(Alpha) General relativity dissatisfies the equivalence principle
- Thread starter zanket
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It need not be. An inertial frame is by definition a frame that is small enough that the tidal force is negligible for the purposes of a given experiment. Then it need not be point-sized. Wikipedia notes that yours is a misconception: “The freely-falling object or laboratory, however, must still be small, so that tidal forces may be neglected. This idealized requirement has been misunderstood. This form of the equivalence principle does not imply that the effects of a gravitational field cannot be measured by observers in free-fall.” Do you disagree with the definition of an inertial frame in the OP, or Wikipedia’s comment here?
It is not splitting hairs to expect your argument to be consistent. You say it’s okay to use non-point-sized measuring equipment to test the SEP. Then by that logic it must be okay to have X and Y at different r-coordinates within a space the size of the measuring equipment. You’re trying to have your cake and eat it too. We are talking about an actual experiment, an experiment that can be performed in principle according to GR.
OK, I thought you were agreeing with me there. You said “clearly the events in X and Y must be compatible”. I agree. You said “But this violates out initial conclusion (which I am beginning to doubt) that the events in X and Y are not the same”. And I agree that it refutes your conclusion.
Sure I understand what the answer will be. What I did not understand is how you reconcile “There is no Lorentz Transformation to be done” with what you said before, “the Lorentz Transformations must be used”. You needed to rectify the contradiction.
OK, so I understand that you think the criterion for determining whether the Lorentz transformation must really be done when r_1 <> r_2 is whether or not the experiment measures the speed of light. If it does, then “There is no Lorentz Transformation to be done”. Otherwise, “the Lorentz Transformations must be used”. So let’s change the experiment:
Let the muon do a local experiment at r_2: measure the time needed for a cesium-133 atom to perform 9,192,631,770 complete oscillations (the definition of a second). Let the ground-based Earth observer do the same local experiment at r_1. How would you apply a Lorentz transformation to make those frames comparable so you can compare the outcomes of the experiments?
Also I repeat a question: Can you give me an example of a local experiment in the muon’s or Earth observer’s frame for which the Lorentz transforms must really be used in order for measurements of like experiments in two frames at different r-coordinates to be the same?
Regardless what the correspondence principle says or what QM predicts, both are irrelevant here because neither is part of GR or the SEP. The thread topic involves GR’s adherence to the SEP, that’s all, and not any theory or principle outside of that scope. For example, GR predicts that the center of a black hole is a physical singularity. That contradicts QM, which predicts that nothing can be confined to a point. So do our black holes discussed in this thread have a central singularity or not? Answer: They unquestionably do, because we are talking about GR’s predictions here, and not QM’s.
I can ignore your responses (like irrelevant QM stuff) but I cannot ignore your relevant challenges.
Not necessarily. As you said, “if I can show your argument is not consistent, then there is no need in continuing this discussion”. Or if two of your responses depend on a third, I need only show that the third is wrong to show they are all wrong. Rest assured I am attacking your case efficiently. For example, your “Lorentz transformation” argument is all but dead, which in turn refutes your r_1 <> r_2 argument.
The first time you made your claim here was in the post of yours that I’m responding to now, with “your reference frame is not defined at one point in space-time”. Prior to that you talked about “one point” only in terms of events. I just searched your posts to verify that. I did not have three chances. And I refuted it the first time I saw it.
There is no flaw. The conclusion that “this violates out initial conclusion (which I am beginning to doubt) that the events in X and Y are not the same” is correct, except “out” should be “my”, as in yours.
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Do you agree that, in the lab in X, one end of the rope is falling below the horizon whereas the other end is stationary with respect to the horizon? If so, do you agree that the rope is being stretched that way?
Do you agree that there isn’t anything mentioned in the OP that would necessarily break the rope in Y?
Hi zanket,
We know that it must stretch and continuing stretching for as long as the rocket accelerates, just as in X. In both cases, we know that it must eventually break because no rope can stretch indefinitely (if it could, then the rope in X need not break either.)
So why must it stretch and continue stretching?
Because the rope at any given point doesn't "know" what the rocket is doing until some time later. This means that the speed of the rocket in Y is always greater than the speed of any part of the rope in Y. The speed of the rope in Y will be greatest at the rocket, and least at the other end.
That's a valid point. If the rope breaks, then it breaks at the rocket, not below the horizon. So, whether or not a horizon is present, the forces on the end of the rope attached to the rocket will be the same, according to the way the thought experiment was specified. The presence or absence of the horizon therefore makes no difference, as far as I can see.
Ah, like Bell’s Spaceship Paradox! I know all about that (I even wrote a thread here about it), but it just didn’t occur to me that it’d be a factor here. I agree with you. If the rope is to never break then it must be accelerated at a higher rate at the dangling end then at the rocket end. Good job Pete, and Thanks! (And I'm not saying that just because of the Alpha rules!)
However, I disagree with JamesR that the “presence or absence of the horizon therefore makes no difference”. I believe this issue is a complication that obscures the difference between X and Y and that can be overcome by changing the thought experiment. I think I have a solution that will stymie y’all with its relative simplicity. I could post it here, but the idea is different enough that the thread would become messy. I’d like to start fresh with a new thread, and in the OP there better forestall common objections that arose, like about inertial frames crossing the horizon. I would wrap up remaining issues in this thread with BenTheMan, or he could migrate. (It might take me a few days anyway to write a new OP.) Do you have a problem with that?
Sounds good to me.
I confess I haven't followed all strands of the thread right through. When Alpha threads get complicated like this (with a number of interweaved discussions happening at once), I'll be inclined to let it go unless someone raises an objection to a post or sidetrack in the rules thread, or if I spot some flagrant abuse, blatant diversionary tactics, or other obvious issues.
I confess I haven't followed all strands of the thread right through. When Alpha threads get complicated like this (with a number of interweaved discussions happening at once), I'll be inclined to let it go unless someone raises an objection to a post or sidetrack in the rules thread, or if I spot some flagrant abuse, blatant diversionary tactics, or other obvious issues.
Zanket: maybe you could talk about a laboratory on a very tall tower above the surface of a planet. There's a reel of rope unwinding, with the end of the rope dropping towards the ground. Because of the gradient that has to exist for gravity to exist, the gravity closer to the ground is greater. So the rope snaps before it snaps in an identical laboratory accelerating through free space.
I think that's a good way to handle it.
Well, if the gravity closer to the ground is significantly greater, then there's a significant tidal force, and my thought experiment depends on a negligible tidal force. But if you think you can overcome that, start a thread!
First,a few things.
No I'm sorry. This is not a fesable experiment to preform.
Certainly not. I don't understand how you can expect two frames to not to be related by a Lorentz transformation.
Sigh. The speed of light is the same in all frames. There is no Lorentz Transformation to be done because the speed of light is the same in all frames. What other way can I say this? If you had said "measure anything else except the speed of light", I would have explained that the Lorentz transformations in Special Relativity reduce to the simple length contractions and time dialations, etc.
Time dialation.
Pion decay, or the example you listed above. I have been quite careful in the past to describe the pion decay experiment and how the Lorentz transformations were to be used.
I guess I disagree with Wikipedia's definition. The bit that you have cut and paste from Wikipedia is sufficiently vague to support your point. For example, what is "this form of the SEP", and is it consistent with the other derfinitions that we have been using?
Here is the paradox I was pointing out, that I think you missed. (Which is also a consequence of Wikipedia's definition.)
If the only requirement for reference frames is that the tidal forces are negligible, then,if X and Y are at similar curvatures, and they both have negligible tidal forces, then one can define a frame S that contains X and Y. Then the physics at X and Y is the same, by virtue of being contained in a single reference frame. The clear conclusion is that the rope doesn't break in either case (or it breaks in both cases).
If you accept that S can be defined such that it contains X and Y (which you have already told me that you did), then the presence of a horizon makes no difference.
If the presence of a horizon makes a difference, i.e. if you think the rope breaks in one frame and not the other, then there is no frame S that contains X and Y. But X and Y have similar curvatures and no tidal forces, so this disagrees with your (and Wikipedia's) definition.
Is this clear?
No I'm sorry. This is not a fesable experiment to preform.
Certainly not. I don't understand how you can expect two frames to not to be related by a Lorentz transformation.
Sigh. The speed of light is the same in all frames. There is no Lorentz Transformation to be done because the speed of light is the same in all frames. What other way can I say this? If you had said "measure anything else except the speed of light", I would have explained that the Lorentz transformations in Special Relativity reduce to the simple length contractions and time dialations, etc.
Time dialation.
Pion decay, or the example you listed above. I have been quite careful in the past to describe the pion decay experiment and how the Lorentz transformations were to be used.
I guess I disagree with Wikipedia's definition. The bit that you have cut and paste from Wikipedia is sufficiently vague to support your point. For example, what is "this form of the SEP", and is it consistent with the other derfinitions that we have been using?
Here is the paradox I was pointing out, that I think you missed. (Which is also a consequence of Wikipedia's definition.)
If the only requirement for reference frames is that the tidal forces are negligible, then,if X and Y are at similar curvatures, and they both have negligible tidal forces, then one can define a frame S that contains X and Y. Then the physics at X and Y is the same, by virtue of being contained in a single reference frame. The clear conclusion is that the rope doesn't break in either case (or it breaks in both cases).
If you accept that S can be defined such that it contains X and Y (which you have already told me that you did), then the presence of a horizon makes no difference.
If the presence of a horizon makes a difference, i.e. if you think the rope breaks in one frame and not the other, then there is no frame S that contains X and Y. But X and Y have similar curvatures and no tidal forces, so this disagrees with your (and Wikipedia's) definition.
Is this clear?
BenTheMan,
The speed of light is the same in all inertial frames when measured with local cesium clocks in those frames, correct? The speed of light is the same in co-moving inertial frames, correct?
There's an important subtlety here. Imagine you have a metre ruler and you consider yourself to be at rest. You measure the speed of light to be 300,000km/s. Now you accelerate to what you consider to be .99c holding your metre ruler transverse to your motion to avoid length contraction. You measure the speed of light again. Again the result is 300,000km/s. But you can calculate the time dilation, which means your seconds are now seven times longer than they were. You can then calculate the speed of light to be one seventh of what it was. So it's too simplistic to say the speed of light is the same in all frames.
If you measure the speed of light at rest or at .99c and get the same result, then that would mean the speed of light has nothing to do with you but instead is more a property of spacetime.
Time dilation is calculated between your frame at rest and .99c, relative to each other. This does not have an effect on the speed of light.
Let me rephrase it:
You measure the speed of light c1 = 300,000 km/s1 while at rest, then accelerate to .99c and measure it again as c2 = 300,000 km/s2. You're using a ruler held transverse to your motion to avoid length contraction effects, so a kilometre in both frames is the same. Speed = distance / time or c = km/s, so if s1 <> s2, then despite the measurement, c1 <> c2.
Like I said, it's a subtlety. It's related to the way "time goes slower" in a "gravitational field".
You measure the speed of light c1 = 300,000 km/s1 while at rest, then accelerate to .99c and measure it again as c2 = 300,000 km/s2. You're using a ruler held transverse to your motion to avoid length contraction effects, so a kilometre in both frames is the same. Speed = distance / time or c = km/s, so if s1 <> s2, then despite the measurement, c1 <> c2.
Like I said, it's a subtlety. It's related to the way "time goes slower" in a "gravitational field".