Relativistic Kinematics of Particle Interactions Introduction

Home , Collision, Invariant mass, Particle decay, Protonium, Rest frame

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Relativistic Kinematics of Particle Interactions

byW von Schlipp e, March2002

1. Notation; 4-vectors, covariant and contravariant comp onents, metric tensor, invariants.

2. Lorentz transformation; frequently used reference frames: Lab frame, centre-of-mass

frame; Minkowski metric, rapidity.

3. Two-b o dy decays.

4. Three-b o dy decays.

5. Particle collisions.

6. Elastic collisions.

7. Inelastic collisions: quasi-elastic collisions, particle creation.

8. Deep inelastic scattering.

9. Phase space integrals.

Intro duction

These notes are intended to provide a summary of the essentials of relativistic kinematics

of particle reactions. A basic familiarity with the sp ecial theory of relativity is assumed. Most

derivations are omitted: it is assumed that the interested reader will b e able to verify the results,

which usually requires no more than elementary algebra. Only the phase space calculations

are done in some detail since we recognise that they are frequently a bit of a struggle. For a

deep er study of this sub ject the reader should consult the monograph on particle kinematics

byByckling and Ka jantie.

Section 1 sets the scene with an intro duction of the notation used here. Although other

notations and conventions are used elsewhere, I present only one version which I b elieveto

b e the one most frequently encountered in the literature on particle physics, notably in such

widely used textb o oks as Relativistic Quantum Mechanics by Bjorken and Drell and in the

b o oks listed in the bibliography.

This is followed in section 2 by a brief discussion of the Lorentz transformation. This can

b e dealt with fairly brie y b ecause Lorentz transformations play no ma jor part in the kind of

calculations characteristic for the rest of this b o ok: central to most relativistic calculations is

the notion of Lorentz invariants. This is so b ecause all observables can b e expressed in terms of

invariants. The transformations b etween di erent reference frames are usually straight forward

for the invariants of the problem without involving the use of the Lorentz transformation

formul.

In the following sections the kinematics of particle reactions is develop ed b eginning with the

simplest case, that of two-b o dy decays of unstable particles, and culminating in multiparticle

kinematics which is of paramount imp ortance for the study of particle reactions at mo dern high

energy accelerators and colliders. The notes conclude with a fairly detailed discussion of phase

space calculations.

The most exhaustive treatment of the sub ject of relativistic kinematics is given in the

monograph byByckling and Ka jantie [1 ]. The notation and other conventions used here are 1

those of [2 ]. A summary of formulas of relativistic kinematics can b e found in the Particle Data

Tables [3],

1. Notation and units

The covariantcomponents of 4-vectors are denoted by sup erscript Greek letters , etc.

Thus we write x , =0; 1; 2; 3, such that

0 1 2 3

x = ct; x = x; x = y; x = z

The contravariantcomponents of 4-vectors are lab eled by subscripts:

x = x ;x = x; x = y; x = z

0 1 2 3

The scalar pro duct of two 4-vectors is given by

0 0 1 1 2 2 3 3

x y = x y = x y x y x y x y

where the notation x y implies summation over .

The minus signs in the scalar pro duct signify that the 4-vector space is not Euclidian. This

is also expressed byintro ducing the metric tensor g with diagonal elements (1; 1; 1; 1)

and zero o -diagonal elements:

g = diag(1; 1; 1; 1)

The contravariantelements of the metric tensor coincide with its covariant elements, i.e. we

havealsog = diag(1; 1; 1; 1)

Using the summation convention one can checkby explicit calculation that

x = g x and x = g x

This pro cedure is referred to as the lowering and raising of the indices.

Using the metric tensor one can express the scalar pro duct in the following equivalent forms:

x y = g x y = g x y

The invariant square of a 4-vector is given by

x = x x

2 2 2

If x > 0 the 4-vector is said to b e time-like,ifx =0 itis light-like, and if x < 0itis

space-like.

In the following we shall use units de ned by c = 1, where c is the sp eed of light. This

is convenient in the kind of calculations characteristic of relativistic kinematics, b ecause all

expressions must then b e homogeneous in energies, momenta and masses, which all havethe

same dimension. Velo cities do not o ccur very often in these calculations, but one must remember

that particle velo cities are dimensionless and do not exceed 1. Thus the relativistic factor of

a particle of velo city v is written as

1=2

= 1 v (1)

The units of energy, momentum and mass most frequently used in elementary particle

6 9

physics are the Mega-electron Volt(1MeV=10 eV), Giga-electron Volt (1 GeV = 10 eV)

c.f. P.A.M. Dirac, The Principles of Quantum Mechanics, 4th edition, OUP, 1958, p. 254 2

and Tera-electron Volt (1 TeV = 10 eV). One should rememb er that the electron mass is

approximately 0.5 MeV, the proton mass is nearly 1 GeV (a more accurate value is 0.94 GeV),

and the highest energy accelerators in op eration accelerate protons to ab out 1 TeV.

2. Lorentz transformation.

The Lorentz transformation b etween two inertial frames is of the form

x ! x = a x (2)

The invariance of the scalar pro duct of two4-vectors implies that

a a = (3)

where is the Kronecker symb ol.

If the axes of the two frames coincide at time t = t = 0 and the frames move with relative

velo city ~v ,then

00 0

x = (x + vx )

0 0

x = (x + vx ) (4)

~x = ~x

where v = j~v j, x is the comp onentof x in the direction of ~v , i.e. the mo dulus of ~x =(~x ~v )~v=v ,

jj jj

2 1=2

~x = ~x ~x and =(1 v ) .Thus in this particular case the Lorentz transformation is

de ned by three parameters, namely the three comp onents of the velo city ~v .

If in addition the two frames are rotated with resp ect to each other, then the transfor-

mation matrix dep ends also on the three Euler angles. Therefore the most general Lorentz

transformation dep ends on six parameters.

The inverse Lorentz transformation from the primed to the unprimed frame is obtained by

replacing in Eq. (4) the primed by the unprimed co ordinates and vice versa, and changing the

sign of v , i.e.

0 00 0 0 00 0

x = (x vx ) x = (x vx ) ~x = ~x (5)

jj jj ?

Energy-momentum 4-vector.

Like time, so also energy is not invariant in relativity, but rather transforms under Lorentz

transformation like the zeroth comp onentofa 4-vector, the other three comp onents b eing the

ordinary 3-vector momentum.

Let us denote the energy-momentum 4-vector of a particle of mass m byp, i.e.

p=(p ;~p) (6)

Frequently we shall denote the zeroth comp onent p , i.e. the energy,by E .Thus the invariant

square of p is

2 2 2 2

p = p p = E p~ = m (7)

The velo city ~v of the particle is de ned by

~p = m~v (8)

From Eqs. (1), (7) and (8) we nd the useful relations

= E=m (9) 3

and

~v = p=E~ (10)

The energy-momentum 4-vector p transforms like the space-time 4-vector x (c.f. Eq. (4)):

00 0

p = (p + vp )

0 0

p = (p + vp ) (11)

p~ = p~

Minkowski metric; rapidity.

An alternative description of the Lorentz transformation (4) is based on the Minkowski

trickofintro ducing an imaginary time x = it. There is now no need to havetwo kinds of

vectors, such as the contravariantandcovariantvectors considered ab ove. Instead we denote

the 4-vector comp onents of space-time by

x = x; x = y; x = z; x = it:

1 2 3 4

and the Lorentz invariant square b ecomes

2 2 2 2

x + x + x + x =invariant:

1 2 3 4

The transformation, which leaves this expression invariant, is a rotation. Thus, for instance, a

rotation in the (x ;x ) plane, leaving x and x unchanged, is of the form

1 4 2 3

0 0

x = x cos x sin ; x = x sin + x cos

1 4 1 4

1 4

Of course, no mathematical trick gets us o the ho ok: nature demands space-time to b e

non-Euclidian. Going back from the world of an imaginary time to the real world wemust

write our rotation as

0 0

x = x cos it sin ; it = x sin + it cos

and using the identities

i sin =sinhi ; cos =coshi

and setting y = i we recover a completely real form of the Lorentz transformation, viz

0 0

x = x cosh y + t sinh y; t = x sinh y + t cosh y

For this to b e equivalent with Eq. (4) wemust demand that y b e real, and hence is

imaginary.Thus the price to pay for a familiar Euclidian form of the rotation in the (x ;x )

1 4

plane is an imaginary angle of rotation. The real quantity y de ned ab ove is called the rapidity

of the transformation; it is related to the relativevelo city v between the two frames and to the

relativistic factor by

cosh y = ; sinh y = v

hence v =tanhy .

We can similarly write the Lorentz transformation of the 4-momentum as

0 0

p = p cosh y + E sinh y; E = p sinh y + E cosh y

In the ultra relativistic case, when the particle mass can b e neglected and wehave E = p,

we get the useful formula

y =ln(E =E ) (12)

ur 4

3. Two-b o dy decays of unstable particles.

The simplest kind of particle reaction is the two-b o dy decay of unstable particles. Awell

known example from nuclear physics is the alpha decayofheavy nuclei. In particle physics one

observes, for instance, decays of charged pions or kaons into muons and neutrinos, or decays of

neutral kaons into pairs of pions, etc. The unstable particle is the mother particle and its decay

pro ducts are the daughter particles.

Consider the decay of a particle of mass M which is initially at rest. Then its 4-momentum

is P = (M; 0; 0; 0). This reference frame is called the centre-of mass frame (CMS). Denote the 4-

momenta of the two daughter particles byp and p :p =(E ;~p ), p =(E ;~p ). 4-momentum

1 1 2 2

1 2 1 2

conservation requires that

P=p +p (13)

1 2

and hence p~ = ~p .We can therefore omit the subscript on the particle momenta and hence

2 1

energy conservation takes on the form

q q

2 2

E + E = + p + + p = M (14) m m

1 2

Solving this equation for p weget

2 2 2 2

(m m ) ][M (m + m ) ] (15) [M p =

1 2 1 2

An immediate consequence of Eq. (14) is that

M m + m (16)

1 2

i.e. a particle can decay only if its mass exceeds the sum of the masses of its decay pro ducts.

Conversely, if some particle has a mass that exceeds the masses of two other particles, then this

particle is unstable and decays, unless the decay is forbidden by some conservation law, such

as conservation of charge or of angular momentum etc.

Another p oint to note is that the momenta of the daughter particles and hence also their

energies are xed by the masses of the three particles. In the next section we shall see that

there is no equivalent statement in the case of three-b o dy decays: the momenta of the daughter

particles in three-b o dy decays can takeonanyvalue from zero to some maximum, and it is

only the maximum momentum that is xed by the masses of the particles.

Let us complete our calculation by deriving the formul for the energies of the daughter

particles. This is straightforward if we b egin from the energy conservation formula (14) and

2 2 2

express E in terms of E ,viz. E = , and then solvefor E to get + m m E

2 1 2 1

2 1 1

2 2 2

E = M + m m (17)

1 2

and similarly

2 2 2

M + m m (18) E =

2 1

We also note that there is no preferred direction in which the daughter particles travel (the

decayissaidtobeisotropic), but if the direction of one of the particles is chosen (e.g. by

the p ositioning of a detector), then the direction of the second particle is xed by momentum

conservation: the daughter particles are traveling back-to-back in the rest frame of the mother

particle.

Frequently one observes that the masses of the twodaughter particles are equal, for instance

in the decay of a neutral kaon into a pair of pions. In this case the previous formul simplify: the 5

1 1

2 2

energies of the daughter particles are E = E = M and the momenta are p = M 4m ,

1 2

2 2

if we denote the common mass of the daughter particles by m.

Of interest is also the two-b o dy decay of unstable particles in ight. For instance, high

energy b eams of muons or of neutrinos are pro duced in accelerators by letting the internal

b eam of protons impinge on a target of metal (thin foils or wires are used in practice) to

pro duce pions and kaons, which are then steered in a vacuum tub e in which they decayinto

muons and neutrinos. Other cases of great interest are the decays of very short-lived reaction

pro ducts of high energy collisions, such as, for instance, the decays of B mesons or of D mesons,

which are copiously pro duced in mo dern high energy colliders. To illustrate the imp ortance of

a prop er discussion of such in- ightdecays suce it to say that this is frequently the only way

to measure the mass of a neutral particle.

Thus wenowhave the following 4-momenta of the three particles: for the mother particles

we write P = (E; 0; 0;p), and for the daughter particles wehavep =(E ;~p ;p )andp =

1 1? 1z

1 2

(E ;~p ;p ). This means that wehavechosen the z axis along the direction of ightofthe

2 2? 2z

mother particle. The immediate result of this is that by momentum conservation the (two-

dimensional) transverse momentum vectors are equal in magnitude and opp osite in sign:

~p ~p = ~p (19)

? 1? 2?

The energies and the z comp onents of the particle momenta are related to those in the CMS

by a Lorentz b o ost with a b o ost velo city equal to the sp eed of the mother particle. We lab el

the kinematical variables in the CMS with asterisks and write the Lorentz transformation of

particle 1 in the form of

E = (E + vp )

1 1z

p = (p + vE ) (20)

1z 1

~p = ~p

and similarly for particle 2. Here v = p=E and = E=M . This completely solves the problem

in principle. We can now, for instance, nd the angles whichthetwo daughter particles make

with the z axis and with each other as functions of the momentum of the mother particle

(Exercise 1). But it is also of interest to approach the problem in a di erentway, without using

the Lorentz transformation, starting from energy-momentum conservation:

q q

2 2 2 2

E = E + E = m m + p + p + (21)

1 2

1 2 1 2

~p = ~p + p~ (22)

1 2

2 2

Thus, replacing in the energy conservation equation, p by(~p p~ ) we get an equation with un-

known momentum p and angle between ~p and the z axis. Solving for p is a straightforward

1 1 1 1

if lengthy calculation (Exercise 2). In the end weget

2 2 2

p sin M p m (M + m m )p cos 2E

1 1

1 2

(23) p =

2 2

2(M + p sin )

2 2

This result is of interest in the following sense. Realityof p demands that (M p

2 2

m p sin ) 0. This condition is satis ed for all angles if Mp =m p>1. In this case the

1 1 1

lower sign must b e rejected since otherwise wewould get unphysical negativevalues of p for

>=2. On the other hand, if Mp =m p<1, then there is a maximum value of ,given by

1 1 1

sin = Mp =m p. Now b oth signs mustbekept: for eachvalue of < there are

1max 1 1 1max 6

twovalues of p , and corresp ondingly also twovalues of p .Several examples illustratingthis

1 2

are given in Exercise 3.

Exercise 1: Show that the LAB angle that daughter particle 1 makes with the direction of

ight of the mother particle in a two-b o dy decay is related to the CMS angle by the following

equation:

sin

(24) tan =

) + cos (v=v

1 1

where v is the the LAB velo city of the mother particle and v is the CMS velo cityofthe

daughter particle.

Exercise 2: DeriveEq. (23).

Exercise 3: Consider the following decay pro cesses:

0 +

(i) K ! ,

(ii) ! p

(iii) ! 2 .

In each case assume that the mother particle has a LAB energy of 1 GeV. Find the max-

imum Lab angles and the corresp onding LAB momenta for all decay pro ducts. Show that

the minimum op ening angle b etween the decay pro ducts corresp onds to a CMS angle of 90

with the line of ight of the mother particle and calculate the corresp onding LAB momenta.

Assuming that one of the daughter particles makes an angle half the maximum angle, nd

the corresp onding twomomenta and the two momenta and LAB angles of the other daughter

particle.

4. Three-b o dy decays; Dalitz plot.

Consider the decay of a mother particle of mass M into three particles of masses m , m and

1 2

m . Denote their 4-momenta byP,p ,p and p ,respectively. Energy-momentum conservation

1 2 3

is expressed by

P=p +p +p (25)

1 2 3

De ne the following invariants:

2 2

s = P = M

2 2

s = (P p ) =(p +p )

1 2 3

2 2

s = (P p ) =(p +p ) (26)

2 3 1

2 2

s = (P p ) =(p +p )

3 1 2

Invariant s is a trivial invariant, trivial in the sense that it is constant and thus has no

dynamical signi cance. To understand the physical signi cance of invariant s consider the

s is the invariant mass of the second part of its de nition: s =(p +p ) . This shows that

1 1

2 3

p p

s and subsystem of particles 2 and 3. Similarly s are the invariant masses of subsystems

2 3

(3,1) and (1,2), resp ectively.

The three invariants s , s and s are not indep endent: it follows from their de nitions

1 2 3

together with 4-momentum conservation that

2 2 2 2

s + s + s = M + m + m + m (27)

1 2 3

therefore we could alternatively write s instead of s , with similar notation for s and s .

23 1 2 3 7

Kinematical limits

In the case of three-particle decayanimportant question arises as to the limits of the

kinematical variables. We shall see that this question also arises in collision problems. Therefore

the present derivation is of more general use. It is for this reason that we discuss it in greater

detail than most other derivations in these notes.

The space spanned byany set of indep endent kinematical variables is called phase space.

Therefore we can alternatively say that we are deriving the b oundaries of phase space.

Consider the decay pro cess in the rest frame of the mother particle (CMS). Here wehave

2 2

s = M + m 2ME (28)

1 1

2 2

with E = m + p , where p is the CMS momentum of particle 1. Thus E m , and hence

1 1 1 1

1 1

max s =(M m ) (29)

1 1

To nd min s weevaluate s in the rest frame of subsystem (2,3). Let us denote all frame

1 1

dep endent kinematical variables in this frame by a little zero ab ove the symbol. Thus weget

2 2 2

s =(p +p ) =( + ) (m + m ) (30)

E E

1 2 3 2 3

2 3

with similar formul for s and s . In summary,we get the following limits of the invariants

2 3

s , s ,ands :

1 2 3

i h

2 2

s 2 (m + m ) ; (M m )

1 2 3 1

h i

2 2

s 2 (m + m ) ; (M m ) (31)

2 3 1 2

h i

2 2

s 2 (m + m ) ; (M m )

3 1 2 3

However not the entire cub e de ned by Eq. (31) is kinematically accessible. To nd the limits

2 2

of s ,say, for xed s 2 [(m + m ) ; (M m ) ]itisconvenient to consider the Jackson frame

2 1 2 3 1

~ ~ ~

p p p

, S23. This frame is de ned by = . Momentum conservation then implies that =

3 2 1

and it follows that

0 1

r r

2 2

@ A

p p

) = s =( + + (32) M m

E E

1 1

weget Solving for

h ih i

1 1

2 2

= s (M m ) s ;M ;m (33) s (M + m )

1 1 1 1 1

1 1

4s 4s

1 1

Similarlywe get from

(34) + s =(p +p ) =

E E

3 2 1

2 3

2 2

p p

the corresp onding expression for and :

2 3

2 2

p p

(35) ;m s ;m = =

3 2 2 3

Now consider the invariant s :

2 2 2

p p

cos (36) s =(p +p ) = m + m +2

E E

3 1 1 3 2

1 3

such a frame is called Jackson frame; this particular Jackson frame will b e denoted S23.

4 2 2 2

The kinematical function (x; y ; z ) is de ned by (x; y ; z )=x + y + z 2xy 2yz 2zx 8

~ ~

p p

where is the angle b etween and .Taking into account Eqs. (33) and (35), we see that

1 3

s dep ends only on if s is xed. It follows that s max s and s min s corresp ond

2 1 2+ 2 2 2

to = and = 0, resp ectively, i.e.

2 2

p p

s = m + m +2 (37)

E E

1 2 3 1 3

1 3

in terms of s , i.e. if we write and If we express

E E

3 1 1

1 1

2 2 2

s s m s + m = = m , (38)

p p

E E

1 1 1 3

1 3 2

2 s 2 s

1 1

then wegets explicitly as a function of s :

2 1

h i

2 2 2 1=2 2 1=2 2 2 2 2

s s m s m + m (s ;s;m ) (s ;m ;m ) (39) s = m + m +

1 1 1 1 2

1 2 3 1 2 3 1 3

The curvede nedby Eq. (39) is the b oundary of the Dalitz plot in the (s ;s ) plane.

1 2

Of interest are also the maximum values of the three-momenta of the daughter particles in the

rest frame of the mother particle. From Eq. (28) we see that max E , and hence also max p ,

1 1

corresp onds to min s , and a straight forward calculation gives the result

2 2 2 2

p = [M (m + m + m ) ][M (m + m m ) ] (40)

1max 1 2 3 2 3 1

and we get the similar expressions for p and p by the cyclic replacement of the sub-

2max 3max

scripts: 1 ! 2, 2 ! 3 and 3 ! 1.

In the particular case when one of the daughter particles is massless, such as for instance in

+ 0 +

K decay: K ! ,we nd that the maximum momenta of the two massive daughter

particles are equal and greater than the maximum momentum of the massless daughter particle.

If twodaughter particles are massless, suchasinmuon decay: ! e , then the maximum

momenta of all daughter particles coincide.

5. Particle collisions

Centre-of-mass frame and Lab oratory frame.

The total 4-momentum p of a system of n particles with 4-momenta p ; p ; ::: ;p is

1 2

given by

+ ::: +p +p p =p

2 1

The centre-of-mass frame (CMS) of the system is de ned as the frame in which the total three-

momentum of the system is equal to nought. Lab eling CMS variables by asterisks, we express

this de nition by the following equation:

p~ = p~ + p~ + ::: + p~ =0 (41)

1 2 n

In particle physics the lab oratory frame (LAB frame) of the system is de ned as that

reference frame in which one of the initial particles is at rest. This particle is called the target

particle, the other b eing the b eam particle or incident particle. The CMS and LAB frames are

illustrated for the reaction a + b ! c + d by the kinematical diagrams in Fig. 1.

We use the convention that the b eam particle is incident along the p ositive z axis. Its

momentum and energy are denoted p and E , resp ectively. Then its 4-momentum is

LAB LAB

given by

p =(E ; 0; 0;p ) (42)

LAB LAB

1 9

c(p )

b(m )

a(p ) a(p ) b(p )

1 1 2

- - z -

d(p ) d(p )

4 4

(b) (a)

Figure 1: Kinematical diagram: (a) CMS; (b) LAB.

and the 4-momentum of the target particle is given by

p =(m ; 0; 0; 0)

where wehave arbitrarily denoted the incident particle as particle 1 and the target particle as

particle 2. is the scattering angle and is the recoil angle. In the CMS the scattering angle

is .

The total 4-momentum of the initial particles is

p=p +p (43)

1 2

The invariant square of p is usually denoted by s. In the CMS frame wehave ~p + ~p =0 and

1 2

hence

s =(E + E ) (44)

1 2

i.e. s is the total CMS energy of the system.

We also note that ~p + ~p = 0 implies that we can drop the subscripts on the magnitudes

1 2

of the three-momenta and write

q q

2 2

E = m + p ; E = m + p (45)

1 2

Substituting these expressions in Eq. (44) and solving for p weget

io ih nh

1=2

2 2

(46) s (m + m ) s (m m ) p =

1 2 1 2

2 s

If particles c and d are di erentfrom a and b, then wemust distinguish b etween the initial

CMS momentum p ,given by Eq. (46), and the nal CMS momentum p , given by

i f

nh ih io

1=2

2 2

p = (47) s (m m ) s (m + m )

3 4 3 4

2 s

In the LAB frame s is given by

2 2

s =p p = m + m +2m E (48)

2 LAB

1 2

This is one of the Mandelstam variables; in section 6 we shall de ne the full set of three Mandelstam

variables. 10

from whichwe get the useful relation

2 2

E =(s m m )=2m (49)

LAB 2

1 2

2 2

and hence, substituting E = and solving for p ,we get p + m

LAB LAB

LAB

nh ih io

1=2

1 1

2 2 2 2

s (m m ) s (m + m ) (s; m ;m ) (50) = p =

1 2 1 2 LAB

1 2

2m 2m

2 2

Comparing Eq. (50) with (46) we ndthat

p = p (51)

LAB

and hence

s (52) E =(m + m E )=

2 LAB

1;2 1;2

It is interesting to note that for ultra-relativistic particles, i.e. when we can neglect all

masses in comparison with the particle energies, we get E = E , which means that wecan

1 2

drop the subscript of the CMS energies, and hence we get from Eqs. (48) and (52) the simple

relation

E m E (53)

2 LAB

thus the CMS energy grows only like the square-ro ot of the LAB energy. In particle collisions

only the centre-of-mass energy can b e converted into the reaction pro ducts of the nal state.

Therefore we see from Eq. (53) that in xed-target exp eriments at high energies most of the

b eam energy is lost in kinetic energy of the particle system. This limitation is overcome by

the construction of colliders, i.e. accelerators with two b eams moving in opp osite directions: in

the case of particles of equal masses in the two b eams, such an arrangement ensures that the

combined b eam energy is the centre-of-mass energy and no energy is lost in the form of overall

kinetic energy of the reaction pro ducts. This can b e illustrated strikingly by the example of

the creation of Z b osons in electron-p ositron annihilations. In an e e collider weneedtwo

b eams of half the Z b oson mass, i.e. of approximately 45.5 GeV each. The equivalent LAB

energy of a p ositron b eam impinging on target electrons is 8:3 10 GeV, i.e. an energy which

is unattainable in the foreseeable future.

Let us use our results to nd the LAB velo cityofthecentre of mass of the initial particles.

This is also the b o ost velo city of the Lorentz transformation from the CMS to the LAB frame.

We note that the Lorentz transformation Eq. (20) applies to any particle. Therefore, if weadd

the rst of Eqs. (20) to the corresp onding equation for particle 2, i.e. for the target particle,

then weget

E + E E + m = [E + E + v (p + p )]

1 2 LAB 2 cm cm

1 2 1z 2z

and hence with (44) and recalling that p + p =0,weget

1z 2z

=(E + m )= s

cm LAB 2

and nally also the LAB velo city of the centre of mass

v = p =(E + m )

cm LAB LAB 2

Consider the particular case of equal masses. Then in the nonrelativistic limit weexpect

intuitively that the centre of mass is moving with half the sp eed of the incident particle. The

relativisticformula simpli es to

E m

LAB

v =

E + m

LAB 11

where wehavesetm = m = m . Then, with p m,weget

1 2 LAB

p p

LAB

1 v =

2m 4m

or, since the LAB velo city v of the incident particle is v = p = m and =

LAB LAB LAB LAB LAB

1=2 4 2

, ) we get, neglecting terms of order v 1=(1 v

LAB LAB

1 1

v = v v 1+

cm LAB

LAB

2 4

and we see that the second term in the brackets can b e neglected in the nonrelativisticlimit.

Thus wehave found the exp ected result.

We can use the expressions for v and to write down the Lorentz transformation from

cm cm

the LAB frame to the centre-of-mass frame for the energy of particle 1 using Eq. (20):

s E = (E v p )=(m + m E )=

cm LAB cm LAB 2 LAB

1 1

which repro duces the result Eq. (52). The corresp onding transformation formula for the

momentum repro duces Eq. (51).

To carry out the Lorentz transformation of the target particle from the LAB frame to the

CMS we substitute its LAB energy E = m and LAB momentum p = 0 in Eq. (20),

2 LAB 2 2 LAB

hence

E = m = m (m + E )= s

cm 2 2 2 LAB

in agreement with Eq. (52).

6. Elastic Collisions

In the case of elastic collisions we denote the 4-momenta of the incident particles p and p ,

1 2

and the 4-momenta of the scattered particles p and p ,such that

3 4

2 2 2 2 2 2

p =p = m and p =p = m (54)

3 1 1 4 2 2

4-momentum conservation is expressed by

p +p =p +p (55)

1 2 3 4

Consider the invariants which can b e constructed from the 4-momenta p ;:::;p . Such

1 4

invariants are of the form p p ;i;j=1; 2; 3; 4, and hence there are sixteen invariants. Four

i j

2 2

of these are of the form p = m , i.e. they are constants without any dynamical contents.

i i

They are therefore referred to as trivial invariants. Of the remaining twelveinvariants there

are only six di erent ones as a result of the symmetry p p =p p . This leaves us with six

i j j i

invariants, which are further constrained by the 4-momentum conservation (55). Thus there

remain two indep endentinvariants. However rather than working with twoinvariants it is

frequently convenient to use three invariants with one constraint. The most widely used choice

of suchinvariants are the Mandelstam variables, which are de ned by the following equations:

2 2

s = (p +p ) =(p +p )

1 2 3 4

2 2

t = (p p ) =(p p ) (56)

1 3 2 4

2 2

u = (p p ) =(p p )

1 4 2 3

Since only two of the Mandelstam variables are indep endent there exists one relation b etween

them, namely

2 2

s + t + u =2m +2m (57)

1 2 12

The variable t has a simple meaning in the CMS where one has E = E , and hence

1 3

2 2

t = (~p ~p ) = 2p (1 cos ) (58)

1 3

Thus up to a sign, t is the squared momentum transfer in the CMS. For this reason t is

generally and not very accurately referred to as the 4-momentum transfer. An imp ortant result

of Eq. (58) is that in elastic scattering t is always negative except at = 0 (forward scattering)

where t =0.

Interesting is also the signi cance of t in the LAB frame. Here wehave

t =(p p ) =2m (m E )=2m T (59)

2 2 4 2 4

2 4

where wehave denoted the LAB kinetic energy of particle 4 - the recoil particle - by T , i.e.

T = E m .

4 4 2

7. Inelastic collisions

If the particles in the nal state are di erent from the particles in the initial state then the

collision is said to b e inelastic. Examples of inelastic collisions are the creation of additional

pions in pion-proton collisions,

+ + +

+ p ! + p + + (60)

or the annihilation of an electron-p ositron pair into a muon pair,

+ +

e + e ! + (61)

Atvery high energies where many particles are created one frequently measures only one

of the nal state particles, for instance a fast outgoing electron in electron-proton collisions,

with all other particles remaining unobserved. This is referred to as inclusive col lision, and the

corresp onding reaction equation is written in the form

e + p ! e + X (62)

where X denotes any system of nal state particles.

Let us denote the 4-momenta of the initial particles p and p , and the 4-momenta of the

1 2

nal state particles p ; p ;::: ;p . Energy and momentum conservation are expressed by

3 4 n

p +p =p +p + ::: +p (63)

1 2 3 4 n

Let us calculate the minimum LAB energy needed for this reaction. This energy is called

the threshold energy. Let particle 1 b e the incident particle and particle 2 the target particle,

then in the LAB frame

p =(E ; 0; 0;p ) p =(m ; 0; 0; 0) (64)

LAB LAB 2

1 2

and hence

2 2 2

s =(p +p ) = m + m +2m E (65)

2 LAB

1 2

By 4-momentum conservation wehave also s =(p +p + ::: +p ) . We make use of the

3 4 n

invariance of s and evaluate the latter expression in the CMS; then, byEq. (63), wehave

2 2

s =(E + E + ::: + E ) (m + m + ::: + m ) (66)

3 4 n

2 2

= where in the last step wehaveusedE m . + p~ m

i i

i 13

Thus the threshold CMS energy is

s = m + m + ::: + m (67) E =

min 3 4 n

thr

and hence with Eq. (65) and setting M = m + m + ::: + m

3 4 n

h i

thr 2 2 2

E = M m m (68)

LAB 1 2

or, if weintro duce the LAB kinetic energy T = E m ,

LAB LAB 1

h i

thr 2 2

T = M (m + m ) =2m (69)

1 2 2

LAB

Equation (67) has an imp ortantphysical interpretation: at threshold the nal state particles

are at rest relativetoeach other. Expressed di erently this means that there is no internal

motion in the system of nal state particles. This implies that they move together with equal

velo city in the LAB frame.

+ + +

Let us calculate the threshold kinetic energy for the reaction + p ! + p + +

using the approximate mass values m =940MeV, m = m = m = 140 MeV:

i h

thr 2 2

T = (m +3m ) (m + m )

p p

LAB

2m 1

(2m +4m )(2m )=2m 1+ =

2m m

p p

= 2 140 (1 + 2 140=940) = 363:4MeV (70)

The kinetic energy of the incident pion is the only source of energy to create the additional

mass of the nal state, whichis2m . But our result shows clearly that wemust supply more

kinetic energy than the additional mass. Qualitatively this result can b e understo o d from our

discussion of Eq. (67): part of the LAB kinetic energy is converted into the rest energy of the

additional particles, and more kinetic energy is needed to impart LAB kinetic energy to the

created particles.

Quasi-elastic collisions.

A particular case of inelastic collisions are the quasi-elastic collisions, i.e. 2 ! 2 b o dy reactions,

in which the two particles in the nal state are di erent from the initial particles. The reaction

+ +

e + e ! + is a typical example of such reactions. Elastic collisions are a particular

case of quasi-elastic collisions.

To describ e 2 ! 2 b o dy reactions we shall again use the Mandelstam variables s, t and u.

In this case Eq. (57) must b e replaced by the more general relation

2 2 2 2

s + t + u = m + m + m + m (71)

1 2 3 4

As b efore wehave in the CMS ~p + ~p = ~p + ~p ,but

1 2 3 4

j~p j6= j~p j (72)

1 3

Let us therefore distinguish the magnitudes of the CMS momenta by a prime on the CMS

momentum in the nal state. Thus

j (73) j = j~p j and p = j~p j = j~p p = j~p

4 3 2 1

This statement has to b e mo di ed if there are massless particles in the nal state. 14

Equation (46) remains valid for p ; to express p in terms of s wemust replace in Eq. (46)

m by m and m by m ,thus

1 3 2 4

nh ih io

1=2

2 2 0

) (74) s (m m ) s (m + m ) p =

3 4 3 4

2 s

Consider the reaction

+ p ! K +

with the pion incident on the target proton. The threshold energy for pro ducing the kaon

(m =0:494 GeV ) and the (m =1:189 GeV) is 1.03 GeV by Eq. (68). Assuming a

LAB energy of the incident pion of 1.5 GeV weget

2 2 2 2 2

s = m + m +2m E =0:14 +0:94 +2 0:94 1:5=3:71 GeV

p LAB

and hence by Eqs. (46) and (74) wehave

p =0:727 GeV and p =0:438 GeV

i.e. the CMS momenta of the nal state particles are less than the CMS momenta of the initial

particles. Qualitatively this result is obvious: the reaction pro ducts are heavier than the initial

particles, and therefore part of the initial kinetic energy must b e converted into mass.

Conversely one exp ects the CMS momentum of the nal state to b e greater than the CMS

momentum of the initial state if the nal state particles are lighter than the initial state particles.

This can b e checked for the example of the annihilation of a proton-antiproton pair into a pair

of pions. It is interesting to note that this reaction can take place at zero relative momentum

of the proton and antiproton; even under this condition the pions are created with highly

relativistic momenta. Exp erimentally the annihilation at rest of proton-antiproton pairs can

b e realized for instance in the low-energy antiproton ring LEAR at CERN, where antiproton

b eams of extremely low momenta can b e pro duced. These low-energy antiprotons can lose

practically their entire kinetic energy in a liquid hydrogen target b efore b eing captured bya

hydrogen atom to form a protonium atom and nally annihilate with the proton.

Without derivation we collect here some useful formulas relating CMS and LAB variables to

the Mandelstam invariants.

CMS variables :

2 2 2 2

s + m m s + m m

2 1 1 2

p p

; E = ; (75) E =

2 1

2 2 s s

2 2 2 2

s + m m s + m m

4 3 3 4

p p

; E = ; (76) E =

4 3

s s 2 2

i ih h

2 2

p = p = ; (77) s (m + m ) s (m m )

1 2 1 2

1 2

2 s

h ih i

2 2

s (m m ) s (m + m ) ; (78) = = p p

3 4 3 4

4 3

s 2

Denoting the CMS angle b etween ~p and p~ by ,wehave

3 1

cos =1 (t t)=2p p =1 (u u )=2p p (79)

0 0

1 3 1 4

where

2 2 2 2

t = m + m 2(E E p p )= m + m 2(E E p p )

1 3 1 3 1 3 2 4 2 4 2 4 15

and

2 2 2 2

u = m + m 2(E E + p p )= m + m 2(E E + p p )

1 4 1 4 1 4 2 3 2 3 2 3

:by de nition wehave in the LAB p~ = 0, hence LAB variables

2 2

s m m

1 2

; E = m (80) E =

2 2 1

2 2 2 2

m + m u m + m t

2 3 2 4

E = ; E = (81)

3 4

2m 2m

2 2

the pro duction angle of particle 3, i.e. the angle made by p~ with the z axis, is given by

2 2

+2E E m t m

1 3

3 1

(82) cos =

2p p

1 3

and similarly

2 2

u m m +2E E

1 4

cos = (83)

2p p

1 4

8. Deep inelastic scattering.

In lepton-hadron scattering at suciently high energies one nds a large numb er of hadrons

in the nal state: this is deep inelastic scattering (DIS). The multiplicity of the hadronic system

varies eventbyevent. The reaction equation for electron-proton DIS is written as

e + p ! e + X (84)

where X stands for the hadronic system with an arbitrary numb er of particles. A generic

diagram depicting the DIS pro cess is shown in Fig. 2.

e(k ) e(k )

(q )

X (P )

p(P )

Figure 2: Generic diagram of deep inelastic scattering.

To describ e the DIS reaction kinematics we denote the 4-momentum of the incoming electron

byk =(E; 0; 0;k), that of the target proton by P and those of the scattered electron and of the

0 0

hadronic system byk and P , resp ectively. The exchanged virtual photon has 4-momentum

q=k k . 4-momentum conservation demands

0 0

k+P = k +P (85)

2 0 2

2 2

and wehave the mass-shell conditions k =k = m and P = m . Since energies characteristic

e p

of DIS are at least of several GeV, the electron mass can b e safely set equal to zero. Then we

2 2 0

get for the square of the 4-momentum transfer q =(k k ) = 2EE (1 cos ), and we see

that q 0, i.e. the exchanged photon is space-like. 16

2 0 2

The invariant W = P is variable b ecause of the variable multiplicity of particles in the

hadronic system, each of which can have an arbitrary kinetic energy up to some maximum

value. Therefore the complete kinematics of DIS is determined by three indep endentinvariants

rather than twoaswe are used to in elastic collisions. A natural choice of one of these invariants

is the square of the total CMS energy S ,

2 2

S = (k+P) = m +2k P (86)

which is de ned by the b eam energy.

The second invariant is usually chosen to b e the negative square of 4-momentum transfer:

2 2 2 0 2

Q = q = (k k ) =4EE sin (87)

The third indep endentinvariant can b e taken to b e W or alternatively one of the dimensionless

variables

x = (88)

2P q

P q

y = (89)

k P

where q = k k .

The variable y has a simple physical meaning in the target rest frame where P = (m ; 0; 0; 0),

0 0 0

k=(E ; 0; 0;E ), and k =(E ;~p ), hence y =1 E =E , i.e. y is the relative

LAB LAB LAB

LAB 3 LAB

energy loss of the electron in the LAB frame.

The invariant x is the Bjorken scaling variable or simply Bjorken-x.Itwas rst recognised

as an imp ortantvariable of DIS by J.D. Bjorken who predicted the prop erty of scaling in DIS

whichwas subsequently con rmed exp erimentally.

Interesting is the expression of S in terms of the b eam energies. In xed target DIS we

have the electron or muon b eam with 4-momentum k = (E; 0; 0;E) and the proton target with

P=(m ; 0; 0; 0), hence

S = m +2m E

whereas in an electron-proton collider like HERA wehave 4-momenta P = (E ; 0; 0;E ) and

p p

k=(E ; 0; 0; E ) and hence

e e

S =4E E

e p

Other useful relations b etween the various kinematical variables are the following:

Q = xy S (90)

and

2 2 2

W = m + Q (1=x 1) (91)

where in the latter formula wehavekept the proton mass in order to indicate that the threshold

of W corresp onds to elastic scattering.

Within the framework of the parton mo del, DIS pro ceeds by the exchange of a photon or

intermediate vector b oson with only one of the quarks in the proton. This is shown in the

diagram in Fig. 3.

The electron-quark collision is elastic. As a result of this collision the struck quark acquires

a sucientmomentum to break away from the rest of the proton as far as the colour force

allows it to travel. At this stage some of the binding energy is converted into the creation of a

quark-antiquark pair from the vacuum; the antiquark combines with the original quark into a

meson, leaving b ehind a quark which can give rise to the creation of another quark-antiquark 17

e(k ) e(k )

(q )

q (p )

q (p)

X (P )

p(P )

Figure 3: Parton mo del diagram of deep inelastic scattering.

pair. This pro cess, called fragmentation,continues until the remaining energy drops b elowthe

threshold for the creation of another pair. Thus, as a result of fragmentation, several mesons

are created whichtravel roughly in the direction of the struck quark. Such a system of mesons,

or more generally of hadrons, is called a jet. The residue of the proton is a highly unstable

system: it has lost a quark, absorb ed a quark presumably of the wrong sort that is left over

from the fragmentation, and has absorb ed a fraction of the energy transferred from the electron.

Therefore it breaks up into several hadrons.

The elastic electron-quark collision is the hard subprocess of DIS. If we think of the incoming

electron and proton as travelling in opp osite directions, then the quark carries a fraction of the

proton momentum. At a suciently high momentum, where the proton mass is negligible, the

energy of the quark is the same fraction of the proton energy. It turns out that this fraction is

identical with the Bjorken-x de ned ab ove. Denoting the 4-momentum of the incoming quark

by p wehave therefore

p = xP

Denoting the invariant(k + p) by s, which is the squared CMS energy of the subpro cess, we

have therefore also

s = xS (92)

This, together with the de nition of Q ,shows that the two indep endentinvariants that control

the kinematics of the subpro cess are x and Q .

The rst DIS exp eriments were carried out in 1967 at the Stanford 2-mile linear electron

accelerator with electron b eams of up to 20 GeV and hydrogen targets at rest, giving a CMS

energy of ab out 6 GeV. Subsequent xed target exp eriments were done in other lab oratories,

notably at the CERN SPS with muon b eams of up to nearly 300 GeV and hence of CMS

energies up to ab out 25 GeV. The range of energies available for DIS was extended byan

order of magnitude when in 1992 the electron-proton collider HERA came into op eration at

the DESY lab oratory in Hamburg. In this collider the electrons are accelerated up to nearly

30 GeV and the protons up to 820 GeV, giving a CMS energy of 314 GeV. Theoretically the

2 5 2

corresp onding values of Q go up to ab out 10 GeV .

An imp ortant to ol to study the structure of the nucleon is also deep inelastic scattering with

neutrinos as b eam particles. The kinematics is identical with the one describ ed ab ove, but one

must bare in mind that the exchanged particle in neutrino-DIS is an intermediate vector b oson,

either the W or the Z .

9. Phase Space Integrals.

One step in the evaluation of a decay rate or of a cross section is the evaluation of phase

space integrals. This step is logically part of kinematics, the dynamics of the pro cess b eing

contained in the matrix element. 18

The di erential decay rate of a particle of 4-momentum P = (E; P )into n particles of

4-momenta p ,p , :::,p is given by

1 2 n

(2 )

2 3n

jMj d (P; p ; p ;::: ; p ) (93) d=

1 2 n

2 3n

where jMj is the Lorentz invariant square of the matrix elementand d (P;p ; p ;::: ; p )is

1 2 n

the Lorentz invariant di erential phase space factor. The usual task of phase space calculations

is to integrate over all unobserved degrees of freedom.

Consider the simplest case, namely two-b o dy decay. Then the Lorentz invariant di erential

phase space factor is

3 3

d p d p

2 1

6 4

(94) d (P; p ; p )= (P p p )

1 2 1 2

3 3

(2 ) 2E (2 ) 2E

1 2

where P = (E; P ), p =(E ;~p ) and p =(E ;~p )arethe4-momenta of the mother particle

1 1 2 2

1 2

and the twodaughter particles, resp ectively, and the four dimensional delta function ensures

energy-momentum conservation.

Thus, in the present case, if wewant to nd the total decay rate, then wemust integrate

the di erential decayrate

(2 )

d= jMj d (P; p ; p ) (95)

1 2

over all comp onents of p and p . Here jMj is the spin-averaged square of the matrix element,

1 2

whichinthecaseoftwo-b o dy decay is a constant. Therefore the integrations can b e carried

out exactly. Integrating over one of the three-momenta removes the three dimensional delta

function (P p~ ~p ), hence

1 2

d p

3 6

d (P;p ; p )=(2 ) (E E E )

2 1 2

1 2

2 2 2

where E = p (P p~ ) + m + m .Nowwe exploit the Lorentz invariance of and E =

1 1 2

1 2 1

by nishing the integration in the \frame of convenience", which in this case is the rest

frame of the mother particle (CMS), where P =0 and E = m, hence

d p

3 6

d (CMS)= (2 ) (m E E )

2 1 2

2 2

Here E retains its previous form and E = . Since the integrand has no angular + m p

1 2

2 1

dep endence, the integration over the full solid angle gives a factor of 4 , leaving us with the

integration of the mo dulus of ~p . This is conveniently converted into an integration over E ,

1 1

2 2 2

using E = p + m , hence p dp = E dE . The nal integration over E is accomplished by

1 1 1 1 1

1 1 1

setting the argument of the delta function equal to f (E ) and using the identity

0 1

(f (E )) = jf (E )j (E E )

1 10 1 10

where E is the ro ot of f (E )=0.Thus nally we get

10 1

1CMS

(2 ) m

2 2 2 2 1=2

where the CMS momentum is p = f[m (m m ) ][m (m + m ) ]g =2m.

1 CMS 1 2 1 2 19

Example: pion decay ! .

The spin averaged mo dulus squared of the matrix element of pion decay is in the lowest order

of p erturbation theory given by

2 2 2

jMj =4G f m (p k) (96)

where G is the Fermi constant, f is the pion decayconstant, m is the muon mass, p and k

are the four-momenta of the muon and neutrino, resp ectively, and we assume the neutrino to

have zero mass. The di erential decay rate of this pro cess is given by

(2 )

d= jMj d (P; p; k)

where P is the four-momentum of the pion and

3 3

d p d k

6 4

d (P; p; k) = (P p k)

3 3

(2 ) 2E (2 ) 2!

where E and ! are the energies of the muon and the neutrino, resp ectively.We get the total

decayratebyintegrating over the momenta ~p and k . Let us do this in two steps. First we

integrate over the neutrino momentum:

3 3 4

d p d k (2 )

3 3

jMj (E E ! ) (P ~p k ) d =

3 3

2m (2 ) 2E (2 ) 2!

d p 1

jMj (E E ! ) =

32 m E!

then weintegrate over the p olar angle and azimuth of the muon. Todothiswe represent

3 2

d p as p dp d , where d =d d cos ,andwe use EdE = pdp,thus

1 pdE d

d == jMj (E E ! )

32 m !

2 2

Nowwe note that the matrix element is constant. Indeed, wehaveP =(p+k) = m +2p k,

2 2 2

but also P = m , hence p k=(1=2)(m m ), and the matrix elementisgiven by

2 2 2 2 2

jMj =2G f m (m m )

and therefore the integrand has no angular dep endence and integration over gives just a

factor of 4 .Thus

1 pdE

d= jMj (E E ! )

8m !

Now, since momentum conservation has already b een imp osed byintegration over k , the energies

2 2

E m . Let us denote the E and ! are no longer indep endent: wehave ! = k = p =

argument of the remaining function by f (E ): f (E )=E E ! , or, working from nowon

in the pion rest frame, f (E )=m E ! . Then, using the identity

0 1

(f (E )) = jf (E )j (E E )

0 0

we get

jMj p =

m . Using once more momentum conservation, wealsohave p = k = where p = E

0 0 0

2 2

! =(m m )=2m . Finally we get for the pion decay rate

2 2

G f

= (97) m m 1

8 m

Nowthedecay rate is related to the mean life by=h=tau. Therefore we can nd the

pion decayconstant f from Eq. (97) by substituting the known values of the Planck constant

hbar ,theFermi constant G and the muon and pion masses.

Example: Muon decay ! e .

In this example wehave three nal state particles, two of which are massless. Eachof

the outgoing particles can have an energy of up to one half of the muon rest energy, i.e.

approximetely up to 50 MeV, which is an ultrarelativistic energy for the electron. Atmomenta

close to zero, the electron is of course nonrelativistic, but at such energies the approximation

whichwe are going to use, where the matrix element is calculated in the lowest order of

p erturbation theory,isnotvery accurate and one must include radiative corrections. This

is outside the scop e of these notes. We shall therefore carry out the calculations with an

ultrarelativistic electron over the entire phase space, and hence set the electron mass equal to

zero.

The reaction equation of muon decayis

(p) ! e(p )+ (k)+ (k )

with the following four-momenta in the muon rest frame:

0 0 0

0 0

~ ~

p=(m; 0); p =(E ; p ); k= (!; k ); k =(! ; k )

The di erential decay rate is given by

d= jMj d (p; p ; k ; k )

where in lowest order p erturbation theory wehave

2 0

jMj =64G (p k )(k p )

0 0 0

= m! Now, p k ,butk p involves the energies ! and E and the angle b etween k and p .

This is inconvenient. Therefore we use four-momentum conservation to transform this scalar

pro duct:

0 0 2 2 2 0

2k p =(k+p) =(p k ) = m 2m!

hence

2 2 0 0

jMj =64G m ! (m=2 ! ) (98)

and in this form jMj dep ends only on the energy ! ; therefore we need not consider it until

we get to the integration over ! .

) in detail, wehave therefore Writing the Lorentz invariant phase space factor d (p;p ; k; k

for the di erential decayrate

3 0 3 3 0

d k d k d p 1

4 4 0

jMj (2 ) (p p k k ) d=

3 0 3 3 0

2m (2 ) 2E (2 ) 2! (2 ) 2! 21

and we b egin the calculation byintegrating over the muon-neutrino three-momentum k ;weget

d k 1

4 0 0 0

(p p k k ) = (m E ! ! )

! !

0 0 0 0 0 0 0

~ ~ ~

~ ~

+2! E cos , where is the angle b etween p and k . + E ! with ! = jk j = jk + p j =

3 0

Next weintegrate over k .To do this we represent d k in p olar co ordinates with the p olar

3 0 0 0

axis along p ,whichiskept xed at this stage, i.e. d k =2! d! d cos , where the factor 2

comes from the integration over the azimuth. First the integration over cos :

3 0

1 d p d cos

0 0 0 0

d= jMj ! d! (m ! ! E )

4 0

(4 ) m E !

Let us put x =cos and denote the argumentofthe function by f (x), i.e.

0 0

0 0 0 0

+ E +2! E x f (x)= m ! E !

and pro ceed by using the standard formula

0 1

(f (x)) = jf (x )j (x x )

0 0

where x is de ned by f (x )=0 and f (x ) is the derivativeof f (x)at x = x :

0 0 0 0

0 0 0

f (x )=(E ! =! )

0 0

with ! = ! (x ), and hence

0 0

dx dx

= (x x ) (f (x))

0 0

! E !

which gives us immediately

3 0

1 d p

d! d= jMj

(4 ) m

and wemust rememb er that the function gives a nonzero value only if 1 x 1or,

equivalently,if ! ! ! , where

0 0

0 0 0 0

! = ! + E 2! E = j! E j

0 0

and since by virtue of the function wehave ! = m ! E , this condition can b e cast in

the form of

0 0 0 0 0 0

j! E j m ! E ! + E

0 0

or, if weadd(! + E ) and divide by2,

1 m

0 0 0 0 0 0

(j! E j + ! + E ) ! + E (99)

2 2

0 0

The left-hand side of this inequality is the greater of ! and E , and hence the left-hand

inequality is equivalenttotwo inequalities:

0 0

E m=2 and ! m=2

0 0

which, together with the right-hand inequality, de nes a triangle in the (E ;!) plane over

0 0

whichtheintegration over ! and E has to b e taken, see Fig. 4. 22

m=2

- @

m=2

Figure 4: The shaded region is the domain of integration (99)

0 0

The integral over ! go es from (m=2 E )tom=2, and wemust now rememb er that the

matrix element dep ends on ! .Thus

2 2 0 3 0 2

m=2

m 16G m 4E d p 64G m

0 0 0 0 0

! ! d! = 1 d= 4E dE

4 4

(4 ) 2 (4 ) 3m

m=2E E

3 0

where in the last step wehave also represented d p in p olar co ordinates and done the angular

integrations, whichhave yielded a factor of 4 . We can rewrite our result in the form of the

energy distribution of the emitted electron:

2 2 0

d 16 m 4E

= E (100) 1

0 3

dE (4 ) 3m

This result is very close to the exp erimental data.

Nowwe can also get the total decayrateby carrying out the nal integration over E from

zero to m=2, hence

2 5

G m

= (101)

192

and hence the muon lifetime:

192h h

= =

2 5

G m

This formula can b e used to get a value for the Fermi constant G from the accurately known

values ofh and of the muon mass and lifetime. The result is remarkably close to the result

obtained from similar calculations for nuclear decay. The small di erences are accounted for

by radiative corrections. However, we shall not enter into these considerations, rememb ering

that the purp ose of this exercise was an illustration of the phase space integration in the case

of three-b o dy decay.

The derivation of the three-particle phase space given here followed the most direct route.

One can nd other, more ingenious metho ds in the literature. For instance, in Part 2 of

the \Relativistic Quantum Mechanics"by E.M. Lifshits and L.P. Pitaevsky, the following

alternative derivation is presented.

M. Bardon et al., Phys. Rev. Lett. 14 (1965) 449; the deviation is partly due to radiative corrections and

partly due to exp erimental acceptance.

this is Vol. IV of the monograph on theoretical physics by Landau and Lifshits; the derivation mentioned

here is given in Chapter XVI, section 146 of the Russian 1971 edition 23

Without spin summation, the mo dulus squared matrix elementis

2 2

k jMj =32G (p m a ) (p m a ) k

e e e

where p and p are the four-momenta of the electron and muon, resp ectively, m and m are

e e

their masses, a and a are their p olarization vectors, and k and k are the four-momenta of

e e

the electron and muon neutrino, resp ectively. The 4-vector comp onents are lab eled by and

. The di erential decay rateisgiven by

3 3 3 2

d p d k d k jMj

e e

4 4

d=(2 ) (p + k + k p )

e e

3 3 3

2m (2 ) 2E (2 ) 2! (2 ) 2!

e e

where E , ! and ! are the energies of the electron, electron neutrino and muon neutrino,

e e

resp ectively.

Integration over the neutrino momenta is accomplished by taking the following integral:

3 3

d k d k

I = k k (k + k q )

! !

where q = p p . The only dynamical variable, on which this symmetric tensor dep ends, is q ;

therefore the integral must b e of the form of

I = Aq g + Bq q

where A and B are numerical factors. Contracting I with g we get

2 2

Iq (4A + B )q =

and contracting with q q weget

2 2 2 2

(A + B )(q ) = I (q )

where

3 3

d k d k

I = (k + k q )

! !

3 3

~ ~

Integration over d k removes the three-dimensional function (k + k q ), leaving us with

the integral

d k

I = (! + ! ! )

! !

~ ~

which is done in the CMS of the two neutrinos, where k + k = 0 and ! = ! , and there is no

e e

angular dep endence, so the angular integration gives a factor of 4 , leaving us with the integral

I =4 (2! ! ) d! =2

e e

and hence

(q g +2q q ) I =

and the nal result is found by straight forward calculations.

For n particle decays the phase space integrals can b e done recursively in the CMS:

1 d p

2 2 1=2

([E 2EE + m ] ) (E )=

n1 n n

(2 ) 2E

n 24

As wehave seen, the three-particle phase space integral can b e evaluated analytically.For n>3

one has to resort to numerical integration which is usually done by Monte Carlo metho ds.

Phase Space Calculations in the Case of Collision Pro cesses.

Phase space calculations are equally imp ortant in collision pro cesses. Here one must b ear

in mind that the matrix element is not constanteven in the simplest case of 2 ! 2 collisions.

However the structure of the spin-averaged square of the matrix elementofa two-b o dy collision

is suciently simple for analytical calculation to b e feasible. For inelastic 2 ! n particle

collisions with n 3 one resorts to numerical integration.

For 2 ! 2 b o dy collisions, i.e. elastic in quasielastic collisions, the observed quantityisthe

di erential cross section; it is given bytheformula

jMj

4 6

d =(2 ) d (p +p ;p ; p ) (102)

1 2 3 4

2 2

(p p ) m m is the Lorentz invariant ux of the incoming particles of masses where F =4

1 2

m and m and 4-momenta p and p , and the 4-momenta of the outgoing particles are denoted

1 2

p and p . In the LAB frame the ux factor takes the form of

3 4

F =4m p

2 LAB

where, as b efore, m is the mass of the target partcle. In the CMS wehave

F =4p s

CMS

Consider the particular example of electron-muon elastic scattering in the LAB frame. Then

the spin-averaged matrix element is given by

" #

4 2

8e q

2 2 0 2

jMj = cos 2M EE sin

4 2

q 2 2M 2

where M is the muon mass, E and E are the initial and nal electron energies, is the

2 0

is the 4-momentum transfer, and wehave neglected the scattering angle, q = 4EE sin

electron mass.

Note that jMj dep ends on the observable scattering angle . Therefore, b earing in mind

that wewanttointegrate over all unobserved degrees of freedom, wemust from the b eginning

plan the phase space integrations suchas not to integrate over , except p ossibly as a very last

step if wewant to nd the total cross section.

The Lorentz invarianttwo-b o dy phase space factor is

3 3

d p d p

4 3

6 (4)

d = (p +p p p )

3 4 1 2

3 3

(2 ) 2E (2 ) 2E

3 4

and wehave to carry out four of the six integrations to get rid of the four-dimensional function.

We b egin byintegrating over the 3-momentum ~p ; this removes the three-dimensional function

(3)

(~p + p~ p~ p~ ) and leaves us with

3 4 1 2

p dp d

d = (E + E E E )

2 3 4 1 2

3 2

[2(2 ) ] E E

3 4

where wehave expressed the three-dimensional di erential d p in p olar co ordinates with d =

sin dd, where and are the p olar angle and azimuth, resp ectively. 25

At this p ointwechange the notation to that used previously for the LAB frame by setting

q q

2 0 2 0 2 2

~ ~

k ; E = E = k ; E = M; and E = E = E = ~p + M

3 2 4 1

hence

0 0

E dE 1

0 3

d (E + E E M ) d =

4 2

4(2 ) E

We note that our integration over ~p has already enforced momentum conservation. There-

fore wehave

0 0 0 2 2 0 2 0

~ ~

~p = k k ; hence ~p = E + E 2EE cos

Now consider the argument of the function. Let us denote it by f (E ), i.e.

0 0

0 2 2

f (E )=E + ~p + M E M

Its zero corresp onds to energy conservation. Toevaluate the integral over E we rewrite the

function in the form of

df (E )

0 0 0

(f (E )= (E E )

0 0 0 0

where E isthezeroof f (E ). Di erentiating f (E ) with resp ect to E weget

0 0 0

E E cos E + E E cos df (E )

=1+ =

dE E E

4 4

or, applying energy conservation, E + E = E + M ,

M + E (1 cos ) ME df (E )

= =

0 0

dE E E E

4 4

2 0 0

where in the last step wehave used q = 2EE (1 cos )=2M (E E ). Thus nally we

have

E E

0 0 0

(f (E )) = (E E )

Nowwe can do the integral over E and get

0 2

1 1 E

d = d

6 2

4(2 ) (4 ) ME

where in the nal expression wehave dropp ed the subscript of E , whichisnow redundant.

Putting our results for the matrix element, the phase space factor and the ux factor

together, we get for the di erential cross section of elastic electron-muon scattering the following

expression:

! !

0 2 2

E q d

2 2

= cos sin (103)

d E 2 2M 2

4E sin

LAB

where = e =4 is the ne structure constant.

References

[1] E. Byckling and K. Ka jantie, \Particle Kinematics", Wiley, New York, 1973.

[2] J.D. Bjorken and S.D. Drell, \Relativistic Quantium Mechanics", McGraw-Hill, New York,

1964.

[3] \Review of Particle Physics", Particle Data Group, Europ ean Physical Journal C 15 (2000)

1-878.

Popularly known as the Particle Data Tables, this review is published every other year; it

is available on the website http://p dg.lbl.gov 26