Does Special Relativity Predict Zero Acceleration in Free Fall?

[PeterDonis]

the second postulate of SR implies that light cannot bend in vacuum, as measured with a classical inertial frame.

Then I'm confused; how is an optical accelerometer supposed to register a nonzero reading in free fall?

 

[stevendaryl]

Then I'm confused; how is an optical accelerometer supposed to register a nonzero reading in free fall?

I think that Harry is saying that SR predicts no bending of light by gravity. Therefore, the prediction is that light will appear to bend upward inside a falling elevator car (the elevator is falling, but light is not).

 

[PeterDonis]

I think that Harry is saying that SR predicts no bending of light by gravity. Therefore, the prediction is that light will appear to bend upward inside a falling elevator car (the elevator is falling, but light is not).

That doesn't make sense either, because the prediction that the elevator is falling, if by that we mean falling with respect to the Earth, can't be made in SR; there's no way to model the gravitational field of the Earth in SR. So it isn't that SR predicts no bending of light by gravity; it's that SR can't model gravity at all, it doesn't even yield a prediction, just a "reply hazy, ask again later".

If instead you say the elevator is "falling" with reference to a uniformly accelerated observer in flat spacetime, then SR *does* predict light bending--for the uniformly accelerated observer, *not* for the observer in the elevator (since the latter is obviously at rest in a global inertial frame, and there is no light bending in such a frame), as I said before. (Of course, then the issue is that SR alone predicts only half the light bending that is actually observed.)

 

[Samshorn]

There is another translation of Langevin's paper by B. L. Sykes (published in Scientia, 1973, v. 108). Online freely available on:
http://amshistorica.unibo.it/diglib.php?inv=7&int_ptnum=108&term_ptnum=302

The relevant passage reads:
p. 297: To do this, our traveller would need only to agree to being shut up inside a projectile that the Earth would launch at a velocity sufficiently close to that of light, but still less than it, which is physically possible, arranging for an encounter with, say, a star to take place at the end of one year in the lifetime of the traveller and to send him back towards the Earth at the same velocity.

Thanks for that link. This translation seems more consistent with the original Wiki translation, since it doesn't attribute the turn-around to the encounter, it simply says we can arrange to send him to a distant star, say, and then back to Earth at the same speed. Only in the revised translation on Wikipedia by Harald88 do we find wording that seems to have been crafted to suggest a gravitational slingshot mechanism, although of course there is no actual mention of gravitation in any of the translations, nor in the original French. And for good reason, since a gravitational slingshot of a near light speed object would be a contradiction in the context of special relativity (since it would obviously entail superluminal speed at the perigee).

Even if it is a mis-attribution, for the purposes of this thread "Langevin scenario" is a gravitational twins paradox. The translation doesn't eliminate the concept, only possibly correct the misattribution.

I think the subject of this thread is free fall acceleration in the context of special relativity, particularly the scenario involving a gravitational slingshot of a projectile that travels to a star at nearly the speed of light. This is inherently a historical subject, since we don't normally deal with gravity in the context of special relativity today. It's also inherently self-contradictory, since (as Langevin himself observed) the projectile cannot exceed the speed of light according to special relativity, and yet a gravitational slingshot in this context would require the projectile to be moving significantly faster than light at the perigee (and the "star" would have to have a superluminal escape velocity at the surface, and couldn't precisely reverse the projectile's direction in any case).

The OP cites Langevin's 1911 article for this scenario, and if this was true then Langevin was guilty of a rather glaring self-contradiction and error in reasoning. Is it appropriate to attribute this blunder to Langevin (who was a pretty smart guy)? Well, when we read the paper in question, we find no mention at all of gravitation. The idea that Langevin had in mind a gravitational slingshot is based entirely on the fact that he said the projectile travels to a star and then returns. But when he mentions the star as the destination point, he says "for example", implying that the turn-around point need not be a star, and of course he does not state the mechanism for the turn-around, any more than he explains how the projectile was originally accelerated to near light speed when departing the Earth.

The only arguably damning evidence against Langevin is the current Wiki translation (as revised in 2010 by a wikipedia editor named Harald88), which is worded to indicate that the encounter with the star is what sends the projectile back. If that translation is accurate, it might suggest (although certainly wouldn't prove) that Langevin really did have that erroneous idea in mind. However, in two other translations the wording is more neutral. Also, the fact remains that in none of the translations does Langevin actually mention gravity. So, I'd be inclined to give him the benefit of the doubt, and attribute the error in reasoning to some of Langevin's modern day readers, rather than to Langevin himself.

 

[Dale]

So, I'd be inclined to give him the benefit of the doubt, and attribute the error in reasoning to some of Langevin's modern day readers, rather than to Langevin himself.

That is fine by me. I have never read anything by him and have no opinion about his reasoning. "Langevin scenario" nothing more than a convenient shorthand for "gravitational slingshot turn around twins paradox scenario". No judgement of Langevin is implied by me. But if you find the association objectionable then I can just say "GST" for "gravitational slingshot twins".

 

[pervect]

That doesn't make sense either, because the prediction that the elevator is falling, if by that we mean falling with respect to the Earth, can't be made in SR; there's no way to model the gravitational field of the Earth in SR. So it isn't that SR predicts no bending of light by gravity; it's that SR can't model gravity at all, it doesn't even yield a prediction, just a "reply hazy, ask again later".

If instead you say the elevator is "falling" with reference to a uniformly accelerated observer in flat spacetime, then SR *does* predict light bending--for the uniformly accelerated observer, *not* for the observer in the elevator (since the latter is obviously at rest in a global inertial frame, and there is no light bending in such a frame), as I said before. (Of course, then the issue is that SR alone predicts only half the light bending that is actually observed.)

SR should correctly predict the amount of bending in an accelerating elevator. It's basically a flat space-time situation, so it shouldn't have any problem with it. (I haven't done a detailed calculation, but I don't see how it could possibly come out otherwise).

The problem arises in applying the results to the GR situation.

I'd have to think carefully before I placed the blame for the discrepancy. I don't think one can calculate it light bending correctly without actually knowing the GR field equations (I might be wrong).

We occasionally have had arguments over the best "explanation" for light bending as well. As I recall, I thought most people were convinced that the "extra deflection" of light in GR could be blamed on spatial curvature alone - one argument is that it's only sensitive to the PPN parameter gamma. But I'm not sure if everyone got convinced by this.

 

[PeterDonis]

SR should correctly predict the amount of bending in an accelerating elevator.

Yes, and this calculation, by the equivalence principle, should also predict bending measured by, for example, an accelerated observer standing at rest on the surface of a planet like the Earth, or "hovering" above the surface of a star like the Sun. But this calculation will be "local"; there is no way to calculate, using SR, the light bending in a "global" scenario such as light from a distant star grazing the Sun and arriving at a telescope on Earth.

 

[harrylin]

There is another translation of Langevin's paper by B. L. Sykes (published in Scientia, 1973, v. 108). Online freely available on:
http://amshistorica.unibo.it/diglib.php?inv=7&int_ptnum=108&term_ptnum=302

The relevant passage reads:
p. 297: To do this, our traveller would need only to agree to being shut up inside a projectile that the Earth would launch at a velocity sufficiently close to that of light, but still less than it, which is physically possible, arranging for an encounter with, say, a star to take place at the end of one year in the lifetime of the traveller and to send him back towards the Earth at the same velocity.

(Il suffirait pour cela que notre voyageur consente à s’enfermer dans un projectile que la Terre lancerait avec une vitesse suffisamment voisine de celle de la lumière, quoique inférieure, ce qui est physiquement possible, en s’arrangeant pour qu’une rencontre, avec une étoile par exemple, se produise au bout d’une année de la vie du voyageur et le renvoie vers la Terre avec la même vitesse.)

Thanks Histspec I did not know that another translation already existence. Nice!
The English there is a bit fuzzy though, for it is unclear what is sending him back in that translation, while Langevin explains by what means all this could be possible in principle. I now checked with two natively speaking French colleagues that the wiki translation is correct on that point (and the one by Sykes inaccurate). I was pretty sure of that but it's always good to double-check.

 

[harrylin]

I think that Harry is saying that SR predicts no bending of light by gravity. [..]

Yes that's the consequence of the second postulate, as Einstein also explained.

 

[PeterDonis]

Yes that's the consequence of the second postulate, as Einstein also explained.

No, that's not what Einstein explained. You yourself quoted him as saying that SR cannot be used in cases where the effects of gravity cannot be neglected. Light bending by the Sun--i.e., in a global context, where light comes in from infinity, passes close to a gravitating mass like the Sun, and then goes back out to infinity with some change in angle, not a local context like an accelerating elevator (see my post #148)--is a case in which the effects of gravity cannot be neglected. So SR cannot make a prediction in this case; that's not the same as SR predicting no bending.

 

[harrylin]

No, that's not what Einstein explained. [..]

I thought that he was clear, but apparently not... it's like Newton's mechanics which can fail at very high speeds because it has no limit speed; that's not the same as saying that Newton's mechanics makes no predictions at the speed of light. And I think that Einstein's explanations about the relationship between light speed and light bending according to SR vs GR have been sufficiently clarified on this forum in the past. But if not, then it's a good topic to discuss.

 

[Dale]

I thought that he was clear, but apparently not... it's like Newton's mechanics which can fail at very high speeds because it has no limit speed; that's not the same as saying that Newton's mechanics makes no predictions at the speed of light.

It is more like Newtonian mechanics with the strong and weak nuclear forces. It doesn't have any mechanism for making predictions where the nuclear forces are important. All you can do is ignore them.

In any case, forgetting the analogies, the direct point is that SR can only be used where "we are able to disregard the influences of gravitational fields on the phenomena". That is not the case for the GST, where the errors from neglecting gravity are large.

 

[PeterDonis]

it's like Newton's mechanics which can fail at very high speeds because it has no limit speed; that's not the same as saying that Newton's mechanics makes no predictions at the speed of light.

Newtonian mechanics makes predictions at the speed of light because "speed" is a meaningful concept in Newtonian mechanics; it just doesn't have a limiting speed. SR can't make predictions at all in the presence of gravity because "gravity" is not even a meaningful concept in SR; the presence of gravity (more precisely, the presence of tidal gravity, but the field of any real gravitating mass, like the Earth, includes tidal gravity, so it amounts to the same thing) violates the fundamental assumptions upon which SR is built.

 

[Dale]

Let's not ask people to disclose their usernames on other sites please.

IMO, the translation is irrelevant to the physics, it only serves in the discussion of the attribution of the GST to Langevin. The GST is a legitimate scenario to analyze regardless of attribution.

 

[Samshorn]

The question is what SR predicts that an accelerometer in free-fall will read.

I think the question is ill-posed, because special relativity isn't a theory of gravity, and hence it makes no predictions about the effects of gravity. Presumably the OP meant to ask what is predicted by some theory of gravity that is formulated in the context of, and consistent with, special relativity. But even this clarification doesn't really help, because there is no such theory of gravity. Needless to say, Newtonian gravity is not consistent with special relativity, so we can't use that as a basis for answering the question.

Probably the closest thing to a viable theory of gravity that is (nominally) consistent with special relativity is Nordstrom's second theory in 1913. This theory arose from Nordstrom's attempt to devise, in the most direct way, a Lorentz invariant theory of gravity, compatible with special relativity. In its final form (to which Einstein and Laue actually made significant contributions in their discussions with Nordstrom) it was regarded as logically unobjectionable. Einstein considered it the only serious competitor to general relativity. However, Nordstrom's theory predicts no deflection of light, and it gives the wrong sign for the precession of orbits, so it is ruled out empirically. Also, the final version of Nordstrom's theory was only formalistically consistent with special relativity, in the sense that the posited underlying Minkowski framework was strictly un-observable. The effective observable metric of spacetime in Nordstrom's theory is curved, so it is now regarded as the first metrical theory of gravity, and it serves as another example of how every attempt to model gravity in a self-consistent way inevitably "breaks" special relativity.

It might also be worth mentioning that, just as Nordstrom's theory predicts no deflection of light, it would also be problematic to claim that it could "turn around" a near-light-speed projectile, so it defeats the OP's original claim about a twins scenario using gravity for the turn-around. (I haven't worked it out, but since it is a metrical theory of gravity in which test particles follow geodesics, I assume the trajectory of a near-light-speed particle would be similar to the trajectory of a light-speed pulse.)

Overall, I suspect the OP was misled by Langevin's 1911 paper, especially the part where Langevin claims (naively, at best) that if an observer and an electric charge were both floating freely inside a sealed capsule in free fall in a uniform gravitational field, the observer (according to Langevin) would detect radiation from the co-moving electric charge (because he says it is absolutely accelerating). Langevin says this "fact" proves the existence of an "ether" (although he doesn't say why). But of course that "fact" violates the strong equivalence principle, and is certainly not in evidence. He also considers tandem acceleration from other causes, and makes the same assertion, but, again, his naive assumptions in this area are known to have been wrong - although I suppose he can be forgiven, since even today there are endless threads on this forum arguing about whether a co-moving observer will detect radiation from a free-falling charge. It's hard for people to understand that the presence or absence of "radiation" is not an absolute fact (especially for people steeped in the Faraday-Maxwell field interpretation, rather than the Ampere-Weber particle interpretation).

 

[PeterDonis]

I think the question is ill-posed, because special relativity isn't a theory of gravity, and hence it makes no predictions about the effects of gravity.

This can't be quite right, because the concept of free fall makes sense in the absence of gravity, as does the concept of proper acceleration and the idea of an accelerometer. But I agree that any prediction SR makes about accelerometer readings (as with any prediction of SR in general) can only be made on the assumption that gravity is negligible.

 

[Nugatory]

I agree that any prediction SR makes about accelerometer readings (as with any prediction of SR in general) can only be made on the assumption that [STRIKE]gravity is negligible[/STRIKE] any tidal effects from gravity are negligible.

In the interests of precise wording for the next person who wanders into this thread, I will suggest the correction above - but that's an editorial correction not a disagreement.

 

[Samshorn]

... the concept of free fall makes sense in the absence of gravity...

The subject of the thread is "Free fall acceleration in SR", which obviously refers to gravity (see the words "fall" and "acceleration"), so I don't quite understand your criticism, unless you are claiming that the very term "free fall" necessarily implies a state indistinguishable from uniform motion in flat spacetime. Along these lines, we could simply define "free fall" as a state of motion in which every kind of accelerometer reads zero. But then the answer to the OP's question would be tautological regardless of the theoretical context (although it would shift the question to whether "free-fall" actually exists in certain circumstances).

But clearly the OP didn't want to assume that tautology, he wanted to know if any "accelerometer" could measure or detect, from within a sealed projectile, the gravitational "acceleration" of the projectile as it loops around a star (for example). He cites Langevin's 1911 paper, in which (as described in my previous post) Langevin claims the answer is yes, by the radiation emitted by a charged particle floating inside the capsule. Now, we know this violates the strong equivalence principle, but that is something which is open to question (especially in 1911), so the answer depends on what theory of gravity we apply. Likewise Nordstrom's theory of gravity has no light deflection (and doesn't strictly satisfy the strong equivalence principle) so it could lead to detectability of gravitational acceleration relative to its posited Minkowskian background.

So, as I said, the question is ill-posed, because the answer to the question (what does an accelerometer read in gravitational free fall, in the context of special relativity) depends on what theory of gravity we choose, and there is no theory of gravity that is both consistent with special relativity and empirically viable.

 

[PeterDonis]

The subject of the thread is "Free fall acceleration in SR", which obviously refers to gravity (see the words "fall" and "acceleration")

Huh? Doesn't "SR" automatically exclude gravity? (Or more precisely tidal gravity, as Nugatory pointed out.)

I interpreted the OP's question to mean, what does SR predict that an ideal accelerometer will read? That question can be formulated perfectly well in flat spacetime, without gravity. (Yes, I realize that's not the question the OP actually wanted answered; see further comments below.)

unless you are claiming that the very term "free fall" necessarily implies a state indistinguishable from uniform motion in flat spacetime.

Yes, I would say that (with the qualifier that the statement should say "locally indistinguishable"), but the question the OP asked was whether an accelerometer in free fall would read zero. That translates easily to asking whether an accelerometer at rest in some inertial frame in flat spacetime would read zero.

clearly the OP didn't want to assume that tautology, he wanted to know if any "accelerometer" could measure or detect, from within a sealed projectile, the gravitational "acceleration" of the projectile as it loops around a star (for example).

Yes, I know, but the OP also assumed, incorrectly IMO, that this question can even be formulated within SR, so he did not ask the question you just described; he asked what SR predicts that an accelerometer in free fall would read. The whole point is that the latter question *can* be formulated within SR (since it can be formulated in flat spacetime, as I said above), but the former question, which is the one the OP really wanted answered, cannot.

So the full response to the OP, IMO, can't just be that "SR can't answer your question"; it has to also distinguish the question the OP actually asked, which SR *can* answer, from the question the OP really wanted answered, which SR cannot answer.

 

[Samshorn]

Huh? Doesn't "SR" automatically exclude gravity? (Or more precisely tidal gravity, as Nugatory pointed out.)

As discussed previously, one can formulate a nominally viable theory of gravity within the context of special relativity (i.e., a Lorentz invariant theory in flat Minkowski spacetime), and in fact this is precisely what several people did between 1905 and 1915, most notably Nordstrom's second theory of 1913. The thing that "automatically excludes" gravity from being operationally modeled within special relativity is the strong equivalence principle, but that is not an apriori fact, it is part of a theory of gravity. If your point is that theories like Nordstrom's (and indeed global special relativity itself) are empirically falsified (at least in operational terms), then I would agree, but the OP was asking for the prediction of an admittedly falsified theory, i.e., he was asking for the predictions within a certain archaic framework (special relativity), recognizing that those predictions are not necessarily in agreement with our best current theory (general relativity).

I interpreted the OP's question to mean, what does SR predict that an ideal accelerometer will read? That question can be formulated perfectly well in flat spacetime, without gravity.

The OP was obviously asking about gravitational free-fall, and even if we restrict this to homogeneous fields it can be converted to a question about accelerating coordinate systems in special relativity only by assuming the strong equivalence principle, which is part of a theory of gravitation. In other words, either you are disavowing any relevance of your answer to gravitational free-fall (in which case, it's pointless), or else you're assuming that physics in a homogeneous gravitational field is indistinguishable from physics in a uniformly accelerating system of coordinates, which enables you to answer a question about gravitational free-fall (neglecting tidal effects) by converting it to a question about an accelerating coordinate system in special relativity. But by doing this you are smuggling in a crucial part of a theory of gravity, to claim the equivalence. If you say you're not answering a question about gravitational free-fall, you're just answering a question about accelerating reference frames in special relativity, then I would say you aren't addressing the question that the OP was obviously asking.

Yes, I realize that's not the question the OP actually wanted answered...

I see.

...but the question the OP asked was whether an accelerometer in free fall would read zero. That translates easily to asking whether an accelerometer at rest in some inertial frame in flat spacetime would read zero.

My point is that this "translation" (if it has any relevance to the topic of discussion, i.e., gravitational free-fall) assumes the strong equivalence principle, which is part of a theory of gravitation, and one that is inconsistent with special relativity. Obviously we can answer what an accelerometer will do according to general relativity, but that isn't the question that was asked. So it comes back to the fact that the question is ill-posed, because to answer it (even to answer it in the manner that you have described) requires some theory of gravity - unless you are just completely disavowing any relevance of your answer to gravitational free-fall, in which case it isn't responsive to the question posed.

So the full response to the OP, IMO, can't just be that "SR can't answer your question"; it has to also distinguish the question the OP actually asked, which SR *can* answer, from the question the OP really wanted answered, which SR cannot answer.

I think we have a different perception of the question the OP actually asked. To me it seemed (and seems) abundantly clear that the subject was gravitational free-fall (arising, as it did, from the gravitational slingshot scenario), and clearly his question can't be answered in the absence of some theory of gravity. You, on the other hand, think he actually asked what an accelerometer would read when moving inertially in flat spacetime. I don't think that's a plausible reading of his question.

To satisfy both interpretations, maybe the best answer to the OP would be: If you are asking what special relativity predicts for the reading of an ideal accelerometer moving inertially in flat spacetime, the answer is zero (essentially by definition of an ideal accelerometer). If you are asking what an accelerometer in gravitational free-fall would read according to special relativity, the answer is that special relativity is not a theory of gravity, so it does not provide an answer. Furthermore, there is no empirically viable theory of gravity consistent with (global) special relativity - at least not in operational terms.

 

[PeterDonis]

As discussed previously, one can formulate a nominally viable theory of gravity within the context of special relativity (i.e., a Lorentz invariant theory in flat Minkowski spacetime), and in fact this is precisely what several people did between 1905 and 1915, most notably Nordstrom's second theory of 1913.

I probably need to read more about these theories, but here's the basic issue I see with trying to formulate any theory within the framework of SR that can model tidal gravity: a basic assumption of SR, the one that allows you to construct global inertial frames, is that if you take two objects at rest with respect to each other, and both of them move inertially--i.e., no forces act on them--then they will stay at rest with respect to each other. Tidal gravity violates this assumption.

The thing that "automatically excludes" gravity from being operationally modeled within special relativity is the strong equivalence principle

I'm not sure I agree; I think it's the impossibility of constructing global inertial frames in the presence of tidal gravity. See above.

The OP was obviously asking about gravitational free-fall, and even if we restrict this to homogeneous fields

Which eliminates the "gravitational slingshot turnaround" scenario, since that requires an inhomogeneous field centered on the star at the turnaround point. Since that scenario was what originally prompted the OP to start this thread, I'm not sure how relevant a discussion of homogeneous fields is.

To put this another way: if you're trying to model the gravitational slingshot turnaround scenario in SR, you're basically assuming that the trajectory of the traveling twin, who slingshots around the distant star at his turnaround, can be modeled in some global inertial frame. But in any such frame, this twin's trajectory has a huge path curvature at the turnaround point. You can make the impact of this on the twin's elapsed proper time arbitrarily small (by making the turnaround a smaller and smaller part of the total trip), so you can get a reasonably accurate answer for the twin's elapsed proper time using SR. But you can't get any kind of accurate prediction of the twin's proper acceleration using SR, because of the large path curvature at the turnaround point. You need a curved spacetime to get the right answer for proper acceleration in this scenario (i.e., zero).

In this context, the OP's question is really something like this: does the large path curvature of the traveling twin at turnaround, in the above SR model of the slingshot scenario, really *have* to correspond to SR predicting (incorrectly) a large nonzero reading on an accelerometer for the traveling twin at turnaround? Is there any way to graft some kind of "gravity" onto SR that would allow SR to somehow predict a zero reading on an accelerometer even on a path with a large ("apparent"?) path curvature in a global inertial frame?

The answer to this, it seems to me, is "no"; but that's not because of the SEP. It's because there's no way to model tidal gravity in SR, as I said above. Put another way, even if we could "simulate" a homogeneous gravitational field with acceleration in SR (which is where the SEP would come in), that won't suffice in the slingshot scenario; as I said above, you need an inhomogeneous field, centered on the turnaround star, and that means you need tidal gravity to be present, which can't be done in SR. In other words, there's no way to even model any kind of "force of gravity" within SR that could somehow "cancel" the path curvature of the traveling twin at turnaround and allow a prediction of zero accelerometer reading.

I think we have a different perception of the question the OP actually asked.

Yes, I think so; and I also wasn't as clear as I should have been on my interpretation of the context of the OP's question. Hopefully the above helps to clarify that.

 

[Samshorn]

...a basic assumption of SR, the one that allows you to construct global inertial frames, is that if you take two objects at rest with respect to each other, and both of them move inertially--i.e., no forces act on them--then they will stay at rest with respect to each other. Tidal gravity violates this assumption.

I think your reasoning tacitly relies on general relativity and the strong equivalence principle, because you are thinking of gravitational free-fall as force-free inertial motion. You say two objects are moving inertially - subject to no forces - in a gravitational field, and since they drift together or apart due to tidal effects you say this contradicts global inertial frames. But you could also conceivably model gravity as a force in Minkowski space, so that objects in gravitational free-fall are not in inertial motion, and hence this doesn't automatically rule out global inertial frames. (There are "tidal effects" for electromagnetism too, but this doesn't make EM incompatible with special relativity.)

Note that even with a force theory of gravity, an accelerometer that works by mechanical means would still be expected to read zero in free-fall, provided that the force of gravity obeys at least the weak equivalence principle, i.e., gravitational mass is proportional to inertial mass (as in Newtonian gravity), so all parts of the accelerometer are accelerated in tandem. What people tried to do soon after special relativity was to develop a Lorentz invariant version of Newton's gravitational force, analogous to the electromagnetic force (which certainly isn't incompatible with special relativity). But it turns out there are difficulties with this simple approach when applied to gravity. (Strangely reminiscent of the difficulties in trying to quantize gravity today...) Also, with just the weak equivalence principle, one might imagine electrical or optical accelerometers that might reveal the "actual" acceleration, depending on our theory of gravitation.

You can't get any kind of accurate prediction of the twin's proper acceleration using SR, because of the large path curvature at the turnaround point. You need a curved spacetime to get the right answer for proper acceleration in this scenario (i.e., zero).

I don't think the OP asked for "the right answer", he asked what special relativity would predict. As noted above, one would get a prediction of zero for accelerometer readings even for a Newtonian-style force theory of gravity, provided only that the force accelerates all parts of the accelerometer in tandem. This, I suspect, is what the OP was driving at. Indeed the default theory of gravity works exactly this way... the problem is that the only known "force" models of gravity that are nominally consistent with special relativity are ruled out by experiment. (You can also argue that they are only formalistically consistent with special relativity, but still...)

In this context, the OP's question is really something like this: does the large path curvature of the traveling twin at turnaround, in the above SR model of the slingshot scenario, really *have* to correspond to SR predicting (incorrectly) a large nonzero reading on an accelerometer for the traveling twin at turnaround?

Hmm... I'm not aware of any sense in which special relativity makes any such prediction (let alone that it must make such a prediction), at least not for any theory of gravity in which the gravitational mass is proportional to inertial mass, so all the components of the accelerometer are accelerated in tandem. I suppose someone could dream up a crazy theory of gravity that didn't obey this weak equivalence principle, in which case an accelerometer would give a non-zero reading, but it would be pretty weird. But maybe when you say special relativity predicts a non-zero reading you are referring to some other kind of accelerometer, perhaps one using electromagnetic or optical features, and a theory of gravity that violates the strong equivalence principle so that these features would reveal the acceleration? That all seems pretty far-fetched to me. And yet you're talking about it as if it's the default SR prediction... and all the while saying that SR even makes a prediction for the accelerometer readings in a gravitational turn-around in the absence of a theory of gravity... I can't follow your reasoning at all.

Is there any way to graft some kind of "gravity" onto SR that would allow SR to somehow predict a zero reading on an accelerometer even on a path with a large ("apparent"?) path curvature in a global inertial frame?

Again, this seems puzzling, because special relativity is not a theory of gravity at all, so to get any kind of gravitational turn-around we obviously need to posit SOME theory of gravity. Whatever theory we propose, if it obeys the weak equivalence principle, any mechanical accelerometer will obviously read zero. So it would take a very weird theory (grossly inconsistent with observation) to give a non-zero reading.

The answer to this, it seems to me, is "no"; but that's not because of the SEP. It's because there's no way to model tidal gravity in SR, as I said above.

That's where we fundamentally differ, because the existence of tidal effects for a physical force don't automatically invalidate global inertial frames. (See electromagnetism.) And our mechanical accelerometers will read zero provided the force satisfies at least the WEP. But, again, to make any kind of "prediction" for what an accelerometer in gravitational free-fall will read, we need to specify a theory of gravity (and then we confront the fact that there is no viable theory of gravity consistent with special relativity, at least operationally).

 

[PeterDonis]

I think your reasoning tacitly relies on general relativity and the strong equivalence principle, because you are thinking of gravitational free-fall as force-free inertial motion. You say two objects are moving inertially - subject to no forces - in a gravitational field, and since they drift together or apart due to tidal effects you say this contradicts global inertial frames. But you could also conceivably model gravity as a force in Minkowski space, so that objects in gravitational free-fall are not in inertial motion, and hence this doesn't automatically rule out global inertial frames. (There are "tidal effects" for electromagnetism too, but this doesn't make EM incompatible with special relativity.)

Hmm. I'll have to think about this. See further comments below.

What people tried to do soon after special relativity was to develop a Lorentz invariant version of Newton's gravitational force, analogous to the electromagnetic force (which certainly isn't incompatible with special relativity). But it turns out there are difficulties with this simple approach when applied to gravity. (Strangely reminiscent of the difficulties in trying to quantize gravity today...)

Can you give any references? This is not an area I'm familiar with, but it looks interesting.

But maybe when you say special relativity predicts a non-zero reading you are referring to some other kind of accelerometer, perhaps one using electromagnetic or optical features, and a theory of gravity that violates the strong equivalence principle so that these features would reveal the acceleration? That all seems pretty far-fetched to me. And yet you're talking about it as if it's the default SR prediction... and all the while saying that SR even makes a prediction for the accelerometer readings in a gravitational turn-around in the absence of a theory of gravity... I can't follow your reasoning at all.

What I'm saying is much simpler than all this. I'm saying that, if we just assume a global inertial frame in which the stay-at-home twin is at rest, and in which the turnaround star is also at rest, and look at the traveling twin's trajectory in this global inertial frame, that trajectory obviously has a large path curvature at the turnaround point. The question of how the traveling twin can have this path curvature if no "force" except "gravity" is acting on it is a different question.

Regarding what I quoted at the beginning of this post, I don't see how I am assuming the SEP in order to set up the global inertial frame I just described. I certainly don't see that I have to assume the SEP in order to set up a scenario where there is a global inertial frame in which both the stay-at-home twin and the turnaround star are at rest. And given such a global inertial frame, the claim that the traveling twin's path curvature must be large at the turnaround seems straightforward too, without requiring the SEP.

Now, suppose that the turnaround "star" has negligible mass, so that gravity is indeed completely absent from the scenario. Instead, the "star" induces the traveling twin's turnaround by, say, sticking out an extremely strong cable that catches the twin's ship and swings it around like a real slingshot (as opposed to a "gravitational slingshot"), releasing it in the opposite direction (back towards the stay-at-home twin). It seems obvious here that the large path curvature of the traveling twin's trajectory at this point, in the global inertial frame, will correspond to a large nonzero reading on the traveling twin's accelerometer. In other words, the force exerted by this slingshot would not obey the WEP. But that's because this slingshot only acts on the ship itself; it doesn't act directly on things inside the ship, like the twin himself and his accelerometer.

If I'm understanding what I quoted at the top of this post correctly, you are saying that a theory of SR + gravity could model gravity as exerting a force which, unlike the above slingshot, would obey the WEP, by exerting its force on everything--the ship, the twin, and his accelerometer--in the same way. This would allow the accelerometer to read zero at the turnaround even though the path curvature of the trajectory, in the global inertial frame, is still large and nonzero. In other words, the presence of gravity, in this theory, would break the connection between path curvature and nonzero accelerometer readings. Or at least, one would have to find some other kind of "accelerometer" if one wanted to detect the nonzero path curvature due to the gravitational force.

 

[Samshorn]

Can you give any references? This is not an area I'm familiar with, but it looks interesting.

I agree it's an interesting topic. Here are a couple of excellent papers by John Norton, describing the development of gravitation theories in the years leading up to 1915, during which Einstein was basically alone in his conviction that the right theory would have to depart from the framework of special relativity:

www.pitt.edu/~jdnorton/papers/Nordstroem.pdf
Einstein, Nordstrom, and the Early Demise of Scalar Lorentz-Covariant Theories of Gravitation

www.pitt.edu/~jdnorton/papers/einstein-nordstroem-HGR3.pdf
Einstein and Nordstrom, Some Lesser Known Thought Experiments in Gravitation

 

[Wes Tausend]

This is a spin-off of a parallel discussion, starting from:
https://www.physicsforums.com/showthread.php?p=4281037#post4281037

The question is what SR predicts that an accelerometer in free-fall will read. This issue may be simply due to different people using a different meaning of "SR", but it could have a deeper cause.

A basic reference for this discussion:
Einstein 1905, http://www.fourmilab.ch/etexts/einstein/specrel/www/
and another one for context:
Langevin 1911, http://en.wikisource.org/wiki/The_Evolution_of_Space_and_Time


SR uses the inertial frames of classical mechanics; in my opinion it's obvious that SR doesn't predict that an accelerometer in free fall will indicate a large acceleration. That conflicts with the known laws of physics, even of classical mechanics.

Arguments in favor of both opinions may help to clarify this issue.

harrylin,

I my opinion... free-fall must be defined as not feeling, nor experiencing any acceleration. The instant one shuts off a rocket engine, gravity from all objects in the universe are free to exert their respective "pull" and a person, or object, in free-fall neither feels, nor can measure any acceleration at all, no matter what tool used (if placed at the center of the ship's gravity). Within the confines of the twin's free-falling ship, an enclosed source/destination of light does not bend laterally, nor does it red shift longitudinally, which is not true in any accelerating ship. The ship becomes a more perfect laboratory to observe this than one can experience on earth, which is the beauty of space-stations. Barring Tidal Effects, a ship in free-fall will find an accelerometer useless. (Still, a small, negligable amount of gravity is always also exerted by the matter within the ship and the observer himself. There is no perfect, gravity-free laboratory.)

Perhaps the best way to picture simple free-fall is to imagine that an object is in rest "suspension" above earth, and only mother Earth is rushing up to meet it. The object has no internal indication of motion, or gravitational tug, whatsoever. Such a suspended object apparently has no internal gravitational field (and one might also think, likely no gravitons if such mysterious particles do indeed exist).

As an example of simplest free-fall, if one twin were to leave his brother and earth, he would accelerate up at a greater uniform velocity than his twin, and in a most simple sense, merely cut the engines and instantaneously free fall also at a greater uniform velocity, until such a time as he would have to decelerate (experience acceleration again) to prevent crashing. The traveling twin would then theoretically be younger than his brother by some degree, purely due to moving relatively faster as time passed by independently for both. The most important point, in my opinion, is that he would be younger, not directly because he experienced the accelerations, but merely because he ended up traveling at a different uniform velocity than the other twin. The only thing that acceleration has contributed, is the fact that it changed thee relative uniform velocities of the two men for a period at which their clocks ran at different rates and only one man felt the excess acceleration which indicates he traveled at the greater speed. However, one can readily see that an innocuous acceleration must always be part of a change of speeds, the speed difference being the key. (see "Does a clock's acceleration affect its timing rate?" below)

Since acceleration has no direct effect on time, a powered loop back, or non-accelerative slingshot (no motor assist) around a planetary body, will not modify, nor affect, the clocks within the spaceship. Only different relative speeds due to any type of acceleration will affect clocks, and then only the accelerated clock(s).

All the ongoing non-relative (time-changing) speeds, under acceleration at the time or not, are merely an added combination of clock-slowing speeds, a percentile greater than the slower speed of the non-accelerated twin on earth. We have no way of speeding clocks up, only slowing them down.

An interesting similar event takes place in this revealing excerpt from Wikipedia on Micro-Gravity (free-fall):
"Floating" objects in a spacecraft in LEO are actually in independent orbits around the Earth. If two objects are placed side-by-side (relative to their direction of motion) they will be orbiting the Earth in different orbital planes. Since all orbital planes pass through the center of the earth, any two orbital planes intersect along a line. Therefore two objects placed side-by-side (at any distance apart) will come together after one quarter of a revolution. If they are placed so they miss each other, they will oscillate past each other, with the same period as the orbit. This corresponds to an inward acceleration of 0.17 μg per meter horizontal distance from the center."
It is not a long page and is a good read.

However note, if the two free-fall "floating" objects in the above paragraph were accelerometers, they would measure nothing. They are merely following their respective "equal-to flat space-time" orbits over the exact center of the Earth in curved space-time. Absolutely all successful orbits must be centered with one another thusly, a point to contemplate.

...

I thought someone mentioned students. I think, that if I were to present this scenario to a student, as to how one twin ended up younger, that I would take the route that Feynman did in his lecture in Six Not So Easy Pieces, Chapter Four, section 4-2 (The Twin Paradox). The presentation was good and I particularily liked the summation of the last paragraph, first sentence:
"So the way to state the rule is to say that the man who has felt the accelerations, who has seen things fall against the walls, and so on, is the one who would be the younger; that is the difference between them in the absolute sense, and it is certainly correct."

To establish lesson fundamentals, I would proceed thusly:
"SR is Special because it specifically excludes gravity or any other non-uniform motion to establish a fundamental baseline. Otherwise Einstein might have merely called it Relativity. He later specifically named General Relativity, General, as to note that it now included gravity along with ordinary SR. SR never, ever has anything to do with gravity or any other acceleration. Any quantity calculation, any quality that deals with an acceleration, falls promptly in the realm of GR, not SR.

The reason Einstein started with SR was that an explanation was needed to describe the silver lining he recognized after the heart-breaking news that Michelson–Morley failed to find Absolute Rest (or Motion) in the certain linear, uniform motion of earth. Motion is certain because Earth was known to be at least moving around the sun, not the center of the universe. Orbital motion is always uniform because no side-acceleration can ever be felt; the Earth is in free-fall as is a non-powered GST slingshot. As a matter of fact, the slightly elipse orbit of Earth around the sun changes speed without any measureable slingshot acceleration."

References
Does a clock's acceleration affect its timing rate?
Micro-g environment
Six Not So Easy Pieces

Wes
...

 

[harrylin]

Newtonian mechanics makes predictions at the speed of light because "speed" is a meaningful concept in Newtonian mechanics; it just doesn't have a limiting speed. SR can't make predictions at all in the presence of gravity because "gravity" is not even a meaningful concept in SR; the presence of gravity (more precisely, the presence of tidal gravity, but the field of any real gravitating mass, like the Earth, includes tidal gravity, so it amounts to the same thing) violates the fundamental assumptions upon which SR is built.

It would make no sense to modify a postulate that doesn't apply and SR is commonly used in the presence of gravity; evidently this is a topic to discuss, although it's not the topic of this thread.

 

[harrylin]

[..] Probably the closest thing to a viable theory of gravity that is (nominally) consistent with special relativity is Nordstrom's second theory in 1913. This theory arose from Nordstrom's attempt to devise, in the most direct way, a Lorentz invariant theory of gravity, compatible with special relativity. [..] Nordstrom's theory predicts no deflection of light [..]

Thanks, that's a constructive take of the question. I'm not familiar with that theory, but of course such a theory can not predict a deflection of light in vacuum wrt a Newtonian reference system. Once more: in contrast to GR, SR based theories of gravitation erroneously predict that a truly optical accelerometer in free fall towards a star will measure its free-fall acceleration.

[..] every attempt to model gravity in a self-consistent way inevitably "breaks" special relativity.

That was suggested earlier in this thread, but it's an interesting point indeed!

[..] Overall, I suspect the OP was misled by Langevin's 1911 paper, especially the part where Langevin claims (naively, at best) that if an observer and an electric charge were both floating freely inside a sealed capsule in free fall in a uniform gravitational field, the observer (according to Langevin) would detect radiation from the co-moving electric charge (because he says it is absolutely accelerating). Langevin says this "fact" proves the existence of an "ether" (although he doesn't say why). But of course that "fact" violates the strong equivalence principle [..]

No, that confounds SR with GR - SR has no equivalence principle. Mixing up theories is a persistent problem in this thread...

He also considers tandem acceleration from other causes, and makes the same assertion, but, again, his naive assumptions in this area are known to have been wrong - although I suppose he can be forgiven, since even today there are endless threads on this forum arguing about whether a co-moving observer will detect radiation from a free-falling charge. It's hard for people to understand that the presence or absence of "radiation" is not an absolute fact (especially for people steeped in the Faraday-Maxwell field interpretation, rather than the Ampere-Weber particle interpretation).

Yes, especially when considering that SR uses Maxwell's field interpretation. It goes to show that one should never have blind faith in a theory.

I don't think the OP asked for "the right answer", he asked what special relativity would predict. As noted above, one would get a prediction of zero for accelerometer readings even for a Newtonian-style force theory of gravity, provided only that the force accelerates all parts of the accelerometer in tandem. This, I suspect, is what the OP was driving at. Indeed the default theory of gravity works exactly this way... the problem is that the only known "force" models of gravity that are nominally consistent with special relativity are ruled out by experiment.

Yes indeed; and we all know that SR was ruled out by experiment, that's why GR successfully replaced it.

 

[Dale]

SR is commonly used in the presence of gravity

But only when the errors from neglecting gravity are negligible, as in the MMX, not the GST.

In the GST you can take the accelerometer reading as a given, in which case neglecting gravity gives a large error in the path. Alternatively you can take the path as a given, in which case neglecting gravity gives a large error in the accelerometer reading. Either way the error from neglecting gravity is large so SR cannot be used in the GST.

 

[Samshorn]

SR based theories of gravitation erroneously predict that a truly optical accelerometer in free fall towards a star will measure its free-fall acceleration.

That's a problematic claim, because the concept of "SR based theory of gravitation" is inherently ill-defined and counter-factual. It would be more accurate to say that for any theory of gravity that does not satisfy the strong equivalence principle there would be ways of locally detecting gravitational acceleration - which is essentially true by definition.

One can't really claim that an "SR based theory of gravitation" could not possibly satisfy (operationally) the strong equivalence principle. In fact, there is a field interpretation of general relativity in which the curved metric is just an "effective" metric, on top of the "true" (but unobservable) flat Minkowski metric of special relativity. (This is similar to how special relativity can be interpreted in a Lorentzian sense, by invoking a metaphysically defined sense of "truth", adding strictly unobservable elements to the theory.) So, according to this interpretation, general relativity is actually a "SR based theory of gravitation". Nordstrom's theory is in the same boat, with an unobservable flat background, the only difference being it is a scalar theory ("spin 0") rather than a tensor theory ("spin 2"). But one can argue that a scalar theory - in which gravity couples with mass but doesn't couple with electromagnetic energy (no light deflection) - is inherently inconsistent with the mass-energy equivalence of special relativity. This is essentially how Einstein argued that a tensor theory was required, and hence general relativity is the closest possible theory to special relativity that incorporates gravity in a logically coherent way, preserving as far as possible (but not farther) the principles of special relativity. And of course if you are someone who enjoys imposing arbitrary metaphysics on top of your physics, you can always claim that general relativity perfectly preserves special relativity even globally (albeit in an unobservable way).

No, that confounds SR with GR - SR has no equivalence principle. Mixing up theories is a persistent problem in this thread...

I'd say the persistent problem in this thread is the mistaken idea that there is a unique theory of gravity consistent with special relativity, so that questions about accelerometer readings in "gravitational free-fall in SR" have a well-defined answer. They don't, because the idea of a "SR based theory of gravitation" is too vague and ambiguous to have any definite meaning, and of course it is ultimately a counter-factual proposition (unless you are talking metaphysics, in which case you can believe whatever you want).

Yes especially when considering that SR uses Maxwell's field interpretation.

The independence of the speed of light from the speed of its source was suggested by the success of Maxwell's equations and the wave model of light, but by the time he wrote the EMB paper in 1905 Einstein had already finished his paper on the photo-electric effect, in which he highlights the "particle" attributes of light, and the inadequacy of Maxwell's equations to account for these attributes. He already knew "Maxwell's equations could not claim unlimited validity". This is why he specifically avoided basing special relativity on Maxwell's equations, and on the wave model of light, and this enabled special relativity to survive the quantum revolution, in which the wave model of light was superceded by the "neither wave nor particle" model of quantum electrodynamics that replaced Maxwell's equations. The relativity postulate was suggested by particle models, and the light postulate was suggested by wave models, and the two were "only apparently irreconcilable". In a sense the whole purpose of the EMB paper was to provide a framework within which it is possible to reconcile both the particle and the wave attributes of phenomena.

 

[Dale]

I'd say the persistent problem in this thread is the mistaken idea that there is a unique theory of gravity consistent with special relativity, so that questions about accelerometer readings in "gravitational free-fall in SR" have a well-defined answer.

And none of the "SR gravity" theories are consistent with observation (AFAIK). Because of that, all you can justifiably do is use SR in situations where the errors arising from neglecting gravity are small.

 

[Samshorn]

And none of the "SR gravity" theories are consistent with observation (AFAIK). Because of that, all you can justifiably do is use SR in situations where the errors arising from neglecting gravity are small.

Yes, and lacking a theory of gravity it's difficult to know with certainty what the errors might be, especially in circumstances where gravity plays an essential role. For the slingshot twins scenario one might think that, since we can make the coasting part of the journey as long as we want, we could ensure the gravitational effect is negligible, but that's not true, because if we posit a theory of gravity that doesn't deflect electromagnetic energy at all, we can't assume it would deflect a near-light-speed projectile very much either. To produce a gravitational slingshot and yet have no deflection of light, we would have to postulate a theory of gravity that couples with mass but not with other forms of energy, which is inconsistent with mass-energy equivalence of special relativity, to which we are supposedly adhering. The only way out would be to completely renounce any claim to empirical viability, and propose a theory of gravity in which hot objects fall differently than cold objects, etc. But if we renounce empirical constraints, then we're free to propose all kinds of crazy theories such as a theory in which gravity has no effect at all, and hence can't turn around any projectile. This is perfectly consistent with special relativity, it's just an empirical failure as a gravitational theory... but so is every other gravitational theory that doesn't satisfy the equivalence principle, which is the most precisely verified principle in physics.

 

[Dale]

the equivalence principle, which is the most precisely verified principle in physics.

This is interesting. Do you mean this literally or is this hyperbole?

 

[Samshorn]

This is interesting. Do you mean this literally or is this hyperbole?

I meant it literally. One often hears that quantum field theory is the most precisely confirmed theory in all of physics, citing for example the computed value of the magnetic moment of an electron using Feynman diagrams up to 8th order, being accurate to within about 1 part in a billion. But the equivalence principle, or at least the equality of inertial and gravitational mass, has been established to 1 part in about 100 billion - and that was in 1964. Needless to say, that doesn't prove the equality is EXACT (as it must be in general relativity), but it sure doesn't give much reason to doubt it.

 

[Dale]

the equivalence principle, or at least the equality of inertial and gravitational mass, has been established to 1 part in about 100 billion - and that was in 1964.

Thanks, I was unaware of that! Do you have a reference, or at least the name of the experiment? It seems like one I should know, but don't.

 

[Samshorn]

Thanks, I was unaware of that! Do you have a reference, or at least the name of the experiment? It seems like one I should know, but don't.

One reference is a refinement of the classic Eotvos experiment, where they achieved precision of 1 part in 100 billion:

P. G. Roll, R. Krotkov, R. H. Dicke, Annals of Physics, 26, 442, 1964.

In 1971 Braginski improved this result by another order of magnitude:

Braginskiǐ, V. B.; Panov, V. I., Soviet Journal of Experimental and Theoretical Physics, Vol. 34, p.463.

I think more recently the precision has been improved still further, but don't have references.