Equations for length contraction

  • Thread starter JDude13
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  • #1
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So I was bored and decided to find some simple equations to deal with relativistic geometric edges and angles at an angle to the direction of movement.
[tex]L=L_0\sqrt{(\frac{\cos(\Delta\theta_0)}{\gamma})^2+(1-\cos(\Delta\theta_0)^2)^2}[/tex]

[tex]\Delta\theta=\tan^{-1}(\gamma\frac{\sqrt{1-\cos(\Delta\theta_0)^2}}{\cos(\Delta\theta_0)}[/tex]
Where
[tex]L[/tex] is the relativistic length of the edge
[tex]L_0[/tex] is the rest length of the edge

[tex]\Delta\theta[/tex] is the “relativistic angle” between the edge and the direction of motion.
[tex]\Delta\theta_0[/tex] is the “rest angle” between the edge and the direction of motion.
[tex]\gamma[/tex] is the Lorentz factor of the object, [tex]\frac{1}{\sqrt{1-\beta^2}}[/tex]

Tell me what you think.
 
Last edited:

Answers and Replies

  • #2
238
6
So I was bored and decided to find some simple equations to deal with relativistic geometric edges and angles at an angle to the direction of movement.
[tex]L=L_0\sqrt{(\frac{\cos(\Delta\theta_0)}{\gamma})^2+(1-\cos(\Delta\theta_0)^2)^2}[/tex]

[tex]\Delta\theta=\tan^{-1}(\gamma\frac{\sqrt{1-\cos(\Delta\theta_0)^2}}{\cos(\Delta\theta_0)}[/tex]
Where
[tex]L[/tex] is the relativistic length of the edge
[tex]L_0[/tex] is the rest length of the edge


[tex]\Delta\theta[/tex] is the “relativistic angle” between the edge and the direction of motion.
[tex]\Delta\theta_0[/tex] is the “rest angle” between the edge and the direction of motion.
[tex]\gamma[/tex] is the Lorentz factor of the object, [tex]\frac{1}{\sqrt{1-\beta^2}}[/tex]


Tell me what you think.
I get division by zero when trying to calc the relativistic angle for a rest angle of 90 degrees. Is the second formula incomplete?
 
  • #3
95
0
I get division by zero when trying to calc the relativistic angle for a rest angle of 90 degrees. Is the second formula incomplete?
Common sense dictates that an edge at 90° to the direction of motion will remain at 90°.
[tex]\tan90=undef.[/tex]
and
[tex]\frac{\sqrt{1-\cos90^2}}{\cos90}=undef.[/tex]
So the equation becomes unclear at 90° but we can safely assume that at 90° the angle remains unchanged.
 

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