We may choose a non-rotating earth as our reference frame and ask ourselves: how about the planetary and stellar motions. A star at a distance of 10 million light years would turn around the earth in 24h with a velocity of 10^18 m/s.

A friend once told me that actually articles have been published delving into this problem, e.g. to prove that fictitious forces emerge from the choice of such a bizar reference frame that ensure that the earth is still (somewhat) flattened at it's poles.


1) Does anybody know of such a publication?

2) I know that even such speeds of 10^18 m/s are not in contradiction with relativity because a limiting velocity only exists for exchange of information, which apparantly does not occur.
Still: could anybody explain why such bizar velocities are allowed?

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    $\begingroup$ I don't know enough to expand this into an answer, but it seems like you're describing the ideas around Mach's principle: en.wikipedia.org/wiki/Mach%27s_principle $\endgroup$ – j.c. Nov 11 '10 at 23:29

Velocities in General Relativity can only be compared at a point, where local tangent planes coincide. Talking about the velocities of far-away stars in any sort of absolute sense is an empty question. Saying 'the coordinate velocity of Andromeda is 10^huge m/s' is, in a sense, not a statement about physics, but rather about your coordinate system. In order to get a meaningful prediction, you would have to devise an experiment whereby you compare the two velocities--say, andromeda sends the earth a light signal at a preassigned 100 Hz. An Earth-based observer then measures the redshift for the light signal, and then uses that to decide their relative velocities.

  • $\begingroup$ I must agree. Any idea how to prove that the earth becomes flattened at the poles? That's actual observation. $\endgroup$ – Gerard Nov 12 '10 at 19:54
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    $\begingroup$ How fancy an answer do you want? The five-cent answer is that you get coriolis forces by choosing a non-inertial reference frame. The five dollar answer is that if you make a coordinate transformation on \phi such that it becomes a coorotating coordinate with the Earth, you induce a time-dependence in g_{phi phi}, which is also proportional to theta, so you get new Christoffel symbols that are zero on the poles, but nonzero at the equator. $\endgroup$ – Jerry Schirmer Nov 14 '10 at 16:49
  • $\begingroup$ And after thinking a little more--the thrust of what I just said is right, but the details are a little off--the coordinate change will induce a time independent term proportional to $\sin (\theta)$ in $g_{t \phi}$, and THIS term will modify the geodesic equation in such a way that you see forces at the equator and not the poles. $\endgroup$ – Jerry Schirmer Nov 14 '10 at 17:27

1) In this reference frame it is obvious that Earth is flattened at the poles. There is a centrifugal force pushing out away from the axis of rotation with magnitude $\omega^2 r$, where $r$ is the distance from the axis. If you're at the poles, $r = 0$ and you aren't pushed out at all. If you're at the equator $r = R_{earth}$ and you're pushed out a lot.

2) (Not a direct answer) Rotation in special relativity is tricky. For example, the ratio of the circumference of a disk to its diameter is not $\pi$. See http://en.wikipedia.org/wiki/Ehrenfest_paradox

  • $\begingroup$ In this reference frame there is nó earth rotation so one of the problems is to prove that the rest of the universe causes a fictitious centrifugal force on earth - you may not simply assume it exists, not in this reference frame. $\endgroup$ – Gerard Nov 12 '10 at 19:48
  • $\begingroup$ @Gerard : In this reference frame, the rotation is replaced by inertial force (centrifugal force and Coriolis force). So the centrifugal force only exists in the reference frame where the Earth does not move. In the geocentric reference frame, where the Earth does rotate, you do not need to add centrifugal force to explain the the Earth flatness, since the rotation induces the centripetal acceleration of the equator which explains the flatness. $\endgroup$ – Frédéric Grosshans Nov 29 '10 at 16:31

You already answered yourself: this is because the principle of casualty does not prohibit such motion.

Tangential motion and moving away from another object with superluminal velocity is not prohibited because in such processes the casualty cannot brake. What is prohibited is approaching another object with superluminal velocity.

The only reason why we do not see objects moving away from each other at superluminal speeds is because moving from some object at v>c usually means approaching another object with v>c, and this is prohibited. But in the case of the expansion of Universe it is possible to move away from one object without approaching another, and that's why superluminal growth of distances between two galaxies separated by big distances is possible.

Even more, due to expansion of universe any light rays going from us (or any other atationary observer) have velocity somewhat greater than c and any light rays going toward us have velocity smaller than c. This of course means also that any two stationary observers move away from each other.

There is only one case where one can imagine superluminal approaching of two objects: when both are under the horizon of a black hole. When an object approaches a black hole, he reaches superluminal tangential velocity at ergosphere (which is outside the black hole horizon so the object can return) and the radial velocity reaches c exactly at the horizon. Mathematically this means that inside the horizon the object should have radial velocity greater than c. But this is not the case: once object reached c (or near c) at the horizon, the time stops for him and he remains there until the black hole explodes (finally evaporates).


Planetary motions in such a frame would be depicted properly by the deferent an epicycle model of Ptolemy:


So, the question is actually answered by Ptolemy. For more reference, look up details in the Almagest! :-)

  • $\begingroup$ This doesn't address either of the 2 questions in Gerard's question. $\endgroup$ – j.c. Nov 12 '10 at 14:31
  • $\begingroup$ You're right. Maybe it should be a comment rather than an answer. When I begun thinking about this kind of motions, Pholemy's theory came to my mind. Should I delete the answer and post a comment? $\endgroup$ – asanlua Nov 12 '10 at 14:40

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