Why is gravity a fictitious force?

[Demystifier]

Then you are doing quantum field theory, not classical GR. This is the classical GR forum, so here, we use the GR definition of force.

There is a classical version of this. Formulate gravity as a spin-2 field on a Minkowski background. Develop a perturbative method of classical scattering computation, which leads to classical Feynman diagrams (differing from the quantum Feynman diagrams by having no loop diagrams). The internal lines of Feynman diagrams contain Green functions (which is just a classical name for the mathematical object called propagator in quantum physics). Then you can say that the force is represented by this Green function in the classical Feynman diagram.

 

[PeterDonis]

There is a classical version of this.

This can only represent "gravitational fields" on the chosen background. It can't, for example, represent a black hole spacetime, because the global topology of that spacetime is incompatible with the Minkowski background.

Also, classically it's a weird way of doing things, because the background spacetime is unobservable--the "field" appears as a change in the spacetime geometry itself. With other QFTs, such as the Standard Model, that's not the case.

 

[renormalize]

This can only represent "gravitational fields" on the chosen background. It can't, for example, represent a black hole spacetime, because the global topology of that spacetime is incompatible with the Minkowski background.

How do you reconcile this claim with the actual published literature on perturbative quantum gravity?
For example, here are two recent papers from Phys Rev Letters that independently sum spin-2 diagrams to all orders on flat spacetime, both of which successfully arrive at the exact Schwarzschild metric:

https://journals.aps.org/prl/pdf/10.1103/PhysRevLett.132.251603:

https://journals.aps.org/prl/pdf/10.1103/PhysRevLett.133.111601:

 

[PeterDonis]

here are two recent papers from Phys Rev Letters that independently sum spin-2 diagrams to all orders on flat spacetime, both of which successfully arrive at the exact Schwarzschild metric

"The exact Schwarzschild metric" is not the same as "a black hole spacetime".

The issue here isn't the local solution; it's been known since the 1960s that the spin-2 field theory gives you the Einstein Field Equation of General Relativity. For a paper published in 2024 to show this for just one solution of the EFE is hardly a breakthrough.

The issue is global topology. The global topology of Minkowski spacetime is $R^4$. That will also be true of a curved spacetime describing, for example, a Schwarzschild vacuum region surrounding a spherically symmetric planet or star. So you can get such a global solution by putting a spin-2 field on Minkowski spacetime. (Of course inside the planet or star the stress-energy tensor is nonzero so you have to deal with the matter fields as well as the spin-2 field.)

But the global topology of a black hole spacetime is $R^2 \times S^2$, which is not the same as $R^4$. So there is no way to get such a global solution by putting a spin-2 field on Minkowski spacetime.

 

[Demystifier]

The difference between Schwarzschild metric and black hole spacetime is in the interpretation of the interior behind the horizon. But we don't know what happens behind the horizon and there are theoretical indications (e.g. related to the information paradox) that the usual GR picture of physics in the interior might be wrong. So it seems reasonable to focus only on the black hole exterior, and not too close to the horizon, where there is no difference between Schwarzschild metric and black hole spacetime. (I know, I know, this remark would be more meaningful in the BSM subforum, but one cannot always strictly separate such discussions.)

 

[PeterDonis]

The difference between Schwarzschild metric and black hole spacetime is in the interpretation of the interior behind the horizon.

There is no interpretation involved in the statement I made about global topology.

Whether the actual, physical objects we call "black holes" actually have that spacetime geometry and its associated topology is a different question. If they don't--if they're something more like a Bardeen black hole, say (which we've had previous PF threads on), which has no singularity and no true event horizon at all, and whose global topology is $R^4$--then of course there's no issue with modeling them with a spin-2 field on a Miinkowski background (although in such solutions the stress-energy tensor is again nonzero so you need to deal with matter fields--and ones that violate energy conditions at that).

But even that is not a matter of "interpretation of the interior". It's simply a matter of what solution of the EFE actually describes the objects. Once we pick a solution, there's no "interpretation" required to say what it predicts.

 

[PeterDonis]

it seems reasonable to focus only on the black hole exterior, and not too close to the horizon, where there is no difference between Schwarzschild metric and black hole spacetime.

Yes, there is--if we only focus on the exterior, we're not using a "black hole spacetime" at all. We're just using one particular piece of a spacetime whose full global structure we're leaving unspecified--but which we're assuming must have global topology $R^4$ if we're going to claim that our spin-2 field theory will end up being able to model it. Which means we're excluding an actual black hole spacetime, which doesn't have the right global topology.

 

[Demystifier]

There is no interpretation involved in the statement I made about global topology.

I'm not sure about that. Suppose that someone tells you that the Schwarzschild $g_{\mu\nu}(t,r,\theta,\varphi)$ is not a metric tensor, but just some tensor on a 4-dimensional manifold with unspecified metric. Would that influence your conclusions about the global topology of the manifold? If yes, I would call it a different interpretation.

 

[renormalize]

But the global topology of a black hole spacetime is $R^2 \times S^2$, which is not the same as $R^4$. So there is no way to get such a global solution by putting a spin-2 field on Minkowski spacetime.

That's of course a good point. But what empirical evidence exists that demonstrates that we don't really live in $R^4\,$? It's been shown that spin-2 quantum field theory sums to Schwarzschild outside the event horizon, but perhaps it sums to something different at and inside the horizon. To me it's exciting that work like I cited above suggests that physicists are getting close to actually predicting what gravity looks like outside a point mass if it arises, not from geometry, but rather from nonlinear spin-2 QFT on flat spacetime.

 

[PeterDonis]

Suppose that someone tells you that the Schwarzschild $g_{\mu\nu}(t,r,\theta,\varphi)$ is not a metric tensor, but just some tensor on a 4-dimensional manifold with unspecified metric.

What would that even mean? Until you tell me what this is supposed to model physically, how can I possibly say anything about it?

 

[PeterDonis]

what empirical evidence exists that demonstrates that we don't really live in $R^4\,$?

None, of course. I've already said that the objects we currently call "black holes" might not actually be described by that particular global spacetime geometry and topology. And I've already referred to at least one solution, the Bardeen black hole (the "black hole" is of course a misnomer in this case), which has global topology $R^4$ and doesn't raise any issue.

It's been shown that spin-2 quantum field theory sums to Schwarzschild

Yes.

outside the event horizon

No, the local solution in the paper you described is valid in any local patch of spacetime which meets the conditions of vacuum and spherical symmetry. Again, that's been known since the 1960s; it's not a new discovery that that 2024 paper made. Locally, there is no way of telling whether the patch of spacetime in question is outside or inside an event horizon, or even whether the global spacetime it's a patch of has an event horizon at all.

perhaps it sums to something different at and inside the horizon.

Nope. See above.

To me it's exciting that work like I cited above suggests that physicists are getting close to actually predicting what gravity looks like outside a point mass if it arises, not from geometry, but rather from nonlinear spin-2 QFT on flat spacetime.

Um, I guess you missed the part where I said that everything in that paper has been known since the 1960s? It's not telling anyone familiar with the field anything they haven't already known for decades.

Here are some more things that have been known since the 1960s:

The nonlinear spin-2 QFT is not renormalizable. (The ultimate reason for this is that the coupling constant in the theory, Newton's gravitational constant, is not dimensionless.)

That QFT is best viewed as an effective theory overlaid on some other underlying physics (string theory, loop quantum gravity, or whatever). In this respect, of course, it's no different from the Standard Model, or indeed from any QFT physicists use.

We don't know whether there is any physical regime where that QFT is actually relevant--meaning it works as an effective theory, and the quantum effects it predicts, which are not predicted by classical GR, are significant. It might well be that by the time we get to a regime where classical GR no longer works, the spin-2 QFT also no longer works--because gravity is strong enough that the other underlying physics has already taken over. If, for example, it turns out that classical GR works down to the Planck scale, it would seem highly unlikely that the spin-2 QFT would have any usefulness, because by the Planck scale we would expect it too to break down and some other underlying physics to take over.

 

[Demystifier]

What would that even mean? Until you tell me what this is supposed to model physically, how can I possibly say anything about it?

Your very question proves me right, because you ask about the physical meaning, namely about the interpretation. But I meant it as an exercise in differential geometry, that has nothing to do with physics.

 

[martinbn]

I'm not sure about that. Suppose that someone tells you that the Schwarzschild $g_{\mu\nu}(t,r,\theta,\varphi)$ is not a metric tensor, but just some tensor on a 4-dimensional manifold with unspecified metric. Would that influence your conclusions about the global topology of the manifold? If yes, I would call it a different interpretation.

The topology comes with the manifold, no matter what tensors you consider. There is no room for interpretations.

 

[Dale]

one cannot always strictly separate such discussions

Sure one can. One doesn’t need to bring in irrelevant topics to hijack a thread. This is utterly irrelevant to the discussion here because none of this reformulation stuff changes the fact that the force of gravity is not locally measurable.

 

[Demystifier]

The topology comes with the manifold, no matter what tensors you consider. There is no room for interpretations.

I see your point, but in GR one usually thinks backwards. One first finds an analytic expression for the metric tensor (usually as a solution of the Einstein equation), and then tries to reconstruct the global topology of the manifold. It seems to me that this reconstruction is somewhat ambiguous, i.e., amenable to different interpretations. For instance, should one take the maximal analytic extension of the Kerr metric or not? Moreover, it seems to me that this ambiguity is even more explicit if the tensor is not interpreted as a metric tensor. I'm not sure how to make my claims precise, but I hope that someone sees what I'm trying to say.

See also this post

Graduate Post in thread 'Feynman's bug (hot plate) in generality'

Mar 23, 2026

My take on the problem is that locally, the "hot plate" analogy works fine, but on a global scale, I don't believe adding in the "heat" that distorts the length of "rulers" on the plate is able to change the global topology of the plate. I.e., for instance, it's connectdness or it's compactness, for instance.

Yes, but I think the right question is this: Can general relativity (GR) "change" topology? Or more precisely, is GR a theory of the spacetime topology?

I think it isn't. GR is a local theory, in that sense it is not much different from the theory of "hot plate". When you...

• Demystifier

where I explained similar ideas on simpler examples.

 

[PAllen]

Tidal gravity can be measured--it's not "fictitious"--but in GR it is not a force. A force is something that causes nonzero proper acceleration. Tidal gravity does not. Tidal gravity is geodesic deviation--the behavior of objects in free fall.

Ok, so if I stretch a rubber band over 2 pegs, so it is stretched, you would claim you must consider that there is no force anywhere, as opposed to balanced forces which sum to 0 at each point. In the case of my simple tidal detector, there a force exerted by the spring on the ball. There must be a force exerted against the spring for conservation of momentum (tidal gravity provides it). Or are you going to discard conservation of momentum at such a small scale in GR? As has been agreed, none of this frame dependent, unlike fictitious forces.

 

[martinbn]

I see your point, but in GR one usually thinks backwards. One first finds an analytic expression for the metric tensor (usually as a solution of the Einstein equation), and then tries to reconstruct the global topology of the manifold. It seems to me that this reconstruction is somewhat ambiguous, i.e., amenable to different interpretations. For instance, should one take the maximal analytic extension of the Kerr metric or not? Moreover, it seems to me that this ambiguity is even more explicit if the tensor is not interpreted as a metric tensor. I'm not sure how to make my claims precise, but I hope that someone sees what I'm trying to say.

See also this post

Graduate Post in thread 'Feynman's bug (hot plate) in generality'

Mar 23, 2026

My take on the problem is that locally, the "hot plate" analogy works fine, but on a global scale, I don't believe adding in the "heat" that distorts the length of "rulers" on the plate is able to change the global topology of the plate. I.e., for instance, it's connectdness or it's compactness, for instance.

Yes, but I think the right question is this: Can general relativity (GR) "change" topology? Or more precisely, is GR a theory of the spacetime topology?

I think it isn't. GR is a local theory, in that sense it is not much different from the theory of "hot plate". When you...

• Demystifier

where I explained similar ideas on simpler examples.

What do yoy mean by "in GR one usually.."? I would say that in GR one usually doesn't do that!

 

[Demystifier]

What do yoy mean by "in GR one usually.."? I would say that in GR one usually doesn't do that!

At least sometimes one does. For definiteness, let us consider the example of the Kerr metric.

 

[PeterDonis]

One first finds an analytic expression for the metric tensor (usually as a solution of the Einstein equation), and then tries to reconstruct the global topology of the manifold.

So how is your hypothetical in which we say that a particular tensor is not the metric tensor relevant to this at all?

It seems to me that this reconstruction is somewhat ambiguous, i.e., amenable to different interpretations.

No, it isn't. You said you posed your question as a problem in differential geometry, not physics. A metric tensor is part of differential geometry. So is finding the maximal analytic extension of a particular manifold with a particular metric tensor. No interpretation involved at all.

should one take the maximal analytic extension of the Kerr metric or not?

If you're asking this as a math question, what does it even mean? Math is math.

If you're asking this as a physics question, then of course the answer is that it depends on whether we think the maximal analytic extension models something that is physically relevant or not. (Most physicists appear to think not.)

If you seriously don't see the obvious difference between these two things, I'm not sure what else I could say to explain it.

 

[PeterDonis]

I stretch a rubber band over 2 pegs, so it is stretched, you would claim you must consider that there is no force anywhere, as opposed to balanced forces which sum to 0 at each point

No, I would make no such claim.

The stretched rubber band has a nonzero stress and strain in it. This is physically measurable and tells us that the band is stretched, not unstretched. Obviously this is due to internal forces within the band.

Since you're stretching the band over 2 pegs, this example has nothing to do with tidal gravity anyway. My claim was that tidal gravity, i.e., geodesic deviation, is not, strictly speaking, a force. But of course an object that is in a curved spacetime will have stresses and strains induced in it that would not be there if the spacetime were flat, and informally we say these are due to tidal gravity.

 

[PeterDonis]

it seems to me that this ambiguity is even more explicit if the tensor is not interpreted as a metric tensor.

I fail to see how this is relevant to the discussion at all. Please note that, as I said in post #109, a metric tensor is a differential geometry concept, independent of any physical interpretation.

 

[PeterDonis]

In the case of my simple tidal detector, there a force exerted by the spring on the ball. There must be a force exerted against the spring for conservation of momentum

Ok.

(tidal gravity provides it).

No, the ball provides it. The spring pulls (or pushes) on the ball, and the ball pulls (or pushes) on the spring. These are electromagnetic forces between the atoms, not tidal gravity.

 

[PeterDonis]

See also this post

Where you said:

the maximal analytic extension of the metric does not need to be the physical one.

You're misdescribing this.

The maximal analytic extension tells us the maximal global manifold that has the same local solution everywhere. For example, the maximal analytic extension of the Schwarzschild metric has the same local metric everywhere--the unique vacuum, spherically symmetric metric.

When we say that, for example, the "physical" metric of a spherically symmetric collapse of a star to a black hole, as in the idealized 1939 Oppenheimer-Snyder model, is not the same everywhere as the maximal analytic extension of the Schwarzschild metric, what we mean is that globally, this spacetime is not vacuum everywhere. There's a region occupied by the collapsing matter which is not vacuum, i.e., the local metric is not the Schwarzschild metric. It's a different metric (it turns out to be a portion of a closed collapsing FLRW metric). In other words, the local solution changes in the matter region, and the global solution is two regions, one a portion of the maximal analytic extension of the vacuum Schwarzschild metric (but not all of it), the other a portion of the maximal analytic extension of the closed collapsing FLRW metric (but not all of it), with the two regions matched properly at the boundary between them.

What you can't do, at least not according to how GR is done by those who work with it, is just declare by fiat that no, the local solution doesn't change, the metric is the vacuum Schwarzschild metric everywhere, but the global manifold is something different from the maximal analytic extension of that metric. The only way to get a different global manifold from that, according to how GR is actually done, is to change the local solution somewhere--to patch together regions having different local solutions, because something changes from region to region (usually the stress-energy tensor).

 

[PAllen]

Where you said:


You're misdescribing this.

The maximal analytic extension tells us the maximal global manifold that has the same local solution everywhere. For example, the maximal analytic extension of the Schwarzschild metric has the same local metric everywhere--the unique vacuum, spherically symmetric metric.

When we say that, for example, the "physical" metric of a spherically symmetric collapse of a star to a black hole, as in the idealized 1939 Oppenheimer-Snyder model, is not the same everywhere as the maximal analytic extension of the Schwarzschild metric, what we mean is that globally, this spacetime is not vacuum everywhere. There's a region occupied by the collapsing matter which is not vacuum, i.e., the local metric is not the Schwarzschild metric. It's a different metric (it turns out to be a portion of a closed collapsing FLRW metric). In other words, the local solution changes in the matter region, and the global solution is two regions, one a portion of the maximal analytic extension of the vacuum Schwarzschild metric (but not all of it), the other a portion of the maximal analytic extension of the closed collapsing FLRW metric (but not all of it), with the two regions matched properly at the boundary between them.

What you can't do, at least not according to how GR is done by those who work with it, is just declare by fiat that no, the local solution doesn't change, the metric is the vacuum Schwarzschild metric everywhere, but the global manifold is something different from the maximal analytic extension of that metric. The only way to get a different global manifold from that, according to how GR is actually done, is to change the local solution somewhere--to patch together regions having different local solutions, because something changes from region to region (usually the stress-energy tensor).

There can be two topologically different manifolds with the same metric everywhere, neither of which is incomplete in any way. A pure Riemannian example is the flat Torus vs the flat plane. This is where analytic extension comes in, to provide uniqueness. In GR, Birkhoffs theorem is quasi local. Without the assumption of analytic continuation, there are many different topologies that are everywhere vacuum SC metric. They are made by various smooth (but not analytic) glueings of pieces of Kruskal. Long ago I linked a math paper that demonstrated perhaps a dozen wildly different topologies that were vacuum SC everywhere.

 

[Demystifier]

because something changes from region to region (usually the stress-energy tensor).

What else can change, if not the stress-energy tensor?

 

[Dale]

What else can change, if not the stress-energy tensor?

The boundary conditions.