special relativity christopher r. prior accelerator science and technology centre rutherford...
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Special Relativity
Christopher R. Prior
Accelerator Science and Technology Centre
Rutherford Appleton Laboratory, U.K.
Fellow and Tutor in Mathematics
Trinity College, Oxford
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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
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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
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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
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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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