Relativistic effects

High Energy Astrophysics 2009

Beaming hypothesis 3C279 CGRO EGRET F>100MeV

couples e e

make tophotonsray -Xwith

interact they because excape,

cannotray - 60 if but..

1000

-

3

+

≥

≥=

γ

σ

γ

γ

lcRm

Ll

e

T

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δ =1/γ (1- β cosθ )

•Δt obs = Δt/δ

•light aberration ν obs = δ ν

Iv /ν 3 is relativistic invariant Iνobs = δ 3Iν

Fν obs = Fν δ 3+α Fbol obs = Fbolδ 4

ΔFobs

Δtobs

= δ 5 ΔF

Δt

l =Lxσ T

Rmec 3 R ~ δcΔt l = δ−5 Lobs

Δtobs

σ T

mec4

Doppler boosting

The compactness problem

t ~ 1-10 ms

Compact sources R0 c t ~ 3 107 cm

•Cosmological sources (D~3 Gpc) L~ fD2~1052

erg/s

~ p R0nσT~σTL/4R0c ~1015

~ 1 MeV ~ 1 MeV

1 2≥ (mec2)2 γγγγ e e++ee--

p fraction of photons above the thereshold of pair production

p fraction of photons above the thereshold of pair production

Optical depth ((γγγγ e e++ee--) ) >>1

σTT==6.25x10-

25cm2

σTT==6.25x10-

25cm2

R of our galaxy ~ 30 kpc: extragalactic objects

Implications: The fireball

Relativistic motion: A plasma of e+ e- , a Fireball, which expands and accelerate to relativistic velocities, optical depth reduced by relativistic expansion with Lorentz factor

How can the photons escape the source?

The reasons:

1. In the comoving frame below the thereshold for pair production ’=/ ’1 ’2≤ (mec2)2

2. Number of photons above the threshold reduced by 2( -1) (~2 high-energy photon index);

3. Emitting region has a size of 2R0

γγ reduced by a factor 2+2 6..

γγ < 1 for ≥100

€

=1

1− (v /c)2

Special relativityThe two basic postulates:

1. The relativity principle: the laws of physics are the same for all inertial observers

2. The constancy of the speed of light: c is the same for all inertial observer, independent of their velocity or motion relative to the source of light.

Two main insights at the base of the theory (Einsten, Poincare, Lorentz):1. Electromagnetism: electromagnetic waves in terms of motion of an aether. But,

if there is an aether, there would be a preferred observational frame of reference. So if all physical laws should hold for observers in all inertial frame one should abandon the idea of an aether (and introduce the idea of a field).

2. A new concept of space and time. Although space and time are different, they can no longer be considered independent, the quantities one measures depend on the speed at which he/she is traveling.

Implications of the constancy of c

It is possible to derive all special relativity from these two postulates. And the two postulate together tells that Newton’s laws are incomplete. If one accepts the two postulate, there is no choice, but to replace Newton’s law with new rules that are consistent with them. The new rules must resemble the Newton’s ones when applied to objects moving slowly.

In Newtonian mechanics speeds are simply added together. But if the speed of light is constant a beam launched by a train running at 0.5 c, runs always at c. Quite a shocking results! How can it be? The explanation is in the new concept of space and time. If they are linked, it is only a combination of the two which appears the same to two observers who move relative to one another.

Time dilatationClock of a woman on a rocket

Same clock seen by a men at rest v

d c

s

€

t0 = Nd /c

s = vt1 = vt /N

ct = N(d2 + s2)1/ 2 = N(d2 + v 2t 2 /N 2)1/ 2

t 2 = N 2d2 /c 2 + v 2t 2 /c 2 = t0 + β 2t 2

t = t0

1

(1− β 2)1/ 2= γt0 γ =

1

(1− β 2)1/ 2>1

assume β = 3 /2 = 0.866 then t = 2t0

so to travel 4 light years the woman aged 2.3 years and the man 4.6 years

How could it be?

Lorentz contractionFor the man at rest the star is indeed 4 light years and the rocket takes 4.6 yr to reach it. But the woman on the rocket thinks she is at rest and the star is rushing by her at 0.866c. Because of the Lorentz contraction, what is 4 light year for the man at rest is 2 light years for the women on the rocket. Both man and woman are right, according to SR any inertial observer has the right to consider himself at rest. If the length of a rod in its rest frame is L0 and if the rod moves along its length at speed v with respect to an observer, than the observer sees a length L=L0/γ

0 L0 non moving rod t0‘

t0

L moving rod

0 t’

v(t-t’) vt’

€

A photon leaves one end at time 0, strikes the other at time

to' L0 = ct0

' . Then it reflects and reaches the original end at

time t0 L0 = c(t0 − t0' ). Adding : 2L0 = ct0

A photon leaves one end at time 0, strikes the other at time t ' .

In the interim the other end has moved a distance vt ' .

Then, L + vt ' = ct ' .Then it reflect and return at the original end

at time t. In the interim the original end has moved a distance

v(t − t '). Therefore, L − v(t − t ') = c(t − t ' ) 2L = (c 2 − v 2)t /c

Lorentz contraction

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2L = (c 2 + v 2)t /c 2L0 = ct0 t = γt0

2L = (c 2 + v 2)γt0 /c = (c 2 + v 2)γ2L0 /c 2

L = L0

(1− β 2)

(1− β 2)1/ 2= L0

1

γ

Lorentz transformations

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x',y ',z' t ' of an event in S' must have a one - to - one relationship with

x, y,z, t of the same event measured in S.

y = y' z = z'

From the translational symmetry of space and time follows that the relationship

between x' and x and t' and t must be linear :

x'= Ax + Bt

t'= Cx + Dt

Lorentz transformations

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2) The origin of the reference frame S' has coordinate x'= 0 and moves with

velocity v relative to S, so that x = vt.

x'= Ax + Bt = 0 = Avt + Bt B = −vA

x'= A(x − vt)

3) The origin of the reference frame S has coordinate x = 0 and moves with

velocity − v relative to the frame S', so that x'= −vt'

x'= A(x − vt) − vt'= A(x − vt) = −vAt

t'= Cx + Dt = Dt = At ⇒ A = D

t'= A(Ex + t) where E = C / A and lets call A = γ

4) A combination of two Lorents transformation must be a Lorentz transformation.

S' moves with velocity v1 relative to S and S'' moves with velocity v2 relative to S

x' '= γ v 2(x'−v2t'), x'= γ v1(x − v1t)

t' '= γ v 2(Ev2x '+t '), t '= γ v1(Ev1x + t)

x' '= γ v 2γ v1[(1− Ev1v2)x − (v1 + v2)t]

t' '= γ v 2γ v1[(Ev1 + Ev 2)x + (1− Ev 2v1)t]

Lorentz transformations

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For a general Lorentz transformation, the coefficients in front of x and t in 3) must be equal

and the equations in 4) must satisfy this requirement. Therefore :

1− E v1v2 =1− Ev 2v1 ⇒v2

Ev2

=v1

E v1

The left - hand side depend on v2 only and the right - hand side depend on v1 only. This eq. can

be satisfied only if v / Ev is a constant a independent on v, i.e. Ev = v /a

Substituting this in 3) one has : x '= γ v (x − vt), t '= γ v (xv /a + t)

5) Let us make a Lorentz transformation from S to S' with velocity v and then back from S' to S

with velocity − v. The eq. are similar to the eq. in 4) :

x = γ−v (x '+vt'), x'= γ v (x − vt)

t = γ−v (−x'v /a + t '), t '= γ v (xv /a + t)

substituting x' and t' from the first eq. in the second eq. one finds :

x = γ−vγ v (1+ v 2 /a)x, t = γ−vγ v (1+ v 2 /a)t

these eq. must be valid for any x and t, so : γ−vγ v =1

1+ v 2 /aBecause of space symmetry γ v must depend only on | v | and not by its direction, so γ−v = γ v :

γ v =1

(1+ v 2 /a)1/ 2

Lorents transformations

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6) Substituting γ =1

(1+ v 2 /a)1/ 2 in 4) one finds :

x'=x − vt

(1+ v 2 /a)1/ 2; t '=

xv /a + t

(1+ v 2 /a)1/ 2 putting a = −c 2 we have the usual L. transformations :

x'=x − vt

(1− v 2 /c 2)1/ 2; t '=

−xv /c 2 + t

(1− v 2 /c 2)1/ 2

7) The Lorentz transformation preserves the space - time interval :

(ct')2 − (x')2 = (ct)2 − (x)2

'Lorentz invariant'

8) If a = ∞ the eq. in 6) produce the Galileo transformations:

x'= x − vt, t '= t

Doppler effectThe effect is the combination of both relativistic time dilation and time retardation. Consider a a source of radiation which emits one period of radiation over the time t it takes to move from P1 to P2

€

If ωem is the emitted circular frequency of the radiation in the rest frame, then Δt'=2π

ωem

, and the time

between the two events in the observer is Δt = γΔt'=2π

ωem

. However, this is not the observed time

between the events, because there is a time difference involved in radiation emitted from P1 and P2.

If D is the distance from P2 to the observer, t1 is the time of emission of rad. from P1 and t2 is the

time of emission of radiation from P2, then the times of reception t1obs and t2

obs are :

t1obs = t1 +

D + vΔt cosθ

c; t2

obs = t2 +D

c

Doppler effect

€

The period of the pulse received in the observer frame is :

t2obs − t1

obs = t2 +D

c

⎛

⎝ ⎜

⎞

⎠ ⎟= t1 +

D + vΔt cosθ

c

⎛

⎝ ⎜

⎞

⎠ ⎟= (t2 − t1) −

v

cΔt cosθ = Δt 1−

v

ccosθ

⎛

⎝ ⎜

⎞

⎠ ⎟

therefore :

2π

ωobs

= γ2π

ωem

1−v

ccosθ

⎛

⎝ ⎜

⎞

⎠ ⎟;

ωobs

ωem

=1

γ 1−v

ccosθ

⎛

⎝ ⎜

⎞

⎠ ⎟=

ν obs

ν em

=1

1+ z

The factor γ is a pure relativistic effect, the factor (1− β cosθ) is the result of time retardation.

δ =1

γ 1−v

ccosθ

⎛

⎝ ⎜

⎞

⎠ ⎟ is the Doppler factor

Four dimensional space-time

Indices are raised and lowered with :

Representation of Lorentz transformations

For the special case of a Lorentz transformation involving a boost along the x-axis:

Lorentz invarianceUnder Lorentz transformations, the dot product of two four vectors, xy = xμyμ is preserved. In terms of Λ, this means:

Our first LORENTZ INVARIANT is the PROPER TIME of an event, which is just the square root of the scalar product of the space-time 4-vector with itself:

€

cτ = xμ x μ = c 2t 2 −r x •

r x

€

The 4 - momentum is defined by :

pμ = m0uμ = [γmc,γmv i]

pμ pμ =E 2

c 2−

r p •

r p =

E 2

c 2− p2

This Lorents invariant is c 4 times the square of the

rest mass of the object whose momentum we are scritinizing

E 2 = p2c 2 + m2c 4

Phase space number density

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nphase =dN

VVphase

dN

dxdydzdpxdpydpz

=n

dpxdpydpz

dpx '= dpx, dpy '= dpy, dpz '= γ (dpz − vdE)

For arbitrary E the product of these three is in general complicate. This is because we are

considering the transformation of a 3D volume in a 4D space. However, the problem can

be avoided by using 3D slices of 4D phase space. This can be dobe by setting E = const.

and thus dE = 0. If this is the case : Vphase ' = γVphase

n =dN

dV=

dN

dxdydz, dx'= dx, dy'= dx, dz'= γ (dz − vdt)

Since all particles are considered at the same time, dt = 0 and :

n'=dN '

dx 'dy'dz'=

dN

dxdydzγ=

n

γ

Therefore :

nphase '=dN '

V 'Vphase '=

dN

V 'Vphase '= nphase

Energy density and specific intensity

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Energy density within d3 p = hνnphased3 p = Energy density per unit frequency and unit

solid angle

Uν dνdΩ = hνnphased3 p = hνnphase p2dpdΩ

However, Uν =Iν

c where Iν is the specific intensity. Therefore :

Iν

cdνdΩ = hνnphase p2dp = hνnphase

hv

c

⎛

⎝ ⎜

⎞

⎠ ⎟2

dhv

c

⎛

⎝ ⎜

⎞

⎠ ⎟

nphase ∝Iν

v 3⇒ Lorentz invariant

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δ =1/γ (1- β cosθ ) γ = (1− β 2)−1/ 2

β = angle between velocity and LOS

Observed transverse v. of emitting blob va = β ac

True velocity v = βc

β a =β sinθ

1− β cosθ

if β > 0.7 superluminal motion!!

β a max = γ 2 −1 if cosθ = β or sinθ = γ −1

This implies a γ min = β a2 +1 if β a = 5 γ > 5.1!!

βc

θδ

BeamingBeaming relativistico relativistico - Moto superluminaleMoto superluminale

Per θ = arccos(β) si ottiene:

Quindi se β ~1:

c

Rtt 11

'1 +=

c

Rtt 2

2'2 +=

)(1

)( 12' RR

ctt −+=

tcRR ⋅−=− )(cos12 θβ} )]cos(1[)( ' θβ−= tt

tcxxx ⋅=−= )(sin12 θβ

θβθββ

cos1

sin

)(

1

)(

1'' −

=Δ

Δ

Δ

Δ=

Δ

Δ=

t

t

t

x

ct

x

cobs

γββ

ββ =−

=21

)( MAXobs

1)( >=γβ MAXobs