Equations for length contraction

[JDude13]

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.
L=L_0\sqrt{(\frac{\cos(\Delta\theta_0)}{\gamma})^2+(1-\cos(\Delta\theta_0)^2)^2}

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

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

Tell me what you think.

 

[MikeLizzi]

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.
L=L_0\sqrt{(\frac{\cos(\Delta\theta_0)}{\gamma})^2+(1-\cos(\Delta\theta_0)^2)^2}

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


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


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?

 

[JDude13]

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°.
\tan90=undef.
and
\frac{\sqrt{1-\cos90^2}}{\cos90}=undef.
So the equation becomes unclear at 90° but we can safely assume that at 90° the angle remains unchanged.