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  Only vacuum is possible in the large $D$ limit of General Relativity?

+ 2 like - 0 dislike

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

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


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

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


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


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

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

$8\pi T =  \dfrac{D-2}{2} (R_0 - R)$

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.)

asked May 4, 2017 in Theoretical Physics by Dvij D.C. (40 points) [ no revision ]

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