Free fall acceleration in SR


by harrylin
Tags: acceleration, free fall, special relativity
harrylin
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#73
Feb28-13, 08:57 AM
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Quote Quote by PeterDonis View Post
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
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#74
Feb28-13, 09:04 AM
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Quote Quote by harrylin View Post
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
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#75
Feb28-13, 09:07 AM
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Quote Quote by Mentz114 View Post
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
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#76
Feb28-13, 09:18 AM
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Quote Quote by harrylin
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
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#77
Feb28-13, 09:22 AM
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Quote Quote by harrylin View Post
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
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#78
Feb28-13, 09:31 AM
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Quote Quote by Mentz114 View Post
I refer to this.
I referred there to:
Quote Quote by DaleSpam View Post
[..] 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
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#79
Feb28-13, 09:35 AM
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Quote Quote by harrylin View Post
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
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#80
Feb28-13, 09:36 AM
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Quote Quote by PeterDonis View Post
[..] 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
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#81
Feb28-13, 09:42 AM
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Quote Quote by stevendaryl View Post
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
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#82
Feb28-13, 10:01 AM
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Quote Quote by harrylin View Post
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:

[itex]m \dfrac{d^2 x^\mu}{d \tau^2} = F^\mu[/itex]

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
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#83
Feb28-13, 10:32 AM
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Quote Quote by stevendaryl View Post
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
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#84
Feb28-13, 11:13 AM
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Quote Quote by harrylin View Post
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
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#85
Feb28-13, 11:18 AM
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Quote Quote by stevendaryl View Post
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
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#86
Feb28-13, 11:24 AM
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Quote Quote by stevendaryl View Post
The second approach is to treat gravity as an external force and use SR's equations of motion for such a force:

[itex]m \dfrac{d^2 x^\mu}{d \tau^2} = F^\mu[/itex]

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
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Feb28-13, 11:54 AM
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Quote Quote by PeterDonis View Post
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
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Feb28-13, 12:03 PM
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Quote Quote by PeterDonis View Post
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
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Feb28-13, 12:20 PM
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Quote Quote by georgir View Post
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
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Feb28-13, 12:29 PM
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Quote Quote by stevendaryl View Post
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.

Quote Quote by stevendaryl View Post
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.

Quote Quote by stevendaryl View Post
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.

Quote Quote by stevendaryl View Post
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.


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