Forest Relativity Notes

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Lorentz Transformations

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

ForestRelativityLorentzFrame.jpg

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

4- vector notation

The 4-vector notation is given as

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

where

[math] x^0 \equiv ct[/math]
[math]x^1 \equiv x[/math]
[math]x^2\equiv y[/math]
[math]x^3\equiv z[/math]
[math]\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]\beta = \frac{v}{c} = \frac{pc}{E}[/math]
[math]\gamma = \frac{1}{\sqrt{1 -\beta^2}} = \frac{E_{tot}}{mc^2}[/math]
NOTE
It is common in particle physics to define [math] c \equiv 1[/math] making [math]\gamma = \frac{E}{m}[/math] where [math]m[/math] is in units of [math]\frac{\mbox{MeV}}{\mbox{c}^2}[/math]
example
[math]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]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]\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]\sum[/math] symbols and assumes it to exist whenever there is a repeated subscript and uperscript
ie; [math]x^{\mu^{\prime}} = \Lambda_{\nu}^{\mu} x^{\nu}[/math]
in the example above the[math] \nu[/math] symbol is repeated thereby indicating a summation over [math]\nu[/math].

Momentum 4-vector

[math]p^{\mu} \equiv (\frac{E}{c} , \vec{p})[/math]
[math]p_{\mu} \equiv (\frac{E}{c} , -\vec{p})[/math]
[math]p_{\mu}p^{\mu} = \frac{E^2}{c^2} - p^2 \equiv E^2 - p^2 = m^2[/math]


Trig Method

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

[math]\sin (\alpha) \equiv \beta \equiv \frac{v}{c}[/math]
[math]\cos(\alpha) \equiv \frac{1}{\gamma} \equiv \sqrt{1 - \beta^2}[/math]
The Matrix form pf the tranformation looks like
[math]\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]\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]\Lambda^{-1}[/math]

Proper Time and Length

Proper Time

Proper Time [math]\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]\gamma \Tau[/math]

Note
since [math]\gamma \gt 1[/math] you expect the Proper time interval to be the smallest

Proper Length

Proper Length[math] (c\Tau)[/math]
The length of an object in the object's rest frame.

Invariant Length

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

:[math]p_{\mu}p^{\mu}(1)= \frac{E^2(1)}{c^2} - p^2(1) \equiv E^2 - p^2 = m^2(1)[/math]
p_2^2 = m_2^2 : similarly

Let [math]E_{21}[/math] = Energy of particle #2 as observed by particle #1