# Simple reasoning that the equivalence principle suggests curvature

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• luinthoron
In summary, Dale wrote a helpful article that discusses how to introduce students to the concepts of general relativity. He suggests using simple examples that connect the principles of the equivalence principle and curvature of spacetime. In this example, the student is looking for an example of spacetime curvature that is easy to understand. Every book that the student has read talks about light going up a gravity well, the Pound-Rebka experiment. This experiment was not testing the equivalence principle. It was testing gravitational time dilation. There is a relationship between the two, but I don't think it's what you think it is.
luinthoron
TL;DR Summary
I am looking for simple examples showing that the equivalence principle implies or at least suggests spacetime curvature (i. e. nonvanishing nonconstant components of the metric tensor).
I am a high-school teacher and a PhD. student. I am looking for ways to introduce my students to GR. In my treatment, I speak about the equivalence principle and later about curvature in general and consequently that of spacetime. I am missing a connection of these two parts that would be understandable to a high-schooler. Every book that I found talks about light going up a gravity well, the Pound-Rebka experiment. That is all fine, but I don't like that light is now, in my opinion, unnecessarily involved. I would like a simple dynamic example that would show that equivalence principle implies that the spacetime interval (which my students know from Special Relativity) now contains nonvanishing nonconstant coefficients like on a curved surface. I have been trying some examples with a falling observer flying near two stationary observers in a gravitational field but I can't get the reasoning right. I also don't want to use the common argument about some general coordinate transformation from the Local Inertial Frame to a general one, as that would be, again in my opinion, an unnecessary complication for the students. Could someone please suggest such a thought experiment to fill the gap between the EP and curvature of spacetime?

vanhees71
Non-constant non-vanishing components of the metric do not imply curvature. The most straightforward example being polar coordinates in the 2D Euclidean plane. Curvature is somewhat more involved.

vanhees71
luinthoron said:
I am looking for simple examples showing that the equivalence principle implies or at least suggests spacetime curvature
This doesn't make sense, since the point of the equivalence principle is that, whether a spacetime is globally curved or not, locally, i.e., in a small enough patch of spacetime, you can treat it as flat and ignore the effects of curvature.

luinthoron said:
Every book that I found talks about light going up a gravity well, the Pound-Rebka experiment.
This experiment was not testing the equivalence principle. It was testing gravitational time dilation. There is a relationship between the two, but I don't think it's what you think it is.

Einstein's original argument for why there should be gravitational time dilation was indeed based on the equivalence principle: his argument was basically that we should expect time dilation between the top and bottom of an accelerating "elevator" (nowadays we would probably say "rocket ship") in flat spacetime, and therefore, by the equivalence principle, we should also expect time dilation between the top and bottom of a similar "elevator" (or rocket ship) sitting at rest in a gravitational field. But that argument also means that the presence of that time dilation, in and of itself, does not tell you that spacetime is curved. It can't, because that time dilation is also present in flat spacetime. This is an example of why you can't use the equivalence principle to detect the presence of spacetime curvature.

luinthoron said:
I would like a simple dynamic example that would show that equivalence principle implies that the spacetime interval (which my students know from Special Relativity) now contains nonvanishing nonconstant coefficients like on a curved surface.
If you are trying to teach your students about curved spacetime, you will have to teach them about the difference between "the spacetime interval" (an invariant) and "the components of the metric tensor" (which depend on your choice of coordinates). The two are not the same, but you are treating them as the same. I don't think that approach will work.

In the example discussed above, about time dilation in an accelerating rocket ship in flat spacetime, you can set up coordinates in which the rocket ship is at rest, and an observer at rest inside the ship feels acceleration. (The technical term for this is "proper acceleration".) These coordinates are called "Rindler coordinates", and the metric tensor in these coordinates has a "nonvanishing nonconstant coefficent", the metric coefficient ##g_{00}##, which now depends on "height" (the spatial coordinate in the direction of the acceleration). But these coordinates are still describing flat spacetime. The geometry of spacetime, which includes all "spacetime intervals", is the same. So this is an example of why you cannot use the presence of "nonvanishing nonconstant coefficients" in the metric tensor to tell that spacetime is curved: you can have the same thing in flat spacetime.

The correct way to think of spacetime curvature is as tidal gravity. (Note that this is how the Insights article that @Dale wrote and referred you to talks about it.) The equivalence principle, by definition, cannot tell you anything about tidal gravity, since by definition it only applies to a patch of spacetime that is small enough that no tidal gravity is observable.

For "dynamic examples" of tidal gravity in action, which should be accessible to your students if they understand how Newtonian gravity works, consider these two scenarios:

(1) Two apples, both momentarily at rest relative to each other and Earth, along the same radial line but at different altitudes. Both apples will start falling towards Earth, but the lower apple will have a greater acceleration towards Earth, so the two apples will gradually diverge.

(2) Two apples, both momentarily at rest relative to each other Earth, at the same altitude but separated tangentially. Both apples will have the same magnitude of acceleration towards Earth, but in different directions (since both are accelerating towards the center of the Earth), so the two apples will gradually converge.

Note that neither of these scenarios can be analyzed using the equivalence principle, because any region of spacetime large enough to cover both apples and include their divergence or convergence will be too large for the equivalence principle to apply.

cianfa72, pasala, vanhees71 and 3 others
luinthoron said:
Summary:: I am looking for simple examples showing that the equivalence principle implies or at least suggests spacetime curvature (i. e. nonvanishing nonconstant components of the metric tensor).

I am a high-school teacher and a PhD. student. I am looking for ways to introduce my students to GR. In my treatment, I speak about the equivalence principle and later about curvature in general and consequently that of spacetime. I am missing a connection of these two parts that would be understandable to a high-schooler. Every book that I found talks about light going up a gravity well, the Pound-Rebka experiment. That is all fine, but I don't like that light is now, in my opinion, unnecessarily involved.

Unfortunately, while I recall seeing that approach used in at least one textbook (MTW), it's wrong, as others have mentioned. One could carry out a Pound-Rebka type experiment on an accelerating spaceship in the flat space-time of special relativity, where there is no curvature. So a positive result on the Pound-Rebka experiment is not sufficient to show the existence of curvature.

My favorite approach to spatial curvature is the sphere - rather than talk about parallel transport at a high school level, I'd talk about the sum of the angles of a triangle being greater than 180 degrees to attempt to explain the idea of curvature. Unfortunately, this approach won't work for space-time curvature, because the sum of the angles approach just doesn't generalize :(. The idea of drawing a space-time diagram on a curved surface, such as a sphere, rather than a flat plane is a valid approach, but the sum-of-angles idea doesn't survive the translation from Euclidean geometry to Lorentzian geometry.

I've seen it argued that having two geodesics cross each other demonstrates that a geometry is curved, (for instance a self-published book by a PF poster, "Realtivity for Poets" by Ben Crowell), but there's a counterexample for that argument as well - a cylinder :(.

pervect said:
while I recall seeing that approach used in at least one textbook (MTW), it's wrong
What do you think is wrong about MTW's treatment of the Pound-Rebka experiment? They talk about it as a test of gravitational redshift, but AFAIK they don't claim that the results of a single such experiment, considered in isolation, show spacetime curvature. (There is one caveat to this; see next paragraph below.) The results of multiple such experiments, done at different, widely separated locations on Earth, are evidence for spacetime curvature, since the direction of the measured redshift (the "vertical" direction in which the experiment is done) would vary by location, which would not be the case in an accelerated rocket in flat spacetime. And that is the argument that MTW uses (see in particular p. 1058, second paragraph from the bottom).

(MTW does mention their earlier discussion of Schild's argument that gravitational redshift by itself, i.e., a single measurement of it at one location on Earth, implies spacetime curvature; we had a long thread on that some time ago, started by me, because on its face the argument appears to be false for the very reason you give: you can also observe gravitational redshift in an accelerating rocket in flat spacetime. Plenty of subtleties were brought out in that thread, which I won't go into here; but on p. 1058 MTW give the argument I describe above, which makes use of multiple observations of gravitational redshift in different directions, and call it a "reworked version" of Schild's argument. This seems to me to be a tacit admission that the earlier version of Schild's argument, by itself, at least if it is read as arguing that a single measurement of gravitational time dilation, in isolation, implies spacetime curvature, is not really correct.)

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cianfa72
pervect said:
I've seen it argued that having two geodesics cross each other demonstrates that a geometry is curved, (for instance a self-published book by a PF poster, "Realtivity for Poets" by Ben Crowell)
Hm. I haven't read that book by Ben, but I would be surprised if the claim he makes is that simplistic. Obviously just the presence of two crossing geodesics doesn't show curvature.

What shows curvature is two initially parallel geodesics crossing. You will not have that on a cylinder, or any other flat geometry.

PeterDonis said:
What shows curvature is two initially parallel geodesics crossing. You will not have that on a cylinder, or any other flat geometry.
Cone. (Remove apex.)

But this is a result of the global properties, not local.

Orodruin said:
Cone. (Remove apex.)
Which initially parallel geodesics would cross?

PeterDonis said:
Which initially parallel geodesics would cross?
Any as geodesics cross themselves.

Edit: Well, any that are not directed at the apex.

Orodruin said:
Any as geodesics cross themselves.

Edit: Well, any that are not directed at the apex.
I'm afraid I still don't understand. Isn't the cone supposed to be infinitely extended? That is, it's equivalent to a plane with one missing point (the apex). The geodesics are the curves that are straight lines on the plane; those curves don't intersect themselves, or any other straight lines that are parallel to them.

A.T.
PeterDonis said:
I'm afraid I still don't understand. Isn't the cone supposed to be infinitely extended? That is, it's equivalent to a plane with one missing point (the apex). The geodesics are the curves that are straight lines on the plane; those curves don't intersect themselves, or any other straight lines that are parallel to them.
Each time the geodesic makes a turn around the apex its direction changes by an angle related to the apex angle.

luinthoron said:
I am looking for simple examples showing that the equivalence principle implies or at least suggests spacetime curvature (i. e. nonvanishing nonconstant components of the metric tensor).
Curvature is the reason, why the equivalence principle doesn't apply globally, just as a local approximation. In the below animation there is no intrinsic curvature shown (a cone is intrinsically flat), because its effect is negligible locally.

But to reconcile the local situation on one side of the Earth, with the local situation on the other side of the Earth, you must have curvature in between.

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Orodruin said:
Each time the geodesic makes a turn around the apex its direction changes by an angle related to the apex angle.
Yes, but in those space-time analogies (embeddings) you arrive on a new layer after going around the cone, because time is not cyclic.

A.T. said:
Yes, but in those space-time analogies (embeddings) you arrive on a new layer after going around the cone, because time is not cyclic.
Well, that is besides my point, which was that geodesics may intersect themselves depending on the manifold, but that this is more of a result of the global structure of the manifold than it is of curvature.

I think that the argument about curved spacetime is not too difficult to make.

Within a small region of spacetime, there are many ways to set up a coordinate system, which gives 4 numbers, (x,y,z,t) to each point in that region. Newtonian physics would tell us that (if our coordinate system is inertial and Cartesian) then a test particle (one with no forces acting on it) will travel at a constant velocity: If you let ##x(t), y(t), z(t)## be its position as a function of time, and you plot ##x## versus ##t## or ##y## versus ##t## or ##z## versus ##t##, you'll get straight lines.

For some choices of a coordinate system, though, a test particle's coordinate velocity does not remain constant with time but accelerates or changes directions as if acted on by a force. Some examples: a rotating coordinate system has apparent centrifugal and Coriolis forces, and an accelerating coordinate system such as onboard a rocket has g-forces. We can call such apparent forces "inertial forces" (or "fictitious forces") if they can be made to disappear by changing to a different coordinate system. The equivalence principle just states that gravity itself is an inertial force in this sense. You can set up a "free fall" coordinate system such that objects behave as if there were no gravity.

In a small region of "flat" spacetime, you can pick a coordinate system that gets rid of inertial forces everywhere within that region. In "curved" spacetime, you can make the inertial forces vanish at a single point, but then they will be nonzero once you get far enough away from that point.

If you assume (because of the equivalence principle) that gravity is an inertial force, then that means that spacetime is curved near the Earth, because there is no coordinate system in which it is zero everywhere. Near a falling object, you can pick a "free fall" coordinate system in which that object has zero coordinate acceleration. But that coordinate system can't include distant freefalling objects to make it where they are unaccelerated, as well. Objects on opposite sides of the Earth fall in different directions, so you can't set up a coordinate system in which objects at rest remain at rest.

pervect said:
My favorite approach to spatial curvature is the sphere - rather than talk about parallel transport at a high school level, I'd talk about the sum of the angles of a triangle being greater than 180 degrees to attempt to explain the idea of curvature. Unfortunately, this approach won't work for space-time curvature, because the sum of the angles approach just doesn't generalize :(. The idea of drawing a space-time diagram on a curved surface, such as a sphere, rather than a flat plane is a valid approach, but the sum-of-angles idea doesn't survive the translation from Euclidean geometry to Lorentzian geometry.

I think this can be translated to Lorentzian geometry, but only in the special case where all three sides of the triangle are timelike.

First of all we have to remove the dependence on 180°, and express the Euclidean result in the form that one exterior angle is equal to the sum of the opposite interior angles, i.e. in the diagram below, ##\varphi = \theta + \psi ##.

Now, reinterpret the diagram as a spacetime diagram where ##\theta ##, ##\varphi ## and ##\psi ## are rapidities instead of angles. This now represents a variation of the twins paradox, let's call it a triplets paradox, where triplets Alice, Bob and Carol are initially together moving inertially, Alice then starts to move inertially with rapidity ##\theta ## relative to Bob and Carol. Some time later, Carol then starts to move inertially with rapidity ##\varphi ## (greater than ##\theta ##) relative to Bob in a direction directly towards Alice. When Carol catches up with Alice, what will be the rapidity ##\psi ## of Carol relative to Alice? In flat spacetime the answer will be ##\psi = \varphi - \theta ##, but in curved spacetime it won't.

This can be translated into relative velocities ##u = \tanh \theta ##, ##v = \tanh \varphi ## and ##w = \tanh \psi ## (under the convention ##c = 1##), and so the standard velocity composition formula
$$w = \frac{v - u}{1 - vu}$$
will hold in flat spacetime, and generally won't hold around a triangle of timelike geodesics in curved spacetime. The discrepancy will depend on the size of the triangle.

"Relative velocity" in this case means as measured in the locally inertial frame in which one of the two participants is at rest. Alternatively you can define in invariant way by the formula
$$U_\alpha V^\alpha = \frac{1}{\sqrt{1 - v^2}} \quad (=\cosh \varphi)$$
to define the relative velocity ##v ## between two 4-velocities ##\textbf{U} ## and ##\textbf{V} ##.

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cianfa72, pervect and (deleted member)
PeterDonis said:
You will not have that on a cylinder
The article I linked to in post #17 talks about the cylinder case towards the end. However, I'm still not sure that initially parallel geodesics on a cylinder will intersect. (I still haven't worked out the details for initially parallel geodesics even on a cone.) What the article I linked to makes clear is that, on either a cone or a cylinder, even though they are flat geometrically, you can have multiple geodesics between the same pair of points. That's not quite the same thing as initially parallel geodesics intersecting.

vanhees71
Hi

I have not yet read any of the previous replies so what I am saying here may have already been covered.

If we interpret the Equivalence Principle to say that, an observer in a freely falling reference frame (elevator) will not be able to tell if he is in free fall in a gravitational field or if he is in an inertial frame moving at constant velocity then we can look at the following Thought Experiment;

Imagine two observers, Observer one is in a closed box, which is not in a gravitational field but is taken to be an inertial frame of reference. Imagine that observer one has a light source on the left wall of his box, possibly i meter below the top of the box. Now a beam of light is emitted and goes across to the opposite wall. It should strike the opposite wall at a point 1 meter from the top of the box, on that side, and observer one should notice this.

Now imagine that observer Two s box is equipped with an identical light source also emitting a beam of light. However, observer twos box is in free fall in a gravitational field. Now the Equivalence Principal (as I have interpreted it here) tells us that Observer two should see the exact same result as seen by observer one. Hence if observer two sees the beam emitted from a point 1 meter below the top of the box, and strike the opposite wall at a point 1 meter from the top of the box, since, in the time it takes the beam to move across the box will have dropped within the gravitational field, it can only mean that, in the gravitational field, the beam did not move straight across the box but must have followed a curve downwards. In other words, for the reception point of the light beam on the opposite wall to be below the emission point (as seen by someone outside the box and not falling with the box - someone at infinity) the path taken by the light beam must have curved down from the point of emission.

This is a fairly simple bit of reasoning. It would be interesting to hear what others say to this.

Paul

Phinrich said:
This is a fairly simple bit of reasoning. It would be interesting to hear what others say to this.

Paul

Phinrich said:
Hi

I have not yet read any of the previous replies so what I am saying here may have already been covered.

If we interpret the Equivalence Principle to say that, an observer in a freely falling reference frame (elevator) will not be able to tell if he is in free fall in a gravitational field or if he is in an inertial frame moving at constant velocity then we can look at the following Thought Experiment;

Imagine two observers, Observer one is in a closed box, which is not in a gravitational field but is taken to be an inertial frame of reference. Imagine that observer one has a light source on the left wall of his box, possibly i meter below the top of the box. Now a beam of light is emitted and goes across to the opposite wall. It should strike the opposite wall at a point 1 meter from the top of the box, on that side, and observer one should notice this.

Now imagine that observer Two s box is equipped with an identical light source also emitting a beam of light. However, observer twos box is in free fall in a gravitational field. Now the Equivalence Principal (as I have interpreted it here) tells us that Observer two should see the exact same result as seen by observer one. Hence if observer two sees the beam emitted from a point 1 meter below the top of the box, and strike the opposite wall at a point 1 meter from the top of the box, since, in the time it takes the beam to move across the box will have dropped within the gravitational field, it can only mean that, in the gravitational field, the beam did not move straight across the box but must have followed a curve downwards. In other words, for the reception point of the light beam on the opposite wall to be below the emission point (as seen by someone outside the box and not falling with the box - someone at infinity) the path taken by the light beam must have curved down from the point of emission.

This is a fairly simple bit of reasoning. It would be interesting to hear what others say to this.

Paul
That's exactly right. The bending of light by gravity can be predicted this way (although the amount of bending is harder to get right). However, bending of light doesn't by itself show that spacetime is curved, exactly because it's predicted in flat spacetime using a noninertial coordinate system.

Phinrich said:
since, in the time it takes the beam to move across the box will have dropped within the gravitational field, it can only mean that, in the gravitational field, the beam did not move straight across the box but must have followed a curve downwards.
This is not an argument based on the equivalence principle. If you want to use the EP, you have to confine yourself to observations made inside the "elevator". Arguing that the beam "must have followed a curve downwards" is not doing that; the "downwards" is in reference to an external object (the source of the gravitational field).

To correctly deduce light bending using the EP, you need to use an "elevator" that is accelerated. In other words: consider an "elevator" that is accelerating at 1 g in flat spacetime, where by "accelerating at 1 g" we mean that that is the acceleration that is felt by observers at rest inside the elevator, or measured by an accelerometer attached to the elevator. A light beam emitted horizontally in this elevator will hit the opposite wall at a point lower than the point from which it was emitted.

Now consider the equivalent (according to the EP) situation of an "elevator" sitting at rest on the surface of the Earth. Observers at rest in the elevator will feel a 1 g acceleration, and an accelerometer attached to the elevator will measure a 1 g acceleration. Therefore, by the EP, a light beam emitted horizontally in this elevator will also hit the wall at a point lower than the point from which it was emitted, just as in the case above.

As @stevendaryl points out, while this argument is enough to indicate that light bending occurs in a gravitational field, it is not enough to tell you the correct magnitude of the light bending when measured globally. The correct global result for light bending by a spherically symmetric massive object turns out to be twice as large as the result you would calculate from the EP argument above.

stevendaryl said:
That's exactly right.
No, it isn't, for the reason I gave in post #23 just now. You correctly state that you need to use a non-inertial, accelerated frame to apply the EP to deduce light bending in a gravitational field; but @Phinrich didn't do that.

PeterDonis said:
No, it isn't, for the reason I gave in post #23 just now. You correctly state that you need to use a non-inertial, accelerated frame to apply the EP to deduce light bending in a gravitational field; but @Phinrich didn't do that.
Well, it's one additional step. You have a planet with gravity. You have an elevator that is falling in that gravitational field. Consider two different reference frames: one used by people on the surface of the planet, where the ground is at rest and the elevator is falling, and another one used by people inside the elevator, in which the elevator is at rest. In the elevator reference frame, by the equivalence principle, you know that a light beam that starts off perpendicular to one wall of the elevator and travels to the other will arrive at the same height above the floor as when it left. In the planet reference frame, that means the light beam bends down since the elevator drops during the time that the light beam travels from one wall to the other.

Usually, people compare an accelerating elevator to one sitting on the surface of a planet, but you can just as well compare an elevator traveling inertially through empty space to one falling in a gravitational field.

stevendaryl said:
Consider two different reference frames
The first of which is not one that can be used in any argument involving the equivalence principle. But @Phinrich was making an argument using the equivalence principle.

stevendaryl said:
you can just as well compare an elevator traveling inertially through empty space to one falling in a gravitational field.
But you can't use the equivalence principle in such a comparison to make any argument about light bending, since the light bending observed inside both elevators is zero, and that's all you're allowed to take into account if you're making an argument using the equivalence principle. To make an argument for light bending using the EP, you have to use accelerated elevators, because only then will light bending be observable inside the elevators.

Off on a bit of a tangent, I had some thoughts. Given the existence of the structure of General relativity, we can equate the curvature of space-time with the Riemann curvature tensor, and we can demonstrate that instruments, such as gravity gradiometers (for example, the Forward mass detector) can measure some of the components of this curvature. We can also demonstrate that such instruments respond to the presence of nearby masses, but do not respond to accelerated motion of the observer, though there are some interesting subtle features to the calculations involved that I've thought about but have never seen anyone else discuss.

However, demonstrating the necessity of the existence of space-time curvature a priori is a much more difficult task, and it's likely over-ambitious. It's safer to say that we have a theory, General Relativity, which incorporates space-time curavature and which also works well, than to make bold statements about what sort of theories are possible.

Historically, we can say that Einstein was motivated to find a covariant theory of gravity that was consistent with special relativity and with experiment. Nordstrom's 1913 theory of gravity is an interesting example of a theory that comes very close to meeting these goals. Nordstrom's theory is covariant, and is compatible with SR, but it winds up not being compatible with experiment, for instance it predicts zero for light deflection. See for instance https://en.wikipedia.org/w/index.php?title=Nordström's_theory_of_gravitation&oldid=1020198139 which I used to refresh my memory on this interesting theory.

I

Dale
PeterDonis said:
The first of which is not one that can be used in any argument involving the equivalence principle. But @Phinrich was making an argument using the equivalence principle.
I don’t know why you are saying that. That seems completely wrong to me.

stevendaryl said:
I don’t know why you are saying that. That seems completely wrong to me.

It seems like a simple deduction:
1. If light does not bend, as observed inside an elevator floating in gravity-free space, then it does not bend, as observed inside an elevator in freefall near the Earth.
2. If light does not bend as observed in an elevator in freefall near the Earth, then it DOES bend as observed by someone at rest on the surface of the Earth.

Point 1 follows from the equivalence principle. Point 2 is just algebra. If it takes ##\delta t## seconds for light to travel from one side of the elevator to the other, then since the elevator is falling, the elevator will have fallen a small distance during that time. For the light to hit the same distance above the floor, then the light path has to bend down.

stevendaryl said:
It seems like a simple deduction
I'm not saying the argument you make is wrong. I'm just saying that this part...

stevendaryl said:
If light does not bend as observed in an elevator in freefall near the Earth, then it DOES bend as observed by someone at rest on the surface of the Earth.
...is not an argument based on the equivalence principle, since your argument involves "someone at rest on the surface of the Earth" who is not inside the elevator and who is not within the small patch of spacetime in which the EP can be used.

What the EP tells you is that measurements restricted to inside an elevator in free fall can't tell you anything about the bending of light, since such measurements detect no bending of light; but measurements restricted to inside an accelerated elevator can.

stevendaryl said:
Point 2 is just algebra.
No, it's physics, but physics not restricted to the small patch of spacetime within the elevator that the EP is restricted to.

PeterDonis said:
No, it's physics, but physics not restricted to the small patch of spacetime within the elevator that the EP is restricted to.
Here is a paper making a related point.

Ehlers, J., Rindler, W.
Local and Global Light Bending in Einstein's and Other Gravitational Theories.
General Relativity and Gravitation 29, 519–529 (1997).
https://doi.org/10.1023/A:1018843001842

PeterDonis said:
...is not an argument based on the equivalence principle, since your argument involves "someone at rest on the surface of the Earth" who is not inside the elevator and who is not within the small patch of spacetime in which the EP can be used.
This seems silly to me. If you like, you can let the person on the surface of the Earth be inside an elevator. That certainly doesn't make any difference. I'm NOT invoking the equivalence principle to compare the person at rest on the Earth to anybody in a rocket ship. I'm invoking the equivalence principle to compare the person in a freefalling elevator to a person in an elevator floating in space.

It's a two-part argument. The first part involves the equivalence principle. The second part only assumes the falling elevator approximately obeys Newtonian physics, that its height as a function of time is approximately given by ##h = h_0 - \frac{1}{2} g t^2##.

PeterDonis said:
No, it's physics, but physics not restricted to the small patch of spacetime within the elevator that the EP is restricted to.
Your comment seems very weird to me. You're saying that if it all takes place inside an elevator, that's a small enough patch. But if we make it large enough to include the elevator, plus someone standing beside the elevator as it falls, then that's too big to be a single patch?

If you like, we can imagine a building that is 3 stories tall that contains an elevator that is falling. Is a 3-story tall building too big to be a single patch?

The point of saying "a small patch" is that you're in a small enough region that geodesic deviation is negligible. If a 3 story building is too big, let it be a two-story building. Let it be a toy elevator inside a dollhouse.

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