Isn't the Equivalence Principle technically wrong?

Question

Isn't the Equivalence principle not quite true, technically speaking? A gravitational acceleration and an inertial acceleration are communicated very differently. A gravitational acceleration is communicated throughout your body and its stress is spread out in your body, whereas an inertial acceleration is communicated by points of contact and its stress is focused on these points of contact. And the different types of acceleration will also affect the fluid in your inner ears different, which means you will not be able to balance as easily in an acceleration caused by inertia as one caused by gravity.

Am I missing some caveat or rule or disclaimer, that makes these things irrelevant?

For instance in the classic elevator example, the walls of the elevator will communicate the acceleration to you. If the elevator is moving towards your head, the acceleration is communicated through your feet, and due to inertia your body is compressed. Whereas if you are on Earth and step off a building, your feet will fall first, and then the rest of your body, so you get stretched slightly.

And if you had a constant space elevator acceleration to match Earth's gravitational acceleration rate, your skeleton should be under more stress in the elevator because the acceleration has to begin at your feet and then overcome the inertia of your body, and the fluid in your inner ears would not give you the same sense of balance perception as if you are standing still on Earth. If you are standing still on Earth the acceleration is communicated simultaneously throughout your skeleton since your whole body is in Earth's gravity well.

What is it that the way the forces are communicated, and the different stresses they place on your body and skeleton, not considered a violation of the Equivalence principle? What am I missing?

Answer 1

What you are pointing out is in fact one of the key insights of the equivalence principle: if every atom in your body is accelerated in the same fashion then there is no differential acceleration across your body and therefore no force in your body and therefore no way of "feeling" the acceleration or experiencing it in a way that can be distinguished from free fall. Similarly if a scale you are standing on is accelerating at the same rate as you, then you will exert no force on it and you will be weightless. Since gravity results in the same acceleration on all atoms in your body and on any nearby scale (locally, i.e. ignoring tidal effects) freefall in gravity is therefore the same as being in an inertial frame that is not in a gravitational field.

In the case of acceleration, say in a rocket, yes, indeed the force is communicated via the points of contact with the chair or floor or whatever, and this is indeed different from the above description of gravity in free fall or an inertial frame. We call it a non-inertial frame. And indeed the point is that if you are standing on, say, Earth, the floor is similarly exerting a force that is communicated via the points of contact with the floor, and the result is a physical situation identical to the rocket accelerating. We also call this a non-inertial frame: you are not in free-fall, because the Earth/floor is preventing you from being in the inertial reference frame of a freely falling observer.

Answer 2

I'll post my comments here since I've written quite a bit.

It seems like you're mis-matching the scenarios. Consider four scenarios: A1 is "Standing on Earth," A2 is "Standing on the back of the rocket as it accelerates at 9.8 m/s^2," B1 is "Floating in space," B2 is "Free falling in a vacuum." The equivalence principle says A1 is analogous to A2, and it says B1 is analogous to B2. In your post, you seem to compare B1 to A2 and A1 to B2, which are not the analogies.

Now as a follow-up question, you asked,

A question. For the person in free fall, since they are in a not quite symmetric graviational field, since the gravity is slightly stronger below them, their body should be slightly stretched. Which should be different than if they were floating in space. Also, the gravitational field should alter the inertial vectors of their quantum states, as compared to simply floating in space. For these situations to be truly identical, my logic must be wrong. But what is that flaw(s)?

Also, if a person was in free fall above a black hole, they would get spaghetti-fied. What is it about that scenario that breaks it of the analogy with your B1, floating in space?

That's a good observation. I was thinking about including that in my comment, but I ended up leaving it out, because I wasn't sure if it was relevant.

On Earth, you are infinitesimally compressed from the sides (because from a Newtonian framework the force vectors point inward toward the Earth) and additionally you have an uneven strength of gravity at your feet vs your head (for the reasons you point out). This is called the effect of tidal forces.

The way to salvage the equivalence principle is to point out that the equivalence principle holds for vanishingly small regions of spacetime. If you are considering a region of spacetime where tidal forces matter, take a smaller region, and the tidal forces are less noticeable. They're not exactly gone, but a sufficiently small region will make them hard to detect.

Using tidal forces, you can deduce the gravitational field. However, as you "zoom in" into a small region of spacetime, the tidal forces become vanishingly small. On Earth you know the direct effect of gravity on a box: it falls down. However, trying to measure tidal effect on the box would require extraordinarily precise instrumentation. Near a black hole, a person would be spaghettified, yes, but if you "zoom in" to the region of an electron, this effect would no longer hold as strongly. Simply said, tidal forces depend on gravitational strength and length scale, and you can always zoom in to get a smaller length scale.

The equivalence principle, stated more accurately, talks about the similarity of scenarios B1 and B2 as you zoom in on smaller and smaller regions of spacetime.

As a related note, this point of zooming in on a region of spacetime to "get rid of" tidal forces is analogous to the idea of zooming in on a sphere to "get rid of" curvature effects. When you zoom in on a sphere, it looks more and more like a flat Euclidean plane. The curvature effects are never actually gone, just like how the tidal forces are never actually gone, but they become harder and harder to detect as you zoom in. This is the geometric significance of the equivalence principle: as you zoom in to a region of spacetime, it looks more and more like Minkowski space from special relativity.

In fact, there's a mathematical way to state this: given any geodesic of a free-falling body (i.e. worldline of a free-falling body), there exist coordinates for which the metric tensor $g_{\mu\nu}$ is exactly in the form of the Minkowski metric $\eta_{\mu\nu}$ at every point of the geodesic. At points near the geodesic, $g_{\mu\nu}$ is not exactly $\eta_{\mu\nu}$, but it looks more and more like $\eta_{\mu\nu}$ in the limit as you approach a point of the geodesic. These coordinates are called Fermi normal coordinates.

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Regarding the question in your title, I would say that the "technical version of the equivalence principle" is technically correct, but the "intuitive version that people talk about" is technically incorrect. It was always meant to refer to the technical version, but when you explain it intuitively, you have to brush some things under the rug.

Answer 3

In essence you are suggesting that an established scientific theory is not valid. The proper way to prove you are right would be to propose an experiment that could be performed in an elevator car, at rest on Earth and then in a constant state of acceleration in space and would give different results. One experiment would be sufficient, and since it is obviously difficult to perform, a well described thought experiment would be a useful first step.

On the question of a human body standing on points on the floor of the elevator car, the forces from the floor to the feet would conventionally be described as "forces" in the dynamic case and "reactions" in the gravitation case, whereas the masses of all the particles of the body would be described as "masses" or "inertias" in the dynamic case but forces or loads (masses multiplied by g) in the dynamic case, but I suggest that these would merely be matters of semantics and that every force, pressure, stress and strain in the body would be exactly the same in both cases (all other things being equal). If, for example you suggest that a vertical spring standing on the floor of the elevator car would compress in different ways in either case then you should describe numerically what those differences would be and why.

I don't know what evidence you have that the fluids in the inner ear would behave differently in either case. Again the onus is on you to provide that evidence.

When you say "the acceleration has to begin at your feet and then overcome the inertia of your body", this implies a non-steady-state condition, but the constantly accelerating lift car is a steady-state condition. There may be some changes, perhaps waves through the body at the onset of the acceleration but after that I believe the situation would be identical to the gravitation one.

Answer 4

Any loose part of our body has an acceleration of $g$ at the surface of the earth. The fluid of our inner ears only stay where it is because the structure of the ears act as a container. The ears are supported by the head, that is supported by the skeleton. The whole body doesn't fall because it is supported by the floor.

The same thing happens inside an accelerated rocket in the outer space. The important point is the experimental fact that all bodies, independent of density or form, fall with the same acceleration close to the earth surface in the vacuum. And the fact that inside an accelerated frame all loose bodies have also the same acceleration with respect to the body frame.