I was thinking of a universe that has 3 spatial dimensions and one temporal dimension, and in which the spacetime around massive objects is curved. However the geometry of spacetime near a massive body is different from the geometry described by the General Relativity of our universe. When spacetime curvature is negligible, you can use special relativity, just as you would use special relativity in our universe. Also position invariance applies to this universe, meaning that the laws of physics of this universe are the same for all positions. And direction invariance applies to this universe as well, meaning that the laws of physics apply to all directions.

Could this universe be self consistent?

  • 1
    $\begingroup$ Given that there are many paradoxes about our own universe that cause one to question if OUR universe is self consistent, I am not sure what the issue is about your universe being self consistent? $\endgroup$ – Justin Thyme the Second Jun 2 at 1:43
  • 1
    $\begingroup$ Historically physicists have modeled many different space/time curvature configurations You can chose any you want - its your Universe $\endgroup$ – Mon Jun 2 at 1:54
  • $\begingroup$ Special relativity has nothing to do with spacetime curvature $\endgroup$ – L.Dutch - Reinstate Monica Jun 2 at 4:22
  • $\begingroup$ What is the role of temporal dimension? You can introduce as many dimensions as you desire which I think some string theorists may have gotten too excited, but no general relativity being mentioned. $\endgroup$ – user6760 Jun 2 at 8:17
  • 3
    $\begingroup$ @L.Dutch-ReinstateMonica Yes it does, or at least in does in the sense that spacetime curvature affects how accurate special relativity is as an approximation to general relativity, which is what the question is saying. To be more precise about it, if our spacetime manifold has zero curvature everywhere, then we can find a global reference frame covering the whole manifold that has a minkowski metric. If there is curvature, we can only do so locally. $\endgroup$ – el duderino Jun 2 at 15:58

Yes, Absolutely!

Not only can you do such a thing, but in fact boatloads of alternate metric theories of gravity have been proposed both before and after Einstein came out with GR. For instance, the first metric theory of gravity was actually due to Nordström in 1913. Let's go over what exactly the equations governing GR look like, and then we can dig into a few alternate theories that might be similar to what you're looking for.

Review of General Relativity

GR certainly has a formidable reputation, but hopefully I can give names to enough of the most fundamental concepts that you can follow along with my discussion. Now, the key equation that describes how spacetime evolves in GR is the Einstein Field Equation:

$$R_{\mu \nu} - \frac{1}{2}g_{\mu \nu}R = 8\pi T_{\mu \nu} - \Lambda g_{\mu \nu}$$


  • $g_{\mu \nu}$ is the metric of spacetime, which encodes information about how distances and orientations change as you travel through spacetime. The $\mu$ and $\nu$ are integer indices ranging from 0 to 3-- one for each spatial/temporal dimension. So this object actually has 16 entries, although it turns out most of them aren't independent. When the indices are in the upper position, it means you're talking about the inverse of the metric in a matrix inverse sense.
  • $R_{\mu \nu}$ is the Ricci curvature tensor, which contains information about curvature of spacetime. More specifically, it is the trace of the more general Riemann tensor $R_{\alpha \beta \gamma \delta}$: $R_{\alpha \beta} = R^\mu_{\alpha \mu \beta} = R^0_{\alpha 0 \beta} + R^1_{\alpha 1 \beta} + R^2_{\alpha 2 \beta} + R^3_{\alpha 3 \beta}$. The Riemann tensor contains all information about spacetime curvature, and it's more or less the second derivative of the metric. Note the convention that repeated indices are summed over. The indices in the upper position are raised by the metric: $R^\alpha_{\beta \gamma \delta} = g^{\alpha \mu} R_{\mu \beta \gamma \delta}$.
  • $R$ is the Ricci scalar, which is the trace of the Ricci tensor: $R=R^\alpha_\alpha$. Basically, R is the simplest scalar that gives information about spacetime curvature, a fact we will come back to several times.
  • $T_{\mu \nu}$ is the stress energy tensor, which tells you about how matter and energy is distributed through space.
  • $\Lambda$ is the cosmological constant and can be thought of as a negative energy density permeating all of space.

I know that's a lot of stuff to remember if you haven't seen this before, but even having a cursory understanding of what these symbols mean makes it possible to have a much more meaningful discussion about alternate theories of gravity. The main takeaways from this are:

  1. The left side of the equation tells you about the curvature of space and the right side tells you about energy and matter in space, which is why we say that matter curves space.
  2. The fields governing gravity, namely $g_{\mu \nu}$ and by extension $R_{\mu \nu}$, are real tensors with two indices. This is what people mean when they say gravity is a spin 2 theory and is a large portion of why we have so much trouble integrating it with quantum mechanics. The renormalization procedures used in QM don't play nicely with spin 2 theories.

Nordström's Theory of Gravity

As I said before, Nordström beat Einstein to the punch by 2 years in developing the first metric theory of gravity. In contrast to the Einstein field equations, Nordströms theory was governed by the two equations (using units where $G=c=1$):

$$R=24 \pi T $$ $$ C_{\alpha \beta \gamma \delta} =0$$

Here, $T$ is the trace of the stress energy tensor, and $C_{\alpha \beta \gamma \delta}$ is the Weyl tensor, which is essentially the traceless part of the Riemann tensor that isn't captured by the Ricci tensor. The second condition is know as conformal flatness and is the same thing as requiring that the metric can be written as

$$g_{\mu\nu} = \phi^2 \eta_{\mu \nu}$$

Where $\eta_{\mu \nu}$ is the metric from special relativity that we all know and love, and $\phi$ is a non-zero scalar function over all of spacetime.

Coming from the Einstein field equations, we can see how nice these equations look-- there are way fewer variables that show up, and the ones that do are all scalars. In this theory, gravity would be a spin 0 force, which are easier to integrate into QM. It also satisfies your conditions perfectly-- it reduces to special relativity in an empty universe and to Newtonian gravity in the weak field limit, and properly predicts gravitational redshift.

So if it's such a great theory, why don't we all sarcastically call people Nordström when they're acting stupid? Well, unfortunately it just doesn't describe the universe that we inhabit. It predicts that there should be no gravitational lensing, and that orbits should precess in the wrong direction at the wrong speed, among other inaccuracies. However, it's a completely consistent mathematical theory that reduces to what we expect to see for classical phenomena, so it would be a great theory to use to describe a fictional universe you're building.

Another Approach for Alternate Metric Theories of Gravity

There is another way to come up with what you're looking for, if you want more options. Similar to particle physics and classical mechanics, instead of starting with the field equations we can start with a mathematical expression called the Lagrangian and and derive the field equations from it. I'm not gonna get into the details of how this is done, but all you really need to know is that such a thing exists and completely determines how the theory behaves. Now, for general relativity, the Lagrangian is

$$\mathcal{L} = \frac{1}{16\pi}(R-2\Lambda) + \mathcal{L_M}$$

Where $\mathcal{L}_M$ is the Lagrangian describing other matter and radiation fields. So we can see that from this perspective, General relativity is the simplest metric theory of gravity we can create, since it's action in an empty universe is just the Ricci scalar $R$, which is the simplest scalar related to spacetime curvature that we can construct.

However, we can construct other Lagrangians and look at the theory they create-- our only real requirement is that the Lagrangian be a scalar (ie be made out of tensor/scalar quantities and have no unpaired indices) so that the field equations work for any reference frame. So for instance, we can explore Lagrangians like

$$\mathcal{L} = \frac{1}{16\pi}(R+a\nabla^\mu R \nabla_\mu R -2\Lambda) + \mathcal{L_M}$$


$$\mathcal{L} = \frac{1}{16\pi}(R+bR_{\alpha\beta\gamma\delta}R^{\alpha\beta\gamma\delta} -2\Lambda) + \mathcal{L_M}$$

where $a$ and $b$ are coupling constants and $\nabla_{\mu}$ is a tensorial operator known as the covariant derivative. I personally don't know too much about the consequences of these theories aside from that as $a,b \rightarrow 0$ they become GR, but there has been papers published on them before so that might be a good place to look if you can manage to parse them.

| improve this answer | |
  • $\begingroup$ I absolutely love the 'cooks fudge factor variable constant' concept introduced into the last two equations - the 'coupling constants'. Basically, you can put any term into an equation you want, provided that as it approaches zero, it completely self-eliminates itself. Assume a term, any term, and then zero it out. $\endgroup$ – Justin Thyme the Second Jun 3 at 1:16
  • $\begingroup$ So would Nordström's Theory of Gravity, be related to the Lagrangian you showed, or is it independent of the Lagrangian you gave? $\endgroup$ – Anders Gustafson Jun 3 at 10:35
  • $\begingroup$ @AndersGustafson Nordstrom's theory of gravity can be derived from a Lagrangian, but it doesn't have the same form as the ones I showed. This is because the ones I showed assume our degrees of freedom are the independent components of $g_{\mu\nu}$, and that all these components can be varied independently. This isn't true for Nordstrom's gravity, since the off diagonal components of $g$ are forced to be zero and all the diagonal components are proportional to each other. So when we write the Lagrangian, we must take this into account (which naturally leads to writing it in terms of $\phi$). $\endgroup$ – el duderino Jun 3 at 14:38

Well you would use general relativity if you really wanted to. However even in a curved universe in the really tiny parts of space and time there would be no curvature even though the whole universe is curved. So special relativity could be used as long it is not used over long distances.

About the universe being self-consistent, well sort but sort of not. The laws of physics will be the same,yes, however the way you apply the laws of physics will change depending on where you are in space. So in shorter regions of space you would use special relativity but over long distances or stretches of space in your universe you would use general relativity.

| improve this answer | |
  • $\begingroup$ '...really tiny parts of space and time there would be no curvature...' So at the quantum level, is there a curvature? And therefore in entanglement, is there any curvature? At both 'ends' of a Planck unit, can there be curvature? Can a Planck unit be curved? How can two points define a curve if there is no possibility of another point in the middle? At the quantum level, doesn't GR=SR universally? At what scale can the universe 'curve'? $\endgroup$ – Justin Thyme the Second Jun 3 at 14:39
  • $\begingroup$ Well by small I do not mean quantum. I actually mean like few light years or slightly less. Space time would not be curved and this is if the object is away from a massive object. Well whenever space time is curved Planck lengths are curved because Planck lengths make up space. There is a point in the middle but it is as if it exists everywhere in the Planck length. In the quantum world GR and SR break down. Well technically the universe does curve in the Planck length but for you to do that you have to deal with scales around a few light years and this too has to be far away from other mass. $\endgroup$ – Roghan Arun Jun 3 at 17:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.