I was debating a geocentrist online who said that Einstein and a bunch of other physicists admitted a geocentric framework is valid. I replied that this was technically correct, but if you wanted to adopt a framework where the Earth is stationary and non-rotating, you would need to introduce gravity fields throughout the entire universe which pushed everything else around.

Right after saying this, I worried that I might be spreading a misconception. I think it makes sense in light of Einstein's equivalence principle. I know this applies to uniform acceleration, but any acceleration could be considered as a bunch of tiny uniform accelerations stitched together, right? Maybe I'm missing something. Would what I described above be a valid way of thinking about a geocentric reference frame in General Relativity?

Edit: just to further clarify what I'm asking, I essentially want to know if all fictitious forces in any non-inertial reference frame can be described completely by introducing new gravity fields in GR. I changed the title of the question to reflect this.

  • $\begingroup$ AFAIK it's an unsolved problem in GR whether the rotation is truly absolute or whether spinning the whole universe the other way around would get you the same measurements. $\endgroup$
    – g s
    Commented Aug 22, 2023 at 2:41
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    $\begingroup$ Or what "spinning the whole universe the other way around" even means ;) $\endgroup$
    – m4r35n357
    Commented Aug 22, 2023 at 8:11
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    $\begingroup$ It is in fact true that a geocentric framework is allowed; this is simply mathematics. All the gymnastics with tensor fields etc are precisely because we want the laws of physics not to depend on your coordinate system, whether it is flat Euclidean or somehow curved. Even in Newtonian physics you can place Earth at the centre, but you just make it a lot harder to understand the motions of the planets. $\endgroup$
    – j4nd3r53n
    Commented Aug 22, 2023 at 9:13
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    $\begingroup$ @j4nd3r53n I am not convinced people who haven't tried it appreciate just how hard "non-inertial" analysis is, even in Newtonian systems, e.g. en.wikipedia.org/wiki/Coriolis_force. Nice nick BTW ;) $\endgroup$
    – m4r35n357
    Commented Aug 22, 2023 at 10:33
  • $\begingroup$ @m4r35n357 Well, in principle, it's simple enough; just tensors and Christoffel symbols ;-) The problem is doing it in practise. $\endgroup$
    – j4nd3r53n
    Commented Aug 22, 2023 at 12:47

1 Answer 1


Fictitious forces give all bodies the same acceleration, irrespective of their mass. The same is true, so we believe, for test point particles in any gravity field. One can formally remove the distinction between the two forces and describe motions influenced by both gravity and fictitious forces as geodesic ("free") motions in some appropriately curved spacetime, where both fictitious forces and gravity forces together are described by some single total appropriate tensor field (e.g. the metric tensor).

This is a formal trick, with important role in historical development of GR and other theories of gravity, but of questionable relevance to the question of whether fictitious forces are as physically real as gravity. Consider that in order to describe motion of a distant star (but also the Sun) in Earth's frame, we have to introduce curved spacetime due to rotation of the Earth that extends all the way to the star, and which makes it go in circles around the Earth, with immense speed. And this speed and corresponding acceleration on the circular orbit gets greater the more distant the object is. This is exactly the opposite of what we expect of physical influences, including real gravity - to get weaker with increasing distance.

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    $\begingroup$ FWIW, in the fixed Earth frame, a body at 27.482 au is rotating faster than light. That's smaller than the orbit of Neptune (~30 au). $\endgroup$
    – PM 2Ring
    Commented Aug 22, 2023 at 2:06

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