Special Relativity

Christopher R. Prior

Accelerator Science and Technology Centre

Rutherford Appleton Laboratory, U.K.

Fellow and Tutor in Mathematics

Trinity College, Oxford

2

Overview• The principle of special relativity• Lorentz transformation and consequences• Space-time• 4-vectors: position, velocity, momentum, invariants,

covariance.• Derivation of E=mc2

• Examples of the use of 4-vectors• Inter-relation between and , momentum and energy• An accelerator problem in relativity• Relativistic particle dynamics• Lagrangian and Hamiltonian Formulation• Radiation from an Accelerating Charge• Photons and wave 4-vector• Motion faster than speed of light

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Reading• W. Rindler: Introduction to Special Relativity (OUP

1991)• D. Lawden: An Introduction to Tensor Calculus and

Relativity• N.M.J. Woodhouse: Special Relativity (Springer

2002)• A.P. French: Special Relativity, MIT Introductory

Physics Series (Nelson Thomes)• Misner, Thorne and Wheeler: Relativity• C. Prior: Special Relativity, CERN Accelerator

School (Zeegse)

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Historical background• Groundwork of Special Relativity laid by Lorentz in

studies of electrodynamics, with crucial concepts contributed by Einstein to place the theory on a consistent footing.

• Maxwell’s equations (1863) attempted to explain electromagnetism and optics through wave theory

– light propagates with speed c = 3108 m/s in “ether” but

with different speeds in other frames

– the ether exists solely for the transport of e/m waves

– Maxwell’s equations not invariant under Galilean transformations

– To avoid setting e/m apart from classical mechanics, assume

• light has speed c only in frames where source is at rest• the ether has a small interaction with matter and is carried

along with astronomical objects

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Contradicted by:

• Aberration of star light (small shift in apparent positions of distant stars)

• Fizeau’s 1859 experiments on velocity of light in liquids

• Michelson-Morley 1907 experiment to detect motion of the earth through ether

• Suggestion: perhaps material objects contract in the direction of their motion

L v =L01−v2

c2 1

2

This was the last gasp of ether advocates and the germ This was the last gasp of ether advocates and the germ of Special Relativity led by Lorentz, Minkowski and of Special Relativity led by Lorentz, Minkowski and

Einstein.Einstein.

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The Principle of Special Relativity

• A frame in which particles under no forces move with constant velocity is inertial.

• Consider relations between inertial frames where measuring apparatus (rulers, clocks) can be transferred from one to another: related frames.

• Assume:– Behaviour of apparatus transferred from F to F' is independent

of mode of transfer

– Apparatus transferred from F to F', then from F' to F'', agrees with apparatus transferred directly from F to F''.

• The Principle of Special Relativity states that all physical The Principle of Special Relativity states that all physical laws take equivalent forms in related inertial frames, so laws take equivalent forms in related inertial frames, so that we cannot distinguish between the framesthat we cannot distinguish between the frames.

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Simultaneity• Two clocks A and B are synchronised if light rays

emitted at the same time from A and B meet at the mid-point of AB

• Frame F' moving with respect to F. Events simultaneous in F cannot be simultaneous in F'.

• Simultaneity is not absolute but frame dependent.

A BCFrame F

A’’ B’’C’’

A’

B’C’Frame F’

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The Lorentz Transformation• Must be linear to agree with standard Galilean transformation

in low velocity limit• Preserves wave fronts of pulses of light,

• Solution is the Solution is the Lorentz transformationLorentz transformation from frame F from frame F ((t,x,y,z)t,x,y,z) to frame F to frame F''((tt'',x,x'',y,y'',z,z'')) moving with velocity moving with velocity vv along along the the xx-axis:-axis: t '=γ t− vx

c2 x '=γ x−vt

y '= yz '=z

¿ } ¿ } ¿ } ¿

¿ where γ=1− v2

c2 −1

2 ¿

i .e . P≡x2 y2z2−c2 t2 =02

2

2

2

whenever Q≡x ¿ y ¿ z¿ c2t ¿ 0

ct

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Set t '=αtβxx '=γxδty '=εyz '=ςz

Then P=kQ

⇔c2t '2−x' 2− y '2−z ' 2=k c2t2−x2− y2−z2 ⇒ c2 αt βx 2− γxδt 2−ε2 y2−ς2 z2=k c2 t2−x2− y2−z2 Equate coefficients of x,y,z,t.Isotropy of space ⇒k =k v =k ∣v∣=±1Apply some common sense e . g . ε ,ς , k =+1 and not -1

Outline of Derivation

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x'=xv γt γ −1 v⋅xv2

t '=γ tv⋅xc2

General 3D form of Lorentz Transformation:

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Consequences: length contraction

z

x

Frame F

v

Frame F’z’

x’

RodA B

Moving objects appear contracted in the direction of the Moving objects appear contracted in the direction of the motionmotion

Rod AB of length L' fixed in F' at x'A, x'B. What is its length measured in F?

Must measure positions of ends in F at the same time, so events in F are (t,xA) and (t,xB). From Lorentz:

xA' =γ xA−vt xB

' =γ xB−vt L'=xB

' −x A' =γ xB− xA =γLL

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Consequences: time dilation

• Clock in frame F at point with coordinates (x,y,z) at different times tA and tB

• In frame In frame FF'' moving with speed moving with speed v, v, Lorentz Lorentz transformation givestransformation gives

• So:So:

t A' = γ t A− vx

c 2 t B' = γ tB− vx

c2 Δ t '=t B

' − t A' = γ t B− t A = γΔt Δt

Moving clocks appear to run slowMoving clocks appear to run slow

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Schematic Representation of the Lorentz Transformation

Frame F

Frame F

t

x

t

x

L

L

Length contraction L<L

t

x

t

x

tt

Time dilatation: t<tRod at rest in F. Measurement in F at fixed time t, along a line parallel to x-axis

Clock at rest in F. Time difference in F from line parallel to x-axis

Frame F

Frame F

x '= γ x− vt t '=γ t − vx

c2

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v = 0.8c

v = 0.9c

v = 0.99c

v = 0.9999c

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Example: High Speed Train

• Observers A and B at exit and entrance of tunnel say the train is moving, has contracted and has length

• But the tunnel is moving relative to the driver and guard on the train and they say the train is 100 m in length but the tunnel has contracted to 50 m

100γ

=100×1− v2

c2 1

2=100×1− 34

12=50m

All clocks synchronised.

A’s clock and driver’s clock read 0 as front of train emerges from tunnel.

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Moving train length 50m, so driver has still 50m to travel before his clock reads 0. Hence clock reading is

Question 1A’s clock reads zero as the driver exits tunnel. What does B’s clock read when the guard goes in?

−50v

=−100

3c≈−200 ns

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Question 2

What does the guard’s clock read as he goes in?

To the guard, tunnel is only 50m long, so driver is 50m past the exit as guard goes in. Hence clock reading is

50v

=+100

3c≈200 ns

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Question 3

Where is the guard

when his clock

reads 0?

Guard’s clock reads 0 when driver’s clock reads 0, which is as driver exits the tunnel. To guard and driver, tunnel is 50m, so guard is 50m from the entrance in the train’s frame, or 100m in tunnel frame.

So the guard is 100m from the entrance to the tunnel when his clock reads 0.

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F(t,x) is frame of A and B, F'(t',x') is frame of driver and guard.

Question 1A’s clock reads zero as the driver exits tunnel. What does B’s clock read when the guard goes in?

x=γ x'−v t ' t=γ t '−v x '

c2 x'=γ xvt t '=γ tvx

c2 xB=100 , xG

' =100

⇒ t B=1001−γγv

=−50v

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Question 2

What does the guard’s clock read as he goes in?

F(t,x) is frame of A and B, F'(t',x') is frame of driver and guard.

x=γ x'−v t ' t=γ t '−v x '

c2 x'=γ xvt t '=γ tvx

c2 x=100 , xG

' =100

⇒ tG' =100

γ −1γv

=50v

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Question 3

Where is the guard

when his clock

reads 0?

F(t,x) is frame of A and B, F'(t',x') is frame of driver and guard.

x=γ x'−v t ' t=γ t '−v x '

c2 x'=γ xvt t '=γ tvx

c2 xG

' =100 , t G' =0

⇒ x= γ x '=200 mOr 100m from the entrance to the tunnel

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Question 4Where was the driver

when his clock reads

the same as the

guard’s when he

enters the tunnel?F(t,x) is frame of A and B, F'(t',x') is frame of driver and guard.

x=γ x'−v t ' t=γ t '−v x '

c2 x'=γ xvt t '=γ tvx

c2

xD' =0, t D

' =50v

⇒ x=−γv t D' =−100 m

Or 100m beyond the exit to the tunnel

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Example: Cosmic Rays

-mesons are created in the upper atmosphere, 90km from earth. Their half life is =2 s, so they can travel at most 2 10-6c=600m before decaying. So how do more than 50% reach the earth’s surface?

• Mesons see distance contracted by , so

• Earthlings say mesons’ clocks run slow so their half-life is and

• Both giveγvc

=90 kmcτ

=150 , v≈c , γ≈150

vτ≈90γ km

v γτ ≈90 km

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Space-time• An invariant is a quantity that has the same

value in all inertial frames.

• Lorentz transformation is based on invariance of

• 4D space with coordinates (t,x,y,z) is called space-time and the point

is called an event.

• Fundamental invariant (preservation of speed of light):Δs2=c2 Δt2−Δx2−Δy 2− Δz2=c 2 Δt 21− Δx 2 Δy2Δz2

c 2 Δt 2 =c 2 Δt 21− v2

c2 =c2 Δtγ

2

c2 t2− x2 y2 z 2= ct 2−x2

t , x , y , z = t , x

τ=∫ dtγ

is called proper time, time in instantaneous rest frame, an invariant. s=c is called the separation between two events

t

x

Absolute future

Absolute past

Conditional present

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4-VectorsThe Lorentz transformation can be written in matrix form as

t '=γ t−vx

c2 x'=γ x−vt y'= yz'=z

ct '

x '

y '

z ' =γ −γvc

0 0

−γvc

γ 0 0

0 0 1 00 0 0 1

L

ctxyz

Position 4-vector Χ= ct ,x An object made up of 4 elements which transforms like X is called a 4-vector

(analogous to the 3-vector of classical mechanics) X '= LX

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Invariants

c2 t2−x2− y2−z2= ct , x , y , z [1 0 0 00 −1 0 00 0 −1 00 0 0 −1

]ctxyz

=XT gX=X⋅X

A =a0 ,a , B= b0 , b

A⋅A= A T gA=a0 b0−a1b1−a2 b2−a3b3=a0 b0−a⋅b

Basic invariant

Inner product of two 4-vectors

Invariance:

A '⋅A '= LA T g LA =AT LT gL A=AT gA=A⋅A

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4-Vectors in S.R. Mechanics

• Velocity:

• Note invariant

• Momentum

V = dΧdτ

=γdΧdt

=γddt

ct , x =γ c , v

V⋅V =γ2 c2−v 2= c 2−v 2

1−v 2/ c2=c2

P =m0 V = m 0 γ c ,v = mc , p

m=m0 γ is relativistic mass

p=m0 γ v=m v is the 3-momentum

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Example of Transformation:

Addition of Velocities• A particle moves with velocity    in

frame F, so has 4-velocity

• Add velocity by transforming to frame F to get new velocity .

• Lorentz transformation gives

V = γ u c , u u = u x , u y , u z

t ↔γ u, x ↔ γu u

v = v , 0 ,0

γ w=γv γuvγuu x

c2 γ w wx=γ v γuu xvγu γ w w y=γu u y

γ w wz=γuu z

⇒

w x=u xv

1vux

c2

w y=u y

γ v 1vux

c2 wz=

u z

γv 1vux

c2

w

x=γ x'−v t ' t=γ t '−v x '

c2 x'=γ xvt t '=γ tvx

c2

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4-ForceFrom Newton’s 2nd Law expect 4-Force given by

F=dPdτ

=γdPdt

=γddt

mc , p =γ c dmdt

,d pdt

=γ cdmdt

, f Note: 3−force equation f = d p

dt=m0

ddt

γ v

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• Momentum invariant

• Differentiate

Einstein’s Relation

P⋅dPdτ

=0 ⇒ V⋅dPdτ

=0 ⇒ V⋅F =0

⇒ γ c ,v ⋅γ cdmdt

,f =0

⇒ ddt

mc2 −v . f =0

P⋅P=m02 V⋅V =m0

2 c2

Therefore kinetic energy

T=mc2constant =m0 c2 γ −1

= rate force does work = rate of change of kinetic energyv .f

E=mcE=mc2 2 is total energyis total energy

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Basic Quantities used in Accelerator Calculations

γ =1−v 2

c2 −12 = 1− β2

−12

⇒ βγ 2=γ2 v2

c 2=γ2−1 ⇒ β 2=v 2

c2=1−1

γ2

β=vc

v= βcp=mv=m0 γβc

T= m-m0 c2=m0c2 γ−1

Relative velocity

Velocity

Momentum

Kinetic energy

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Velocity v. Energy

T =m0 γ−1 c2

γ =1T

m0 c2

β=1−1

γ2

p=mo c βγ

For v << c , γ=1−v2

c2 −1

2≈112

v2

c2

so T =m0c2 γ−1 ≈12

m0 v2

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Energy/Momentum Invariant

P⋅P=m02 V⋅V =m0

2 c2

E2

c2−p2 ⇒ p2 c2=E2− m0 c2 2=E2−E0

2

¿ E−E0 EE0 ¿T T2 E0

Example: ISIS 800 MeV protons (E0=938 MeV)

=> pc=1.463 GeV

βγ =m0 βγc2

m0c2= pc

E0

=1.56

γ2= βγ 21 ⇒ γ =1.85

β= βγγ

=0. 84

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Relationships between small variations in parameters

E, T, p, , βγ 2=γ2−1

⇒ βγ Δ βγ =γΔγ⇒ βΔ βγ =Δγ 1

1

γ2=1−β2

⇒ 1

γ 3 Δγ =βΔβ 2

Δpp

=Δm0 γβc m0 γβc

=Δ βγ βγ

=1

β 2

Δγγ

=1

β 2

ΔEE

=γ2 Δββ

=γγ1

ΔTT

(exercise)

Note: valid to first order only

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4-Momentum Conservation• Equivalent expression

for 4-momentum

• Invariant

• Classical momentum conservation laws conservation of 4-momentum. Total 3-momentum and total energy are conserved.

P=m0 γ c , v =mc , p = Ec

, p

m02 c2=P⋅P = E2

c2−p2 E2

c2=m02 c2p2

∑particles, i

Pi=constant

⇒ ∑particles, i

Ei and ∑particles,i

pi constant

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A body of mass M disintegrates while at rest into two parts of rest masses M1 and M2. Show that the energies of the parts are given by

E1=c 2 M 2 M 12− M 2

2

2M, E2=c2 M 2−M 1

2 M 22

2M

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SolutionBefore:

P= Mc , 0 After:

P1=E 1

c, p

P2=E2

c,−p

Conservation of 4-momentum:P=P1P2 ⇒ P−P1=P2

⇒ P−P1 ⋅ P− P1 =P2⋅P2

⇒ P⋅P −2P⋅P1P1⋅P1=P2⋅P2

⇒ M 2 c2−2 ME1M 12 c 2=M 2

2 c 2

⇒ E1=M 2 M 1

2−M 22

2M

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Example of use of invariants

• Two particles have equal rest mass m0.

– Frame 1: one particle at rest, total energy is E1.

– Frame 2: centre of mass frame where velocities are equal and opposite, total energy is E2.

Problem: Relate E1 to E2

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Total energy E1

(Fixed target experiment)

Total energy E2

(Colliding beams expt)

P1= E1−m0 c2

c, p P2=m0c , 0

P1= E2

2c, p ' P2= E2

2c,−p '

Invariant: P2⋅P1P2 m0 c ×

E1

c−0 × p=

E2

2c×

E2

c p ' × 0

⇒ 2m 0 c2 E1=E22

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Collider Problem

• In an accelerator, a proton p1 with rest mass m0 collides with an anti-proton p2 (with the same rest mass), producing two particles W1 and W2 with equal mass M0=100m0

– Expt 1: p1 and p2 have equal and opposite velocities in the

lab frame. Find the minimum energy of p2 in order for W1

and W2 to be produced.

– Expt 2: in the rest frame of p1, find the minimum energy

E' of p2 in order for W1 and W2 to be produced.

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Note: same m0, same p mean same E.E2

c2=p2m02 c2

p1

W1

p2

W2

Total 3-momentum is zero before collision and so is zero after impact

4-momenta before collision:

P1=Ec

, p P2= Ec

,−p 4-momenta after collision:

P1=E '

c,q P2= E'

c,−q

Energy conservation E=E > rest energy = M0c2 = 100 m0c2

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p2

p1

W1

W2

Before collision:

P1= m0 c ,0 P2= E'

c, p

Total energy is

E1=E 'm0 c2

Use previous result 2m0c2 E1=E22 to relate E1 to total energy E2 in

C.O.M frame2m 0 c2 E1=E2

2

⇒ 2m 0 c2 E 'm 0c 2 = 2 E 2 200m 0c 2 2⇒ E ' 2×104−1 m0 c2≈20 , 000m0 c2

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4-Acceleration

A = γ γ 3 v⋅ac

,γa γ 3 v⋅ac2 v

A= dVdτ

=γddt

γc , γ v Acceleration = “rate of change of velocity”

1

γ2=1− v⋅v

c2⇒ 1

γ3

dγdt

= v⋅vc2

= v⋅ac 2Use

A = 0,a , A⋅A =−∣a∣2Instantaneous rest-frame

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Radiation from an accelerating charged particle

• Rate of radiation, R, known to be invariant and proportional to in instantaneous rest frame.

• But in instantaneous rest-frame• Deduce

• Rearranged:

∣a∣2

A⋅A=−∣a∣2

R∝ A⋅A=−γ6 v⋅ac

2

1

γ2a2

R=2e2

3c3 γ6 [∣a∣2−a∧v 2

c2 ] Relativistic Larmor Formula

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Motion under constant acceleration; world lines

• Introduce rapidity defined by

• Then

• And

• So constant acceleration satisfies

V =γ c , v =c cosh ρ ,sinh ρ

β= vc

=tanh ρ ⇒ γ= 1

1−β2=cosh ρ

A = dVdτ

=c sinh ρ ,cosh ρ dρdτ

a2=∣a∣2=−A⋅A=c2 dρdτ

2

⇒ dρdτ

= ac

, ρ= aτc

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World line of particle is hyperbolic

c sinh ρ=c sinhaτc

=γν=γdxdt

=dxdτ

⇒ x=x0c2

a coshaτc

−1 cosh ρ=cosh

aτc

=γ =dtdτ

⇒ t=ca

sinhaτc

Particle Pathscosh2 ρ−sinh2 ρ=1

⇒ x−x 0c2

a 2

−c 2 t 2=c4

a2

x≈ x0c2

a 112

a2 τ 2

c2−1 ≈ x01

2aτ 2

t≈ca

×aτc

≈ τ , parabola

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Relativistic Lagrangian and Hamiltonian Formulation

f = d pdt

⇒ m0ddt x

1−v2/c2 1 /2 =− ∂ V∂ x

etc

ddt ∂ L

∂ x = ∂ L∂ x

etc ⇒ ∂ L∂ x

=m0 x

1− v2 /c 2 1 /2 , ∂ L∂ x

=−∂ V∂ x

v 2= x2 y2 z 2

3-force eqn of motion under potential V:

Standard Lagrangian formalism:

Since , deduce

L=−m0 c21− v2

c2 1

2−V =−m0 c2

γ−V

Relativistic Lagrangian

49

H=∑ x ∂ L∂ x

− L=m0

1−v2

c2 1

2

x2 y2 z2 − L

¿ m0 γv2m0c2

γV =m0 γc2V

¿ EV ,

Hamiltonian

total energy

Hamilton’s equations of motion

E2=p2 c2m02 c 4Since

H = c p2 m02 c2

12 V

x= ∂ H∂ px

, p x=− ∂ H∂ x

, etc

50

Photons and Wave 4-Vectors

Monochromatic plane wave:

Phase is number of wave crests

passing an observer, an invariant.

sin ωt−k⋅x

12π

ωt−k⋅x

ωt−k⋅x= ct , x ⋅ ωc

, k

k =wave vector, ∣k∣= 2πλ

; ω=angular frequency =2πν

Position 4-vector, X Wave 4-vector, K

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Relativistic Doppler Shiftω

∣k∣=c

nK = ωc

1,n

For light rays, phase velocity is

So where is a unit vector

v

F F'

'

Lorentz transform (t↔ω/c2, x↔ωn/c):

ω '=γω1−vc

cosθ tan { θ '=sin θ

γ cosθ−vc

¿

Note: transverse Doppler effect even when θ=½

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Motion faster than light• Two rods sliding over each

other. Speed of intersection point is v/sinα, which can be made greater than c.

• Explosion of planetary nebula. Observer sees bright spot spreading out. Light from P arrives t=dα2/2c later.

t= dα 2

2c≈ x

cα2

<< xc

xc

P

Od

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