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.