Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 2 down vote accepted

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.