Relativity for Accelerator
Physicists
Christopher R. Prior
Fellow and Tutor in
Mathematics
Trinity College, Oxford
Accelerator Science and
Technology Centre
Rutherford Appleton
Laboratory, U.K.
Overview
The principle of special relativity
‰ Lorentz transformation and its consequences
‰ 4-vectors: position, velocity, momentum,
invariants. Derivation of E=mc2
‰ Examples of the use of 4-vectors
‰ Inter-relation between β and γ, momentum and
energy
‰ An accelerator problem in relativity
‰
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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)
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Historical background
‰
‰
Groundwork by Lorentz in studies of
electrodynamics, with crucial concepts contributed
by Einstein to place the theory on a consistent
basis.
Maxwell’s equations (1863) attempted to explain
electromagnetism and optics through wave theory
ƒ
ƒ
ƒ
ƒ
ƒ
light propagates with speed c = 3×108 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
And the ether has a small interaction with matter and is carried along with
astronomical objects
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Nonsense! 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 1contract in the
direction of their motion
⎛ v2 ⎞ 2
L(v) = L0⎜⎜1− 2 ⎟⎟
⎝ c⎠
This was the last gasp of ether advocates and the germ of
Special Relativity led by Lorentz, Minkowski and Einstein.
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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
Frame F’
A
C
B
A’
C’
B’
A’’
‰
‰
C’’
B’’
Frame F' moving with respect to F. Events
simultaneous in F cannot be simultaneous in F'.
Simultaneity is not absolute but frame dependent.
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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.
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
laws take equivalent forms in related inertial frames, so
that we cannot distinguish between the frames.
frames
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The Lorentz Transformation
‰
‰
‰
Must be linear to agree with
standard Galilean
transformation in low
velocity limit
Preserves wave fronts of
i.e. P ≡ x 2 + y 2 + z 2 − c 2t 2 = 0
pulses of light,
whenever Q ≡ x ′2 + y′2 + z ′2 − c 2t ′2 = 0
Solution is the Lorentz
transformation from frame
F (t,x,y,z) to frame
F'(t',x',y',z') moving with
velocity v along the x-axis:
⎛ vx ⎞
t' = γ ⎜ t − 2 ⎟
⎝ c ⎠
x' = γ ( x − vt )
y' = y
z' = z
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⎫
⎪
1
2 − 2
⎪⎪
⎛ v ⎞
⎜⎜1- 2 ⎟⎟
where
γ
=
⎬
⎝ c ⎠
⎪
⎪
⎪⎭
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Outline of Derivation
Set
t' = α t + β x
x' = γ x + δ t
y' = ε y
z' = ς z
Then P = kQ
(
⇔ c 2t '2 − x '2 − y '2 − z '2 = k c 2t 2 − x 2 − y 2 − z 2
)
(
⇒ c 2 (α t + β x ) − (γ x + δ t ) − ε 2 y 2 − ς 2 z 2 = k c 2t 2 − x 2 − y 2 − z 2
2
2
)
Equate coefficients of x,y,z,t.
r
r
Isotropy of space ⇒ k = k(v ) = k( v ) = ±1
Apply some common sense (e.g. ε , ς , k = +1 and not - 1)
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Consequences: length contraction
Frame F
z’
z
Frame F’
v
A
Rod
x
B
x’
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:
x′A = γ ( x A − vt )
x′B = γ ( xB − vt )
L′ = x′B − x′A = γ ( xB − x A ) = γ L > L
Moving objects appear contracted in the direction of the motion
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Consequences: time dilatation
‰
Clock in frame F at point with coordinates (x,y,z)
at different times tA and tB
‰
In frame F' moving with speed v, Lorentz
transformation gives
vx ⎞
⎛
t ′A = γ ⎜ t A − 2 ⎟
c ⎠
⎝
‰
So:
vx ⎞
⎛
t ′B = γ ⎜ t B − 2 ⎟
c ⎠
⎝
∆t ′ = t B′ − t ′A = γ (t B − t A ) = γ ∆t > ∆t
Moving clocks appear to run slow
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Schematic Representation of the Lorentz
Transformation
Frame F′
t
Frame F′
t
t′
Frame F
t′
Frame F
∆t
L
L′
∆t′
x′
x′
x
x
Length contraction L<L′
Time dilatation: ∆t<∆t′
Rod 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
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Example: High Speed Train
‰
‰
‰
All clocks synchronised.
Observers A and B at entrance and exit of tunnel say the train is moving, has
contracted and has length
1
1
2
2
⎛ v ⎞
100
⎛ 3⎞ 2
= 100 × ⎜⎜1 − 2 ⎟⎟ = 100 × ⎜1 − ⎟ = 50m
γ
⎝ 4⎠
⎝ c ⎠
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
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A Simple Problem
‰
‰
F=frame of tunnel (and
A,B)
F'=frame of train (and
driver D and guard G)
⎛ vx ⎞
t′ = γ ⎜ t + 2 ⎟
⎝ c ⎠
x′ = γ ( x + vt )
vx′ ⎞
⎛
t = γ ⎜ t′ − 2 ⎟
c ⎠
⎝
x = γ ( x′ − vt ′)
x A = 0 xB = 100
x′D = 0 xG′ = 100
‰
What does B’s clock read
when the guard goes into
the tunnel?
Coincident events (t B , xB ) and (tG′ , xG′ )
xG′ = γ ( xB + vt B ) ⇒ 100 = 2(100 + vt B )
⇒ tB = −
100
c 3
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Simple Problem (continued)
‰
‰
xB = γ ( x'G −vtG′ ) ⇒ 100 = 2(100 − vtG′ )
What does the guard’s
clock read as he goes in?
⇒ tG′ = +
Where is the guard
when his clock reads 0?
Put tG′ = 0 in x = γ ( xG′ − vtG′ )
⇒ x = 2 xG′ = 200 m
100
3c
So the guard is still 100m from
the tunnel entrance when his
clock reads zero.
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Example: π-mesons
‰
‰
Mesons are created in the upper
atmosphere, 90km from earth. Their half
life is τ=2 µs, so they can travel at most
2 ×10-6c=600m before decaying. So how
do more than 50% reach the earth’s
surface undecayed?
Mesons see distance contracted by γ, so
⎛ 90 ⎞
vτ ≈ ⎜⎜ ⎟⎟km
⎝γ ⎠
‰
Earthlings say mesons’ clocks run slow
so their half-life is γτ and
v(γτ ) ≈ 90 km
‰
Both give
βγ =
90 km
= 150, β ≈ 1, γ ≈ 150
cτ
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Invariants
‰
‰
‰
An invariant is a quantity that has the same value in all inertial frames.
Lorentz transformation is based on invariance of
r
c 2t 2 − ( x 2 + y 2 + z 2 ) = (ct ) 2 − x 2
r
Write in terms of the 4-position vector Χ = (ct , x ) as X•X
r
r
If Χ = ( x0 , x ), Υ = ( y0 , y ), define the invariant product
r r
Χ • Υ = x0 y0 − x ⋅ y
‰
Fundamental invariant (preservation of speed of light):
⎛ ∆x 2 + ∆y 2 + ∆z 2 ⎞
⎟⎟
c ∆t − ∆x − ∆y − ∆z = c ∆t ⎜⎜1 −
2
2
c ∆t
⎠
⎝
2
2
2
2
2
2
2
⎛ ∆t ⎞
v2 ⎞
2
2⎛
= c ∆t ⎜⎜1 − 2 ⎟⎟ = c 2 ⎜⎜ ⎟⎟
⎝γ ⎠
⎝ c ⎠
‰
‰
2
τ is the invariant proper time where ∆τ=∆t/γ
τ is time in the rest frame
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4-Vectors
‰
The Lorentz transformation can be written in
matrix form
⎛ vx ⎞
t′ = γ ⎜ t − 2 ⎟
⎝ c ⎠
x′ = γ ( x − vt )
y′ = y
z′ = z
⇒
⎛
⎛ ct ′ ⎞ ⎜ γ
⎜ ⎟ ⎜
⎜ x′ ⎟ ⎜ − γ v
⎜ y′ ⎟ = ⎜ c
⎜ ⎟ ⎜ 0
⎜ z′ ⎟
⎝ ⎠ ⎜
⎝ 0
An object made up of 4 elements which
transforms like X is called a 4-vector
(analogous to the 3-vector of classical
mechanics)
−
γv
c
γ
0
0
⎞
0 0 ⎟⎛ ct ⎞
⎟⎜ ⎟
x⎟
⎜
⎟
0 0
⎟⎜ y ⎟
1 0 ⎟⎜⎜ ⎟⎟
⎟⎝ z ⎠
0 1⎠
r
Position 4-vector Χ = (ct , x )
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4-Vectors in S.R. Mechanics
‰
‰
‰
V=
Velocity:
Note invariant
Momentum
r
r
dΧ
dΧ
d
=γ
= γ (ct , x ) = γ (c, v )
dτ
dt
dt
r2
V •V = γ (c − v ) = c2
2
2
r
r
P = m0V = m0γ (c, v ) = (mc, p)
m = m0γ is relativistic mass
r
r
r
p = m0γ v = mv is the 3 - momentum
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Example of Transformation: Addition of
Velocities
‰
‰
‰
r
A particle moves with velocity u = (u x , u y , u z ) in frame F, so
r
has 4-velocity V = γ u (c, u )
r
Add velocity v = (v,0,0) by transforming to frame F′ to get
r
new velocity w .
r
r
Lorentz transformation gives (t ↔ γ , x ↔ γ u )
ux + v
vγ u u x ⎞
⎛
wx =
γ w = γ v ⎜γ u + 2 ⎟
vu x
c
1+ 2
⎝
⎠
c
γ w wx = γ v (γ u u x + vγ u ) ⇒
uy
wy =
γ w wy = γ u u y
⎛ vu x ⎞
γ
v ⎜1 +
2 ⎟
γ w wz = γ u u z
c ⎠
⎝
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wz =
uz
⎛
⎝
γ v ⎜1 +
vu x ⎞
2 ⎟
c ⎠
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Einstein’s relation
‰
Momentum invariant P • P = m02 (V • V ) = m02c 2
P•
dP
dP
=0 ⇒ V•
=0
dτ
dτ
‰
Differentiate
‰
From Newton’s 2nd Law expect 4-Force given by
r
r
dP
dP
d
⎛ dm dp ⎞
⎛ dm r ⎞
F=
, ⎟ = γ ⎜c
,f⎟
=γ
= γ (mc, p ) = γ ⎜ c
dτ
dt
dt
⎝ dt dt ⎠
⎝ dt
⎠
v r
dP
‰ But V •
= 0 ⇒ V • F = 0 Rate of doing work, v ⋅ f = rate of change
of kinetic energy
dτ
‰
So
r r
d
2
mc − v ⋅ f = 0
dt
( )
Therefore kinetic energy
T = mc 2 + constant = m0 c 2 (γ − 1)
E=mc2 is total energy
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Basic quantities used in Accelerator
calculations
Velocity
β = vc
v = βc
Momentum
p = mv = m0γ β c
Relative velocity
T = (m-m0 )c = m0 c (γ − 1)
2
Kinetic energy
⎛ v2 ⎞
γ = ⎜⎜1 − 2 ⎟⎟
⎝ c ⎠
⇒
(βγ )
2
−
=
1
2
(
= 1− β
γ 2v 2
c
2
2
)
1
2 −2
= γ −1 ⇒
2
1
v2
β = 2 = 1− 2
c
γ
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Velocity as a function of energy
T = m0 (γ − 1)c 2
γ = 1+ T
m0 c 2
β = 1− 1
γ2
p = mo cβγ
−1
⎛ v ⎞
v
, γ = ⎜⎜ 1 − 2 ⎟⎟
For small
c
⎝ c ⎠
1
2
so T = m0 c (γ − 1) ≈ m0 v 2
2
2
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2
1 v2
≈ 1+
2 c2
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Relationships between small variations in
parameters ∆E, ∆T, ∆p, ∆β, ∆γ
⇒
⇒
(βγ )2 = γ 2 − 1
βγ ∆(βγ ) = γ ∆γ
β ∆(βγ ) = ∆γ
1
γ2
⇒
1
γ
3
∆p ∆(m0γ βc) ∆(βγ )
=
=
p
m0γ βc
βγ
(1)
= 1− β 2
1 ∆γ
1 ∆E
= 2
= 2
β γ
β E
2 ∆β
=γ
β
∆γ = β ∆β
(2)
γ
∆T
=
γ +1 T
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(exercise)
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4-Momentum Conservation
‰
‰
‰
(
Equivalent expression for
r
r E r
4-momentum
P = m0γ (c, v ) = (mc, p) =
,p
2
r
E
Invariant m02 c 2 = P • P = 2 − p 2
c
Classical momentum
conservation laws →
conservation of 4momentum. Total 3momentum and total
energy are conserved.
E
2
c2
c
)
r2
=m c + p
2 2
0
∑ P = constant
i
particles, i
r
⇒ ∑ Ei and ∑ pi constant
particles, i
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particles,i
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Example of use of invariants
‰
Two particles have equal rest mass m0.
ƒ Frame 1: one particle at rest, total energy is E1.
ƒ 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)
⎛ E1 − m0 c 2 r ⎞
P1 = ⎜⎜
, p ⎟⎟
c
⎠
⎝
⎛ E2 r ′ ⎞
P1 = ⎜ , p ⎟
⎝ 2c
⎠
P2 = (m0 c,0 )
Total energy E2
(Colliding beams expt)
⎛ E2 r ′ ⎞
P2 = ⎜ ,− p ⎟
⎝ 2c
⎠
Invariant : P2 • (P1 + P2 )
E2 E2
E1
m0 c × − 0 × p =
× + p′ × 0
2c c
c
2m0 c 2 E1 = E22
⇒
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Accelerator 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
ƒ 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.
ƒ 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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2
Note: E
c
2
r
= p 2 + m02 c 2
⇒ same m0, same p mean same E.
Total 3-momentum is zero before
collision and so is zero after impact
W1
p1
p2
4-momenta before collision:
(
r
E
P1 =
,p
c
)
(
r
E
P2 =
,− p
c
4-momenta after collision:
W2
(
r
′
E
P1 =
,q
c
)
(
r
′
E
P2 =
,− q
c
)
)
Energy conservation ⇒ E=E′ > rest energy = M0c2 = 100 m0c2
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W1
Before collision:
(
r
P1 = m0 c,0
)
r
′
E
, p)
P2 = (
c
p1
p2
Total energy is
E1 = E ′ + m0 c
2
W2
Use previous result 2m0c2 E1=E22 to relate E1 to total energy E2 in
C.O.M frame
(
2m0 c 2 E1 = E22
)
(
⇒ 2m0 c E ′ + m0 c = (2 E ) > 200 m0 c
2
⇒
2
(
2
)
)
2 2
E ′ > 2 ×10 4 − 1 m0 c 2 ≈ 20,000 m0 c 2
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