# Equations for length contraction

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.

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

$$\Delta\theta=\tan^{-1}(\gamma\frac{\sqrt{1-\cos(\Delta\theta_0)^2}}{\cos(\Delta\theta_0)}$$
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?

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.