Difference between revisions of "Forest Relativity Notes"

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The Kinetic Energy <math>(\gamma-1)*m_oc^2)</math> is Given by
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The Kinetic Energy (<math>(\gamma-1)m_oc^2</math>) is Given by
  
 
  root [43] (P1.Gamma()-1)*0.511
 
  root [43] (P1.Gamma()-1)*0.511

Latest revision as of 17:39, 28 April 2015

Lorentz Transformations

The picture below represents the relative orientation of two different coordinate systems (S,S) . S is at rest (Lab Frame) and S is moving at a velocity v to the right with respect to frame S.

ForestRelativityLorentzFrame.jpg

The relationship between the coordinate(x,y,z,ct) of an object in frame S to the same object described using the coordinates (x,y,z,ct) in frame S is geven by the Lorentz transformation:

Notation

The relativistic transformation used to transform the coordinates of an object observed in the rest frame S to a moving reference frameS is given by:

xμ=3ν=0Λμνxν

where

x0ct
x1x
x2y
x3z
Λ=[γγβ00γβγ0000100001]
β=vc=pcE
γ=11β2=Etotmc2
NOTE
It is common in particle physics to define c1 making γ=Em where m is in units of MeVc2
example
x0=2ν=0Λ0νxν=Λ00x0+Λ01x1Λ02x2+Λ03x2
ct=γx0γβx1+0x2+0x3=γctγβx=γ(ctβx)
Or in matrix form the tranformation looks like
(ctxyz)=[γγβ00γβγ0000100001](ctxyz)
Note
Einstein's summation convention drops the symbols and assumes it to exist whenever there is a repeated subscript and uperscript
ie; xμ=Λμνxν
in the example above theν symbol is repeated thereby indicating a summation over ν.

Momentum 4-vector

The momentum 4 -vector is denoted as:

pμ(Ec,p)
pμ(Ec,p)
pμpμ=E2c2p2E2p2=m2
Note
There is another convention used for 4-vector notation by Perkins and Koller which goes like this
pμ(p,iE)
pμ(p,iE)

Trig Method

Another way to represent the Lorentz transformation is by using the substitution

sin(α)βvc
cos(α)1γ1β2
The Matrix form of the tranformation looks like
(ctxyz)=[sec(α)tan(α)00tan(α)sec(α)0000100001](ctxyz)
Or the reverse transformation
(ctxyz)=[sec(α)tan(α)00tan(α)sec(α)0000100001](ctxyz)
Notice that you just needed to change the signs for the inverse matrix Λ1

Proper Time and Length

Proper Time

Proper Time \Tau
The time measured in the rest frame of the clock. The time interval is measured at the same x,y,z coordinates because the clock chose is in a frame which is not moving (rest frame).

The time given in any frame (t) = γ\Tau

Note
since γ>1 you expect the Proper time interval to be the smallest

Proper Length

Proper Length(c\Tau)
The length of an object in the object's rest frame.

Invariant Length

Transformation Examples

Decay of a Particle at Rest to 2 Bodies

Consider the decay of the ρ0 meson of mass M at rest into two pions (π+ and π ) of mass m1 and m2 respectively.


File:NeutralRhoMesonDecayDiagram.jpg

The diagram above shows a ρ0 meson at rest in the lab which then decays into two pions of momentum p1 and p2 in the center of momentum frame of the ρ0 meson.

Our goal is to determine the momentum and energy of each pion (E1 & E2) resulting from the decay of the Mother particle.

Conservation of 4-Momentum implies that if Pμ represents the total momentum of the system before the decay then

Pμ=(E,0)=(M,0)=(p1)μ+(p2)μ
0=p1+p2

or

p1=p2

Let

p|p1|=|p2|

Conservation of Energy

Etot=M=E1+E2=m21+p2+m22+p2

solving for p

p=12M[M2(m1m2)2][M2(m1+m2)2]
Mm1+m2 is required to avoid the unphysical condition that the momentum of the particles after a decay would be an imaginary number

Using

p|p1|=|p2|
E21m21=E22m22
E2=E21m21+m22

Combine this with the conservation of energy equation above:

E1+E2=E1+E21m21+m22=M
E1M=E21m21+m22

Square both sides of the above equation

E212ME1+M2=E21m21+m22
E1=M2+m21m222M

Similarly

E2=M2+m22m212M


Note
p1=p2
The daughter particles (pions) from the decay of the Mother particle (ρ) travel in opposite directions with respect to eachother ( ie; they are "back - to -back")
This means that there is no preferential direction for the decay (the particles are distributed isotropically such that they are back-to-back)

Decay of Moving Particle to 2 Bodies (decay in flight)

Pμ=(E,ptot)=(M,ptot)=(p1)μ+(p2)μ

Assuming that the Mother particle is moving along the Z-axis, then the momentum of the daughter particles perpendicular to the Z-axis (transverse components:p1, and p2,) are equal and opposite by conservation of momentum.

p1,p2,

The center of momentum frame is moving such that

βCM=ptotM
γCM=EtotM

A Lorentz transformation of the kinematics for particle 1 between the Center of Momentum (cm) frame and the lab is given by:

E1=γcm(ECM1+βcmpCM1,z)
p1,z=γCM(pCM1,z+βcmECM1)
p1,=pCM

where

ECM1 = Kinetic Energy (not total) of particle 1 in the center of momentum (CM) reference frame
pCM1,z = momentum of particle 1 along the direction of the mother particle in the CM frame
p1, = the component of particle 1's momentum perpendicular to Mother particle's momentum
You can now use the results for E1 and p1=p from the previous section where the Mother particle is at rest to determine the kinematics of the particles in the lab frame given that you know the initial 4-Momentum of the mother particle. You will need to specify the daugher decay angles in the CM frame in order to find the momentum components pz and p.
It can be shown that the lab angle for daughter particle 1 (θ1) is given by
tan(θ1)=sin(θCM1)γCM(βCMβCM1+cos(θCM1))

where

βCM1=p1E=β for daughter particle 1 in CM frame.

One could also find p1 without using the Lorentz transformation. Just use conservation of Energy and Momentum:

Etot=Etot1+Etot2=m21+p21+m22+p22
ptot=p1+p2

Solve the conservation of momentum equation for p22

p22=(ptotp1)2

and substitute the above for p22 in the Conservation of Energy equation above. The dot product gives you the angle between the daughter momentum and the Mother momentum (θ1) as a variable. After a lot of algebra you can show that

p1=(M2+m21m22)ptotcos(θ1)±2EM2p2m21p2totsin2(θ1)2(M2+p2sin2(θ1))
Note
p is the momentum of the two daughter particles in the CM frame which was derived when the Mother particle was at rest. ptot is the momentum of the Mother particle.


In order for a real solution

M2p2m21p2totsin2(θ1)0
Mpm1ptotsin(θ1)


If Mpm1ptot>1 then θ1 can be any angle and the "-" sign possibility in "±" is rejected to avoided negative values for p1 when θ1 > π2.


If Mpm1ptot<1 then the maximum emmission angle for daughter 1 is given by

sin(θmax1)=Mpm1ptot

The "±" is kept because for each θ1<θmax1 there are two possible trajectories for daughter particle 1 and as a result 2 trajectories for daughter particle 2.

Decay of Particle to 3 Bodies (Dalitz plot)

Now lets consider the case where a Mother particle of mass M decays into 3 daughter particles of masses m1, m2, and m3. The 4-mometum conservation is written as

Pμ=(p1)μ+(p2)μ+(p3)μ

The following invariants are defined

s=PμPμ=M2
s1=(Pp1)μ(Pp1)μ=(p2+p3)μ(p2+p3)μ
s2=(Pp2)μ(Pp2)μ=(p3+p1)μ(p3+p1)μ
s3=(Pp3)μ(Pp3)μ=(p1+p2)μ(p1+p2)μ

The invariants s1, s2 and s3 are not independent (the motivation for what is known as a Dalitz plot). Based on the definitions of these invariants and 4-momentum conversation one can show that

s1+s2+s3=(P22Pp1+p21)+(P22Pp2+p22)+(P22Pp3+p23)
=3P22P(p1+p2+p3)+p21+p22+p23
=3P22P2+p21+p22+p23
=P2+p21+p22+p23
=M2+m21+m22+m23


Also Note
s1 is the invariant mass of a subsystem defined by treating daughter particles 2 and 3 as one object. similar interpretations for s2 and s3.


There are limits to the values of the invariant masses s1 - s3

In the Center of Momentum system we have the

s1=M2+m212ME1

because

E1=m21+p21 we expect E1m1

This lead to

s1|max=M2+m212M(m1)=(Mm1)2


TO find the minimum value of s1 we evaluate s1 in the rest frame of the m2, m3 subsystem.

s1=(p2+p3)μ(p2+p3)μ=(Ecm232+Ecm233)2>(m2+m3)2 : In the CM frame E=p2+m2=m

so the limits of s_1 are

(m2+m3)2s1(Mm1)2

Invariant mass Dalitz plot Limits

Analagously you can show the the Min - Max limits for the s1, s2, and s3 invariant masses are given by.

(m3+m1)2s2(Mm2)2
(m1+m2)2s3(Mm3)2

set theory notation is often used to express the above limits as

s1[(m2+m3)2,(Mm1)2]
s2[(m3+m1)2,(Mm2)2]
s2[(m1+m2)2,(Mm3)2]

Dalitz plot boundary curve equation

Suppose we wish to plot a curve which describes the depence of s_1 on s_2 (determine s_2 as a function of s_1). This is refered to as a Dalitz plot in which s2 appears on the x-axis (abscissa) and s1 appears on the y-axis (ordinate). Becuase of the "squares" in the invariant mass, we can have two values of s2 for a given value of s1.

since

s2=(p3+p1)μ(p3+p1)μ

We should look seek equations for p1 and p3 in terms of masses(M,m1,m2,m3) and invariants (s,s1).

In order to find such expressions lets consider the problem as viewed by a reference frame in which p3 = -p2. This reference frame is sometimes referred to as the Jackson Frame (JF).

In the JF frame

PJF=pJF1 : because the other two particles cancel the total momentum of the system is carried by m1.
s1≡=(Pp1)μ(Pp1)μ
=[(E,P)(E1,p1)][(E,P)(E1,p1)]
=[(EJF,PJF)(EJF1,pJF1)][(EJF,PJF)(EJF1,pJF1)]
=(EJFEJF1)2:PJF=pJF1
=(M2+(PJF)2m2+(pJF1)2) : Def of Etot
=(M2+(pJF1)2m2+(pJF1)2)

where

EJF = total energy of Mother particle before it decays
EJF1 = total energy of m1 particle

solving for pJF1

(pJF1)2=14s1[s1(Mm1)2][s1(M+m1)2]


The function λ is defined such that

λ(A,B2,C2)=[A(BC)2][A(B+C)2] : notice the "squared" arguments
=A2+B2+C22AB2BC2CA

since this solution above is so common in relativistic kinematics

(pJF1)2=14s1λ(s1,M2,m21)

Doing an analogous calculation for the other form os s_1

s1=(p2+p3)μ(p2+p3)μ

we will find

pJF2=pJF3=14s1λ(s1,m2,m23)

Lets substitute p1 and p3 into the definition for s2 in the Jackson Frame.

s2=(p3+p1)μ(p3+p1)μ
=(pJF3+pJF1)μ(pJF3+pJF1)μ
=[(EJF1,pJF1)+(EJF3,pJF3)][(EJF1,pJF1)+(EJF3,pJF3)]
=m21+m23+2(EJF1EJF3pJF1pJF2cos(θ13))

where

cos(θ13) = angle between pJF1 and pJF3

pJF1 and pJF2 are given above we just need to figure out what EJF1 and EJF2 are


If we assume s1 is fixed (you generate the Dalitz plot boundary by determining the two values of s2 for a given value of s1) then

EJF1=12s1(M2s1m21)
EJF3=12s1(s1+m23m22)

The only remaining unkown is the angle (α) between math>\vec{p}_1^{JF}</math> and pJF3 which we can treat as either π or 0 to determine the min and max values of s_2 for a given value of 2_1.

s2=m21+m23+2(EJF1EJF3pJF1pJF2cos(θ13))
=m21+m23+2(EJF1EJF3±pJF1pJF2)
=m21+m23+2(12s1(M2s1m21)12s1(s1+m23m22)±pJF1pJF2)

Substituting for pJF1 and pJF2

s2=m21+m23+1s1((M2s1m21)(s1+m23m22)±λ(s1,M2,m21)λ(s1,m22,m23))

The above equation for s2 defines a boundary line of the Dalitz plot. The kinematics of the particles in constrained to the interior of this bounding.

Example:D+s Meson decay

Consider the D+s meson which decays 5\% of the time into three particles; K+Kπ+

M=1969MeV for aD+s Meson
m1=494MeV for a K+
m2=140MeV is a π+
m3=494MeV is a K


Dalitz Plot limits are:

X-axis(s2)
Min = (m3+m1)2=(988MeV)21GeV2
Max = (Mm2)2=(1829MeV)2=3.35GeV2
Y-axis (s1)
Min (m2+m3)2(634MeV)2=0.402GeV2
Max= (Mm1)2=(1475MeV)2=2.2GeV2

The image below is from experiment E698

DalitzPlot Dsplus-to-KKPi.jpg

Notice the dark bands which have been labeled ϕ(1020) and ˉK0(892). These dark bands indicate a tendency for the decay to clump into states with specific masses, namely s2=1.04GeV2(m=1020MeV) and s1=0.796GeV2(m=892MeV2). The isobar model suggests that the D+s decays into two particles and then a third particle. In one case the decay is

D+sϕ0+π+

and then the ϕ decays via

ϕ0K++K

Notice that the dark bands are not infinitely thin but have widths as well. The uncertainty principle (ΔEΔt>) suggest that only a particle with an infinite lifetime can have a finite , well defined, mass. The width of these dark bands can be used to determine the lifetimes of the intermediate ϕ and ˉK0 states.

Also notice that there are 2 "clumps" of darks spots in each dark band. The limits for s2 where based on Min/Max values of the cos(θ13) term. This term tells us how aligned or misaligned the momentum vecors of m1 and m3 (the Kaons)are.

Elastic Scattering

File:ForestRelativity ElasticScateringDiagram.jpg


Given the elastic scattering of 2 particles such that the following properties are

Known
m1 = mass of the incident particle #1
m2 = mass of the target particle (at rest) #2
p1 = momentum of the incident particle #1
θ1 = scattering angle of particle #1

You can show that

p1=B±B24AC2A = Final momentum of scattered particle #1
p2=(p1)22p1p1cos(θ1)+p21 = Final momentum of the target particle
sin(θ2)=p1sin(θ1)p2

where

A=(p21+m21+m2)2p21cos2(θ1)
B=2p1cos(θ1)(m21+m2p21+m21)
C=[m41+(m22m21)(p21+m21)m21m22]

In-Elastic Scattering

File:ForestRelativity InelasticScatDiagram.jpg

List of 4-vectors
qμ(ω,q) = momentum transfered from incident particle to target
kμi(Ei,ki) = initial momentum of incident particle
kμf(Ef,kf) = final momentum of incident particle


qμkμikμf = definition of momentum transfer based on conservation of momentum

Momentum Transfer Squared (Q2)

The momentum transfer squared is given by

qμqμ=(EiEf)2(kikf)(kikf)
=(E2i2EiEf+F2f)(|ki|22|ki||kf|cos(θ)+|kf|2)
E2=|k|2+m2
qμqμ=(|ki|2+m2i2EiEf+|kf|2+m2f)|ki|2|kf|2+2|ki||kf|2cos(θ)
=m2i+m2f2EiEf+2|ki||kf|2cos(θ)


In the case of electron scattering

mi=mf
Ei|ki|
Ef|kf|
mE
q2=2|ki||kf|(1cos(θ))
=4|ki||kf|sin2(θ2)4EiEfsin2(θ2)

ifq2<0 spacelike (scattering)

if q2>0 timelike (free particle)

The Momentum Transfer squared for scattering is define as Q2 such that

Q2=q2

Missing Mass (W)

Consider an inelastic scattering process where the particles have the 4-Momentum vectors defined as

(pμe)(Ei,ki) = initial momentum 4-vector of the incident electron
(pμp)(Mp,0) = initial momentum 4-vector of the target proton
(pμe)(Ef,kf) = final momentum 4-vector of the scattered electron
(pμp)(EX,pX) = final momentum 4-vector of the target proton
W2(E2Xp2X) = mass of scattered proton

Conservation of 4-Momentum

(pe)μ+(pp)μ=(pe)μ+(pp)μ

solve for final proton momentum 4-vector and determine the length

(pp)μ(pp)μ=[(pe)μ+(pp)μ(pe)μ][(pe)μ+(pp)μ(pe)μ]


W2(pp)μ(pp)μ
= [(Ei,ki)+(Mp,0)(Ef,kf)][(Eiki)+(Mp0)(Efkf)]
=(E2ik2i)+(E2fk2f)+M2p+2Mp(EiEf)2(EiEfkikf)
= m2e+m2e+M2p+2Mp(EiEf)2(EiEfkikf)
q2=m2i+m2f2EiEf+2|ki||kf|2cos(θ)


substitution

W2=M2p+2Mp(EiEf)+q2=M2p+2Mp(EiEf)Q2

Tranform to Center of Mass

A relativistic transformation from a rest frame where E,p are given to a frame moving with velocity β is given by
(Epxpypz)=[γγβ00γβγ0000100001](Epxpypz)


The velocity of the center of mass frame for the case of a fixed target scattering event (the target is at rest) is given by

vcmc=β=p1p0=p1E1+m2

where

E1,p1 are the energy and momentum for the incident particle
m2 is the mass of the target particle
β=vc=pcE
γ=11β2=Etotmc2=E1+m2m21+m22+2E1m2


Assume an incident electron of 11 GeV moller scatters

pμ1=(E1,p1)=(sqrt112+0.0005112,11ˆi)=(11.0000000118691,11ˆi)
pμ2=(E2,0)=(0.000511,0)
vcm=m1v1m1+m2


Using ROOT functions

TLorentzVector P1,P2;


set the momentum four vector for an 11 GeV electron

P1.SetPxPyPzE(11,0,0,sqrt(11*11+0.000511*0.000511))
P2.SetPxPyPzE(0,0,0,0.000511)

Check that you get the mass of the electron in Units of GeV


The invariant mass of the particle is given by

root [36] P1.Mag()
(const Double_t)5.11000016012715737e-04

The Beta and Gamma of the Particle are Given by


P1.Beta()
(const Double_t)9.99999998920987565e-01
root [42] P1.Gamma()
(const Double_t)2.15264184050203912e+04


The Kinetic Energy ((γ1)moc2) is Given by

root [43] (P1.Gamma()-1)*0.511
(const double)1.09994888049654201e+04
root [46] (P1.Gamma()-1)*P1.Mag()
(const double)1.09994891496458251e+01


assume you moller scatter off a free electron at rest.

Construct the boost vector


TLorentzVector CMS;
CMS=P1+P2;
P1.Boost(-CMS.BoostVector());
P2.Boost(-CMS.BoostVector());
root [138] P1.Px()
(const Double_t)5.30129176950140391e-02
root [139] P2.Px()
(const Double_t)(-5.30129176949399178e-02)

A final state moller scattering event from GEANT4 has

The scattered electron

P3.SetPxPyPzE(0.433025,-0.858867,10999.6,sqrt(0.433025*0.433025+0.858867*0.858867+10999.6*10999.6+ 0.000511*0.000511))

The moller electron

P4.SetPxPyPzE(-0.433025,0.858867,0.905366,sqrt(0.433025*0.433025+0.858867*0.858867+0.905366*0.905366+ 0.000511*0.000511))


CMS=P3+P4


P3.Boost(-CMS.BoostVector());
root [191] P4.Boost(-CMS.BoostVector());
root [192] P3.Px()
(const Double_t)4.33024999999999993e-01
root [193] P4.Px()
(const Double_t)(-4.33024999999999993e-01)
root [194] P3.Py()
(const Double_t)(-8.58867000000000047e-01)
root [195] P4.Py()
(const Double_t)8.58867000000000047e-01
root [196] P3.Pz()
(const Double_t)4.78021531980484724e+01
root [197] P4.Pz()
(const Double_t)(-4.78021531981362244e+01)
root [198] P3.E()
(const Double_t)4.78118292245973180e+01
root [199] P4.E()
(const Double_t)4.78118292247171723e+01


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