# Lorentz Transformations

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

The relationship between the coordinate of an object in frame to the same object described using the coordinates in frame is geven by the Lorentz transformation:

## Notation

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

where

NOTE
It is common in particle physics to define making where is in units of
example
Or in matrix form the tranformation looks like
Note
Einstein's summation convention drops the symbols and assumes it to exist whenever there is a repeated subscript and uperscript
ie;
in the example above the symbol is repeated thereby indicating a summation over .

## Momentum 4-vector

The momentum 4 -vector is denoted as:

Note
There is another convention used for 4-vector notation by Perkins and Koller which goes like this

## Trig Method

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

The Matrix form of the tranformation looks like
Or the reverse transformation
Notice that you just needed to change the signs for the inverse matrix

# Proper Time and Length

## Proper Time

Proper Time
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) =

Note
since you expect the Proper time interval to be the smallest

## Proper Length

Proper Length
The length of an object in the object's rest frame.

# Transformation Examples

## Decay of a Particle at Rest to 2 Bodies

Consider the decay of the meson of mass at rest into two pions ( and ) of mass and respectively.

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

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

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

or

Let

Conservation of Energy

solving for p

is required to avoid the unphysical condition that the momentum of the particles after a decay would be an imaginary number

Using

Combine this with the conservation of energy equation above:

Square both sides of the above equation

Similarly

Note
The daughter particles (pions) from the decay of the Mother particle 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)

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: and ) are equal and opposite by conservation of momentum.

The center of momentum frame is moving such that

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

where

= Kinetic Energy (not total) of particle 1 in the center of momentum (CM) reference frame
= momentum of particle 1 along the direction of the mother particle in the CM frame
= the component of particle 1's momentum perpendicular to Mother particle's momentum
You can now use the results for and 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 and .
It can be shown that the lab angle for daughter particle 1 () is given by

where

for daughter particle 1 in CM frame.

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

Solve the conservation of momentum equation for

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

Note
is the momentum of the two daughter particles in the CM frame which was derived when the Mother particle was at rest. is the momentum of the Mother particle.

In order for a real solution

If then can be any angle and the "-" sign possibility in "" is rejected to avoided negative values for when > .

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

The "" is kept because for each 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 decays into 3 daughter particles of masses , , and . The 4-mometum conservation is written as

The following invariants are defined

The invariants , and 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

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

There are limits to the values of the invariant masses -

In the Center of Momentum system we have the

because

we expect

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

: In the CM frame

so the limits of s_1 are

### Invariant mass Dalitz plot Limits

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

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

### 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 appears on the x-axis (abscissa) and appears on the y-axis (ordinate). Becuase of the "squares" in the invariant mass, we can have two values of for a given value of .

since

We should look seek equations for and in terms of masses and invariants .

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

In the JF frame

: because the other two particles cancel the total momentum of the system is carried by .
= : Def of
=

where

= total energy of Mother particle before it decays
= total energy of particle

solving for

The function is defined such that

: notice the "squared" arguments

since this solution above is so common in relativistic kinematics

Doing an analogous calculation for the other form os s_1

we will find

Lets substitute and into the definition for in the Jackson Frame.

where

= angle between and

and are given above we just need to figure out what and are

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

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

Substituting for and

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

### Example: Meson decay

Consider the meson which decays 5\% of the time into three particles;

for a Meson
for a
is a
is a

Dalitz Plot limits are:

X-axis
Min =
Max =
Y-axis
Min
Max=

The image below is from experiment E698

Notice the dark bands which have been labeled and . These dark bands indicate a tendency for the decay to clump into states with specific masses, namely and . The isobar model suggests that the decays into two particles and then a third particle. In one case the decay is

and then the decays via

Notice that the dark bands are not infinitely thin but have widths as well. The uncertainty principle () 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 and states.

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

## Elastic Scattering

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

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

You can show that

= Final momentum of scattered particle #1
= Final momentum of the target particle

where

## In-Elastic Scattering

List of 4-vectors
= momentum transfered from incident particle to target
= initial momentum of incident particle
= final momentum of incident particle

= definition of momentum transfer based on conservation of momentum

### Momentum Transfer Squared

The momentum transfer squared is given by

In the case of electron scattering

if spacelike (scattering)

if timelike (free particle)

The Momentum Transfer squared for scattering is define as such that

### Missing Mass

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

= initial momentum 4-vector of the incident electron
= initial momentum 4-vector of the target proton
= final momentum 4-vector of the scattered electron
= final momentum 4-vector of the target proton
= mass of scattered proton

Conservation of 4-Momentum

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

=
=

substitution

## Tranform to Center of Mass

A relativistic transformation from a rest frame where are given to a frame moving with velocity is given by

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

where

are the energy and momentum for the incident particle
is the mass of the target particle

Assume an incident electron of 11 GeV moller scatters

### 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 () 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