Equivalence Principle: Are there any restrictions

[Jorl17]

Hello all.

I've been thinking a lot about the Equivalence Principle as it has been taught to me: "The effects of the acceleration (vector) a of a referential are indistinguishable of those of a gravitational field (vector) g=-a". The same book has the phrase: "There is no experience that allows us to distinguish if we are in a gravitational field or attached to an accelerated referential".

Now, here's a simple question: Are there any restrictions? Specifically, as I cannot understand from the sentence, are we assuming a constant (vector) a?

Why am I asking that? I'm not pondering if the module of (vector) a is varying, I'm rather thinking about cases where (vector) a changes in direction -- for example, centripetal acceleration in a uniform rotation.

Can I say that even if (vector) a varies over time (absolute time), the full effects can be described by
$$\vec{g}=-\vec{a}$$?

And if your answer is yes, then there is a more specific question: Does this imply that (vector) g varies as well, so that (vector) g isn't equal to (vector) a₀ but a changing acceleration over time?

I hope my question is clear, thanks,

Jorl17

 

[bcrowell]

I'm not pondering if the module of (vector) a is varying, I'm rather thinking about cases where (vector) a changes in direction -- for example, centripetal acceleration in a uniform rotation.

To an observer rotating on a carousel in flat spacetime, the direction of the gravitational field is not changing over time.

One way of looking at it is that the equivalence principle is basically a statement that $\vec{g}$ is not the right thing to study. The distinction between time-varying and constant $\vec{g}$ is uninteresting in GR for the same reason that $\vec{g}$ itself is uninteresting in GR. $\vec{g}$ is not a tensor, and you can make it have any value you like by choosing coordinates. Similarly, $d\vec{g}/dt$ is not a tensor, and you can make it have any value you like by choosing coordinates.

Another way of looking at it is that the e.p. basically says that different test bodies follow the same geodesic when given the same initial position and velocity. From this point of view, it should be clear that the e.p.'s validity isn't dependent on any distinction between constant acceleration and non-constant acceleration. If two bodies "fly in formation," i.e., their world-lines are essentially the same, then that's true regardless of your coordinate system (accelerating at a constant rate, at a changing rate, etc.).

 

[Jorl17]

Thanks, but I don't think that answered my original question (or I simply didn't get it). Does the Equivalence Principle imply a constant vectorial acceleration (direction included ofc)?

 

[bcrowell]

Thanks, but I don't think that answered my original question (or I simply didn't get it). Does the Equivalence Principle imply a constant vectorial acceleration (direction included ofc)?

No, it doesn't. It's valid regardless of whether the acceleration is constant. It isn't even possible to decide whether a particular acceleration is constant. (BTW, I edited my post after posting it but before you replied.)

 

[atyy]

In Newtonian gravity, it does seem that every acceleration could be mimicked by a gravitational field.

However, not every gravitational field can be mimicked by an accelerated frame, except locally. So it requires not so much constant acceleration, but a "constant" gravitational field (I don't know the technical statement for "constant" gravitational field off the top of my head, but it should have something to do with spatial derivatives or tidal forces).

 

[PAllen]

In Newtonian gravity, it does seem that every acceleration could be mimicked by a gravitational field.

What about, e.g. a cylinder spinning on its access. You are inside. Walk all around the outer wall, and force out from center is uniform. Now further imagine the cylinder is very flat (small dimension in direction of its spin access). Now it is known that spherical shell produces no net gravitation on the inside. I think a thin slice of this would preserve the effect. Then it seems no arrangement of mass could produce the pseudo-gravity of a spinning flat disc.

Obviously 'locally' you could mimic this; locally it is just uniform acceleration. Thus, I think the 'locally' restriction is required for gravity mimicking acceleration as well as for acceleration mimicking gravity.

 

[bcrowell]

However, not every gravitational field can be mimicked by an accelerated frame, except locally.

Well, in GR frames don't even exist globally -- they only exist locally.

Now it is known that spherical shell produces no net gravitation on the inside. I think a thin slice of this would preserve the effect.

I don't think it's true for a ring, but anyway I don't see why you need a physical structure here. All you have to do is transform Minkowski space, represented in the usual coordinates, into a rotating frame.

Then it seems no arrangement of mass could produce the pseudo-gravity of a spinning flat disc.

Zero mass everywhere produces this gravitational field.

 

[atyy]

Well, in GR frames don't even exist globally -- they only exist locally.

I was talking about Newtonian gravity (and Newtonian tidal forces).

 

[PAllen]

I don't think it's true for a ring, but anyway I don't see why you need a physical structure here. All you have to do is transform Minkowski space, represented in the usual coordinates, into a rotating frame.


Zero mass everywhere produces this gravitational field.

I think we're talking in different terms. I interpreted atty as saying that fictitious forces from non-inertial motion could be mimicked by gravity in a Newtonian context. I don't believe this. I don't believe an outward radial force around a ring can be reproduced by any mass distribution.

As for the claim about hollow shell theorem, I didn't work it out. I relied on an intuition that what works for 2 sphere in 3-space inverse square law should work for a 1-sphere in 2-space inverse square law.

 

[PAllen]

I think we're talking in different terms. I interpreted atty as saying that fictitious forces from non-inertial motion could be mimicked by gravity in a Newtonian context. I don't believe this. I don't believe an outward radial force around a ring can be reproduced by any mass distribution.

As for the claim about hollow shell theorem, I didn't work it out. I relied on an intuition that what works for 2 sphere in 3-space inverse square law should work for a 1-sphere in 2-space inverse square law.

Well, now I did calculate things, and immediately see that ring and hollow sphere are completely different,as Bcrowell stated. A ring would have zero internal force for inverse r force law; a sphere for and inverse square force law. However, my calculation still validates my main point. The fictitious force along rotating disk is simply proportional to r (distance from center of disk; for a given angular velocity). Any attempt to emulate this with a ring of matter fails, because the force will be a much more complicated function of radial distance from the center of the ring. In fact, after writing down my integrals for this case (and seeing they would be a bear to integrate exactly) I went looking online. I found the page below, with amusing coincidence that my integrals matched those on this page even down to choice of every symbol, except that I was only interested in the plane of the ring, not altitude above it. This page shows nice graphs of the non-linear r dependence for a ring of matter, validating that it cannot reproduce the fictitious forces of a spinning disk.

http://www.mathpages.com/home/kmath402/kmath402.htm