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In two spacetime dimensions the Einstein tensor is identically zero. Therefore if we consider the EFE with a non-zero cosmological constant and a stress-energy tensor as: $$\Lambda g_{\mu\nu} =\alpha T_{\mu\nu}.$$ Then how can we calculate the stress energy tensor for the hyperbolic metric, $$g_{\mu\nu}=\frac{d\mu^{2}+2d\sigma{^2}}{\sigma^2}.$$

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  • $\begingroup$ a small correction: the second expression is of square of the line element $ds^{2}$. The coefficient of the differential in the RHS are the components of the metric tensor. $\endgroup$ Dec 31, 2021 at 12:38

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You pretty much solved the problem already. Just divide by $\alpha$ on both sides and the equation you wrote yields $$T_{\mu\nu}\text{d}x^\mu \text{d}x^\nu = \frac{\Lambda}{\alpha \sigma^2} (\text{d}\mu^2 + 2\text{d}\sigma^2).$$

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  • $\begingroup$ Thanks a lot. Can you please suggest any Lagrangian to derive this stress energy tensor. In my previous post I describe a Lagrangian which generate this S-E tensor but it appears to me as fictitious. $\endgroup$ Dec 31, 2021 at 12:14
  • $\begingroup$ @DebasisMazumdar I can't think of any straightforward way of seeing the Lagrangian. However, I also don't really know whether there is a manner of obtaining a "non-fictitious" Lagrangian for this scenario, since two-dimensional gravity usually doesn't involve any matter fields at all. In fact, I'm not even sure of whether 2D gravity has any applications instead of being just a theoretical laboratory, and if it doesn't then naturally there aren't any "non-fictitious" possibilites $\endgroup$ Dec 31, 2021 at 18:03
  • $\begingroup$ @DebasisMazumdar I just noticed that that your metric isn't even Lorentzian, and hence it certainly can't describe spacetime. $\endgroup$ Jan 1, 2022 at 5:53

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