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:

4- vector notation

The 4-vector notation is given as

[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

[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 pf 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 Particle to 2 Bodies

Consider the decay of the [math]\displaystyle{ \rho_0 }[/math] meson at rest into two pions ([math]\displaystyle{ \pi^+ }[/math] and [math]\displaystyle{ \pi^- }[/math] )

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.

If [math]\displaystyle{ P^{\mu} }[/math] represent 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 which 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}_{\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_{\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_1^2 + p_1^2} }[/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 thee 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 wa 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) > 0 }[/math]

[math]\displaystyle{ \Rightarrow \frac{M p}{m_1 p_{tot}} > \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 = 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].

Phase space

Phase space

In this type of problem, Phase space represents a kinematics range (space) spanned by any set of INDEPENDENT kinematic variables.

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]