Equivalence Principal, Acceleration, Curved Space

[Widdekind]

According to the Equivalence Principal, the effects of Gravity are locally indistinguishable from those of Acceleration.

QUESTION: Since Gravity curves Space, a la the Flamm Paraboloid, does acceleration do the same ?

Acceleration does impose a comparable Time Dilation effect, from the bottom to the top of the typical proverbial "elevator" ...

 

[A.T.]

According to the Equivalence Principal, the effects of Gravity are locally indistinguishable from those of Acceleration.

"Locally" is the keyword here. An uniformly accelerated frame is equivalent to an uniform gravitational field.

QUESTION: Since Gravity curves Space, a la the Flamm Paraboloid, does acceleration do the same ?

Flamm's Paraboloid represents the purely spatial (nor space-time) curvature of the Schwarzschild metric, which is not the metric of an uniform gravitational field. The space-time metric in an uniformly accelerating frame and also in an uniform gravitational field is the http://en.wikipedia.org/wiki/Rindler_coordinates" . I think this space-time doesn't have intrinsic curvature, but is still different from a standard Minkowski space-time. It's not really the curvature that causes the effects(gravity, gravitational time dialtion), but rather any distortion of distances (metric).

 

[DrGreg]

Space-time curvature is responsible for gravitational "tidal effects". It describes how space-time deviates from "uniform acceleration". If there were such a thing as a "uniform gravitational field", it would be indistinguishable from no field at all. That's the equivalence principle. Real gravitational fields aren't uniform, but "locally" they almost are.

"Gravitational time dilation" really has nothing to do with gravity, it's due to the proper acceleration of the observer. You get it for observers hovering a fixed distance above planet, you also get it for accelerating observers in the absence of any gravitational source.

Curvature of space-time is an intrinsic property of space-time and does not depend on the observer or on the choice of coordinates to describe a given space-time.

So in the absence of a gravitational source, there is never any curvature, regardless of observer or coordinate choice.

Also, to be pedantic, the "metric" is also an intrinsic property of space-time and does not depend on the observer or on the choice of coordinates. The "Rindler metric" is therefore misnamed and really ought to called "the Minkowski metric expressed in Rindler coordinates".

 

[A.T.]

"Gravitational time dilation" really has nothing to do with gravity, it's due to the proper acceleration of the observer.

Doesn't a free falling clock in the center of the mass run slower than a free falling clock far away, regardless any proper acceleration?

You get it for observers hovering a fixed distance above planet, you also get it for accelerating observers in the absence of any gravitational source.

I agree that "gravitational time dilation" doesn't imply a "gravitational source". But by "gravity" (as opposed to "gravitation") I usually just mean an inertial force in an accelerated frame, which we sometimes call "the force of gravity". This force alway occurs together with gravitational time dilation.

Also, to be pedantic, the "metric" is also an intrinsic property of space-time and does not depend on the observer or on the choice of coordinates.

This puzzles me. Wouldn't an observer, moving relative to the gravitational source, observe a different, Lorentz-contracted metric? Or given two gravitational sources, moving relative to each other, wouldn't the metric created by their combined influence look different in thieir individual reference frames?

 

[DrGreg]

"Gravitational time dilation" really has nothing to do with gravity, it's due to the proper acceleration of the observer.

Doesn't a free falling clock in the center of the mass run slower than a free falling clock far away, regardless any proper acceleration?

What I said is probably a bit of an oversimplification. I'm thinking in terms of the local approximation in which the equivalence principle is valid. To be honest, I don't know the answer to that specific question.

You get it for observers hovering a fixed distance above planet, you also get it for accelerating observers in the absence of any gravitational source.

I agree that "gravitational time dilation" doesn't imply a "gravitational source". But by "gravity" (as opposed to "gravitation") I usually just mean an inertial force in an accelerated frame, which we sometimes call "the force of gravity". This force alway occurs together with gravitational time dilation.

Yes, in terms of relativistic terminology the "pseudo-gravity" inside an accelerating rocket isn't "pseudo" at all, it really is "gravity".

Also, to be pedantic, the "metric" is also an intrinsic property of space-time and does not depend on the observer or on the choice of coordinates. The "Rindler metric" is therefore misnamed and really ought to called "the Minkowski metric expressed in Rindler coordinates".

This puzzles me. Wouldn't an observer, moving relative to the gravitational source, observe a different, Lorentz-contracted metric? Or given two gravitational sources, moving relative to each other, wouldn't the metric created by their combined influence look different in thieir individual reference frames?

I'm referring here to the metric tensor, the 4x4 matrix of coefficients usually expressed by an equation of the form ds² = ... . Just as with 4-vectors, when you change coordinate systems, the values of the 16 numbers in the matrix will change, but it's still regarded as the "same" tensor, just expressed in a different coordinate system. This is just a question of terminology; the numbers change but it's still the "same entity". And if you think of the space metric inherited from the space-time metric, yes that really does change, as your decomposition of space-time into 3D space + 1D time changes.

To give an example, in flat, Euclidean 2D space, the equations

$$ds^2 = dx^2 + dy^2$$
$$ds^2 = dr^2 + r^2 d\theta^2$$​

both describe the same 2D Euclidean metric.

I am being a little pedantic here, because in practice people often (though technically incorrectly) use the word "metric" to refer to a specific equation in a particular coordinate system.