Chapter 3.4 by Manu Verma

Chapter 3 Relativistic Kinematics

Chapter 3 3.1 Relativistic Transformation 3.2 Relativistic Transformation II 3.3 Four Vector, Space Time 3.4 Kinematics: Basics 3.5 Interactive Exercises

3.4 Kinematics: Phase Space

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Kinematics and Particle Reactions A classical collision, in which object A hits object B producing objects C and D (Fig1). During a collision in classical frame 1) Mass is conserved: mA + mB = mc + mD. 2)Momentum is conserved: pA + pB = pC + pD . 3)Kinetic energy may or may not be conserved.

Fig 1: A collision in which A + B → C + D.

c(pc) a(pa) Beam

b(pb)

target

θ θr d(pd)

Fig. 3

LAB kinematics diagram of particle collision a + b → c + d Dayalbagh Educational Institute

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Definition of LAB frame: the target particle is at rest The particles have masses ma,mb ,mc ,md Particle a is the incident particle Particle b is the target particle: this is initially at rest in the LAB Particle c is the scattered particle Particle d is the recoil particle θ is the LAB scattering angle: θr is the recoil angle c(pc*)

b(pb*)

a(pa*)

d(pd*)

Fig. 2

Kinematics diagram of particle collision in the CMS Definition of CMS: the total momentum of the initial system is zero. It follows by momentum conservation that the total momentum of the final system is also equal to zero. Energy and momentum conservation: E a + Eb = E c + Ed →

pa + → pb = → pc + → pd

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(vectors!)

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E = √(mc2)2 + (pc)2 is the total relativistic energy c = speed of light (in the vacuum!) In particle physics one frequently uses units such that c = 1 (and ħ=1) and then the energy-momentum relation is E = √m2 + p2 Elastic scattering: ma= mc , mb = md Typical problem of particle kinematics (needed by experimentalists): Given the masses of the initial particles and the momentum of the incident particle, find the momenta of the final particles in an elastic collision; also find the LAB recoil angle. The problem is solved by using energy and momentum conservation:

where p is the LAB momentum of a, W = ELAB + mb is the total LAB energy of the initial system.

tan θr = pc sin θ/(p - pc cos θ) Example: elastic pion - proton scattering mπ =140 MeV, mp = 940 MeV,

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Assume: LAB energy of the incident pion E = 1000 MeV Let θ =30 deg Question: what are the momenta and energies of the scattered pion and of the recoil proton: what is the recoil angle? Using the above formulas we get the following answer: Pπ = 866.3 MeV, Eπ = 877.54 MeV PP = 495.2 MeV, 1062.46 MeV θr = 1.07 rad = 61 deg. Note the conservation of energy: initially we had a total energy W = E+mP = 1940 MeV; after the collision we have Eπ + EP. = 877.54+1062.46 = 1940 MeV Our result looks surprising: the proton, which was initially at rest and which was hit by a pion of LAB energy 1000 MeV, has acquired an energy of 1062 MeV! The reason is that these energies are relativistic total energies: they include the rest energy. which in the case of the proton is 940 MeV. More intuitive than the total energy is the kinetic energy (K.E.): this is defined by K.E. = total energy - rest energy: T = E-mc2 We can check the expression for T takes on the familiar form of the nonrelativistic K.E. if the particle velocity is small compared with c: T = √m2c4 + p2c2 - mc2 = mc2 √1 + p2/m2c2 - mc2 and since non-relativistically p/mc <<1 we have √1 + p2/m2c2 ≈ 1 + p2 /2m2c2 Dayalbagh Educational Institute

hence

Tnr ≈ p2/ 2m 7

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In our example of elastic pion - proton scattering, the energy balance, expressed in terms of the K.E.s, is (Tπ +mπ c2) + (Tp +mpc2) = (T'π +mπ c2) + (T'p +mpc2)

i.e. the rest energies cancel and we are left with the balance of K.E.s: Tπ + Tp = T'π + T'p and in our previous numerical example we have: Tπ = 860 MeV, Tp= 0, T'π = 737.54 MeV, T'p =112.46 MeV and we see that only a small part of the initial pion K.E. got Transferred to the proton. But remember: the K.E.s balance only in the case of elastic scattering. In inelastic collisions only the total energies balance!

Inelastic collisions

Example of an inelastic collision: π+ + p → π+ + p + π 0 Here an additional (neutral) pion has been created. A first question is: what LAB K.E. is needed to produce this extra pion? (The smallest energy to produce the extra pion is called threshold energy) The calculation is simple in the CMS. To find the LAB threshold K.E. we then have two ways: 1) do a Lorentz transformation from CMS to LAB: this is difficult or Dayalbagh Educational Institute

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2) use the concept of invariance: that’s the easy way Let us do the calculation in the CMS, then use invariance

CMS threshold energy By definition of the CMS, the total momentum is equal to zero, both before and after the reaction. Before the reaction we have p→π + p→p = 0, hence pπ = pp and we con drop the subscripts on the momenta; the total CMS energy before the collision is therefore (“in” for initial) Usually the square of Ein is denoted by s, i.e. we have

and if we solve for p (Exercise!). then we get

After the reaction we have in the CMS

and the minimum of this corresponds to all final state particles being at rest in the CMS. i.e. Pi = 0, i = l,2,3, hence smin = M2 = {m1 + m2 + m3}2 Dayalbagh Educational Institute

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One can show that the quantity s is given in terms of the LAB energy of the incident particle by the following expression: and this is invariant, meaning that it takes the same value in any reference frame. We can therefore equate the threshold value of s in the LAB with its threshold value in the CMS, hence

Putting in the numbers for our example of production of a neutral pion in a pion-proton collision: π + + p → π + + p + π0 mπ+ = 140MeV, mπ0 = 135 MeV, mp = 940 MeV we get (Exercise!): M = 1215 MeV and for the LAB energy and LAB K.E. of the incident pion (Exercise!)

Thus the LAB threshold K.E. is greater than the extra mass produced. This is so because in the LAB the entire final state system is moving. i.e. it has a K.E. which is taken from the K.E. of the incident particle.

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Use of kinematics in planning an experiment: In a 2-to-2 reaction, the final momenta span a plane, called the reaction plane. By momentum conservation the reaction plane contains the beam axis (axis!). We can therefore position two detectors such as to detect the two final state particles: Detector 1 beam

Fig. 3

Beam direction target

Detector 2

We can do something a bit more clever than just position the detectors at the angles expected for the reaction we want to study: we know also the speed of the particles and therefore the time it takes them to get to the detectors: Time of Flight (ToF). So we put the detectors at such distances that the ToFs coincide, then use electronics to count only particles which arrive at the detectors in coincidence.

Time of Flight calculation Distance between interaction point (target!) and detector: D Particle velocity.....................................................: v hence ToF ..................................................... : t = D/v But velocities are relativistic in typical particle physics experiments! Dayalbagh Educational Institute

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The relativistic relation between momentum and velocity is p= γmv where γ = 1/√1-(v/ c)2

(“re1ativistic γ factor”)

The relativistic relation for the energy is E = γmc2 hence

v/c = pc/E

this is the particle velocity “in units of c”; one says for instance: “the particle is travelling at a speed of 0.9 of the speed of light’'. Example: π+ p → π+ p with ELAB = 1000 MeV Note that we have plenty of energy to produce additional pions: to produce just a single (neutral) pion we need only 305MeV: you can work out for yourselves how many pions can be produced if the beam energy is 1000 MeV (Exercise!). So the design of our experiment should be such as to count all elastic collisions and not to count the inelastic ones. We have previously worked out for a scattering angle of 30 deg: pπ = 866.3 MeV/c; Eπ =877.54 MeV pp = 495.2 MeV/c; Ep = 1062.46 MeV and θr =61 deg. Hence vπ = 866.3/877.54 = 0.987 c and vp = 0.466 c

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Why Electron Volts for Masses? When one scatters a beam of relativistic particles from a target, some of the kinematic energy in the beam can appear as mass → new particles produced Collision

Before Beam

After

Beam New particles

Target

Fig 4

Target

So it is convenient to measure masses in same units you use for energies

E2 Total Relativistic energy

p 2 c2 + m2 c4

Momentum Velocity of Light E2

Mass sometimes called “Rest Mass”

p 2 c 2 + m2 c 4

For a particle with no mass (!) (Photon - γ ; Neutrino – υ) E=p.c p = E/c = [eV]/c

Unit of momentum

A massless particle having an energy of 1eV, has a momentum of 1eV/c for a massive particle at rest ( in its rest frame) Dayalbagh Educational Institute

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Total Relativistic energy

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c2 mass

M = E/c2 =

[eV]/c2

Unit of mass If a particle has a mass of 1ev/c2, when at rest it has a total relativistic energy of 1eV

p 2 c2 + m 2 c4

(1)

Subatomic physics use units where c=1

p 2 + m2

(â&#x2020;&#x2019; sometimes confusing. I will use 'c' explicitly.) Should be comfortable with either Each term of (1) has dimensions [E]2 Example, for a particle Mass = 1000 MeV/c2 Momentum = 1000 MeV/c E2 = 106 MeV2/c2 .c2 + 106 MeV2/c4 .c4 So

E2 = 2.106 MeV2 The particle has a total relativistic energy,

E = â&#x2C6;&#x161;2 . 103 MeV

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Example 1 Two lumps of plaster cine, each of mass m, collide head-on at â&#x2026;&#x201D;c (Figure 1). They stick together. What is the mass M of the final composite lump?

Solution: Conservation of energy says E1 + E2 = EM. Conservation of momentum says p1 + p2 = pM . In this case, conservation of momentum is trivial: p1 = -p2 , so the final lump is at rest (which was obvious from the start). The initial energies are equal, so conservation of energy yields.

Fig.1 Collision of two equal masses (Example 1).

Conclusion M = 6/ď&#x192;&#x201E;. Notice that this is greater than the sum of the initial masses; in heated collisions kinetic energy is converted into rest energy, so the mass increases. Dayalbagh Educational Institute

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Example 2 A particle of mass M, initially at rest, decays into two pieces, each of mass m (Figure 2). What is the speed of each piece as it flies off?

Solution: This is, of course, the reverse of the process in Example 1 Conservation of momentum just says that the two lumps fly off in opposite directions at equal speeds. Conservation of energy requires that

This answer makes no sense unless M exceeds 2m: M = 2m is the threshold for the process M â&#x2020;&#x2019; 2m to occur. The deuteron, for example, is below the threshold for decay into proton plus neutron (md =1875.6 Mev/c2; mp + mn = 1877.9 MeV/c2) and therefore is stable. A deuteron can be pulled apart, but only by pumping enough energy into the system to make up the difference. (If it puzzles you that a bound state of p and n should weigh less than the sum of its parts, the point is that the binding energy of the deuteron - which, like all internal energy, is reflected in its rest mass - is negative. Indeed, for any stable bound state the binding energy must be negative; if the composite particle weighs more than the sum of its constituents, it will spontaneously disintegrate. Dayalbagh Educational Institute

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Example 3 A pion at rest decays into a muon plus a neutrino (Figure 3). Question: What is the speed of the muon?

Fig. 2 A particle decays into two equal pieces. (Example 2).

Fig. 3 Decay of the charged pion (Example 3.3).

Solution: Conservation of energy requires Eπ = Eμ + Eυ. Conservation of momentum gives pπ = pμ + pυ; but pπ = 0, so pμ = -pυ. Thus the muon and the neutrino fly off back-toback, with equal and opposite momenta. Rule 1: To get the energy of a particle, when you know its momentum (or vice versa), use the invariant E2 — p2c2 = m2c4 Dayalbagh Educational Institute

(2) 17

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In the present case, then: Eπ = mπc2

Eυ =|pυ|c = |pμ|c Putting these into the equation for conservation of energy, we have or Solving for |pμ|,

Meanwhile, the energy of the muon is

Once we know the energy and momentum of a particle, it is easy to find its velocity. If E = γ mc2 and p = γ mv, dividing gives p/E = v/c2 Rule 2: If you know the energy and momentum of a particle, and you want to determine its velocity, use v = pc2 / E

(3)

So the answer to our problem is

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Putting in the actual masses, I get vμ = 0.271c. There is nothing wrong with that calculation; it was a straightforward and systematic exploitation, of the conservation laws. Rule 3: A method to get the energy and momentum of the muon by using four-vector notation. Conservation of energy and momentum requires pπ = pμ + pυ ,

pυ = pπ - pμ

Taking the scalar product of each side with itself, we obtain But

Therefore from which E, follows immediately. By the same token

pμ = pπ - pυ

Squaring yields But Eυ =|pυ|c = |pμ|c, so

We call this the center-of-momentum (CM) frame, because in this system the total (three-vector) momentum is zero. Dayalbagh Educational Institute

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Example 4 The Bevatron at Berkeley was built with the idea of producing tons, by the reaction p + p â&#x2020;&#x2019; p + p + p + p. That is, a high-energy proton strikes a proton at rest, creating (in addition to the original particles) a protonantiproton pair. Question: What is the threshold energy for this reaction (i.e. the minimum energy of the incident proton)? In the laboratory the process looks like Figure 4 a; in the CM frame, it looks like Figure 8 b. Now, what is the condition for threshold?

Fig. 4 p + p â&#x2020;&#x2019; p + p + p + p . (a) In the lab frame; (b) in the CM frame. Dayalbagh Educational Institute

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Just barely enough incident energy to

create the two extra particles. In the lab frame, it is hard to see how we would formulate this condition, but in the CM it is easy: all four final particles must be at rest, with nothing ‘wasted’ in the form of kinetic energy. (We can’t have that in the lab frame, of course, since conservation of momentum requires that there be some residual motion.) Let pμTOT be the total energy- momentum four-vector in the lab; it is conserved, so it doesn’t matter whether we evaluate it before or after the collision. We'll do it before:

where E and p are the energy and momentum of the incident proton, and m is the proton mass. Let pμTOT’ be the total energy-momentum four-vector in the CM. Again, we can evaluate it before or after the collision; this time we’ll do it after:

since (at threshold) all four particles are at rest. Now pμTOT ≠ pμTOT' obviously, but the invariant products pμ'TOT pμTOT' and pμTOT pμTOT are equal:

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Using the standard invariant E2 â&#x20AC;&#x201D; p2c2 = m2c4 to eliminate p2. and solving for E, we find E = 7mc 2 Evidently, the incident proton must carry a kinetic energy, for this process to occur. (And in fact the first antiprotons were discovered when the machine reached about 6000 MeV.) The distinction between a conserved quantity and an invariant quantity. Energy is conserved - the same value after the collision as before - but it is not invariant. Mass is invariant - the same in all inertial systems â&#x20AC;&#x201C; but it is not conserved. Some quantities are both invariant and conserved. (e.g. electric charge); many are neither (speed, for instance). Rule 5: If a problem seems lengthy , in the lab frame, try analyzing it in the CM system. Even if youâ&#x20AC;&#x2122;re dealing with something more complicated than a collision of two identical particles, the CM (in which pTOT = 0) is still a useful reference frame, for in this system conservation of momentum is trivial: zero before, zero after. But you might wonder whether there is always a CM frame.

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In other words, given a swarm of particles with masses m1, m2, m3 and velocities v1, v2' v3,â&#x20AC;Ś, does there necessarily exist an inertial system in which their total (three-vector) momentum is zero? The answer is yes, I will prove it by finding the velocity of that frame and demonstrating that this velocity is less than c. The total energy and momentum in the lab frame (S) are (4) Since pÎźTOT is a four-vector, we can use the Lorentz transformations to get the momentum in system S', moving in the direction of pTOT with speed v

In particular this momentum is zero if v is chosen such that

Now, the length of the sum of three-vectors cannot exceed the sum of their lengths (this geometrically evident fact is known as the triangle inequality), so

and since vi < c, we can be sure that v < c.

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Thus the CM system always exists, and its velocity relative to the lab frame is given by (5) It seems odd, looking back at the answer to Example (5) that it takes an incident kinetic energy six times the proton rest energy to produce a p/p pair. i.e. only creating 2mc2 of new rest energy. This illustrates the inefficiency of scattering off a stationary target; conservation of momentum forces you to waste a lot of energy as kinetic energy in the final state. Suppose we could have fired the two protons at one another, making the laboratory itself the CM system. Then it would suffice to give each proton a kinetic energy of only mc2, one-sixth of what the stationarytarget experiment requires. This realization led, in the early 1970s, to the development of colliding-beam machines (see Figure 9). Today, virtually every new machine in high-energy physics is a collider.

Example 5 Suppose two identical particles, each with mass m and kinetic energy T, collide head-on. Question: What is their relative kinetic energy, T' (i.e. the kinetic energy of one in the rest system of the other)?

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Fig. 5 Two experimental arrangements: (a) Colliding beams; (b) fixed target.

Solution: There are many ways to do this one. A quick method is to write down the total four-momentum in the CM and in the lab

set (pTOT)2 = (pTOT')2:

use Equation 2 to eliminate p' 2E2 = mc2 (E’ + mc2 ) and express the answer in terms of T = = E - mc2 and T' = E’ - mc2 (6) The classical answer would have been T' = 4T, to which this reduces when T « mc2. (In the rest system of B, A has, classically, twice the velocity, and hence four times as much kinetic energy, as in the CM.) Now, a factor of 4 is some benefit, to be sure, but the relativistic gain can be greater by far.

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Colliding electrons with a laboratory kinetic energy of 1 GeV, for example, would have a relative kinetic energy of 4000 GeV!

Example 6 A pion traveling at speed v decays into a muon and a neutrino If the neutrino emerges at 90o to the original pion direction , at what angle does the Îź come off.

Solution 6 The 4 vector momentum conservation implies

or Here

but

But

So,

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Also we know that

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Example 7:

Suppose that a Φ meson of energy 2000 MeV decays to a pair of charged kaons as in the diagram below. Some properties of Φ and K mesons can be found in the table at the bottom of this question.

Q1(a)Draw a Feynman diagram of the decay (at the quark level). What force is responsible for this decay

A1(a)

The process proceeds via the strong force. The gluon in the above diagram is optional- there are many ways one could draw in virtual particle(s) in the diagram. Q1(b)Calculate the velocity of the Φ before it decays. A1(b)One way to do this is to calculate the gamma factor. Since E = γmc2 Dayalbagh Educational Institute

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we have

Now

and so i.e. since β = υ/c, the Φ is travelling at 86% of the speed of light. Q1(c)Write down the components of the 4-momentum of the Φ, A1(c)The 4-vector of the Φ is PΦ = (E, pxc, pyc, pzc). The Φ is traveling in the x direction so py = pz = 0. We know that E = 2000 MeV. The easiest way to get px is to note that in this case, px = p and that p = βE/c. So in this case px = βE/c = 0.860 × 2000 MeV/c = 1720MeV/c Hence the 4-vector of the Φ is PΦ = (2000 MeV, 1720 MeV, 0, 0) Note that it was not necessary to explicitly substitute for c anywhere, if we stick to what amounts to “energy equivalent” values. Q1(d)Calculate the angle θ at which the kaons emerge. Particle

Quark Content

Mass (MeV/c2)

Spin(ħ)

1020

K+

494

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A1(d)Call the K+ kaon 1 and the K- kaon 2. To get the angle θ of the kaons with respect to the x axis, we note that by energy and momentum conservation P Φ = P1 + P2 i.e. EΦ = E1 + E 2 pxΦ = px1 + px2 = p1 cos θ + p2 cos θ pyΦ = py1 + py2 = p1 sin θ − p2 sin θ = 0 pzΦ = pz1 + pz2 = 0 From the equation for the y component it is obvious that P1 = P2, i.e. the kaons have equal magnitude of momentum. This means that they also have equal energy E1 = E2 since they have equal masses and for any particle Thus, from the energy equation, E1 = E2 = 0.5EΦ = 1000 MeV. Now Now we use the equation for the x component to get

Thus the kaons emerge at an angle of around cos−1(0.989) = 8.5° to the x axis.

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