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RelativisticKinematics
U. Blumenschein
LorentzTransformations
Four Vectors
Energy andMomentum
Collisions
Relativistic Kinematics
U. Blumenschein
School of Physics and AstrophysicsQueen Mary University of London
EPP, SPA6306
RelativisticKinematics
U. Blumenschein
LorentzTransformations
Four Vectors
Energy andMomentum
Collisions
Outline
Lorentz Transformations
Four Vectors
Energy and Momentum
Collisions
RelativisticKinematics
U. Blumenschein
LorentzTransformations
Four Vectors
Energy andMomentum
Collisions
Lorentz TransformationsI According to Special Theory of Relativity: the laws of
physics are equally valid in all inertial reference systemsI An inertial system is one in which Newton’s first law
(the law of inertia) is obeyed: objects keep moving instraight line at constant speeds unless acted upon by aforce.
I Example: inertial systems S and S’ with S’ moving atuniform velocity v
S S’
y y
z z’
x
x’
v
RelativisticKinematics
U. Blumenschein
LorentzTransformations
Four Vectors
Energy andMomentum
Collisions
Lorentz TransformationsAssuming an event at coordinates (x ,y ,z) and time t in S,what are the corresponding space-time coordinates in S’?
i x ′ = γ(x − vt)
ii y ′ = y
iii z ′ = z
iv t ′ = γ(t − vc2 x)
where γ = 1√1−v2/c2
and β = vc . Note γ > 1 and
asymptotic at v = c .
The inverse transformations from S’ to S are obtainedchanging the sign of v .
RelativisticKinematics
U. Blumenschein
LorentzTransformations
Four Vectors
Energy andMomentum
Collisions
Lorentz Transformations
γ = 1√1−v2/c2
and β = vc .
RelativisticKinematics
U. Blumenschein
LorentzTransformations
Four Vectors
Energy andMomentum
Collisions
Time Dilation and length contraction
The Lorentz transformations have several interestingconsequences:
I Time dilation: Time difference of a clock resting in Smeasured from S’: ∆t ′ = t1′ − t2′ =γ(t1− vx
c2 − t2 + vxc2 ) = γ(t1− t2) ⇒: a clock resting in
S seems to be running slower from the perspective of anobserver in S’
I Application: Life time of a particle moving withrelativistic velocities is increased by a factor of γ:τ ′ = γτ
I length contraction: The length component s of anobject resting in S in the dimension parallel to thevelocity of S’ is measured by an observer in S’ asshorter by a factor of γ: s ′ = 1
γ s
RelativisticKinematics
U. Blumenschein
LorentzTransformations
Four Vectors
Energy andMomentum
Collisions
Velocity addition
I If a particle moves along x ′ in S’ with velocity u′, whatis its velocity u in S?
I Using the Lorentz transformations, we know it travels adistance ∆x = γ(∆x ′ + v∆t ′) in a time∆t = γ(∆t ′ + (v/c2)∆x ′), so:
I u = ∆x∆t = ∆x′+v∆t′
∆t′+(v/c2)∆x′ = (∆x′/∆t′)+v1+(v/c2)∆x′/∆t′ = u′+v
1+(v/c2)u′
I Note that if u′ = c then u = c the speed of light isthe same in all inertial systems.
RelativisticKinematics
U. Blumenschein
LorentzTransformations
Four Vectors
Energy andMomentum
Collisions
Four VectorsGlobal notation for space and time variables
Introducing some simplifying notation using a position-timefour vector, xµ, µ = 0, 1, 2, 3, as follows:
I x0 = ct x1 = x x2 = y x3 = z
The Lorentz transformations then take the form:
I x0′= γ(x0 − βx1)
I x1′= γ(x1 − βx0)
I x2′= x2
I x3′= x3
where β = vc .
RelativisticKinematics
U. Blumenschein
LorentzTransformations
Four Vectors
Energy andMomentum
Collisions
Lorentz transformations in a matrix form
This can be written in a matrix form, i.e.
I xµ′
=∑3
ν=0 Λµνxν (µ = 0, 1, 2, 3)
I where Λ is defined as:
Λ =
γ −γβ 0 0−γβ γ 0 0
0 0 1 00 0 0 1
thus xµ
′= Λµνxν
I Λ can have a different form if the motion is not along x,but the matrix form is the same.
RelativisticKinematics
U. Blumenschein
LorentzTransformations
Four Vectors
Energy andMomentum
Collisions
More about four Vectors
I It is useful to define::I contravariant four vector: xµ =̂ (x0, x1, x2, x3)I covariant four-vector: xµ ≡ gµνx
ν =̂(x0,−x1,−x2,−x3)
I with gνµ is defined as:
g =
1 0 0 00 −1 0 00 0 −1 00 0 0 −1
RelativisticKinematics
U. Blumenschein
LorentzTransformations
Four Vectors
Energy andMomentum
Collisions
Scalar product and lorentz invariants
I We can now introduce a scalar product:
a · b ≡3∑
µ=0
aµbµ = a0b0 − a1b1 − a2b2 − a3b3
I We introduce the sum convention: Implicit sum overtwo equal indices: aµb
µ ≡∑3
µ=0 aµbµ
I For bµ = aµ we obtain the squared length of a 4-vector:
I = aµaµ = a0a0 − a1a1 − a2a2 − a3a3
I The length of a 4-vector is Lorentz invariant, i.e.aµa
µ = a′µa′µ
I We can differentiate:I aµa
µ > 0, aµ is timelike, normal massive particleI aµa
µ < 0, aµ is spacelikeI aµa
µ = 0, aµ is lightlike, photon or massless particle
RelativisticKinematics
U. Blumenschein
LorentzTransformations
Four Vectors
Energy andMomentum
Collisions
Typical frames of reference
I In particle physics we use different frames of referencethat make specific computations simpler.
I In the Laboratory frame the laboratory is taken to be atrest.
I In the Centre of Mass (CM)-frame the observer is takento be travelling with the same speed as the centre ofmass of a system
I In the CM-frame the total momentum is zero.I Particles travel with equal and opposite momenta.I In general, CM frame and LAB frame are not the same
RelativisticKinematics
U. Blumenschein
LorentzTransformations
Four Vectors
Energy andMomentum
Collisions
The 4-Momentum
I Simple 4-velocity: vµ = dxµ
dt = (c , v) = (c, vx , vy , vz)
I We define the 4-momentum aspµ = mγvµ = mγ(c , vx , vy , vz). The 3-momentum preduces to the classic momentum for v << c .
I With the definition of the relativitistic energy asE ≡ γmc2, we can write the momentum as:pµ = (E
c , px , py , pz)
I Its scalar product is: pµ · pµ = E2
c2 − p2 = m2c2. It is aninvariant and corresponds to the rest mass of the objectwith the specified 4-momentum: invariant mass.
I The invariant mass M of the 4-momentum sum of twoparticles A and B: M2 = (pA + pB)µ(pA + pB)µ gives usthe mass of a potential mother particle which hasdecayed into A and B.
RelativisticKinematics
U. Blumenschein
LorentzTransformations
Four Vectors
Energy andMomentum
Collisions
Classical and relativistic energy
Let’s expand the energy (E = γmc2 = mc2√
1−v2c2) in a Taylor
series to understand the relation between the classical andrelativistic energy:E = mc2(1 + 1
2v2
c2 + 38
v4
c4 + ...) = mc2 + 12mv2 + 3
8v4
c2 + ...
I Rest energy: mc2. Doesn’t exist in classical mechanics.
I Relativistic kinetic energy:mc2(γ − 1) = 1
2mv2 + 38
v4
c2 + ...
Special case: massless particles m=0
I Momentum p=0, kinetic energy T=0, no force (F=ma)can be sustained in classical physics.
I However, in the relativistic case if v=c fromE2
c2 − p2 = m2c2 we get E =| p | c.
I From quantum dynamics: E = ~ν
RelativisticKinematics
U. Blumenschein
LorentzTransformations
Four Vectors
Energy andMomentum
Collisions
Summary of useful kinematic relations
I Velocity and Lorentz factor:I β = v
cI γ = 1√
1−v2/c2
I 4-vectors and scalar product:I xµ =̂ (x0, x1, x2, x3)I xµ ≡ gµνx
ν =̂ (x0,−x1,−x2,−x3)I xµx
µ = (x0)2 − (x1)2 − (x2)2 − (x3)2
I 4-momentum: pµ = (E/c , px , py , pz)I Main energy-momentum relation: E 2 = p2c2 + m2c4
I Invariant mass (or center-of-mass energy) of A and B:M =
√(pA + pB)µ(pA + pB)µ
I Relation between β, γ energy and momentum:I E = γmc2 ⇒ γ = E
mc2
I p = βγmc ⇒ βγ = pmc
I ⇒ β = pcE
RelativisticKinematics
U. Blumenschein
LorentzTransformations
Four Vectors
Energy andMomentum
Collisions
Application: 2-body decayI Particle A, at rest decays into two particles B and C
(A→ B + C ). What are the energies of the daughterparticles?
PµA = Pµ
B + ePµC ⇒ Pµ
C = PµA − Pµ
B
Square both sides:
P2C = P2
A + P2B − 2Pµ
APBµ
Use the definition for the squared 4-momentum:
P2C = m2
Cc2, P2A = mAc2, P2
B = m2Bc2
and for the mixed product:
PµAPBµ = EAEB/c
2 − pA · pB
In this case we have particle A at rest:
pA = 0, EA = mAc2.
RelativisticKinematics
U. Blumenschein
LorentzTransformations
Four Vectors
Energy andMomentum
Collisions
Application: 2-body decay
We can now substitute in the equation above:
m2Cc2 = m2
Ac2 + m2Bc2 − 2(mAEB)
To get EB we can rearrange as:
2mAEB = (m2A + m2
B −m2C )c2
thus getting:
EB =m2
A + m2B −m2
C
2mAc2
Similar for particle C:
EC =m2
A + m2C −m2
B
2mAc2
⇒ the energies of the daughter particles are determined bythe masses of the mother and of the daughter particles.
RelativisticKinematics
U. Blumenschein
LorentzTransformations
Four Vectors
Energy andMomentum
Collisions
Energy and momentum in collisions
I Are Total energy, total momentum, mass and kinemticenergy conserved?
I Energy is conserved: EA + EB = EC + ED
I Momentum is conserved: pA + pB = pC + pD
I Kinetic energy and rest mass may or may not beconserved.
I All four components of the energy-momentum fourvector are conserved: pµA + pµB = pµC + pµD
I Special case elastic collision: kinetic energy and restmass conserved
RelativisticKinematics
U. Blumenschein
LorentzTransformations
Four Vectors
Energy andMomentum
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Centre of Mass Energy
I We consider a collision of two particles A and B in thelaboratory frame.
I Invariant mass of the system:P2 = pµ.p
µ = (EA/c + EB/c)2 − (pA + pB)2
= m2Ac2 + m2
Bc2 + 2EAEB/c2 − 2pA.pB
I As this is a Lorentz-invariant it has the same value inthe CM frame, where the two momenta of A and B areof equal size and in opposite direction.
P2 = P2CM = (E ∗A/c + E ∗B/c)2 − (p∗A + p∗B)2
= (E ∗A/c + E ∗B/c)2 = E ∗/c2
Where * denotes energies and momenta in the CMframe.
I It follows that the Invariant mass of two collidingparticles is equivalent to their center-of-mass energy.
RelativisticKinematics
U. Blumenschein
LorentzTransformations
Four Vectors
Energy andMomentum
Collisions
Example: Target at Rest
I Using(E ∗/c)2 = P2 = m2
Ac2 + m2Bc2 + 2EAEB/c
2 − 2pA.pB
I If particle B is at rest we can set pB = 0 andEB = mBc2 resulting in:(E ∗)2 = P2 = m2
Ac2 + m2Bc2 + 2EAmB
I If particle A is highly relativistic with EA >> mAc2 andEA >> mBc2, then:E ∗ =
√2EAmB
I Example: A E = 450 GeV proton hitting a stationaryproton target (mB = 0.938GeV /c)E ∗ =
√2× 450× 0.938 = 24GeV
I Only 24GeV is available to produce new particles. Theremaining energy boosts the centre-of-mass system.
RelativisticKinematics
U. Blumenschein
LorentzTransformations
Four Vectors
Energy andMomentum
Collisions
Example: Colliding Beams
I Using(E ∗/c)2 = P2 = m2
Ac2 + m2Bc2 + 2EAEB/c
2 − 2pA.pB
I If particle A and B collide head-on: −2pApB = 2pApB
I If they are highly relativistic with EA >> mAc2 andEB >> mBc2 then: 2pApB
∼= 2EAEB
I (E ∗)2 = 4EAEB
I If in addition EA = EB : (E ∗)2 = 4E 2A ⇒ E ∗ = 2EA
I Example: A E = 450 GeV proton beam colliding with aE = 450 GeV proton beam results in 900GeV beingavailable to the interaction.