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#1 |
Muse
Join Date: Jul 2018
Posts: 959
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Is the frame dragging part of the Special Relativity?
Hi all,
is the frame dragging embedded in the Special Relativity? ![]() The platform observer at x=0cs, t=0s sends a light across, the blue arrow up, 90 degree up at the top part of the figure. The train observer at x'=0cs', t'=0s' sends a light across, the red arrow up, 90 degree up at the top part of the figure. This happens when they are aligned, x=x'=0 and t=t'=0. Now frame dragging happens. The red arrow up is frame dragged across the platform frame. The red arrow up appears straight up for the train observer but as the blue arrow to the right from the platform x=0cs origin position. The blue arrow up is frame dragged across the train frame in the reciprocal way. The blue arrow up appears straight up for the platform observer but as the red arrow to the left from the train x'=0cs' origin position. Your thoughts, SDG |
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#2 |
Illuminator
Join Date: Aug 2007
Posts: 4,127
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Frame dragging is predicted by GR not SR and isn't what you just described.
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#3 |
Muse
Join Date: Jul 2018
Posts: 959
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Would you have a better name for what I described?
SDG |
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#4 |
Penultimate Amazing
Join Date: Aug 2007
Location: Hong Kong
Posts: 48,530
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"special relativity"?
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#5 |
Penultimate Amazing
Join Date: Mar 2009
Posts: 17,683
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Yeah, it's not frame dragging. It's two different frames of reference, moving relative to each other. It's the bog standard premise and whole point of SR.
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#6 |
Penultimate Amazing
Join Date: Mar 2009
Posts: 17,683
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But to answer what it is, basically in Newtonian mechanics, gravity only depends on the distance between two bodies. If you're talking about how a moon will move around a planet, only their masses and distance matter. How fast the planet spins, or whether it even spins at all, it plays no role whatsoever.
In GR, it turns out that it matters. It's pretty much lost in the decimals, but it matters. Basically if the Earth were to spin ten times faster, the moon's orbit would be a little different. Again, infinitesimally, but still... And basically that's all there is to it. It matters if something creating a gravity well spins or not. That's what we call frame dragging. |
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Which part of "Ph'nglui mglw'nafh Cthulhu R'lyeh wgah'nagl fhtagn" don't you understand? |
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#7 |
Penultimate Amazing
Join Date: Aug 2007
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What I think is neat is that the Sun-Mercury system is massy enough, and energetic enough, that we could actually find the frame-dragging phenomenon "in the decimals", as far back as 1859. It took GR to explain this measurable deviation from the Newtonian prediction.
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#8 |
Penultimate Amazing
Join Date: Jun 2003
Posts: 47,892
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Nope. The perihelion precession of Mercury's orbit is indeed observably affected by general relativity, but this is mostly not frame dragging. The frame dragging component is 0.002 arcseconds per century, whereas the non-rotating GR contribution (the component which would exist if the sun didn't rotate at all) is about 43 arcseconds per century (source). The overall GR contribution was observable back in 1859, but the frame dragging component was not. Even today the error margin on Mercury's perihelion precession is much larger (0.65 arcsec/century) than the frame dragging contribution.
Frame dragging is very difficult to observe, and wasn't experimentally confirmed until Gravity Probe B. |
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"As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law |
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#9 |
Muse
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#10 |
Penultimate Amazing
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#11 |
Penultimate Amazing
Join Date: Jun 2003
Posts: 47,892
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Which premise? You have described a scenario, not a premise. "The speed of light is constant in all inertial frames" is a premise.
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"As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law |
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#12 |
Muse
Join Date: Jul 2018
Posts: 959
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The scenario is based on that premise and it is easy to show that it is the case.
Having said that, what the train observer considers as 'c' the platform observer sees 'c/2'. This is reciprocal when we switch frames.
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If any light round trip is time dilated then the twin paradox is not resolved. Each twin has own light round trips as own proper time clocks and they are time dilated from the other moving frame. That's the contradiction in your replies. Jano |
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#13 |
The Clarity Is Devastating
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#14 |
Penultimate Amazing
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"As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law |
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#15 |
Muse
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#16 |
Penultimate Amazing
Join Date: Mar 2009
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If we're talking SR, we're talking inertial frames, i.e., they don't accelerate at all in respect to each other. And yes, the whole idea of SR is that there is no preferred frame. Any frame behaves just like any other frame, and is just as good as any other frame.
The twin paradox though is actually fairly easy to solve, even while sticking strictly to SR. The key is simply that there are not two frames of reference, but THREE. One for the guy sitting at home, one for the ship as it's flying away, and one for the ship as it's coming back. The solution is basically just to notice that one of the guys is switching reference frames. As for what it's good for, well, just like any other physics it's good when you need to calculate anything where SR applies, really. |
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#17 |
Penultimate Amazing
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Which part of "Ph'nglui mglw'nafh Cthulhu R'lyeh wgah'nagl fhtagn" don't you understand? |
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#18 |
Muse
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#19 |
Muse
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#20 |
Penultimate Amazing
Join Date: Mar 2009
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To understand why both see it as c, you need to also understand that clocks are not synchronized in the two reference frames. When you calculate how fast something moves, basically you measure when it passes by some kind of points, then basically do a v=(x2-x1)/(t2-t1). And in the other frame of reference that would be v'=(x'2-x'1)/(t'2-t'1).
When you do the Lorentz transformations for BOTH space and time, what one sees as c, the other also sees as exactly c. Not c/2, not anything else than exactly c. But basically here's an illustration I've used before about how not only intervals are different in different frames, but even two events being synchronous doesn't carry over to other frames. Let's say we have a 10m long hangar which is basically a straight tunnel with force fields at both ends that can be opened or closed. They're currently open. Let's say you're sitting at the controls of those doors, while I'm flying a 20m long spaceship at 0.866c through that hangar. Well, from your point of view, my ship is only 10m long, because I chose the speed so the transformation works that way. So there is one moment when, as seen from your frame, my ship fits snugly inside your hangar. I mean, both are 10m long in your frame of reference. So for an instant you could close both doors simultaneously, with my 20m ship fitting perfectly inside your 10m hangar. Well, now from my perspective as I'm sitting in the captain's chair on that ship, my ship is still 20m long, and it's your hangar that is half the length. So it's only 5m long. There's no way in hell my 20m ship will ever fit neatly inside your 5m hangar, as seen from my frame. So what happens from my point of view when you close the doors? Well, from my point of view the two force fields don't close simultaneously. First the one in front closes for an instant in front of the tip of my ship, then it opens, my ship keeps passing through, and then the one behind closes right behind my ship for an instant. Basically, when they said that worse things happen at sea... yeah, even worse happen in relativity ![]() But yeah, that's the kind of effects you have to contend with, in order to understand why what looks like c to one of them, also looks like c to the other. |
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Which part of "Ph'nglui mglw'nafh Cthulhu R'lyeh wgah'nagl fhtagn" don't you understand? |
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#21 |
Penultimate Amazing
Join Date: Mar 2009
Posts: 17,683
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To use a cliched quote, "You keep using that word. I do not think it means what you think it means."
![]() As a couple of us have been trying to tell you, the Lorentz transformations are NOT the same as frame dragging. They don't have ANYTHING to do with what is actually called frame dragging. Just because one frame moves in relation to the other (by being dragged or otherwise), doesn't mean you can call it frame dragging. Jargon terms can mean something very different than just the sum of the words. So anyway, it's not frame dragging unless you're calculating gravity around a massive rotating body (and I mean, as in, you'd have to be pretty damn close to a rotating black hole or at least neutron star to even need to apply that correction,) but the normal Lorentz transformations still apply anyway. Well, if you get the obviously wrong result, that should be your clue that you just disproved your premises via an ad absurdum disproof. If you mean if we want a wrong calculation that is based on not even starting to understand SR, yeah, that's not what we want ![]() |
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Which part of "Ph'nglui mglw'nafh Cthulhu R'lyeh wgah'nagl fhtagn" don't you understand? |
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#22 |
Penultimate Amazing
Join Date: Jan 2003
Location: Yokohama, Japan
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Maybe you already know this, but it's a good for anyone.
The Planet That Wasn't By Isaac Asimov Reading this taught me that the planet Vulcan wasn't originally something invented by the writers of Star Trek. It was something that astronomers hypothesized to explain why Mercury's orbit was so weird and didn't conform to the Newtonian prediction. So they spent a lot of time looking for this planet they called Vulcan that didn't actually exist. |
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A fool thinks himself to be wise, but a wise man knows himself to be a fool. William Shakespeare |
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#23 |
Muse
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#24 |
Muse
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#25 |
Philosopher
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We don't want good, sound arguments. We want arguments that sound good. |
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#26 |
Muse
Join Date: Jul 2018
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The blue arrow up from the platform observer is 90 degrees to the velocity vector of the train.
The train moves along x axis and the up direction is in y axis. The x, y are platform axes. The red arrow up moves along y' axis that is 90 degree to x' axis. The blue arrow up in platform frame is observed as the red arrow to the left in the train frame. |
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#27 |
The Clarity Is Devastating
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#28 |
Penultimate Amazing
Join Date: Oct 2012
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Wait! We have two clocks;
Clock A: at rest Clock B: moving at, say, 20%C As Clock A passes Clock B, we observe that both clocks are exactly sycnhronized. Clock A then proceeds on at 20%C and completes a large circle, traveling at 20%C all the way, never accelerating or decelerating. After a period of time, it passes Clock B again. Are the clocks still synchronized? |
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#29 |
Muse
Join Date: Jul 2018
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Agreed, but the photons are comoving with the train frame.
Yes, moving at an angle, this is aberration of the light. To make things easier, I will not call it frame dragging but comoving. The effect is obvious the final light speed is c, the y component is c/2 and the x component is v, based on the comoving frame velocity. c^2 = (c/2)^2 + v^2 |
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#30 |
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#31 |
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#32 |
Philosopher
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#33 |
Penultimate Amazing
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#34 |
Penultimate Amazing
Join Date: Mar 2009
Posts: 17,683
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What meaning does "comoving" even convey there?
What you actually have, as the very minimum to measure a speed, is that in a frame of reference something passed through point {x1, y1, z1} at moment t1, and then through point {x2, y2, z2} at moment t2. You take the distance, divide by t2-t1, get the speed. Meanwhile in another frame, let's call it prime, it passed through point {x'1, y'1, z'1} at moment t'1, and then through point {x'2, y'2, z'2} at moment t'2. You take the distance, divide by t'2-t'1, get the speed. Whether you're talking SR or plain old Newton, the body whose motion you're measuring is simply passing through those points at those times. Thinking about it as "comoving" with anything isn't necessary or bringing you any bit of extra information at all. Because it doesn't matter if the thing you're measuring is walking inside the train, or on the ground, or in a second train. It changes nothing. It still passes through those coordinates at that time. You still just need to transform the coordinates and times, do the division, and that's it. As long as you keep it inside one frame, and it's still SR, sure, Pythagoras still applies when you project any vector into orthogonal components. Not sure what it's supposed to illustrate, though. The reason both still measure exactly c for that scenario, even though one sees the light travelling diagonally over a larger distance, is that their clocks aren't synchronized between the two frames. The guy in the train sees the light moving only half the distance that the guy on the ground sees, but also in only half the time that the guy on the ground measures. The guy on the ground sees the light moving diagonally twice the distance that the guy on the train sees, but in twice the time that the guy in the train measured. So while it's mathematically correct, I'm not seeing anything particularly enlightening there. I'm not seeing what information is calling it "comoving" supposed to convey. Not really, no. In fact, that's kinda the whole point. |
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#35 |
Penultimate Amazing
Join Date: Mar 2009
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As for the 'twins paradox', instead of thinking it as one ship that goes from 0 to a point at distance r, and then back, which involves acceleration somewhere in between, you can think of it as two ships which move at constant speed and never accelerate. One starts from the origin and keeps going in the positive direction of the x axis at constant speed. One starts from 2r and goes towards the origin. When they pass each other by at point r, they synchronize their clocks. When the second one passes the origin by, it compares its clock to the one of the guy who never moved from the origin.
That version is perfectly compatible with SR, since what we have is three perfectly inertial frames. And then it turns out it's no paradox at all. |
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#36 |
Muse
Join Date: Jul 2018
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The enlightening part is that the time dilation determines the v_y.
This example is generic, it applies to any two moving inertial frames. The illustration is the blue arrow up traveling 1s in the platform frame and it takes 2s' in the train frame. This is reciprocal to the first solution, this is the paradox and this is the foundation for the twin paradox. The SR is reciprocal, paradoxical and it is a problem. |
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#37 |
Muse
Join Date: Jul 2018
Posts: 959
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No it is not because the twins do not meet again.
![]() The reversal of the frames at r that you described is generating time discontinuation at the stay at home twin. It is seen as the jump from A to B. That's not realistic. The time does not jump like that. Is this some kind of magic? |
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#38 |
The Clarity Is Devastating
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The equation c^2 = (c/2)^2 + v^2 is valid if and only if v = .866c (or v = -.866c). That is to say, it can be solved for v, and doing so merely recapitulates your initially assumed conditions. What is the "aberration?" Are you talking about the direction of travel of the light beams (assuming they're narrow focused beams, not just omnidirectional light flashes)? If the train observer aligns his beam emitter (stationary in his reference frame) perpendicular to the train in his reference frame, and the platform observer aligns his beam emitter (stationary in his reference frame) perpendicular to the platform in his reference frame, both observers will observe the beams diverging once they're emitted. If there are detectors arranged along the center line of the train and stationary with respect to the train, the beam from the train's emitter will be detected at a detector positioned perpendicular to the train's emitter, while the beam from the platform's emitter will be detected at a detector farther to the rear of the train. Both observers will agree which detector detects which beam. Both observers will agree that the beams diverge. What they will disagree on is how the emitters are aimed. The platform observer won't agree that the train's emitter is perpendicular to the train (it'll seem to be aimed at a "forward" angle). The train observer won't agree that the platform's emitter is perpendicular to the platform (it'll seem to be aimed at a "backward" angle). That's because to be aimed in a direction other than parallel to the x axis, the emitters have to have a length along the y axis. The Lorentz transformation thus induces a rotation of the direction of the emitters. (But... but... but... the Lorentz transformation doesn't affect the y coordinate! It says right there, y' = y! But since the emitter has length, light takes time to travel from the back of the emitter to the front, and in that time, as observed from a relatively moving frame, the x position of the emitter changes and is affected by the Lorentz transformation. That's observed as a change in the "aim" direction. Such complications are why typical elementary SR thought experiments usually involve long thin moving components like rockets and trains, and flashes of light rather than aimed beams.) |
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#39 |
The Clarity Is Devastating
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The stay at home twin experiences no time discontinuity. The only discontinuity is in what the traveling twin observes is happening to the stay at home twin.
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It's the same kind of magic that allows the traveling twin at r to instantaneously reverse direction in the first place, requiring infinite acceleration and infinite energy. (In the "no acceleration" version, where there are two travelers and they simply cross paths and transfer clock information at r, neither of the travelers observes any discontinuity for the stay at home twin. They merely disagree with one another about what time it is back on Earth at the moment of the exchange... which is expected, since they're observing Earth from different inertial reference frames.) |
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#40 |
Muse
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