Equivalence Principle Misunderstanding?

[EskWIRED]

I've seen a lot of statements regarding Einstein's equivalence principle.

Many formulate it to say that no experiment can distinguish between a reference frame in a gravitational field and an accelerating reference frame.

But - isn't is true that in a gravitational field, tidal effects are present, while in an accelerating frame, no such tidal effects re present?

Isn't this a dead giveaway as to which phenomenon is being observed?

Is the problem merely a sloppy description of the equivalence principle?

Is a more precise formulation that one cannot distinguish between the two ONLY in the context of an experiment involving a single fundamental particle?

If so, ISTM that the principle is very limited in its applicability. If so, does this limitation have any profound implications?

Am I being pedantic? Just complaining about fuzzy descriptions I've read in the popular literature?

Where can I read more precise and accurate descriptions of the principle, its limitations and its implications?

 

[HallsofIvy]

The equivalence principal says the you cannot distinguish locally between an acceleration and a force. Tidal effects are not local.

 

[A.T.]

Many formulate it to say that no experiment can distinguish between a reference frame in a gravitational field and an accelerating reference frame.

An uniform gravitational field.

 

[yuiop]

I would say it is useful in the same way that Newtonian physics is useful. If we are being very fussy, Newtonian physics is just an approximation of relativity and only accurate for situations where nothing is moving and maybe not even then, but it is still useful for everyday calculations. Einstein used the equivalence principle as a guiding light in formulating GR. He knew if he was on the right tracks that, the equations of GR should reduce to those of SR, in a local (enough) region, just as should reduce to Newtonian physics in the weak field limit. I tend to agree that many popular descriptions over emphasize the equivalence aspect and under emphasize the limitations. Something that might be worth investigating further in your studies, is that the "local" aspect applies to space and time.

 

[yuiop]

An uniform gravitational field.

Just curious how a uniform gravitational field is defined. Is it one where the proper acceleration is the same at any height?

 

[EskWIRED]

I will learn more about the meaning of "local".

Thanks guys.

 

[bahamagreen]

Einstein, Minkowski, et al define "local" to mean infinity close, the distance radius from a point being "local" when that radius is infinitely small, approaching zero.

It makes the math work, but "local" becomes smaller than any discernible measure, smaller than atoms, smaller than the Planck length, smaller than super strings... infinitely so. For practical purposes there is no "local", for theoretical mathematical purposes, there is, but it is infinity small.

 

[yuiop]

For practical purposes there is no "local", for theoretical mathematical purposes, there is, but it is infinity small.

Can we not use a definition of local "for practical purposes" that is based on an acceptable margin of error? For example if we decide to define local as the region within which SR calculations are accurate to within 0.1% then this "local region" would be quite large in a weak field, but significantly smaller in a strong field. Would that work?

By analogy, a Newtonian calculation is in error for any any non zero velocity, but for practical purposes the margin of error in most everyday situations is so small, especially when the practical limitations of measuring equipment is taken into account, that the errors are negligible?

 

[PeterDonis]

Einstein, Minkowski, et al define "local" to mean infinity close, the distance radius from a point being "local" when that radius is infinitely small, approaching zero.

This is not the correct definition of "local" for the purpose that we're using the term here. The correct definition is that a "local" region of spacetime is a region centered on some particular event of interest, which is small enough that it can be considered flat to within the desired accuracy--i.e., within the local region, the effects of curvature are too small to measure at the desired accuracy. Tidal effects are effects of curvature, and so, as HallsOfIvy said, they are not local; they are not measurable in a local region of spacetime.

The reason this definition works is that one can actually find such "local" regions of spacetime. In other words, we can find local regions in which we can treat spacetime as flat and find the equivalence principle to hold--we can set up local inertial coordinates around particular events and find that, within those coordinates, objects freely falling under gravity (like falling rocks) behave like inertial objects in flat spacetime and objects at rest in the gravitational field (like observers standing at rest on the surface of the Earth) behave like accelerated objects in flat spacetime (such as an observer standing on the floor of a rocket with thrust equal to 1 g acceleration)--while the region is still small enough that tidal effects are negligible.

Edit: I see yuiop posted along the same lines; I am basically saying that what he proposes *is* the standard definition of "local" in GR.

 

[DrGreg]

Just curious how a uniform gravitational field is defined. Is it one where the proper acceleration is the same at any height?

Well, I guess if you want the statement "no experiment can distinguish between a reference frame in a uniform gravitational field and an accelerating reference frame" to be exactly true (and not just approximately true locally to first order) then a uniform gravitational field has to be one where the variation with height is the same as that in Rindler coordinates. Slightly odd use of the word "uniform", but there you go.

The alternative approach is not to worry about uniformity and say that the equivalence principle is true only "locally" i.e. as a first-order approximation, i.e. ignore variation proportional to squared distance or smaller.

 

[rbj]

sorry, being an electrical engineer, i don't know all the words...

Well, I guess if you want the statement "no experiment can distinguish between a reference frame in a uniform gravitational field and an accelerating reference frame" to be exactly true (and not just approximately true locally to first order) then a uniform gravitational field has to be one where the variation with height is the same as that in Rindler coordinates. Slightly odd use of the word "uniform", but there you go.

do you mean that a uniform gravitational field has to be constant with variation in height? Dunno Rindler coordinates.

thanx,

r b-j

 

[DrGreg]

do you mean that a uniform gravitational field has to be constant with variation in height? Dunno Rindler coordinates.

Rindler coordinates apply to a rigid rocket (in the absence of gravity) undergoing proper acceleration that is constant over time but is inversely proportional to height. The rocket is at rest relative to the coordinates.

 

[rbj]

Rindler coordinates apply to a rigid rocket (in the absence of gravity) undergoing proper acceleration that is constant over time but is inversely proportional to height.

how is it constant in time, when it is not constant in height and the height changes with time?

 

[atyy]

Where can I read more precise and accurate descriptions of the principle, its limitations and its implications?

The equivalence principal says the you cannot distinguish locally between an acceleration and a force. Tidal effects are not local.

The alternative approach is not to worry about uniformity and say that the equivalence principle is true only "locally" i.e. as a first-order approximation, i.e. ignore variation proportional to squared distance or smaller.

A useful discussion of the above points of view are in http://www.pma.caltech.edu/Courses/ph136/yr2006/0424.1.K.pdf , section 24.7: "In this conclusion the word local is crucial: The equivalence principle is strictly valid only at the spatial origin of a local Lorentz frame; and, correspondingly, it is in danger of failure for any law of physics that cannot be formulated solely in terms of quantities which reside at the spatial origin"

 

[DrGreg]

how is it constant in time, when it is not constant in height and the height changes with time?

Sorry, I should have explained that height is being measured relative to a fixed point on the rocket*. Remember an accelerating rocket without gravity is equivalent to a "stationary" rocket with gravity (e.g. on the Earth's surface).


_______
*Or, rather, to be more precise, a point that is a fixed distance c^2/a underneath a part of the rocket undergoing proper acceleration a, as measured by the rocket.

 

[rbj]

Sorry, I should have explained that height is being measured relative to a fixed point on the rocket. Remember an accelerating rocket without gravity is equivalent to a "stationary" rocket with gravity (e.g. on the Earth's surface).

i understand that. i am (now) trying to understand what the distance c^2/a is about. i don't get where this value comes from.

 

[Dale]

i understand that. i am (now) trying to understand what the distance c^2/a is about. i don't get where this value comes from.

Remember that the path of a uniformly (proper) accelerating object is a hyperbola in spacetime. That distance is essentially the "radius" of that hyperbola. A smaller radius of curvature corresponds to a higher acceleration.

 

[yuiop]

i understand that. i am (now) trying to understand what the distance c^2/a is about. i don't get where this value comes from.

For a rocket undergoing Born rigid acceleration, such that the proper length of the rocket remains constant, the proper acceleration (a) on any given floor of the rocket is equal to c²/r, where r is the constant distance measured inside the rocket from the hypothetical Rindler horizon. If the proper acceleration measured on one floor is a₁ and the proper acceleration measured on a higher floor is a₂ then the fixed ruler distance between those floors, as measured on board the rocket, is (c²/a₂ - c²/a₁). The radar distance between floors will be different and will also depend upon whether it measured from above or below.

That explains a little more about what c²/a is, but not where it comes from. Dalespam's answer that the it is the constant hyperbolic radius of the rocket's path gives some geometrical insight into that.

 

[EskWIRED]

Remember that the path of a uniformly (proper) accelerating object is a hyperbola in spacetime. That distance is essentially the "radius" of that hyperbola. A smaller radius of curvature corresponds to a higher acceleration.

I'm having trouble with that. Or more specifically, why that is the case.

Is it because of the mass of the object itself? I notice that you did not specify that the object is in a gravitational field.

Is your statement true in intergalactic space where gravitational fields are so weak that they can be ignored? If so, again, is the phenomenon due to the gravity attributable to the object itself?

Or is the phenomenon due to the curvature of the universe itself, and a corollary of the fact that spacetime is everywhere curved?

Or is it because, due to the equivalence principle, the high acceleration is equivalent to the gravitational field produced by a large amount of mass?

Or ... ?

 

[yuiop]

I'm having trouble with that. Or more specifically, why that is the case.

Is it because of the mass of the object itself? I notice that you did not specify that the object is in a gravitational field.

Is your statement true in intergalactic space where gravitational fields are so weak that they can be ignored? If so, again, is the phenomenon due to the gravity attributable to the object itself?

Or is the phenomenon due to the curvature of the universe itself, and a corollary of the fact that spacetime is everywhere curved?

Dalespam was talking about a rocket artificially accelerated in flat space, far away from any massive gravitational objects. The proper acceleration and redshift of signals measured inside the artificially accelerated rocket is equivalent locally (in a very small region) to a stationary rocket sitting on a massive gravitational object.

In a real gravitational field, the vertical distance will differ significantly from c^2/a over any non local height difference and that is one of the limitations of the equivalence principle.

You added this after I posted:

Or is it because, due to the equivalence principle, the high acceleration is equivalent to the gravitational field produced by a large amount of mass?

This is more like it ;)

 

[yuiop]

Well, I guess if you want the statement "no experiment can distinguish between a reference frame in a uniform gravitational field and an accelerating reference frame" to be exactly true (and not just approximately true locally to first order) then a uniform gravitational field has to be one where the variation with height is the same as that in Rindler coordinates.

I am a little bit bothered by this statement because as far as I am aware there is no gravitational field (even in principle) where the measurements would be exactly the same as in an accelerating reference frame. For example an infinitely long massive cylinder or an an infinitely flat massive slab would still not be exactly equivalent to an accelerating reference frame over non local distances. I may be wrong and would be interested in any thoughts on this.

Secondly, can we produce an array of accelerating rockets that can exactly duplicate the gravitational field around a massive spherical object, e.g. all accelerating away from a central point so that they look like an expanding sphere. I think the answer is no. Over any non local distance the differences would be obvious.

 

[DrGreg]

I am a little bit bothered by this statement because as far as I am aware there is no gravitational field (even in principle) where the measurements would be exactly the same as in an accelerating reference frame. For example an infinitely long massive cylinder or an an infinitely flat massive slab would still not be exactly equivalent to an accelerating reference frame over non local distances. I may be wrong and would be interested in any thoughts on this.

Secondly, can we produce an array of accelerating rockets that can exactly duplicate the gravitational field around a massive spherical object, e.g. all accelerating away from a central point so that they look like an expanding sphere. I think the answer is no. Over any non local distance the differences would be obvious.

I think everything you say is correct, so my opinion is that there's no point in trying to work out circumstances when the equivalence principle is exactly true, and accept that it's only ever locally true as an approximation.

Bear in mind that wherever you go in the Universe, although you may find places where the effects of gravity are negligible, you'll never (according to theory) find somewhere where it's mathematically zero.

 

[Dale]

as far as I am aware there is no gravitational field (even in principle) where the measurements would be exactly the same as in an accelerating reference frame.

You might want to check out this paper:
http://cdsweb.cern.ch/record/328471/files/9706071.pdf

I'm still digesting it, but it seems relevant.

 

[atyy]

At the origin of Fermi normal coordinates, acceleration (like Rindler coordinates) produces first order deviations (non-zero Christoffel symbols) from flat coordinates, while tidal gravity produces second order deviations (zero Christoffel symbols, non-zero Christoffel symbol derivatives) from flat coordinates.

 

[yuiop]

There seems to a hard line extremist "local" faction in this thread that agrees with the definition that" 'local' becomes smaller than any discernible measure, smaller than atoms, smaller than the Planck length, smaller than super strings... infinitely so". Mathematically this is correct if we expect the equivalence principle to hold exactly, but perhaps is too extreme for practical purposes? For example, if we apply the same exacting standards to SR, then SR is only valid for particles with zero mass (and energy) because any non zero mass introduces a gravitational field that distorts the ideal flat spacetime that SR assumes.

Can we afford to be that demanding in our expectations of accuracy, in a universe where the uncertainty principle tells us that we cannot even simultaneously know the exact momentum and exact position of a particle at any given instant?

 

[Austin0]

Well, I guess if you want the statement "no experiment can distinguish between a reference frame in a uniform gravitational field and an accelerating reference frame" to be exactly true (and not just approximately true locally to first order) then a uniform gravitational field has to be one where the variation with height is the same as that in Rindler coordinates. Slightly odd use of the word "uniform", but there you go.

The alternative approach is not to worry about uniformity and say that the equivalence principle is true only "locally" i.e. as a first-order approximation, i.e. ignore variation proportional to squared distance or smaller.

Hi I have related question. Were the Rindler coordinates developed in GR from the equations there and then exported to an accelerating system in SR or is it the other way around?
I thought I read that the gamma relation to potential altitude was derived from calculations in the context of SR and were part of the process of formulating GR?.
Thanks

 

[atyy]

There seems to a hard line extremist "local" faction in this thread that agrees with the definition that" 'local' becomes smaller than any discernible measure, smaller than atoms, smaller than the Planck length, smaller than super strings... infinitely so". Mathematically this is correct if we expect the equivalence principle to hold exactly, but perhaps is too extreme for practical purposes? For example, if we apply the same exacting standards to SR, then SR is only valid for particles with zero mass (and energy) because any non zero mass introduces a gravitational field that distorts the ideal flat spacetime that SR assumes.

Can we afford to be that demanding in our expectations of accuracy, in a universe where the uncertainty principle tells us that we cannot even simultaneously know the exact momentum and exact position of a particle at any given instant?

It's exact in the theory, but of course (unless you are a believer in humanity having the final theory, as some physicists dream) the theory is not exact with respect to reality.

 

[A.T.]

You might want to check out this paper:
http://cdsweb.cern.ch/record/328471/files/9706071.pdf

I'm still digesting it, but it seems relevant.

Thanks. In the abstract they also state that the "Rindler solution" represents an uniform gravitational field. How to reconcile this with the fact pointed out by DrGreg that the proper acceleration varies with height in Rindler coordinates?

 

[Dale]

Thanks. In the abstract they also state that the "Rindler solution" represents an uniform gravitational field. How to reconcile this with the fact pointed out by DrGreg that the proper acceleration varies with height in Rindler coordinates?

I think that uniform means that the field corresponds to the field which would be considered uniform historically, I.e. In Newtonian gravity. They didn't attempt to reconcile it here, merely presented it as an accepted meaning.

 

[DrGreg]

Hi I have related question. Were the Rindler coordinates developed in GR from the equations there and then exported to an accelerating system in SR or is it the other way around?
I thought I read that the gamma relation to potential altitude was derived from calculations in the context of SR and were part of the process of formulating GR?.
Thanks

I don't know the historical order of development, but Rindler coordinates apply only in SR (i.e. in flat spacetime, i.e. in the absence of gravity). The mathematical relationship between Rindler coordinates and the standard Minkowski inertial coordinates in SR has many similarities between the mathematical relationship between Schwarzschild coordinates and Kruskal–Szekeres coordinates in GR.

 

[yuiop]

Hi I have related question. Were the Rindler coordinates developed in GR from the equations there and then exported to an accelerating system in SR or is it the other way around?
I thought I read that the gamma relation to potential altitude was derived from calculations in the context of SR and were part of the process of formulating GR?.
Thanks

I did a internet search but failed to find a single reference to when Rindler actually introduced his coordinate system. If they had been developed before Einstein introduced GR in 1915, they would have provided a clue to expect a coordinate singularity or event horizon for a black hole in the GR solutions. Personally I find it fascinating that SR and the equivalence principle predicts event horizons. Unfortunately, the role of Rindler coordinates in the development of GR does not seem to get much of a mention in the historical records (or I am looking in the wrong places).

 

[yuiop]

I think that uniform means that the field corresponds to the field which would be considered uniform historically, I.e. In Newtonian gravity. They didn't attempt to reconcile it here, merely presented it as an accepted meaning.

I have been wondering if a uniform field implies a field in which neighbouring vertical lines (as measured by plumb bobs) are parallel?

 

[DrGreg]

I did a internet search but failed to find a single reference to when Rindler actually introduced his coordinate system. If they had been developed before Einstein introduced GR in 1915, they would have provided a clue to expect a coordinate singularity or event horizon for a black hole in the GR solutions. Personally I find it fascinating that SR and the equivalence principle predicts event horizons. Unfortunately, the role of Rindler coordinates in the development of GR does not seem to get much of a mention in the historical records (or I am looking in the wrong places).

Rindler coordinates are named after Wolfgang Rindler who wasn't even born until 1924, but (if the entirely unsourced Wikipedia article is to be believed(?)) was the person who invented the term "event horizon".

 

[atyy]

I'm uncertain about terminology - I believe Rindler in his GR textbook says that Rindler coordinates are not a "uniform" "gravitational field".