Derive Einstein's field equations in one spatial and one time dimensions?

Question

Can someone derive Einstein's general relativity field equation for one spatial dimension and one time dimension from the beginning? I think this will help many beginners to get a feel and understand details and effects of spacetime curvature easily.

I am expecting an explanation starting from equivalence principle or anything related. Something like this https://www.youtube.com/watch?v=pES_tNZJm3Q but simplified to a single spatial dimension for a beginner to be able to mess with and get a feel of the model.

For example, I want to be able to describe things like, If this is the space time, what would happen if I add a big mass at (0,8). What would happen if I add 2 big masses at (0,6) and (0,10). How would the space time curve. What are the world lines of these objects. How do they change with the mass and their position.

Answer 1

The Einstein field equations may be derived using the action principle from the action,

$$S = \frac{1}{2\kappa^2}\int d^Dx \, \sqrt{|g|} \, \mathcal R,$$

potentially supplemented by a cosmological constant term, or a matter Lagrangian with other fields if coupling gravity to another theory. The Einstein field equations follow from the variation with respect to $g^{\mu\nu}$ and at no point does one assume $D=4$, so the derivation for $D= 2$ is the same.

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The Atiyah-Singer index theorem applied to the de Rham complex for a manifold $\mathcal M$ reads,

$$\chi(\mathcal M) = \int_{\mathcal M} e(T\mathcal M)$$

where $\chi$ is the Euler characteristic, a topological invariant and $e(T\mathcal M)$ is the Euler class of the tangent bundle of $\mathcal M$. In $D= 2$, this integral reduces to the Einstein-Hilbert action, up to constants and thus gravity in $D=2$ is classically purely topological.

Since $S$ becomes topological, $\frac{\delta S}{\delta g^{\mu\nu} }=0$ which implies stress-energy $T_{\mu\nu} = 0$ vanishes. Solutions are manifolds, of varying genus, otherwise they are seen as the same system by the action, due to the homeomorphism invariance.

Answer 2

Due to various topological constraints, the Einstein field equation in 2 dimensions are almost trivial. That is, any matter content will go with any metric, except if you decide to add a cosmological term, in which case the equations become

$$\Lambda g_{\mu\nu} = T_{\mu\nu}$$

This is due to the Hilbert action being a constant term due to the Gauss Bonnet theorem. $$S_H = \int dt dx R(x,t) = C$$

Its variation will therefore always be 0 no matter the metric. The addition of a cosmological term will give you the above equation.

Answer 3

As explained in previous answers the Einstein--Hilbert (EH) action is topological in two dimensions $$ \int d^2 x \sqrt{g}\, R = 4\pi\chi $$ where $\chi$ is the Euler characteristics. Equivalently this can be seen by the fact that the symmetries of the Ricci tensor $R_{\mu\nu}$ imply that $$ R_{\mu\nu} = \frac{R}{2}\, g_{\mu\nu} $$ and thus that the Einstein tensor is identically zero $$ G_{\mu\nu} = R_{\mu\nu} - \frac{R}{2}\, g_{\mu\nu} \equiv 0. $$ Another explanation is that the action is invariant under both diffeomorphisms (fixing two components of the metric) and Weyl symmetry (fixing the last component). So all the usual arguments used to infer the Einstein equations in $D = 4$ dimensions (or any $D > 2$, for what matters) cannot be applied in two dimensions.

There has been some attempts to recover the Einstein equation in vacuum but this involves some baroque constructions (see for example Teitelboim '83). Other researchers have preferred to generalize the action by introducing a dilaton (see for example hep-th/9204002 or gr-qc/9309018) or by searching what is the natural generalization of the EH action in 2d (it has been shown that a properly defined limit $\epsilon \to 0$ of $D = 2 + \epsilon$ yields the Liouville action, for example look at hep-th/9303123).

Finally note that $2d$ gravity presents a lot of pathologies. First if one considers that there is just a cosmological constant $$ S_\mu = \mu \int d^2 x \sqrt{g} $$ besides the EH term then the equation of motion reduces to $$ \mu = 0 $$ which has no solution. The same result is found if the Lagrangian contains matter that is invariant under the Weyl symmetry. Finally if one considers matter not invariant under the Weyl symmetry one can still show that many models have no dynamics or even don't exist (this is the topic of a recent paper I have written).

You asked for a simple and intuitive derivations and thus the above comments may look complicated, but I think it is necessary to explain why considering gravity in two dimensions is not a good idea.