I understand that the closer something travels to the speed of light, that time will stretch by a factor, and distance will compress by the same factor.

My question is, if something travels in a circle, close to the speed of light, what does the distance of the journey look like to them? They measure that the trip took them 10 minutes. And an outside observer says that the journey took 20 minutes, and the outside observer measured that they did, for example, 1000 laps of a circle circumference 1000 km.

So if the plane had a distance trip counter, what would it read? And if they were looking out of the window, would the circle still look like it had a circumference of 1000km?


Circular motion in special relativity is somewhat tricky: Note that for circular motion, the acceleration in the spaceship travelling in a circle is not zero, so the spaceship is not in a single frame of inertia.

Here is an interesting thought: Distances perpendicular to the direction of motion are not subject to contraction. Hence, if the observes on earth see the spaceship going on a circle with radius $R$, then in their frame of inertia the spaceship is always a distance $R$ away from the earth. Since the line from earth to spaceship is perpendicular to the direction of flight, the people on the spaceship will also believe that they are always a distance $R$ away from earth, so they will also fly on a circle.

They will, nonetheless, experience a different circumference! The best way to solve this is to consider a polygon with $N$ sides and then let $N$ go to infinity. If people on the earth measure each side of the polygon as $L_0/N$ where $L_0$ is the circumference of the polygon in the earth's frame of inertia, then the spaceship-people will measure each side to be $L_0/(N\gamma)$. Hence, for $N \rightarrow \infty$, the polygon becomes a circle. Measured from earth, it has circumference $L_0$, but for the spaceship it has circumference $L_0/\gamma$.

This suggests that the spaceship moves through non-Euclidean geometry, because it travels on a circle whose ratio between circumference and diameter is less than $2\pi$. This is a hint that accelerating frames have non-Euclidean geometry, which is excessively treated in General Relativity.

Reference: http://abacus.bates.edu/~msemon/WortelMalinSemon.pdf

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    $\begingroup$ Thanks for your answer. So in the case of the polygon, the numbers of sides would experience the contraction and therefore, so would the circumference. So I would be right in saying that from the view of the plane, the circumference would seem shorter than the observer would perceive, but as you said the radius would remain constant. $\endgroup$ – Juddling Apr 14 '11 at 23:44
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    $\begingroup$ Yes. The circumference gets smaller ($\gamma$ here is the relativistic contraction factor $1/\sqrt{1-v/c}$). It still is a circle and it still has the same radius. This can only happen if the space the plane travels in is not the classical Euclidean space. Indeed, the fact that the plane is accelerating all the time warps its space. $\endgroup$ – Lagerbaer Apr 14 '11 at 23:54
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    $\begingroup$ +1 for a good, clear answer. I would just like to add that in one of my 4th year particle physics lectures, the Professor calculated the circumference of the circle travelled by a proton circling the LHC in the frame of the proton (as if one were sitting on it.) At about 7TeV, he found that rather than the 27km circumference as measured in the lab frame (i.e. standing next to the LHC,) the proton was in fact circling a circumference of about 3m!! Unfortunately I don't has the calculation to hand but it shouldn't be too difficult to replicate. $\endgroup$ – qftme Apr 15 '11 at 13:51
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    $\begingroup$ As a final teaser: what do you think would happen if one were to increase the energy of the proton (and hence also the speed), such that the circumference becomes less than the size of the proton? $\endgroup$ – qftme Apr 15 '11 at 13:54

You seem to be forgetting that the distance (ie. the laps) will contract in length as the rocket ship approaches the speed of light. From the perspective of the pilot, EACH lap that he takes will be shorter in distance than the one that is seen from the point of view of the bystander. In order for this to be balanced out, time will dilate for the pilot.

  • $\begingroup$ That's exactly what I said. Each lap that he takes will be shorter in distance. $\endgroup$ – Alex Bratchuli May 12 '11 at 3:30
  • $\begingroup$ Well put Alex. This is the simplest explanation for the twin paradox too. The fact that the pilot is accelerating (non-inertially), in either the OP's case or in the twin paradox, really doesn't matter. $\endgroup$ – finbot May 14 '11 at 16:05

Everyone, including Einstein, assumes that length contraction only occurs in the direction of motion. This is wrong. The Ehrenfest paradox is not a paradox because length contraction occurs unilaterally in every direction. We just can't perceive the perpendicular length contraction directly. To do so, we need a ruler with invariant length. We have one. It's called light.

Imagine a spaceship which is 50 light years from Earth, traveling towards earth at 0.5c. Now imagine a star, 100 light years from both the spaceship and earth, goes supernovae.

From Earth's perspective the spaceship arrives at earth, the same time they see the star explode. From the spaceships perspective, the journey takes 86.6 years. From the spaceships perspective, the star is 86.6 light years away, even though it's perpendicular to the direction of motion.

The implications of this is the the Michelson and Morley experiment, as well as all subsequent experiments, are flawed, as they rely on the assumption that length contraction only occurs in the direction of motion.


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