Is equivalence principal really true?

[mritunjay]

Is it possible to eliminate Gravity completely even locally as equivalence principal says? I think this is not possible,since in a curved spacetime, Riemann tensor cann't be made zero in any frame. Then in what sense equivalence principal(which says that for a freely falling observer in a gravity field, laws of physics hold locally) is valid?
To me it seems that it is more of a order of magnitude business. Upto the order one cann't distinguish gravity from acceleration, laws of physics will hold upto that order.
the statement of equality of inertial and gravitational mass is, I think okey. But the alternative statement of equivalence principal (laws holding good locally in freely falling frame due to indistinguishable nature of gravity and acceleration) is not making sense to me? what are your opinians?

 

[atyy]

Yes, you are right.

 

[dx]

Curvature is a second order effect. Just like a smooth surface is locally like Euclidean space, a general relativistic spacetime is locally like Minkowski spacetime.

More precisely, for any given point in spacetime, one can always find a coordinate system in which the metric components at that point are diag(-1,1,1,1), and the first derivatives of the components are zero.

 

[Al68]

Are we missing the point of the equivalence principle? Of course there is no existing gravitational field that meets the strict conditions for equivalence, but that's not the point.

In the example given by Einstein, with the box in deep space being pulled by a rope, an observer could perform no local experiment to distinguish between proper acceleration in a uniform gravitational field and acceleration in the absence of one. That seems to be still true hypothetically even if no perfectly uniform gravitational field actually exists.

And a uniform gravitational field would be undetectable locally for an observer in freefall. That still seems to be true hypothetically, even if no such field actually exists. Perfectly inertial motion doesn't exist in reality either, but we still consider the concept important, and the equations correct in principle.

 

[atyy]

"More exactly, it says that a freely falling observer will not experience gravity except through the tidal forces, or equivalently, the curvature components." Susskind and Lindesay, http://books.google.com/books?id=cx...principle+string+theory&source=gbs_navlinks_s

 

[nutgeb]

As I've mentioned before, Einstein said that the equivalence principle should be applied only to a truly uniform gravitational field, not to one with tidal characteristics. Of course the uniform gravitational field is a fictional construct, it does not exist in the real world. Einstein felt that reducing the local inertial frame to its infinitesimal limit was not helpful because the accelerations would correspondingly be reduced to infinitesimal magnitudes.

 

[atyy]

As I've mentioned before, Einstein said that the equivalence principle should be applied only to a truly uniform gravitational field, not to one with tidal characteristics. Of course the uniform gravitational field is a fictional construct, it does not exist in the real world. Einstein felt that reducing the local inertial frame to its infinitesimal limit was not helpful because the accelerations would correspondingly be reduced to infinitesimal magnitudes.

Uniform field = Minkowski space?

 

[nutgeb]

No, uniform field meaning spacetime curvature that has no tidal characteristics -- e.g. it does not vary with distance from the source, and is not angular toward a point source.

An infinite gravitating plane would produce an infinite field of perfectly uniform strength, orthogonal to the planar surface.

 

[atyy]

No, uniform field meaning spacetime curvature that has no tidal characteristics -- e.g. it does not vary with distance from the source, and is not angular toward a point source.

An infinite gravitating plane would produce an infinite field of perfectly uniform strength, orthogonal to the planar surface.

Is this a solution of GR?

 

[mritunjay]

But, for a uniform gravity field case there exist a tranformation in which Riemann tensor becomes zero everywhere and hence manifold is flat for a uniform gravity field. (In short curvature is zero for a uniform gravity field)

 

[dx]

What do you mean by a uniform gravitational field? If you mean that there exists a coordinate system covering spacetime such that the metric components are constant, then yes that does mean that the curvature is zero, since that statement is the exact condition for spacetime being a flat minkowski space. A world with a 'uniform gravitational field' is exactly the same as a world with zero gravitational field, or any other value.

 

[nutgeb]

I don't think so. I don't think you can transform away a uniform acceleration. It's not like velocity. Acceleration is a very real effect -- you will fall to the floor of your spaceship.

 

[dx]

Uniform acceleration (of all bodies) can always be transformed away, and that is precisely the insight that allows us to regard particles in a gravitational field as 'free particles' moving on 'straight lines' or geodesics.

 

[Al68]

I don't think so. I don't think you can transform away a uniform acceleration. It's not like velocity. Acceleration is a very real effect -- you will fall to the floor of your spaceship.

It's not the proper acceleration that is transformed away, it's the (uniform) gravitational field.

Coordinate acceleration is frame dependent and can be transformed away easily.

 

[atyy]

The Ehlers and Rindler reference at the end does not formulate the EP in GR using uniform fields - that is a Newtonian conception. For comparison, the EP also holds in Newtonian gravity, and the theory can indeed be formulated geometrically as Newton-Cartan theory, as suggested by the Newtonian EP. http://www.aei.mpg.de/einsteinOnline/en/spotlights/equivalence_deflection/index.html

"The principle of equivalence has a local character. The mentioned equivalence is only valid as far as the measurements does not reveal a possible curvature of space." http://arxiv.org/abs/0806.0464

 

[nutgeb]

It's not the proper acceleration that is transformed away, it's the (uniform) gravitational field.

Coordinate acceleration is frame dependent and can be transformed away easily.

Well you sound like you know more about it than I do, so you may be right. But what does it mean to "transform away" the uniform acceleration when you always fall to the floor of your spaceship in stead of floating in freefall? It is obvious within this closed laboratory that some kind of acceleration is occurring, which may be either kinetic or gravitational. The observed effect of gravity in the closed laboratory will never 'go away.' So I'm wondering if were losing the forest for the trees here.

A uniform gravitational field means there are no acceleration gradients. It doesn't mean there's no gravitational acceleration.

 

[Al68]

Well you sound like you know more about it than I do, so you may be right. But what does it mean to "transform away" the uniform acceleration when you always fall to the floor of your spaceship in stead of floating in freefall? It is obvious within this closed laboratory that some kind of acceleration is occurring, which may be either kinetic or gravitational. The observed effect of gravity in the closed laboratory will never 'go away.' So I'm wondering if were losing the forest for the trees here.

Those effects are the result of proper acceleration and would be identical with or without the presence of a uniform G-field. And proper acceleration can't be transformed away. The G-field can be.

And coordinate acceleration can be. For example, if you're on a rocket thrusting at 1 G proper acceleration, hovering 10 feet above Earth's surface, then the coordinate acceleration is 1 G relative to an inertial frame. But the coordinate acceleration relative to Earth's surface is zero. And if the G-field were uniform, you could not detect locally whether or not the G-field was present. You could not determine locally whether you were near Earth's surface or in deep space. And the G-field could be transformed away simple by switching to an inertial frame in which the coordinate acceleration is 1 G instead of zero.

Likewise if in deep space, a G-field could be "transformed in" by switching from an inertial frame to the accelerated frame of the rocket in which the rocket is stationary and the "G-field" is making things "fall".

 

[atyy]

In GR, accelerations and gravity can both be locally transformed away. However accelerations and gravity are both absolute in that they respecively produce first and second order deviations from flatness, and so can be detected non-locally.

 

[mritunjay]

Well you sound like you know more about it than I do, so you may be right. But what does it mean to "transform away" the uniform acceleration when you always fall to the floor of your spaceship in stead of floating in freefall? It is obvious within this closed laboratory that some kind of acceleration is occurring, which may be either kinetic or gravitational. The observed effect of gravity in the closed laboratory will never 'go away.' So I'm wondering if were losing the forest for the trees here.

A uniform gravitational field means there are no acceleration gradients. It doesn't mean there's no gravitational acceleration.

transforming away uniform gravity field means you go to a frame in which there is no gravity i.e. you observe the first accelerating frame from an inertial frame. for transforming away gravity you don't have to stick to the accelerating frame.

 

[nutgeb]

You guys are correct, I don't know what I was thinking (probably about a spaceship siting on the surface of the plane). Of course a uniform gravitational field can be transformed away. If the spaceship is in freefall (allowing gravity to accelerate it) then the observer inside will feel no gravity.

This of course is the basic idea of the equivalence principle (especially as Einstein wanted it), that a uniform gravitational field can be transformed entirely away, with no tidal effects to perturb it.