relativistic kinematics - 2019.qmplus.qmul.ac.uk

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

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U. Blumenschein

LorentzTransformations

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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

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Lorentz Transformations

γ = 1√1−v2/c2

and β = vc .

RelativisticKinematics

U. Blumenschein

LorentzTransformations

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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

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U. Blumenschein

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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.

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U. Blumenschein

LorentzTransformations

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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

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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.

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U. Blumenschein

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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

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U. Blumenschein

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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

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U. Blumenschein

LorentzTransformations

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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

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U. Blumenschein

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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.

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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

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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

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LorentzTransformations

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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.

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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

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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.

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U. Blumenschein

LorentzTransformations

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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.

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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.