Difference between revisions of "Relativistic Frames of Reference"

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<center><math>c^2=\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></center>
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<center><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></center>
  
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This is possible since the ratios of distance to time are multiples of the same base, i.e. <math>\frac{3\times 10^8\ m}{s}</math>
  
  

Revision as of 04:07, 3 June 2017

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

[math]\vartriangleleft [/math] [math]\triangle [/math] [math]\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.


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


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

[math]\vartriangleleft [/math] [math]\triangle [/math] [math]\vartriangleright [/math]