March 9, 2011 Special Relativity, continued. Lorentz Transform.

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March 9, 2011

Special Relativity, continued

cos1

coscos

c

v

)(

) v(

2v xtt

zz

yy

txx

c

21

1

Lorentz Transform

Stellar Aberration

Discovered by James Bradley in 1728

Bradley was trying to confirm a claimof the detection of stellar parallax,by Hooke, about 50 years earlier

Parallax was reliably measuredfor the first time by Friedrich Wilhelm Bessel in 1838

Refn:A. Stewart: The Discovery of Stellar Aberration, Scientific American,March 1964Term paper by Vernon Dunlap, 2005

Because of the Earth’s motion in its orbit around the Sun, the angle atwhich 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 referenceframe for a short period of time.

Bradley’s Telescope

With Samuel Molyneux, Bradley had master 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 hisAunt’s house. This telescope is now in theGreenwich 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:Bradley’s micrometer

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

410v c

Aberration formula:

coscoscos

)cos1)((cos

cos1

cos'cos

22

2sincoscos (small β) (1)

Let aberration of angle

Then

sinsincoscos

)cos(cos

α is very small, so cosα~1, sinα~α, so

sincoscos (2)

Compare to (1): 2sincoscos

we get sin Since β~10^4 radians 40 arcsec at most

BEAMING

Another very important implication of the aberration formula isrelativistic beaming

cos1

cos'cos

cos

sintan

Suppose 2 That is, consider a photon emitted at

right angles to v in the K’ frame.

Then

1tan

1sin

small is sin ,1 For 1

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 observerMoving source: moves from point 1to point 2 with velocity vEmits a pulse at (1) and at (2)

The difference in arrival times between emission at pt (1) and pt (2) is

where

cos1 tc

dttA

2

t

ω` is the frequency in the source frame.ω is the observed frequency

cos1

2

At

Relativistic Doppler Effect

1

term: relativistic dilation

cos1

1

classicalgeometric term

Transverse Doppler Effect:

cos1

When θ=90 degrees,

Proper Time

Lorentz Invariant = quantity which is the same inertial frames

One such quantity is the proper time

c 2dτ 2 = c 2dt 2 − dx 2 + dy 2 + dz2( )

It is easily shown that under the Lorentz transform

dd

is sometimes called the space-time interval between two eventscd

• dimension : distance

• For events connected by a light signal:

0cd

Space-Time Intervals and Causality

Space-time diagrams can be useful for visualizing the relationshipsbetween events.

ct

x

World line for light

future

past

The lines x=+/ ct representworld lines of light signals passingthrough the origin.

Events in the past are in the regionindicated.Events in the future are in the regionon the top.

Generally, a particle will have some world line in the shaded area

x

ct

The shaded regions here cannotbe reached by an observer whose worldline passes through the origin since toget to them requires velocities > c

Proper time between two events:

222 xct

0222 xct “time-like” interval

22 xct “light-like” interval

0222 xct “space-like” interval

x

ct

x’

ct’

x=ctx’=ct’

Depicting another frame

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

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

VLBI (Very Long BaselineInterferometry) or VLBA

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”.

+ indicates position of X-ray binary source,which is a 14 solar massblack hole. The “blobs”are moving with v = 1.25 c.

GRS 1915+105 Radio Emission

Mirabel & Rodriguez

Most likely explanation of Superluminal Expansion:

vΔtθ

v cosθ Δt

(1)

(2)

v sinθ Δt

Observer

Blob moves from point (1) to point (2)in time Δt, at velocity v

The distance between (1) and (2) is v Δt

However, since the blob is closer to the observer at (2), the apparent time difference is

cos

c

v1tt app

The apparent velocity on the plane of the sky is then

coscv

1

sin v

sin v v

app

app t

t

coscv

1

sin v v

app

v(app)/c

To find the angle at which v(app) is maximum, take the derivative of

coscv

1

sin v v

app

and set it equal to zero, solve for θmax

Result:c

vcos MAX and

1

1sin 2 MAX

then v1

1vv

2

2

MAXWhen 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

zx

yx

xx

ctx

3

2

1

0

Written 3,2,1,0 x

4-vectors can be transformed via multiplication by a 4x4 matrix.

1000

0100

0010

0001

if 0

1,2,3 if 1

0 if 1Or

The Minkowski Metric

Then the invariant s

222222 zyxtcs can be written

3

0

3

0

2

xxs

3

0

3

0

2

xxs

It’s cumbersome to write

So, following Einstein, we adopt the convention that when Greekindices are repeated in an expression, then it is implied that we are summing over the index for 0,1,2,3.

(1)

(1) becomes: xxs 2

Now let’s define xμ – with SUBSCRIPT rather than SUPERSCRIPT.

Covariant4-vector:

zx

yx

xx

ctx

x

3

2

1

0

Contravariant4-vector:

zx

yx

xx

ctx

x

3

2

1

0

More on what this means later.

So we can write

xx

xx

i.e. the Minkowski metric,

can be used to “raise” or “lower” indices.

Note that instead of writing xxs 2

we could write

xxs 2

assume the Minkowski metric.

The Lorentz Transformation

1000

0100

00

00

where

21

1 and

v

c

Notation:

33323130

23222120

13121110

03020100

FFFF

FFFF

FFFF

FFFF

F

Instead of writing the Lorentz transform as

)v

(

)v(

2x

ctt

zz

yy

txx

we can write

xx

z

y

xx

xt

z

y

x

ct

1000

0100

00

00

or

zz

yy

ctxx

xcttc

We can transform an arbitrary 4-vector Aν

AA

Kronecker-δ

Define

for 0

for 1

1000

0100

0010

0001

Note:

(1)

AA

(2) For an arbitrary 4-vector A

Inverse Lorentz Transformation

~

We wrote the Lorentz transformation for CONTRAVARIANT 4-vectors as

xx The L.T. for COVARIANT 4-vectors than can be written as

xx ~where

~

Since xxs 2

is a Lorentz invariant,

~

~ xxxx

xxxx

or Kronecker Delta

General 4-vectors A (contravariant)

Transforms via

AA

Covariant version found by

AA

Minkowski metric

Covariant 4-vectors transform via

AA

~

Lorentz Invariants or SCALARS

Given two 4-vectors BA BA

and

SCALAR PRODUCT

BABA

This is a Lorentz Invariant since

BA

BA

BA

BABA

~

~

Note: AA

can be positive (space-like) zero (null) negative (time-like)

The 4-Velocity

d

dxu

(1) The zeroth component, or time-component, is

ucd

dtc

d

dxu

00 where

2

2

1

1

cu

u

and elocityordinary v theof magnitude uu

Note: γu is NOT the γ in the Lorentz transform which is

dt

dz

dt

dy

dt

dx,,

2

2v1

1

c

The 4-Velocity

d

dxu

(2) The spatial components

iu

ii u

d

dxu

where

2

2

1

1

cu

u

elocityordinary v theu

dt

dz

dt

dy

dt

dx,,

u

cuu So the 4-velocity is

So we had to multiply by to make a 4-vector, i.e. something whose square is a Lorentz invariant.

u

How does transform?u

uu

33

22

101

100

)(

) (

uu

uu

uuu

uuu

so... or

33

22

11

1

)(

)(

uu

uu

ucu

ucc

uu

uu

uuu

uuu

where

2/1

2

2

2/1

2

2

2/1

2

2

v1

1

1

c

c

u

c

u

u

u

where v=velocity between frames

Wave-vector 4-vector

Recall the solution to the E&M Wave equations:

)exp( tirkiE

The phase of the wave must be a Lorentz invariant sinceif E=B=0 at some time and place in one frame, it must alsobe = 0 in any other frame.

k

ck /

Tensors

3.2.1.0

3,2,1,0

T

(1) Definitions zeroth-rank tensor Lorentz scalar first-rank tensor 4-vector second-rank tensor 16 components:

(2) Lorentz Transform of a 2nd rank tensor:

TT

(3) T contravariant tensor

T covariant tensor

related by

TT

transforms via

TT

~

~

(4) Mixed Tensors

one subscript -- covariantone superscript – contra variant

TT

so the Minkowski metric “raises” or “lowers” indices.

TT

(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 arefunctions of the space-time coordinates,

3210 ,,, xxxx

(7) Gradients of Tensor Fields

Given a tensor field, operate on it with

∂∂x μ

for x μ = 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 then

x

is a covariant 4-vector

xWe denote as

,

Example: if T is a second-ranked tensor

xT

, third rank tensor

where T of components the

(8) Divergence of a tensor field

Take the gradient of the tensor field, and then contract.

Example:

Given vectorA Divergence is

,A

Example:

TensorT Divergence is

,T

(9) Symmetric and anti-symmetric tensors

If TT then it is symmetric

If TT 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 xx to

or32103210 ,,,,,, xxxxxxxx by some rule.

A COVARIANT 4-vector, Bα, transforms “like” the basis vector, or

3

3

2

2

1

1

0

0

Bx

xB

x

xB

x

xB

x

xB

Bx

xB

or

A CONTRAVARIANT 4-vector transforms “oppositely” from the basisvector

A

x

xA

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

dx

xd

xd

dxxd

dx

xxdx

xd

xx

cos

cos

cos

cos

But also

so

x’

y’

xy

Peacock gives examples for transformations in normal flat 3-spacefor non-orthogonal axes where

dx

xd

xd

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

AA

Where the Minkowski Matrix is

1000

0100

0010

0001

So the difference isthe sign of the time-likecomponent

Example: Show that xμ=(ct,x,y,z) transforms like a contravariant vector:

xcttc

ctxx

x

33

22

11

00

xx

xx

x

xx

x

xx

x

x

xx

xx

Let’s let 0 'ctx

ctxx

x

00

∂ x μ

∂x1x1 = −γβx

In SR

xx

In GR AgA

tensormetric theg

Gravity treated as curved space.

Of course, this typeof picture is for 2D space,and space is really 3D

Two Equations of Dynamics:

02

2

d

dx

d

dx

d

xd

dxdxgdc 22

where eproper tim

x

g

x

g

x

gg

2

1and

= The Affine Connection, or Christoffel Symbol

For an S.R. observer in an inertial frame:

1000

0100

0010

0001

g

And the equation of motion is simply

02

2

d

xdAcceleration is zero.