JVNY Posted December 10, 2013 Share Posted December 10, 2013 A typical objection to the equivalence principle is that you can tell that you are falling in gravity because you can see that an item falling alongside you compresses horizontally and stretches vertically. However, this results from tidal effects of non-uniform gravity, whereas Einstein stated the equivalence principle very clearly to apply in the case of homogeneous (that is, uniform) gravitational fields. See post here, http://www.scienceforums.net/topic/79255-distinguishing-between-acceleration-and-gravity/?p=772228, and Einstein quotation from 1907: We consider two systems S1 and S2… Let S1 be accelerated in the direction of its X axis, and let g be the (temporally constant) magnitude of that acceleration. S2 shall be at rest, but it shall be located in a homogeneous gravitational field that imparts to all objects an acceleration –g in the direction of the X axis. As far as we know, the physical laws with respect to S1 do not differ from those with respect to S2; this is based on the fact that all bodies are equally accelerated in a gravitational field. At our present state of experience we have thus no reason to assume that the systems S1 and S2 differ from each other in any respect, and in the discussion that follows we shall therefore assume the complete physical equivalence of a gravitational field and a corresponding acceleration of the reference system. Quoted at: http://www.mathpages.com/home/kmath622/kmath622.htm So it seems that the typical objection is wrong, and it may not be possible to identify anything in S2 that is distinguishable. Is there anything that you can identify in S2, given a homogeneous gravitational field, that is distinguishable? Alternatively, if the equivalence principle does not hold true over extended spacetime (but only over small local spacetime), then is the reason because of something identifiable in S1 (the accelerating frame), rather than S2 (the frame at rest in a homogeneous gravitational field)? Consider someone in a really tall Einstein elevator in space (far from any gravitational mass). The elevator is accelerated upward relative to him at a constant magnitude for a very long time before the floor comes up and strikes him. He observes the walls and everything on them racing upward past him and thinks that he is falling down in the elevator. Here are two potentially identifiable things in the upwardly accelerating elevator that could distinguish it from an elevator at rest in a gravitational field. Are they correct? Do they distinguish acceleration from gravity over extended spacetime, such that the equivalence principle can hold only in small extensions of spacetime? 1. It has been suggested that the degree to which light bends in the accelerating frame is less than that which light bends in equivalent gravity. See post here: http://www.scienceforums.net/topic/67141-equivalency-principle/?p=684962. The "falling" observer could theoretically determine this by measuring the light bending and the rate at which his speed relative to the walls grows. Then S1 and S2 would be distinguishable. Does this occur, so would it distinguish the two reference frames? 2. Accelerating the elevator at a constant magnitude raises Bell's spaceship paradox issues. If you accelerate all parts of the elevator at a constant rate (same magnitude over the entire height of the elevator, which is equivalent to length in this context) then the elevator will keep its same height in the "falling" observer's frame rather than length contracting; the observer will notice this (contrary to the elevator at rest in a homogeneous gravitational field, which an observer falling within it should observe as length contracting upward toward him). Also, the accelerating elevator would increase in proper height, and it would tear apart (which the "falling" observer would note). If instead you accelerate the elevator's parts in a Born rigid manner, the bottom will accelerate at a greater rate in the observer's frame than the top, so the elevator should length contract at a greater rate than the elevator at rest described above (which the "falling" observer could presumably measure). And, if the elevator is really tall (such that its proper length is greater than c^2 divided by the proper acceleration of the top end), then even Born rigid acceleration will not prevent the elevator's proper height from expanding, and the elevator will tear apart toward its bottom. The "falling" observer will be able to observe this. Does any of this occur, such that it could distinguish the two reference frames? Link to comment Share on other sites More sharing options...

xyzt Posted December 10, 2013 Share Posted December 10, 2013 (edited) 1. It has been suggested that the degree to which light bends in the accelerating frame is less than that which light bends in equivalent gravity. See post here: http://www.scienceforums.net/topic/67141-equivalency-principle/?p=684962. The "falling" observer could theoretically determine this by measuring the light bending and the rate at which his speed relative to the walls grows. Then S1 and S2 would be distinguishable. Does this occur, so would it distinguish the two reference frame Not clear where this is coming from. The actual statement, verifiable through math, is that the light bending predicted by GR is exactly double the light bending predicted by Newtonian physics ([math]\Theta_{GR}=\frac{4GM}{Rc^2} [/math] vs [math]\Theta_{Newton}=\frac{2GM}{Rc^2}[/math]). So, the IM Egdall guy must have misinterpreted what he read. It is also true that it may be virtually impossible to derive "photon orbits" without the benefit of the GR mathematical machinery, using accelerated frames only. I have shown , in the other thread, how to do something as simple as to derive the timing of the radial motion using accelerated frames. The results are not identical with the ones derived using GR.2. Accelerating the elevator at a constant magnitude raises Bell's spaceship paradox issues. If you accelerate all parts of the elevator at a constant rate (same magnitude over the entire height of the elevator, which is equivalent to length in this context) then the elevator will keep its same height in the "falling" observer's frame rather than length contracting; the observer will notice this (contrary to the elevator at rest in a homogeneous gravitational field, which an observer falling within it should observe as length contracting upward toward him). Also, the accelerating elevator would increase in proper height, and it would tear apart (which the "falling" observer would note). I do not understand your fixation with length contraction. In GR, calculating the length of an object in motion wrt to an observer is a complicated problem that cannot be reduced to calculating length contraction. Length contraction is an artifact of the Lorentz transforms, while it is true that the spacetime is locally Lorentzian , you can't really use length contraction. Edited December 10, 2013 by xyzt Link to comment Share on other sites More sharing options...

ajb Posted December 10, 2013 Share Posted December 10, 2013 Alternatively, if the equivalence principle does not hold true over extended spacetime (but only over small local spacetime)... This is how I understand it. The equivalence principal holds on "small enough" regions of space-time where we see so tidal effects. It is a local statement not a global one. It is related to the fact that we can employ Riemann normal coordinates. Link to comment Share on other sites More sharing options...

Toffo Posted December 10, 2013 Share Posted December 10, 2013 1. It has been suggested that the degree to which light bends in the accelerating frame is less than that which light bends in equivalent gravity. See post here: http://www.scienceforums.net/topic/67141-equivalency-principle/?p=684962. The "falling" observer could theoretically determine this by measuring the light bending and the rate at which his speed relative to the walls grows. Then S1 and S2 would be distinguishable. Does this occur, so would it distinguish the two reference frames? Let me think ... The length contraction motion that is slowing down! That must be the solution. A bullet shot across an accelerating rocket: One might think that length contraction would cause the bullet to hit "wrong" place. But the bullet has some initial length contraction motion velocity ... and the target will lose some length contraction motion velocity during the flight of the bullet. Bullet lands where it's supposed to. Same thing with laser beam. Link to comment Share on other sites More sharing options...

Toffo Posted December 10, 2013 Share Posted December 10, 2013 If instead you accelerate the elevator's parts in a Born rigid manner, the bottom will accelerate at a greater rate in the observer's frame than the top, so the elevator should length contract at a greater rate than the elevator at rest described above (which the "falling" observer could presumably measure). And, if the elevator is really tall (such that its proper length is greater than c^2 divided by the proper acceleration of the top end), then even Born rigid acceleration will not prevent the elevator's proper height from expanding, and the elevator will tear apart toward its bottom. The "falling" observer will be able to observe this. Does any of this occur, such that it could distinguish the two reference frames? Equivalently: It's impossible for a very long rocket to accelerate rapidly. And it's impossible for a very tall rocket to stand on the ground in a strong gravity field, because the ground can not exist, because matter is pulled into a singularity in the place where the ground is supposed to be. Link to comment Share on other sites More sharing options...

JVNY Posted December 10, 2013 Author Share Posted December 10, 2013 Thanks for all of the responses. To follow up: ajb, do you have thoughts on Einstein's counterfactual thought experiment? Assume that gravity is homogeneous (1g whether you are here, 100 km higher, or 100km to the right, etc.). Then would that make it globally impossible to tell that you are in gravity? xyzt, I don't think that IM Egdall is disputing that GR gets the bending of light right. He suggests that the amount by which light bends in the elevator is wrong -- light does not bend enough in an accelerating elevator, so an occupant of the accelerating elevator can tell that he is not in a gravitational field. The problem with the form of the equivalence principle as quoted above is not that an elevator occupant can observe anything unusual when the elevator is at rest in a gravitational field and he is falling downward; it is that he can observe something unusual when he is in free float and the elevator is being accelerated upward. Length contraction is important in this case because the observer can measure the distance between (a) where the light started out horizontally at one wall, and (b) where it strikes the opposite wall after bending. If the walls are not only moving upward (which gives the observer the sense that he is falling), but also length contracting upward, the length contraction will have an independent effect on where the light hits the opposite wall. This might cause the light to strike the wall at a different place than it does in the case of an elevator at rest in a homogeneous gravitational field. Again, this is no argument about GR. Everyone agrees that GR shows the correct amount of light bending in the case of gravity. The question is, does the accelerating elevator frame get the wrong amount of bending? If it gets the wrong amount of bending, then the broad equivalence principle quoted above is not globally correct. But the reason that it is not globally correct is not because of something that the observer measures due to gravity. It is because of something that the observer measures due to acceleration. The other examples I give of length contraction or proper length expansion are just other ways that an observer in free float can tell that an elevator in which he lives is being accelerated. It should be possible to determine whether the light flash hits the opposite wall in the same place in an elevator at rest in a gravitational field as in an elevator undergoing constant acceleration, at least in the case of actual (non-uniform) gravity (because the GR formulas describe actual gravity). I don't know whether it is possible to compare the results of homogeneous gravity versus constant elevator acceleration. Toffo, let's push the question a bit further. Are you using homogeneous gravity or actual gravity? Also, say the observer carries his own meter sticks. Does the light bend the same amount in his meter sticks in both the gravitational field and the accelerating elevator? If the light hits the same place on the wall in both cases, it might be because of the offsetting effects you describe, which means that the observer could use his own meter sticks to determine that light bent a different amount in the two cases, so the two frames are not equivalent. It would be great to be able to make the equivalence principle true globally by restricting it to the case of homogeneous gravity. But it seems that the accelerating reference frame has a lot of unique features that an observer can actually identify, which would mean that you cannot make the principle globally true that way. It also raises the issue about how to define locality in the more narrow definition of the equivalence principle that ajb utilizes. It is not enough to restrict your measurements to a small enough region of spacetime to avoid being able to measure tidal effects. You also have to restrict yourself to a small enough region of spacetime to avoid being able to measure the effects of acceleration on the elevator. You have to restrict to the lesser of the two. Although I do not know which requires a smaller area. Link to comment Share on other sites More sharing options...

xyzt Posted December 10, 2013 Share Posted December 10, 2013 (edited) xyzt, I don't think that IM Egdall is disputing that GR gets the bending of light right. He suggests that the amount by which light bends in the elevator is wrong -- light does not bend enough in an accelerating elevator, so an occupant of the accelerating elevator can tell that he is not in a gravitational field. The problem with the form of the equivalence principle as quoted above is not that an elevator occupant can observe anything unusual when the elevator is at rest in a gravitational field and he is falling downward; it is that he can observe something unusual when he is in free float and the elevator is being accelerated upward. I disagree, he clearly mixes up things when he says: But the amount of bending is only half what one would measure for a light beam in an elevator at rest in a gravitational field of the same magnitude. Length contraction is important in this case because the observer can measure the distance between (a) where the light started out horizontally at one wall, and (b) where it strikes the opposite wall after bending. If the walls are not only moving upward (which gives the observer the sense that he is falling), but also length contracting upward, the length contraction will have an independent effect on where the light hits the opposite wall. This might cause the light to strike the wall at a different place than it does in the case of an elevator at rest in a homogeneous gravitational field. This is a very bad way of attacking the issue , as I said, length contraction is an effect that can not be easily translated from SR into GR (and , more importantly, it shouldn't). The GR mathematical machinery is more than capable of dealing with all the effects by itself, no need to resorting to mixing in length contraction. This is not to say that the effects are not physical, it is just to say that you need to stay withing the confines of the GR math, no point in mixing in SR. Rindler does it in a few places in his book and it looks very ugly (most likely it is incorrect as well). Again, this is no argument about GR. Everyone agrees that GR shows the correct amount of light bending in the case of gravity. The question is, does the accelerating elevator frame get the wrong amount of bending? If it gets the wrong amount of bending, then the broad equivalence principle quoted above is not globally correct. But the reason that it is not globally correct is not because of something that the observer measures due to gravity. It is because of something that the observer measures due to acceleration. The other examples I give of length contraction or proper length expansion are just other ways that an observer in free float can tell that an elevator in which he lives is being accelerated. I think so. I did not do the calculations for the amount of bending of the transverse ray of light but I once did the calculations for the transit time for a ray of light moving radially (up and down the elevator) and , though the symbolic formulas were a little different, when I did a Taylor expansion, they agreed to the third term. Considering that the third term is [math](\frac{gh}{c^2})^2[/math], this is an amazing precision. Edited December 10, 2013 by xyzt 1 Link to comment Share on other sites More sharing options...

ajb Posted December 10, 2013 Share Posted December 10, 2013 ajb, do you have thoughts on Einstein's counterfactual thought experiment? Assume that gravity is homogeneous (1g whether you are here, 100 km higher, or 100km to the right, etc.). Then would that make it globally impossible to tell that you are in gravity? The equivalence principal is a local thing. Einstein's elevator is small enough and does not fall far enough for you to notice any difference in the gravitational field. This is an idealised situation, but it is a good approximation for weak gravitational fields and small enough objects. Link to comment Share on other sites More sharing options...

Toffo Posted December 11, 2013 Share Posted December 11, 2013 (edited) Toffo, let's push the question a bit further. Are you using homogeneous gravity or actual gravity? Also, say the observer carries his own meter sticks. Does the light bend the same amount in his meter sticks in both the gravitational field and the accelerating elevator? If the light hits the same place on the wall in both cases, it might be because of the offsetting effects you describe, which means that the observer could use his own meter sticks to determine that light bent a different amount in the two cases, so the two frames are not equivalent. Of course there is a homogeneous pseudo gravity field inside an accelerating rocket. And of course the measuring devices are owned by the person that is measuring with the devices Does the light bend the same amount according to an observer's meter sticks in both the gravitational field and the accelerating elevator? That's the question. Maybe we should calculate the effect of my offsetting effect. My offsetting effect can be expressed like this: A slow bullet spends its flying time in a gravity field that is smaller on the average than the gravity field that is effecting a faster bullet, because the gravity field decreases with time. It was an outside observer that observed the bullets and the rocket, and the observer imagined a gravity field effecting the bullets, that observer is the observer that observes a weakening imaginary gravity field. (I should mention that the rocket accelerates with constant proper acceleration.) It takes time t for the rocket's speedometer needle to go from speed x to speed y, according to a clock inside the rocket. Outside observer says the time is t*gamma , according to his clock. (gamma is the relativistic factor of change) So we see: acceleration = initial acceleration/gamma ... And here end my mathematiclal skills. Edited December 11, 2013 by Toffo Link to comment Share on other sites More sharing options...

JVNY Posted December 11, 2013 Author Share Posted December 11, 2013 Thanks. Hopefully someone with full GR math skills can help us Is it possible to run the actual numbers and compare them? Here is a stab at the numbers for acceleration. Assume an elevator that is 1LY wide, at rest in an inertial frame. Light is flashed into the elevator horizontally at time t=0. Simultaneously, the elevator begins to accelerate upward at 1g, which I believe is 1.0326 LY/Y/Y. In the inertial frame, the flash takes 1Y to strike the opposite wall. In 1Y, an object accelerating at constant proper velocity 1.0326 LY/Y/Y travels 0.4236 LY in the inertial frame. So an observer in free float in the elevator would observe that the light was bent, with a total 0.4236 movement downward for 1.0 movement across. Is this a fair way to interpret the accelerating elevator in the equivalence principle? Can anyone confirm or correct those numbers? If they are correct, how do they compare to what GR shows to be the amount of bending in a homogeneous 1g gravitational field? It turns out that there was already a thread covering the question of the amount of bending, although it does not seem to have reached consensus. It is here: http://www.scienceforums.net/topic/64305-why-does-light-bend-during-acceleration/ Link to comment Share on other sites More sharing options...

xyzt Posted December 11, 2013 Share Posted December 11, 2013 (edited) Thanks. Hopefully someone with full GR math skills can help us Is it possible to run the actual numbers and compare them? Here is a stab at the numbers for acceleration. Assume an elevator that is 1LY wide, at rest in an inertial frame. Light is flashed into the elevator horizontally at time t=0. Simultaneously, the elevator begins to accelerate upward at 1g, which I believe is 1.0326 LY/Y/Y. In the inertial frame, the flash takes 1Y to strike the opposite wall. In 1Y, an object accelerating at constant proper velocity 1.0326 LY/Y/Y travels 0.4236 LY in the inertial frame. So an observer in free float in the elevator would observe that the light was bent, with a total 0.4236 movement downward for 1.0 movement across. Is this a fair way to interpret the accelerating elevator in the equivalence principle? Can anyone confirm or correct those numbers? If they are correct, how do they compare to what GR shows to be the amount of bending in a homogeneous 1g gravitational field? It turns out that there was already a thread covering the question of the amount of bending, although it does not seem to have reached consensus. It is here: http://www.scienceforums.net/topic/64305-why-does-light-bend-during-acceleration/ Firstly, you will never get any relevant answer if you persist in reducing everything to numerical calculations, you NEED to LEARN how to use fully symbolic calculations. Secondly, the transverse case is quite difficult, I can post the radial case and you are going to glean how difficult it is to deal with "photon orbits". Thirdly, you can wait for your Rindler book to arrive and read the appropriate chapter on "photon orbits". The chapter gives the "easier" treatment, based on the metric. You can use that to try to produce the more difficult solution, based on the equations of hyperbolic motion. Edited December 11, 2013 by xyzt Link to comment Share on other sites More sharing options...

Toffo Posted December 12, 2013 Share Posted December 12, 2013 Assume an elevator that is 1LY wide, at rest in an inertial frame. Light is flashed into the elevator horizontally at time t=0. Simultaneously, the elevator begins to accelerate upward at 1g, which I believe is 1.0326 LY/Y/Y. In the inertial frame, the flash takes 1Y to strike the opposite wall. In 1Y, an object accelerating at constant proper velocity 1.0326 LY/Y/Y travels 0.4236 LY in the inertial frame. So an observer in free float in the elevator would observe that the light was bent, with a total 0.4236 movement downward for 1.0 movement across. gamma * 0.4236 is the distance in the elevator frame, if 0.4236 is the distance in the inertial frame. Seems very simple ... but there might be something wrong with even that. Link to comment Share on other sites More sharing options...

md65536 Posted December 16, 2013 Share Posted December 16, 2013 2. Accelerating the elevator at a constant magnitude raises Bell's spaceship paradox issues. If you accelerate all parts of the elevator at a constant rate (same magnitude over the entire height of the elevator, which is equivalent to length in this context) then the elevator will keep its same height in the "falling" observer's frame rather than length contracting; the observer will notice this (contrary to the elevator at rest in a homogeneous gravitational field, which an observer falling within it should observe as length contracting upward toward him). Also, the accelerating elevator would increase in proper height, and it would tear apart (which the "falling" observer would note). Suppose you have an elevator appropriately fixed in a uniform gravitational field. The observer at the "top" is still at a higher gravitational potential. Even though the force of gravity doesn't change between the two observers, there is still work involved in moving a mass from one location to the other. So there is still Doppler shift observed by the two, and their clocks do not remain in sync. So they don't have to observe the other having the same coordinate acceleration as their self. I don't think there is any room here to distinguish between a uniform proper acceleration and uniform gravitational field. Link to comment Share on other sites More sharing options...

xyzt Posted December 16, 2013 Share Posted December 16, 2013 (edited) Suppose you have an elevator appropriately fixed in a uniform gravitational field. The observer at the "top" is still at a higher gravitational potential. Even though the force of gravity doesn't change between the two observers, there is still work involved in moving a mass from one location to the other. You are mixing GR concepts with Newtonian ones, there is no "force of gravity" in GR. Moreover, in the framework of Newtonian physics, your statement is false, the "force of gravity" is different, [math]\frac{GMm}{r_1^2}[/math] for observer 1 and [math]\frac{GMm}{r_2^2}[/math] for observer 2. Since [math]r_1 \ne r_2[/math] the force is different. So there is still Doppler shift observed by the two, and their clocks do not remain in sync. So they don't have to observe the other having the same coordinate acceleration as their self. This is a non-sequitur, the coordinate acceleration has nothing to do with the Doppler shift. Besides, it isn't a Doppler shift to begin with, it is a gravitational shift since the observers are stationary. You keep piling up mistakes. . Edited December 16, 2013 by xyzt Link to comment Share on other sites More sharing options...

md65536 Posted December 16, 2013 Share Posted December 16, 2013 You are mixing GR concepts with Newtonian ones, there is no "force of gravity" in GR. Moreover, in the framework of Newtonian physics, your statement is false, the "force of gravity" is different, [math]\frac{GMm}{r_1^2}[/math] for observer 1 and [math]\frac{GMm}{r_2^2}[/math] for observer 2. Since [math]r_1 \ne r_2[/math] the force is different. This is a non-sequitur, the coordinate acceleration has nothing to do with the Doppler shift. Besides, it isn't a Doppler shift to begin with, it is a gravitational shift since the observers are stationary. You keep piling up mistakes. Right. Gravitational redshift/blueshift is the correct term. I apologize for the huge pile of mistakes. What are [math]r_1[/math] and [math]r_2[/math]? Link to comment Share on other sites More sharing options...

xyzt Posted December 16, 2013 Share Posted December 16, 2013 Right. Gravitational redshift/blueshift is the correct term. I apologize for the huge pile of mistakes. What are [math]r_1[/math] and [math]r_2[/math]? The radial distance , it is the universal law of attraction, no? Link to comment Share on other sites More sharing options...

md65536 Posted December 16, 2013 Share Posted December 16, 2013 (edited) The radial distance , it is the universal law of attraction, no?No, I was talking about a uniform gravitational field. Anyway I might be wrong... Is gravitational redshift associated with a change in gravitational potential, or with change in gravitational field? I think it must be the former, because it's related to conservation of energy? But the wikipedia page http://en.wikipedia.org/wiki/Gravitational_redshift begins with: "... electromagnetic radiation originating from a source that is in a gravitational field is reduced in frequency, or redshifted, when observed in a region of a weaker gravitational field." It also says "frequency of the electromagnetic radiation is reduced in an area of a higher gravitational potential (i.e., equivalently, of lower gravitational field)", and I don't think these are equivalent in a uniform gravitational field, so they must be assuming a gravitational mass and non-uniform field? Edited December 17, 2013 by md65536 Link to comment Share on other sites More sharing options...

xyzt Posted December 17, 2013 Share Posted December 17, 2013 (edited) No, I was talking about a uniform gravitational field. You were talking Newtonian mechanics, i.e [math]F=m \frac{d \Phi}{dr}=\frac{GMm}{r^2}[/math]. Anyway I might be wrong... Is gravitational redshift associated with a change in gravitational potential, or with change in gravitational field? Potential. In Schwarzschild coordinates: [math]\frac{f_1}{f_2}=\sqrt{\frac{1-r_s/r_2}{1-r_s/r_1}}=\sqrt{\frac{1+2 \Phi_2/c^2}{1+2 \Phi_1/c^2}}[/math] I think it must be the former, because it's related to conservation of energy? But the wikipedia page http://en.wikipedia.org/wiki/Gravitational_redshift begins with: "... electromagnetic radiation originating from a source that is in a gravitational field is reduced in frequency, or redshifted, when observed in a region of a weaker gravitational field." It also says "frequency of the electromagnetic radiation is reduced in an area of a higher gravitational potential (i.e., equivalently, of lower gravitational field)", and I don't think these are equivalent in a uniform gravitational field, so they must be assuming a gravitational mass and non-uniform field? The Earth gravitational field is not uniform, it is (approximately) radially symmetric, it can be described by the Schwarzschild metric to a very good approximation. The potential is not constant, [math]\Phi =-\frac{GM}{r}=-\frac{r_s c^2}{2r}[/math] , so the field is not uniform. This explains the [math]\frac{f_1}{f_2}=\sqrt{\frac{1-r_s/r_2}{1-r_s/r_1}}=\sqrt{\frac{1+2 \Phi_2/c^2}{1+2 \Phi_1/c^2}}[/math] above. Edited December 17, 2013 by xyzt 1 Link to comment Share on other sites More sharing options...

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