Although "Spacetime Curvature is Gravity" is the first statement that is told to an infant by the "popular science", I believe that really really misrepresents what GR has to say. I am posting this question to confirm whether my understanding is appropriate or not.

I have the following arguments against "Spacetime Curvature is Gravity":

  1. Technically speaking, we mean gravitational field by gravity in the classical physics, and thus, according to the correspondence principle, the quantity used to describe gravity in a modern theory should, in the classical limit, become the good old (Newtonian) gravitational field. Christoffel symbols are the quantities that do so and not the Reimann curvature tensor. So, how come the curvature of spacetime can be called gravity?

  2. More importantly, the curvature of spacetime can be zero and we can still have non-zero Christoffel symbols. Since the vanishing or non-vanishing nature of Christoffel symbols is what determines whether the laws of Physics take the special relativistic form or not in a given frame, we can most certainly have gravity in the absence of the curvature of spacetime.

  3. It is argued at some places that it is the curvature tensor that determines whether there is any "true gravity" or not. Specifically, if the curvature is zero everywhere in the spacetime, one can always have a transformation that takes her from a generic coordinate system to a global Minkowskian coordinate system - in other words, one can globally gauge-away the gravity. Okay, great - but just the fact that you can gauge away the gravity doesn't mean that it doesn't really exist in those frames as well in which the laws of physics don't have a special relativistic form. One has got to accept in GR that the existence or otherwise of gravity is purely a frame-dependent fact. I think this is the most important insight of general relativity that the existence of gravity itself is a relativistic fact - it is perfectly fine that one frame sees gravity and the other doesn't. The famous "elevator experiment" - where an elevator, in the deep empty space (say, with zero curvature), is getting pulled by some rope - illustrates the very fact that wholeheartedly accepting that gravity is truly existent in the frame of the accelerated elevator is the way to do Physics in the accelerated frames. I agree that one can, in the cases where there is no curvature at all, do away with special relativity by cleverly chosing the inertial frame or by keep on transforming his equations referring to the inertial frame. But there is no other way to do physics in a non-inertial frame without keep referring to an other frame than accepting that gravity exists in non-inertial frames even if the spacetime is Minkowskian. And honestly, if one is willing to assert that just because gravity can be gauged away, gravity doesn't really exist then one should actually say that gravity doesn't exist locally even if the curvature is non-zero. But certainly, such an assertion is absurd.

So, shouldn't we say "Christoffel Symbols are Gravity" rather than "Spacetime Curvature is Gravity"?

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Sep 18 '17 at 15:04

No, we should not say that Christoffel symbols are gravity. The big reason, which really should be enough, is that they are coordinate dependent. One of the main tenets of General Relativity is that coordinates don't matter. Everything physical must be expressible in a coordinate independent and/or tensorial manner.

As I said in the comments, personally I think it's a bit ridiculous to suggest that using polar coordinates somehow brings gravity into the mix, while using Cartesian coordinates does not. The equation for a straight line changes, but you can verify using any number of methods that it's still a straight line. If polar coordinates show gravity, then where is that gravity coming from? What physical object is generating it? There was none in Cartesian coordinates.

But let me address your three points:

  1. It is not true in absolute generality that the Christoffel symbols correspond to the gravitational field, for the reasons I gave above. A gravitational field manifests itself in the Christoffel symbols, but not the other way around. Also remember that even in Newtonian gravity the equivalence principle holds, and one might argue that only tidal forces are measurable for someone in free fall, so that's one argument in favor of the curvature.

  2. Again, even in flat spacetime there are curved coordinates. "Special relativistic" means that the metric is $\eta_{\mu\nu}$ when expressed in locally Lorentzian (i.e. Cartesian) coordinates, not in any coordinates.

  3. This is basically the same as 2, but see the following paragraph.

I think the deep issue is that you are misunderstanding gravity for coordinate acceleration. You actually make a very good point in your number 3 argument, but you draw the wrong conclusion. The lesson of the equivalence principle is not that acceleration is relative and hence gravity is relative. You could take it as the conclusion, but then the word "gravity" is not very useful anymore because it is coordinate-dependent.

Instead, the lesson you should take away is that "gravity" should refer to something that has a physical existence independently of the observer, and that something is tidal forces, precisely because of the equivalence principle. Since coordinate acceleration is relative, the smart thing to do is to make "gravity" mean something that is not relative.

I insist with a very important point: this is not just a matter of definition; physical reality has my back here. I say this because it turns out that every time there are tidal forces, one can identify some physical object (planet, star, whatever) responsible for it. However, sometimes objects seem to not obey the Cartestian-flat-space geodesic equation with no apparent source of gravity nearby. To me it just makes much more sense to say that the thing that always manifests itself near a heavy object is gravity and that the thing that sometimes happens as a result of weird coordinates is not gravity.

  • 2
    $\begingroup$ "every time there are tidal forces, one can identify some physical object" - No. GR admits propagating degrees of freedom. In a complete vacuum spacetime with no sources whatsoever, you can have gravitational waves. $\endgroup$ – Dvij Mankad Sep 16 '17 at 7:39
  • 3
    $\begingroup$ Another thing: GR has a certain degree of gauge-freedom and propagating degrees of freedom. This allows you to set your Christoffel symbol to behave, up to that freedom, the way you want no matter what the $T_{\mu\nu}$ is. Physically, this corresponds to the existence of propagating degrees of freedom and to "being able to choose both accelerated frames as well as different kinds of spatially curvilinear sort of frames".... $\endgroup$ – Dvij Mankad Sep 16 '17 at 7:39
  • 3
    $\begingroup$ ...This is the reason you should not be really talking about what exactly causes the gravity. It is the gauge-freedom and propagating degrees of freedom along with the $T_{\mu\nu}$ that decides whether gravity exists or not. $\endgroup$ – Dvij Mankad Sep 16 '17 at 7:40
  • 2
    $\begingroup$ "One of the main tenets of GR is that the coordinates don't matter." IMO, this is an oversimplification. GR says is that laws of Physics should take a covariant form - we should be able to write the laws in a coordinate independent way. It doesn't mean that the existence or otherwise of any effect cannot be coordinate dependent. If the effect is a geometric fact, for example, curvature, then it should be coordinate independent. But if the effect involves some business with setting gauges then it doesn't need to be coordinate-free. The overall laws are still covariant and that's good enough. $\endgroup$ – Dvij Mankad Sep 16 '17 at 7:50
  • $\begingroup$ Let me ask you about 3D space. Assume it not to be curved, but somebody gives you coordinates with non zero Christoffel symbols. Would you say there is something going on, or that person just made a poor choice. Of course this does not directly extend to spacetime; depending on how different you.view space and time from spacetime. $\endgroup$ – lalala Sep 16 '17 at 9:49

To answer your last question directly, we shouldn't, because at any point, there exists a coordinate transformation that causes the Christoffel symbols to vanish (geodesic coordinates). In other words, an observer following a geodesic is in free fall and doesn't experience any force of gravity.

What the observer DOES notice is geodesic deviation. This is the key thing that differentiates between "gravity" and "coordinates that look like gravity". Regardless of the coordinates, if an observer sees geodesic deviation, there is gravity present. If not, the spacetime is flat. This is another way of saying that in the presence of gravity (geodesic deviation), the Riemann curvature is nonzero.

  • 1
    $\begingroup$ "we shouldn't, because at any point, there exists a coordinate transformation that causes the Christoffel symbols to vanish" - Why is it bad? Isn't it the fundamental tenet of the Equivalence principle that we should always be able to locally gauge-away gravity? You see geodesic deviation and you can be sure that the spacetime is not locally curvature-free. Agreed. But why do you want the curvature to represent gravity is the question in the first place which you seem to sort of circularly bypass. $\endgroup$ – Dvij Mankad Sep 16 '17 at 7:29
  • $\begingroup$ The point I'm trying to make is that gravity = geodesic deviation, rather than gravity = curvature. In Newtownian mechanics, if there's a region of uniform acceleration, say, $10 m/s^2$ as in freshman physics problems, there's no geodesic curvature, and you could argue that you might actually be somewhere out in the vacuum of space accelerating at a constant rate. However, if the acceleration is non-uniform, like for a spherically symmetric body, the tidal forces (geodesic deviation) indicates that the gravity is real. $\endgroup$ – pianyon Sep 16 '17 at 23:42
  • $\begingroup$ In the context of general relativity, the geodesic deviation is mathematically equivalent to an expression involving the curvature, which is why we say that curvature indicates the presence of gravity, or "spacetime curvature is gravity." $\endgroup$ – pianyon Sep 16 '17 at 23:44
  • 1
    $\begingroup$ @Dvij this started as something looking suspiciously resembling a newbie question, but it turns out you are well-aware of the common wisdom that people put in the answers, and are arguing about the notation. Well, the point of notation is to commonly refer to things without spelling out their definitions every time. So I guess my answer is: you are wrong, because most people say that you are :) Or, if you wish, there is no right and wrong; but if you want to share ideas with others you're better off speaking the language. And no language is perfect btw. $\endgroup$ – Prof. Legolasov Sep 18 '17 at 14:04

When you discuss an interaction in nature due to a field, you need to consider two things: what determines the field, and what the field determines. To take the simpler example of electromagnetism to begin, Maxwell's equations show the four-current determines $F^{\mu\nu}$ (up to gauge invariance), while the Lorentz force law determines the four-force on a charged particle.

Now let's consider gravity. Verbally we say matter tells space how to bend (this is the field-determination part) and space tells matter how to move (this is the determined-by-field part). Mathematically, the former is expressible in terms of the Riemann tensor (or more concisely in terms of the Ricci tensor), and the latter in terms of Christoffel symbols. The former only uses tensors; the latter uses non-tensors in terms of which we can write down a formula for the Riemann tensor.

Part 1: $R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}g^{\rho\sigma}R_{\rho\sigma}+\Lambda g_{\mu\nu}=\kappa T_{\mu\nu}$. Part 2: Newton's second law $m\dfrac{d^2 x^i}{dt^2}=f^i$ generalises in special relativity to $m\dfrac{d^2 x^\mu}{d\tau^2}=f^\mu$ and in general relativity to $m\dfrac{d^2 x^\mu}{d\tau^2}=f^\mu-m\Gamma^{\mu}_{\nu\rho}\dfrac{dx^\nu}{d\tau}\dfrac{dx^\rho}{d\tau}$.

  • $\begingroup$ Your answer seems quite intriguing to me. What I understand you are saying is this: "One can say that gravity is Christoffel symbols when described in terms of its effects on a system. But gravity can be called the curvature of spacetime when we want to describe it in terms of how matter determines gravity (to whatever extent it does). But these two notions don't agree with each other up to the extent allowed by gauge freedoms and propagating degrees of freedom." Am I right? $\endgroup$ – Dvij Mankad Sep 16 '17 at 9:54
  • $\begingroup$ @Dvij Whether you define "gravity" in terms of part 1, part 2 or both is semantic, but I'd go with both because you don't have a full theory otherwise, just as you need both parts to understand electromagnetism. But it's wrong to say "these two nations don't agree". As with other physical problems, when you combine both "parts" you get a differential equation, but determining its solution requires initial/boundary conditions. $\endgroup$ – J.G. Sep 16 '17 at 11:14

Old question, but I wanted to contribute.

I think you are correct. You are actually echoing a point of view that was historically held by Einstein himself. In a letter written in 1950 to Max von Laue he stated:

It is true that in the case of the $R^i_{klm}$ [the Riemann curvature tensor components] vanish, so that one could say: “There is no gravitational field present.” However, what characterizes the existence of a gravitational field from the empirical standpoint is the non-vanishing of the $\Gamma^l_{ik}$ [the connection components], not the vanishing of the $R^i_{klm}$. If one does not think intuitively in such a way, one cannot grasp why something like curvature should have anything at all to do with gravitation. In any case, no reasonable person would have hit upon such a thing. The key for understanding the equality of inertial and gravitational mass [the Equivalence Principle] is missing.

Einstein did not see coordinate-dependence as a problem, but as a real physical property of the gravito-inertial field. See also Ron Maimon's answer here.

Of course, only global uniform gravitational fields can be canceled by a choice of coordinates(*). In general, when there is spacetime curvature at least somewhere in the domain, this isn't possible(**). From Einstein's point of view, gravity is not a curvature of space-time, but rather curvature is a manifestation of gravity.

Einstein's point of view seems to have fallen out of fashion these days, since it goes against the idea that "real" quantities should have an absolute existence indipendent of the observer, but in principle there is nothing wrong with it.

See also this paper


I argue that, contrary to folklore, Einstein never really cared for geometrizing the gravitational or (subsequently) the electromagnetic field; indeed, he thought that the very statement that General Relativity geometrizes gravity “is not saying anything at all”. Instead, I shall show that Einstein saw the “unification” of inertia and gravity as one of the major achievements of General Relativity. Interestingly, Einstein did not locate this unification in the field equations but in his interpretation of the geodesic equation, the law of motion of test particles.

And this short monograph on the subject


There exists some confusion, as evidenced in the literature, regarding the nature of the gravitational field in Einstein's General Theory of Relativity. It is argued here the this confusion is a result of a change in interpretation of the gravitational field. Einstein identified the existence of gravity with the inertial motion of accelerating bodies (i.e. bodies in free-fall) whereas contemporary physicists identify the existence of gravity with space-time curvature (i.e. tidal forces). The interpretation of gravity as a curvature in space-time is an interpretation Einstein did not agree with.

(*) It should also be pointed out that Einstein considered inertial forces not as "pseudo-forces", but as real gravitational "forces" originating from the circumstance that, in certain frames, the distant masses in the universe accelerate relative to the observer; this induces these "forces" in a similar way moving relative to a charge induces magnetic forces.

Consequently, the fact that this (uniform, given the distance and distribution of the sources) part of the gravitational field can be removed by using a different coordinate system would stem from the circumstance that in this new frame the distant masses are not accelerating, and so only their (locally negligible) tidal forces are detectable, in principle.

(**) An example where it is clear that there is a gravitational field present but no local curvature (i.e, tidal forces) can be found in Newtonian gravity: A uniform density spherical mass with a non-concentric spherical cavity inside it. There is a uniform gravitational field inside the cavity but no tidal forces. Clearly, there will be tidal forces very close and inside the walls of the cavity, but not locally inside the cavity itself. I would expect that, in the appropriate limit, this result carries over to general relativity.


Not the answer you're looking for? Browse other questions tagged or ask your own question.