March 21, 2011
Turn in HW 6
Pick up HW 7: Due Monday, March 28
Midterm 2: Friday, April 1
4-vectors and Tensors
Four Vectors
x,y,z and t can be formed into a 4-dimensional vector with
components
0
x  ct
x1  x
x  y
2
x3  z
Written


x 
0
,
1
,
2
,
3
4-vectors can be transformed via multiplication by a 4x4 matrix.
The Minkowski Metric
1
0
  
0

0
0 0 0

1 0 0
0 1 0

0 0 1
Or



1if
0


1if
1,2
0if

Then the invariant s
s


c
t

x

y

z
2 2
2222
can be written
3
3
s



2


0

0

xx

It’s cumbersome to write
3
3
s



2


0

0

xx

So, following Einstein, we adopt the convention that when Greek
indices are repeated in an expression, then it is implied that we
are summing over the index for 0,1,2,3.
(1) becomes:


s
xx
2
(1)
Now let’s define xμ – with SUBSCRIPT rather than SUPERSCRIPT.
Covariant
4-vector:
x
Contravariant
4-vector:
x

x 0   ct
x 0   ct
x1  x
x1  x
x2  y
x3  z
More on what this means later.
x  y
2
x3  z
x  x

So we can write
x  x







i.e. the Minkowski metric,
can be used to “raise” or “lower” indices.
2

s

x
x
Note that instead of writing
we could write

s
xx

2
assume the Minkowski metric.
The Lorentz Transformation
  
 


 
0
0

0
0
where
0
0
1
0
0
0

0

1
v
1

 and
 2
c
1


Notation:
F
00
F
 10
F


F
20

30
F
F

01 F
02 F
03

F
11 F
12 F
13


F
21 F
22 F
23

F
31 F
32 F
33

Instead of writing the Lorentz transform as
x   ( x  vt )
y  y
z  z
v
t    (t  2 x )
c
we can write




x

x

0 0
ct
t

x
 


 


x
 




0
0


x


x


 


0

y
  y 
0 10


 

0 01
0

z  z 
or
ct   ct   x
x  x   ct
y  y
z  z
We can transform an arbitrary 4-vector Aν




A
A





0for
1
for

 


Kronecker-δ
Define
1
0
 
0

0
Note:
(1)
0 0 0
1 0 0

0 1 0

0 0 1
 



 
(2) For an arbitrary 4-vector

A

A

A



Inverse Lorentz Transformation
~ 

We wrote the Lorentz transformation for CONTRAVARIANT 4-vectors as




x

x

The L.T. for COVARIANT 4-vectors than can be written as
~

x


x

Since

where

~

 




 
s
xx
 is a Lorentz invariant,
xx xx
  ~


 x x x x
 ~

or
 
Kronecker Delta
2

General 4-vectors
Transforms via
A
(contravariant)
 

A


A


A
A

Covariant version found by
Minkowski metric
Covariant 4-vectors transform via
~


A


A

Lorentz Invariants or SCALARS

A  A
Given two 4-vectors

and
B  B
SCALAR PRODUCT




A

A
BB

This is a Lorentz Invariant since

A


B  

 

~
A  B

~ 
 A B
 
  A B

 A B

Note:
A A
can be positive (space-like)
zero
(null)
negative (time-like)
The 4-Velocity

dx
u 
d

(1) The zeroth component, or time-component, is
0
dx
dt
0
u 
c 
c

d
 d
 u
and
where
u 
1
u2
1 2
c

u

u

mag
of
ord
the
el
dxdydz
, ,
dt dt dt
Note: γu is NOT the γ in the Lorentz transform which is

1
v2
1 2
c

dx
u 
d

The 4-Velocity
(2) The spatial components
i
dx
i
i
u 
uu
d

So the 4-velocity is
c
uuu


where
u 
1
u2
1 2
c

u

ord
the
el
So we had to multiply by u to make a 4-vector,
i.e. something whose square is a Lorentz invariant.
dxdydz
, ,
dt dt dt



u

u

How does
so...
u
transform?
u0   (u0   u1)
u1   (u0 u1)
u2  u2
u3  u3
where


u
u
2


u
 1 
2

c


u
 1 

c


v
   1  2
c

2
2
2











or
uc(cu uu )
1
1

uu (cu uu )
uu2 uu2
uu3 uu3
1/ 2
1 / 2
1 / 2
where v=velocity between frames
1
Wave-vector 4-vector
Recall the solution to the E&M Wave equations:



E

exp(
i
k

r

it
)
The phase of the wave must be a Lorentz invariant since
if E=B=0 at some time and place in one frame, it must also
be = 0 in any other frame.
/c


k 
 k 



Tensors
(1) Definitions
zeroth-rank tensor
first-rank tensor
second-rank tensor
Lorentz scalar
4-vector
16 components:
(2) Lorentz Transform of a 2nd rank tensor:





T


T


T 
  0,1,2,3
  0.1.2.3
(3)
T 
T
contravariant tensor
covariant tensor
related by



T

T



transforms via
~
~



T



T




(4) Mixed Tensors
one subscript -- covariant
one superscript – contra variant




T
T


 
T
 T

so the Minkowski metric “raises” or “lowers” indices.
(5) Higher order tensors (more indices)
T
T



etc
(6) Contraction of Tensors
Repeating an index implies a summation over that index.
 result is a tensor of rank = original rank - 2
Example:

T
is the contraction of
 
T
(sum over nu)
(7) Tensor Fields
A tensor field is a tensor whose components are
functions of the space-time coordinates,
0 1 2 3
x
,x
,x
,x
(7) Gradients of Tensor Fields
Given a tensor field, operate on it with

x 
for x   x 0, x1, x 2, x 3
to get a tensor field of 1 higher rank, i.e. with a new index

Example:
if


scal ar
then

x 
We denote

x 
is a covariant 4-vector
as
,
Example:
if

T
is a second-ranked tensor




T , 

x
where

third rank tensor



co
the
o
T
m
f
(8) Divergence of a tensor field
Take the gradient of the tensor field, and then contract.
Example:

A
Divergence is
A ,

Divergence is
T,
Given vector
Example:
Tensor
T
(9) Symmetric and anti-symmetric tensors
If
If
 
T
T


T

T
then it is symmetric
then it is anti-symmetric
COVARIANT v. CONTRAVARIANT 4-vectors
Refn: Jackson E&M p. 533
Peacock: Cosmological Physics
Suppose you have a coordinate transformation which relates
or


x
to
x
0
1
2
30
1
2
3




x
,
x
,
x
,
x

x
,
x
,
x
,
x
by some rule.
A COVARIANT 4-vector, Bα, transforms “like” the basis vector, or
0
1
2
3


x

x

x

x

x
or

  B
B

B

B

B

B
B

0
1
2
3






 
 
 


x
x
x
x

x
A CONTRAVARIANT 4-vector transforms “oppositely” from the basis
vector

x 
  A
A

x

For “NORMAL” 3-space, transformations between e.g. Cartesian
coordinates with orthogonal axes and “flat” space 
NO DISTINCTION
Example: Rotation of x-axis by angle θ
y’
y
x
x’
But also
so
x 
dx
dx
x 
dx
dx
dx
dx
x cos 
 cos 
x  cos 
 cos 
dx

dx
Peacock gives examples for transformations in normal flat 3-space
for non-orthogonal axes where
dx dx

dx dx
Now in SR, we add ct and consider 4-vectors.
However, we consider only inertial reference frames:
- no acceleration
- space is FLAT
So COVARIANT and CONTRAVARIANT 4-vectors differ by

A


A
 
Where the Minkowski Matrix is
1
0
  
0

0
0 0 0
1 0 0

0 1 0

0 0 1
So the difference is
the sign of the time-like
component
Example: Show that xμ=(ct,x,y,z) transforms like a contravariant vector:
xx
ct
ctct

x



x

  x
x

x
x
 0 
 1 
 2 
 3

x
x
x
x
 0x 1x 2x 3x

x

x

x

x
Let’s let
0


x
ct
'
x 0
x ct
0
x
x  1
x  x
1
x
In SR
In GR

x

x
 

A
g
A


g

th
metric
ten
e
so

Gravity treated as curved space.
Of course, this type
of picture is for 2D space,
and space is really 3D
Two Equations of Dynamics:
2


d
x dx
dx



0

2
d

d
d


22


c
d

g
d
d
x
 x
where
and


proper
e
tim

g

g



g
1











g



2

x 
x 
x

= The Affine Connection, or Christoffel Symbol
For an S.R. observer in an inertial frame:
1
0

g


0

0
0
1
0
0
0
0
1
0
0
0

0

1
And the equation of motion is simply
2 
dx
0
2
d
Acceleration is zero.
Covariance of Electromagnetic Phenomena
Covariance of Electromagnetic Phenomena
4-current and 4-potentials
Define the 4-current
where

j

c

j 


j
 

= 3-vector, current
= charge density
Recall the equation for consersvation of charge:
In tensor notation

j ,0

 


j0

t
Let’s look at all this another way:
Consider a volume element (cube) with dimensions
containing N electrons
Charge in the cube = N e
Charge density
Ne
0  3
l0
Suppose in the K’ frame, the charges are at rest, so that the
current
j0 0
What is the current in the K frame? Assume motion with velocity v,
parallel to the x-direction
l0
l0
l0

The volume in K will be
l0 1 

v 2
c
 l0  l0  l0
1 
length contraction
in one direction
The number of electrons in the volume must be the same in both
the K and K’ frames. Thus,

Ne
l0 1 
3

v 2
c

0
1 

v 2
c
3

v 2
c
Similarly, for the current:
j  v
Nev

l0
3
1 

v 2
c


0 v
1 

v 2
c
Now, in analogy to the expression for proper time:



2
22
2222
c

c
t

x

y

z
we can write


c 2 2  jx  jy  jz  c 20
2
2
2
2
The transformation equations are:

jx 
j x  v
1- 

v 2
c
jy  j y

jz  j z
  
  vj x /c 2
1- 

v 2
c
4-Potential
Recall the vector potential
which satisfied

A scalar potential
and

2

1  A 4 
2
A 2 2  j
c t
c
1 2
2
 2 2 4
c t
and the Lorentz guage
Define the 4-potential
 1


A


0
c
t
 
 
    Ax 

A       
 A  Ay 
A 
 z

 1


A


0
c
t
can be written

2

1A
4 
2
A 2 2 
j
c t
c
1 2
2
 2 2 4
c t

A,
0
become

4 

,
A ,
 j
c
4-vector

Wh
about
E
an
B
?
at
d
Recall that E and B are related to A and φ by


BA


1A
E

c t
E and B have 6 independent components, so we’ll write
the Electro-magnetic force as an anti-symmetric 2nd rank tensor
with 6 independent components:
F

A
,

A
,





F

A
,

A
,





Can show that


E
E
 0 E
x 
y 
z


0 B
B
E
x
z 
y
F



E

B
0

B
z
x
 y

E B B 0 
y
x
 z

We can re-write Maxwell’s Equations

E4

 1E 4

B
 j
c t c

B0

 1B
E
0
c t
4

F
j
 
c
,
F

F

F

0

,


,


,

More Notation:
Instead of writing
F

F

F

0

,


,


,

Write
F
0

,

where [ ] means: permute indices
even permutations  + sign
odd permutations  - sign
E.G.:
A
A

A





Transformation of E and B:
Lorentz transform Fμν
Let
~
 ~


F




F


vvelocity
between
frame
K'
and
K

 v

c
 


E
component
of
E
,B
parallel
to
v
|,
|B
| |
 

E
,B
component
perp.
to
v




 
 
E
E
B
B
||
||
||
||

   
  
E

E


B
B

B


E




 

 

or, for v = velocity in x-direction

Ex  Ex

Ey   Ey  vBz 
Ez   Ez  vBy 


Bx  Bx
vE z 

By    By  2 
c 

vE y 


B z    B z  2 
c 


NOTE:
The concept of a pure electric field (B=0) or a pure magnetic
field (E=0) is NOT a Lorentz invariant.
if B=0 in one frame, in general
0 in other frames

E
0and B

What is invariant?
(1)
F F
i.e.
(2)

r2 r 2
2 B E



2 2 2 2is an invariant


B

E

B

E
r r
det F  E  B

so


2

 is an invariant


E

B

E

B
Fields of a Uniformly Moving Charge
If we consider a charge q at rest in the K’ frame,
the E and B fields are
qx
Ex  3
r
 qy
Ey  3
r
 qz
Ez  3
r

where

Bx  0

By  0

Bz  0
r  x   y   z 
3
2
2
2 3/2
Transform to frame K
We’ll skip the derivation: see R&L p.130-132
The field of a moving charge is the expression we derived from
the Lienard-Wiechart potential:
r 
E  q


r2
r r
n   1  

r3 2

 R



We’ll consider some implications 


 


1

n

Consider the following special case:
(1) charged particle at x=y=z=0 at t=0
v = (v,0,0) uniform velocity
(2) Observer at x = z = 0 and y = b sees
Ex  
Ey 
qvt
 v t
2 2 2
b

2 3/2
qvb
 v t
2 2 2
b
Ez  0
Bx By 0
Bz Ey

2 3/2
What do Ex(t), Ey(t) look like?
Ex(t) and Ey(t)
(1) max(Ey) >> max (Ex) particularly for gamma >>1
(2) Ey, Bz strong only for 2t0
(3)
max(
Ey)
max(
Ex)

t0 
b
v
As particle goes faster, γ increases,
E-field points in y-direction more
Ex(t) and Ey(t)
(4) The observer sees a pulse of radiation, of duration
(5) When gamma >>1, β~1 so

(6) To get the spectrum that the observer sees, take the
B
E

E
z
y
y
fourier transform of E(t)  E(w)
We can already guess the answer 

2b
v
The spectrum will be
where
ˆ()
E
dW ˆ 2

c
E
(

)
dAd

is the fourier transform
1
Eˆ ( ) 
E y (t)e it dt

2
qb  2 2 2
2 3 / 2 it

 v t  b  e dt


2 
The integral can be written in terms of the modified Bessel function
of order one, K1



qb

b

ˆ


E
(

)

K
1



b
v

v

v

The spectrum cuts off for
~
v
b
Rybicki & Lightman give expressions for
dW
d
and some approximate analytic forms -- Eqns 4.74, 4.75
see Numerical Recipes
Bessel Functions
Bessel functions are useful for solving differential equations
for systems with cylindrical symmetry
Bessel Fn. of the First Kind
Bessel Fn. of the Second Kind
Jn(z), n integer
Yn(z), n integer
Jn, Yn are linearly independent solutions of
2
d
y
dy
2
2
2
z

z

z

n
y  0

2
dz
dz

Modified Bessel Functions:
I
(
z
),
K
(
z
)
ar
solu
e
to
n
n
2
d
y
dy
2
2
2
z

z

z

n
y 0


2
dz
dz
