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

Kinematics in Center of Momentum Frame

In this example we would like to determine the energy of particle #2 as seen from a referenve frame attached to particle # 1, ie; particle #1's rest frame. Our list of Lorentz invariants is

$\left ( p^1\right )_{\mu}\left (p^1 \left )^{\mu}= \frac{E^2(1)}{c^2} - p^2(1) \equiv E^2 - p^2 = m^2(1)$= Lorentz invariant

p_2^2 = m_2^2 : similarly

Let $E_{21}$ = Energy of particle #2 as observed by particle #1