Relativistic Frames of Reference: Difference between revisions

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Revision as of 01:47, 6 June 2017

[math]\displaystyle{ \textbf{\underline{Navigation}} }[/math]

[math]\displaystyle{ \vartriangleleft }[/math] [math]\displaystyle{ \triangle }[/math] [math]\displaystyle{ \vartriangleright }[/math]

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


Galilean Frames of Reference
Figure 2.1: Primed reference frame moving in the z direction with velocity v.


In the rest frame of v=0

[math]\displaystyle{ \begin{cases} t= t' \\ x=x' \\ y=y' \\ z=z'+vt' \end{cases} }[/math]

While conversely, from the rest frame of v'=0

[math]\displaystyle{ \begin{cases} t'= t \\ x'=x \\ y'=y \\ z'=z-vt \end{cases} }[/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]\displaystyle{ speed=\frac{\Delta Distance}{\Delta Time} }[/math]


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


where

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


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


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



We can rewrite this as


[math]\displaystyle{ \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]\displaystyle{ (\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]\displaystyle{ 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]\displaystyle{ \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]\displaystyle{ ds^2 }[/math] when the change is infinitesimal


[math]\displaystyle{ ds^2\equiv c^2 dt^{'2}-dx^{'2}-dy^{'2}-dz^{'2}= c^2 dt^{2}-dx^2-dy^2-dz^2 }[/math]


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.

[math]\displaystyle{ ds^2\equiv c^2 dt^{'2}-dx^{'2}-dy^{'2}-dz^{'2}= c^2 dt^{2}-dx^2-dy^2-dz^2 }[/math]


[math]\displaystyle{ s^2\equiv c^2 dt^{'2}-dr^{'2}= c^2 dt^{2}-dr^2 }[/math]


[math]\displaystyle{ s^2\equiv (c^2 -v^{'2})dt^{'2}= (c^2 -v^2)dt^{2} }[/math]


From the rest frame of v'=0


[math]\displaystyle{ s^2\equiv c^2 dt^{'2}= (c^2 -v^2)dt^{2} }[/math]


[math]\displaystyle{ \Rightarrow dt^{'2}= (1-\frac{v^2}{c^2 })dt^{2} }[/math]


[math]\displaystyle{ \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} }[/math]


[math]\displaystyle{ \Rightarrow \begin{cases} t'=\frac{1}{\gamma} t\\ \\ t=\gamma t' \end{cases} }[/math]


Assuming motion is only along the z direction


[math]\displaystyle{ z \equiv ct\ \ \ \ z'\equiv ct' }[/math]



Substituting these changes into the Galilean transformations


[math]\displaystyle{ \underline{\textbf{Galilean\ Transformations}}\quad \underline{\textbf{Lorentz\ Transformations}} }[/math]
[math]\displaystyle{ v=0\Rightarrow \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} }[/math]


[math]\displaystyle{ v'=0\Rightarrow \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} }[/math]


[math]\displaystyle{ \textbf{\underline{Navigation}} }[/math]

[math]\displaystyle{ \vartriangleleft }[/math] [math]\displaystyle{ \triangle }[/math] [math]\displaystyle{ \vartriangleright }[/math]