Forest Relativity Notes

Lorentz Transformations

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

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

4- vector notation

The 4-vector notation is given as

$x^{\mu^{\prime}} = \sum_{\nu=0}^3 \Lambda_{\nu}^{\mu} x^{\nu}$

where

$x^0 \equiv ct$

$x^1 \equiv x$

$x^2\equiv y$

$x^3\equiv z$

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

$\beta = \frac{v}{c} = \frac{pc}{E}$

$\gamma = \frac{1}{\sqrt{1 -\beta^2}} = \frac{E_{tot}}{mc^2}$

NOTE

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

example

$x^{0^{\prime}} = \sum_{\nu=0}^2 \Lambda_{\nu}^0 x^{\nu} = \Lambda_0^0 x^0 + \Lambda_1^0 x^1 \Lambda_2^0 x^2 + \Lambda_3^0 x^2$

$ct^{\prime}= \gamma x^0 - \gamma \beta x^1 + 0 x^2 + 0 x^3 = \gamma ct - \gamma \beta x = \gamma(ct -\beta x)$

Or in matrix form the tranformation looks like

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

Note

Einstein's summation convention drops the $\sum$ symbols and assumes it to exist whenever there is a repeated subscript and uperscript

ie; $x^{\mu^{\prime}} = \Lambda_{\nu}^{\mu} x^{\nu}$

in the example above the$\nu$ symbol is repeated thereby indicating a summation over $\nu$.

Momentum 4-vector

$p^{\mu} \equiv (\frac{E}{c} , \vec{p})$

$p_{\mu} \equiv (\frac{E}{c} , -\vec{p})$

$p_{\mu}p^{\mu} = \frac{E^2}{c^2} - p^2 \equiv E^2 - p^2 = m^2$

Note

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

$p^{\mu} \equiv (\vec{p},iE)$

$p_{\mu} \equiv (\vec{p},iE)$

Trig Method

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

$\sin (\alpha) \equiv \beta \equiv \frac{v}{c}$

$\cos(\alpha) \equiv \frac{1}{\gamma} \equiv \sqrt{1 - \beta^2}$

The Matrix form pf the tranformation looks like

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

Or the reverse transformation

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

Notice that you just needed to change the signs for the inverse matrix $\Lambda^{-1}$

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) = $\gamma \Tau$

Note

since $\gamma \gt 1$ 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.

Transformation Examples

Decay of Particle to 2 Bodies

Consider the decay of the $\rho_0$ meson at rest into two pions ($\pi^+$ and $\pi^-$ )

File:NeutralRhoMesonDecayDiagram.jpg

The diagram above shows a $\rho_0$ meson at rest in the lab which then decays into two pions of momentum $p_1$ and $p_2$ in the center of momentum frame of the $\rho_0$ meson.

If $P^{\mu}$ represent the total momentum of the system before the decay then

$P^{\mu} = (E,0) =(M,0) = \left ( p_1 \right )^{\mu} + \left ( p_2 \right)^{\mu}$

$\Rightarrow 0 = \vec{p}_1 + \vec{p}_2$

or

$\vec{p}_1 = - \vec{p}_2$

Let

$p \equiv |\vec{p}_1 | = |\vec{p}_2 |$

Conservation of Energy

$\Rightarrow E_{tot} = M = E_1 + E_2 = \sqrt{m_1^2 + p^2} + \sqrt{m_2^2 + p^2}$

solving for p

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

$\Rightarrow M \ge m_1 + m_2$ is required to avoid the unphysical condition that the momentum of the particles after a decay would be an imaginary number

Using

$p \equiv |\vec{p}_1 | = |\vec{p}_2 |$

$E_1^2 - m_1^2 = E_2^2 - m_2^2$

$\Rightarrow E_2 = \sqrt{E_1^2 - m_1^2 + m_2^2}$

Combine this with the conservation of energy equation above:

$E_1 + E_2 = E_1 + \sqrt{E_1^2 - m_1^2 + m_2^2} = M$

$\Rightarrow E_1 - M = \sqrt{E_1^2 - m_1^2 + m_2^2}$

Square both sides of the above equation

$E_1^2 -2ME_1 + M^2 = E_1^2- m_1^2 + m_2^2$

$\Rightarrow E_1 = \frac{M^2+m_1^2-m_2^2}{2M}$

Similarly

$E_2 = \frac{M^2+m_2^2-m^2_1}{2M}$

Note

$\vec{p}_1 = -\vec{p}_2$

$\Rightarrow$ The daughter particles (pions) from the decay of the Mother particle $(\rho)$ 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)

$P^{\mu} = (E,\vec{p}_{tot}) =(M,\vec{p}_{tot}) = \left ( p_1 \right )^{\mu} + \left ( p_2 \right)^{\mu}$

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:$\vec{p}_{1,\perp}$ and $\vec{p}_{2,\perp}$) are equal and opposite by conservation of momentum.

$\vec{p}_{\perp}\equiv -\vec{p}_{2,\perp}$