While studying general relativity, I noticed that much of the theory could easily be generalized from a $(3,1)$-dimensional spacetime to an $(n,1)$-dimensional spacetime without any changes. So, is this the case for all results that we can derive in general relativity? Or are there results which depend crucially on spacetime having $3$ spatial dimensions?
2 Answers
I'll first discuss some quite physical differences, which might already be enough for your interests. Afterwards, I'll also mention how mathematically things can be sort of unique in four dimensions.
Physical Predictions
There are results that depend on the number of spatial dimensions. For example, the Riemann tensor has $\frac{d^2 (d^2-1)}{12}$ independent components in $d=n+1$ spatial dimensions. This implies that
- in $3+1$ dimensions, it has $20$ independent components. $10$ in the Ricci tensor, $10$ in the Weyl tensor;
- in $2+1$ dimensions, it has $6$ independent components. All of them are in the Ricci tensor, since the Weyl tensor vanishes on manifolds of dimension less than $4$;
- in $1+1$ dimensions, it has a single independent component. It corresponds to the Ricci scalar.
The Einstein equations read $$R_{ab} - \frac{1}{2}Rg_{ab} = 8 \pi T_{ab}.$$ If we contract with the metric on both sides, we see that \begin{align} g^{ab}R_{ab} - \frac{1}{2}Rg^{ab} g_{ab} &= 8 \pi g^{ab}T_{ab}, \\ R - \frac{d}{2}R &= 8 \pi T, \\ \left(1 - \frac{d}{2}\right)R &= 8 \pi T, \\ R &= \frac{16 \pi}{2 - d} T, \\ \end{align} where I defined $T = g^{ab}T_{ab}$ and used the fact that $g^{ab}g_{ab} = d$ in $d$ dimensions. Notice that, for $d=2=1+1$, the last step is a division by zero, so I'll treat it separately. As for the other dimensions, a consequence of this expression is that we can rewrite the Einstein equations as $$R_{ab} = 8\pi\left(T_{ab} + \frac{T}{2-d} g_{ab}\right).$$
Let us consider what happens in vacuum: $T_{ab} = 0$. In all dimensions but $d=2$, we get $R_{ab} = 0$. If $d=4$, we can still have curvature because only $10$ components of the Riemann tensor are on the Ricci tensor, so there can be non-vanishing curvature. This is what allows the Earth to orbit the Sun: even though the Earth is in vacuum, spacetime is still curved.
In $d=3$, things are not like that. The Einstein equations are telling us that the Ricci tensor vanishes, but all of the Riemann tensor's independent components are in the Ricci tensor. Hence, in $d=3$, there is no curvature in vacuum. Gravity is no longer long-range in this sense, it is only present where matter is present.
Let us now consider $d=2$. In this case, we actually have that $R_{ab} = \frac{1}{2}R g_{ab}$ always. The Einstein tensor always vanishes. Hence, the Einstein equations actually only imply that the stress-energy tensor must vanish, which is wildly different from what we get in other dimensions.
In short, the dimensionality of spacetime can have quite some impact on the physical predictions of General Relativity. Padmanabhan's Gravitation: Foundations and Frontiers has a chapter dedicated to gravity in other numbers of dimensions which might be of your interest. While I only mentioned gravity in lower dimensions, it also covers gravity in higher dimensions.
Four Dimensions are Hard (Topologically Speaking)
Differential topology is quite difficult in four dimensions. For example, if your space if topologically equal (i.e., homeomorphic) to $\mathbb{R}^n$, then it surely has the same differential structure as $\mathbb{R}^n$ (i.e., they are diffeomorphic). Unless $n=4$. In this case, there are infinitely many possible different differential structures, known as exotic $\mathbb{R}^4$'s. This is just one example of how things can get rough. In short, techniques that apply to large dimensions fail in $d=4$, and so do the techniques that apply to low dimensions. As this review by C. Manolescu puts it, "This makes it [$d=4$] the most challenging dimension to study".
I'm not an expert in $d=4$ topology and this paragraph pretty much used everything I know about it, but my point is that not only do physical predictions depend on the dimensionality of spacetime, but even more general properties of differential geometry depend highly on the dimensionality of the manifold and can get quite complicated in the particular case of $d=4$.
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$\begingroup$ Huh, didn't know differential geometry was so complicated in $d=4$. And to think that that is what we reside in. Such a coincidence. $\endgroup$ Commented Mar 21, 2022 at 21:12
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$\begingroup$ Also, are there Physical Predictions which differ between $d=4$ and $d>4$? That was actually what I wanted to ask, but I forgot that the situation was so different in $d=2, 3$. $\endgroup$ Commented Mar 21, 2022 at 21:14
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$\begingroup$ @IshanDeo I'm quite certain there are physical predictions that will differ in more dimensions, but I'm not really familiar with them. Notice the number of independent components of the Ricci tensor is $\frac{d^2 + d}{2}$, i.e., quadratic in $d$, while the number of components of the Riemann tensor is quartic. Hence, in large dimensions most components of curvature are due to the Weyl tensor, which does not occur on the Einstein equations. This suggests the vacuum solutions to be much richer $\endgroup$ Commented Mar 21, 2022 at 21:56
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$\begingroup$ In fact, Wikipedia says that in $d=5$ one can get "black rings", which would be analogues to black holes, but with horizons that are not necessarily homeomorphic to a sphere $\endgroup$ Commented Mar 21, 2022 at 21:58
Fun fact: The familiar 3+1 dimensions are the only number of dimensions in which stable planetary orbits exist.
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$\begingroup$ Oh, wow! Do you have any links you can recommend where I could find more information about this? $\endgroup$ Commented Mar 22, 2022 at 15:45
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2$\begingroup$ @JohnMadden See physics.stackexchange.com/q/50142/64170 $\endgroup$ Commented Mar 22, 2022 at 18:41
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1$\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ Commented Mar 25, 2022 at 11:02
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$\begingroup$ @JohnRennie How do you mean this does not provide an answer to the question? The OP asks for results in GR that depend uniquely spacetime having 3+1 dimensions. This answer gives one (for humankind very crucial) example of such a result. Its short nature is not in any way meant as a critique of the question. If this was put as comment, moderators would, rightly, be complaining that answers should not be put in comments. $\endgroup$– TimRiasCommented Mar 25, 2022 at 22:35