# Theories that respect one form of the equivalence principle but violate another

This extract is taken from Sean Carroll's textbook. It talks about a theory that respects the Weak Equivalence Principle but violates the Einstein Equivalence Principle.

we could imagine a theory of gravity in which freely falling particles begin to rotate as they moved through a gravitational field. Then they could fall along the same paths as they would in an accelerated frame (thereby satisfying the WEP), but you could nevertheless detect the existence of the gravitational field (in violation of the EEP). Such theories seem contrived, but there is no law of nature that forbids them.

Now, the Weak Equivalence Principle has any one of the following forms:

1. the inertial mass is equal to the gravitational mass
2. there exists a preferred class of trajectories through spacetime, known as inertial or freely-falling trajectories, on which unaccelerated particles travel - where unaccelerated means "subject only to gravity"
3. the motion of freely-falling particles are the same in a gravitational field and a uniformly accelerated frame, in small enough regions of spacetime

and the Einstein Equivalence Principle has the following form:

In small enough regions of spacetime, the laws of physics reduce to those of special relativity; it is impossible to detect the existence of a gravitational field by means of local experiments.

In the theory mentioned in the extract, what does it mean for the particles to fall along the same paths as they would in an accelerated frame and how does that satisfy the WEP?

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## Answers and Replies

PeterDonis
Mentor
what does it mean for the particles to fall along the same paths as they would in an accelerated frame and how does that satisfy the WEP?

Imagine that you were in an rocket accelerating at 1 g (meaning an object at rest in the rocket would feel weight equivalent to a 1 g acceleration) in empty space, and you were standing on the "floor" of the rocket (the "floor" here would actually be the end of the rocket where the exhaust was being ejected), and you dropped an object. It would accelerate straight downward (towards the floor) at 1 g. And this behavior would be indistinguishable from the behavior of an object dropped by a person standing in an enclosed room on the surface of the Earth, feeling weight equivalent to a 1 g acceleration. A frame of reference in which the rocket, or the enclosed room, is at rest would be an accelerated frame, and the paths of the objects in both cases, relative to that frame, would be the same. This would satisfy the WEP.

spaghetti3451
Thanks for the explanation!

I now understand that the free-falling particles in both the accelerated frame (with acceleration ##\textbf{a}=-\textbf{g}## relative to some inertial frame (inertial as defined in special relativity)) and the gravitational field (with field strength ##\textbf{g}##) fall downwards in a helical motion. This respects the WEP.

But how could we nevertheless detect the existence of the gravitational field?

PeterDonis
Mentor
I now understand that the free-falling particles in both the accelerated frame (with acceleration ##\textbf{a}=-\textbf{g}## relative to some inertial frame (inertial as defined in special relativity)) and the gravitational field (with field strength ##\textbf{g}##) fall downwards in a helical motion.

Huh? That's not what I said. I said they will both fall straight downward.

how could we nevertheless detect the existence of the gravitational field?

To test the hypothetical theory of gravity Carroll describes, you would look for whether the free-falling particles started to spin as they fell. If they did, they must be falling in a gravitational field (in this hypothetical theory); if they didn't, they must be falling in an accelerating rocket in flat spacetime.

Of course, in actual experiments, we see no difference--the free-falling particles don't spin in either case. That tells us that hypothetical theories of the sort Carroll describes are ruled out.

To test the hypothetical theory of gravity Carroll describes, you would look for whether the free-falling particles started to spin as they fell. If they did, they must be falling in a gravitational field (in this hypothetical theory); if they didn't, they must be falling in an accelerating rocket in flat spacetime.

Of course, in actual experiments, we see no difference--the free-falling particles don't spin in either case. That tells us that hypothetical theories of the sort Carroll describes are ruled out.

Well, the Weak Equivalence Principle states that the motion of freely-falling particles are the same in a gravitational field and a uniformly accelerated frame, in small enough regions of spacetime.

Does this not mean that, if WEP is to be satisfied, then if the particle rotates as it falls freely in a gravitational field (in the hypothetical theory), it also ought to rotate as it falls freely in a uniformly accelerated frame in flat spacetime?

stevendaryl
Staff Emeritus
As I understand it, a particle with intrinsic angular momentum, such as an electron, will experience a very tiny spin-dependent force due to the rotation of a massive planet. This effect is analogous to the force between two magnetic dipoles in electromagnetism. If you take this tiny force into account (I think it's called the Lense-Thirring effect), then the path of an electron in freefall will not be a geodesic. So it seems that General Relativity does not actually satisfy the strong equivalence principle. Even in a very small region of spacetime, the motion of an electron would be different in the presence of a massive spinning planet than it would in flat spacetime.

Maybe somebody who knows more than I do can clarify this: Does GR satisfy the strong equivalence principle, when you take into account spin?

PeterDonis
Mentor
Does this not mean that, if WEP is to be satisfied, then if the particle rotates as it falls freely in a gravitational field (in the hypothetical theory), it also ought to rotate as it falls freely in a uniformly accelerated frame in flat spacetime?

No. The WEP only talks about the particle's trajectory. It does not talk about any other degrees of freedom of the particle. Rotation is a separate degree of freedom from the particle's trajectory, so the WEP doesn't say anything about it.

PeterDonis
Mentor
If you take this tiny force into account (I think it's called the Lense-Thirring effect), then the path of an electron in freefall will not be a geodesic.

I don't think this is correct. A particle with spin will still follow a geodesic; it will just be a different geodesic than the one a particle with no spin, but all other attributes the same, would follow.

Also, I don't think the Lense-Thirring effect (which Gravity Probe B was testing for) is limited to particles with spin; it's due to the angular momentum of the massive object, not any angular momentum of the test object, and is present for spinless test objects. But I believe there is an extra effect something like spin-orbit coupling, over and above the usual Lense-Thirring effect, which comes into play for particles with spin orbiting a spinning massive object.

An electron moving in an electromagnetic field will not follow a geodesic; but that's because the EM field is exerting a force on it which results in it having nonzero proper acceleration. The Lense-Thirring effect and its relatives (see above) is a "gravitational" effect and does not cause nonzero proper acceleration. At least, as I understand it.

Also, AFAIK there is a different effect due to spacetime curvature which can cause nonzero proper acceleration; it arises from different parts of an asymmetrical object interacting with spacetime curvature (tidal gravity) in different ways. But an object like this is not a "test object", and all versions of the EP only apply to test objects (objects that do not cause any spacetime curvature and have no internal structure). So the EP doesn't apply to this scenario.

Does GR satisfy the strong equivalence principle, when you take into account spin?

AFAIK, yes, as above.

stevendaryl
Staff Emeritus
I don't think this is correct. A particle with spin will still follow a geodesic; it will just be a different geodesic than the one a particle with no spin, but all other attributes the same, would follow.

I suppose that depends on how you define "geodesic", but if the path is spin-dependent, then it certainly is not a geodesic in the sense of extremizing $\int d\lambda \sqrt{g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda}}$. But the main point is that it violates the equivalence principle, in that two electrons floating in empty space will not experience spin-dependent deviation, while the same two electrons in freefall will.

PeterDonis
Mentor
if the path is spin-dependent, then it certainly is not a geodesic in the sense of extremizing ##\int d\lambda \sqrt{g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda}}##.

Yes, I see what you mean. I'll have to look up the actual math on this; I might be misremembering since it's been quite a while since I looked at it.

PeterDonis
Mentor
Ok, a quick look at the Gravity Probe B website reminded me of at least one thing I misremembered. The Lense-Thirring effect does not change the trajectory of an object; it causes the spin axis of a spinning test object ("gyroscope") orbiting a spinning massive object to precess. (More precisely, it adds an additional precession to the geodetic precession that is present for a gyroscope orbiting any massive object, spinning or not.) But the precession is not observable locally; you have to have a distant reference in a fixed direction (such as a distant star) to measure it. So the L-T effect does not violate the EP. (Also it does not affect spinless objects, so I misremembered that too. )

Here is the Gravity Probe B website's page on the L-T precession (it also discusses other relevant effects like geodetic precession):

https://einstein.stanford.edu/SPACETIME/spacetime4.html

There is another issue involved here as well, to do with the effect of frame dragging on radial trajectories, but I'll save that for a separate post when I have more time.

stevendaryl
Staff Emeritus
Ok, a quick look at the Gravity Probe B website reminded me of at least one thing I misremembered. The Lense-Thirring effect does not change the trajectory of an object; it causes the spin axis of a spinning test object ("gyroscope") orbiting a spinning massive object to precess.

Hmm. I misunderstood, then. Let's consider two massive circular rotating disks. What I thought was the case was that the force between the disks was different, depending on whether the disks were rotating in the same direction or opposite direction.

In Wikipedia: (https://en.wikipedia.org/wiki/Gravitoelectromagnetism#Higher-order_effects)
For instance, if two wheels are spun on a common axis, the mutual gravitational attraction between the two wheels will be greater if they spin in opposite directions than in the same direction. This can be expressed as an attractive or repulsive gravitomagnetic component.

PeterDonis
Mentor
Let's consider two massive circular rotating disks. What I thought was the case was that the force between the disks was different, depending on whether the disks were rotating in the same direction or opposite direction.

Yes, this is true. But neither disk is a test object, since both are producing significant spacetime curvature; one way of thinking of the "force" produced is as an interaction between the spacetime curvatures produced by the two objects. What I said about the trajectory not being affected by the L-T effect was only intended to apply to test objects. Sorry if that wasn't clear.

Even before considering rotation of the test objects or of the planet below them, in that case, aren't there are some little things to account for already?

- a pair of test objects released at rest with respect to one another not subject to an external gravitational influence (deep space) will accelerate toward each other

- a pair of test objects released at rest with respect to one another subject to an external gravitational influence (planet) will fall toward the planet's center of mass, so as to approach each other, and will further approach each other due to mutual gravitation

- a pair of test objects released at rest with respect to one another subject to an acceleration (elevator) will have parallel vectors due to that acceleration, and also approach each other due to mutual gravitation

The test objects' net mutual approach seems different in these three cases, least for deep space, most for planet, elevator in the middle (same as deep space up to relativistic energy increase from elevator perspective)?
If the gravitational case is not a planet with a center of mass but a gravitational field without convergence, won't the source center of mass be either at an infinite distance or an infinite plane of some thickness at a finite distance... both impossible (and identical in effect to the accelerating elevator)?

In the case of the planet, if the test objects are connected by a rigid rod, does the resistance to the various path components acting to laterally displace the test objects toward each other influence their vertical trajectories (does gravity care whether it's efforts to bring the test object's paths together toward the center of the planet's mass are being effective or wasted)?

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timmdeeg
Gold Member
To test the hypothetical theory of gravity Carroll describes, you would look for whether the free-falling particles started to spin as they fell. If they did, they must be falling in a gravitational field (in this hypothetical theory); if they didn't, they must be falling in an accelerating rocket in flat spacetime.

Of course, in actual experiments, we see no difference--the free-falling particles don't spin in either case. That tells us that hypothetical theories of the sort Carroll describes are ruled out.
http://arxiv.org/pdf/1206.7093v1.pdf
According to Mathisson a spinning particle in a gravitational field performs a helical motion. Shouldn't this allow us to distinguish acceleration from gravity?

PAllen
http://arxiv.org/pdf/1206.7093v1.pdf
According to Mathisson a spinning particle in a gravitational field performs a helical motion. Shouldn't this allow us to distinguish acceleration from gravity?
This effect is not inconsistent with the principle of equivalence. In reading the paper, note the discussion from the bottom of page 2 to the top of page 3. There is a minimum size of spinning body in GR if you don't want local violations of SR (FTL flows; captured in the dominant energy condition). This means that a spinning body is not really a test particle, as it is interacting with the curvature over a finite size. This is made more explicit in the following pater, which cites the above:

http://arxiv.org/abs/1501.04879

where (see the top of page 4), the effect is noted as precisely due to curvature over scale of the body - parallel transport around a closed loop not preserving the vector. Thus, this effect is due to the relative motion of parts of the body interacting with the tidal gravity over the scale of the body. This makes it irrelevant for evaluating the equivalence principle, which explicitly applies only on scales where tidal gravity can be ignored.

PeterDonis
stevendaryl
Staff Emeritus
This effect is not inconsistent with the principle of equivalence. In reading the paper, note the discussion from the bottom of page 2 to the top of page 3. There is a minimum size of spinning body in GR if you don't want local violations of SR (FTL flows; captured in the dominant energy condition). This means that a spinning body is not really a test particle, as it is interacting with the curvature over a finite size. This is made more explicit in the following pater, which cites the above:

http://arxiv.org/abs/1501.04879

where (see the top of page 4), the effect is notes as precisely due to curvature over scale of the body - parallel transport around a closed loop not preserving the vector. Thus, this effect is due to the relative motion of parts of the body interacting with the tidal gravity over the scale of the body. This makes it irrelevant for evaluating the equivalence principle, which explicitly applies only on scales where tidal gravity can be ignored.

Yes, I thought about that--even if a spinning body as a whole doesn't follow geodesics, it could be that the little bits of matter that make it up could satisfy the equivalence principle. But what about particles with intrinsic spin?

PAllen
Yes, I thought about that--even if a spinning body as a whole doesn't follow geodesics, it could be that the little bits of matter that make it up could satisfy the equivalence principle. But what about particles with intrinsic spin?
In classical GR, there can be no such thing. Any spin rate for radius below some size implies FTL motion. Also note that the second paper notes that a spinning 'particle' will follow a geodesic if the there is no curvature detectable over a closed loop of characteristic size. That is, even with a spinning particle idealization (which this paper uses), the effect is only due to interaction with tidal gravity - thus irrelevant to equivalence principle.

stevendaryl
Staff Emeritus
In classical GR, there can be no such thing. Any spin rate for radius below some size implies FTL motion. Also note that the second paper notes that a spinning 'particle' will follow a geodesic if the there is no curvature detectable over a closed loop of characteristic size. That is, even with a spinning particle idealization (which this paper uses), the effect is only due to interaction with tidal gravity - thus irrelevant to equivalence principle.

My question isn't really about GR, but about people's best guess as to whatever extension of GR would be needed to allow particles such as electrons that have intrinsic spin. If it's really a point-particle, then I would think that tidal forces couldn't be important.

PAllen
My question isn't really about GR, but about people's best guess as to whatever extension of GR would be needed to allow particles such as electrons that have intrinsic spin. If it's really a point-particle, then I would think that tidal forces couldn't be important.
The paper in my post #16 suggests that a particle with intrinsic spin does detect tidal gravity within a characteristic radius determined by its spin parameter; and if tidal gravity is minimal within that radius, the particle would act no differently than a non-spinning particle. That is consistent with the tests of falling neutrons - they behave no dfferently than expected for a particle without spin (due to weakness of tidal gravity near earth). In particular, these experiments have shown that the free fall is unaffected by the direction of the neutron's spin axis.

PeterDonis
Mentor
a pair of test objects released at rest with respect to one another not subject to an external gravitational influence (deep space) will accelerate toward each other

Strictly speaking, a "test object" is an object that does not produce any gravitation, so a pair of test objects released at rest with respect to one another, with no external gravitational influence, will just stay at rest relative to each other forever.

What I think you mean is a pair of gravitating objects, with no other gravitating objects present. But these objects will produce spacetime curvature, which already complicates the analysis. See below.

a pair of test objects released at rest with respect to one another subject to an external gravitational influence (planet) will fall toward the planet's center of mass, so as to approach each other, and will further approach each other due to mutual gravitation

Strictly speaking, we don't know this. You are assuming that we can take the solution to the above case, with two "small" gravitating objects (small compared to the third one that is providing the external gravitational influence in this case), and just add it to the solution for test objects (i.e,. true test objects that produce no gravity) in the field of the external gravitational influence (the planet), to get the solution for the case you describe. But GR is nonlinear, so you can't do this in the general case.

Even if you assume that all fields are sufficiently weak that they superpose linearly (the nonlinearities are not measurable), you still have complications if you're trying to apply the equivalence principle. See below.

a pair of test objects released at rest with respect to one another subject to an acceleration (elevator) will have parallel vectors due to that acceleration, and also approach each other due to mutual gravitation

And if you are looking at a large enough piece of spacetime that you can see the difference between all these effects, you are looking at a large enough piece of spacetime for tidal gravity to be significant, which means the equivalence principle does not apply. In a scenario where you have two "small" gravitating objects, the only way to apply the EP is to look at a small enough piece of spacetime that the "force" each object exerts on the other is negligible over the distance and time you are looking at. None of your three scenarios above meet that condition, so none of them are suitable for applying the EP. You would have to look at the motion of a true "test object" (producing negligible gravity) over a piece of spacetime small enough that an observer accelerating to remain at rest relative to one of the "small" gravitating objects would see the test object "accelerating" towards that same "small" gravitating object at the appropriate acceleration. And over that small piece of spacetime, the effect would be exactly the same as that of being in an elevator accelerating with the same proper acceleration in flat spacetime.

timmdeeg
Gold Member
This effect is not inconsistent with the principle of equivalence. In reading the paper, note the discussion from the bottom of page 2 to the top of page 3. There is a minimum size of spinning body in GR if you don't want local violations of SR (FTL flows; captured in the dominant energy condition). This means that a spinning body is not really a test particle, as it is interacting with the curvature over a finite size.
Ah, I see. Thanks for clarifying.

Buzz Bloom
Gold Member
It occurs to me that there should be in principle a very small but observable/measurable difference between (1) being in a box accelerating at 1 g far away from any gravitating body, and (2) being in the same box on the surface of the earth. The actual in the earth bound box, a falling object at the top of the box has a slightly smaller acceleration than at the bottom of the box. Therefore the ratio of the time to fall a given fraction of the height of the box to the time to fall the full height will be slightly more in the earth bound box than in the distant accelerating box. I don't have the time now to do the exact calculation for say a 10 meter box, but a rough mental calculation suggests the difference in these ratios should be measurable with current technology.

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PAllen
It occurs to me that there should be in principle a very small bu observable/measurable difference between (1) being in a box accelerating at 1 g far away from any gravitating body, and (2) being in the same box on the surface of the earth. The actual in the earth bound box, a falling object at the top of the box has a slightly smaller acceleration than at the bottom of the box. Therefore the ratio of the time to fall a given fraction of the height of the box to the time to fall the full height will be slightly more in the earth bound box than in the distant accelerating box. I don't have the time now to do the exact calculation for say a 10 meter box, but a rough mental calculation suggests the difference in these ratios should be measurable with current technology.

What you describe is called tidal gravity. The principle of equivalence is accurate exactly to the extent tidal gravity can be ignored. This is a common non-argument against the principle of equivalence. A more challenging one involves charged bodies, a topic on which there is sitll debate and on which the opinion of as great a physicist as Richard Feynman is now the minority opinion. The majority opinion is that a charged body falls slightly differently in an accelerating rocket than in a gravitational field, because the former does not radiate while the latter does. Philosophically, the principle of equivalence is still rescued by the observation the radiation of a free falling charge (in gravity) is due to the interaction of the charge with its 'far field' [equivalently, the absence of global inertial frames in GR], and this global aspect makes the (theoretically tiny, never observed) effect not relevant to the equivalence principle. Further, it is believed that if a charged body is encased in a conducting shell of opposite charge, preventing the exterior from detecting any EM field, the composite will fall exactly as a neutral body. This is supported by the very sensitive tests of free fall of ordinary matter, which consists of intense charges arranged so as not to have a net external EM field.

So far as I know, the only way tidal gravity can be detected with current technology over a distance like 10 meters is with super sensitive accelerometers (e.g. SQUID based). The practical error rates for timing a fall are overwhelmingly larger than this effect (irrespective of clock sensitivity). Similarly, even with the super clocks that detect gravitational time dilation over a few feet in altitude change cannot yet distinguish potential change given uniform acceleration assumption versus tidal assumption over an altitude change as small as 10 meters.

maline, Buzz Bloom and PeterDonis
Buzz Bloom
Gold Member
So far as I know, the only way tidal gravity can be detected with current technology over a distance like 10 meters is with super sensitive accelerometers (e.g. SQUID based). The practical error rates for timing a fall are overwhelmingly larger than this effect (irrespective of clock sensitivity).
Hi Paul:

Thank you for your post.

I am not an electronics engineer or an instrument designer, but as a thought experiment it seems plausible to me that the critical component needed is a pulse generator, say 1 ghz, feeding a binary bit counter with a reasonable memory, say 64 bits. This in effect becomes the core of a very precise digital stop watch with a precision of 1 ns, and a capacity way in excess of what is needed. I envisage a vertical tube about 10 meter long with a little extra room at top and bottom. At the top of the tube there is a devise that will drop a small spherical ball when signaled to do so. At the bottom is some device that will softly collect the ball without injuring it. This bottom ball catcher will move the ball back to the top of the tube where it will be readied for another drop.
All of this equipment is inside the tube that has been evacuated of any as to achieve a very high vacuum that will cause so little friction as the ball falls that the friction effect can be ignored.

At both 2.5 meters and 10 meters from the top there is a device that emits a narrow beam of light horizontally. Next to each of these light emitter is a devise to detect the reflection of the light as the ball passes by vertically. When the reflected light is detected, a signal is sent to the "stop watch" to record the time. When the "drop the ball" signal (originated by the experimenter) is received by the device which will drop the ball, another signal then is sent to the stop watch to record the time. Each trial results in three distinguishable times being recorded: (1) drop, (2) ball at 2.5 meters, and (3) ball at 10 meters. I recognize that these three times need to be adjusted with respect to time it takes the signals to travel from each of the sources to the stop watch. These signal travel times can be measured by preliminary tests and wired into the software that will perform calculations after each trial so that these times can be subtracted from the three respective times recorded by the stop watch.

BTW, the 2.5 meters and 10 meters need not be very precise for the purpose of the thought experiment. However, it will help in keeping the description of calculations clear to use these precise values. I assume that the same test equipment would be used for two environments: (a) on the surface of the Earth, and (b) in an accelerating rocket far away from any gravitating body.

I have calculated the ratio of the time for the ball to fall 2.5 meters and 10 meters for both environments. I do not currently have available these results, but I will include the details of my calculations in a later post. For the present, I will summarize from memory. With respect to environment (b), the ratio of the time to fall 2.5 meters to the time to fall 10 meters is exactly 0.5 independent of the acceleration. With respect to (a), this ratio is very roughly 10-7 larger, which seems to me to be a value within the range of measurement and calculation with respect to the thought experiment I have described.

I apologize for not presenting the calculations here. I hope I will be able to post them tomorrow, and if I have made an error, I will much appreciate someone explaining my error to me.

Regards,
Buzz

PAllen
Buzz,

I did one of my standard big O() computations of your scenario and agree with your results. However, there is subtle flaw in the assumptions of your experiment. If you are going to probe this precision, you need to be aware that exact uniform acceleration of a 10 meter body is impossible in special relativity. A rocket undergoing uniform thrust 'far from everything', maintaining is size and shape, will experience slight less acceleration at the top than the bottom. This difference is described by Rindler metric. The upshot is that special relativity predicts a linear change in acceleration with distance that cancels the linear component of the Taylor expansion of the exact function for earth. This means that to detect tidal gravity (distinction from what SR predicts for a uniformly thrust rocket) this way, you need precision better than 1 part in 1014. To understand this issue , read about Born rigidity and the Rindler metric in wikipedia, as a start.

Realizing this, I need to correct that I don't think even SQUID accelerometers could make the distinction over 10 meters (between Rindler and earth metric). It is much easier to detect tidal gravity by change in direction of acceleration over lateral distance since SR predicts no such effect at all for an accelerating platform. My comments about the theoretical difference in falling charges are unaffected by this since SR predicts no radiation in the case of the charge free falling in a rocket because it moving inertially in globally flat spacetime. Unfortunately, for this theoretical effect has never been detected at all due to limitations in how much net charge can be placed on a body, and how small the predicted radiation is (and also its characteristics make it hard to detect).

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Buzz Bloom
PAllen
Buzz,

I did one of my standard big O() computations of your scenario and agree with your results. However, there is subtle flaw in the assumptions of your experiment. If you are going to probe this precision, you need to be aware that exact uniform acceleration of a 10 meter body is impossible in special relativity. A rocket undergoing uniform thrust 'far from everything', maintaining is size and shape, will experience slight less acceleration at the top than the bottom. This difference is described by Rindler metric. The upshot is that special relativity predicts a linear change in acceleration with distance that cancels the linear component of the Taylor expansion of the exact function for earth. This means that to detect tidal gravity (distinction from what SR predicts for a uniformly thrust rocket) this way, you need precision better than 1 part in 1014. To understand this issue , read about Born rigidity and the Rindler metric in wikipedia, as a start.

Realizing this, I need to correct that I don't think even SQUID accelerometers could make the distinction over 10 meters (between Rindler and earth metric). It is much easier to detect tidal gravity by change in direction of acceleration over lateral distance since SR predicts no such effect at all for an accelerating platform. My comments about the theoretical difference in falling charges are unaffected by this since SR predicts no radiation in the case of the charge free falling in a rocket because it moving inertially in globally flat spacetime. Unfortunately, for this theoretical effect has never been detected at all due to limitations in how much net charge can be placed on a body, and how small the predicted radiation is (and also its characteristics make it hard to detect).
The above was written in a rush after a return from a Saturday night out. No surprise an error needs corrections. The SR predicted gradient of acceleration with distance from a starting point with specified acceleration is much smaller than what would occur in earth's gravity. As a result, Buzz's experiment should conceptually work (and also, almost all of the acceleration change with altitude measured by very sensitive accelerometer is due to tidal gravity, not the SR effect). It is true that gravitational time dilation measurements over small distances are only measuring a local SR effect to achievable sensitivity because they measure potential difference; the potential difference change in prediction for SR vs. GR is ballpark one part in 107; the time dilation over 10 meters is only measurable to an accuracy much less than that.

Thus, any issue with the experiment as away of detecting tidal gravity are practical, rather than conceptual. One that comes to mind is that varition in position of the ball's drop point on the order of 1 part in 107 would wipe out the effect. This may be quite achievable, but I don't know such issues of currently achievable experimental limits. Also, as a practical matter, beam interruption rather than reflection would likely work better.

Buzz Bloom
I was not trying to confound things and am happy to be informed about the test objects not being gravitational sources and the need for avoiding tidal effects.

What I was trying to show was that in the elevator the objects fall parallel but in the planet scenario the objects fall non-parallel each going towards the planet's center of mass, therefore approaching each other because those lines of motion converge to a point at the center of mass of the planet.

That deviation from parallel fall is small in the planet scenario (where the source center is far enough to minimize tidal effects), but how close and massive could that center be without producing tidal effects? Will the tidal effects always appear before the fall lines become noticeably non-parallel?

PAllen
I was not trying to confound things and am happy to be informed about the test objects not being gravitational sources and the need for avoiding tidal effects.

What I was trying to show was that in the elevator the objects fall parallel but in the planet scenario the objects fall non-parallel each going towards the planet's center of mass, therefore approaching each other because those lines of motion converge to a point at the center of mass of the planet.

That deviation from parallel fall is small in the planet scenario (where the source center is far enough to minimize tidal effects), but how close and massive could that center be without producing tidal effects? Will the tidal effects always appear before the fall lines become noticeably non-parallel?
That change of distance between geodesics is tidal gravity, thus not relevant to the principle of equivalence. Geometrically, it results from spacetime curvature and the principle of equivalence applies only to the extend spacetime curvature can be ignored.

[edit: one nice description of the affect of GR curvature in a vacuum is that a ball of test particles in free fall becomes - in the free fall fame - elliptical while preserving volume. In the case of such ball dropped near earth, in the direction parallel to earth surface, the ball will squeeze as the particles paths converge; the ball will stretch in the other direction as particles starting closer to earth fall faster. These effects (exceedingly small near earth), then define the extent to which you cannot rely on the principle of equivalence.]

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Buzz Bloom
Gold Member
Thus, any issue with the experiment as away of detecting tidal gravity are practical, rather than conceptual. One that comes to mind is that varition in position of the ball's drop point on the order of 1 part in 107 would wipe out the effect. This may be quite achievable, but I don't know such issues of currently achievable experimental limits. Also, as a practical matter, beam interruption rather than reflection would likely work better.
Hi Paul:

Thanks much for your posts. I particularly thank you for your suggestion to use beam interruption. With that in mind I would also make two further "improvements". In addition to the correction for the time for signals to travel to the timer from the drop device and the two beam interruption detectors, there should also be a correction for the time it takes light to travel from center of the tube to the interrupt detectors. Also two signals should be send from each interruption detector: one when the beam is interrupted, and another when the beam is detected again. The mean of the two times would be a better measurement of when the ball passes a beam than just the time the beam interruption is detected. Actually I think the geometric mean would be better than the linear mean, but I have not done the math to calculate the difference.

Regarding the variation of the dropping position, in principle this can be corrected for, and also to as great a precision as desired, by running a lot of trials and taking averages over the trials. I haven't considered what assumptions should be used about the variance of the dropping position, so I can't calculate the number of trials needed to compensate for this variance. My guess is it might well be an impractically large number.

Here are my calculations regarding the expected difference between the Earth based measurement and the rocket based measurement.
x is the distance in meters the ball has dropped in t seconds.
x0 is the value of x at the drop point.
x1 is the value of x at the upper beam interruption detector. x1 = 2.5.
x2 is the value of x at the lower beam interruption detector. x2 = 10.
r is the distance between the center of the Earth and the drop point. I assume r = 6,371,010 meters.​
g is the assumed Earth acceleration (in m/s2) at the top of the tube. I assume
g = 9.78033.
For the rocket case,
s = 0.5 g t2
t = √(2 s /g)​
Therefore the time ratio (if measured precisely enough) is exactly t(2.5)/t(10) = 0.5, for any acceleration. Using the thought experiment equipment, the measurements would be:
t(10) = 1,430,007,263 ns
t(2.5) = 715,003,631 ns
The ratio = 0.499999999650351.
The following is for t he Earth case.
h is the increase per meter in accleration below the drop point.
h = 0.00000307026116885822 1/s2
is calculated from
g+h = g (r/r-1)2 .​
The calculation of t is:
d2x/dx2 = g + hx
Boundary conditions: x(0) = 0, dx(0)/dt = 0
x = (g/h) (cosh (t√h) - 1)
t = (1/√h) acosh (1 + hx/g)​
In particular, the measurements would be:
t(10) = 1.43000688973752 ns
t(2.5) = 0.715003585178547 ns
The ratio is 0.500000098251275.​

Regards,
Buzz