Relativistic Covariance and Kinematics

3

Galilean Transformation/Relativity• Galilean transformation is used to transform between

the coordinates of two inertial frames of reference which differ only by constant relative motion within the constructs of Newtonian physics.

Newton’s law is invariant under the Galilean transformation.

However, Maxwell’e equations are not invariant under the Galilean transformation.

• Lorentz transformation is the result of attempts by Lorentz and others to explain how the speed of light was observed to be independent of the reference frame, and to understand the symmetries of the Maxwell’s equations.

x

y

z

x′ O

v

O ′

y′

z′

K K ′

t = t ′ = 0same origin at

x′ = x −vty′ = yz′ = zt′ = t

4

* Review of Lorentz Transformations *• Postulates in the special theory of relativity

(1) The laws of nature are the same in two frames of reference in uniform relative motion with no rotation.

(2) The speed of light is c in all such frames.

• space-time event: an event that takes place at a location in space and time.

• Derivation of Lorentz transforms:

If a pulse of light is emitted at the origin at t = 0, each observer will see an expanding sphere centered on his own origin. Therefore, we have the equations of the expanding sphere in each frame.

Since space is assumed to be homogeneous, the transformation must be linear.

We note that the origin of is a point that moves with speed as seen in K. Its location in K is given by . Therefore, we have

x2 + y2 + z2 − c2t 2 = 0, x′2 + y′2 + z′2 − c2t′2 = 0

x′ = a1x + a2t, y′ = y, z′ = z, t′ = b1x + b2t

x = vtK ′ (x′ = 0)

a2a1

= −v

x′ = a1(x −vt)y′ = yz′ = zt′ = b1x + b2t

(1)

(2)

v

5

Substitute Eqs. (2) into Eq. (1):

(Note: we didn’t assume that )

Therefore, the following equations should be satisfied.

Finally, we obtain the Lorentz transformation (and its inverse):

a12 (x −vt)2 + y2 + z2 − c2 (b1x + b2t)2 = x2 + y2 + z2 − c2t 2

(a12 − c2b1

2 )x2 − 2(a12v + c2b1b2 )xt + (a12v2 − c2b22 )t 2 + y2 + z2 = x2 + y2 + z2 − c2t 2

a12 − c2b1

2 = 1(a1

2v + c2b1b2 ) = 0a12v2 − c2b22 = −c2

(a)(b)(c)

(a) b12 = a1

2 −1c2

(c) b22 = 1+ v2

c2a12

(b) a14v2 = −c4b1

2b22 = c2a1

2 +v2a14 − c2 −v2a12

→ a1 =1

1− v2

c2

≡ γ

(a) b1 = γ , (c) b2 = − v

c2γ

x′ = γ (x −vt)y′ = yz′ = z

t′ = γ t − vc2x⎛

⎝⎜⎞⎠⎟

where γ ≡ 1

1− v2

c2

= 1− β 2( )−1/2 ; β ≡ vc

Lorentz factor 1≤ γ < ∞ ; 0 ≤ β <1

x = γ (x′ +vt′)y = y′z = z′

t = γ t′ + vc2x′⎛

⎝⎜⎞⎠⎟

x′2 + y′2 + z′2 − c2t′2 = x2 + y2 + z2 − c2t 2

x′2 + y′2 + z′2 − c2t′2 = 0

6

Length Contraction / Time Dilation• Length contraction (Lorentz-Fitzgerald contraction): Suppose a rigid rod of length

is carried at rest in . What is the length as measured in K? The positions of the ends of the rod are marked at the same time in K.

Therefore, the rod appears shorter by a factor in K.

If both carry rods (of the same length when compared at rest) each thinks the other’s rod has shrunk!

It would appear to that the two ends of the moving stick were not marked at the same time by the other observer (in K).

• Time dilation: Suppose a clock at rest at the origin of measures off a time interval . What is the time interval measured in K? Note that the clock is at rest at the origin of so that .

The time interval has increased by a factor , so that the moving clock appears to have slowed down.

Time dilation is detected in the increased half-lives of unstable particles moving rapidly in an accelerator or in the cosmic-ray flux.

L0 = x2 ′ − x1′K ′

L0 = x2 ′ − x1′ = γ (x2 − x1) = γ LL = L0 /γ

1/γ

K ′

K ′ T0 = t2 ′ − t1′K ′

x2 ′ = x1′ = 0 T = t2 − t1 = γ (t2 ′ − t1′) = γ T0T = γ T0

γ

7

Transformation of Velocities• Simultaneity is relative: Simultaneous events at two different spatial points in the primed frame is

not simultaneous in the unprimed frame.

• If a point has a velocity in frame , what is its velocity in frame K. Writing Lorentz transformations for differentials

u′ K ′ u

dx = γ (dx′ +vdt′), dy = dy′, dz = dz′

dt = γ dt′ + vc2dx′⎛

⎝⎜⎞⎠⎟

ux =dxdt

= γ (dx′ +vdt′)γ (dt′ +vdx′ / c2 )

= ux ′ +v1+vux ′ / c2

uy =dydt

= dy′γ (dt′ +vdx′ / c2 )

=uy ′

γ (1+vux ′ / c2 )

uz =dzdt

=uz ′

γ (1+vux ′ / c2 )

Review of Lomtz Tnurrfonnations 109

that his own two clocks at x 1 and x2 are synchronized. K’ will object to this, since according to his observations the two clocks in K are not synchro- nized at all.

In both the time-dilation and length-contraction effects we can see the powerful role played by the questions of synchronization of clocks and of the whole concept of simultaneity. Many of the apparent contradictions of special relativity are simply a result of the relativity of simultaneity between two events separated in space.

3. Transformation of Velocities

If a point has velocity u’ in frame K‘, what is its velocity u in frame K (Fig. 4.2)? Writing Lorentz transformations for differentials [cf. Eqs. (4.2)]

dx=y(dx’+udt ’ ) , &=&’

dz = dz‘,

We then have the relations

u; + u - =-= dx y(dx’+udt’) - dt y (dt ’+udx’ /c* ) 1 +uu; /c2 ’

a;

4 !v = y( 1 + uu;/c’) ’

y( 1 + uu:/c’) . u, =

(4.5a)

(4.5b)

(4.52)

K

Figure 4.2 Lorentt tramformation of wlocitks.

u =u ′ +v

1+vu ′ / c2

u⊥ = u⊥ ′γ (1+vu ′ / c2 )

or

8

• Aberration formula: the directions of the velocities in the two frames are related by

• Aberration of lightFor the case of light:

Using the identity,

The aberration formula can be written as:

tanθ = u⊥

u= u⊥ ′γ (u ′ +v)

= u′sinθ′γ (u′cosθ′ +v) where u′ ≡ u′ .

u′ = c

tanθ = sinθ′

γ (cosθ′ +v / c)= sinθ′γ (cosθ′ + β )

cosθ = γ (cosθ′ +v / c)

γ 2 (cosθ′ +v / c)2 + sin2θ′= cosθ′ + β1+ β cosθ′

sinθ = sinθ′

γ 2 (cosθ′ +v / c)2 + sin2θ′= sinθ′γ 1+ β cosθ′( )

tanθ2= sinθ1+ cosθ

tan θ2

⎛⎝⎜

⎞⎠⎟ =

(1 /γ )sinθ′1+ β cosθ′ + cosθ′ + β

= (1 /γ )sinθ′(1+ β )(1+ cosθ′)

tan θ2

⎛⎝⎜

⎞⎠⎟ =

1− β1+ β

⎛⎝⎜

⎞⎠⎟

1/2

tan θ′2

⎛⎝⎜

⎞⎠⎟ → θ <θ′

9

• Beaming (“headlight”) effect:If photons are emitted isotropically in , then half will have and half .

Consider a photon emitted at right angles to in . Then we have

For highly relativistic speeds, , becomes small:

Therefore, in frame K, photons are concentrated in the forward direction, with half of them lying within a cone of half-angle . Very few photons will be emitted .

K ′ θ′ < π / 2 θ′ > π / 2

v K ′

sinθb =1γ, cosθb = β, or tan θb

2⎛⎝⎜

⎞⎠⎟ =

1− β1+ β

⎛⎝⎜

⎞⎠⎟

1/2

γ 1 θb

θb ~1γ

Review of Lotentz Tmnsformatioav 11 1

h ’ h

Figuw 4.3 frrune K’.

Relativistic beaming of mdiation emitted isotmpically in the rest

4. Doppler Effect

We have seen that any periodic phenomenon in the moving frame K’ will appear to have a longer period by a factor y when viewed by local observers in frame K. If, on the other hand, we measure the arrival times of pulses or other indications of the periodic phenomenon that propagate with the velocity of light, then there will be an additional effect on the observed period due to the delay times for light propagation. The joint effect is called the Doppler effect.

In the rest frame of the observer K imagine that the moving source emits one period of radiation as it moves from point 1 to point 2 at velocity u. If the frequency of the radiation in the rest frame of the source is o’ then the time taken to move from point 1 to point 2 in the observer’s frame is given by the time-dilation effect:

Now consider Fig. 4.4 and note I = o h t and d = v At cose. The difference in arrival times AtA of the radiation emitted at 1 and 2 is equal to At minus the time taken for radiation to propagate a distance d. Thus we have

Therefore, the observed frequency w will be

277 w’ w= - = (4.1 1)

This is the relativistic Doppler formula. The factor y - ’ is purely a relativistic effect, whereas the 1 -(u/c)cosB factor appears even classi- cally. One distinction between the classical and relativistic points of view should be mentioned, however. The classical Doppler effect (say, for sound

K ′ K

1/γ θ 1/γ

beam half-angle:

10

11 2 Relativistic C o o a h and Kinematics

Observer

Figwe 4.4 Geometty for the Doppler effect.

waves) requires knowledge not only of the relative velocity between source and observer but also the velocities of source and observer relative to the medium (say, air) carrying the waves. The relativistic formula has no reference to an underlying medium for the propagation of light, and only the relative velocity of source and observer appears.

We can also write the Doppler formula as

(4.12a)

It is easy to show that the inverse of this is

(4.12b)

5. Proper Time

Although intervals of space and time are separately subject to Lorentz transformation and thus have differing values in differing frames of reference, there are some quantities that are the same in all Lorentz frames. An important such Lorentz invariant is the quantity dr defined by

C’ dT2 = C’ dt2 - ( dx2 + 4’ + dz’). (4.13)

• In the rest frame of the observer K, imagine that the moving source emits one period of radiation as it moves from point 1 to point 2 at velocity .

Let frequency of the radiation in the rest frame of the source = . Then the time taken to move from point 1 to point 2 in the observer’s frame is given by the time-dilation effect:

Difference in arrival times of the radiation emitted at 1 and 2:

Therefore, the observed frequency will be

Note appears even classically. The factor is purely a relativistic effect.

Transverse (or second-order) Doppler effect :

Doppler Effect

v

ω′(K ′)

Δt = Δt′γ = 2πω′

γ

ΔtA

ΔtA = Δt − dc= Δt 1− β cosθ( )

ω

ω = 2πΔtA

= ω′γ (1− β cosθ )

, or ωω′

= 1γ (1− β cosθ )

1− β cosθ γ −1

ωω′

= 1γ≤1 at θ = π / 2

11

• Beam half-angle:

• Angle for null Doppler shift:

Relativistic Doppler effect can yield redshift even as a source approaches.

• Note

sinθb = γ−1

ωω′

= 1γ (1− β cosθn )

= 1

→ cosθn =1−γ −1

β= 1−γ −1

1+ γ −1

⎛⎝⎜

⎞⎠⎟

1/2

θb ≤θn

cosθn =1−γ −1

1+ γ −1

⎛⎝⎜

⎞⎠⎟

1/2

≈1− 1γ

for γ 1

1− θn2

2≈1− 1

γ

∴ θn ≈2γ

≈ 2θb

0.0 0.2 0.4 0.6 0.8 1.0β

0102030405060708090

angl

e (d

eg)

θb

θn

100 101 102 103

γ

10-1

100

101

102

angl

e (d

eg) θb

θn

ωω′

>1 (θ <θn )ωω′

<1 (θ >θn ) �

source

b

n

vredshift

blueshift

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Lorentz Invariant• Lorentz invariant: A quantity (scalar) that remains unchanged by a Lorentz transform is said to

be a “Lorentz invariant.”

• Proper distance: Since all events are subject to the same transformation, the space-time “interval” between two event is also invariant.

This is the spatial distance between two events occurring at the same time. This is called the proper distance between the two points.

• Proper time (interval):

This measures time intervals between events occurring at the same spatial location

If the coordinate differentials refer to the position of the origin of another reference frame traveling with velocity , then

This is the time dilation formula in which is the time interval measured by the frame in motion.

ds2 ≡ dx2 + dy2 + dz2 − c2dt 2

(dx = dy = dz = 0)

v dτ = dt 1− β 2( )1/2 = dt /γdτ

x′2 + y′2 + z′2 − c2t′2 = γ 2 (x − βct)2 + y2 + z2 −γ 2 (ct − βx)2

= γ 2 (1− β 2 )x2 + y2 + z2 + γ 2 (β 2c2 − c2 )t 2

= x2 + y2 + z2 − c2t 2

c2dτ 2 ≡ −ds2 = c2dt 2 − (dx2 + dy2 + dz2 )

13

* Four-Vectors *• Four-vector: Invariant in 3D rotations:

By analogy, the invariance of the space-time interval suggests to define a vector in 4D space (4 dimensional space-time vector or four-vector). The quantities define coordinates of an event in space-time.

• Minkowski space: Space-time is not a Euclidean space; it is called Minkowski space.

Minkowski metric:

• Summation convention:

The invariant can now be written in terms of the Minkowski metric:

In any single term containing a Greek index repeated twice (between contravariant and covariant indices), a summation is implied over that index (originated by Einstein). This index is often called a dummy index.

x ≡ xµ =

x0

x1

x2

x3

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

=

ctxyz

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

= ctx

⎛⎝⎜

⎞⎠⎟

dx2 + dy2 + dz2

xµ (µ = 0,1,2,3)

ηµν =ηµν =

−1 0 0 00 1 0 00 0 1 00 0 0 1

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

ηµν =ηνµ

Note that this metric is symmetric:

s2 =µ=0

3

∑ν=0

3

∑ηµν xµxν

s2 =ηµν xµxν ηµµx

µNote that is regarded as meaningless.

Contravariant components

14

• Contravariant/Covariant components

They are related by

The metric can be used to raise or lower indices.

• Lorentz transform (corresponding to a boot along the x axis)

Any arbitrary Lorentz transformation can be written in the above form, since the spatial 3D rotation necessary to align the x axes before and after the boost are also of linear form.

xµ =

x0

x1

x2

x3

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

=

ctxyz

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

xµ =

x0x1x2x3

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

=

−ctxyz

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

contravariant components:

(superscripted)

covariant components:(subscripted)

xµ =ηµν xν , xµ =ηµν xν

s2 = xµxµ

Λµν =

γ −βγ 0 0−βγ γ 0 00 0 1 00 0 0 1

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

x′µ = Λµν x

ν

Λµν =

∂x′µ

∂xν

transformation matrix: Lorentz transformation:

x′µ =∂x′µ

∂xνxν ,

∂A

∂x′µ=

∂xν

∂x′µ∂A

∂xν

The components of a position (velocity etc.) vector contra-vary with a change of basis vectors to compensate. Transformation rules between the following two vector components are inverse. This is the basic idea of “contravariant” and “covariant.”

15

• Conditions for the Lorentz transformation:

From the invariance of , we must have

This can be true for arbitrary only if

Taking determinants yields

Proper Lorentz transformations (to keep the right-handness), which rules out reflections.

Isochronous Lorentz transformations (to ensure that the sense of flow of time is the same in frames)

• Lorentz transformation of the covariant component

s2

ηµν xµxν =ηστ x′

σ x′τ =ηστΛσµΛ

τν x

µxν

xµ

ηµν = ΛσµΛ

τνηστ or equivalently η = ΛTηΛ in matrix form

det Λ = ±1

det Λ = 1

Λ00 ≥1

′xµ =ηµτ x′τ =ηµτΛ

τσ x

σ =ηµτΛτση

σν xν

∴ ′xµ = Λ µνxν where Λ µ

ν≡ ηµτΛ

τση

σν

Λ µν=∂ ′xµ∂xν

16

• From the invariance of :

• For any arbitrary contravariant components,

• Note that

• Inverse transform

s2 = xµxµ

′xσ ′xσ = Λσ

ν xν Λσ

µxµ = Λσ

ν Λσµxν xµ

∴Λσν Λσ

µ= δ µ

ν

δ µν =

1 0 0 00 1 0 00 0 1 00 0 0 1

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

where we have introducedthe Kronecker delta

Qµ = δ µνQ

ν

Identity matrix

ηµσησν = δµν

∴Λσµ= Λ−1( )µσ

Λσ

µ× x′σ = Λσ

ν xν( ) → xµ = Λσ

µx′σ note: Λσ

µ= Λ−1( )µσ

17

Other Four-vectors• Four-vector:

• Consider two four-vectors

Therefore, the scalar product of any two four-vectors is a Lorentz invariant or scalar. In particular, the “square” of a four vector is an invariant. Thus, our starting point, the invariance of , is seen to be a general property of four-vectors.

• Note

Aµ =ηµνAν

A′µ = ΛµνA

ν

Aµ =ηµνAν

′Aµ = Λ µνAν

contravariant covariant

A →

A and

B

A′µ ′Bµ = Λµ

ν Λ µσAνBσ = δ σ

νAνBσ = AνBν →

A ⋅B = AµBµ = A′

µ ′Bµ

s2 = xµxµ

A ⋅A > 0 → spacelike four-vector= 0 → light-like (or null) four-vector< 0 → time-like four-vector

A0 → time componentAi → space-components (ordinary three-vector)

A ⋅B = −A0B0 +A ⋅B = −A0B0 + AiBi (i = 1,2,3)

18

Four-velocityThe (infinitesimally small) difference between the coordinates of two events is also a four-vector. Dividing by the proper time yields a four-vector, the four-velocity:

Transformation of the four-velocity:

U µ ≡ dxµ

dτU 0 = cdt

dτ= cγ u

U i = dxi

dτ= γ uu

i

γ u ≡ 1− u2 / c2( )−1/2 , u ≡ dxdt

U = γ u

cu

⎛⎝⎜

⎞⎠⎟

or where

U ⋅U =U µUµ = − γ uc( )2 + γ uu( )2 = −c2length of the four-velocity :

′U 0 = γ (U 0 − βU1)′U 1 = γ (−βU 0 +U1)′U 2 =U 2

′U 3 =U 3

γ ′u c = γ (cγ u − βγ uu1)

γ ′u ′u 1 = γ (−βcγ u + γ uu1)

γ ′u ′u 2 = γ uu2

γ ′u ′u 3 = γ uu3

γ ′u = γγ u (1−vu1 / c2 )γ ′u ′u 1 = γγ u (u

1 −v)

′u 1 = u1 −v1−vu1 / c2

γ ′u = γγ u 1−vu1

c2⎛⎝⎜

⎞⎠⎟

velocity component:

speed:

This is the previously derived formula.

19

Momentum and Energy• Four-momentum of a particle with a mass m0 is defined by

• In the nonrelativistic limit,

Therefore, we interpret as the total energy of the particle.

The quantity is interpreted as the rest energy of the particle.

Then,

Since , we obtain

• Photons are massless, but we can still define

Pµ ≡ m0Uµ

P0 = m0cγ v

Pi = γ vm0v

P0c = m0c

2γ = m0c2 1− v2

c2⎛⎝⎜

⎞⎠⎟

−1/2

= m0c2 + 12m0v2 +

E ≡ P0c = γ vm0c2

m0c2

p ≡γ vm0v, Pµ = (E / c, p)

P2 = −m0

2c2 = − E2

c2+ p 2

E2 = m02c4 + c2 p 2

U 2 = −c2

→P2 = 0Pµ = (E / c, p), E = p c

20

Wavenumber vector and frequency• Quantum relations:

We can define four wavenumber vector:

Then, we obtain an invariant:

Therefore, the phase of the plane wave is an invariant.

• Transform for (Doppler formula):

E = hν = ωp = E / c = k

k ≡ 1

P = ω

c, k⎛

⎝⎜⎞⎠⎟

ω = 2πνk = 2π / λ

⎛

⎝⎜⎞

⎠⎟

k ⋅ x = kµx

µ = k ⋅x −ωt

k ⋅k = k 2 −ω 2 / c2 = 0

Note that it’s a null vector:

k

′k 0 = γ (k0 − βk1)′k 1 = γ (−βk0 + k1)′k 2 = k2

′k 3 = k 3

′ω = γ (ω − βck1) =ωγ 1− v

ccosθ⎛

⎝⎜⎞⎠⎟

k1 = (ω / c)cosθ

21

* Tensor Analysis *• Definition:

zeroth-rank tensor : Lorentz invariant (scalar)

first-rank tensor : four-vector

second-rank tensor:

• Covariant components and mixed components:

• Transformation rules:

• symmetric tensor = a tensor that is invariant under a permutation of its indices.

• antisymmetric tensor : if it alternates sign when any two indices of the subset are interchanged.

′T µν = ΛµσΛ

ντT

στ

′x µ = Λµν x

ν

s′ = s

Tµν =ηµσηντTστ Tµ

ν =ηµσTσνT µ

ν =ηντTµτ

′Tµν =ηµαηνβ ′T αβ

=ηµαηνβΛαγ Λ

βδT

γδ

=ηµαηνβΛαγ Λ

βδη

γσηδτTστ

= Λ µσΛν

τTστ

νµ′T =ηνα ′T µα

=ηναΛµσΛ

αδT

σδ

=ηναΛµσΛ

αδη

δτT στ

= Λµσ Λν

τT σ

τ

′Tµν =ηµα ′T αν

=ηµαΛαβΛ

ντT

βτ

=ηµαΛαβΛ

ντη

βσTστ

= Λ µβΛν

τTστ

T µν = T νµ

T µν = −T νµ

22

• Examples of the second-rank tensors

A product of two vectors:

The Minkowski metric:

The Kronecker-delta:

AµBν

′A µ ′B ν = ΛµσΛ

ντA

σBτ

ηµν

δ µν

23

• Higher-rank tensors

- Addition:

- Multiplication:

- Raising and Lowering Indices: The metric can be used to change contravariant indices into covariant ones, and vice versa, by the processes of raising and lowering.

- Contraction:

- Gradients of Tensor Fields: A tensor field is a tensor that is a function of the spacetime coordinates in Cartesian coordinate systems. The gradient operation acting on such a field produces a tensor field of on higher rank with μ as a new covariant index.

- Invariance of form or Lorentz covariance or covariance: A fundamental property of a tensor equation is that if it is true in one Lorentz frame, then it is true in all Lorentz frames. Covariance plays a powerful role in helping decide what the proper equations of physics are.

Aµ + Bµ , Fµν +Gµν

AµBν , FµνGστ

AµBν → AµBµ

T µνσ → T µν

ν

scalar

vector

ν′T µν = ΛµαΛ

νβ Λν

τT αβ

τ = Λµαδ

τβT

αβτ = Λµ

αTαβ

β

∂/ ∂xµ ≡ ∂µ

λ → ∂λ∂xµ ≡ ∂µλ ≡ λ,µ vector (gradient) Aµ → ∂Aµ

∂xµ ≡ ∂µAµ ≡ Aµ ,µ scalar (divergence)

24

Mathematical Formulae• Gamma function

• Euler-Mascheroni constant

• Modified Bessel function of the second kind

Kn (x) ≈− ln(x / 2)−γ if n = 0

Γ(n)2

2x

⎛⎝⎜

⎞⎠⎟n

if n > 0

⎧

⎨⎪

⎩⎪

Kn (x) ≈π2xe− x 1+ (4n

2 −1)8x

⎡

⎣⎢

⎤

⎦⎥

(2) x≫ n2 −1/ 4

(1) 0 < x < n +1

γ ≡ limn→∞

1k− ln(n)

k=1

n

∑⎛⎝⎜⎞⎠⎟= − e− x ln xdx

0

∞

∫ = 0.577215664901532

Kn (x) ≡Γ(n +1/ 2)(2x)n

πcost

t 2 + x2( )n+1/2dt

0

∞

∫

Γ(x) ≡ t x−1e− t dt0

∞

∫ , Γ(x) = (x −1)!= (x −1)Γ(x − 2), Γ(3 / 2) = 12Γ(1 / 2) = π

2

Kn−1(x)− Kn+1(x) = − 2nxKn (x)

Kn−1(x)+ Kn+1(x) = −2 ′Kn (x)

Recurrence formulae

xKn2 (x)dx = 1

2x2 Kn

2 (x)− Kn−1(x)Kn+1(x)⎡⎣ ⎤⎦∫= −xKn−1(x)Kn (x)+

12x2 Kn

2 (x)− Kn−12 (x)⎡⎣ ⎤⎦

Integral formula

2

[Covariance of Electromagnetic Phenomena]• Equation of charge conservation:

The above equation can be written as a tensor equation,

if the four-current is defined by

• Note that the Jacobian (determinant) of the transformation from to is simply the determinant of , which is unity. Therefore, the four-volume element is an invariant.

Since is the zeroth component of the four-current, the charge element within a three-volume element is an invariant.

It is also an empirical fact that e is invariant.

∂ρ∂t

+∇⋅ j = 0

∂ jµ

∂xµ = 0, jµ ,µ = 0 or ∂µ jµ = 0

jµ =ρcj

⎛

⎝⎜

⎞

⎠⎟

xµ ′xµΛ

d ′x0d ′x1d ′x2d ′x3 = detΛ dx0dx1dx2dx3 = dx0dx1dx2dx3

ρ

de = ρdx1dx2dx3

jµ =−ρcj

⎛

⎝⎜

⎞

⎠⎟

∂µ≡∂

∂xµ = ∂c∂t

, ∂∂x, ∂∂y, ∂∂z

⎛⎝⎜

⎞⎠⎟

∂µ≡ ∂∂xµ

= − ∂c∂t

, ∂∂x, ∂∂y, ∂∂z

⎛⎝⎜

⎞⎠⎟

3

• The set of vector and scalar wave equations in the Lorentz gauge is

If we define the four-potential

then the wave equations can be written as the tensor equations

• The Lorentz gauge should be preserved under Lorentz transformations.

∇2A − 1c2

∂2A∂t 2

= − 4πcj

∇2φ − 1c2

∂2φ∂t 2

= − 4πc

ρc( )

Aµ = φA

⎛

⎝⎜⎞

⎠⎟, Aµ =

−φA

⎛

⎝⎜⎞

⎠⎟

∂2Aµ

∂xν ∂xν= − 4π

cjµ , ∂ν ∂

ν Aµ = − 4πcjµ or A,ν

µ ,ν = − 4πcjµ

∇⋅A + 1c∂φ∂t

= 0 → ∂Aµ

∂xµ = 0 or Aµ ,µ = 0

! ≡ ∂2

∂xν ∂xν→ !Aµ = − 4π

cjµd’Alembertian operator:

4

• Electromagnetic field tensor:

The fields are expressed in terms of the potentials as

The x components of the electric and magnetic fields are explicitly

These equations imply that the electric and magnetic fields, six components in all, are the elements of a second-rank, antisymmetric field-strength tensor, because a rank two antisymmetric tensor has exactly six independent components.

E = −∇φ − 1c∂A∂t

B = ∇×A

Ex = − 1c∂Ax

∂t− ∂φ∂x

= ∂0A1 − ∂1A0( )

Bx =∂Az∂y

−∂Ay

∂z= ∂2A3 − ∂3A2( )

Fµν ≡ ∂µAν − ∂ν Aµ =

0 Ex Ey Ez

−Ex 0 Bz −By

−Ey −Bz 0 Bx

−Ez By −Bx 0

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

F0i = Ei

Fi0 = −Ei

F12 = −F21 = B3,!

5

covariant field-strength tensor

• The two Maxwell equations containing sources (inhomogeneous equations):

Fµν =ηµαηνβFαβ

F0i =η0αηiβFαβ = −F0i

Fi0 =ηiαη0βFαβ = −Fi0

Fij =ηiαη jβFαβ = Fij

∇⋅E = 4πρ

∇×B− 1c∂E∂t

= 4πcj

∂i Eii=1

3

∑ = 4πcj0

∂2B3 − ∂3B2 − ∂0E1 =4πcj1

− ∂i Fi0

i=1

3

∑ = 4πcj0

−∂0F01 − ∂2F

21 − ∂3F31 = 4π

cj1

∂µFµν = − 4π

cjν or ∂ν F

µν = 4πcjµ ∂µFµν = − 4π

cjν or ∂ν Fµν =

4πcjµ

Fµν ≡ ∂µAν − ∂ν Aµ =

0 −Ex −Ey −Ez

Ex 0 Bz −By

Ey −Bz 0 Bx

Ez By −Bx 0

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

F0i = −Ei

Fi0 = Ei

F12 = −F21 = B3,!

6

• The conservation of charge easily follows from the above equation and the asymmetric property.

• The “internal” Maxwell equations (homogeneous equations):

The equation can be written concisely as , where [ ] around indices denote all permutations of indices, with even permutation contributing with a positive sign and odd permutation with a negative sign, for example,

∇⋅B = 0

∇×E+ 1c∂B∂t

= 0

∂i Bii=1

3

∑ = 0

∂2E3 − ∂3E2 + ∂0B1 = 0

∂1F23 + ∂2F

31 + ∂3F12 = 0

∂2F30 + ∂3F

20 + ∂0F23 = 0

∂µFνσ + ∂ν F

σµ + ∂σ Fµν = 0 or ∂µFνσ + ∂ν Fσµ + ∂

σ Fµν = 0

F[µν ,σ ] = 0 or F[µν ,σ ] = 0

Fµν = ∂µAν − ∂ν Aµ = A[ν ,µ ]

∂ν jν = − c

4π∂ν ∂µF

µν = 0

∂µ∂ν Fµν = −∂ν ∂µF

νµ = −∂µ∂ν Fµν → ∂µ∂ν F

µν = 0

7

- Transformation of Electromagnetic Fields• Since is a second-rank tensor, its components transform in the usual way:

For a pure boost along the x-axis:

• In general,

The concept of a pure electric or pure magnetic is not Lorentz invariant.

Fµν

F′µν = ∂x′µ

∂xα∂x′ν

∂xβFαβ = Λµ

αΛνβF

αβ

Λµν =

γ −βγ 0 0−βγ γ 0 00 0 1 00 0 0 1

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

Ex′ = F′01 = Λ0

0Λ11F

01 + Λ01Λ

10F

10 = γ 2Ex − β2γ 2Ex = Ex

Ey′ = F′02 = Λ0

0Λ22F

02 + Λ01Λ

22F

12 = γ Ey − βγ Bz

Ez′ = F′03 = Λ0

0Λ33F

03 + Λ01Λ

33F

13 = γ Ez + βγ By

Bx′ = F′23 = Λ2

2Λ33F

23 = Bx

By′ = F′31 = Λ3

3 Λ10F

30 − Λ11F

31( ) = βγ Ez + γ By

Bz′ = F′12 = Λ1

0Λ22F

02 + Λ11Λ

22F

12 = −βγ Ey + γ Bz

′E|| = E||′E⊥ = γ E⊥ + β ×B( )

′B|| = B||′B⊥ = γ B⊥ − β ×E( )

8

• Lorenz invariants:

dot product of F with itself or “square” of F:

determinant of F:

FµνFµν = F0iF0ii=1

3

∑ + Fi0Fi0i=1

3

∑ + FijFiji≠ j∑ = 2 B2 −E2( )

det F = E ⋅B( )2

9

[Relativistic Mechanics and the Lorentz Four-Force]• We can define a four-acceleration in exactly the same way as we obtained the four-velocity.

Note that the four-acceleration and four-velocity are orthogonal:

• We can also define the four-force from the Lorentz force, so as to obtain a relativistic form of Newton’s equation.

Since , the Lorentz four-force should involve (1) the electromagnetic field tensor and (2) the four-velocity and should also be (3) a four-vector and (4) proportional to the charge of the particle. Therefore, the simplest possibility is

aµ

aµ ≡ dUµ

dτ

!a ⋅!U ≡ dU

µ

dτUµ =

12ddτ

U µUµ( ) = 12ddτ

−c2( ) = 0

Fµ

Fµ ≡ m0aµ = dP

µ

dτ

!F = d

!Pdτ

= γ d!Pdt

= γ 1cdEdt, dpdt

⎛⎝⎜

⎞⎠⎟

FLorentz = q E+ 1cv ×B( )⎡

⎣⎢⎤⎦⎥

FµLorentz =

qcFµνUν

10

• Let’s check to see if it is indeed what we want.

Therefore, we obtained the desired expression for the four-Lorentz force.

• Note that the four-force is always orthogonal to the four-velocity:

It implies that every four-force must have some velocity dependence.

For the Lorentz four-force, in particular, we find

because is antisymmetric and is symmetric.

F1Lorentz =qcF1νUν =

qcF10 (−γ c)+ F12γ v2 + F13γ v3( )

= qcγ E1c + B3v2 − B2v3( )

dpdt

= q E+ 1cv ×B⎛

⎝⎜⎞⎠⎟

F0

Lorentz =qcF0νUν =

qc

Eiγ vii=1

3

∑ = qcγ (E ⋅v) dE

dt= qE ⋅v : conservation of energy

The rate of change of particle energy is the mechanical work done on the particle by the field.

!F ⋅!U = m0 (

!a ⋅!U ) = 0

!FLorentz ⋅

!U = q

cFµνUµUν = 0,

Fµν UµUν

11

[Fields of a Uniformly Moving Charge]• Let’s find the fields of a charge moving with constant velocity along the x axis. In the rest frame

of the particle the fields are

Since , we obtain

E′ = ′Ex , ′Ey , ′Ez( ) = q′r 3 x′, y′,z′( )

B′ = 0,0,0( )

′r = ′x 2 + ′y 2 + ′z 2( )1/2where

E|| = ′E||E⊥ = γ ′E⊥ − β ×B′( )

B|| = ′B||B⊥ = γ ′B⊥ + β ×E′( )

inverse transformation of the previous one:Ex =

qx′′r 3

Ey = γqy′′r 3

Ez = γqz′′r 3

Bx = 0

By = −γβ qz′′r 3

Bz = γβqy′′r 3

x′ = γ x −vt( ), y′ = y, z′ = z

Ex = γq x −vt( )

r3

Ey = γqyr3

Ez = γqzr3

Bx = 0

By = −γβ qzr3

Bz = γβqyr3

r = x −vt( )2 + y2 + z2⎡⎣ ⎤⎦1/2

where

Is this equivalent to the fields givenby the Lienard-Wiechert potentials?

v

12

- Velocity field from the retarded potential• For simplicity, assume .

Let us first find where the retarded position of the particle is. vt vtret

R

n

x, y( )

x

yz = 0

tret ≡ t − R / c

R2 = x −vtret( )2 + y2 = x + βR( )2 + y2

n = x + βR( )R

x! + yRy! = x

R+ β⎛

⎝⎜⎞⎠⎟ x! + y

Ry!

(1− β 2 )2R2 − 2xβR − x 2 − y2 = 0

R2 − 2xγ 2βR −γ 2 x 2 + y2( ) = 0

E = Ex ,Ey ,Ez( ) = γ qr3

x −vt, y,z( )

= γ qγ 2x 2 + y2( )3/2

x , y,0( ) x ≡ x −vtwhere

R = γ 2βx ± γ 4β 2x 2 + γ 2 x 2 + y2( )⎡⎣ ⎤⎦1/2

= γ 2βx ± γ γ 2β 2x 2 + x 2 + y2( )⎡⎣ ⎤⎦1/2

= γ 2βx ± γ γ 2x 2 + y2( )1/2

R

n− β = x

Rx! + y

Ry!(1)

E = γ qγ 2x 2 + y2( )3/2

n− β( )

(2) γ 2x 2 + y2( )1/2 = R −γ 2βxγ

= Rγ 1γ 2 −

βxR

⎛⎝⎜

⎞⎠⎟

= Rγ 1− β 2 − βxR

⎛⎝⎜

⎞⎠⎟

= Rγ 1− β xR+ β⎛

⎝⎜⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

= Rγ 1− n ⋅ β( ) = Rγκ

positive solution → R = γ 2βx + γ γ 2x 2 + y2( )1/2 ∴ E = qn− β( )γ 2κ 3R2

= qn− β( ) 1− β 2( )

κ 3R2: velocity field

13

- Time-dependence of the electric field at a point• Let us choose the field point to be at .

This involves no loss in generality. Then,

The field of a highly relativistic charge

appears to be a pulse of radiation traveling

in the same direction as the charge and confined

to the tranverse plane.

vt

b

0,b,0( )

x

y

v

E

B

z

(0,b,0)

Ex = − qγ vt(γ 2v2t 2 + b2 )3/2

= − qb2

γ vt / b(γ 2v2t 2 /b2 +1)3/2

Ey =qγ b

(γ 2v2t 2 + b2 )3/2= qγb2

1(γ 2v2t 2 /b2 +1)3/2

Ez = 0

Bx = 0By = 0Bz = βEy

t =

1

2

b

γ v

Max Ex =2

33/2q

b2

-3 -2 -1 0 1 2 3

0

1

2

3Max Ey = γq

b2

t ≈

b

γ v

As γ ≫1 → Ex ≪ Ey

14

- Spectrum of the pulse• Spectrum of this pulse of virtual radiation.

This integral can be done in terms of the modified Bessel function:

Thus the spectrum is

E(ω ) = 12π

Ey (t)eiωt dt∫

= qγ b2π

γ 2v2t 2 + b2( )−3/2 eiωt dt−∞

∞

∫ = qγ b2π

γ 2v2t 2 + b2( )−3/2 eiωt + e− iωt( )dt0

∞

∫

= qγ bπ

γ 2v2t 2 + b2( )−3/2 cosωt dt0

∞

∫

Kn (x) ≡Γ(n +1/ 2)(2x)n

πcost

t 2 + x2( )n+1/2dt

0

∞

∫ Γ(3 / 2) = 12Γ(1 / 2) = π

2Gamma function:

E(ω ) = qγ bπ

γ 2v2

ω 2

⎛⎝⎜

⎞⎠⎟

−3/21ω

ω 2t 2 + b2ω 2

γ 2v2⎛⎝⎜

⎞⎠⎟

−3/2

cosωt dωt0

∞

∫

= qπbv

bωγ v

K1bωγ v

⎛⎝⎜

⎞⎠⎟

dWdAdω

= c E(ω )2= q2

π 2b2v2bωγ v

⎛⎝⎜

⎞⎠⎟

2

K12 bω

γ v⎛⎝⎜

⎞⎠⎟

15

The spectrum starts cut off for .

• Total energy per unit frequency range is obtained by

The lower limit has been chosen as some minimum distance, such that the approximation of the field by means of classical electrodynamics and a point charge is valid.

ω > γ v /b

Δω ~ 1

Δt~ γ v /b

Fields of a Unifonry Moving cluvge 135

Figure 4.8 Area e&ment perpedicuku to the Oelocity of a moving pmti'e&.

are (1) bmin = radius of ion, if field is that of an ion and (2) b---A/rnc = Compton wavelength of particle. The integral is now

where

This integral can be done in terms of Bessel functions

dW 2q2c -- - __ [ x K , ( x ) K , ( x ) - f x 2 ( K : ( x ) - K ~ ( x ) > ] . (4.74b) did nu2

Two limiting forms occur when w is small, w<<yu/b,,,, and when w is large, w>>yu/bmln:

__ dW = --exp( q2c - --), 2wbmin w>- YU

dw 2 v 2 bmn

(4.75a)

(4.75b)

These forms can be derived approximately by direct integration of xK:(x) , using the asymptotic results K l ( x ) - 1 /x, x<< 1, and K l ( x ) - ( n / 2 x ) ' / 2 e - x ,

x>> 1.

dWdω

= 2π dWdAdω

bdbbmin

bmax∫

dWdω

= 2q2c

πv2yK1

2 (y)dyx

∞

∫

= 2q2c

πv2xK0 (x)K1(x)−

12x2 K1

2 (x)− K02 (x)( )⎡

⎣⎢⎤⎦⎥

y ≡ ωb

γ v, and x ≡ ωbmin

γ vwhere

16

• Two limiting cases:

(1)ω ≪ γ v

bmin(x≪1),

xK0 (x)K1(x)−12x2 K1

2 (x)− K02 (x)( )

≈ x − ln(x / 2)−γ( ) 1x− x

2

21x2

− ln(x / 2)+ γ( )2⎡⎣⎢

⎤⎦⎥

≈ ln 2xe−(γ +1/2)⎡

⎣⎢⎤⎦⎥

= ln 0.68x

⎛⎝⎜

⎞⎠⎟

(2)ω ≫ γ v

bmin(x≫1),

xK0 (x)K1(x)−12x2 K1

2 (x)− K02 (x)( )

≈ x π2x

e−2 x − 12x2 π2x

e−2 x 38x

⎛⎝⎜

⎞⎠⎟2

− 18x

⎛⎝⎜

⎞⎠⎟2⎡

⎣⎢

⎤

⎦⎥

= π4e−2 x

dWdω

= 2q2c

πv2ln 0.68 γ v

ωbmin

⎛⎝⎜

⎞⎠⎟

dWdω

= q2c2v2

exp − 2ωbminγ v

⎛⎝⎜

⎞⎠⎟

17

[Emission from Relativistic Particles]• Total emitted power:

Imagine an instantaneous rest frame K’, such that the particle has zero velocity at a certain time. We can then calculate the radiation emitted by use of the dipole (Larmor) formula.

Suppose that the particle emits a total amount of energy in this frame in time . The momentum of this radiation is zero, , because the emission is symmetrical in the frame.

The energy in a frame K moving with velocity w.r.t. the particle is:

The time interval is simply

The total power emitted in frames K and K’ are given by

Thus the total emitted power is a Lorentz invariant for any emitter that emits with front-back symmetry in its instantaneous rest frame.

dW ′ dt′dp′ = 0

−v

dW = γ dW ′ dE = cdP0 = c !Λ0µd ′P µ = c !Λ0

0d ′P 0 = γ d ′Edt

dt = γ dt′

P = dWdt, P′ = dW ′

dt′

P = P′

18

• the Larmor formula in covariant form:Recall that , and because in the instantaneous rest frame of the particle, we have

Therefore,

• Expression of P in terms of the three-vector acceleration

Recall

P′ = 2q2

3c3a′ 2

!a ⋅!U = 0

!U = (c,0)

′a0 = 0 → a′ 2 = ′ak ′a k = ′aµ ′a µ = !a ⋅ !a

P = 2q

2

3c3!a ⋅ !a

dt = γ dt′ + vc2dx′!

⎛⎝⎜

⎞⎠⎟

u! =′u! +v

1+v ′u! / c2

u⊥ = ′u⊥

γ (1+v ′u! / c2 )

σ ≡ (1+v ′u! / c2 )

dt = γ dt′σ

u! =′u! +vσ

u⊥ = ′u⊥

γσ

dt = γ dt′σ

du! =d ′u!σ

−′u! +vσ 2

vc2d ′u!

=d ′u!σ 2 1− v2

c2⎛⎝⎜

⎞⎠⎟=

d ′u!γ 2σ 2

du⊥ = d ′u⊥

γσ− ′u⊥

γσ 2vc2d ′u!

= 1γσ 2 σdu⊥ ′ −

vu⊥ ′c2

du! ′⎛⎝⎜

⎞⎠⎟

19

Hence,

In an instantaneous rest frame of a particle,

Thus we can write

a|| =du!dt

= 1γ 3σ 3

d ′u!dt′

a⊥ = du⊥

dt= 1γ 2σ 3 σ d ′u⊥

dt′− v ′u⊥

c2d ′u!dt′

⎛⎝⎜

⎞⎠⎟

a|| =1

γ 3σ 3 ′a||

a⊥ = 1γ 2σ 3 σ ′a⊥ −

v ′u⊥

c2′a||

⎛⎝⎜

⎞⎠⎟

σ ≡ 1+

v ′u!c2

⎛⎝⎜

⎞⎠⎟where

Transformation of three-vector acceleration:

′u|| = ′u⊥ = 0, σ = 1

′a|| = γ3a||

′a⊥ = γ 2a⊥

P = 2q

2

3c3a′ 2 = 2q

2

3c3′a!2 + ′a⊥

2( ) P = 2q

2

3c3γ 4 γ 2a!

2 + a⊥2( )

tan ′θa ≡

′a⊥′a!= 1γa⊥a!

= 1γtanθaNote

20

• Angular Distribution of Emitted and Received Power

In the instantaneous rest frame of the particle, let us consider an amount of energy dW’ that is emitted into the solid angle (see the above figure).

Recall

I# Relarioistic Covariance and Klnemotics

It is convenient to express P in terms of the three-vector acceleration d2x/dr2 rather than in terms of the four-vector acceleration d2xp/dr2 . It can easily be shown (see Problem 4.3) that if K‘ is an instantaneous rest frame of a particle, then

a;l= Y 3a,,,

a; = y 2 a,.

(4.9 1 a)

(4.91b)

Thus we can write

(4.92)

Angular Distribution of Emitted and Received Power

In the instantaneous rest frame of the particle, let us consider an amount of energy dW‘ that is emitted into the solid angle dSt’=sinB‘dB‘d@’ about the direction at angle 8‘ to the x’ axis (see Fig. 4.9). It is convenient to introduce the notations

Since energy and momentum form a four-vector, the transformation of the energy of the radiation is,

d W = y ( d W’ + u dP:) = y ( 1 + Bp’) dW’. (4.93)

Figurn 4.9 Lorentz tmfonnation of the angular dist&utim of emitted power.

dΩ′ = sinθ′dθ′dφ′

µ ≡ cosθ → dΩ = dµdφ µ′ ≡ cosθ′→ dΩ′ = dµ′dφ′

d ′px = p cosθ′dpx = p cosθ

cosθ = cosθ′ + β1+ β cosθ′

→ µ = µ′ + β1+ βµ′

, or inverse µ′ = µ − β1− βµ

dµ = dµ′1+ βµ′

− µ′ + β1+ βµ′( )2

βdµ′dµ = dµ′

γ 2 1+ βµ′( )2, dµ = γ 2 1− βµ( )2 dµ′

dΩ = dΩ′γ 2 1+ βµ′( )2

, dΩ = γ 2 1− βµ( )2 dΩ′

dφ′ = dφ

Note:

21

• Power

Recall that energy and momentum form a four-vector

In the rest frame, the power emitted in a unit time interval is

However, in the observer’s frame, there are two possible choices for the time interval to calculate the power.

(1) :

This is the time interval during which the emission occurs. With this choice we obtain the emitted power.

(2) :

This is the time interval of the radiation as received by a stationary observer in K. With this choice we obtain the received power.

dP′dΩ′

≡ dW ′dt′dΩ′

dt = γ dt′

dtA = γ (1− βµ)dt′, or dtA = γ−1(1+ βµ′)−1dt′

dWdΩ

= γ 3 1+ βµ′( )3 dW ′dΩ′

, dWdΩ

= γ −3 1− βµ( )−3 dW ′dΩ′

Pµ = Ec,p⎛

⎝⎜⎞⎠⎟ , and p = E

c dW = γ dW ′ +vd ′px( ) = γ 1+ βµ′( )dW ′

∴ dW = γ 1+ βµ′( )dW ′, dW = γ −1 1− βµ( )−1 dW ′

22

• Thus we obtain the two results:

is the power actually measured by an observer. It has the expected symmetry property of yielding the inver transformation by interchanging primed and unprimed variables, along with a change of sign of β.

is used in the discussion of emission coefficient.

In practice, the distinction between emitted and received power is often not important, since they are equal in an average sense for stationary distributions of particles.

• Beaming effect:

If the radiation if isotropic in the particle’s frame, then the angular distribution in the observer’s frame will be highly peaked in the forward direction for highly relativistic velocities.

The factor is sharply peaked near with an angular scale of order .

dPedΩ

= γ 2 1+ βµ′( )3 dP′dΩ′

= γ −4 1− βµ( )−3 dP′dΩ′

dPrdΩ

= γ 4 1+ βµ′( )4 dP′dΩ′

= γ −4 1− βµ( )−4 dP′dΩ′

Pr

Pe

γ −4 1− βµ( )−4

γ −4 1− βµ( )−4 ≈ γ −4 1− 1− 12γ 2

⎛⎝⎜

⎞⎠⎟1− θ

2

2⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

−4

= γ −4 12γ 2 +

θ 2

2⎛⎝⎜

⎞⎠⎟

−4

= 2γ1+ γ 2θ 2

⎛⎝⎜

⎞⎠⎟

4

θ ≈ 0 1/γ

23

• Dipole emission from a slowly moving particle

Using and , we obtain

(1) Acceleration parallel to velocity:

(2) Acceleration perpendicular to velocity:

dP′dΩ′

= q2 ′a 2

4πc3sin2Θ′ Θ′ = the angle between the acceleration and

the direction of emission.

′a|| = γ3a||, ′a⊥ = γ 2a⊥

dPrdΩ

= γ −4 1− βµ( )−4 dP′dΩ′

dPrdΩ

= q2

4πc3γ 2a!

2 + a⊥2( )

1− βµ( )4sin2Θ′ To use this formula, we must

relate to the angles in K.Θ′

Θ′ = θ′, a⊥ = 0

sin2Θ′ = 1− µ′2 = 1− µ − β1− βµ

⎛⎝⎜

⎞⎠⎟

2

= 1− µ2

γ 2 1− βµ( )2

dP!dΩ

=q2a!

2

4πc31− µ2

1− βµ( )6

Emission from Relativistic Partick 143

Figure 4.10 Geometry for dipole emission from a particle instantaneously at &?St.

To use this formula we must relate 0' to the angles in the frame K. This is difficult in the general case, so we work out the angular distribution of the received power for special cases:

1-Acceleration /I to Velocity. Here 0'= 6' so that

(4.100)

where we have used Eq. (4.94). Substituting Eq. (4.100) into Eq. (4.99) with a, =0, we obtain

(4.101)

2-Acceleration I to Velocity. Here cos 0' = sin 6' cos +', so that

Thus we have the result

(4.102)

cosΘ′ = sinθ′cosφ′, a! = 0

sin2Θ′ = 1−1− µ2( )cos2φγ 2 1− βµ( )2

dP⊥dΩ

= q2a⊥2

4πc31

1− βµ( )41−

1− µ2( )cos2φγ 2 1− βµ( )2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

x

y

z

(when a is in y-direction in the above figure)

24

(3) In general

• In the extreme relativistic limit, the radiation becomes

strongly peaked in the forward direction.

“ I )

Figutv 4. I la Dipole mdiation pattern for patii& at mst.

(11)

Figwp 4.llb Angular dktribution of mdiatim emitted by a partic& with parollel accelerariosl and wlocity.

( r ) Figutv 4 .11~ Same as a

(11)

Figwp 4.11d Angular distribution of mdhtion emitted by a particle with perpendicular acceleration and wlocity.

3-Extreme Relativistic Limit. When y>> I , the quantity (1 - Pp) in the denominators becomes small in the forward direction, and the radiation becomes strongly peaked in this direction. Using the same arguments as before, we obtain

“ I )

Figutv 4. I la Dipole mdiation pattern for patii& at mst.

(11)

Figwp 4.llb Angular dktribution of mdiatim emitted by a partic& with parollel accelerariosl and wlocity.

( r ) Figutv 4 .11~ Same as a

(11)

Figwp 4.11d Angular distribution of mdhtion emitted by a particle with perpendicular acceleration and wlocity.

3-Extreme Relativistic Limit. When y>> I , the quantity (1 - Pp) in the denominators becomes small in the forward direction, and the radiation becomes strongly peaked in this direction. Using the same arguments as before, we obtain

“ I )

Figutv 4. I la Dipole mdiation pattern for patii& at mst.

(11)

Figwp 4.llb Angular dktribution of mdiatim emitted by a partic& with parollel accelerariosl and wlocity.

( r ) Figutv 4 .11~ Same as a

(11)

Figwp 4.11d Angular distribution of mdhtion emitted by a particle with perpendicular acceleration and wlocity.

3-Extreme Relativistic Limit. When y>> I , the quantity (1 - Pp) in the denominators becomes small in the forward direction, and the radiation becomes strongly peaked in this direction. Using the same arguments as before, we obtain

“ I )

Figutv 4. I la Dipole mdiation pattern for patii& at mst.

(11)

Figwp 4.llb Angular dktribution of mdiatim emitted by a partic& with parollel accelerariosl and wlocity.

( r ) Figutv 4 .11~ Same as a

(11)

Figwp 4.11d Angular distribution of mdhtion emitted by a particle with perpendicular acceleration and wlocity.

3-Extreme Relativistic Limit. When y>> I , the quantity (1 - Pp) in the denominators becomes small in the forward direction, and the radiation becomes strongly peaked in this direction. Using the same arguments as before, we obtain

x′

y′

z′

′φaφ′

′θaθ′

a

vΘ′

Ω′cosΘ′ = µ′ ′µa + 1− ′µ 2( )1/2 1− ′µa

2( )1/2 cos φ′ − ′φa( )

See Eq. (219) in Chadrasekhar (1960)

particle’s rest frame: observer’s frame:

parallel acceleration:

perpendicular acceleration:

25

[Invariant Phase Volumes and Specific Intensity]• Phase volume

Consider a group of particles that occupy a slight spread in position and in momentum at a particular time. In a rest frame comoving with the particles, they occupy a spatial volume element and a momentum volume element.

In the observer’s frame,

We note that because the velocities are near zero in the comoving frame and the energy is quadratic in velocity. Therefore, we have

This contains no reference to particle mass, and therefore it has applicability to photons.

The phase space density

is an invariant, since the number of particles within the phase volume element is a countable quantity and itself invariant.

d 3x′ = dx′dy′dz′d 3p′ = d ′pxd ′pyd ′pz d ′V ≡ d 3x′d 3p′ = dx′dy′dz′d ′pxd ′pyd ′pz

phase volume in the comoving frame:

dx = γ −1dx′, dy = dy′, dz = dz′dpx = γ d ′px + βd ′P0( ), dpy = d ′py , dpz = d ′pz

d ′P0 = 0 +O(d ′px2 )

dpx = γ d ′pxz d ′V ≡ d 3x′d 3p′ = d 3x d 3p ≡ dVand : Lorentz invariant

f ≡ dN

dV

26

• Specific Intensity and Source FunctionDefinition of the energy density per unit solid angle per frequency range.

Since we find that

Because the source function occurs in the transfer equation as the difference , the source function must have the same transformation properties as the intensity.

• Optical Depth, Absorption Coefficient and Emission CoefficientThe optical depth must be an invariant, since gives the fraction of photons passing through the material, and this involves simple counting.

hν fp2dpdΩ =Uν (Ω)dΩdν

p = hν / c and Uν (Ω) = Iν / cIνν 3 = Lorentz invariant

Iν − Sµ

e−τ

τ = Lorentz invariant

27

• Absorption Coefficient and Emission CoefficientConsider the optical depth in two frames:

Then, the optical depth is

Note that is proportional to the y component of the photon four-momentum .

Both are the same in both frames, being perpendicular to the motion. Therefore, we have

Finally, we obtain the transformation of the emission coefficient from the definition of the source function:

rmMricurt Phase vobnes atid S ~ C $ C Intensity 147

Figurn 4.12 Tmfonnation of a moving, absorbing medium

To find the transformation of absorption coefficient we imagine material in frame K streaming with velocity u between two planes parallel to the x axis. Let K‘ be the rest frame of the material. (See Fig. 4.12). The optical depth T along the ray must be an invariant, since e-‘ gives the fraction of photons passing through the material, and this involves simple counting. Thus we have the result

vayy = Lorentz invariant. I -- 1% 7 = - -

sine vsine

The transformation of sin8 can be found by noting that vsin8 is simply proportional to the y component of the photon four-momentum k,. But both k, and I are the same in both frames, being perpendicular to the motion. Therefore

vg = Lorentz invariant. (4.112)

Finally we find the transformation of the emission coefficient j , = aYS, from Eqs. (4.1 1 1) and (4.1 12):

J Y - = Lorentz invariant. Y 2

(4.1 13)

Another derivation of Eq. (4.1 13) can be based on Eq. (4.97a). The emission coefficient can be written as

(4.1 14)

where n is the density of emitters (particles/cm’). Now, from Eq. (4.12b) we have dv = dv’y( 1 + Pp’), and also n = yn’ by Lorentz contraction along the motion. Thus we have

τ = lαν

sinθ= lν sinθ

ναν = Lorentz invariant

ν sinθ

!k = ω

c, k⎛

⎝⎜⎞⎠⎟

ky and l

ναν = Lorentz invariant

jνν 2 = Lorentz invariant

Sν ≡ jν /αν

28