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Popular books suggest that for an observer in the Einstein lift the following situations are equivalent:

1, the lift hangs motionless (relative to the Earth) on a cable in the gravitational field of the Earth

2, there is no gravity but the lift is accelerated up

The natural question then is:

"If I stand on the Earth and feel its gravity does it mean that there is no gravity in fact and the Earth is expanding with acceleration?"

I know the short answer (no) but as a devil's advocate I would appreciate a longer version.

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The answer you have selected as the best, is wrong. The geodesics equation of GR holds trivially for an object standing at the surface of the Earth (and so it cannot be said that such object is not following a geodesic), since $dx^{i}/ds=0$ and $\Gamma ^{0} _{00}=0$ (the Schwarzschild metric is static, no $x^{0}$-dependent terms). And the Earth rotation doesn't change this in any significant way. –  Eduardo Guerras Valera Nov 25 '12 at 22:48
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3 Answers 3

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I can't improve on the two answers already posted, but I'd like to offer a slightly different viewpoint and claim that the Earth is indeed accelerating outwards.

An object is only not accelerating if it's following a geodesic, so if you are not following a geodesic you are by definition accelerating. The surface of the Earth is not following a geodesic, because an object following a geodesic would (from our perspective) accelerate towards the core, and therefore the surface of the Earth must be accelerating outwards.

We tend to intuitively think of accelerating as accelerating away from some spacetime frame, and indeed in SR this is a reasonable way to think of things. However it's also possible to think of the frame accelerating away from us, and in GR the two are equivalent. In an SR approach you'd argue that the surface of the Earth can't be accelerating outwards because if it was the volume of the Earth would be changing. However when you stand on the surface of the Earth the geodesics passing through you are curving away from your trajectory. In this sense spacetime is accelerating away from you.

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+1 for the twist. –  Chris White Nov 15 '12 at 5:50
    
Wait a moment! That is wrong! A non-moving object DOES follow the geodesics equation trivially, because $dx^{i}/ds$ equals zero, so it is not accelerating, by your own definition. Your statement is only valid for unconstraint particles. The geodesics equation of GR holds too for objects that are not moving due to (non holonomic in this case) constraints. Therefore, objects standing on the surface of the Earth are not accelerating, even in your definition. –  Eduardo Guerras Valera Nov 17 '12 at 17:21
    
I mean when $i$ denotes the spacial coordinates. Then, only $ \Gamma ^{0} _{00} $ needs to be checked in the geodesics equation, but it is zero because there is no $x^{0}$-dependent terms in the Schwarzschild metric. –  Eduardo Guerras Valera Nov 17 '12 at 21:05
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What the elevator story illustrates (and the meaning of the equivalence principle) is that you cannot distinguish $locally$ between an accelerated system and a gravitational field. Einstein uses too another example in the Princeton lectures in 1921: in a rotating platform, you would feel a centrifugal force near the outer edge. As with you at the surface of the Earth, the force at the point where you are is the same as a gravitational pull or the inertia in an accelerated elevator. But only in a small region and time interval around your position when you are there. There is no way you can emulate the whole 360deg radial outwards centrifugal field with gravitating mass. Conversely, there is no way to emulate the static inwards radial gravitational field of the whole Earth for a long time by means of a simple accelerated movement.

In more formal words, the equivalence principle allows you to change the coordinate system between any two non-inertial observers (you standing on the Earth surface, and somebody in an accelerating rocket far from the Earth) as if they were inertial, and physics equations shall hold equally well in both ones, but only in a small region and time interval around them. It makes no sense for the whole Earth, and that is why you are correctly detecting a paradox.

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The issue is local versus global measurements. If you are confined to the lift, then indeed you could never tell the difference between accelerating away from a point or being stationary in a gravitational field.

However, if you had other people in other lifts around the planet, and everyone was allowed to communicate, you would quickly see that the distance between observers is not changing. You just need to take measurements over large enough separations to find the contradiction. After all, if people were accelerating away from the center of the Earth at $9.8\ \mathrm{m}/\mathrm{s}^2$, we would notice it pretty soon.

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