skippy1729

## Main Question or Discussion Point

Suppose I set up two labs, Alice on earth and Bob in a region of intergalactic space (or maybe in a void) where the Riemann tensor is zero. I accelerate Bob's lab at 32.2 ft/sec/sec. Both measure the Riemann tensor using identical equipment. Alice will get a nonzero value for some components, Bob will get zero for all components. My understanding of the equivalence principle is that they should get identical results for all classical experiments.

Thanks for any clarification. Skippy

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Matterwave
Gold Member
Why will Bob get zero for all components? If he is in an accelerated frame, he should no longer be in Euclidean space.

From my understanding (and it is a very humble one), Bob should arrive at the Riemann Curvature Tensor identical to Alice. Wasn't this the whole point of the equivalence principle? To say that an accelerated reference frame is in fact the same as one acted on by some force we call gravity?

It would be easier to imagine a woman on a disc which is rotating. She draws a circle, and tries to measure pi. She won't get 3.1415... because her circle is going to be distorted by the acceleration. So, to us on the outside, we say "hey you're in an accelerated reference frame, that's why you can't draw right and you don't get the right value of pi", but to her she says "No, I'm in a space that has a force of gravity, and this force of gravity is messing with my space".

This is how I understand this, but I may be wrong.

Ich
My understanding of the equivalence principle is that they should get identical results for all classical experiments.
...as long as curvature effects can be neglected. That's the equivalence principle.

Matterwave
Gold Member
How can you neglect curvature effects? I thought curvature was the entire new law of gravitation.

If curvature effects are different in the two frames, then objects would take different paths in the two frames (since the geodesics are different), and it would be simple to tell the two apart.

Again, I'm no expert so I'm asking a question, I'm not trying to argue here.

skippy1729
Why will Bob get zero for all components? If he is in an accelerated frame, he should no longer be in Euclidean space.
In a given domain, if a tensor is zero a particular coordinate system it will transform to zero in any other coordinate system.

Matterwave
Gold Member
That's true, I suppose. I will study this in more detail and get back to you in a few years. :P

Ich
How can you neglect curvature effects? I thought curvature was the entire new law of gravitation.
The EFE connect energy to curvature, that's true. But curvature isn't "gravitation", it's rather akin to Newtonian tidal acceleration.
Gravitational acceleration cannot be discerned from "normal" acceleration, in fact both are locally the same. That's the equivalence principle.
Curvature is the change of gravitational acceleration with position or time (not strictly true, but ok to illustrate the point), and can be neglected locally.
So you give Alice and Bob small instruments, eg accelerometers, and both measure the same. Give the large instruments capable of detecting accurately the difference from one position to another, they measure different things.

George Jones
Staff Emeritus
Gold Member
The equivalence principle is a somewhat vague and imprecise concept. The treatment that I like best links the equivalence principle with a model that takes spacetime to be a differentiable manifold for which the tangent space at each event is a Minkowski vector space. See from the bottom of page 7 to the end of page 13 in

for a readable exposition of this.

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Ich posted:
So you give Alice and Bob small instruments, eg accelerometers, and both measure the same. Give the large instruments capable of detecting accurately the difference from one position to another, they measure different things
I believe this is accurate but have never quite understood why. Just what are the different things measured that restricts this to "local" equivalent measures? Thanks.

skippy1729
The equivalence principle is a somewhat vague and imprecise concept. The treatment that I like best links the equivalence principle with a model that takes spacetime to be a differentiable manifold for which the tangent space at each event is a Minkowski vector space. See from the bottom of page 7 to the end of page 13 in

for a readable exposition of this.
If all it means is that the tangent space at a point is a flat Minkowski Space the Equivalence Principle
would seem to be a trivial consequence of general covariance. Any generally covariant metric theory of gravity (with correct signature) would automatically satisfy the Equivalence Principle since covariant derivatives reduce to partial derivatives in the tangent space.

If this is correct, it seems the Equivalence Principle has no physical content (independent of general covariance).

still confused, Skippy

Matterwave
Gold Member
So, if I read this correctly, all the Equivalence principle says is that we live on manifolds?

Ich
If this is correct, it seems the Equivalence Principle has no physical content (independent of general covariance).
Well, it means that gravity can be described by a metric theory, thus that gravitation is not a force, that there is only one simplest theory - GR -, how it looks in detail, with astonishing consequences.
If that's "no physical content", I suggest you try to unlearn it, re-derive GR and see how trivial it is.

atyy
If all it means is that the tangent space at a point is a flat Minkowski Space the Equivalence Principle
would seem to be a trivial consequence of general covariance. Any generally covariant metric theory of gravity (with correct signature) would automatically satisfy the Equivalence Principle since covariant derivatives reduce to partial derivatives in the tangent space.

If this is correct, it seems the Equivalence Principle has no physical content (independent of general covariance).

still confused, Skippy
Yes, within GR, the EP requires both
(i) locally, the metric looks Minkowskian
(ii) the local laws of physics are the same as those in special relativity

Because of semi-riemannian geometry, we always get (i), but (ii) is not necessarily true (what are the "local" laws of physics?).

Try chapter 24, section 7 "Curvature coupling delicacies" of Blandford and Thorne's http://www.pma.caltech.edu/Courses/ph136/yr2006/text.html.

Also, Carroll's http://nedwww.ipac.caltech.edu/level5/March01/Carroll3/Carroll4.html "From the modern point of view, we do not expect the EEP to be rigorously true."

The EP does not imply GR - but it does imply a geometric formulation. Since Newton's gravity obeys an EP, it has a geometric formulation as Newton-Cartan theory. There are also consistent relativistic theories of gravity that obey the EP and are not GR, such as Nordstrom's theory. Thus the EP is a constraint on theories of gravity, but is not a sufficiently strong constraint to imply GR.

BTW, general covariance has no physical content either. All theories can be cast in generally covariant form, ie. all theories give the same predictions if you use a diffeomorphism to act on everything. Some theories obey the EP, and of those theories, what distinguishes GR is that it has (a) "no prior geometry" or that the predictions of the theory are the same if you use a diffeomorphism to act on the dynamical fields and (b) something about no higher than second derivatives in the action - I don't know the exact statement.

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Al68
I think the key word may be "uniform". Uniform acceleration is the same whether it is due to gravity or applied force. It's the concept of acceleration that is equivalent. A rocket accelerating at 1 G might be increasing speed relative to earth, or hovering 10 feet off the ground. It's a 1 G accelerated reference frame either way. Any difference in experimental results is due to the non-uniformity of the gravitational field, not due to a conceptual difference between gravity and applied force.

I've been looking at this issue in a little detail recently, and some aspects of the opening post are incorrect. The metric for a uniform gravitational field can be written as

ds^2 = -(1+gz)^2 dt^2 + dx^2 + dy^2 + dz^2

(e.g. Principle of Equivalence, F. Rohrlich, Ann. Phys. 22, 169-191, (1963), page 173 ). If you measure the curvature for this, you'll find it's flat, and all the terms in the Riemann tensor are zero. A little algebra shows that this metric, however, is the same as an accelerated frame in Minkowski space - the two are equivalent (i.e. the EP).

What this uniform gravitational field is missing are the tidal forces, which are what the non-zero Riemann terms tell you.

DrGreg
Gold Member
I've been looking at this issue in a little detail recently, and some aspects of the opening post are incorrect. The metric for a uniform gravitational field can be written as

ds^2 = -(1+gz)^2 dt^2 + dx^2 + dy^2 + dz^2

(e.g. Principle of Equivalence, F. Rohrlich, Ann. Phys. 22, 169-191, (1963), page 173 ). If you measure the curvature for this, you'll find it's flat, and all the terms in the Riemann tensor are zero. A little algebra shows that this metric, however, is the same as an accelerated frame in Minkowski space - the two are equivalent (i.e. the EP).

What this uniform gravitational field is missing are the tidal forces, which are what the non-zero Riemann terms tell you.
But that doesn't disagree with post #1 at all! (Apart from the poster's question about the EP.)

Bob has the Rindler metric you wrote above and measures zero curvature. Alice is in Earth's non-uniform gravitational field and measures non-zero curvature.

Point taken - but the key point is that the "uniform" gravitational field has a zero Riemann tensor, just the same as uniform acceleration in Minkowski space, and the surface of the Earth is not a uniform gravitational field.