Forest Relativity Notes

Lorentz Transformations

The picture below represents the relative orientation of two different coordinate systems [math]\displaystyle{ (S, S^{\prime}) }[/math] . [math]\displaystyle{ S }[/math] is at rest (Lab Frame) and [math]\displaystyle{ S^{\prime} }[/math] is moving at a velocity v to the right with respect to frame [math]\displaystyle{ S }[/math].

The relationship between the coordinate[math]\displaystyle{ (x,y,z,ct) }[/math] of an object in frame [math]\displaystyle{ S }[/math] to the same object described using the coordinates [math]\displaystyle{ (x^{\prime},y^{\prime},z^{\prime},ct^{\prime}) }[/math] in frame [math]\displaystyle{ S^{\prime} }[/math] is geven by the Lorentz transformation:

Notation

The relativistic transformation used to transform the coordinates of an object observed in the rest frame [math]\displaystyle{ S }[/math] to a moving reference frame[math]\displaystyle{ S^{\prime} }[/math] is given by:

[math]\displaystyle{ x^{\mu^{\prime}} = \sum_{\nu=0}^3 \Lambda_{\nu}^{\mu} x^{\nu} }[/math]

where

[math]\displaystyle{ x^0 \equiv ct }[/math]

[math]\displaystyle{ x^1 \equiv x }[/math]

[math]\displaystyle{ x^2\equiv y }[/math]

[math]\displaystyle{ x^3\equiv z }[/math]

[math]\displaystyle{ \Lambda = \left [ \begin{matrix} \gamma & -\gamma \beta & 0 & 0 \\ -\gamma \beta & \gamma &0 &0 \\ 0 &0 &1 &0 \\ 0 &0 &0 &1\end{matrix} \right ] }[/math]

[math]\displaystyle{ \beta = \frac{v}{c} = \frac{pc}{E} }[/math]

[math]\displaystyle{ \gamma = \frac{1}{\sqrt{1 -\beta^2}} = \frac{E_{tot}}{mc^2} }[/math]

NOTE

It is common in particle physics to define [math]\displaystyle{ c \equiv 1 }[/math] making [math]\displaystyle{ \gamma = \frac{E}{m} }[/math] where [math]\displaystyle{ m }[/math] is in units of [math]\displaystyle{ \frac{\mbox{MeV}}{\mbox{c}^2} }[/math]

example

[math]\displaystyle{ x^{0^{\prime}} = \sum_{\nu=0}^2 \Lambda_{\nu}^0 x^{\nu} = \Lambda_0^0 x^0 + \Lambda_1^0 x^1 \Lambda_2^0 x^2 + \Lambda_3^0 x^2 }[/math]

[math]\displaystyle{ ct^{\prime}= \gamma x^0 - \gamma \beta x^1 + 0 x^2 + 0 x^3 = \gamma ct - \gamma \beta x = \gamma(ct -\beta x) }[/math]

Or in matrix form the tranformation looks like

[math]\displaystyle{ \left ( \begin{matrix} ct^{\prime} \\ x^{\prime} \\ y^{\prime} \\ z^{\prime} \end{matrix} \right )= \left [ \begin{matrix} \gamma & -\gamma \beta & 0 & 0 \\ -\gamma \beta & \gamma &0 &0 \\ 0 &0 &1 &0 \\ 0 &0 &0 &1\end{matrix} \right ] \left ( \begin{matrix} ct \\ x \\ y \\ z \end{matrix} \right ) }[/math]

Note

Einstein's summation convention drops the [math]\displaystyle{ \sum }[/math] symbols and assumes it to exist whenever there is a repeated subscript and uperscript

ie; [math]\displaystyle{ x^{\mu^{\prime}} = \Lambda_{\nu}^{\mu} x^{\nu} }[/math]

in the example above the[math]\displaystyle{ \nu }[/math] symbol is repeated thereby indicating a summation over [math]\displaystyle{ \nu }[/math].

Momentum 4-vector

The momentum 4 -vector is denoted as:

[math]\displaystyle{ p^{\mu} \equiv (\frac{E}{c} , \vec{p}) }[/math]

[math]\displaystyle{ p_{\mu} \equiv (\frac{E}{c} , -\vec{p}) }[/math]

[math]\displaystyle{ p_{\mu}p^{\mu} = \frac{E^2}{c^2} - p^2 \equiv E^2 - p^2 = m^2 }[/math]

Note

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

[math]\displaystyle{ p^{\mu} \equiv (\vec{p},iE) }[/math]

[math]\displaystyle{ p_{\mu} \equiv (\vec{p},iE) }[/math]

Trig Method

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

[math]\displaystyle{ \sin (\alpha) \equiv \beta \equiv \frac{v}{c} }[/math]

[math]\displaystyle{ \cos(\alpha) \equiv \frac{1}{\gamma} \equiv \sqrt{1 - \beta^2} }[/math]

The Matrix form of the tranformation looks like

[math]\displaystyle{ \left ( \begin{matrix} ct^{\prime} \\ x^{\prime} \\ y^{\prime} \\ z^{\prime} \end{matrix} \right )= \left [ \begin{matrix} \sec(\alpha) & -\tan(\alpha) & 0 & 0 \\ -\tan(\alpha) & \sec(\alpha) &0 &0 \\ 0 &0 &1 &0 \\ 0 &0 &0 &1\end{matrix} \right ] \left ( \begin{matrix} ct \\ x \\ y \\ z \end{matrix} \right ) }[/math]

Or the reverse transformation

[math]\displaystyle{ \left ( \begin{matrix} ct \\ x \\ y \\ z \end{matrix}\right )= \left [ \begin{matrix} \sec(\alpha) & \tan(\alpha) & 0 & 0 \\ \tan(\alpha) & \sec(\alpha) &0 &0 \\ 0 &0 &1 &0 \\ 0 &0 &0 &1\end{matrix} \right ] \left ( \begin{matrix} ct^{\prime} \\ x^{\prime} \\ y^{\prime} \\ z^{\prime} \end{matrix} \right ) }[/math]

Notice that you just needed to change the signs for the inverse matrix [math]\displaystyle{ \Lambda^{-1} }[/math]

Proper Time and Length

Proper Time

Proper Time [math]\displaystyle{ \Tau }[/math]

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) = [math]\displaystyle{ \gamma \Tau }[/math]

Note

since [math]\displaystyle{ \gamma > 1 }[/math] you expect the Proper time interval to be the smallest

Proper Length

Proper Length[math]\displaystyle{ (c\Tau) }[/math]

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 [math]\displaystyle{ \rho_0 }[/math] meson of mass [math]\displaystyle{ M }[/math] at rest into two pions ([math]\displaystyle{ \pi^+ }[/math] and [math]\displaystyle{ \pi^- }[/math] ) of mass [math]\displaystyle{ m_1 }[/math] and [math]\displaystyle{ m_2 }[/math] respectively.

File:NeutralRhoMesonDecayDiagram.jpg

The diagram above shows a [math]\displaystyle{ \rho_0 }[/math] meson at rest in the lab which then decays into two pions of momentum [math]\displaystyle{ p_1 }[/math] and [math]\displaystyle{ p_2 }[/math] in the center of momentum frame of the [math]\displaystyle{ \rho_0 }[/math] meson.

Our goal is to determine the momentum and energy of each pion ([math]\displaystyle{ E_1 }[/math] & [math]\displaystyle{ E_2 }[/math]) resulting from the decay of the Mother particle.

Conservation of 4-Momentum implies that if [math]\displaystyle{ P^{\mu} }[/math] represents the total momentum of the system before the decay then

[math]\displaystyle{ P^{\mu} = (E,0) =(M,0) = \left ( p_1 \right )^{\mu} + \left ( p_2 \right)^{\mu} }[/math]

[math]\displaystyle{ \Rightarrow 0 = \vec{p}_1 + \vec{p}_2 }[/math]

or

[math]\displaystyle{ \vec{p}_1 = - \vec{p}_2 }[/math]

Let

[math]\displaystyle{ p \equiv |\vec{p}_1 | = |\vec{p}_2 | }[/math]

Conservation of Energy

[math]\displaystyle{ \Rightarrow E_{tot} = M = E_1 + E_2 = \sqrt{m_1^2 + p^2} + \sqrt{m_2^2 + p^2} }[/math]

solving for p

[math]\displaystyle{ \Rightarrow p = \frac{1}{2M} \sqrt{[M^2 - (m_1-m_2)^2][M^2-(m_1+m_2)^2]} }[/math]

[math]\displaystyle{ \Rightarrow M \ge m_1 + m_2 }[/math] is required to avoid the unphysical condition that the momentum of the particles after a decay would be an imaginary number

Using

[math]\displaystyle{ p \equiv |\vec{p}_1 | = |\vec{p}_2 | }[/math]

[math]\displaystyle{ E_1^2 - m_1^2 = E_2^2 - m_2^2 }[/math]

[math]\displaystyle{ \Rightarrow E_2 = \sqrt{E_1^2 - m_1^2 + m_2^2} }[/math]

Combine this with the conservation of energy equation above:

[math]\displaystyle{ E_1 + E_2 = E_1 + \sqrt{E_1^2 - m_1^2 + m_2^2} = M }[/math]

[math]\displaystyle{ \Rightarrow E_1 - M = \sqrt{E_1^2 - m_1^2 + m_2^2} }[/math]

Square both sides of the above equation

[math]\displaystyle{ E_1^2 -2ME_1 + M^2 = E_1^2- m_1^2 + m_2^2 }[/math]

[math]\displaystyle{ \Rightarrow E_1 = \frac{M^2+m_1^2-m_2^2}{2M} }[/math]

Similarly

[math]\displaystyle{ E_2 = \frac{M^2+m_2^2-m^2_1}{2M} }[/math]

Note

[math]\displaystyle{ \vec{p}_1 = -\vec{p}_2 }[/math]

[math]\displaystyle{ \Rightarrow }[/math] The daughter particles (pions) from the decay of the Mother particle [math]\displaystyle{ (\rho) }[/math] 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)

[math]\displaystyle{ P^{\mu} = (E,\vec{p}_{tot}) =(M,\vec{p}_{tot}) = \left ( p_1 \right )^{\mu} + \left ( p_2 \right)^{\mu} }[/math]

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:[math]\displaystyle{ \vec{p}_{1,\perp} }[/math] and [math]\displaystyle{ \vec{p}_{2,\perp} }[/math]) are equal and opposite by conservation of momentum.

[math]\displaystyle{ \vec{p}_{1,\perp}\equiv -\vec{p}_{2,\perp} }[/math]

The center of momentum frame is moving such that

[math]\displaystyle{ \beta_{CM} = \frac{p_{tot}}{M} }[/math]

[math]\displaystyle{ \gamma_{CM} = \frac{E_{tot}}{M} }[/math]

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

[math]\displaystyle{ E_1 = \gamma_{cm}(E_1^{CM} + \beta_{cm}p_{1,z}^{CM}) }[/math]

[math]\displaystyle{ p_{1,z} = \gamma_{CM}(p_{1,z}^{CM} + \beta_{cm} E_1^{CM}) }[/math]

[math]\displaystyle{ p_{1,\perp} = p_{\perp}^{CM} }[/math]

where

[math]\displaystyle{ E_1^{CM} }[/math] = Kinetic Energy (not total) of particle 1 in the center of momentum (CM) reference frame

[math]\displaystyle{ p_{1,z}^{CM} }[/math] = momentum of particle 1 along the direction of the mother particle in the CM frame

[math]\displaystyle{ p_{1,\perp} }[/math] = the component of particle 1's momentum perpendicular to Mother particle's momentum

You can now use the results for [math]\displaystyle{ E_1 }[/math] and [math]\displaystyle{ p_1=p }[/math] 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 [math]\displaystyle{ p_z }[/math] and [math]\displaystyle{ p_{\perp} }[/math].

It can be shown that the lab angle for daughter particle 1 ([math]\displaystyle{ \theta_1 }[/math]) is given by

[math]\displaystyle{ \tan(\theta_1) = \frac{\sin(\theta_1^{CM})}{\gamma_{CM}\left (\frac{\beta_{CM}}{\beta_1^{CM}} + \cos(\theta_1^{CM}) \right )} }[/math]

where

[math]\displaystyle{ \beta_1^{CM} = \frac{p_1}{E} = \beta }[/math] for daughter particle 1 in CM frame.

One could also find [math]\displaystyle{ \vec{p}_1 }[/math] without using the Lorentz transformation. Just use conservation of Energy and Momentum:

[math]\displaystyle{ E_{tot} = E_1^{tot} + E_2^{tot} = \sqrt{m_1^2 + p_1^2} + \sqrt{m_2^2 + p_2^2} }[/math]

[math]\displaystyle{ \vec{p_{tot}} = \vec{p}_1 + \vec{p}_2 }[/math]

Solve the conservation of momentum equation for [math]\displaystyle{ p_2^2 }[/math]

[math]\displaystyle{ p_2^2 = (\vec{p_{tot}}- \vec{p}_1)^2 }[/math]

and substitute the above for [math]\displaystyle{ p_2^2 }[/math] in the Conservation of Energy equation above. The dot product gives you the angle between the daughter momentum and the Mother momentum ([math]\displaystyle{ \theta_1 }[/math]) as a variable. After a lot of algebra you can show that

[math]\displaystyle{ p_1 = \frac{\left ( M^2 + m_1^2 -m_2^2 \right ) p_{tot} \cos(\theta_1) \pm 2E\sqrt{M^2p^2 - m_1^2p^2_{tot} \sin^2(\theta_1)}}{2 \left( M^2 + p^2 \sin^2(\theta_1)\right )} }[/math]

Note

[math]\displaystyle{ p }[/math] is the momentum of the two daughter particles in the CM frame which was derived when the Mother particle was at rest. [math]\displaystyle{ p_{tot} }[/math] is the momentum of the Mother particle.

In order for a real solution

[math]\displaystyle{ M^2p^2 - m_1^2p^2_{tot} \sin^2(\theta_1) \ge 0 }[/math]

[math]\displaystyle{ \Rightarrow \frac{M p}{m_1 p_{tot}} \ge \sin(\theta_1) }[/math]

If [math]\displaystyle{ \frac{M p}{m_1 p_{tot}} > 1 }[/math] then [math]\displaystyle{ \theta_1 }[/math] can be any angle and the "-" sign possibility in "[math]\displaystyle{ \pm }[/math]" is rejected to avoided negative values for [math]\displaystyle{ p_1 }[/math] when [math]\displaystyle{ \theta_1 }[/math] > [math]\displaystyle{ \frac{\pi}{2} }[/math].

If [math]\displaystyle{ \frac{M p}{m_1 p_{tot}} < 1 }[/math] then the maximum emmission angle for daughter 1 is given by

[math]\displaystyle{ \sin(\theta_1^{max}) = \frac{M p}{m_1 p_{tot}} }[/math]

The "[math]\displaystyle{ \pm }[/math]" is kept because for each [math]\displaystyle{ \theta_1 < \theta_1^{max} }[/math] 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 [math]\displaystyle{ M }[/math] decays into 3 daughter particles of masses [math]\displaystyle{ m_1 }[/math], [math]\displaystyle{ m_2 }[/math], and [math]\displaystyle{ m_3 }[/math]. The 4-mometum conservation is written as

[math]\displaystyle{ P^{\mu} = \left ( p_1 \right )^{\mu} +\left ( p_2 \right )^{\mu} +\left ( p_3 \right )^{\mu} }[/math]

The following invariants are defined

[math]\displaystyle{ s = P_{\mu}P^{\mu} = M^2 }[/math]

[math]\displaystyle{ s_1 = \left (P -p_1 \right)_{\mu}\left (P -p_1 \right)^{\mu}=\left (p_2 + p_3 \right)_{\mu}\left (p_2 + p_3 \right)^{\mu} }[/math]

[math]\displaystyle{ s_2 = \left (P -p_2 \right)_{\mu}\left (P -p_2 \right)^{\mu}=\left (p_3 + p_1 \right)_{\mu}\left (p_3 + p_1 \right)^{\mu} }[/math]

[math]\displaystyle{ s_3 = \left (P -p_3 \right)_{\mu}\left (P -p_3 \right)^{\mu}=\left (p_1 + p_2 \right)_{\mu}\left (p_1 + p_2 \right)^{\mu} }[/math]

The invariants [math]\displaystyle{ s_1 }[/math], [math]\displaystyle{ s_2 }[/math] and [math]\displaystyle{ s_3 }[/math] 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

[math]\displaystyle{ s_1 + s_2 + s_3 = (P^2 -2Pp_1 +p_1^2) +(P^2 -2Pp_2 +p_2^2) + (P^2 -2Pp_3 +p_3^2) }[/math]

[math]\displaystyle{ =3P^2 -2P(p_1+p_2+p_3) +p_1^2+p_2^2+p_3^2 }[/math]

[math]\displaystyle{ =3P^2 -2P^2 +p_1^2+p_2^2+p_3^2 }[/math]

[math]\displaystyle{ =P^2 +p_1^2+p_2^2+p_3^2 }[/math]

[math]\displaystyle{ =M^2 + m_1^2 + m_2^2 +m_3^2 }[/math]

Also Note

[math]\displaystyle{ \sqrt{s_1} }[/math] is the invariant mass of a subsystem defined by treating daughter particles 2 and 3 as one object. similar interpretations for [math]\displaystyle{ \sqrt{s_2} }[/math] and [math]\displaystyle{ \sqrt{s_3} }[/math].

There are limits to the values of the invariant masses [math]\displaystyle{ s_1 }[/math] - [math]\displaystyle{ s_3 }[/math]

In the Center of Momentum system we have the

[math]\displaystyle{ s_1 = M^2 +m_1^2 - 2ME_1 }[/math]

because

[math]\displaystyle{ E_1 = m_1^2 + p_1^2 }[/math] we expect [math]\displaystyle{ E_1 \ge m_1 }[/math]

This lead to

[math]\displaystyle{ \left . s_1 \right |_{max} = M^2 +m_1^2 - 2M(m_1) = (M-m_1)^2 }[/math]

TO find the minimum value of [math]\displaystyle{ s_1 }[/math] we evaluate [math]\displaystyle{ s_1 }[/math] in the rest frame of the [math]\displaystyle{ m_2 }[/math], [math]\displaystyle{ m_3 }[/math] subsystem.

[math]\displaystyle{ s_1 = \left (p_2 + p_3 \right)_{\mu}\left (p_2 + p_3 \right)^{\mu}= (E_2^{cm2-3} +E_3^{cm2-3})^2 > (m_2 + m_3)^2 }[/math] : In the CM frame [math]\displaystyle{ E = \sqrt{p^2 +m^2} = m }[/math]

so the limits of s_1 are

[math]\displaystyle{ (m_2 + m_3)^2 \le s_1 \le (M-m_1)^2 }[/math]

Invariant mass Dalitz plot Limits

Analagously you can show the the Min - Max limits for the [math]\displaystyle{ s_1 }[/math], [math]\displaystyle{ s_2 }[/math], and [math]\displaystyle{ s_3 }[/math] invariant masses are given by.

[math]\displaystyle{ (m_3 + m_1)^2 \le s_2 \le (M-m_2)^2 }[/math]

[math]\displaystyle{ (m_1 + m_2)^2 \le s_3 \le (M-m_3)^2 }[/math]

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

[math]\displaystyle{ s_1 \in \left [ (m_2 + m_3)^2 , (M-m_1)^2 \right ] }[/math]

[math]\displaystyle{ s_2 \in \left [ (m_3 + m_1)^2 , (M-m_2)^2 \right ] }[/math]

[math]\displaystyle{ s_2 \in \left [ (m_1 + m_2)^2, (M-m_3)^2 \right ] }[/math]

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 [math]\displaystyle{ s_2 }[/math] appears on the x-axis (abscissa) and [math]\displaystyle{ s_1 }[/math] appears on the y-axis (ordinate). Becuase of the "squares" in the invariant mass, we can have two values of [math]\displaystyle{ s_2 }[/math] for a given value of [math]\displaystyle{ s_1 }[/math].

since

[math]\displaystyle{ s_2 = \left (p_3 + p_1 \right)_{\mu}\left (p_3 + p_1 \right)^{\mu} }[/math]

We should look seek equations for [math]\displaystyle{ p_1 }[/math] and [math]\displaystyle{ p_3 }[/math] in terms of masses[math]\displaystyle{ (M,m_1,m_2,m_3) }[/math] and invariants [math]\displaystyle{ (s,s_1) }[/math].

In order to find such expressions lets consider the problem as viewed by a reference frame in which [math]\displaystyle{ \vec{p}_3 }[/math] = -[math]\displaystyle{ \vec{p}_2 }[/math]. This reference frame is sometimes referred to as the Jackson Frame (JF).

In the JF frame

[math]\displaystyle{ \vec{P}^{JF} = \vec{p}_1^{JF} }[/math] : because the other two particles cancel the total momentum of the system is carried by [math]\displaystyle{ m_1 }[/math].

[math]\displaystyle{ s_1 \equiv = \left (P -p_1 \right)_{\mu}\left (P -p_1 \right)^{\mu} }[/math]

[math]\displaystyle{ =\left [ (E,-\vec{P}) - (E_1,-\vec{p_1})\right ]\left [ (E,\vec{P}) - (E_1,\vec{p}_1)\right ] }[/math]

[math]\displaystyle{ =\left [ (E^{JF},-\vec{P}^{JF}) - (E_1^{JF},-\vec{p}_1^{JF})\right ]\left [ (E^{JF},\vec{P}^{JF}) - (E_1^{JF},\vec{p}_1^{JF})\right ] }[/math]

[math]\displaystyle{ = (E^{JF}-E_1^{JF})^2 : \vec{P}^{JF}=\vec{p}_1^{JF} }[/math]

=[math]\displaystyle{ \left( \sqrt{M^2+(P^{JF})^2} - \sqrt{m^2+(p_1^{JF})^2}\right) }[/math] : Def of [math]\displaystyle{ E_{tot} }[/math]

=[math]\displaystyle{ \left( \sqrt{M^2+(p_1^{JF})^2} - \sqrt{m^2+(p_1^{JF})^2}\right) }[/math]

where

[math]\displaystyle{ E^{JF} }[/math] = total energy of Mother particle before it decays

[math]\displaystyle{ E_1^{JF} }[/math] = total energy of [math]\displaystyle{ m_1 }[/math] particle

solving for [math]\displaystyle{ p_1^{JF} }[/math]

[math]\displaystyle{ \left( p_1^{JF}\right )^2 = \frac{1}{4s_1} \left [ s_1 -(M-m_1)^2\right ]\left [ s_1 -(M+m_1)^2\right ] }[/math]

The function [math]\displaystyle{ \lambda }[/math] is defined such that

[math]\displaystyle{ \lambda(A,B^2,C^2) = \left [ A -(B-C)^2\right ]\left [ A -(B+C)^2\right ] }[/math] : notice the "squared" arguments

[math]\displaystyle{ =A^2 + B^2 + C^2 -2AB -2BC -2CA }[/math]

since this solution above is so common in relativistic kinematics

[math]\displaystyle{ \Rightarrow \left( p_1^{JF}\right )^2 = \frac{1}{4s_1}\lambda(s_1,M^2,m_1^2) }[/math]

Doing an analogous calculation for the other form os s_1

[math]\displaystyle{ s_1 =\left (p_2 + p_3 \right)_{\mu}\left (p_2 + p_3 \right)^{\mu} }[/math]

we will find

[math]\displaystyle{ p_2^{JF} = p_3^{JF} = \frac{1}{4s_1}\lambda(s_1,m^2,m_3^2) }[/math]

Lets substitute [math]\displaystyle{ p_1 }[/math] and [math]\displaystyle{ p_3 }[/math] into the definition for [math]\displaystyle{ s_2 }[/math] in the Jackson Frame.

[math]\displaystyle{ s_2 = \left (p_3 + p_1 \right)_{\mu}\left (p_3 + p_1 \right)^{\mu} }[/math]

[math]\displaystyle{ =\left (p_3^{JF} + p_1^{JF} \right)_{\mu}\left (p_3^{JF} + p_1^{JF} \right)^{\mu} }[/math]

[math]\displaystyle{ =\left [ (E_1^{JF},-\vec{p}_1^{JF}) + (E_3^{JF},-\vec{p}_3^{JF})\right]\left [ (E_1^{JF},\vec{p}_1^{JF}) + (E_3^{JF},\vec{p}_3^{JF})\right] }[/math]

[math]\displaystyle{ = m_1^2 +m_3^2 + 2\left ( E_1^{JF} E_3^{JF} - p_1^{JF} p_2^{JF} \cos(\theta_{13})\right ) }[/math]

where

[math]\displaystyle{ \cos(\theta_{13}) }[/math] = angle between [math]\displaystyle{ \vec{p}_1^{JF} }[/math] and [math]\displaystyle{ \vec{p}_3^{JF} }[/math]

[math]\displaystyle{ p_1^{JF} }[/math] and [math]\displaystyle{ p_2^{JF} }[/math] are given above we just need to figure out what [math]\displaystyle{ E_1^{JF} }[/math] and [math]\displaystyle{ E_2^{JF} }[/math] are

If we assume [math]\displaystyle{ s_1 }[/math] is fixed (you generate the Dalitz plot boundary by determining the two values of [math]\displaystyle{ s_2 }[/math] for a given value of [math]\displaystyle{ s_1 }[/math]) then

[math]\displaystyle{ E_1^{JF} = \frac{1}{2\sqrt{s_1}} (M^2 - s_1 -m_1^2) }[/math]

[math]\displaystyle{ E_3^{JF} = \frac{1}{2\sqrt{s_1}} (s_1 + m_3^2 -m_2^2) }[/math]

The only remaining unkown is the angle [math]\displaystyle{ (\alpha) }[/math] between math>\vec{p}_1^{JF}</math> and [math]\displaystyle{ \vec{p}_3^{JF} }[/math] which we can treat as either [math]\displaystyle{ \pi }[/math] or 0 to determine the min and max values of s_2 for a given value of 2_1.

[math]\displaystyle{ s_2= m_1^2 +m_3^2 + 2\left ( E_1^{JF} E_3^{JF} - p_1^{JF} p_2^{JF} \cos(\theta_{13})\right ) }[/math]

[math]\displaystyle{ = m_1^2 +m_3^2 + 2\left ( E_1^{JF} E_3^{JF} \pm p_1^{JF} p_2^{JF} \right ) }[/math]

[math]\displaystyle{ = m_1^2 +m_3^2 + 2\left ( \frac{1}{2\sqrt{s_1}} (M^2 - s_1 -m_1^2) \frac{1}{2\sqrt{s_1}} (s_1 + m_3^2 -m_2^2) \pm p_1^{JF} p_2^{JF} \right ) }[/math]

Substituting for [math]\displaystyle{ p_1^{JF} }[/math] and [math]\displaystyle{ p_2^{JF} }[/math]

[math]\displaystyle{ s_2=m_1^2 +m_3^2 + \frac{1}{s_1} \left ((M^2 - s_1 -m_1^2)(s_1 + m_3^2 -m_2^2) \pm \sqrt{\lambda(s_1,M^2,m_1^2) \lambda(s_1,m_2^2,m_3^2)}\right ) }[/math]

The above equation for [math]\displaystyle{ s_2 }[/math] defines a boundary line of the Dalitz plot. The kinematics of the particles in constrained to the interior of this bounding.

Example:[math]\displaystyle{ D_s^+ }[/math] Meson decay

Consider the [math]\displaystyle{ D_s^+ }[/math] meson which decays [math]\displaystyle{ \sim }[/math] 5\% of the time into three particles; [math]\displaystyle{ K^+K^-\pi^+ }[/math]

[math]\displaystyle{ M =1969 MeV }[/math] for a[math]\displaystyle{ D_s^+ }[/math] Meson

[math]\displaystyle{ m_1 = 494 MeV }[/math] for a [math]\displaystyle{ K^+ }[/math]

[math]\displaystyle{ m_2 = 140 MeV }[/math] is a [math]\displaystyle{ \pi^+ }[/math]

[math]\displaystyle{ m_3=494 MeV }[/math] is a [math]\displaystyle{ K^- }[/math]

Dalitz Plot limits are:

X-axis[math]\displaystyle{ (s_2) }[/math]

Min = [math]\displaystyle{ (m_3 + m_1)^2 = (988 MeV)^2 \; \sim 1 GeV^2 }[/math]

Max = [math]\displaystyle{ (M-m_2)^2 = (1829 MeV)^2 = 3.35 GeV^2 }[/math]

Y-axis [math]\displaystyle{ (s_1) }[/math]

Min [math]\displaystyle{ (m_2+m_3)^2 \; \sim (634 MeV)^2 = 0.402 GeV^2 }[/math]

Max= [math]\displaystyle{ (M-m_1)^2 = (1475 MeV)^2 = 2.2 GeV^2 }[/math]

The image below is from experiment E698

Notice the dark bands which have been labeled [math]\displaystyle{ \phi(1020) }[/math] and [math]\displaystyle{ \bar{K}^{*0}(892) }[/math]. These dark bands indicate a tendency for the decay to clump into states with specific masses, namely [math]\displaystyle{ s_2=1.04 GeV^2 (\Rightarrow m=1020 MeV) }[/math] and [math]\displaystyle{ s_1=0.796 GeV^2 (\Rightarrow m= 892 MeV^2) }[/math]. The isobar model suggests that the [math]\displaystyle{ D_s^+ }[/math] decays into two particles and then a third particle. In one case the decay is

[math]\displaystyle{ D_s^+ \rightarrow \phi^0 + \pi^+ }[/math]

and then the [math]\displaystyle{ \phi }[/math] decays via

[math]\displaystyle{ \phi^0 \rightarrow K^+ + K^- }[/math]

Notice that the dark bands are not infinitely thin but have widths as well. The uncertainty principle ([math]\displaystyle{ \Delta E \Delta t > \hbar }[/math]) 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 [math]\displaystyle{ \phi }[/math] and [math]\displaystyle{ \bar{K}^{*0} }[/math] states.

Also notice that there are 2 "clumps" of darks spots in each dark band. The limits for [math]\displaystyle{ s_2 }[/math] where based on Min/Max values of the [math]\displaystyle{ \cos(\theta_{13}) }[/math] term. This term tells us how aligned or misaligned the momentum vecors of [math]\displaystyle{ m_1 }[/math] and [math]\displaystyle{ m_3 }[/math] (the Kaons)are.

Elastic Scattering

File:ForestRelativity ElasticScateringDiagram.jpg

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

Known

[math]\displaystyle{ m_1 }[/math] = mass of the incident particle #1

[math]\displaystyle{ m_2 }[/math] = mass of the target particle (at rest) #2

[math]\displaystyle{ p_1 }[/math] = momentum of the incident particle #1

[math]\displaystyle{ \theta_1 }[/math] = scattering angle of particle #1

You can show that

[math]\displaystyle{ p_1^{\prime} = \frac{-B \pm \sqrt{B^2 -4AC}}{2A} }[/math] = Final momentum of scattered particle #1

[math]\displaystyle{ p_2^{\prime} = \left ( p_1^{\prime} \right )^2 -2p_1p_1^{\prime}\cos(\theta_1) + p_1^2 }[/math] = Final momentum of the target particle

[math]\displaystyle{ \sin (\theta_2) = - \frac{p_1^{\prime} \sin(\theta_1)}{p_2^{\prime}} }[/math]

where

[math]\displaystyle{ A = \left ( \sqrt{p_1^2 +m_1^2} +m_2 \right )^2 -p_1^2 \cos^2(\theta_1) }[/math]

[math]\displaystyle{ B = -2p_1 \cos(\theta_1) \left ( m_1^2 + m_2 \sqrt{p_1^2 + m_1^2}\right ) }[/math]

[math]\displaystyle{ C = - \left [ m_1^4 + (m_2^2 -m_1^2)(p_1^2 + m_1^2) -m_1^2m_2^2\right ] }[/math]

In-Elastic Scattering

File:ForestRelativity InelasticScatDiagram.jpg

List of 4-vectors

[math]\displaystyle{ q^{\mu} \equiv ( \omega , \vec{q} ) }[/math] = momentum transfered from incident particle to target

[math]\displaystyle{ k_i^{\mu} \equiv (E_i, \vec{k}_i ) }[/math] = initial momentum of incident particle

[math]\displaystyle{ k_f^{\mu} \equiv (E_f, \vec{k}_f ) }[/math] = final momentum of incident particle

[math]\displaystyle{ q^{\mu} \equiv k_i^{\mu} - k_f^{\mu} }[/math] = definition of momentum transfer based on conservation of momentum

Momentum Transfer Squared [math]\displaystyle{ (Q^2) }[/math]

The momentum transfer squared is given by

[math]\displaystyle{ q_{\mu}q^{\mu} = (E_i - E_f)^2 -(\vec{k}_i - \vec{k}_f ) (\vec{k}_i - \vec{k}_f ) }[/math]

[math]\displaystyle{ =(E_i^2 -2E_iE_f +F_f^2) - ( |\vec{k}_i |^2 - 2 |\vec{k}_i| |\vec{k}_f| \cos(\theta) + | k_f |^2) }[/math]

[math]\displaystyle{ E^2 = | \vec{k} |^2 +m^2 }[/math]

[math]\displaystyle{ \Rightarrow q_{\mu}q^{\mu} = \left ( | \vec{k}_i |^2 +m_i^2 - 2 E_i E_f +| \vec{k}_f |^2 +m_f^2\right ) - | \vec{k}_i |^2 - | \vec{k}_f |^2 + 2 | \vec{k}_i | | \vec{k}_f |^2 \cos(\theta) }[/math]

[math]\displaystyle{ = m_i^2 + m_f^2 - 2 E_i E_f + 2| \vec{k}_i | | \vec{k}_f |^2 \cos(\theta) }[/math]

In the case of electron scattering

[math]\displaystyle{ m_i = m_f }[/math]

[math]\displaystyle{ E_i \sim | \vec{k}_i | }[/math]

[math]\displaystyle{ E_f \sim | \vec{k}_f | }[/math]

[math]\displaystyle{ m \ll E }[/math]

[math]\displaystyle{ \Rightarrow q^2 = -2 | \vec{k}_i | | \vec{k}_f | \left ( 1 - \cos (\theta) \right ) }[/math]

[math]\displaystyle{ = -4 | \vec{k}_i | | \vec{k}_f | \sin^2(\frac{\theta}{2}) \equiv -4 E_i E_f \sin^2(\frac{\theta}{2}) }[/math]

if[math]\displaystyle{ q^2 < 0 \Rightarrow }[/math] spacelike (scattering)

if [math]\displaystyle{ q^2 > 0 \Rightarrow }[/math] timelike (free particle)

The Momentum Transfer squared for scattering is define as [math]\displaystyle{ Q^2 }[/math] such that

[math]\displaystyle{ Q^2 = -q^2 }[/math]

Missing Mass [math]\displaystyle{ (W) }[/math]

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

[math]\displaystyle{ \left ( p_e^{\mu} \right ) \equiv (E_i, \vec{k}_i) }[/math] = initial momentum 4-vector of the incident electron

[math]\displaystyle{ \left ( p_p^{\mu} \right ) \equiv (M_p, 0) }[/math] = initial momentum 4-vector of the target proton

[math]\displaystyle{ \left ( p_e^{\mu} \right )^{\prime} \equiv (E_f, \vec{k}_f) }[/math] = final momentum 4-vector of the scattered electron

[math]\displaystyle{ \left ( p_p^{\mu} \right )^{\prime} \equiv (E_X, \vec{p}_X) }[/math] = final momentum 4-vector of the target proton

[math]\displaystyle{ W^2 \equiv \left (E_X^2 -p_X^2 \right ) }[/math] = mass of scattered proton

Conservation of 4-Momentum

[math]\displaystyle{ \left ( p_e \right )^{\mu} + \left ( p_p \right )^{\mu} = \left ( p_e^{\prime} \right )^{\mu} + \left ( p_p^{\prime} \right )^{\mu} }[/math]

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

[math]\displaystyle{ \left ( p_p^{\prime} \right )_{\mu} \left ( p_p^{\prime} \right )^{\mu} = \left [ \left ( p_e \right )^{\mu} + \left ( p_p \right )^{\mu} - \left ( p_e^{\prime} \right )^{\mu}\right ] \left [ \left ( p_e \right )_{\mu} + \left ( p_p \right )_{\mu} - \left ( p_e^{\prime} \right )_{\mu}\right ] }[/math]

[math]\displaystyle{ W^2 \equiv \left ( p_p^{\prime} \right )_{\mu} \left ( p_p^{\prime} \right )^{\mu} }[/math]

= [math]\displaystyle{ \left [ (E_i, \vec{k}_i) + (M_p, 0) - (E_f, \vec{k}_f) \right ]\left [ \left ( {E_i \atop \vec{k}_i }\right ) + \left ( {M_p \atop 0 }\right ) - \left ( {E_f \atop \vec{k}_f }\right )\right ] }[/math]

[math]\displaystyle{ = (E_i^2 - k_i^2 ) + (E_f^2 - k_f^2 ) + M_p^2 + 2 M_p(E_i - E_f) - 2(E_iE_f - \vec{k}_i \cdot \vec{k}_f ) }[/math]

= [math]\displaystyle{ m_e^2 + m_e^2 +M_p^2 + 2 M_p(E_i - E_f) - 2(E_iE_f - \vec{k}_i \cdot \vec{k}_f ) }[/math]

[math]\displaystyle{ q^2= m_i^2 + m_f^2 - 2 E_i E_f + 2| \vec{k}_i | | \vec{k}_f |^2 \cos(\theta) }[/math]

substitution

[math]\displaystyle{ W^2 = M_p^2 + 2M_p (E_i -E_f) + q^2 = M_p^2 + 2 M_p(E_i - E_f) -Q^2 }[/math]

Tranform to Center of Mass

A relativistic transformation from a rest frame where [math]\displaystyle{ E,\vec{p} }[/math] are given to a frame moving with velocity [math]\displaystyle{ \beta }[/math] is given by

[math]\displaystyle{ \left ( \begin{matrix} E^{\prime} \\ p^{\prime}_x \\ p^{\prime}_y \\ p^{\prime}_z \end{matrix} \right )= \left [ \begin{matrix} \gamma & -\gamma \beta & 0 & 0 \\ -\gamma \beta & \gamma &0 &0 \\ 0 &0 &1 &0 \\ 0 &0 &0 &1\end{matrix} \right ] \left ( \begin{matrix} E \\ p_x \\ p_y \\ p_z \end{matrix} \right ) }[/math]

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

[math]\displaystyle{ \vec v_{cm} c = \vec \beta = \frac{\vec{p_1}}{p^0} = \frac{\vec p_1}{E_1+m_2} }[/math]

where

[math]\displaystyle{ E_1,\vec p_1 }[/math] are the energy and momentum for the incident particle

[math]\displaystyle{ m_2 }[/math] is the mass of the target particle

[math]\displaystyle{ \beta = \frac{v}{c} = \frac{pc}{E} }[/math]

[math]\displaystyle{ \gamma = \frac{1}{\sqrt{1 -\beta^2}} = \frac{E_{tot}}{mc^2} = \frac{E_1 + m_2}{\sqrt{m_1^2+m_2^2+2 E_1m_2}} }[/math]

Assume an incident electron of 11 GeV moller scatters

[math]\displaystyle{ p_1^{\mu} = ( E_1, \vec p_1) = (sqrt{11^2+0.000511^2, 11 \hat i} )= (11.0000000118691,11 \hat i) }[/math]

[math]\displaystyle{ p_2^{\mu} = ( E_2,\vec 0) = (0.000511,0) }[/math]

[math]\displaystyle{ v_{cm} = \frac{m_1 v_1}{m_1+m_2} }[/math]

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 ([math]\displaystyle{ (\gamma-1)m_oc^2 }[/math]) 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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