Does Special Relativity Predict Zero Acceleration in Free Fall?

[Dale]

Instead, your formula for what according to SR an accelerometer in free fall will read was not supported by your references and as for me, I will need some time to search more specific references myself. I won't respond anymore to such personal attacks but discuss such references and basic derivations.

How is pointing out your failure to provide a reference a "personal attack"? I am attacking your argument as being speculative, not you personally. You shouldn't take that as a personal attack.

My formula was supported by the references.

The references on the four-acceleration gave the formula in terms of the four-velocity. The references on the four-velocity gave the four-velocity in terms of the worldline. To obtain my formula, simply substitute the four-velocity into the four-acceleration. I can provide a reference on substitution too if needed.

The references on the four-acceleration also explained the relationship to proper acceleration. In the other thread I provided references showing that the proper acceleration is the acceleration measured by an accelerometer. I didnt feel the need to repeat those here.

Bottom line, my formula is correct, and well supported by references. Yours is neither. You can choose to ignore the facts, but that doesn't change them.

Meanwhile I think that the participants to this thread are not a bad sample of "mainstream" opinion, and the opinions are divided.

No, a poll on PF might constitute mainstream public opinion, but does not qualify as a mainstream scientific reference. Particularly not in terms of the PF rule against overly-speculative posts.

 

[Dale]

This thread was to discuss the several times repeated claim by Dalespam that according to SR an accelerometer in free fall will have a large reading.

My claim, as I stated in post 2 and repeated multiple times is: "That question cannot even be addressed by SR since SR does not handle gravitation. You cannot generally establish a self-consistent SR inertial frame in the presence of gravity. This question is outside the domain of applicability of SR."

As I have mentioned several times, all you can do in SR is to neglect gravity. When you do neglect gravity then you introduce errors in your analysis. Whether or not the use of SR is justified depends on the magnitude of those errors.

The formula used to give the SR prediction has been provided. Applying that formula to the Langevin scenario, those errors are large because the SR prediction is a high proper acceleration and the real accelerometer would measure 0.

Any time I mentioned a high accelerometer reading for free-fall it was specifically in reference to this, neglecting gravity for the Langevin scenario. If you treat the Langevin scenario using SR then you use the referenced formula for accelerometers and get an erroneously high prediction.

 

[Samshorn]

Any time I mentioned a high accelerometer reading for free-fall it was specifically in reference to this, neglecting gravity for the Langevin scenario. If you treat the Langevin scenario using SR then you use the referenced formula for accelerometers and get an erroneously high prediction.

What is the "Langevin scenario"? Are you referring to a twin scenario where the turn-around is achieved by looping around a distant star under the effect of the star's gravitation? If so, Langevin never mentioned any such thing. (Granted, another poster claimed the gravitational turn-around scenario was in Langevin's 1911 paper, but when I asked him to point to where in that paper the gravitational turn-around was discussed, he admitted there was no such thing in that paper, so it was just a mis-attribution.)

 

[PeterDonis]

What is the "Langevin scenario"? Are you referring to a twin scenario where the turn-around is achieved by looping around a distant star under the effect of the star's gravitation? If so, Langevin never mentioned any such thing.

Yes, that is what we're referring to. Even if Langevin didn't actually come up with it, it's named after him now (at least for this thread and its relatives).

 

[georgir]

This thread is a mess...

DaleSpam, you still have not addressed my last edit. So I will repeat

your quote again:

[..] SR predicts a very large accelerometer reading during the turnaround, and real free falling accelerometers read 0.

If there were noticeable tidal forces, then a real free falling accelerometer would also not read 0.

You have two separate cases:
a) tidal gravity is negligible, SR predicts 0 and real experiments show 0;
b) tidal gravity is significant, SR predicts >0, irrelevant how accurate, and real experiments also show >0
So your quote is still wrong no matter how you look at it.

PeterDonis, I just don't know where to start with your posts, I'll leave it for another day.

 

[Dale]

You have two separate cases:
a) tidal gravity is negligible, SR predicts 0 and real experiments show 0;
b) tidal gravity is significant, SR predicts >0, irrelevant how accurate, and real experiments also show >0

This is a false dichotomy and, in particular, it is not true in the case under consideration. Real accelerometers would still show 0 in the gravitational twin scenario, but the SR prediction is >0.

 

[PeterDonis]

For example, suppose they are both charged, with the same charge/mass ratio, and you are accelerating them upward by using a uniform electric field. Then the spring will be neither stretched nor compressed.

Really? Think carefully about the Bell Spaceship Paradox before you answer.

 

[georgir]

This is a false dichotomy and, in particular, it is not true in the case under consideration. Real accelerometers would still show 0 in the gravitational twin scenario, but the SR prediction is >0.

Again, no. SR will not predict >0. It predicts 0 without tidal effects. Whatever formula you think you have for an accelerometer's reading in SR is clearly wrong. There can be no change in the distance between any components of an accelerometer in its proper frame according to SR if the acting forces are uniform, that means zero reading.

 

[Dale]

Whatever formula you think you have for an accelerometer's reading in SR is clearly wrong.

The formula is the correct one for SR. If you disagree then please post the equation you believe to be the correct one for SR and provide your references.

 

[stevendaryl]

Note that you said "laws of motion", *not* "law of gravity". Yes, SR is a more accurate version of the laws of motion, which takes into account the finite speed of light and its consequences. But it's *not* a replacement for the law of gravity.

This doesn't necessarily mean that none of the scenarios you mention can be analyzed using SR. For example, if I'm calculating the stress on a beam, I can do the calculation in a local inertial frame, which means I can use SR if for some reason I need that level of accuracy. But note that in doing this, I am not adding a "force of gravity" to SR. I am using the equivalence principle to mimic the local effects of "gravity" with acceleration: I am basically modeling the beam as being inside an accelerating rocket and assuming that the results will apply.

Well, the exact equations of motion for a test particle in curved spacetime in an arbitrary coordinate system are:

$m (\dfrac{d^2 x^\mu}{d \tau^2}+ \Gamma^\mu_{\nu \lambda} \dfrac{d x^\nu}{d \tau} \dfrac{d x^\lambda}{d \tau}) = F^\mu$

Treating gravity like a force simply amounts to moving the $\Gamma^\mu_{\nu \lambda}$ term to the right side:

$F_{eff}^\mu = F^\mu - m \Gamma^\mu_{\nu \lambda} \dfrac{d x^\nu}{d \tau} \dfrac{d x^\lambda}{d \tau}$

Then it looks just like SR with an effective force $F_{eff}^\mu$

$m \dfrac{d^2 x^\mu}{d \tau^2} = F_{eff}^\mu$

The problem with this approach is that $F_{eff}^\mu$ doesn't obey the third law; there is no equal and opposite force. That's why I said that the problem with considering gravity a force is in correctly handling the evolution of the forces. If you assume that the particle has a negligible effect on the field, then I don't think there's any problem.

 

[stevendaryl]

Really? Think carefully about the Bell Spaceship Paradox before you answer.

You're certainly right about that, but that's a higher-order effect. It is not noticeable until the spaceship reaches relativistic speeds.

Anyway, the point holds, that if both ends are accelerated, then the amount of stretching is not an accurate measure of how much you are accelerating. To get zero stretching, I suppose you would have to have the electric field get weaker with height in just the right way.

 

[PeterDonis]

Well, the exact equations of motion for a test particle in curved spacetime in an arbitrary coordinate system are:

Remember I said we're using SR; we're working in a local inertial frame. In such a frame the Christoffel symbols are zero; there is no "gravitational field".

If you want to use non-inertial coordinates instead, with the Christoffel symbols present, you can, sure; for example, you could use Rindler coordinates for the accelerating rocket, which is basically equivalent to using Schwarzschild coordinates on the Earth's surface (provided you are working with a small enough range of altitudes that tidal effects can be ignored). I'm not sure this makes the analysis any easier, but you could do it.

The problem with this approach is that $F_{eff}^\mu$ doesn't obey the third law; there is no equal and opposite force. That's why I said that the problem with considering gravity a force is in correctly handling the evolution of the forces. If you assume that the particle has a negligible effect on the field, then I don't think there's any problem.

I don't follow this. The problem with $F_{eff}^\mu$ is not a problem about the particle affecting the field; it's about the "force" on the particle not obeying Newton's third law. That problem is independent of the particle's effect as a source of the field.

 

[stevendaryl]

Remember I said we're using SR; we're working in a local inertial frame. In such a frame the Christoffel symbols are zero; there is no "gravitational field".

I'm saying that treating the gravitational field as a force, and using SR is equivalent to the exact GR treatment, if you don't worry about the evolution/propagation of the force. In other words, I'm saying that one can use SR with an effective force and get good agreement with the exact GR equations.

 

[stevendaryl]

I don't follow this. The problem with $F_{eff}^\mu$ is not a problem about the particle affecting the field; it's about the "force" on the particle not obeying Newton's third law. That problem is independent of the particle's effect as a source of the field.

In SR, there is no action at a distance; objects are only affected by fields, not by other objects, and they only affect fields, not other objects. So the equivalent of the third law is that when a field affects an object, the object affects the field, as well, so as to conserve momentum, when you include the momentum of the field as well as particles.

The problem (or I should say, one of the problems) with regarding fictitious forces as real forces is that they act on particles, but are not acted upon by the particles.

 

[PeterDonis]

I'm saying that treating the gravitational field as a force, and using SR is equivalent to the exact GR treatment, if you don't worry about the evolution/propagation of the force. In other words, I'm saying that one can use SR with an effective force and get good agreement with the exact GR equations.

If by this you mean that you can use curvilinear coordinates in flat spacetime (e.g., Rindler coordinates), and then interpret the Christoffel symbols in those curvilinear coordinates as the "gravitational field", then yes, I agree.

But if you mean that you can *add* a "force of gravity" to SR as something extra, over and above just using curvilinear coordinates, then no, I don't agree. Doing that destroys the physical predictiveness of the theory.

For example, what does proper acceleration--path curvature of a worldline--mean in this "SR + gravity" theory? I know we've gone back and forth about what different types of accelerometers would read, but in standard SR and GR, a key element of the physical interpretation of the theory is that proper acceleration, path curvature of a worldline, is a direct physical observable; there is *some* device, call it an "ideal accelerometer", that measures it. A given worldline in the presence of gravity has *different* path curvature according to "SR + gravity" than it does according to GR, because "SR + gravity" still uses flat spacetime; so "SR + gravity" can't possibly get good agreement with the physical predictions of GR with regard to path curvature.

 

[georgir]

The formula is the correct one for SR. If you disagree then please post the equation you believe to be the correct one for SR and provide your references.

I can post a formula for an object's acceleration. It will probably be the same as your formula (where is this original thread that you posted it anyway, I have yet to see it)

That is not a formula for an accelerometer's reading.

Applied to each component of an accelerometer, it will give the same result. So the relative difference between each component will be zero. Now that is a formula for an accelerometer's reading. "0"

 

[Dale]

I can post a formula for an object's acceleration.

That is not a formula for an accelerometer's reading.

You appear to be unaware of the difference between proper acceleration and coordinate acceleration. See here:
http://en.wikipedia.org/wiki/Proper_acceleration

 

[georgir]

You appear to be unaware of the difference between proper acceleration and coordinate acceleration. See here:
http://en.wikipedia.org/wiki/Proper_acceleration

That definition is for the GR case, not for SR. For SR you still define proper acceleration as the acceleration of an object in a reference frame where it is (at the particular instant) at rest. But there it is not always measurable by an accelerometer.

Even in the GR case it would not be measurable if it were caused by a uniform force other than Gravity.

 

[stevendaryl]

If by this you mean that you can use curvilinear coordinates in flat spacetime (e.g., Rindler coordinates), and then interpret the Christoffel symbols in those curvilinear coordinates as the "gravitational field", then yes, I agree.

But if you mean that you can *add* a "force of gravity" to SR as something extra, over and above just using curvilinear coordinates, then no, I don't agree. Doing that destroys the physical predictiveness of the theory.

I don't understand why you would say that. In GR, the equations of motion for a test particle of mass $m$ in curved spacetime when acted upon by a non-gravitational interaction force $F^\mu_{ng}$ are:

$m (\dfrac{d^2 x^\mu}{d \tau^2}+ \Gamma^\mu_{\nu \lambda} \dfrac{d x^\nu}{d \tau} \dfrac{d x^\lambda}{d \tau} ) = F_{ng}^\mu$

Now, if we define $F^\mu_{grav}$ via:
$F^\mu_{grav} = -m \Gamma^\mu_{\nu \lambda} \dfrac{d x^\nu}{d \tau} \dfrac{d x^\lambda}{d \tau}$, then we have:

$m \dfrac{d^2 x^\mu}{d \tau^2} = F^\mu_{grav} + F^\mu_{ng}$

What you seem to be saying is that the exact same equation makes a correct prediction if we got to it by starting with GR, but it makes an incorrect prediction if we started with SR and added $F^\mu_{grav}$ by hand. I don't understand that.

 

[Dale]

That definition is for the GR case, not for SR.

The difference between SR and GR is simply that you neglect gravity in SR. In the absence of gravity, the definition of proper acceleration and its relationship to an accelerometer is the same in SR and GR.

For SR you still define proper acceleration as the acceleration of an object in a reference frame where it is (at the particular instant) at rest.

Reference please.

 

[stevendaryl]

For example, what does proper acceleration--path curvature of a worldline--mean in this "SR + gravity" theory? I know we've gone back and forth about what different types of accelerometers would read, but in standard SR and GR, a key element of the physical interpretation of the theory is that proper acceleration, path curvature of a worldline, is a direct physical observable; there is *some* device, call it an "ideal accelerometer", that measures it. A given worldline in the presence of gravity has *different* path curvature according to "SR + gravity" than it does according to GR, because "SR + gravity" still uses flat spacetime; so "SR + gravity" can't possibly get good agreement with the physical predictions of GR with regard to path curvature.

After thinking about it, I'm not convinced that you are right, but I am convinced that it's much more complicated than I was thinking it was. The problem is that, as you say, things like lengths and proper times and proper accelerations are assumed to be measurable in most SR type thought experiments, while they are no longer measurable if you assume a universal "gravitational force". The notion of proper time in the theory SR + gravity will not be the same as the GR notion of proper time, and similarly for length measurements and proper acceleration measurements. That makes the comparison of "SR + gravity" with experiment exceedingly difficult.

 

[Dale]

I am convinced that it's much more complicated than I was thinking it was.

I agree. If it were straightforward then it would have been done more than a century ago and GR would probably never have been developed in the first place.

Personally, I stand by the assertion that all you can do with gravity in SR is neglect it. I may be wrong in that, but it will take someone smarter than Einstein to show it.

 

[PeterDonis]

The problem is that, as you say, things like lengths and proper times and proper accelerations are assumed to be measurable in most SR type thought experiments, while they are no longer measurable if you assume a universal "gravitational force". The notion of proper time in the theory SR + gravity will not be the same as the GR notion of proper time, and similarly for length measurements and proper acceleration measurements.

Yes, exactly, this is what I was getting at.

 

[harrylin]

As remarked earlier (post #82), it depends a lot on one's exact definitions of "special relativity", "accelerometer", etc.
In a quick search I found no other literature on this topic, so I just summarize with the few "authorative" literature references that I already had. For completeness I include again also Langevin, as apparently is necessary:

What is the "Langevin scenario"? Are you referring to a twin scenario where the turn-around is achieved by looping around a distant star under the effect of the star's gravitation? If so, Langevin never mentioned any such thing. (Granted, another poster claimed the gravitational turn-around scenario was in Langevin's 1911 paper, but when I asked him to point to where in that paper the gravitational turn-around was discussed, he admitted there was no such thing in that paper, so it was just a mis-attribution.)

- Einstein 1905:
Theory from two postulates.
http://www.fourmilab.ch/etexts/einstein/specrel/www/

- Langevin 1911
"it is sufficient that our traveler consents to be locked in a projectile that would be launched from Earth with a velocity sufficiently close to that of light but lower, which is physically possible, while arranging an encounter with, for example, a star that happens after one year of the traveler's life, and which sends him back to Earth with the same velocity. "
http://en.wikisource.org/wiki/The_Evolution_of_Space_and_Time

He thus gives the prediction of SR for a turnaround using a gravitational sling.

- Einstein 1916:
"The Special Relativity Theory does not differ from the classical mechanics through the assumption of [the first] postulate, but only through the postulate of the constancy of light-velocity in vacuum "
http://en.wikisource.org/wiki/The_F..._1._Remarks_on_the_Special_Relativity_Theory.

- Einstein 1916:
"the special theory of relativity cannot claim an unlimited domain of validity; its results hold only so long as we are able to disregard the influences of gravitational fields on the phenomena".
http://en.wikisource.org/wiki/Relat...nces_from_the_General_Principle_of_Relativity

I note that Einstein similarly did not claim that SR has no results in the presence of gravitation; it can be used as long as the ignored effects are relatively small - as is always the case with theories. I think that most of us agree that this is the case in principle with Langevin's scenario which assumes a relatively short turn-around (see for example post #85).
And with "the phenomena" he obviously meant (or should have meant) the "relativistic" effects and not all phenomena of the experiment (just think of MMX which in full uses the orbit of the Earth with the Sun as gravitational sling!).

Taking, as he did himself, Einstein's postulates as basis and definition for SR, I see no reason for a change of the classical prediction that a perfect mechanical accelerometer will indicate null in free fall: neither postulate seems to impose a difference in gravitational acceleration between the different parts of a standard mechanical accelerometer.
To elaborate with the simple basic case of a silicon accelerometer in free fall (and in agreement with post #113): as gravitation cannot be shielded, the attractive force of a far away body on one Si atom of the accelerometer must be equal to the attractive force on another Si atom (any basic textbook suffices!).

In contrast, it follows directly from the second postulate that a non-mechanical (fully optical) accelerometer would indicate free-fall acceleration (contrary to GR and experience and explained in posts #50 and #72).

 

[Dale]

Einstein 1916:
"the special theory of relativity cannot claim an unlimited domain of validity; its results hold only so long as we are able to disregard the influences of gravitational fields on the phenomena".
http://en.wikisource.org/wiki/Relat...nces_from_the_General_Principle_of_Relativity

I note that Einstein similarly did not claim that SR has no results in the presence of gravitation; it can be used as long as the ignored effects are relatively small - as is always the case with theories. I think that most of us agree that this is the case in principle with Langevin's scenario which assumes a relatively short turn-around

I don't agree. A large error for a small amount of time is not the same as a small error.

Einstein's quote is correct, but the Langevin scenario is one where we are not "able to disregard the influences of gravitational fields on the phenomena", IMO. Not only are the accelerometer readings erroneous during the turnaround if you disregard gravitational fields, but also if you disregard the gravitational fields then the path itself is erroneous since it is gravitational fields which cause the turnaround. By disregarding the influences of gravitational fields you get two large errors, one of which is a large error for a small amount of time, and the other is a large error for a large amount of time.

 

[harrylin]

Indeed, also the full MMX makes in its conception use of the gravitational field for measurements at different velocities*.

Thanks everyone for the comments, I think that this topic has been sufficiently discussed now.

* it is just possible that the resultant velocity at the time of the observations was small though the chances are much against it. The experiment will therefore be repeated at intervals of three months, and thus all uncertainty will be avoided - http://en.wikisource.org/wiki/On_the_Relative_Motion_of_the_Earth_and_the_Luminiferous_Ether

 

[Dale]

I don't know what you think the relevance of that is. The analysis you quote was using the aether wind theory which was current at the time. It was not a SR analysis.

In any case, the errors in the MMX are small, not the Langevin scenario.

 

[PeterDonis]

In contrast, it follows directly from the second postulate that a non-mechanical (fully optical) accelerometer would indicate free-fall acceleration (contrary to GR and experience and explained in posts #50 and #72).

I thought I had already responded to the substance of this, but it doesn't look like I have. You appear to be claiming that light bending would make a freely falling optical accelerometer register a nonzero reading. But there is no light bending in a local inertial frame; there is only light bending in an accelerated frame. If I am in a freely falling elevator and watch light bounding back and forth between mirrors on the walls, I won't measure any bending; but someone standing on the ground watching the experiment will measure bending of the same light beams. So an optical accelerometer using light bending as a measure of acceleration should only measure actual proper acceleration; it should read zero in free fall.

 

[harrylin]

Still a few more clarifications then:

I don't know what you think the relevance of that is. The analysis you quote was using the aether wind theory which was current at the time. It was not a SR analysis. [..]

Most people and textbooks hold that SR can be used to analyse MMX, despite its reliance on the gravitational swing of the Sun to enable measurements at considerably different velocities. I agree with that.

[..] You appear to be claiming that light bending would make a freely falling optical accelerometer register a nonzero reading. [...]

Certainly not. Instead, and as Einstein elaborated in his 1916 papers, the second postulate of SR implies that light cannot bend in vacuum, as measured with a classical inertial frame. That was modified with GR.

 

[Dale]

Most people and textbooks hold that SR can be used to analyse MMX, despite its reliance on the gravitational swing of the Sun to enable measurements at considerably different velocities. I agree with that.

I agree that the MMX can be analyzed using SR also, but the quote you used earlier (from the MMX paper itself) is irrelevant since it refers to an aether analysis, not an SR analysis. The reason that the MMX can be analyzed using SR is that the interference fringes are not sensitive to the gravitational effects in the experiment. It has nothing to do with changing velocities over the course of the year, there simply is no measurable effect at any velocity.

The same is not true of the Langevin version of the twins paradox. In the twins paradox the key to resolving the scenario is to identify that there is an asymmetry between the twins. This asymmetry is either proper acceleration measured by an accelerometer or spacetime curvature. The error produced by neglecting gravity is therefore important to Langevin where it was not important to MMX.

 

[Samshorn]

"For this it is sufficient that our traveler consents to be locked in a projectile that would be launched from Earth with a velocity sufficiently close to that of light but lower, which is physically possible, while arranging an encounter with, for example, a star that happens after one year of the traveler's life, and which sends him back to Earth with the same velocity."

He thus gives the prediction of SR for a turnaround using a gravitational sling.

One might surmise that this is what Langevin had in mind, especially from this translation, but he doesn’t actually mention gravitation – and for good reason: No star (as Langevin knew them) would have been sufficient to accomplish such a turn-around. In order for a projectile moving at nearly the speed of light to be turned around gravitationally, it would need to pass within a distance where the escape velocity is comparable to the speed of light, which is not possible for any star that Langevin could have known about – unless you think he was assuming the existence of something like John Michell’s "black holes", from which even light could not escape, but this would have been (to his contemporaries) even more fantastical than the twins paradox that he was trying to illustrate, especially in the context of special relativity (with which Michell didn’t have to contend). A Newtonian slingshot would require the projectile to speed up to superluminal speed at the perigee, which Langevin knew was impossible, so he really wasn't in a position to reconcile a gravitational mechanism with special relativity. Also no Newtonian hyperbolic or parabolic trajectory could precisely turn the projectile around, although it could come close. (Ironically, in general relativity it actually is possible to send a particle back in exactly the same direction as its approach, but not in Newtonian gravity.)

I’d also be careful with the English translation, especially the words "and which sends him back", which don't seem to come from a simple literal translation of the French. In fact, looking back at the original version of the Wiki translation you cited, that specific phrase was translated differently. It originally said

“For this it is sufficient that our traveler consents to be locked in a projectile that would launched from Earth with a velocity sufficiently close to that of light, although lower so that it is physically possible, then arranging an encounter with, for example, a star that happens after one year of life, and then the traveler returns to Earth with the same velocity.”

This is even less suggestive of gravitational sling mechanism. However, the words were changed from this to the current version (the one phrased to suggest that the destination star somehow “sends” the projectile back) by a wikipedia editor named Harald88. I don’t know which version of the translation is more accurate.

But even accepting the revised translation of Harald88, which seems worded to promote the idea that Langevin was talking about a gravitational slingshot, the fact remains that Langevin didn't specify the mechanics of how the projectile is accelerated from Earth, nor how it is decelerated at the star. The same method (e.g., rockets) could have been used for both. If he really had in mind a gravitational slingshot (which is only circumstantially suggested by his mentioning of a star “for example” as the destination point), then at the very least he was being coy about the fantastical kind of “star” that would be required to accomplish a gravitational turn-around of an object traveling at near light speed.

It’s also strange that he says "meeting with a star for example...". Why does he say "for example"? If he was describing a gravitational slingshot, what else could it be other than a star (or rather, a black hole)? This suggests he was just using a star as an arbitrary destination and turn-around point. All he really said is, the traveler accelerates (by unspecified means) from Earth out to a distant star (for example) at high speed, then turns around (by unspecified means) and returns at the same speed. Readers who realized that a gravitational slingshot could not possibly accomplish the turn-around (for any star known at the time) would have been less likely to assume that he was claiming a gravitational slingshot as the mechanism.

 

[Histspec]

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.)

 

[Dale]

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.

 

[stevendaryl]

“For this it is sufficient that our traveler consents to be locked in a projectile that would launched from Earth with a velocity sufficiently close to that of light, although lower so that it is physically possible, then arranging an encounter with, for example, a star that happens after one year of life, and then the traveler returns to Earth with the same velocity.”

This is even less suggestive of gravitational sling mechanism.

I can't imagine why it would be important that there be a distant star in the first place, if the star is not supposed to play a role in turning the rocket around. Without the star, how is this any different from the usual twin paradox?

 

[Histspec]

I can't imagine why it would be important that there be a distant star in the first place, if the star is not supposed to play a role in turning the rocket around. Without the star, how is this any different from the usual twin paradox?

Well, it simply could play the role of the travel destination, at which the traveler turns around. This doesn't necessarily imply a gravitational influence as the cause of the turnaround. However, even if Langevin thought that a gravitational turnaround can be fully described with SR, it is irrelevant from a modern perspective. Because in 1911 the relation between gravitation and SR was widely unknown, only some speculative, tentative models by Einstein and Abraham were published.