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.

[math]t= t'[/math] [math]x=x'[/math] [math]y=y'[/math] [math]z=z'+vt[/math]

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:

[math]speed=\frac{\Delta Distance}{\Delta Time}[/math]

[math]c=\frac{\Delta d}{\Delta t}[/math]

where

[math]c=3\times 10^8\ m/s[/math]

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

[math]c=\frac{\sqrt{(\Delta x)^2+(\Delta y)^2+(\Delta z)^2}}{\Delta t}[/math]

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

[math]c=\frac{\sqrt{(\Delta x')^2+(\Delta y')^2+(\Delta z')^2}}{\Delta t'}[/math]

We can rewrite this as

[math]\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}[/math]

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 [math](\frac{3\times 10^8\ m}{s})^2[/math]. 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

[math]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[/math]

[math]\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[/math]

This quantity is known as the time space interval [math]ds^2[/math] when the change is infinitesimal

[math]ds^2\equiv c^2 d t^{'2}-(d x')^2-(d y')^2-(d z')^2= c^2 d t^{2}-(d x)^2-(d y)^2-(d z)^2[/math]

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