# Is general relativity only applicable where the curvature is small?

In Eddington Finkelstein Coordinates: Lecture By Emil Akhmedov at 7:10 (transcript below the video) the instructor mentions in passing that on general grounds, Einstein's theory of general relativity can only be expected to apply where the curvature is small, but he doesn't go into further detail.

Why is that? Is there a physical reason why we should expect that?

EDIT What he says is this (slightly adapted from the transcript):

... this tensor measures the strength of tidal forces, and they become enormous as $r$ goes to zero. That's the reason that this metric is applicable only beyond this point. Moreover, one can say that Einstein's general theory of relativity is not applicable as $r$ goes to zero anymore, because higher terms and powers of curvature are becoming relevant. So the curvature becomes strong. And Einstein's theory on general grounds can be expected to be applicable only if curvature is small, sufficiently small.

He seems to be saying that the Schwarzschild geometry ceases to be applicable for $r\to 0$ not only because $R_{\mu\nu\alpha\beta}R^{\mu\nu\alpha\beta} \to \infty$, but also because more generally the theory is only applicable when the curvature is small. The fact that he speaks about higher powers of the curvature seems to indicate that general relativity is just a first order approximation of a more general theory.

EDIT 2

I asked on the course forum, where I only got two views but one of them was the lecturer's. Essentially what he was alluding to was what @gj255 already said in the comments:

It is an assumption, if you think about it more carefully. We have assumed the simplest invariant - $R$ - as the Lagrangian density, but why not $R^2$ e.g.? If you look at this, you will realize that $R^2$, or even higher powers of Ricci scalar or tensor or Riemann tensor, become more relevant, as $r\to 0$, because then curvature tends to infinity...

To give some context: in the derivation of the Einstein equations in this course, an invariant Lagrangian density depending on the metric was constructed. The simplest possible term, $\sqrt{|g|}$ cannot account for any dynamics. Adding the term $\sqrt{|g|}R$ we get dynamical equations of motion and the relative coefficient between these terms is the cosmological constant. This is where he stopped, but apparently there is no a reason why we should, and apparently we should expect non-zero higher order contributions.

• In the context given, the comment does not make too much sense. GR is only an approximation once we include quantum effects, but he doesn't seem to be talking about that. Once we include "quantum effects" (whatever that means), then we have $$\text{large curvature}\leftrightarrow\text{large energies}\leftrightarrow\text{quantum effects}.$$ GR cannot describe the singularity of the black hole because that's not even a part of the spacetime manifold. (There are ways of including it, but strictly speaking that's not GR.) – Ryan Unger Jun 16 '17 at 20:04
• Minor comment to the post (v1): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. Also Emil deserves the credit. – Qmechanic Jun 16 '17 at 20:20
• I don't agree with the comment in the video that G.R. is only applicable when the curvature is small. In very rough terms, if "curvature" is small, I..e, negligible, then, you don't have G.R. at all, your spacetime is Minkowski, and you are in the realm of special relativity. What I think he meant, and it is sort of clear, is that he is computing the Kretschmann scalar, $K = R^{abcd}R_{abcd}$, which for Schwarzschild is $~1/r^6$, so, G.R. breaks down when $r \to 0$. This was my reading of it. – Dr. Ikjyot Singh Kohli Jun 16 '17 at 20:42
• @IkjyotSinghKohli I don't think he means to say "negligible", but maybe that its square is negligible. I edited to include his text, in which he seems to say that there is a reason in addition to the Kretschmann scalar diverging. – doetoe Jun 16 '17 at 21:35
• I think the point the lecturer was trying to get across is that the Einstein-Hilbert action, $\mathcal{L} = R$, is in some sense the lowest-order gravitational action you can write down. Just as Hooke's Law, $F = k x$, is only the first term in a Taylor expansion of the force-extension relation of an elastic medium, and breaks down for large forces, so we might expect that the Einstein-Hilbert action is corrected by other terms, such as $R_{\mu \nu \rho \sigma}R^{\mu \nu \rho \sigma}$, at large curvatures. This has nothing to do with quantum mechanics. – gj255 Jun 16 '17 at 23:52

Maybe you should for a moment forget the narrative of the lecture and consider the following:

Einstein constructed relativity as a more or less "unique" theory of gravity. However, this theory has been robustly proven to exhibit singularities such as the one you see at $r=0$ in the Schwarzschild black hole. Many people don't take singularities seriously enough; when a singularity occurs, the theory has no prediction whatsoever for the behaviour at that point. In other words, the "one and only" Einstein's theory of gravity tells you that it cannot, even in principle, apply in the curvature singularity in the center of the black hole and is thus necessarily an incomplete theory.

This is an extremely clear statement on which the vast majority of theorists agree. What they don't agree on, though, is any kind of resolution or even the bare framing in which an alternate theory should be developed.

Let me sketch what kind of problems we have even with framing how Einstein's gravity should be changed. The curvature tensor $R^{\mu}_{\;\nu\kappa\lambda}$ can be understood as a $4 \times 4\times 4\times 4$ table of numbers (of dimension length$^{-2}$) and some of them blow up near the singularity in a black hole. In principle you could try to define a transition to some modification of relativity when any of the components of $R^{\mu}_{\;\nu\kappa\lambda}$ grow beyond a certain number, lets say one over Planck length squared $1/l_p^2$.

But this is not a coordinate-covariant statement. You could make a funny coordinate transformation even in a weak gravitational field and the components of $R^{\mu}_{\;\nu\kappa\lambda}$ may look large. Or you can make coordinate transform near the black hole singularity and make $R^{\mu}_{\;\nu\kappa\lambda}$ components arbitrarily small when arbitrarily close to the singularity.

What relativists usually do is that they show that coordinate invariants such as the Kretschmann scalar $K=R^{\mu\nu\kappa \lambda} R_{\mu \nu \kappa \lambda}$ blow up. And that truly is a valid, coordinate-independent criterion. But when you go into more complicated cases such as the spinning black hole, you will find e.g. the Kretschmann invariant to be zero arbitrarily close to the central singularity (see e.g. Cherubini et al. 2003). This is because $R^{\mu\nu\kappa\lambda}R_{\mu\nu\kappa\lambda}$ has the character of $F^{\mu\nu}F_{\mu\nu} = E^2 - B^2$ in electromagnetism and becomes zero at points where the gravitomagnetic effects balance the gravitoelectric ones. When you dig around some more you find out that there is really no true "scalar strength" of Einstein's gravity, very much in analogy to the fact that there is no "scalar strength" of the electromagnetic field.

So, there is no easy way to classify what exactly do we expect from the corrections to Einstein's gravity, and this is also the part of the conundrum of contemporary theoretical physics. In other words, the phrase about "small curvature" or "corrections at higher curvature" will be mentioned in many relativity lectures but it will never be given any concrete development because it does not have one, at least not one supported by a wide consensus.

• Good point, smallness only is a meaningful concept when we are dealing with coordinate independent quantities, like $R$ or $K$. – doetoe Jun 17 '17 at 10:15

The quotation is misleading, because "small" is meaningless unless you specify "small relative to what." He really means that general relativity can only be a good approximate description when the curvature is finite, i.e. bounded above. The curvature becomes arbitrarily large near a singularity, so we expect any quantum effects and/or higher-derivative corrections to the Einstein-Hilbert action to kick in near there, regardless of how small the length threshold is where they become significant.

• Strictly speaking you are right, but as I am sure you are aware, this is a very common manner of speech, even in mathematics. Usually it means either "sufficiently small", or "equal to first order". In neither case there is any absolute meaning attached to it. It could be that indeed he does mean that it only works when the curvature is finite, but from the way he says it seems that he says that there is an additional reason. Maybe the second part of your answer, that there must exist a quantum theory of gravity of which general relativity can only be a good approximation when R is small. – doetoe Jun 16 '17 at 21:59
• @doetoe Yes, I suspect he's thinking of quantum corrections that presumably kick in when the curvature scale approaches the Planck scale. – tparker Jun 16 '17 at 22:08
• @doetoe I'm sure he doesn't just mean "accurate to first order" though, because that corresponds to linearized GR, and we have good (though circumstantial) observational evidence that GR holds well beyond the linearized regime - for example, we have very strong observation evidence for the existence of black holes. – tparker Jun 16 '17 at 22:10
• Even if we didn't know about quantum mechanics, we would know that GR had problems at singularities. The problems become especially acute at a naked singularity, because you can't predict what will come out of it. – user4552 Jun 16 '17 at 22:11

But, I still don't understand what "small" curvature means. For example, take the FLRW metric for any $k = -1,0,1$, the Ricci scalar is:

$R = -6\left[\frac{\ddot{a}}{a} + \left(\frac{\dot{a}}{a}\right)^2 + \frac{k}{a^2}\right]$

The curvature is $< \infty$ as long as practically $a(t)$ does not go to zero, but this is a cosmological singularity anyways, where quantum gravity is expected to become relevant (unless you like Conformal Cyclic Cosmology a la Penrose, but that is a different story :) ), so $R < \infty$ in this case for all practical purposes, even if it is large, which means the scale factor can be small. G.R. is certainly valid in the region where $R < \infty$, i.e., as long as $a(t)$ is bounded. So, I don't know what exactly the lecturer is referring to.

• Yes, I guess you are right that general relativity itself doesn't break down at large R, only that for reasons beyond GR it is assumed/known that GR doesn't correctly describe nature anymore. – doetoe Jun 16 '17 at 22:03
• Hi. could you please explain this comment a bit further? – Dr. Ikjyot Singh Kohli Jun 17 '17 at 17:20
• I just meant to say that it is not general relativity as devised by Einstein that breaks down for large $R$, but that it may be that it ceases to provide a correct description of nature at large $R$. – doetoe Jun 18 '17 at 10:21