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There's an observer in a closed room without windows under an influence of gravity force. Can he determine what is the source of gravity - whether it's a spinning motion, acceleration or huge mass object?

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Locally, no. But since you can measure it in a nonzero volume, you could theoretically tell apart rotating or otherwise accelerating reference frame from the gravity of (point/spherical) masses. If you had enough space to fly the ([GOCE probe], you could deduce the mass distribution in surrounding objects... – dominecf Feb 29 at 20:51
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Not 'locally', where 'locally' means doing experiments over a sufficiently small region. Experiments done over larger regions can distinguish things.

What this means more formally is that spacetime is well-approximated by Minkowski spacetime in sufficiently small neighbourhoods, but can be distinguished from it on larger scales.

As an example, it would be very hard to distinguish sitting in a lift on Earth from sitting in an ordinary-sized lift being uniformly accelerated, and impossible to do so in the limit as the size of the lift goes to zero. However if the lift was 100 miles on a side it would be much easier (measure the direction that freely-falling objects move in, relative to the lift, at one edge of the floor and compare it with that at the other, for instance).

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The other thing you'd want to do in your 100 mile long lift is look for a horizon, since if it were curved just right, measuring the path of falling objects relative to the floor could be misleading. – Todd Wilcox Mar 1 at 0:31

LIGO is an example of an observer in a closed room. LIGO has seen a gravitational signal deduced to be from two inspiralling black holes. However, LIGO is non-local. If LIGO were vanishingly small (ie: the arm lengths L were made infinitely small so as to make a truly local measurement) the mirror motion dx due to the gravitational strain would be too small to detect (dx=L* strain). So, LIGO is not an example of determining the source of gravity from a "local" measurement, but does show it can be done by a measurement over an extended region.

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This is a really good example! – tfb Feb 29 at 22:09
More important, int he case of LIGO is not its size, but the fact that the LIGO experiment happens over a finite time. "sufficiently small" means "sufficiently small" in all four dimensions. – Jerry Schirmer Feb 29 at 22:25
@Jerry Yes, good point, as the LIGO arms shorten, the travel time of a light wave fronts along the two arms also shortens, and it is the difference in these travel times the interference pattern registers. So, as you correctly infer, a LIGO measurement is not local in either space or time. It would only become local if the arm lengths and light travel time shrunk to zero. – Gary Godfrey Mar 1 at 22:19

I will touch upon a very practical (not theoretical) aspect of the question.

Let us take the case of gravity vs uniform acceleration of a man sized lift, say between 4 and 10 feet tall, 4 feet wide and 4 feet deep. Let us ignore the spinning motion for this example

The assumption is uniform acceleration. However, uniform acceleration means a constant force at all the times. Think of any practical way of providing a constant force. Practicality is the key here.

Now, put an object on a sensitive weighing scale, and jump on the floor of the lift. In case of a constant force, there will be a momentary change in the weight of the object. In case of the gravity, either that change will not be there, or it will be much less than that of the constant force.

So, by bouncing inside the thing, it would be possible to figure out what it is.

In order for the lift to give same effect in case of a jump inside, it has to be as heavy as the planet. But in that case, It will cause enough gravity anyway.

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If I understand it correctly, you exert some varying force to the walls of the box to determine its mass. But this experiment does not tell apart a finite-weight box in a gravitational field from a finite-weight box pulled by some accelerating force. Thus, your jumping experiment does not provide any information on the origin of the constant acceleration. – dominecf Feb 29 at 22:24
@dominecf: No, just jump inside the lift, the lift acceleration will slow down due to the reaction and the object lying on the scale (on floor of lift) will loose weight momentarily due to decreased acceleration of the lift floor (till the acceleration is caught up again). It is like, you move forward in a little boat, the boat moves backward a little. If you do same in a big ship, the ship does not move backward. That is why, I am saying it is a practical aspect, not theoretical. – kpv Feb 29 at 22:28
But this is what I had in my mind - you can determine a small boat from a big ship. By jumping and observing the walls movement, you can not tell whether they are accelerating at constant rate, or pulled by some external object's gravity. – dominecf Feb 29 at 22:31
When you jump inside an accelerating lift, the acceleration of the lift is bound to slow down for all practical purposes. Because, you exert a force on the lift floor. You can also understand it in terms of conservation of momentum. What you are weighing, is a third object that is lying on a scale on the floor, before the jump, during the jump, and after the jump. That scale will show less reading for a moment. It appears pretty simple. Like someone jumping on a flex deck/patio, gives other person a feeling of going little low. – kpv Feb 29 at 22:36
What I am saying is constant acceleration for a third object is not practically possible, if you jump on the floor of the lift. It is only possible if it is gravity, due to enormous size of the planet. In any other case, you can momentarily slow it down for another object inside the lift, by jumping on the lift. Please move to a chat if still not clear. – kpv Feb 29 at 22:52

protected by Qmechanic Feb 29 at 22:33

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