# General “relativity” without special relativity

I was wondering if there could be a way of using the principle of equivalence and develop a theory of warped space-time without using special relativity, ie, without assuming any maximum possible speed. Would such a theory be possible in principle? Or does GR specifically depend on the absoluteness of $$c$$ in order to make sense as a theory of gravity with warped space-time? Any thoughts?

PD: I have seen this post, but for what I understand, the question (and answers) are concerned with the possibility of having GR as we know it, instead of a modified GR, without a limit speed (which is what I’m asking).

• – G. Smith Aug 27 at 19:33
• Interesting! Do you know if the predictions of the theory match exactly those of Newton? And would it also match the case of GR with $c\to\infty$? – Alan Rapoport Aug 27 at 19:38
• So the central point of special relativity is that when I move with acceleration $\vec a$, I see clocks at coordinate $\vec r$ tick at a rate of $1+\vec a\cdot\vec r/c^2,$ faster “ahead of me” and slower “behind me” assuming I am oriented in the direction I am accelerating. This is for example what makes gravitational time dilation happen, and it predicts an event horizon behind me where $\vec a \cdot \vec r = -c^2$ and clocks no longer appear to tick. Are you explicitly looking to omit gravitational time dilation and black holes from your approach? – CR Drost Aug 27 at 19:41
• I am aware that Newton-Cartan theory exists, but I haven’t looked into its predictions. – G. Smith Aug 27 at 19:48
• @CRDrost Making $c\to\infty$ (ie, no SR) implies, for example, that the Schwarzschild radius goes to $0$, so there would be no BHs in the theory. And in the same way there would be no time dilation. But I'm not sure whether $c\to\infty$ abolishes any sort of curvature, and if so, how gravity comes up. – Alan Rapoport Aug 27 at 19:49

I was wondering if there could be a way of using the principle of equivalence and develop a theory of warped space-time without using special relativity, ie, without assuming any maximum possible speed. Would such a theory be possible in principle?

I would say that the answer is basically no, although it may be necessary to highlight the relevant assumptions more clearly.

The existence of a maximum speed is not an assumption in special relativity. Depending on what axiomatization you use for SR, there could be an assumption that there's an invariant speed (Einstein's 1905 axiomatization), which is a different thing. Actually Einstein's axiomatization is probably not one that anyone would pick today. You just see it in popularizationa dn undergraduate textbooks because it's become a traditional way of presenting it. For other styles of presentation, see, e.g., Pal, "Nothing but relativity," https://arxiv.org/abs/physics/0302045 , or Bertel Laurent, Introduction to spacetime: a first course on relativity.

The machinery of general relativity, in the formalism that is most often used, depends on the existence of a metric. Without a metric, you can't do things like raising and lower indices. In particular, you need this metric to be nondegenerate. If the equivalence principle holds, then spacetime is locally flat, and SR applies locally, where you can choose coordinates such that this metric has the form $$\operatorname{diag}(1,-1,-1,-1)$$.

The Pal paper explains why, under certain reasonable basic assumptions, a flat spacetime has to be either Galilean or Lorentzian. If you want spacetime to be locally Galilean, then you don't have a metric (or you could think of the metric as being degenerate or split into separate time and space parts). This is not compatible with the machinery of GR as outlined above.

• Yes, instead of assuming a maximum speed it would probably be better to say that we have a pseudo-Riemannian metric which is locally Minkowski. So in my case, no SR would mean to have a degenerate metric split into time and space, which is locally Galilean. But that would be the case of the Newton-Cartan Theory as mentioned by G. Smith, right? Wouldn't all the machinery work anyway in this case? What I don't know is if it would be the same as GR with $c\to\infty$, or if it would predict the same results as Newtonian mechanics. – Alan Rapoport Aug 27 at 21:29
• No, the machinery won't work in that case, for the reasons described in the answer. – Ben Crowell Aug 27 at 23:10
• @AlanRapoport What splits the metric in two is exactly the infinite speed of light. Also gravity is caused by a gradient of the time dilation or equivalently by a gradient of the speed of light. So there is no gravity with the infinite speed of light. This is why gravity is given as a postulated force in Newtonian mechanics. – safesphere Aug 28 at 0:13
• @AlanRapoport While you can envision a curved dual metric, only spatial part of it can be curved. There can be no curvature in the 1D temporal metric. And spatial curvature does not cause gravity, only temporal curvature does (see: en.wikipedia.org/wiki/Ellis_wormhole). So again, no gravity with the infinite speed of light. There also is a commonly overlooked case of the zero speed of light. It applies in areas of the infinite time dilation (such as at the horizon) and freezes all local relative motion (as in frame dragging, linear or rotational). – safesphere Aug 28 at 0:26

There's a really nice version of Newtonian gravity that talks about curvature in space-time and I think it might be interesting to you.

It begins with the idea that, since all things fall at the same rate, this might be explained by a curved space. However, when you try to get the equations, they fail: the geodesic equation looks like $$\frac{dv^i}{dt}=-\Gamma^{i}_{jk}v^kv^j$$ Where i,j,k, go from 1 to 3, since they represent only time direction.

Why do they fail? Well, as you can see, the RHS clearly depends on the velocity, while Newtonian gravity does not. However, there's a really nice way to go around this. What if, instead of space, we worked with space-time? In this case, our velocity vector would have components $$(1, v_x,v_y,v_z)$$ instead of just the last three. And that one could really come in handy. Using it, we can make the geodesic equation look like the newtonian one just by making $$\Gamma^i_{jk}=0$$ for $$j=0$$ or $$k=0$$.

In that case, it would just look like: $$\frac{dv^i}{dt}=-\Gamma^{i}_{00}v^0v^0=-\Gamma^{i}_{00}$$ (where i goes from 0 to 3 this time)

Which is way more sensible now.