```March 9, 2011
Special Relativity, continued
Lorentz Transform
x   ( x  v t )
y  y
z  z
t    (t  cv2 x )

cos


cos


1


cos


1
12
v

c
Stellar Aberration
Discovered by James Bradley in 1728
Bradley was trying to confirm a claim
of the detection of stellar parallax,
by Hooke, about 50 years earlier
Parallax was reliably measured
for the first time by
Friedrich Wilhelm Bessel in 1838
Refn:
A. Stewart: The Discovery of
Stellar Aberration, Scientific American,
March 1964
Term paper by Vernon Dunlap, 2005
Because of the Earth’s motion in its orbit around the Sun, the angle at
which you must point a telescope at a star changes
A stationary telescope
Telescope moving at velocity v
Analogy of running in the rain
As the Earth moves around the Sun, it carries us through a
succession of reference frames, each of which is an inertial reference
frame for a short period of time.
clockmaker George Graham (1675 – 1751) build
a transit telescope with a micrometer which
allowed Bradley to line up a star with
cross-hairs and measure its position WRT zenith
to an accuracy of 0.25 arcsec.
Note parallax for the nearest stars is
~ 1 arcsec or less, so he would not have been
able to measure parallax.
Bradley chose a star near the zenith to minimize
the effects of atmospheric refraction.
.
The first telescope was over 2 stories high,
attached to his chimney, for stability. He later
made a more accurate telescope at his
Aunt’s house. This telescope is now in the
Greenwich Observatory museum.
Bradley reported his results by writing a letter to the
Astronomer Royal, Edmund Halley.
Later, Brandley became the 3rd Astronomer Royal.
Vern Dunlap sent this picture from the Greenwich Observatory:
In 1727-1728 Bradley measured the star gamma-Draconis.
Note scale
Is ~40 arcsec reasonable?
The orbital velocity of the Earth is about v = 30 km/s
v 4
 10
c
Aberration formula:
cos



cos

'
1

cos


(cos

)(
1

cos

)
2
2

cos

cos

cos

(small β)


2
(1)

cos

cos

sin
Let





Then


angl
of
abe
 

cos

cos


sin

sin


cos
cos(

)



α is very small, so cosα~1, sinα~α, so
(2)

cos

cos

sin
Compare to (1):
we get


2

cos

cos

sin



sin

Since β~10^4 radians  40 arcsec
at most
BEAMING
Another very important implication of the aberration formula is
relativistic beaming

sin

tan




cos


cos



cos

'
1


cos

Suppose
Then
2
tan

1

That is, consider a photon emitted at
right angles to v in the K’ frame.
s in 

1

1
F

1
,
or
si
i
sma
s
n  
So if you have photons being emitted isotropically in the source
frame, they appear concentrated in the forward direction.
The Doppler Effect
When considering the arrival times of pulses (e.g. light waves)
we must consider
- time dilation
- geometrical effect from light travel time
K: rest frame observer
Moving source: moves from point 1
to point 2 with velocity v
Emits a pulse at (1) and at (2)
The difference in arrival times between emission
at pt (1) and pt (2) is
d
t A  t   t 1  cos  
c
where
2
t 

ω` is the frequency in the source frame.
ω is the observed frequency
2



t A  1   cos 
1

term: relativistic dilation
1
1   cos
classical
geometric term
Relativistic Doppler Effect
Transverse Doppler Effect:


 1  cos 
When θ=90 degrees,



Proper Time
Lorentz Invariant = quantity which is the same inertial frames
One such quantity is the proper time

c 2d 2  c 2dt2  dx2  dy2  dz2 
It is easily shown that under the Lorentz transform

d  d 
cd 
is sometimes called the space-time interval between two events
• dimension : distance
• For events connected by a light signal:
cd   0
Space-Time Intervals and Causality
Space-time diagrams can be useful for visualizing the relationships
between events.
ct
future
x
past
The lines x=+/ ct represent
world lines of light signals passing
through the origin.
Events in the past are in the region
indicated.
Events in the future are in the region
on the top.
World line for light
Generally, a particle will have some world line in the shaded area
ct
The shaded regions here cannot
be reached by an observer whose world
line passes through the origin since to
get to them requires velocities > c
x
Proper time between two events:
   ct  x
2
   ct  x
2
2
0
2
ct   x
2
2
“time-like” interval
“light-like” interval
2
   ct  x
2
2
2
0
“space-like” interval
2
ct
ct’
Depicting another frame
x’
x
x=ct
x’=ct’
In 2D
Superluminal Expansion
Rybicki & Lightman Problem 4.8
- One of the niftiest examples of Special Relativity in astronomy is the
observation that in some radio galaxies and quasars, and Galactic
black holes, in the very core, blobs of radio emission appear to
move superluminally, i.e. at v>>c.
- When you look in cm-wave radio
emission, e.g. with the VLA,
they appear to have radio jets
emanating from a central core
and
ending in large lobes.
DRAGN = double-lobed radio-loud active galactic nucleus
Superluminal expansion
VLBI (Very Long Baseline
Interferometry) or VLBA
Proper motion
μ=1.20 ± 0.03 marcsec/yr
 v(apparent)=8.0 ± 0.2 c
μ=0.76 ± 0.05 marcsec/yr
 v(apparent)=5.1 ± 0.3 c
Another example:
M 87
HST WFPC2 Observations of optical emission from jet,
over course of 5 years:
v(apparent) = 6c
Birreta et al
Recently, superluminal motions have been seen in Galactic jets,
associated with stellar-mass black holes in the Milky Way –
“micro-quasars”.
GRS 1915+105 Radio Emission
+ indicates position
of X-ray binary source,
which is a 14 solar mass
black hole. The “blobs”
are moving with
v = 1.25 c.
Mirabel & Rodriguez
Most likely explanation of Superluminal Expansion:
(1)
v cosθ Δt
Blob moves from point (1) to point (2)
in time Δt, at velocity v
θ
vΔt
The distance between (1) and (2) is
v Δt
(2)
v sinθ Δt
However, since the blob is closer to the
observer at (2), the apparent time
difference is
Observer
t app
 v

 t 1  cos 
 c

The apparent velocity on the plane of the sky is then
v app 
v t sin 

t app
v sin 
v
1    cos
c
v app
v sin 

v
1    cos
c
v(app)/c
To find the angle at which v(app) is maximum, take the derivative of
v app
v sin 

v
1    cos
c
and set it equal to zero, solve for θmax
Result:
then
cos  MAX
v MAX
v

c
and
v 1  2

 v
2
1 
sin  MAX  1   
2
1

When γ>>1,
then v(max) >> v
Special Relativity:
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
1
0
0
0

0
0

1
0
0
1
0
   1 if     0
Or
 1 if     1,2,3
0 if   
Then the invariant s
s  c t  x  y  z
2
2 2
2
2
can be written
3
s 
2
 0
3
  x


0
 
x
2
3
s 
It’s cumbersome to write
2
 0
3
  x


0
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   x x
2
 
x
(1)
Now let’s define xμ – with SUBSCRIPT rather than SUPERSCRIPT.
Covariant
4-vector:
x
Contravariant
4-vector:
x0  ct
x1  x
x2  y
x3  z
More on what this means later.
x
x 0   ct
x1  x
x2  y
x3  z
So we can write

x    x


x   x
  
i.e. the Minkowski metric,

can be used to “raise” or “lower” indices.
s 2   x x
Note that instead of writing
we could write

s  x x
2
assume the Minkowski metric.
The Lorentz Transformation


where
 
  

 0

 0
 

0
0
0
0
1
0
0
0
0

1
v
1

and  
c
1  2
Notation:
 F00
F
F   10
 F20

 F30
F01
F02
F11
F21
F12
F22
F31
F32
F03 

F13 
F23 

F33 
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
or
 

0
0
0
0
1
0
0 ct   t   x 
0  x    x  x 


0  y  
y
  

1  z  
z

ct   ct   x
x  x   ct
y  y
z  z
We can transform an arbitrary 4-vector Aν



A    A
Kronecker-δ

Define
 
Note:
(1)
1
0

0

0

0
1
0
0
   
(2) For an arbitrary 4-vector
0
0
1
0
0
0
0

1


A
A    A



 1 for   

 0 for   
~ 

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

s  x x
2
or
where
~ 


     
is a Lorentz invariant,
x x  x  x

 ~

  x   x  x x
 ~
   
Kronecker Delta
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 B B
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 u
d
d
and
where
u 
1
u2
1 2
c

u  u  magnitudeof theordinary velocity
dx dy dz
, ,
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  theordinary velocity
So we had to multiply by u to make a 4-vector,
i.e. something whose square is a Lorentz invariant.
dx dy dz
, ,
dt dt dt

How does
so...
u

u0   (u 0   u1 )
u1   ( u 0  u1 )
u2  u 2
 u
or
 uc   (c u    u u )
1
1
1

 uu   ( c u   u u )
 u  u 3   u u 3
2


u
 1 
2

c


u
 1  2

c

2
u

2
2

 u u   u u
u 3  u 3
where


u   u
transform?

v 
  1  2 
c 

2








1 / 2
1 / 2
1 / 2
where v=velocity between frames
Wave-vector 4-vector
Recall the solution to the E&M Wave equations:
 
E  exp(ik  r  it )
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 
contravariant tensor
T
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
x ,x ,x ,x
3
(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
  scalar

x 
We denote

x 
then
is a covariant 4-vector
as
,
Example:
T
if

is a second-ranked tensor


T ,  
x

where


third rank tensor
 thecomponentsof T

(8) Divergence of a tensor field
Take the gradient of the tensor field, and then contract.
Example:
A
Given vector


Divergence is
A ,
Divergence is
T  ,
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
x  to x
Suppose you have a coordinate transformation which relates
or
x0 , x1, x 2 , x3  x0 , x1, x2 , x3
by some rule.
A COVARIANT 4-vector, Bα, transforms “like” the basis vector, or
x 0
x1
x 2
x 3
B   B0   B1   B2   B3
x
x
x
x
or
x 
B   B
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  x cos
dx
 cos
dx
x  x cos
dx
 cos
dx
dx dx

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
1
0
0
0
0
1
0
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
x  0 x  1 x  2 x  3
 0 x  1 x  2 x  3 x
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   themetrictensor
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


c d  g dx dx
2
where
and
2
  proper time
1   g  g  g 
  g      
2
x
x
 x





= The Affine Connection, or Christoffel Symbol
For an S.R. observer in an inertial frame:
g  
 1
0

0

0
0
1
0
0
0
0
1
0
0
0
0

1
And the equation of motion is simply
2

d x
0
2
d
Acceleration is zero.
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