Relativistic Frames of Reference

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Relativistic Frames of Reference

From the Galilean description of motion for a frame of reference moving relative to another frame considered stationary we know that

Figure 2.1: Primed reference frame moving in the z direction with velocity v.

In the rest frame of v=0

$\begin{cases} t= t' \\ x=x' \\ y=y' \\ z=z'+vt' \end{cases}$

Using Einstein's Theory of Relativity, we know that the speed of light is a constant, c, for all reference frames. In the unprimed frame, from the definition of speed:

$speed=\frac{\Delta Distance}{\Delta Time}$

$c=\frac{\Delta d}{\Delta t}$

where

$c=3\times 10^8\ m/s$

Using the distance equation in a Cartesian coordinate system, the equation for the speed of light becomes

$c=\frac{\sqrt{\Delta x^2+\Delta y^2+\Delta z^2}}{t}$

Following the postulate of Special Relativity, this implies for the primed frame

$c=\frac{\sqrt{\Delta x^{'2}+\Delta y^{'2}+\Delta z^{'2}}}{t}$

We can rewrite this as

$\frac{\Delta x^{'2}+\Delta y^{'2}+\Delta z^{'2}}{\Delta t^{'2}}= c^2=\frac{\Delta x^2+\Delta y^2+\Delta z^2}{\Delta t^2}$

This is possible since the ratios of distance to time are multiples of the same base, i.e. the square of the speed of light $(\frac{3\times 10^8\ m}{s})^2$. Therefore for the relative change in the time in one frame, the distance must change by the same factor to maintain the same constant. With this we can write

$c^2 \Delta t^{'2}=\Delta x^{'2}+\Delta y^{'2}+\Delta z^{'2}\ \ \ \ \ c^2 \Delta t^{2}=\Delta x^2+\Delta y^2+\Delta z^2$

$\Rightarrow c^2 \Delta t^{'2}-\Delta x^{'2}-\Delta y^{'2}-\Delta z^{'2}= c^2 \Delta t^{2}-\Delta x^2-\Delta y^2-\Delta z^2$

This quantity is known as the time space interval $ds^2$ when the change is infinitesimal

$s^2\equiv c^2 dt^{'2}-dx^{'2}-dy^{'2}-dz^{'2}= c^2 dt^{2}-dx^2-dy^2-dz^2$

Since the speed of light is a constant for all frames of reference, this allows the space time interval to also be invariant for inertial frames.

$s^2\equiv c^2 dt^{'2}-dx^{'2}-dy^{'2}-dz^{'2}= c^2 dt^{2}-dx^2-dy^2-dz^2$

$s^2\equiv c^2 dt^{'2}-dr^{'2}= c^2 dt^{2}-dr^2$

$s^2\equiv (c^2 -v^{'2})dt^{'2}= (c^2 -v^2)dt^{2}$

From the rest frame of v'=0

$s^2\equiv c^2 dt^{'2}= (c^2 -v^2)dt^{2}$

$\Rightarrow dt^{'2}= (1-\frac{v^2}{c^2 })dt^{2}$

$\Rightarrow \begin{cases} dt'= \sqrt{1-\frac{v^2}{c^2 }}dt=\frac{1}{\gamma} dt\\ \\ dt= \frac{1}{\sqrt{1-\frac{v^2}{c^2 }}}dt'=\gamma dt' \end{cases}$

$\Rightarrow \begin{cases} t'=\frac{1}{\gamma} t\\ \\ t=\gamma t' \end{cases}$

Assuming motion is only along the z direction

$z \equiv ct\ \ \ \ z'\equiv ct'$

Substituting these changes into the Galilean transformations

$\underline{\textbf{Galilean\ Transformations}}\quad \underline{\textbf{Lorentz\ Transformations}}$ $\begin{cases} t= t' \\ x=x' \\ y=y' \\ z=z'+vt' \end{cases} \qquad \Rightarrow \qquad \quad \begin{cases} t=\gamma (t'+vz'/c^2) \\ x=x' \\ y=y' \\ z=\gamma (z'+vt') \end{cases}$

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