# Limit on velocity in Minkowski Spacetime geometry

Let A be a rocket moving with velocity v.

Then the slope of its worldline in a spacetime diagram is given by c/v.

Since it is a slope, c/v = tan(theta) for some theta > 45 and theta < 90.

Does this impose a mathematical limit on v?

If so what is it?

As in, we know tan(89.9999999999) = 572957795131.

And c = 299792458.

Using tan(89.9999999999) as our limit of precision, the smallest v we can use is:

c/v = tan(89.9999999999)

=> 299792458 / v = 572957795131

Therefore, v = 1911.18 m/s

What is the smallest non zero value of v? Is there a limit on this?

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When I see a numerical argument like this, I tune out. If there's some point to be made here, it can be made in the language of algebra. – Ben Crowell May 9 '13 at 2:26

Since a worldline along the time axis on Minkowski diagram is at rest, it is more intuitive to measure angles from that axis instead, as then 'slope' is (space)/(time), i.e., a velocity. Then we have the trigonometric relationship: $$\frac{v}{c} = \tanh\alpha$$ where Minkowski spacetime follows hyperbolic trigonometry because of the sign difference in the Minwkoski metric/distance formula compared to Euclidean metric/Pythagorean theorem.

The hyperbolic angle $\alpha$ can be any real number, and limit it imposes on speed under this restriction of real numbers is that $|v|<c$.

A lot of STR formula become rather intuitive in this form, e.g., Lorentz transformation is just a rotation with hyperbolic trigonometry, and the velocity addition formula is: $$\begin{eqnarray*}u\oplus v = \frac{u+v}{1+uv/c^2} &\Longleftrightarrow& \tanh(\alpha+\beta) = \frac{\tanh\alpha+\tanh\beta}{1+\tanh\alpha\tanh\beta}\text{,}\end{eqnarray*}$$ and so forth.

Note that in Euclidean space, the corresponding question is 'if you have three lines intersecting at a point, and the first makes a slope $m$ with the second, while the second makes a slope $l$ with the third, what slope does the first line make with the third?', and the answer to that also follows that pattern of the normal tangent addition formula.

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Why do you stop your largest angle with ten 9s after the decimal point? If you added more of them, then you'd get a smaller bound for the velocity. And you keep adding 9s ad infinitum and you'll "eventually" reach $89.\bar{9}=90$. So eventually, you'll see that the velocity could be arbitrarily small. This just means that the worldline can be vertical... and that describes a particle at rest, with $v=0$ in that frame.

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