# Can there be general relativity without special relativity?

Can General Relativity be correct if Special Relativity is incorrect?

• I think this is difficult to answer because we don't have clear definitions. E.g., scalar-tensor theories might or might not be considered to be GR. You can have GR with one of the matter fields being a preferred, universal timelike vector field, and that would establish a preferred frame, which you could say is inconsistent with SR -- depending on what you mean by SR.
– user4552
Apr 27, 2018 at 16:35
• If SR is incorrect then much more than GR is at stake. QFT relies on SR too. So SR being shown to be wrong would mean a serious crisis in physics. Apr 27, 2018 at 20:04
• Can there be calculus without linear algebra? Mar 13, 2021 at 3:19
• This question seems underspecified. Can special relativity be wrong while GR is right? Sure. SR is incorrect as a spacetime model. Our spacetime is not flat. It exhibits curvature in many ways. On the other hand, GR predicts that spacetime looks locally like Minkowski spacetime, so in that sense SR can't be wrong if GR is correct. Mar 13, 2021 at 3:56

No, special relativity is a special domain of general relativity, hence the name. Special relativity is the limit of general relativity when gravitational effects become negligible. You can't have a theory that's completely correct but has a wrong limit (I use limit here not in the strict mathematical sense, but as like a "domain").

• I'm not sure I agree with the idea that "You can't have a theory that's completely correct but has a wrong limit". As an example, take a scalar field with a potential, $V(\phi)=-\frac{1}{2}\mu^2\phi^2+\frac{1}{4!}\lambda \phi^4$. This potential has two stable minima at $\phi=\pm \sqrt{6} m \lambda^{-1/2}$. In the limit $\lambda\rightarrow 0$, the minima are pushed to infinity, so the potential is unbounded below. Therefore the theory is unstable. I am not sure if this limit is "wrong", but the limit leads to an unphysical theory (if you take the theory "literally" and not an effective theory) Mar 13, 2021 at 3:18

No, SR (Special Relativity) is an intrinsic essential component of GR (General Relativity).

GR says that in perfectly flat spacetime mechanics behaves exactly according to the rules of SR as the answer of enumaris states.

GR also says that stress-energy causes spacetime curvature, but if you chop a region of curved spacetime into sufficiently small subregions then the curvature in each subregion becomes negligible and we can then use SR inside the subregion. To be more precise, as the size of a subregion approaches zero, so does its curvature, and so the rules of SR are valid locally.

This is similar to how we can chop the curved surface of the Earth into a set of flat maps which we can put into an atlas. Each page of the atlas tells us what's in that region, what the scale of that page is, and how that page connects to the other pages in the atlas.

So, roughly speaking, GR tells us how to construct an atlas of curved spacetime so that we can use SR locally on any atlas page. GR gives us a lot of freedom in how we chop up spacetime, and in what coordinate systems we use for our atlas pages.

(Unfortunately, this gets very difficult to do exactly when the spacetime curvature isn't simple or symmetrical: even the general two body problem in GR has no analytic solution).

TL;DR: GR says that SR is always valid locally, but in regions of high curvature the locale may be very small.

• I would be careful with the statement "as the size of a subregion approaches zero, so does its curvature..." It's a bit imprecise. It sounds like it's suggesting that curvature is zero at a point when in fact the Riemann curvature tensor (or w/e other curvature tensors) are tensors which are defined for all points on a manifold (indeed, the limiting to a point is intrinsic to the Riemann curvature's very definition). It's like if you zoom in on a 1-D curve infinitely it looks flat, but its second derivatives exist a each point and it's still curved. Apr 20, 2018 at 15:57
• @enumaris Very good point. My answer presumes that the reader has some familiarity with limit processes and calculus, and so they know that when you zoom into a well-behaved curve it starts to look flat. Apr 20, 2018 at 18:46

I think this question depends a bit on what you mean. My general philosophy is summarized in Isaac Asimov's essay "The Relativity of Wrong" (The Skeptical Inquirer, Fall 1989, Vol. 14, No. 1, Pp. 35-44, web link):

... when people thought the earth was flat, they were wrong. When people thought the earth was spherical, they were wrong. But if you think that thinking the earth is spherical is just as wrong as thinking the earth is flat, then your view is wronger than both of them put together.

Let me explain what I mean.

General Relativity is defined via Einstein's equations $$$$G_{\mu\nu} + \Lambda g_{\mu\nu} = \frac{8 \pi G_N}{c^4} T_{\mu\nu}$$$$ where $$G_{\mu\nu}$$ is the Einstein tensor, $$g_{\mu\nu}$$ is the spacetime metric, $$T_{\mu\nu}$$ is the stress-energy tensor, $$G_N$$ is Newton's constant, $$c$$ is the speed of light, and $$\Lambda$$ is the cosmological constant.

We can consider a Universe with no stress-energy for any matter ($$T_{\mu\nu}=0$$), and with no cosmological constant, $$\Lambda=0$$. In this case, Einstein's equations are $$G_{\mu\nu}=0$$, which can solved by $$g_{\mu\nu}=\eta_{\mu\nu}$$, where $$\eta_{\mu\nu}$$ is a $$4\times 4$$ tensor with components $$(-c^2,1,1,1)$$ on the diagonal and zero elsewhere.$$^\star$$

Now... one quite strict definition of "special relativity" would be that the spacetime must have translations, rotations, and Lorentz boosts as exact symmetries. This is the Universe taught in special relativity courses; the twin paradox assumes such a Universe for example. In this strict sense, special relativity is a solution of Einstein's equations, and is therefore necessarily contained inside of general relativity.

However, you might say it is pretty bizarre physically to consider a Universe with no matter at all. Furthermore we now know the Universe does have a non-zero cosmological constant (or at least behaves in a way that is indistinguishable from a non-zero cosmological constant as far as we can tell) -- and special relativity in the strict sense (meaning $$g_{\mu\nu}=\eta_{\mu\nu}$$ is a solution) is actually not an exact solution of Einstein's equations for non-zero $$\Lambda$$. So in a certain sense, special relativity (in the strict sense defined above) is not correct for our Universe, since $$\Lambda$$ is non-zero.

Nevertheless, there is yet another and deep sense in which special relativity is contained in general relativity. Locally around any spacetime point, we can always choose normal coordinates $$$$g_{\mu\nu}(x) = \eta_{\mu\nu} - \frac{1}{3} R_{\mu\sigma\nu\tau}(0) x^\sigma x^\tau + O(|x|^3)$$$$ This is in fact a statement about geometry, and has nothing to do with Einstein's equations. What we need here is to assume spacetime geometry can be described as a Lorentzian manifold (loosely speaking, we assume have a smooth geometry of spacetime). General relativity certainly does make this assumption, and so at any spacetime point we can always locally describe the metric using the metric of special relativity. Therefore there is a sense in which special relativity is an extremely good approximation, for "small enough" distances, near any point for every solution of general relativity.

This way of looking at things is actually quite similar to the Asimov quote in the beginning. Here's how I would modify the quote so it applies to this situation, modifications in bold ("absolute space and absolute time" here is the idea of Newton that was overturned by special relativity):

When people thought that there was an absolute space and absolute time, they were wrong. When people thought that the Universe was described by special relativity, they were wrong. But if you think that thinking there is an absolute space and absolute time is just as wrong as thinking that special relativity is exactly correct, then your view is wronger than both of them put together.

$$^\star$$ In fact this is not the unique solution, since (a) you can do coordinate transformations but more importantly (b) you can have gravitational waves propagating -- we will ignore this.)

From a theoretical point of view, one may argue that it depends on how one defines the words GR & SR, and how abstract one wants to be. For starters one may generalize to other than 4 spacetime dimensions. Moreover, the local spacetime model could in principle be different from Minkowski spacetime. Instead of Einstein gravity, one may consider e.g. conformal gravity, SUGRA, Newton-Cartan gravity, etc.

• The terms GR and SR are fixed and not really up for debate. Generalising GR or SR to higher dimensions is still GR or SR unless dimensional reduction comes into play and you get Kaluza-Klien theories. Mar 13, 2021 at 5:36

As far as I know, and I'm not an expert or a physicist, Special Relativity developed the principle of Relativity which stipulates mainly two rules:

1. All events follow the same laws of physics in every frame of reference.
2. The independence of the speed of light, which means that light is always moving at a constant value (in the vacuum, at 299792458 m/s), even if one is in an inertial frame and another accelerating.

General Relativity also follows this principle, because the speed of light is always the same in vacuum for all observers, it doesn't change. This might be incorrect, but I think that also GR without SR it would not work. We can take the ISS as an example: We know that the gravity becomes weaker the further that you go from the source of the gravitational field; in our case, the ISS is further from the earth, our source of gravity, than us. Because of the time dilation due to gravity, the further that you go from the source, the faster that time passes, so we might guess that in the ISS the time for the astronauts goes faster than for us. But we need to remember that the ISS is moving, and is moving so fast, fast enough for the relativity effects to be non-negligible, and we know that the fastest and object is moving, the time dilation increase due to the Lorentz factor, and the time gets slower. So having this two assumptions, the SR effects are bigger in the ISS than the GR effects, so the time there gets slower than for us, and the astronauts in the ISS are getting older slowly.

I wouldn't have thought so.

One solution to the vacuum equations of GR without cosmological constant is Minkowski space, this is the space of Special Relativity. Thus we see that General Relativity generalises Special Relativity

And we also see why Einstein opted for the word 'General' when he named this theory - actually this is more to do with general covariance, the theory of which is category theory, at least in one of its senses, it's called functoriality there.

To have GR without SR would mean that we require some physical rule that rules out Special Relativity as a solution. We don't have such a physical rule.

Where gravity is weak the metric of spacetime is flat and SR holds.

For example,

As $$r \to \infty,$$

The Schwarzschild Metric of GR $$\to$$

The flat metric of SR in spherical coordinates:

$$ds^2 = c^2 dt^2 − dr^2 − r^2 dθ^2 − r^2 sin^2 θdφ^2$$

"GR without SR" $$\to$$ There is never no gravity or weak gravity.

No, there can't be general theory of relativity without its special part. Why, what is relative in general theory, length and time interval. From where relative space and time came, from relative speed. But in general theory, there is no more need of relative motion but absolute motion and gravity is just frame of reference as different coordinate frames.

Why there is need of spacetime, because physicists thought for a long that what provide interaction between distant bodies. Space is not vacuum for general theory of relativity to stand. And big bang provide force necessary for motion and expansion is part of it.