# Equivalence Principle Misunderstanding?

1. May 25, 2012

### 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?

2. May 25, 2012

### HallsofIvy

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

3. May 25, 2012

### A.T.

An uniform gravitational field.

4. May 25, 2012

### 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.

5. May 25, 2012

### yuiop

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

6. May 25, 2012

### EskWIRED

Thanks guys.

7. May 25, 2012

### 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.

8. May 25, 2012

### yuiop

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?

9. May 25, 2012

### Staff: Mentor

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.

10. May 25, 2012

### DrGreg

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.

11. May 25, 2012

### rbj

sorry, being an electrical engineer, i dunno all the words...

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

thanx,

r b-j

12. May 25, 2012

### DrGreg

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.

13. May 25, 2012

### rbj

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

14. May 25, 2012

### atyy

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"

15. May 25, 2012

### DrGreg

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.

16. May 26, 2012

### rbj

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.

17. May 26, 2012

### Staff: Mentor

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.

18. May 26, 2012

### yuiop

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.

19. May 26, 2012

### EskWIRED

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 ... ?

20. May 26, 2012

### yuiop

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:
This is more like it ;)

Last edited: May 26, 2012