The equivalence principle and the curvature of spacetime

[geordief]

Suppose one is an observer in Einstein's famous sealed portacabin in empty space and the living space is accelerating at a regular rate in the direction away from the "floor". (The room is a cube of 1x1x1 unit measurement)

Now ,under the equivalence principle the scenario is very ,very similar to that of an observer whose room is not under acceleration but rather subject to the gravitational influence of a massive body positioned in the opposite  direction to the earlier acceleration. 

So ,as I see things the acceleration obtaining in the first scenario should be causing a curvature of spacetime in more or less exactly (ie minus tidal effects) the same way as occurs in conditions of gravity.

If I have understood this right so far ,could I get a little help so as to set up measuring equipment in my accelerating sealed room so that I can measure the curvature in spacetime in a region of the room  and see how the spatial measurement change vis s vis the temporal measurements?

Suppose I have an observer on the floor and place her ,with a clock at the x,y =0,0 point  of that floor (or x,y,z=0,0,0 of the room) can  I place another clock at ,say 1,1,1 and  measure the distance  and times both with and  without acceleration?

Will this setup allow me to  observe the curvature of spacetime in the room?

[MigL]

Not exactly sure what you are after, but IIRC, Einstein used an elevator with holes on either side through which  light shines, either stationary in a gravitational field, or accelerating upward. The fact that there is a deflection from the horizontal for the accelerating elevator then implies that the stationary one in a gravitational field must also experience that deflection due to the equivalence principle.
Geodesics, or allowed paths in spacetime, are an indication and measure of space-time curvature.
In GR, timelike geodesics describe the motion of free falling massive test particles. Convergence of the test particles ( moving closer together as they fall ) would indicate a gravitational field with a central point source ( CoM tidal effect ).
Lightlike geodesics for massless particles ( or null geodesics ) describe the 'curvature' experienced by the light beam as it travels from one side of the elevator to the other. It would deflect downward, whether the elevator is accelerating, or in equivalent gravity.

If Markus logs on tonight, I'm sure he can give you a much more 'in depth' explanation.
 

[Markus Hanke]

10 hours ago, geordief said:

So ,as I see things the acceleration obtaining in the first scenario should be causing a curvature of spacetime in more or less exactly (ie minus tidal effects) the same way as occurs in conditions of gravity.

No, this is a common misunderstanding. Spacetime in a uniformly accelerated reference frame in otherwise empty space is perfectly flat - acceleration is not a source of spacetime curvature. Only distributions of energy-momentum (like planets, stars,...) are. Physically speaking, the defining characteristic of a curved spacetime is the presence of tidal gravity - however, if the reference frame in question is small enough, these tidal effects become negligibly small, which is why in a small local area only uniform acceleration looks like a uniform gravitational field. This is just the equivalence principle, and this is somewhat akin to the surface of Earth looking flat so long as you only look at a small enough section of it.

Anyway, the upshot is that a uniformly accelerated frame in otherwise empty space has no spacetime curvature (the Riemann tensor vanishes).

[MigL]

1 hour ago, Markus Hanke said:

Spacetime in a uniformly accelerated reference frame in otherwise empty space is perfectly flat

While that is obviously true, what then is the cause of the curved trajectories ?
Is a geodesic not an indication of curvature ?

[geordief]

2 hours ago, Markus Hanke said:

No, this is a common misunderstanding. Spacetime in a uniformly accelerated reference frame in otherwise empty space is perfectly flat - acceleration is not a source of spacetime curvature. Only distributions of energy-momentum (like planets, stars,...) are. Physically speaking, the defining characteristic of a curved spacetime is the presence of tidal gravity - however, if the reference frame in question is small enough, these tidal effects become negligibly small, which is why in a small local area only uniform acceleration looks like a uniform gravitational field. This is just the equivalence principle, and this is somewhat akin to the surface of Earth looking flat so long as you only look at a small enough section of it.

Anyway, the upshot is that a uniformly accelerated frame in otherwise empty space has no spacetime curvature (the Riemann tensor vanishes).

Suppose the observer at x,y,z=0,0,0  in the accelerated room  sends a pair of "parallel" beams in the direction of 1,1,1  will she adjudge them to  meet  in the distance or remain parallel? 

 

And the same question for the room in the gravitational field (either from a small source such as a black hole) or an infinitely  large source)  

 

Feel flattered to have graduated to the level of   this "common misunderstanding"😣

Edited September 23, 2020 by geordief

[joigus]

2 hours ago, Markus Hanke said:

No, this is a common misunderstanding. Spacetime in a uniformly accelerated reference frame in otherwise empty space is perfectly flat - acceleration is not a source of spacetime curvature. Only distributions of energy-momentum (like planets, stars,...) are. Physically speaking, the defining characteristic of a curved spacetime is the presence of tidal gravity - however, if the reference frame in question is small enough, these tidal effects become negligibly small, which is why in a small local area only uniform acceleration looks like a uniform gravitational field. This is just the equivalence principle, and this is somewhat akin to the surface of Earth looking flat so long as you only look at a small enough section of it.

Anyway, the upshot is that a uniformly accelerated frame in otherwise empty space has no spacetime curvature (the Riemann tensor vanishes).

Just re-phrasing Markus: Any acceleration field you can remove globally by changing the coordinate system does not betray curvature.

Acceleration fields can be compensated locally, but curvature persists.

Y-a-t'il de la curvature? Cherchez la T mu nu!

Except for point-like sources, which are just a fantasy.

[geordief]

1 hour ago, joigus said:

Just re-phrasing Markus: Any acceleration field you can remove globally by changing the coordinate system does not betray curvature.

Acceleration fields can be compensated locally, but curvature persists.

Y-a-t'il de la curvature? Cherchez la T mu nu!

Except for point-like sources, which are just a fantasy.

You are not saying ,are you that there is any point in the uniformly accelerating room where my two parallel beams of light can  be judged by an observer to be Euclideanly parallel?

 

I thought such parallelness was only a feature of  spacetime devoid of gravity.

[joigus]

2 minutes ago, geordief said:

You are not saying ,are you that there is any point in the uniformly accelerating room where my two parallel beams of light can  be judged by an observer to be Euclideanly parallel?

 

I thought such parallelness was only a feature of  spacetime devoid of gravity.

The point is subtle. The equivalence principle tells you that locally, gravity is indistinguishable from acceleration. But gravity (curvature) goes the next step. Curvature gives you a criterion to judge whether your "field" comes from real presence of energy-momentum (its source, what generates it) or is just an effect of your coordinate system (equivalent to "state of motion", but not just that, but also scale factors, ie., prescriptions for measuring lengths and times, etc.).

In your uniformly accelerated room no tidal forces would appear, because there is no curvature. Picture, if you will, a "naively curved" surface, like a cylinder. A cylinder looks curved to you only because you're looking at it from an embedding 3-dimensional space. But the cylinder is really flat, so geodesics (whether null or not) would be comparable to straight lines that only the 3-d observer would see as curves. In reality, inner observers would never see any trajectories separate. Nor would the observers inside the accelerated room appreciate any separation between photon trajectories. There is Doppler effect, yes, but that's not curvature. Curvature (deviation of geodesics) is equivalent to tidal forces, not just accelerations.

[Markus Hanke]

2 hours ago, MigL said:

While that is obviously true, what then is the cause of the curved trajectories ?
Is a geodesic not an indication of curvature ?

That is just the point - an accelerated frame does not trace out a geodesic in spacetime. 
In the original example, you have an enclosed box under uniform acceleration, so the box itself traces out a world line that is not a geodesic. If you now place a test particle into the interior of the box, and release it, then that test particle will be in free fall, and will thus trace out a geodesic (until it makes contact with one of the walls). The fact that the test particle’s spatial trajectory is curved with respect to the box is precisely due to the difference in the nature of their world lines - that’s one way to look at it, anyway.

1 hour ago, geordief said:

Suppose the observer at x,y,z=0,0,0  in the accelerated room  sends a pair of "parallel" beams in the direction of 1,1,1  will she adjudge them to  meet  in the distance or remain parallel? 

They will remain parallel, if there is only acceleration, but no sources of gravity - this is a flat spacetime.

1 hour ago, geordief said:

And the same question for the room in the gravitational field (either from a small source such as a black hole) or an infinitely  large source)  

If there are sources of gravity, then spacetime is curved, and they will not remain parallel. However, iff the room is small enough, then their geodesic deviation will be so small as to be negligible - in which case these two scenarios become equivalent. But this is true only if the room is small enough.

2 hours ago, geordief said:

Feel flattered to have graduated to the level of   this "common misunderstanding"😣

It wasn’t meant as an insult, please don’t take it personally. It really is a very common misunderstanding - and one which I myself have fallen afoul of in the past (and I made an utter fool of myself on some forum trying to argue the point).

[geordief]

11 minutes ago, joigus said:

The point is subtle. The equivalence principle tells you that locally, gravity is indistinguishable from acceleration. But gravity (curvature) goes the next step. Curvature gives you a criterion to judge whether your "field" comes from real presence of energy-momentum (its source, what generates it) or is just an effect of your coordinate system (equivalent to "state of motion", but not just that, but also scale factors, ie., prescriptions for measuring lengths and times, etc.).

In your uniformly accelerated room no tidal forces would appear, because there is no curvature. Picture, if you will, a "naively curved" surface, like a cylinder. A cylinder looks curved to you only because you're looking at it from an embedding 3-dimensional space. But the cylinder is really flat, so geodesics (whether null or not) would be comparable to straight lines that only the 3-d observer would see as curves. In reality, inner observers would never see any trajectories separate. Nor would the observers inside the accelerated room appreciate any separation between photon trajectories. There is Doppler effect, yes, but that's not curvature. Curvature (deviation of geodesics) is equivalent to tidal forces, not just accelerations.

Is that because the sources of gravity  (the sources of the gravitational accelerations)  are multi-directional?

 

I f the gravity field was infinitely large there would be no curvature?

7 minutes ago, Markus Hanke said:

 

It wasn’t meant as an insult, please don’t take it personally. It really is a very common misunderstanding - and one which I myself have fallen afoul of in the past (and I made an utter fool of myself on some forum trying to argue the point).

I just meant it as a joke at my own expense

Edited September 23, 2020 by geordief

[Markus Hanke]

1 minute ago, geordief said:

Is that because the sources of gravity  (the sources of the accelerations)  are multi-directional?

It’s like longitudinal lines on the surface of the Earth - the closer you get to the pole, the less the distance between these lines becomes (and vice versa).

5 minutes ago, geordief said:

I f the gravity field was infinitely large there would be no curvature?

It’s rather the other way around - if you only look at a small enough region of the field, then it will appear almost flat.

[joigus]

1 minute ago, geordief said:

Is that because the sources of gravity  (the sources of the accelerations)  are multi-directional?

It's rather that localised sources (matter) give rise to spread around gravitational fields (the little g's, the acceleration 3-fields, althought they're not invariant or covariant objects). Because they spread, they change from place to place.

As Markus says, the misunderstanding is common. I also have made the same mistake. Even after one learns it, sometimes one forgets.

-----------------------------

Ignorable ramblings:

Something very peculiar happens with GR. In most other field theories, you have sources; and field divergences are identified with sources by means of field equations, so that the fields kind of "pour out" from the source. Schematically:

divergence(field) = (C)x(source)

C=coupling

In GR, both gravitational field and sources are divergence-less. And one is identified with the other. I've no idea of what the Einstein tensor represents from a purely geometrical POV. The particular combination of second derivatives of the metric that the Einstein tensor represents is built in such a way that it does not "pour out" from anywhere. Neither does energy-momentum.

GR is a completely different scheme of things:

field = (C)x(source)

and

divergence(field) = 0

divergence(source) = 0

Where the field is G, the Einstein tensor (not the little g's), this very peculiar combination that does not represent all the degrees of freedom coded in the curvature, but only certain sums of it.

[MigL]

Thanks for the clarification guys.
Its always a good day when you learn something new.

[Markus Hanke]

1 minute ago, joigus said:

I've no idea of what the Einstein tensor represents from a purely geometrical POV.

When you pass a unit time-like future oriented vector to both slots of the Einstein tensor, you get a scalar that is just precisely the average (!) Gaussian curvature in the spatial directions within a small neighbourhood. So the Einstein tensor is a measure of average curvature around an event. In vacuum, geodesics will diverge in some directions and converge in others, in such a way that the average balances to exactly zero - hence the vanishing of the Einstein tensor in vacuum.

6 minutes ago, joigus said:

this very peculiar combination that does not represent all the degrees of freedom coded in the curvature, but only certain sums of it.

Indeed - it’s average Gaussian curvature, and thus it captures only a particular subset of overall Riemann curvature.

[joigus]

3 minutes ago, Markus Hanke said:

So the Einstein tensor is a measure of average curvature around an event. In vacuum, geodesics will diverge in some directions and converge in others, in such a way that the average balances to exactly zero - hence the vanishing of the Einstein tensor in vacuum.

I nod in awe, because that's awesome.