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RelativisticKinematics
Phase space considerations: discussing a system of many particles we can separate two aspects of the description. The dynamics of the single interaction and the kinematics. When many particles are present the latter tend to dominate the behavior of the system.
The extreme case is a gas of particles and thermodynamics. Here the form of laws is genera and dos not depend of the detail of the collision between two particles !
Reference: Hagedorn, relativistic Kinematics
Phasespace
Let us consider one of the processes we have encountered in one of the previous lessons:
p+p p+p+π++π-+π0
and let use consider some detail.
The initial state |i> is defined by the type of particle and by their momenta that are eg : beam momentum and zero, for an experiment carried out on a fixed target.
Moreover the beam direction is defined say along the z axis.
even if we have fixed the type of final state (i.e. its particle content) , there is an infinite choice of possible final states, depending on how the momenta of final particles are oriented in the space and on what are their intensity. According to QM each final state will be realized proportional to a probability density
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P(i→ f )∝ p1' ⋅ ⋅ ⋅ pn
' S p1 p22
Spectra
Generally we are not going to study the very specific detail, it is a too much complex situation and we look at same simple distribution, eg the momentum spectrum of the π+. This means that we INTEGRATE on all other, non relevant variables.
In a formal way, lets restrict our analysis to a subset F of the possible final states and let’s ask what is the probability that the reaction goes to this subset (eg. momentum of the π+ between 1.2 and 1.3 GeV) we have
S contains dynamical information and kinematical information. Let’s separate them explicitly
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P(i→ F)∝ f Sf ∈F∑ i 2
Probability
In our example the integrals over F are on all possible values of particle momenta except for the π+ where the momentum is limited in the range 1.2 to 1.3 GeV and its angles are in all possible directions. €
P(i→ f )∝ d4∫ p1' d4∫ p1
' ⋅ ⋅ ⋅ d4∫ pn'
δ 4 p1' + p2
' + .....+ pn' − p1 − p2( )
δ(pi' 2
i=1
n
∏ −mi' 2)
S(p1' p2' ⋅ ⋅ ⋅ pn
' | p1p2)
Statisticaltheory
We call momentum-phase-space –factor the integral we have considered above without the function S that describes the dynamics (analogy with gas theory)
and we obtain
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RF = d4∫ p1' d4∫ p1
' ⋅ ⋅ ⋅ d4∫ pn' ⋅ δ 4 p1
' + p2' + .....+ pn
' − p1 − p2( ) ⋅
δ(pi' 2
i=1
n
∏ −mi' 2)
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P(i→ F)∝ SFRF
More generally we define:
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Rn (P,m1,m2,⋅⋅,mn ) = d4∫ p1 d4∫ p2 ⋅ ⋅ ⋅ d4∫ pn ⋅
δ 4 p j − Pj=1
n
∑
⋅ δ(pi
2
i=1
n
∏ −mi2)
Simpleapplications
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R1 = ∫ d3p12 ⋅ E1
δ 4 (m) =δ(E −m)2 ⋅m
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R2 = ∫ d3p12 ⋅ E1
d3p22 ⋅ E2
δ 4 (M;m1,m2) = πp*
Mwhere
p*
MassDistribution(1)
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P
= pii=1
∑
Pn− = pii= +1
n
∑
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M
2 = P
2
Mn−2 = Pn−
2
Take n particles and divide them into two groups 1…. and +1….n we ask what is the mass distribution of the group of particles 1… or what is the probability that the mass lies between M2 and M2 + dM2
Their momenta and masses are:
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Massdistributions(2)
Using
one gets
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δ 4 p j − Pj=1
n
∑
= dP
⋅∫ δ 4 P − P
− p j
j= +1
n
∑
δ 4 p j − Pj=1
∑
€
Rn (P,m1,m2,⋅⋅,mn ) =
d4 p j∫ ×δ 4 P − P− p j
j= +1
n
∑
⋅ δ(pi
2
i= +1
n
∏ −mi2)
d4 p jd4∫ P×δ 4 P
− p j
j=1
∑
⋅ δ(pi
2
i=1
∏ −mi2)
Massdistributions(3)
adding
one gets
€
1= δ(P
2
0
∞
∫ −M 2)dM 2
€
Rn (P,m1,m2,⋅⋅,mn ) = dM 2∫
d4 p j∫ d4P×δ 4 P − P
− p j
j= +1
n
∑
⋅ δ(pi
2
i= +1
n
∏ −mi2) ⋅ δ(P
2 −M 2)
d4 p j∫ ×δ 4 P− p j
j=1
∑
⋅ δ(pi
2
i=1
∏ −mi2)
Massdistributions(3)
and finally
from which follows the mass distribution
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P(M)dM 2 =Rn−+1(P;M,m+1 ⋅ ⋅mn ) × R (P;m1 ⋅ ⋅ ⋅m )
Rn (P;m1 ⋅ ⋅ ⋅mn )dM 2
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Rn (P;m1 ⋅ ⋅ ⋅mn ) = dM 2∫ ⋅ Rn−+1(P;M,m+1 ⋅ ⋅mn ) × R (P;m1 ⋅ ⋅ ⋅m )
Discoveryoftheω
with of 1.6 GeV . In bubble chamber
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p
They collected 2500 events with 4 prong and test the compatibility of the reaction doing the missing mass hypothesis. 800 events do not fit the hypothesis that the reaction is closed without π0 production.
They plot the invariant mass combinations:
Discoveryoftheω
phase space curve
Deviation from phase space= dynamic effects= new particle
Massdistributionofaresonance
Take a particle of rest mass M and lifetime τ=1/Γ . Its wave function in its reference frame is
€
ψ(t) = e− tΓ ⋅ e− iMt
The wave function in term of energy is the Fourier transform
€
ϕ(t) = dt0
∞
∫ ⋅ e− tΓ ⋅ e− iMt ⋅eiEt= 1i(E −M) + Γ
€
1i(E −M) + Γ
2=
1(E −M)2 + Γ2
And the mass distribution is given by:
Massoftheωmeson
Different possibilities for BKG subtraction
BW or BW convoluted with Gaussian (resolution) to get the resonance parameters.
Mass is easily determined (syst. error. on the absolute scale) The width measurement requires that Γ is not too small compared to resolution
If it is a pure BW the width at ½ maximum is 2 Γ.
Threebodyphasespace
We compute it in the rest frame of the three bodies
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R3(M;m1,m2,m3) =
d3 p12E1
∫ d3 p22E2
∫ d3 p32E3
⋅ δ 4∫ (P − p1 − p2 − p3)
We integrate the three component of p3 using the 3 spatial component of the δ. This implies
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E32 = (p1 + p2)
2 + m32
Threebodyphasespace
3
2
1 This angle cannot be chosen independently from the rest
• Fix the direction of 1.
• Integrate 2 around this direction:
• Integrate along all possible directions of 1 :
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d3 p2 → 2π ⋅ p22 ⋅ dcosϑ
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d3 p1→ 4π ⋅ p12
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R3(M;m1,m2,m3) =
8π 2 dp12E1∫ dp2
2E2
dcosϑ2E3
δ∫∫ (M − E1 − E2 − E3)
Threebodyphasespace
Changing variables (p1 p2 cosθ) (E1 E2 E3 ) one gets
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R3(M;m1,m2,m3) = π 2 dE1dE2∫ dE3 ⋅ δ(M − E1 − E2 − E3)
The phase-space density of events in any pair of c.o.m. energies is flat !
Exercise: show that the phase space density is also flat in in any pair of s12,s13,s23 where sij is the square of the invariant mass of the pair.
Solution
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s12 = (P − p3)2 = M 2 + m3
2 − 2ME3
s12 + s13 + s23 = M 2 −m12 −m2
2 −m32
1
2 3
1 2 3
E3 min, s12 max E3 max, s12 min
DalitzTriangle(1)
2
3
1
ρ φ
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T1 =Q3(1+ ρ(− 1
2cosφ − 3
2sinφ)
T2 =Q3(1+ ρ(− 1
2cosφ +
32sinφ)
T3 =Q3(1+ ρcosφ)
Assume equal masses and define Q=M-3m
Ti=Ei-m is the kinetic energy. T is measured as the distance from the side
DalitzTriangle(2)
2
3
1
ρ φ
The position in the plane ensure energy conservation. However momentum conservation is not guaranteed and not all position in the plane are allowed. For example the corners are not allowed.
and the boundary is given by p3=p2+p1
There must exist a boundary curve that delimits the allowed region. Since we are on the boundary these events must correspond to a special kinematic case: It is when the particles are on a straight line: ne balancing the momentum of the other two.
In the region 0<φ<2⁄3π we have T3>T2>T1
DalitzTriangle(3)
2
3
1
ρ φ
This is a difficult equation to solve in the general case because p=√(T2+2mT) and the equation becomes transcendent.
More simple are two limiting cases: m=0 and m>>T. The general case lies in between.
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m = 0T3 = T2 + T11+ ρcosφ = 2 − ρcosφ2ρcosϕ =1
ρcosϕ =12
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T3 =Q3(1+
12) =
Q2
DalitzTriangle(4)
2
3
1
ρ φ
Non relativistic case:
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p3 = p1 + p2p32 = p1
2 + p22 + 2p1p2
T3 = T1 + T2 + 2 T2T1
with some tedious algebra
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ρ =1
DalitzTriangle(5)
2
3
1
ρ φ
Not relativistic case
Relativistic case
All other solutions will be in between
