March 28, 2011

HW 7 due Wed.Midterm #2 on Monday April 4Today: Relativistic Mechanics

Radiation in S.R.

Fields of a Uniformly Moving Charge

If we consider a charge q at rest in the K’ frame, the E and B fields are

3

3

3

r

zqE

r

yqE

r

xqE

z

y

x

0

0

0

z

y

x

B

B

B

where

€

r 3 = ′ x 2 + ′ y 2 + ′ z 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:

€

rE = q

r n −

r β ( ) 1−

r β 2( )

r κ 3R2

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

n1

We’ll consider some implications

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 = −qvγt

γ 2v2t 2 + b2( )

3 / 2

Ey =qvb

γ 2v2t 2 + b2( )

3 / 2

E z = 0

yz

yx

EB

BB

0

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)

v0 b

t

)max(

)max(

x

y

E

EAs 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 fourier transform of E(t) E(w)

We can already guess the answer

v

2

b

yyz EEB

The spectrum will be 2

)(ˆ

EcdAd

dW

where )(ˆ E is the fourier transform

€

ˆ E (ω) =1

2πEy (t)e iωtdt∫

=qγb

2πγ 2v2t 2 + b2

( )−3 / 2

e−iωtdt−∞

∞

∫

The integral can be written in terms of the modified Bessel functionof order one, K1

vvv)(ˆ

1

b

Kb

b

qE

The spectrum cuts off for

b

v~

Rybicki & Lightman give expressions for d

dW

and some approximate analytic forms -- Eqns 4.74, 4.75

Bessel Functions see Numerical Recipes

Bessel functions are useful for solving differential equations for systems with cylindrical symmetry

Bessel Fn. of the First Kind Jn(z), n integerBessel Fn. of the Second Kind Yn(z), n integer

Jn, Yn are linearly independent solutions of

€

z2 d2y

dz2+ z

dy

dz+ z2 − n2y( ) = 0

Modified Bessel Functions:

tosolutions are )(),( zKzI nn

€

z2 d2y

dz2+ z

dy

dz− z2 − n2

( )y = 0

Relativistic Mechanics

4-momentum Ump 0

where m0 = rest mass, i.e. the mass in the inertial reference frame in which the particle is at rest.

For particles with mass:

p

cEp

/where v0V

mp

22420

2

220

pccmE

cmpp

For photons: kpE

,

k

cp

/

Can show that p and E transform in the following way: (v in x-direction)

x

zz

yy

xx

pEE

pp

pp

cpp

v

vE2

where

2/1

2

2v1

c

Relativistic Mass

2

2

0

cv

1

m

m

For what value of v/c will the relativistic mass exceed its rest massby a given fraction f?

0

0

m

mmf

0.001 0.0140.01 0.140.1 0.421.0 0.8710 0.994100 0.999

At high beta, m >> m0

The invariant quantity is

€

E02 = m0c

2

E 2 − cp( )2

= E02or

since 2222 pcEcpE

where

v0

20

mp

cmE

4-acceleration

d

dUa

Newton’s Second Law

4-Force F

F=ma

d

dp

amF

0

For the Lorentz force

The corresponding 4-vector is

€

rF = e

r E + 1

c

r v ×

r B [ ]

UF

c

eF

So the equation of motion for a charge is

UF

cm

ea

0

UF

c

e

d

dp

BEc

E

c

e

c

BBE

BBE

BBE

EEE

c

e

UFc

e

d

dp

z

y

x

xyz

xzy

xzx

zyx

v

v

v

v

v

0

0

0

0

UU

vv

1000

0100

0010

0001

cc

BEc

E

c

eUF

c

e

d

dp

v

v

Does this make sense?

repeat

0

Ec

ed

dW

cd

dp

v

10

W = particle energy

sov Ee

dt

dWOK!

t, not tau

3,2,1

Bv

1

cEe

d

dp

so for μ=1

€

dpx

dt= e Ex +

r v ×

r B ( )

x[ ]

similarly for y,z

Emission from Accelerated Relativistic Particles

Basic Idea:

(i) We know how to calculate the emission from an accelerating particle if <v> << c Larmor result

(ii) We can transform to an inertial frame, K’, in which the particle is instantaneously at rest calculate the radiation field transform back to lab frame, K

Consider a particle with change q in its instantaneous rest frame K’

Let tddW in time emittedenergy total'

0pd

momentum=0 since the particle is at rest

In K frame:

tddt

WddW

Ptd

Wd

dt

dWP

So power

so emitted energy / time =P is a Lorentz invariant

In rest frame, the Larmor result:

23

2

3

2u

c

qP velocityu

or2

3

2

3

2a

c

qP

onacceleratia

In terms of components of the acceleration parallel and perpendicularto the direction of motion between frames, recall that we derived:

aa

aa2

||3

||

provided the particle is instantaneously at rest in K’

€

P = P =2q2

3c 3

r ′ a

2

=2q2

3c 3′ a ||2 + ′ a ⊥

2( )

=2q2

3c 3γ 4 a⊥

2 + γ 2a||2

( )

Angular Distribution of Radiation:

In the rest frame K’, consider emission of energy dW’ into solid angle dΩ’

How do dW’ and dΩ’ transform?

dW’

Since energy and momentrum form a 4-vector, dW transforms like dt

xpdWddW v

For photons, c

W

c

hp

so if we let cos

WddW 1

d Let

ddd

cos

ddd

cos

aberration formula

1

22 1

d

dso

Since dd

22 1

d

d

so...

d

Wd

d

dW 33 1

Now of course we are interested in the power, i.e. the energy/time:

d

Pd

ddt

dW 1

'

'

so how does dt’ transform?

There are two possibilities for relating dt and dt’

(i) tddt This is the interval as seen in K frame emitted power Pe

so

€

dPe

dΩ= γ 2 1+ β ′ μ ( )

3 d ′ P

d ′ Ω

can also write

€

dPe

dΩ=

1

γ 4 1− βμ( )3

d ′ P

d ′ Ω

(ii) 1tddt c.f. Doppler formula

This is better – it’s what you would actually measure as a stationary observer in K

received power Pr

d

Pd

d

dPr 44 1

or

d

Pd

d

dPr44 1

1

We’ll adopt this

Suppose the acceleration of the particle is caused by a forcehaving components

FF and ||with respect to the particle’s velocity (see R&L problem 4.14)

||

||

2||

||||

v

Ftd

pd

xdc

td

Edc

pd

dt

dpF

since dE’ =dx’ =0

and

F

td

pd

xdc

td

pddt

dpF

1

1

€

P =2

3

q2

c 3γ 4 a⊥

2 + γ 2a||2

( )

P =2

3

q2

c 3

1

m2F||

2 + γ 2F⊥2

( )So F

||F

is more effective than

in producing radiation

Beaming As β 1, photons which are isotropic in the rest frame are “beamed” forward

In K:

€

=cosθ ≈1 −θ 2

2

β = 1−1

γ 2

⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 2

=1−1

2γ 2

1

γ 4 1 − βμ( )4 ≈

2γ

1+ γ 2θ 2

⎛

⎝ ⎜

⎞

⎠ ⎟

4

so

so

d

Pd

d

dP4

221

2

Strongly peaked at θ=0

1

~

What happens to a dipole?

Recall that is the rest frame of the emitting particle, Larmor’s result had a dipole angular distribution

2

2

22

sin4 c

aq

d

Pd

in K’

angle between the acceleration and thedirection of emission

writing aaa

||2

24

22||

2

3

2

sin14

aa

c

q

d

dP

Working out the result is messy; see R&L Eqn. 4.101 and 4.103

Qualitative Picture

Transformation of the Equation of Radiative Transfer

How does the specific intensity Iν transform?

Recall the radiation density

ddvolenergyu /// We showed that

c

Iu so volumeenergydd

c

I/

Now, the energy per volume is also

€

f hν p2dpdΩ where f = phase space density of photons

f = # photons / dV’ where dV’ = dx’dy’dz’ dpx’ dpy’ dpz’

How does the phase space volume transform?

zyxzyx pdpdpddpdpdp

zdydxddxdydz

1

VddV so phase space volume is an invariant

But since f = Number of photons/dV

f is a Lorentz invariant

OK:

€

Iνc

dν dΩ = f hν p2dpdΩ

ddc

hfdd

c

I 3

43

sincec

hp

Invariant Lorentz 3

I

What about the source function Sν ?

Recall SI

ds

dI

so S transforms like I

Invariant Lorentz 3

S

Optical depth

e gives fraction of photons absorbed

so Invariant Lorentz

Absorption coefficient

K

θ

v

l

yk

l

c

l

l

2

sin

1

sin

l =distance perp. to v

2

kc

Both l and ky are perpendicular to v transform the same way

yk

lis Lorentz invariant

So Invariant Lorentz

Emission coefficient jν

Sj

Invariant Lorentz is

Invariant Lorentz is

3

S

So Invariant Lorentz 2

j

So, to solve a radiative transfer problem:

• set up problem in emitter rest frame• solve the equation of radiative transfer• transform the result

etc. ,I