Does Special Relativity Predict Zero Acceleration in Free Fall?

[georgir]

In fact I simply meant 3-acceleration, as we only need to examine the problem in a single frame; but it boils down to the same for the prediction.

Well, 3-acceleration due to gravity can not be the same for all objects. It has to depend on their velocity... otherwise we may get to accelerate something above c by accident ;)
Edit: you could say that the "proper acceleration" caused by gravity is the same, but that's just saying 4-acceleration is the same.

 

[Dale]

Neither of us knows such a reference

I am glad that you realize that you are just making stuff up. I would remind you about the rule against speculation and personal theories.

but we made contrary claims about what we think SR predicts for that case.

The difference between your claim and mine is that mine represents the mainstream understanding of SR and yours is unsupported speculation. They are not just two claims with equal validity.

 

[Dale]

Somehow intuitively, I'd expect the same answer for SR, by extension. The problem is the definition of a global uniform force in that context, though.

The problem is that there is no SR theory of gravity. All you can do in SR is neglect gravity.

Please, do not post speculations on what a SR theory of gravity would be like, unless you can produce a mainstream reference for such a theory. This forum is for discussion of mainstream scientific theories only, not development of new ones.

 

[Dale]

in SR gravity is a force which accelerates all matter equally.

No, gravity is not a force which can be treated in SR. Please provide your reference, if you cannot then stop speculating.

 

[georgir]

The equivalence principle + Huygen's principle

I'm sure in freefall light rays will seem to be straight, so then from a static external view point light rays should appear curved. How exactly we can get that in SR though... no clue.

I don't think Huygen's principle is any use though... "its like every point that a wave reaches becumes like a new wave source"... doesn't tell you anything really. Not about how exactly the wave "reaches" any new points from any "source".

In the presence of gravity, or maybe any uniform force field (if we ever discover other kinds at all) it could be like the wave is in a flowing medium instead of a fixed one.

 

[georgir]

No, gravity is not a force which can be treated in SR. Please provide your reference, if you cannot then stop speculating.

ok... change
in SR gravity is a force which accelerates all matter equally.
to
in Classical Physics gravity is a force which accelerates all matter equally.

And just extend that to SR in the most natural way you can.

 

[Dale]

ok... change
in SR gravity is a force which accelerates all matter equally.
to
in Classical Physics gravity is a force which accelerates all matter equally.

And just extend that to SR in the most natural way you can.

If it were that easy then it would have been done more than a century ago, before the development of GR.

Again, stop speculating about gravity in SR and provide any references if you are not speculating.

 

[Fredrik]

ok... change
in SR gravity is a force which accelerates all matter equally.
to
in Classical Physics gravity is a force which accelerates all matter equally.

And just extend that to SR in the most natural way you can.

I think the only way to discuss something that resembles gravity in the context of SR is to consider a uniformly accelerating reference frame. For example, there was recently a thread that discussed the question of what happens to a submarine moving at a relativistic speed under the influence of gravity. The only way to even begin the discussion was to interpret the question as "what happens to a submarine that's moving at a relativistic speed in a huge water tank that's being uniformly accelerated".

 

[georgir]

Ok, DaleSpam, now you are arguing that SR does not and can not say anything about gravity or freefall...
How does that fit with your words quoted in the original post?

SR predicts a very large accelerometer reading during the turnaround

 

[Dale]

Ok, DaleSpam, now you are arguing that SR does not and can not say anything about gravity or freefall...
How does that fit with your words quoted in the original post?

Good question. I have maintained from the beginning that SR cannot model gravity. All you can do is neglect it. So, when you neglect gravity (as you must in SR) and apply the proper SR formula to the Langevin scenario you get an erroneously high accelerometer reading during the turnaround. Indicating that the Langevin scenario is outside the domain of applicability of SR.

 

[georgir]

I think the only way to discuss something that resembles gravity in the context of SR is to consider a uniformly accelerating reference frame. For example, there was recently a thread that discussed the question of what happens to a submarine moving at a relativistic speed under the influence of gravity. The only way to even begin the discussion was to interpret the question as "what happens to a submarine that's moving at a relativistic speed in a huge water tank that's being uniformly accelerated".

When you say "uniformly accelerating reference frame" it is not quite obvious what you mean. Do you mean the actual objects are being accelerated, or just the imaginary measurement reference frame... The two are actually related, I guess. In an accelerated reference frame it would appear that there is a force accelerating the objects.

Yes, when I (and from what I gather, harrylin too) think of Gravity (or a sufficient approximation) in SR, this is pretty much the kind of force I think of.

 

[georgir]

Good question. I have maintained from the beginning that SR cannot model gravity. All you can do is neglect it. So, when you neglect gravity (as you must in SR) and apply the proper SR formula to the Langevin scenario you get an erroneously high accelerometer reading during the turnaround. Indicating that the Langevin scenario is outside the domain of applicability of SR.

Apologies, I do not know the context of that original discussion. Nor do I know what that Langevin scenario is all about. But what happens if instead of neglecting gravity, you replace it with a force field producing locally uniform 4-acceleration?

EDIT: Nevermind, I actually know what happens. What happens is you derive GR, as this is how Einstein came up with it. And get the same 0 predicted reading.

 

[harrylin]

I am glad that you realize that you are just making stuff up. I would remind you about the rule against speculation and personal theories. [..]

No, and I was talking about us, not about only me. Do you claim that you were just speculating and presenting your personal theories?

 

[Dale]

Do you claim that you were just speculating and presenting your personal theories?

I am not speculating. I have references to support my formula; you have none to support yours. My forumla is the correct and well-recognized one; yours is simply made up. I am glad that you recognize that yours is speculative, even if you don't realize yet that mine is not.

 

[Dale]

Apologies, I do not know the context of that original discussion. Nor do I know what that Langevin scenario is all about.

No worries. In short, the Langevin scenario is a standard twins-paradox scenario except that instead of firing a rocket to turn around the traveling twin swings around a star using the star's gravity for the turnaround.

But what happens if instead of neglecting gravity, you replace it with a force field producing locally uniform 4-acceleration?

If you do that you wind up with a theory of gravity without tidal effects.

 

[georgir]

DaleSpam, are you perhaps confusing
"SR predicts a very large accelerometer reading during the turnaround"
with
"SR predicts a very large acceleration of the accelerometer during the turnaround"
?

The two are not the same. The accelerometer may be accelerating and read 0, because all of its components are accelerating completely uniformly.
Anyway... enough with me posting without the full context. I'll be silent until I get more info.
[edit: posted before seeing your last reply, feel free to disregard]

 

[georgir]

If you do that you wind up with a theory of gravity without tidal effects.

Bingo. This is the key to it... tidal effects can always be neglected in a small enough locality, and if we do neglect them, we all agree... zero reading.

The only way an "accelerometer" would detect a freefall is if it was large enough to detect tidal effects. I'd not even call that freefall though.
[edit: the same applies to the classical Newtonian case too, btw]

edit 2: 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.

 

[Fredrik]

When you say "uniformly accelerating reference frame" it is not quite obvious what you mean. Do you mean the actual objects are being accelerated, or just the imaginary measurement reference frame... The two are actually related, I guess. In an accelerated reference frame it would appear that there is a force accelerating the objects.

Yes, when I (and from what I gather, harrylin too) think of Gravity (or a sufficient approximation) in SR, this is pretty much the kind of force I think of.

I mean that the coordinate system is accelerated, so that the coordinate acceleration (d²x/dt²) of an objectively non-accelerating object is non-zero. You could then define the "force" in this coordinate system by F=mx''(t). Note however that if you make the same definition in an inertial coordinate system, the "force" will be 0.

It seems pointless to even mention a "force" in this scenario. Why not just let S be a Rindler coordinate system and describe the motion of some other object in terms of the S coordinates?

To call this force "gravity" is pretty odd in my opinion. It's like using the term "gravity" for the force you feel when you slam the breaks of your car.

 

[georgir]

To call this force "gravity" is pretty odd in my opinion. It's like using the term "gravity" for the force you feel when you slam the breaks of your car.

We don't call that force Gravity. We do the opposite, we claim that Gravity can be viewed as such a force, at least in a small enough locality where it appears uniform
[edit: by "we", I refer to me, myself and Einstein, in his equivalence principle ;)]

 

[harrylin]

I am not speculating. I have references to support my formula; you have none to support yours. My forumla is the correct and well-recognized one; yours is simply made up. I am glad that you recognize that yours is speculative, even if you don't realize yet that mine is not.

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

 

[PeterDonis]

To the contrary, such repeated claims were the reason for this thread; see post #1 (which is non-exhaustive).

Those weren't claims that SR predicts a large accelerometer reading in free fall; they were claims that SR did not predict that the worldline of the traveling twin in the Langevin scenario was a free fall worldline. Which, as I noted in my last post, it doesn't.

I meant it in the sense of unobstructed; motion in under influence of gravity is not "inertial" in SR.

No, motion under the influence of gravity cannot be analyzed in SR. That is the point of my statement that SR does not predict that the traveling twin's worldline in the Langevin scenario is a free-fall worldline. That doesn't mean motion under gravity is not inertial in SR; it means SR gives the wrong answer for motion under gravity, which means that motion under gravity can't be analyzed using SR.

See also posts #26 and #28.

That's talking about an ambiguity in what Einstein said--when he said "Newton's Laws hold good in an inertial frame", did he mean to include the Newtonian law of gravity or not? But I'm not talking about what Einstein said. I'm talking about what actually happens when you try to model the Langevin twin scenario in SR. You get a wrong prediction about the traveling twin. There's no way to finesse that by saying "well, gravity isn't inertial in SR"; you still get a wrong prediction.

 

[harrylin]

Well, 3-acceleration due to gravity can not be the same for all objects. It has to depend on their velocity... otherwise we may get to accelerate something above c by accident ;)
Edit: you could say that the "proper acceleration" caused by gravity is the same, but that's just saying 4-acceleration is the same.

An accelerometer is an instrument in which all parts are attached; at constant read-out all parts follow the same trajectory.

[..] I don't think Huygen's principle is any use though... "its like every point that a wave reaches becumes like a new wave source"... doesn't tell you anything really. Not about how exactly the wave "reaches" any new points from any "source". [..]

The same principle that was still valid in GR tells you in SR that since the speed gradient of a light wave in free space is zero, light in free space cannot bend according to SR.
Here's a mini course: http://en.wikipedia.org/wiki/Huygens–Fresnel_principle

 

[harrylin]

Those ["SR predicts a very large accelerometer reading during the turnaround, and real free falling accelerometers read 0"] weren't claims that SR predicts a large accelerometer reading in free fall [..]

See also the full discussion in the other thread; no further comment needed.

 

[Mentz114]

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

Since you're now invoking a democratic principle, I'd like to register that Dalespam's posts are completely recognizable and familiar to me. They follow from the analysis of local inertial frames in GR, which you clearly know nothing about. If gravity is present, the LIF have a limited spatial extent, in which errors are acceptable. In strong curvature this extent gets smaller. As Dalespam has correctly said, you can only mix gravity with SR if the errors are small enough.

To you, I register no votes. Nil points. Your fundamentalism is dull and incomprehensible.
What is your point ?

 

[harrylin]

you're now invoking a democratic principle

No I'm not, it was Dalespam who invoked mainstream opinion...

They follow from the analysis of local inertial frames in GR,

SR is the topic here, not GR!

As Dalespam has correctly said, you can only mix gravity with SR if the errors are small enough.

Indeed, and I have correctly said the same; everyone agrees on that.

[..] Your fundamentalism is dull and incomprehensible.
What is your point ?

I really wonder what is your point... If you have useful input such as a derivation to offer, or a good reference then please contribute; else please stay away.

 

[Mentz114]

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

I refer to this. Your belief that opinions are divided is a delusion.
The rest of your reply is as irrelevant and pointless as the rest of your posts.

I'll post anywhere I want to. You don't own this thread.

Your questions raised in the first post have been answered. Or do you claim otherwise ?

 

[PeterDonis]

See also the full discussion in the other thread; no further comment needed.

No further comment needed because you agree that you have been misinterpreting what others have been saying? The discussion in the other thread is the same as the one here: people are telling you that SR predicts that the traveling twin's worldline in the Langevin scenario is *not* a free-fall worldline. Nobody was claiming that SR predicts a large accelerometer reading in free fall. That's your misinterpretation of what others were saying.

 

[harrylin]

I refer to this.

I referred there to:

[..] mine represents the mainstream understanding of SR [..].

Your belief that opinions are divided is a delusion.

For the record, one may look at posts #1, #4, #31, #46.

I'll post anywhere I want to. You don't own this thread.

That's true; as you're definitely trolling, you're the first person that I'll now put on my Ignore list here.

 

[stevendaryl]

No I'm not, it was Dalespam who invoked mainstream opinion...

SR is the topic here, not GR!

Indeed, and I have correctly said the same; everyone agrees on that.

I really wonder what is your point... If you have useful input such as a derivation to offer, or a good reference then please contribute; else please stay away.

I'm having trouble following exactly what the disagreement is about, here. Is the question: Can SR be used to compute elapsed times for trips that involve gravity? Without invoking a theory of gravity, we would have to make a guess about what the effect of gravity is on the results. I'm sure that we can make such a guess that would allow a sensible result to be computed. But what exactly is the point of this exercise? I'm confused about that.

 

[harrylin]

[..] The discussion in the other thread is the same as the one here [..] Nobody was claiming that SR predicts a large accelerometer reading in free fall. That's your misinterpretation of what others were saying.

It's good to see that you agree with me concerning the topic here; however I don't take it for granted that you can look into the head of other people.

 

[harrylin]

I'm having trouble following exactly what the disagreement is about, here. Is the question: Can SR be used to compute elapsed times for trips that involve gravity? Without invoking a theory of gravity, we would have to make a guess about what the effect of gravity is on the results. I'm sure that we can make such a guess that would allow a sensible result to be computed. But what exactly is the point of this exercise? I'm confused about that.

I agree with you, and such a computation would make an interesting topic, which is however not the topic of this thread. 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. It could be, as peterdonis thinks, that that claim was merely a poor phrasing; if so, hopefully Dalespam will clarify that.
In the course of the discussion the interesting question of a truly optical accelerometer came up; I think to have given a pertinent answer on that, so that I now distinguish between the prediction for a standard, mechanical accelerometer and a truly optical one.

 

[stevendaryl]

Such a computation would make an interesting topic, which is however not the topic of this thread. This thread was to discuss the several times repeated claim by Dalespam that according to SR an accelerometer in free fall will not read zero. It could be, as peterdonis thinks, that that claim was merely a wrong phrasing; if so, no doubt Dalespam will clarify that.

The question--"What does SR predict for an accelerometer reading in freefall?"--is ambiguous and ill-formed in a number of different ways.

First of all, SR is really a theory about physics when gravity is negligible, so it's not clear what it means to ask what SR predicts in a case where gravity cannot be ignored. There are various approaches to doing SR + gravity that would allow an approximate answer. The first approach would be to invoke the equivalence principle, and treat freefall as approximately equivalent to inertial motion. If you're doing that, then the answer is that there would be no nonzero accelerometer reading in freefall.

The second approach is to treat gravity as an external force and use SR's equations of motion for such a force:

m \dfrac{d^2 x^\mu}{d \tau^2} = F^\mu

That's ambiguous, because (without GR) gravity is only known as a force in the sense of Newtonian physics, which isn't sufficient to describe it as a 4-force of the kind that enters in SR equations of motion.

There is yet another ambiguity in the phrase, which is what "accelerometer" means. If we mean a device that would accurately measure accelerations of a rocket in empty space, then we would have to ask whether it would continue to measure accelerations accurately in the presence of gravity. Then there is another ambiguity, which is the meaning of "acceleration". In GR, acceleration usually means relative to local geodesics, and geodesics are influenced by gravity. So for GR, freefall is usually considered zero acceleration, since we identify geodesics with freefall. If you're talking pure SR, then presumably you don't mean acceleration relative to freefall (unless you're invoking the equivalence principle, in which case freefall = inertial). So what is the intended meaning for "acceleration" in SR when gravity is involved?

So is the point of the question to get a technical answer? In that case, the question has to be clarified considerably before an answer is possible. Or is the point of the question to get a feel for how people would interpret the question? In which case, ambiguity is one of the things you're interested in finding out: do people consider it ambiguous, or not?

 

[harrylin]

The question--"What does SR predict for an accelerometer reading in freefall?"--is ambiguous and ill-formed in a number of different ways.[..]

You have some good points - there is number of things that I thought to be non-ambiguous but that apparently need precision! As I next intend to give a longer commentary complete with literature references (likely some time during the weekend), I'll also include your points in there.
Thanks.

 

[PeterDonis]

I don't take it for granted that you can look into the head of other people.

Neither do I. I'm basing my statement on what they've posted here on PF. If I'm the one misinterpreting them, they're welcome to correct me.

 

[PeterDonis]

Can SR be used to compute elapsed times for trips that involve gravity?

That's one of two questions under discussion. The answer to it is "it depends". For the specific scenario that harrylin brought up, Langevin's version of the twin paradox where the traveling twin swings around a star in free fall to turn around, if the turnaround is short enough compared to the trip as a whole, the error in using SR to compute the traveling twin's elapsed time will be negligible.

But there is a second question, which is, can SR be used to predict the traveling twin's proper acceleration during the turnaround in the Langevin version? The answer to that is no; SR's prediction will be wrong. It will predict that the traveling twin's proper acceleration is nonzero during the turnaround.

 

[PeterDonis]

The second approach is to treat gravity as an external force and use SR's equations of motion for such a force:

m \dfrac{d^2 x^\mu}{d \tau^2} = F^\mu

That's ambiguous

No, it's worse than that. It's incorrect and inconsistent. There's no way to formulate a consistent theory of "SR + gravity" along these lines, and even if it were, the equation you give above obviously gives incorrect predictions (for example, it predicts that astronauts orbiting in the International Space Station will feel weight).

 

[stevendaryl]

No, it's worse than that. It's incorrect and inconsistent. There's no way to formulate a consistent theory of "SR + gravity" along these lines, and even if it were, the equation you give above obviously gives incorrect predictions (for example, it predicts that astronauts orbiting in the International Space Station will feel weight).

Well, that's an interesting result, itself. I'm a little surprised, though.

There is a sense in which it doesn't really matter, because today SR is considered a limiting case of GR, so there really is no good reason for worrying about what SR would predict in the absence of GR. But if we're trying to get into the frame of mind of a physicist living in the decade between the development of SR and the development of GR, then presumably we would have some strategy for dealing with gravity. It's interesting to speculate how someone might approach it.

When it comes to using SR (or Newtonian physics, for that matter) to describe forces, there are (or can be) two different aspects: (1) Describing how "test particles" are affected by the force, and (2) describing how the force itself evolves with time.

The way I understood the incompatibility of SR and gravity was that it was number (2) that caused problems. If you try to model the propagation of the gravitational field along the lines of the electromagnetic field, the resulting theory makes the prediction that "radiation" or a fluctuation in the field carries negative energy. You can't (or at least, I don't know how) have a sensible theory of dynamics if a system can "radiate" negative energy.

But you could make the approximation (which is what people generally do in applying SR or GR to problems such as orbital dynamics) that the gravitational field is approximately static, so we don't need to consider gravitational radiation. In other words, if we just worry about step (1)--the effect of gravity on the motion of "test particles".

I don't see how this approach would predict that people in orbit would feel a weight. The feeling of weight is really about things pressing against other things, such as the floor pressing against your feet. If both your foot and the floor were being acted upon by a force that is proportional to mass, then there would be no pressure of your foot against the floor, and so you wouldn't feel weight.

 

[georgir]

for example, it predicts that astronauts orbiting in the International Space Station will feel weight

How so? It predicts that they will have a proper acceleration, but not that they can feel it or detect it with any device at all.

 

[PeterDonis]

How so? It predicts that they will have a proper acceleration, but not that they can feel it or detect it with any device at all.

To say that SR predicts a proper acceleration but doesn't predict that it's detectable is nonsense; a prediction is a prediction of an observation. SR predicts that astronauts inside the ISS would feel weight, would be able to stand on the "floor" of the station, etc., just as they would inside a rocket with its engine firing. That's what proper acceleration *means*, physically. The mathematical expression is not proper acceleration; it's just how proper acceleration, the physical, detectable phenomenon, is represented in the math.

Not to mention that the proper acceleration would be easy to detect even without looking at the reading on a scale: SR predicts that the astronaut would be able to *stand* on a scale in the ISS, just as it predicts that you would be able to stand on a scale inside the moving chamber in a giant centrifuge floating freely in flat spacetime.

 

[PeterDonis]

There is a sense in which it doesn't really matter, because today SR is considered a limiting case of GR, so there really is no good reason for worrying about what SR would predict in the absence of GR.

True; but you can still, purely as a theoretical exercise, use SR to make (wrong) predictions in situations where gravity is not negligible. See further comments below.

But if we're trying to get into the frame of mind of a physicist living in the decade between the development of SR and the development of GR, then presumably we would have some strategy for dealing with gravity. It's interesting to speculate how someone might approach it.

Interesting as history, perhaps. But not as physics.

The way I understood the incompatibility of SR and gravity was that it was number (2) that caused problems.

No, both (1) and (2) cause problems. SR assumes that there are global inertial frames in which the worldlines of freely falling test objects are straight lines. Try to describe the worldline of a test object orbiting the Earth in an inertial frame; *any* inertial frame. It won't be a straight line. Thus SR predicts that such a worldline is not freely falling. See further comments below.

I don't see how this approach would predict that people in orbit would feel a weight. The feeling of weight is really about things pressing against other things, such as the floor pressing against your feet. If both your foot and the floor were being acted upon by a force that is proportional to mass, then there would be no pressure of your foot against the floor, and so you wouldn't feel weight.

This is how we understand gravity from either a Newtonian viewpoint, or a GR viewpoint, yes. (In the case of GR, this is how we understand its Newtonian approximation, where we try to "translate" GR's statements about curved spacetime into intuitively more palatable statements about gravity as a "force".)

But again, this won't work within the framework of SR, because, once again, the worldline of such an object in an inertial frame is not a straight line. And since we are talking about an astronaut standing on the "floor" of the station, the astronaut and the "floor" will be at slightly different distances from the center (the floor will be a bit further from the center), so the curvature of their paths will be slightly different. That means they will push on each other, i.e., the astronaut will feel weight. Once again, it's the same as if the astronaut were inside the moving chamber of a giant centrifuge floating freely in flat spacetime; do you dispute that SR predicts that such an astronaut will feel weight?

Note that what I am invoking here is *not* tidal gravity. We do not have to assume any difference in the acceleration produced by the "force" between the astronaut and the floor. All that is necessary is the slight difference in distance from the center.

 

[georgir]

To say that SR predicts a proper acceleration but doesn't predict that it's detectable is nonsense; a prediction is a prediction of an observation. SR predicts that astronauts inside the ISS would feel weight, would be able to stand on the "floor" of the station, etc., just as they would inside a rocket with its engine firing. That's what proper acceleration *means*, physically. The mathematical expression is not proper acceleration; it's just how proper acceleration, the physical, detectable phenomenon, is represented in the math.

Not to mention that the proper acceleration would be easy to detect even without looking at the reading on a scale: SR predicts that the astronaut would be able to *stand* on a scale in the ISS, just as it predicts that you would be able to stand on a scale inside the moving chamber in a giant centrifuge floating freely in flat spacetime.

The "floor" is also getting accelerated in exactly the same way as the astronauts are, so the astronauts would not be able to stand on it. The situation is exactly analogous to the Newtonian model.

 

[PeterDonis]

The "floor" is also getting accelerated in exactly the same way as the astronauths are, so the astronauths would not be able to stand on it.

See my response to stevendaryl.

 

[stevendaryl]

To say that SR predicts a proper acceleration but doesn't predict that it's detectable is nonsense; a prediction is a prediction of an observation.

That doesn't seem right to me. As I said in another post, what you "feel" is not acceleration directly, but relative acceleration of the parts of a body.

Take the simple model of an accelerometer, a cubic box with a ball suspended in the center by identical springs connected to each of the 6 sides. If you put the contraption on an accelerating rocket ship, then the position of the ball relative to the walls will shift, and that shift indicates that the box is accelerating.

But now imagine that the box is charged, and so is the ball, and the charge/mass ratio is the same for box and ball. Then use a uniform electric field to accelerate the box. Then to first order, there is no shift in the position of the ball relative to the walls. The accelerometer will not detect acceleration.[/QUOTE]

 

[georgir]

Note that what I am invoking here is *not* tidal gravity. We do not have to assume any difference in the acceleration produced by the "force" between the astronaut and the floor. All that is necessary is the slight difference in distance from the center.

Slight difference in distance means slight difference in acceleration. Exactly tidal forces.
Your analogy with centrifuge is flawed because there the "scale" is fixed to the centrifuge. If it were freefalling together with the astronaut, there would be no weight again.

 

[PeterDonis]

The feeling of weight is really about things pressing against other things, such as the floor pressing against your feet. If both your foot and the floor were being acted upon by a force that is proportional to mass, then there would be no pressure of your foot against the floor, and so you wouldn't feel weight.

I should also note that, on the "SR + a force of gravity" viewpoint you are considering, this argument proves too much: it proves that I should not be able to feel weight on the surface of the Earth. Consider: in an inertial frame centered on the Earth, I and the ground just below me are both motionless. According to SR, that means we should both be freely falling, and should not push on each other at all, so I should not be able to stand on the ground.

 

[stevendaryl]

But again, this won't work within the framework of SR, because, once again, the worldline of such an object in an inertial frame is not a straight line. And since we are talking about an astronaut standing on the "floor" of the station, the astronaut and the "floor" will be at slightly different distances from the center (the floor will be a bit further from the center), so the curvature of their paths will be slightly different. That means they will push on each other, i.e., the astronaut will feel weight. Once again, it's the same as if the astronaut were inside the moving chamber of a giant centrifuge floating freely in flat spacetime; do you dispute that SR predicts that such an astronaut will feel weight?

I think it would be hard to model the feeling of weight realistically, but let's take a drastically simplified model of "feeling acceleration", which is: you have two masses connected by a spring. The spring has an equilibrium length. In this simplified model, "feeling weight" means that the spring is compressed or stretched.

If you put one of the masses on the floor of an accelerating rocket, with the spring holding the other mass up, then the spring will compress. Alternatively, if you attach one mass to the front of the rocket, and let the other mass dangle from the spring, the spring will stretch. So compression or stretching indicates acceleration.

But now, suppose instead of just accelerating one of the masses, you're accelerating both of them: 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. So with this simple model of "feeling acceleration", you don't feel any acceleration when both masses are accelerated together.

So I don't know why you are saying that an astronaut on board a satellite will feel acceleration (or that SR predicts that he will).

 

[stevendaryl]

I should also note that, on the "SR + a force of gravity" viewpoint you are considering, this argument proves too much: it proves that I should not be able to feel weight on the surface of the Earth. Consider: in an inertial frame centered on the Earth, I and the ground just below me are both motionless. According to SR, that means we should both be freely falling, and should not push on each other at all, so I should not be able to stand on the ground.

I don't know why you say that. It's really just like Newtonian physics: there is a force diagram. The floor has an upward force due to the rock underneath it. It has a downward force due to gravity, and a second downward force due to my feet pressing down on it. I have an upward force due to the floor pressing on my feet, and a downward force due to gravity. The total force on me is zero. The total force on the floor is zero.

But even though the total force on me is zero, the downward gravitational force and the upward force due to the floor are not applied at the same point. The upward force applies only to my feet, while the gravitational force applies to all parts of my body. So my feet should accelerate upward relative to the rest of me, which tends to compress me. Of course, my bones act like springs, and provide a resistance force to being compressed, but I will compress a little bit. That little bit of compression is what it feels like to have weight.

 

[PeterDonis]

It's really just like Newtonian physics

But this assumes that "Newtonian physics" of this sort can be consistently modeled in SR.

The floor has an upward force due to the rock underneath it.

Why? I understand why it does in Newtonian physics. But why does it in SR? How do we get this prediction from SR? Both the floor and me are moving in a straight line in an inertial frame. Why should we be pushing on each other at all?

But even though the total force on me is zero, the downward gravitational force and the upward force due to the floor are not applied at the same point. The upward force applies only to my feet, while the gravitational force applies to all parts of my body. So my feet should accelerate upward relative to the rest of me, which tends to compress me. Of course, my bones act like springs, and provide a resistance force to being compressed, but I will compress a little bit. That little bit of compression is what it feels like to have weight.

Is this supposed to be an argument within Newtonian physics? SR? GR?

With respect to Newtonian physics, I don't dispute it, although I haven't tried to think it through in detail.

With respect to GR, I don't think the argument is right (except maybe as a sort of "translation" into the Newtonian approximation), because in a local inertial frame, there is no "force of gravity" at all. There is only the force of the floor pushing on your feet, and causing your worldline to be non-geodesic.

With respect to SR, I can see how this argument could be applied to the astronauts in the ISS--who do not actually feel weight--but I don't see how it can be applied to me standing on the surface of the Earth, because, as I said before, I and the ground below me are both moving on straight lines in an inertial frame, so there's no reason for either of us to push on each other at all.

In other words, SR gets both predictions wrong: it predicts that astronauts in the ISS will feel weight, and your argument helps to explain why--but the prediction is wrong because they actually don't. But SR also predicts that the ground should not push on me, here on the surface of the Earth, because our worldlines are both straight lines in an inertial frame and are therefore freely falling--but the prediction is again wrong because they actually aren't.

 

[Mentz114]

I and the ground below me are both moving on straight lines in an inertial frame

Are you sure ? A geodesic in the Schwarzschild spacetime is falling freely radially. The ground is stopping you from following a geodesic and the accelerometer shows g=9.8 m/s².

I think you probably mean this in some Newtonian way - in which case please disregard it.

 

[PeterDonis]

Are you sure ? A geodesic in the Schwarzschild spacetime is falling freely radially.

Here I'm talking about what SR predicts, not what's actually true. You might need to read back through the discussion for context.