Why is gravity a fictitious force?

[Demystifier]

Both Minkowski spacetime and Schwarzschild spacetime are vacuum spacetimes. They have the same energy-momentum but different boundary conditions. Boundary conditions can also be associated with the topology or symmetry.

It is incorrect to think that boundary conditions are always associated with sources, not just in GR but in other physics also.

Yes, but the Schwarzschild spacetime is unphysical, in the sense that it describes an eternal black hole. A physical black hole is a result of a gravitational collapse, it deviates from Schwarzschild spacetime, and this deviation is associated with a realistic energy-momentum.

 

[PeterDonis]

the maximal analytic extension of Schwarzschild spacetime is unphysical, in the sense that it describes an eternal black hole.

See the bolded addition in the quote above. Without it the statement is false.

A physical black hole is a result of a gravitational collapse, it deviates from the maximal analytic extension of Schwarzschild spacetime, and this deviation is associated with a realistic energy-momentum.

Again see the bolded addition.

I think both additions clarify what you actually intended to say.

 

[Dale]

the Schwarzschild spacetime is unphysical

You missed the point. I was not claiming that either Minkowski spacetime or Schwarzschild spacetime are realistic. They are just clear examples of two distinct spacetimes that have the same stress-energy.

The mathematical fact remains that it is incorrect to think of boundary conditions as being due to sources, not just in GR but generally. Within any given solution region, different boundary conditions do lead to different solutions for the same sources.

While you sometimes can specify the boundary conditions on a small region as being due to external sources, what you are doing then is simply expanding the solution region. You still must have boundary conditions on that expanded region.

And at some point you can no longer expand the solution region, either because you don’t know the sources any more or because the boundary conditions are for the whole universe. In either case you still have boundary conditions.

You cannot avoid boundary conditions and different boundary conditions give different solutions for the same sources.

 

[martinbn]

You cannot avoid boundary conditions and different boundary conditions give different solutions for the same sources.

Just to add that those conditions can be purely geometrical. For example for the initial value problem for the vacuum Einstein equations (no sources), the initial data is the metric restricted to the initial surface and the second fundamental form. Different initial conditions lead to different vacuum solutions, say the Minkowski and the Schwarzschield ones.

 

[PeterDonis]

for the initial value problem for the vacuum Einstein equations (no sources), the initial data is the metric restricted to the initial surface and the second fundamental form. Different initial conditions lead to different vacuum solutions, say the Minkowski and the Schwarzschield ones.

It should be noted that in this formulation, the topology of the initial surface also has to be specified: it's $R^3$ for Minkowski but $R \times S^2$ for Schwarzschild. So "boundary conditions" can include topology as well.

 

[Roberto Pavani]

I read about equivalence principle. I tried to understand Einstein's thought experiment with elevator and I can't understand why we compare elevator in the space and elevator on the surface of the Earth and conclude that gravity is fictitious force.

Why?

Thanks!

Any force is fictitious if, locally and for a point-like object, you can find a reference frame where it vanishes. Gravity satisfies this (free fall). For extended objects the force cannot be fully removed; e.g. tidal forces in gravity remain regardless of your frame.

In practice, since most experiments are small compared to the scale over which gravity varies, gravity can be treated as fictitious to an excellent approximation; much like treating the Earth as flat works well for small-scale measurements.

 

[Sagittarius A-Star]

For extended objects the force cannot be fully removed; e.g. tidal forces in gravity remain regardless of your frame.

That are in GR not gravitational interaction forces but electric forces within the extended object.

 

[Roberto Pavani]

Consider a cloud of non-interacting dust particles in free fall near a black hole: no electric forces at all, yet the cloud stretches (spaghettification). Tidal effects are pure geometry they exist without any internal forces.
Internal EM forces resist the stretching; they don't cause it.

 

[Sagittarius A-Star]

If small bodies follow their geodesics, that is in GR not considered as an interaction force.

Consider in flat spacetime two small moving bodies, that try to follow their (intersecting) geodesics. When they crash into each other, forces appear that are not considered to be gravitational interaction forces.

 

[Roberto Pavani]

I agree that in GR free-falling particles follow geodesics and experience no force by definition. My point was simpler (for undergraduate question): locally, gravity can be transformed away (equivalence principle); globally, curvature cannot.
Whether one calls tidal effects a "force" or "geometry" is terminology; the physical content (measurable increase in separation) is the same.

 

[PeterDonis]

I agree that in GR free-falling particles follow geodesics and experience no force by definition.

Yes, which is why the term "tidal forces", which you used in post #126, is not correct if you are describing the effects of spacetime curvature on geodesics. Tidal "effects" is fine, but these effects are not "forces", as you say.

An extended body moving through a curved spacetime will, in general, experience internal forces due to the fact that not all of its parts are moving on geodesics of the spacetime. But as @Sagittarius A-Star pointed out in post #127, these forces--which are not fictitious, the parts of the body that experience them are not traveling on geodesics--are not "gravity"; they are non-gravitational interactions between the parts of the body that keep the parts from traveling on geodesics.

 

[Roberto Pavani]

Fair point on terminology; "tidal effects" is more precise than "tidal forces". Thanks for the clarification.

 

[DonatasA]

I read about equivalence principle. I tried to understand Einstein's thought experiment with elevator and I can't understand why we compare elevator in the space and elevator on the surface of the Earth and conclude that gravity is fictitious force.

Why?

Thanks!

It is not because it is true, but because it works using same formula. Physics don't need any other reason.
But it does not mean what someone someday will not find out it is wrong or incorrect in some way and fix it or updates it.

 

[Demystifier]

Any force is fictitious if, locally and for a point-like object, you can find a reference frame where it vanishes. Gravity satisfies this (free fall). For extended objects the force cannot be fully removed; e.g. tidal forces in gravity remain regardless of your frame.

In practice, since most experiments are small compared to the scale over which gravity varies, gravity can be treated as fictitious to an excellent approximation; much like treating the Earth as flat works well for small-scale measurements.

There is an interesting similarity with magnetic force. The magnetic force on a point charge is orthogonal to its 3-velocity, so it does no work. But if we have an extended body, typically a magnetic dipole, then magnetic force does work on it.

 

[Roberto Pavani]

The magnetic force on a point charge is orthogonal to its 3-velocity, so it does no work.

Excellent observation. This alone would deserve a dedicated thread (Abraham–Lorentz, radiation reaction, and the energy-balance puzzle for a point charge in pure B), but that's well beyond the undergraduate scope here.

 

[Roberto Pavani]

@Demystifier's observation kept buzzing in my head and produced the following thought:

Gravity is fictitious only to first order. In Riemann normal coordinates at any point $p$, the Christoffel symbols vanish ($\Gamma = 0$), but their derivatives do not ($\partial\Gamma \neq 0$). Since the Riemann tensor is built from $\partial\Gamma$, curvature cannot be transformed away.
"Fictitious" seems to be a first-order statement, not an exact one.

 

[Demystifier]

@Demystifier's observation kept buzzing in my head and produced the following thought:

Gravity is fictitious only to first order. In Riemann normal coordinates at any point $p$, the Christoffel symbols vanish ($\Gamma = 0$), but their derivatives do not ($\partial\Gamma \neq 0$). Since the Riemann tensor is built from $\partial\Gamma$, curvature cannot be transformed away.
"Fictitious" seems to be a first-order statement, not an exact one.

Gravity, of course, is not fictitious because the Riemann curvature is a tensor. But gravity is not the same thing as gravitational force. The claim is that the gravitational force is fictitious, not that gravity is fictitious.

The gravitational force is encoded in the coupling of the gravitational field $g_{\mu\nu}$ with matter. In minimal coupling, the Riemann curvature does not appear. Only the connection appears, through the covariant derivative.

 

[Roberto Pavani]

But gravity is not the same thing as gravitational force.

Fair point. I should have said 'gravitational force', not 'gravity'.
The distinction is precisely the one between Christoffel symbols (eliminable) and Riemann tensor (not).
This is not a minor terminological point: conflating the two is exactly what makes the original question confusing.
Thanks for sharpening it.

 

[DavidMiranda]

So, even if in some frames there is an actual change in momentum, because in others we can make the force vanish, technically it is a byroduct of the choice of frame and not what we would call a real force?

 

[Matterwave]

So, even if in some frames there is an actual change in momentum, because in others we can make the force vanish, technically it is a byroduct of the choice of frame and not what we would call a real force?

In general, we only require that "physical laws look the same" in all *inertial* frames of reference (true for both Galilean and Einsteinian relativity principles). Inertial frames are those moving at constant velocity / no acceleration with respect to each other. "Fictitious" forces appear when we try to write out our physical laws (in this case, the equations of motion) in non-inertial frames.

The question Einstein asked, when he was developing GR was "what really counts as an inertial reference frame?". His happy thought was that the inertial reference frames are the ones in free fall, NOT e.g. the ones at rest relative to the Earth's surface. Such a thought led to his development of his (Einstein's) equivalence principle -- no (non-gravitational) local physical experiment (i.e. before things like geodesic deviation can be assessed) can tell the difference between a freefalling frame in a gravitational field and just a "floating out in space" frame. GR obeys the even stronger "Strong Equivalence Principle" which removes the (non-gravitational) aside.

 

[PeterDonis]

In general, we only require that "physical laws look the same" in all *inertial* frames of reference

In SR, yes. In GR, no--in GR the laws have to look the same in all frames, not just inertial frames.

 

[Matterwave]

In SR, yes. In GR, no--in GR the laws have to look the same in all frames, not just inertial frames.

I put the "physical laws looks the same" in quotes because I realized that was a rather loose statement but it seemed warranted in this thread since the thread is explicitly talking about fictitious forces.

Are you referring to the "diffeomorphism invariance/covariance" of GR here? That we are able to express our physical laws as tensor equations?

 

[PeterDonis]

Are you referring to the "diffeomorphism invariance/covariance" of GR here? That we are able to express our physical laws as tensor equations?

Yes.

 

[Matterwave]

That's fair, and I could definitely be on board with teaching a coordinate independent view as a first-class principle.

It does make talking about fictitious forces a bit difficult though. And TBH I have not seen a rigorous treatment of something like coriolis forces or centrifugal forces in the language of tensors/differential geometry. This is a gap in my knowledge, I admit.

 

[PeterDonis]

It does make talking about fictitious forces a bit difficult though.

No, it doesn't; indeed, it's easier in GR, because in GR, "gravity" is a fictitious force! In other words, a "fictitious force" in GR is just "anything that lay people call a force but which an object doesn't actually feel" (i.e., there's no proper acceleration associated with it, the object is in free fall). So conceptually it's simpler.

TBH I have not seen a rigorous treatment of something like coriolis forces or centrifugal forces in the language of tensors/differential geometry.

Most GR textbooks talk about this, though not all with the same emphasis. See, for example, section 6.6 of MTW, and in particular exercise 6.8 in that section, and Equation 6.25 in that exercise.

In a more detailed treatment, all "fictitious forces" will match up with appropriate connection coefficients (particular combinations of first derivatives of the metric components) in a non-inertial frame.

 

[Matterwave]

No, it doesn't; indeed, it's easier in GR, because in GR, "gravity" is a fictitious force! In other words, a "fictitious force" in GR is just "anything that lay people call a force but which an object doesn't actually feel" (i.e., there's no proper acceleration associated with it, the object is in free fall). So conceptually it's simpler.


Most GR textbooks talk about this, though not all with the same emphasis. See, for example, section 6.6 of MTW, and in particular exercise 6.8 in that section, and Equation 6.25 in that exercise.

In a more detailed treatment, all "fictitious forces" will match up with appropriate connection coefficients (particular combinations of first derivatives of the metric components) in a non-inertial frame.

Well for me it makes it a bit harder :P

But also fair, that's my gap, I should not have implied it for everyone.

Thanks for the references~ another place for me to look into! :D

 

[Roberto Pavani]

A naive question: if the equivalence principle tells us that spacetime is locally flat, and in flat spacetime Newton's third law holds, then when an extended body in free-fall develops internal stresses (EM reaction forces), what is the local "action" they are reacting to?

I understand that globally it's geodesic deviation (Riemann), but locally, in the flat patch, shouldn't there be a paired action for every reaction?

This is essentially why I argued (perhaps incorrectly) that gravity is locally real for extended objects, if there's a reaction, there should be an action.

 

[PeterDonis]

in flat spacetime Newton's third law holds

Not for fictitious forces. For non-fictitious forces, it holds even in curved spacetime. But...

when an extended body in free-fall develops internal stresses (EM reaction forces), what is the local "action" they are reacting to?

Nothing to do with tidal gravity (geodesic deviation). See below.

I understand that globally it's geodesic deviation (Riemann)

Which, since it involves geodesics (free-fall worldlines), is a fictitious force. In other words, the term "tidal forces" is a misnomer; tidal gravity itself (spacetime curvature) is not a force. So no, that is not what EM reaction forces are reacting to. See below.

but locally, in the flat patch

Locally, in a flat patch, there is no geodesic deviation since curvature is ignored. So locally, in a flat patch, there is no such thing as an object developing internal stresses due to spacetime curvature (tidal gravity). You have to look beyond a single local flat patch to model such things. And when you do, the only action-reaction pairs involved are for non-fictitious forces. For example, two adjacent atoms exert equal and opposite EM forces on each other. There is no non-fictitious force anywhere associated with tidal gravity itself.

 

[Dale]

in flat spacetime Newton's third law holds, then when an extended body in free-fall develops internal stresses (EM reaction forces)

I am not sure what kind of scenario you are envisioning here. In flat spacetime there are no tidal effects. So what internal stresses are you talking about? Are you referring to something like a truss built with stressed members?

I understand that globally it's geodesic deviation (Riemann), but locally, in the flat patch, shouldn't there be a paired action for every reaction?

Tidal effects are non-local. You cannot talk about tidal forces and local flatness. They are incompatible.

gravity is locally real for extended objects, if there's a reaction, there should be an action

Locally action and reaction forces are purely mechanical, and they never involve gravitational forces.

It might help to be a bit more specific in the scenario that you are considering here.

 

[Matterwave]

... then when an extended body in free-fall develops internal stresses (EM reaction forces),...

Just to understand your question better, when you bring in "extended body" here, are you envisioning that the body is large enough that one side of the body "feels more gravity" than the other? (Because it's closer to the gravitating object)