Ak  x     ijk  i  R jj
Aj   x 
xi
xi
i, j
i  xi j 
i, j


Aj   x 
   ijk Rii R jj 
Aj   x      ijk  Rkk  Rk k  Rii R jj 
xi
xi
i , j i
j  k  , k 
i , j i
j
• Use the fact that RRT = 1 to rewrite this as
• We already know how the Levi-Civita symbol rotates




Aj   x   det  R   Rkk    A  x   k 
  A  x   k  det  R    ijk  Rkk 
xi
k
i
j k 
• So it turns vectors into pseudovectors
Electricity and Magnetism Under Rotations
• Are the laws of electricity and magnetism invariant under rotations?
• We have to make sensible guesses of how all electromagnetic fields quantities
transform under rotations
• If there is any sense to it, the number
ni  x   ni  x  ni  x 
density of particles is a true scalar field
qi  qi  qi
• And charge is a true scalar
• Therefore, charge density will be a scalar:
  x    qi ni  x 
• Velocity is true vector quantity
i
• Therefore, current is a vector field
J  x    qi ni  x  vi  x 
• If you look at the Lorentz force formula:
i
F  q  E  v  B
• Since F = ma, and a is a true vector, E must be a true vector.
• If B were a true vector, then v  B would be a pseudovector, which is wrong
• So B must be a pseudovector
Maxwell’s Equations and Time Reversal

•
•
•
•
•
•
True scalar
  E   0
J
True vector

E  B  0
E
E, D, P True vector
t
B
B, H, M Pseudovector
Pseudovector

  B   0 J   0 0 E
Let’s look at Maxwell’s Equations in vacuum
t
Coulomb’s Law: both sides are true scalar
B  0
Faraday’s Law: all terms are pseudovector
Ampere’s Law: all terms are true vector
Gauss’s Law for B-fields: left side is pseudoscalar
Summary: All equations are invariant under both proper and improper rotations
•
•
•
•
•
What if we are in a medium?
For example, look at relationship between D, E and P
D   0E  P
This implies that D and P must also be true vectors
B  0  H  M 
Similarly, H and M must be pseudovectors
Everything still works, though some materials may lack certain symmetries
• So far we
have the
following guesses:
Some Other Quantities
• What about scalar and vector potential?

True scalar
• These depend on gauge choice
J
True vector
• Gauge choice could violate rotational invariance
E, D, P True vector
– Axial gauge is defined by A3 = 0, for example
B, H, M Pseudovector
• But the two choices we made are rotationally invariant 
True scalar
• We recall that
E    A t
A
True vector
• This implies that  is a true scalar and A a true vector
u
True scalar
• The latter is consistent with B   A
S, g
True vector
• Energy density is:
u  12  E  D  B  H 
Tij
True tensor
• So it’s a true scalar
• Poynting vector and momentum density S  E  H , g  S c 2
• True vectors
Tij   0 Ei E j  0 H i H j  12  ij  0 E2  0 H 2 
• Maxwell-stress tensor:
• True (rank-2) tensor
Using Symmetries to Solve Problems
• Whenever faced with a new problem, try to identify symmetries of the problem
• Try to choose coordinates that maximize your ability to take advantage of
symmetries
• First, use any translations or proper rotations to argue which coordinates must
be irrelevant
• You may also be able to use these to completely eliminate certain components
of the potential answer
• Then consider one or more improper rotations to restrict it further
– Being careful to remember which fields are pseudovectors!
• Then you can often use simple formulas (like integral versions of Gauss’s Law
or Ampere’s Law) to finish the problem.
– Try to orient your Gaussian surface or Ampere loop so it picks up any
unknown field
Sample Problem 6.3 (1)
A solid infinite cylinder of charge has radius a and uniform
charge density  throughout. Find the electric field everywhere.
• Pick coordinates
– Cylindrical, with the z-axis along the axis of the cylinder
• The problem is translation independent in z-direction
– It must have same answer if z  z + k
• The problem is rotation independent about the z-axis
– It must have same answer if    + k
E  E    ρˆ  E    φˆ  Ez    zˆ
• We now know
• Now let’s try reflection across the z = 0 plane
• This is improper, but E is a true vector
• This reverses z-direction, but leaves  and  unchanged
• So the electric field will be
E  E    ρˆ  E    φˆ  Ez    zˆ
Ez  0
• And therefore
• Similarly, you can reflect across y = 0 plane, reversing  E  0
Sample Problem 6.3 (2)
A solid infinite cylinder of charge has radius a and uniform
charge density  throughout. Find the electric field everywhere.
• We now have
E  E    ρˆ
• Pick an appropriate Gaussian surface
• Gauss’s Law tells us
 r L  q V    0 S E  nˆ da  2 0 rLE
2
E 
• Repeat for a Gaussian surface outside the cylinder
• Combine to a final formula
ρˆ
E
2 0
 r
 2
a r
r  a,
r  a.
r
2 0
Sample Problem 6.4 (1)
A tokamak is in the shape of a rectangular cross-section donut centered on the z-axis, as
sketched in the cutaway view. A total current I is then sent around the rectangular
direction of the tokamak, so it goes up through the hole in the center, across the top,
down on the outside, and then back to the center, equally at all angles. Find B at all
points inside or outside the tokamak.
• Use cylindrical coordinates passing through the middle
• No translation symmetry, but rotation around the z-axis
• All quantities must be independent of 
ˆ  Bz   , z  zˆ
B  B   , z  ρˆ  B   , z  φ
•
•
•
•
•
Current is also symmetric under reflection across the xz-plane
Under this reflection, z and  are unchanged, but  is reversed
B  B ρˆ  B φˆ  Bz zˆ
If B were a true vector, this would imply
But B is a pseudovector, so instead,
B   B ρˆ  B φˆ  Bz zˆ
This implies that
B  0  Bz
Sample Problem 6.4 (2)
A tokamak is in the shape of a rectangular cross-section donut centered on the z-axis, as
sketched in the cutaway view. A total current I is then sent around the rectangular
direction of the tokamak, so it goes up through the hole in the center, across the top,
down on the outside, and then back to the center, equally at all angles. Find B at all
points inside or outside the tokamak.
B  B   , z  φˆ
• Our field is now
• Pick an Ampere loop in the  direction
• For example, to get it in the “cake” of the tokamak,
how about this loop?
• The full current passes through this loop, so
0 I
ˆ

2

B

B

φ

d

 0 I   B  dl
B  in  


2
• A similar loop that is not in the “cake” of the donut can also be drawn
 0 Iφˆ 2 inside
B  out   0
B
0
outside

Electricity and Magnetism and Time Reversal
• Are the laws of electricity and magnetism invariant under time reversal?
• We have to make sensible guesses of how all electromagnetic fields quantities
transform under time reversal
t  t   t
• If there is any sense to it, the number
ni  x, t   ni  x, t   ni  x, t 
density of particles is unchanged under time reversal
• And charge remains unchanged
• Therefore, charge density will be invariant:   x, t     x, t 
• Velocity though, will change sign
J  x, t   J  x, t 
• Therefore, current changes sign
• If you look at the Lorentz force formula:
F  q  E  v  B
• Since F = ma, and a is the second derivative, it will be unchanged
E  x, t   E  x, t 
• Therefore E must be unchanged
• But since v changes sign, B must change sign B  x, t   B  x, t 
Maxwell’s Equations and Time Reversal

•
•
•
•
•
•
++
  E   0
JJ
––

E  B  0
E,
E D, P ++
t
B,
B H, M –

  B   0 J   0 0 E
Let’s look at Maxwell’s Equations in vacuum
t
Coulomb’s Law: both sides are positive
B  0
Faraday’s Law: all terms are positive
Ampere’s Law: all terms are negative
Gauss’s Law for B-fields: left side is negative
Summary: All equations are invariant under time reversal
•
•
•
•
What if we are in a medium?
For example, look at relationship between D, E and P
This implies that D and P must also be positive
Similarly, H and M must also be negative
• So far we
have the
following guesses:
D   0E  P
B  0  H  M 
Some Other Quantities
•
•
•
•
This is getting boring
Sometimes, time reversal can be useful to solve the problem
But often, boundary conditions are not time symmetric
Consider the fields from an arbitrary source in Lorentz gauge:
  x, t  x  x c  3
  x, t   
d x
4 0 x  x
• We explicitly assumed the fields were “caused” by the charges
• This solution is not time symmetric, even if the problem is
• The laws of E and M are time symmetric
• It is a deep question why the universe around us is not

+
J
–
E, D, P +
B, H, M –

A
u
S, g
Tij
+
–
+
–
+