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There is something about Einstein Equivalence Principle that I don't quite get. This is my reasoning:

  1. Equivalence principle $\rightarrow$ locally, acceleration is equivalent to a gravitational field

  2. Forces (which each observer, inertial or not inertial, agree about) causes particles to have a proper acceleration (they don't follow geodesics)

  3. These particles which have a proper acceleration, from their point of view, feel a gravitational field (point one).

Now... is this gravitational field real or fictitious?

If it is real $\rightarrow$ it must depend on mass distribution around the object. So the force which caused the acceleration must be linked to the mass distribution somehow (since these two actions balance themselves so that the object is in equilibrium, from his point of view).

If it is fictitious $\rightarrow$ then it has nothing to do with “real” gravity, which depends on mass and is manifested as the curvature of spacetime. In this case, the Equivalence Principle seems to me just a coincidence which has nothing in common with the geometrized view of curved spacetime.

If my assumptions are correct... which of the two options is true?

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  • $\begingroup$ The equivalence principle was notices by Pholoponus of Alexandrea as far back as the 6th centurey AD when people noticed that falling rocks fell together at the same time regardless of their sizes and masses- see wikipedia article. $\endgroup$
    – Riad
    Commented Dec 19, 2023 at 19:11

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You've somewhat misstated the equivalence principle. It says that the effects of a gravitational field cannot be distinguished from the effects of having an accelerating frame of references. That's different than saying they are equivalent, and it's enough of a difference to break your logic chain at your point #3.

Another route to seeing this is that equivalence principle says that the $m$ in $F=ma$ (inertial mass) is the same as the $m$ in Newton's gravitational law (gravitational mass). It didn't have to be that way in theory, but it is. That does not preclude, however, the existence of other forces.

A bit more, following the comments, by way of explanation. I prefer to think of the equivalence principle as the mathematical statement that $m_{\mathrm{inertial}} = m_{\mathrm{grav}}$, and the (probably more popularly stated) description about indistinguishability between gravitational force and an accelerating frame as a consequence of this equivalence between the two conceptual types of mass. One implies the other though, so I think there's no unique argument for starting with one or the other, other than historical convention.

Note the last statement, which I'll elaborate here: The equivalence between gravitational and inertial mass implies the indistinguishability between a gravitational force and an accelerating frame.

  • If you imagine that you're in a box with no information about what's happening outside of your box, you now try to construct an experiment to determine if your box is in an inertial frame. No problem with this. Hold a pencil out, let it "drop" and see if it heads toward a wall of the box. If it does, then you're in a non-inertial frame.
  • Assume now that it does move toward a wall (meaning that it accelerates since it started at rest in your hand) and try to figure out if you're in a "stationary" box subject to gravity or in an "accelerating" box. Now you're stuck. If the box is not subject to gravity and the frame (e.g. the box) is accelerating due to some other force, then the pencil will move toward the side with acceleration equal to that of the box. If the box is subject to gravity but "stationary" you have $m_{\mathrm{inertial}} a = m_{\mathrm{grav}} g$, where $g$ gives the local strength of the gravitational field (and can be signed to account for direction of the field). But since the two masses are equal, this just gives $a=g$, which is non-informative since you don't have an independent measure of $g$. In either case (or, by extension, any case that includes some elements of both), all you can tell is that the pencil accelerated according to $a$.

The rest of your question about whether the force is "real" or "fictitious" seems to be trying to apply Newtonian reasoning to a relativistic question, and also seems to be based on your subtle misstatement of the principle. A frame exists in a (possibly small) neighborhood of a point but not at a single point. To say that the frame is accelerating is to say that all of the points in the frame are (up to some order) moving rigidly "together" with a single acceleration. That's distinct from looking at individual particles (possibly described in a frame) where each particle has a different relative acceleration. Your question on this part seems to be either comparing different frames (if you build a frame around each particle separately) or confounding the motion of the frame with the motion of the various particles in the frame (if you build one frame around your collection of particles). That's different than what the principle describes.

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  • $\begingroup$ this makes me wonder why it is called "Equivalence" Principle at all... so in this case, the gravitational field they feel is fictitious... does that imply my last deduction ("if it is fictitious --> etc)? $\endgroup$ Commented Sep 11, 2019 at 14:52
  • $\begingroup$ The "equivalence" is between inertial and gravitational mass. $\endgroup$
    – Brick
    Commented Sep 11, 2019 at 14:53
  • $\begingroup$ Your use of the word "fictitious" here is probably not wrong, but it's not especially relativistic in my view. There's no invariant way to describe that - if anything it's an expression of coordinate-dependent concepts that relativity would like to supersede with coordinate-invariant, geometric concepts. $\endgroup$
    – Brick
    Commented Sep 11, 2019 at 14:56
  • $\begingroup$ Ok... but even in this light, I don't see an explanation of the "coincident similarity" between local acceleration effect (due to forces) and local gravitational effects (due to mass-energy). The geometrization of curved spacetime does not attempt to explain this coincidence. Do you agree? $\endgroup$ Commented Sep 11, 2019 at 15:19
  • $\begingroup$ Updated answer with more detail. Writing quickly, so hopefully I got a good balance of precision and detail. $\endgroup$
    – Brick
    Commented Sep 11, 2019 at 16:15
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The argument proposed in the question is rather loosely worded. If we tighten up the language then the issues become clearer.

I will quote the three opening statements, which are loosely worded, and propose a more precisely worded replacement.

  1. Equivalence principle → locally, acceleration is equivalent to a gravitational field

Replace with:

1B. Weak equivalence principle → all entities have the same acceleration under gravity in any given small region.

  1. Forces (which each observer, inertial or not inertial, agree about) causes particles to have a proper acceleration (they don't follow geodesics)

replace with:

2B. (Non-gravitational) forces cause particles to follow non-geodesics. Since all observers agree about geodesics, all will agree that a given worldline is non-geodesic if it is, so they will agree on the presence of a force and consequently a proper acceleration. They will also agree about the invariant magnitude of the acceleration 4-vector but the components of this 4-vector will be frame-dependent.

  1. These particles which have a proper acceleration, from their point of view, feel a gravitational field (point one).

Replace with:

3B. Suppose we attach a frame of reference to a particle whose proper acceleration 4-vector has non-zero magnitude. Any nearby particles in freefall will have 4-acceleration of zero magnitude. Consequently there is a difference in the 4-accelerations of the frame of reference and of particles in freefall. In this reference frame there will be a non-zero three-acceleration of everything in freefall. The term 'gravitational field' may be invoked as a way to refer to the presence and size of this three-acceleration.

Finally, we come to the actual question, which is whether the 'gravitational field' just mentioned is real or fictitious. There is no way to answer that because as a question it is too vague. What one can instead is comment on what is and is not true of this kind of field. First of all, the 3-accelerations we are referring to really are non-zero and they have physical effects. For example, an apple released from a branch of a tree really does have a 3-acceleration relative to the ground and will hit the ground and possibly be bruised. Nothing fictitious about that. On the other hand, one may note that a change of reference frame suffices to make the 3-acceleration of the apple go to zero so it is not an invariant and so not a statement about the apple alone, but rather a way of referring to relative motion between the apple and a chosen reference frame.

Next, if spacetime is flat then merely changing reference frame will not change that fact, so the 'gravitational field' associated with a non-inertial reference frame does not cause spacetime curvature. If you wish to reserve the term 'gravitational field' for the situation of non-zero curvature then you would have to say there is no gravitational field in the Rindler frame (the constantly accelerating frame in flat spacetime) and the effects observed in that frame, such as apples falling to the bottom of an accelerating rocket, should not be called 'gravitational'. That's ok but it does not change the fact that the weak equivalence principle is a useful and insightful observation about the effects of gravity, and indeed it can be used to deduce quantitatively (and correctly!) the curvature of a light ray path under 'gravity' as indicated by the local 3-acceleration, in some given (non-inertial) reference frame, of entities in freefall.

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Now, is this gravitational field real or fictitious?

It's both. If looked from our vantage point of human beings located in space, then the gravitational field and hence force is real. From the vantage of outside of spacetime and viewing it all at once, it is a manifestation of the curvature of spacetime and so not a force.

Which one is the true description? Now some, take the second description as truer because the equations of motion there are simpler. Plus it was discovered by Einstein. But this, to my mind, goes against the philosophical principle of relativity, which says all equivalent descriptions are true. And since the first description is equivalent to the second, when everything is taken into account, then both are true.

It's worth adding that the principle of equivalence means that we shouldn't take one spacetime as the correct description but also, and at the same time, all equivalent ones. In other words, spacetime isn't a single manifold but many of them linked by equivalences. This helps resolve Einstein's famous Hole Argument which showed points don't exist in GR. They don't, you need to keep track of all equivalent points too.

This is made manifestly clearer in Category Theory which is the general theory of covariance - physical and mathematical.

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The equivalence principle was observed far back in the 6 century AD- as reported by John Philoponus of Alexandria- see Wikipedia article. Rocks and other matter were observed to fall from mountains to ground at the same time. Big rocks can break on the way without any effect on their motion. That lead to the conclusion that F/m=g, is a constant gravity-pull regardless of the type, size and weight of matter. If the mass is accelerated to the same value by another force horizontally(normal to gravity), the same force will be needed and the question is why. By Tylor theorem of analysis, if two functions are exactly the same over a small range, they must be the same. This says that gravity must be the equivalent to acceleration and this is what Einstein said in his principle. But this doesn’t really need a curved spacetime to explain, as it can be taken as a fact. What required a bending of space time is the fact that a light beam made of massless things did get affected by gravity which is clearly unexpected. The bending of space time then gives a nice solution to both problems. The acceleration must be due to the bending of the path itself. So that light moves in a straight line and bends only because of the path, and the acceleration by gravity is independent of matter because it is due to the bending of the path of matter. This turned dynamics into kinematics and lead to the phrase by Wheeler; matter tells space how to bend and space tells matter how to move. This of course is only one way to explain these observations and one can invoke conservation of momentum to do the same without the need for path bending. The important issue here is to do with mass itself and gravity is only the description of how mass is behaving which is certainly not fictitious.

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As a side-note to the Equivalence Principle. This principle is often used to "prove" using a Gedanken experiment, that light bends in a gravitational field. Place yourself in an closed elevator. You do not know whether the elevator is at rest on earth or whether it is accelerated in space. As there is no way to distinguish these inside the elevator using an experiment: a ray of light entering the elevator from the side through a hole will seem to bend when the elevator is (really) accelerated. This means that when the elevator is really at rest on earth, the ray should also bend. Thus light bends in a gravitational field.

However there is a caveat here. We make one more hole in the elevator but now in the top. A ray of light passes through and inside we measure the frequency of the light. If the elevator is at rest on earth we measure a constant light frequency, only blue-shifted due to gravity. If the elevator is accelerating (upwards) in space we measure an increasing blue-shift due to the velocity Doppler effect. This means that in principle we are perfectly able to detect whether our elevator is at rest on earth or whether it is accelerating in space (with respect to a distant light).

So at least for this Gedanken experiment (which was used by Einstein to "prove" light bends in a gravitational field) and which was the onset to the Equivalence Principle it is does not "prove" anything at all.

Now to come back to your question: the Equivalence Principle does not tell you whether gravity is real or not. In fact, GR does not tell you, GR is nothing more than applied math. Particle physics likes to explain gravity as a field of gravitons, the mediators of gravity. Other physicists like to explain it as a residual electrostatic field that cannot be shielded (thus as a field of virtual photons). We do not even know from experiments whether gravity acts on particles as small as electrons or neutrons (remember that in GR the particle then should be able to "feel" the very slight difference in curvature from one side of it to the other side, very small indeed). We assume it does. As a last question to you: what do you mean by anything being "real"? Think about that one..

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  • $\begingroup$ You said “If the elevator is at rest on earth we measure a constant light frequency, only blue-shifted due to gravity. If the elevator is accelerating (upwards) in space we measure an increasing blue-shift due to the velocity Doppler effect”. This is wrong. The measurements are the same. $\endgroup$
    – Dale
    Commented Mar 25, 2023 at 12:13
  • $\begingroup$ Blue shift due to gravity is different from blue (red-) shift due to velocity. A photon shot down from a high tower towards earth will gain in energy (blue shift), this is confirmed by experiments. This blue shift is only depending upon the height the photon was shot down and the gravity at earth level. Second, the velocity red shift depends on the velocity of the light source relative to the observer. If this velocity is increasing, the Doppler shift will increase, this is well known from stellar redshift. Realize that there are 2 effects at play: the velocity and acceleration effect. $\endgroup$
    – J.O. Doe
    Commented Mar 26, 2023 at 13:48
  • $\begingroup$ For a nice reference to gravitational redshift see f.i. how GPS works. Redshift by velocity is obvious, see stellar redshifts and determination of stellar velocities w.r.t. earth. One more comment: the redshift in gravity has nothing to do with a change in velocity of light, that's always c. Only its energy $h\nu$ changes. In fact if one says $mc^2 = h\nu$ and one takes into account the change in potential energy of this (hypothetical) photon mass $m = h\nu/c^2$ you'll get exactly the gravitational redshift formula :-) Nice huh? $\endgroup$
    – J.O. Doe
    Commented Mar 26, 2023 at 14:00
  • $\begingroup$ nevertheless your statement is completely false that the equivalence principle fails for the redshift of vertical light. That is the issue why your post was so heavily downvoted. The equivalence principle holds in that and many other tests. Despite your claims that they are different, if you work it out quantitatively using the equivalence principle you get the same result $\endgroup$
    – Dale
    Commented Mar 26, 2023 at 14:41
  • $\begingroup$ Well, see it this way: Two comments. 1: There is never a -sudden- situation in physics. That is, there is never suddenly an earth under your feet. That is not physical (we speak of a -discontinuity- in physics). Neither will there be an earth growing under your feet (increasing gravity/acceleration). In short: in reality we will always be perfectly able to discern both situations: real acceleration or gravity from huge mass. The path towards a physical situation is often more important in theory forming than the final situation. My second argument is below (I'm sorry web master). $\endgroup$
    – J.O. Doe
    Commented Mar 26, 2023 at 16:44

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