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.