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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, enter image description here 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.

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    $\begingroup$ Derive them from what? $\endgroup$
    – ACuriousMind
    Jan 9, 2017 at 16:49
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    $\begingroup$ related: General relativity (gravitation) in time and one spatial dimension $\endgroup$ Jan 9, 2017 at 16:49
  • $\begingroup$ I can't immediately see how changing the standard textbook approach to GR is going to make it any easier to understand. But I can easily imagine that, no offence intended, your suggestion would introduce unnecessary complications. $\endgroup$
    – user140606
    Jan 9, 2017 at 16:55
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    $\begingroup$ If you are trying to get an intuitive picture of the central principle of GR, that is the curvature of spacetime, is it any easier with one dimension than four? $\endgroup$
    – user140606
    Jan 9, 2017 at 17:03
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    $\begingroup$ Indeed, as the answers show, the 1+1 case is a little weird, even though "trivial". I think this kind of subtlety is best met when after you have been messing with the more typical situation for a while. $\endgroup$ Jan 12, 2017 at 13:23

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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.


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.

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    $\begingroup$ I think the last paragraph is a little unclear. It's the fact that $\delta S / \delta g_{\mu \nu} = 0$ that the Einstein equation reduces simply to $0 = T_{\mu \nu}$. There are no more field equations. If one goes through the motions of explicitly varying $S$ with respect to the metric, giving the Einstein tensor, then by comparing with the topological form of the action we find that the Einstein tensor, and hence the Ricci tensor, vanishes. But this argument proves that this holds for all 2D metrics --- stating $R_{\mu \nu} = 0$ as a field equation implies otherwise. $\endgroup$
    – gj255
    Jan 10, 2017 at 1:06
  • $\begingroup$ It is not correct to say that $R_{\mu\nu} = 0$ is the equation of motion since the latter is equal to $R/2\, g_{\mu\nu}$ and since there is no equation (in the absence of matter) the metric, and thus $R$, is arbitrary. $\endgroup$
    – Harold
    Jan 12, 2017 at 14:00
  • $\begingroup$ @Harold See updated answer. $\endgroup$
    – JamalS
    Jan 12, 2017 at 14:10
  • $\begingroup$ The version of AS I am familiar with assumes $\mathcal M$ is compact...interesting spacetimes (in GR) are almost never compact. What conclusions can we make in $D=2$ for $\mathcal M$ noncompact? Furthermore, the only manifold in $D=2$ that can be a spacetime is the torus; it does not make sense to say "varying genus." $\endgroup$
    – Ryan Unger
    Jan 13, 2017 at 13:36
  • $\begingroup$ @0celo7 The Gauss-Bonnet theorem has been shown to hold for any even-dimensional, non-compact hyperbolic manifold of finite volume, and has been shown more generally to hold for non-compact manifolds that are finitely connected with finite volume, as well as quotients of symmetric spaces by arithmetic groups. There are yet further extensions of the Gauss-Bonnet theorem, and presumably I would think this allows one to definitively say two-dimensional gravity is topological, though I will try to find a reference for that. $\endgroup$
    – JamalS
    Jan 13, 2017 at 14:11
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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.

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  • $\begingroup$ "Any matter content will go with any metric" --- I agree that any metric is permitted, but (assuming no cosmological constant) the field equations then give you $T_{\mu \nu} = 0$, no? $\endgroup$
    – gj255
    Jan 9, 2017 at 17:05
  • $\begingroup$ No, the matter content and the metric are completely decoupled, due to this. $\endgroup$
    – Slereah
    Jan 9, 2017 at 17:12
  • $\begingroup$ If I vary the action with respect to the metric, so as to reproduce Einstein's equation, the variation of $S_H$ is zero whilst the variation of the matter action ought to give me $T_{\mu \nu}$, as it usually does. So demanding that my action be invariant under perturbations of the metric gives me the Einstein equation, reading $T_{\mu \nu} = 0$, no? $\endgroup$
    – gj255
    Jan 10, 2017 at 0:58
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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.

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  • $\begingroup$ In the example case of '2 masses on space axis' in the question, I could apply Newton's law of universal gravitation and say they should accelerate towards each other because of the gravitational force. But, as the answers suggest, I cannot apply GR to realise the above 1+1 d case with greater detail and precision? $\endgroup$
    – Makmeksum
    Jan 13, 2017 at 17:08

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