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The Einstein equations with a cosmological constant $\Lambda$ read as:

$R_{{\mu}{\nu}}-\dfrac{1}{2}Rg_{{\mu}{\nu}} + \Lambda g_{{\mu}{\nu}} =8\pi T_{{\mu}{\nu}}$

Therefore,

$R-\dfrac{D}{2}R+D\Lambda=8\pi T$

($D$ is the number of spacetime dimensions.)

Or,

$\Lambda = \dfrac{8\pi T}{D} + \bigg(\dfrac{D-2}{2D}\bigg)R$

Therefore,

$\Lambda = \bigg(\dfrac{D-2}{2D}\bigg)R_0$ where $R_0 := R_{vacuum}$

From the field equations, $R_{{\mu}{\nu}}-\dfrac{1}{2}Rg_{{\mu}{\nu}} + \bigg(\dfrac{D-2}{2D}\bigg)R_0 g_{{\mu}{\nu}} =8\pi T_{{\mu}{\nu}}$

Therefore, $8\pi T = \dfrac{D-2}{2} (R_0 - R)$

Now, in the large $D$ limit, the only way $T$ can be saved from diverging is to make $R$ approach $R_0$. This means that in the large $D$ limit, $R=R_{0}$ everywhere, i.e., the large $D$ limit of General Relativity admits only vacuum solutions.

I am posting this question to confirm if the conclusion I have reached is appropriate because I haven't come across any such claim elsewhere. Also, if this is appropriate then does it denote something interesting or more profound? (Put in other words, can it be associated with some rather known facts or principles?)

PS: I just noticed that in 2-dimensional case (1+1 dimensional case) also, only the vacuum solutions can exist and the cosmological constant also must vanish. (One can verify trivially by putting $D=2$ in the formula for $\Lambda$ and $T$ above.)

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  • $\begingroup$ So you are specifying the scalar curvature $\approx R_0$ but then making a claim about the $R_{\mu \nu}$? $\endgroup$ – AHusain Apr 5 '17 at 20:42
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    $\begingroup$ Well, $T$ being a trace, it's not unexpected that it may scale as $\sim D$, and so there is nothing wrong about it diverging. In other words, we do expect $T$ to diverge in the $D\to\infty$ limit, linearly in $D$. $\endgroup$ – AccidentalFourierTransform Apr 5 '17 at 20:52
  • $\begingroup$ @AHusain I know that one can't generically claim to have obtained $R_{{\mu}{\nu}}$ by only knowing $R$. But, here, since $R=R_{vacuum}$, I think it is safe to say that $R_{{\mu}{\nu}}=(R_{{\mu}{\nu}})_{vacuum}$. In the limiting case of $\Lambda=0$, one can see this pretty clearly as $R_{vacuum}=0$ in that case. $\endgroup$ – Dvij Mankad Apr 5 '17 at 20:52
  • $\begingroup$ @AccidentalFourierTransform But doesn't that create problems? For example, in the static case, it would imply that the energy density also diverges. $\endgroup$ – Dvij Mankad Apr 5 '17 at 20:56
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    $\begingroup$ Side note. You may be interested in arxiv.org/abs/1302.6382 and following papers by Emparan and co devoted to black objects in large D limit. $\endgroup$ – OON May 8 '17 at 13:16
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Yes, higher-dimension does change how matter sources gravity; one particular example is the peeling theorem (see Godazgar, Reall Peeling of the Weyl tensor and gravitational radiation in higher dimensions and references therein) - gravity falls off quicker with growing $D$. But a specific physical model is needed to turn these mathematical statements into any real physical conclusion.

Conclusions such as "only vacuum solutions are allowed in the $D\to \infty$ limit" are meaningless without proper physical context because non-vacuum solutions are certainly allowed for every member of the limiting sequence.

I can only imagine that your statement mean something like the following: "If we have these and these fixed matter densities (numbers of particles, temperatures,...), and these and these scalings of the fundamental constants (remember that both speed of light and Newton's gravitational constant are constants fixed phenomenologically by real physical phenomena and stay hidden in your geometrized units), the solutions will converge to vacuum solutions as $D \to \infty$."

However, even like that you cannot make such statements work without a specific matter model. For instance, the stress energy tensor of any massless particles such as photons is traceless. I.e. there are space-times with $R=R_0$ which are certainly not vacuum ones and the whole argument breaks down.

So, I will just quickly discuss what follows for $R-R_0$ for one particular model, the perfect fluid.


Consider the relativistic perfect fluid, for which the stress energy fluid in the comoving frame reads $$\begin{pmatrix} \varepsilon & 0 & \cdots & 0 \\ 0 & P & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & P \end{pmatrix}$$ The trace is frame-invariant, so we immediately see $T = \varepsilon - (D-1)P$. I.e., at least a part of the tensor scales with $D$, so the trace of the stress-energy tensor seems not to become negligible in the large $D$ limit.

Let's estimate energy/temperature by assuming we are dealing with an ideal gas. (I denote temperature by $\mathcal{T}$ so that it is not confused with trace of stress-energy.) The non-relativistic approximation for the ideal gas holds as long as $D k_B \mathcal{T} \ll m$ ($m$ is the mass of the particles of the gas) but our $D$ will be infinite so we actually can switch directly to the ultrarelativistic limit $D k_B \mathcal{T} \gg m$. for any finite temperature.

In this limit we have $\varepsilon \approx D n k_B\mathcal{T}, P = n k_B T$, where $n$ is the number density of particles. You then see that the trace is simply $T = n k_B\mathcal{T}$ and the thermal contribution to $R-R_0$ falls off as $1/D$.


However, the energy you need to heat the gas to this temperature scales with $D$. I.e., for fixed amount of thermal energy invested per particle and at fixed particle densities in unit volume, the contribution to $R-R_0$ falls of as $1/D^2$.

Consider another example, for a gas at zero temperature ("dust") we have $P=0$ and $\varepsilon = mn$. $mn$ is simply the rest-energy density and its contribution to $R-R_0$ falls off as $1/D$. That is, one can conclude that at least for an ideal gas and a fixed amount of energy per unit volume, $R-R_0$ will always fall off at least as quickly as $1/D$.

To be honest, it is hard to find intuitive arguments for this because it is certainly not that $R-R_0 \to 0$ necessarily means weakened gravity. This is because the full Einstein equations (with an appropriate scaling of the cosmological constant) can still couple to matter in very much the same way. $R-R_0 \to 0$ simply means that the character of gravity changes in large dimension to some sort of conformally coupled gravity. (No surprise here, we know that even in $D=5$ a lot of things change as compared to $D=4$.)

Also, I would still be very careful as to assessing that large-dimensional gravity makes $R-R_0$ vanish without referring to the framework you are working in. One reason is that, depending on how you relate your higher-dimensional gravity to reality, you might have a $\sim D$ factor built into the gravitational constant in front of the stress-energy tensor in the Einstein equations to agree with even the crudest Newtonian phenomenology on some (effective/integrated-out/privileged) 4D hypersurface representing our universe.

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    $\begingroup$ I would like to point out that the cosmological constant and the fine structure constant have dimensions that are dependant on $D$ ($c$ and $\hbar$ don't depend on $D$ and are hidden here. Also : $\Lambda \sim \mathrm{L}^{- 2}$ independently of $D$) : \begin{align}G &\sim \mathrm{L}^{D \,-\, 2}, \\ \alpha &\sim \mathrm{L}^{D \,-\, 4}.\end{align} Thus it's not surprising that these constants may get a $D$-factor "hidden" in them. This $D$-dependance always impressed me, and I don't know what kind of "metaphysics" we could deduce from that aspect. $\endgroup$ – Cham May 8 '17 at 23:35

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