StringJunky Posted June 4, 2017 Posted June 4, 2017 (edited) .... Since we both agree that there is no change in the arms or the wavelength from the grav waves we are back staring at the essential problem I raised in the OP. There is a change in the length of the arms. ...With that in mind, imagine now that you and a friend want to compare how long it takes you to drive to the end of the interferometer arms and back. Just like LIGO's laser light waves, you leave the corner station at exactly the same time, take different paths, and travel at precisely the same speed. You expect to meet up again at the same time. But if a gravitational wave passes while you are on your journey, one of you will end up traveling down the longer arm, and one of you will travel down the shorter arm. Since you're still going the same speed, one of you will take longer to return than the other! https://www.ligo.caltech.edu/page/faq Edited June 4, 2017 by StringJunky
aramis720 Posted June 4, 2017 Author Posted June 4, 2017 There is a change in the length of the arms. Why is there a change in length? Again, a grav wave is defined as being a wave of space, so anything occuping that space will wave in exactly the same way as the grav wave. So why is there a change in length of the arm?
StringJunky Posted June 4, 2017 Posted June 4, 2017 Why is there a change in length? Again, a grav wave is defined as being a wave of space, so anything occuping that space will wave in exactly the same way as the grav wave. So why is there a change in length of the arm? Because that's what nature does when two blackholes merge. The same thing is not happening in two different directions and that is what's measured; as one is contracting the other one is expanding.. What's so hard about that? Things are not 'waving' in the same direction everywhere.
Strange Posted June 4, 2017 Posted June 4, 2017 But it's not true that the time for the propagation of light is changing and thus detectable. What is being detected is a phase difference in the light signals in both arms. The phase difference is due to the difference in light travel time. Since we both agree that there is no change in the arms or the wavelength from the grav waves we are back staring at the essential problem I raised in the OP. We don't agree on that. You seem to have convinced yourself to such an extent that you are not willing to consider that you might be mistaken. Why is there a change in length? Again, a grav wave is defined as being a wave of space, so anything occuping that space will wave in exactly the same way as the grav wave. So why is there a change in length of the arm? You have just explained why there is a change in length: "so anything occuping that space will wave in exactly the same way as the grav wave". The gravitational wave causes a change in space, i.e. length.
geordief Posted June 4, 2017 Posted June 4, 2017 (edited) So the 2 arms of the Ligo interferometer lie at the surface of an expanding sphere centred at the black hole merger when detection is made? If one of the arms happens to be pointing directly into the centre of the sphere is its length unchanged -or is it the other(tangential) arm which would be unchanged in length? They are not both changed in length in complementary ways are they? Edited June 4, 2017 by geordief
aramis720 Posted June 4, 2017 Author Posted June 4, 2017 Because that's what nature does when two blackholes merge. The same thing is not happening in two different directions and that is what's measured; as one is contracting the other one is expanding.. What's so hard about that? Things are not 'waving' in the same direction everywhere. It seems that I'm not communicating well my basic point. Here it is: if a grav wave is defined, as it is, as a wave of space itself, anything occupying that space will wave to EXACTLY the same degree as the grav wave. So what can be used to detect such a wave when anything used to detect the wave must by necessity occupy the same space that is waving? Is this clear now? It doesn't matter what direction the wave is coming from. Any wave direction will wave space itself and anything occupying that space in exactly the same way.
StringJunky Posted June 4, 2017 Posted June 4, 2017 It seems that I'm not communicating well my basic point. Here it is: if a grav wave is defined, as it is, as a wave of space itself, anything occupying that space will wave to EXACTLY the same degree as the grav wave. So what can be used to detect such a wave when anything used to detect the wave must by necessity occupy the same space that is waving? Is this clear now? It doesn't matter what direction the wave is coming from. Any wave direction will wave space itself and anything occupying that space in exactly the same way. The speed of light is constant for all inertial observers and because of that fact the differences can be noted for the travel times in each arm.
aramis720 Posted June 4, 2017 Author Posted June 4, 2017 The phase difference is due to the difference in light travel time. We don't agree on that. You seem to have convinced yourself to such an extent that you are not willing to consider that you might be mistaken. You have just explained why there is a change in length: "so anything occuping that space will wave in exactly the same way as the grav wave". The gravitational wave causes a change in space, i.e. length. You just stated in the previous post that you did agree that neither the arm nor the light wave are being contracted. Here's your quote: "you are correct: the length of the arm and "ruler" (wavelength of light) both change on the same way." Then you stated that even though the arm and the wavelength of light weren't changing, the speed of light must nevertheless change with respect to the interferometer b/c of the special relativity postulate that the speed of light is always constant. I then asked you why you made this conclusion b/c that's not how interferometers work. They work based on comparing phase overlaps. If the arm lengths or wavelengths change the phases won't overlap exactly. So there's no literal time measurement. Could you explain your reasoning further here? I'm always willing to consider that I may be mistaken (in fact, it's almost certain that I am mistaken here b/c of the vast weight of authority against me), yet no one here has answered my questions satisfactorily and there seems to be an ongoing confusion about my question. For example, your last two lines show that you don't understand my point about grav waves distorting space and anything in it. The point of anything in that space that is being waved also waving is that it makes such waves in principle undetectable b/c there is no physical difference that can be detected -- if indeed grav waves are defined as waves of space itself. So what I'm suggesting is the conceptual structure by which physicists are suggesting grav wave detection may in fact be erroneous. The speed of light is constant for all inertial observers and because of that fact the differences can be noted for the travel times in each arm. How? The only measurement is phase overlaps or lack thereof. There's no time measurement at issue here.
Mordred Posted June 5, 2017 Posted June 5, 2017 (edited) How can you have phase variance without time being involved? Seriously we have made every effort to explain the detection methods to you. The part you continuously mistakenly misunderstand is the polarity nature of a GW wave. Calculate the wavelength and then apply a quarter of that to get your required detector length. Then recognize that within that Precise same wavelength you have in essence 4 simultaneous movements. Both -x and +x contract while +y and -y expand. You have an L shape inerferometer it is impossible for both arms to have the same vector components. For the last time if you have length contradiction you MUST have an influence on the time components. It is impossible not to. From the very first link posted by Strange. " This will also cause the laser wavelength in the shortened arm to decrease (blueshift) and the wavelength in the lengthened arm to increase (redshift)" This is two simultaneous movements with different vector components in each arm. Why is that so hard to understand? It is no different than gravitational redshift which relies on SR and GR being correct. Those blue/redshifts alter the frequency of the beams and causes phase shifts as a result. These frequency changes are detectable. Lets put it in electrical terms. You have altered a frequency of light by a frequency of GW. This will induce a propogation delay which results in a phase shift. We don't care about the detector walls. We only care about the phase shifts. Step 1) split a signal with amplitude of x. When you recombine the two signals as they are identical you have complementary waves = constructive interference. The amplitude will be the sum of the two waves. Hit it now with a momentary GW wave. Arm on left contracts and its signal is delayed, signal on the perpinficular DOES NOT undergo the identical affect. It will blueshift not redshift. Its signal also goes out of phase. HOWEVER both signals remain out of phase with each other so you get destructive interference on recombination. The recombined signal will have an amplitude less than the sum of the amplitudes of the two signals. The walls of the detector has nothing to do with the experiment. It is strictly its influence on the beam itself that is involved in the detection. When you have differences in gravitational potentials you gravitational redshift which involves both length and time dilation. Edited June 5, 2017 by Mordred
Strange Posted June 5, 2017 Posted June 5, 2017 It seems that I'm not communicating well my basic point. Here it is: if a grav wave is defined, as it is, as a wave of space itself, anything occupying that space will wave to EXACTLY the same degree as the grav wave. So what can be used to detect such a wave when anything used to detect the wave must by necessity occupy the same space that is waving? Is this clear now? It doesn't matter what direction the wave is coming from. Any wave direction will wave space itself and anything occupying that space in exactly the same way. Yes. Maybe what you are missing is that the two arms are affected differently (because of the nature of the waves). Therefore, the time taken for light to travel each arm will be different (because the speed of light doesn't change and time = distance / velocity), therefore there will be a phase difference between each arm. You just stated in the previous post that you did agree that neither the arm nor the light wave are being contracted. Here's your quote: "you are correct: the length of the arm and "ruler" (wavelength of light) both change on the same way." You seem to have misunderstood. As you can see, I didn't say they are not being contracted (alternately stretched and squeezed, in fact). I said they are both affected by the same amount. So if you tried to use the wavelength of the light to measure the length of the arm, you would measure a difference. But if you use the speed of light, then you will see a difference. The speed of light is not affected. It travels different distances in each arm. Then you stated that even though the arm and the wavelength of light weren't changing, the speed of light must nevertheless change with respect to the interferometer b/c of the special relativity postulate that the speed of light is always constant. Again, I said the speed of light does NOT change. I guess this is why you don't understand the explanations. You seem to be interpreting things to mean the opposite of what is said... How? The only measurement is phase overlaps or lack thereof. There's no time measurement at issue here. Because if light takes a different amount of time to travel one arm than the other, it will have a different phase when it gets back. 1
Lord Antares Posted June 5, 2017 Posted June 5, 2017 Strange is right. Let me ask you something. You are arguing that the experiment might have been faulty. Please explain to me what they measured then? How do you think they overlooked something so simple yet reported results which indicated gravitational waves just as expected? Do you think they made the results up? There you go, as Strange said: http://www.space.com/31913-how-scientists-detected-gravitational-waves-ligo.html The LIGO detectors are in Livingston, Louisiana, and Hanford, Washington. Each one consists of a giant L-shaped structure with arms 2.5 miles (4 kilometers) long. A laser beam shines down each arm from the crux of the "L," and mirrors at the ends of these arms reflect the light back. If the beams from both arms arrive back at the crux at the same time, they cancel each other out, and no signal is produced in the system's light detector. This is the normal, expected scenario. But if one of the beams arrives a bit late, a signal is produced, which could be evidence of a gravitational wave.
tar Posted June 5, 2017 Posted June 5, 2017 Lord Antares, Your quote explains that the interferometer is tuned so the waves cancel each other out. As in no signal. If an arm gets "bumped", contracted by a gravity wave or expanded by a gravity wave, the waves coming from one arm or the other will arrive at the return sensor out of phase for the duration of the distortion. Hence on that day in September, there was a "chirp", a signal where the light merged back from the tuned arms was not, for that moment, in sync. I would imagine, after the wave passed, the instrument went immediately back to the no signal, tuned, in phase condition, or as Mordred suggested, a quarter wavelength out of phase, so as the one arm's sine wave's peak, arrived coincidently with the other arm's sine wave's trough and vice-a-versa. Mordred, you confused me when you said "Step 1) split a signal with amplitude of x. When you recombine the two signals as they are identical you have complementary waves = constructive interference. The amplitude will be the sum of the two waves." However, now after reading through some posts again I think I see where aramis720 and I were getting hung up. Yes, the light waves and the mirrors and the space in the one arm were all getting strained one way or the other together, but that is not why the experiment is faulty, that is why the experiment worked. The only thing that could cause the chirp, once vibrations of the Earth were known not to have contributed, is a distortion of space itself, as if a gravity wave just passed though, squeezing space along one vector, while stretching it along the normal. Regards, TAR Thread, But alas, I am still hung up as to why two black holes spiraling in toward each other would not put out gravity waves with increasing frequency prior the merger. Or why any two orbiting masses do not put out gravity waves. After all as the Earth/moon spins it is sort of like a "lumpy" mass, when taken as one mass. Regards, TAR
aramis720 Posted June 5, 2017 Author Posted June 5, 2017 Yes. Maybe what you are missing is that the two arms are affected differently (because of the nature of the waves). Therefore, the time taken for light to travel each arm will be different (because the speed of light doesn't change and time = distance / velocity), therefore there will be a phase difference between each arm. You seem to have misunderstood. As you can see, I didn't say they are not being contracted (alternately stretched and squeezed, in fact). I said they are both affected by the same amount. So if you tried to use the wavelength of the light to measure the length of the arm, you would measure a difference. But if you use the speed of light, then you will see a difference. The speed of light is not affected. It travels different distances in each arm. Again, I said the speed of light does NOT change. I guess this is why you don't understand the explanations. You seem to be interpreting things to mean the opposite of what is said... Because if light takes a different amount of time to travel one arm than the other, it will have a different phase when it gets back. I could have stated more clearly my summary of your previous statements, I'll agree on that, but my point remains. You had agreed that length and wavelength change to the same degree exactly as the wave, and thus are undetectable, that was my point. You now clarify that because the speed of light remains constant that the interferometer will register a difference. But think again about what is going on with the measurement apparatus. Light does not exist in some other realm than the space we exist in. So if space itself is being distorted so is any light occupying that space. You write:"Because if light takes a different amount of time to travel one arm than the other, it will have a different phase when it gets back." But why would it take a different amount of time to travel one arm? We know, of course, that light is affected by gravity in the same way as mass -- this was the basis for the famous 1919 Eddington experiment looking at the curving of light around the sun during a full eclipse. How can you have phase variance without time being involved? Seriously we have made every effort to explain the detection methods to you. The part you continuously mistakenly misunderstand is the polarity nature of a GW wave. Calculate the wavelength and then apply a quarter of that to get your required detector length. Then recognize that within that Precise same wavelength you have in essence 4 simultaneous movements. Both -x and +x contract while +y and -y expand. You have an L shape inerferometer it is impossible for both arms to have the same vector components. For the last time if you have length contradiction you MUST have an influence on the time components. It is impossible not to. From the very first link posted by Strange. This is two simultaneous movements with different vector components in each arm. Why is that so hard to understand? It is no different than gravitational redshift which relies on SR and GR being correct. Those blue/redshifts alter the frequency of the beams and causes phase shifts as a result. These frequency changes are detectable. Lets put it in electrical terms. You have altered a frequency of light by a frequency of GW. This will induce a propogation delay which results in a phase shift. We don't care about the detector walls. We only care about the phase shifts. Step 1) split a signal with amplitude of x. When you recombine the two signals as they are identical you have complementary waves = constructive interference. The amplitude will be the sum of the two waves. Hit it now with a momentary GW wave. Arm on left contracts and its signal is delayed, signal on the perpinficular DOES NOT undergo the identical affect. It will blueshift not redshift. Its signal also goes out of phase. HOWEVER both signals remain out of phase with each other so you get destructive interference on recombination. The recombined signal will have an amplitude less than the sum of the amplitudes of the two signals. The walls of the detector has nothing to do with the experiment. It is strictly its influence on the beam itself that is involved in the detection. When you have differences in gravitational potentials you gravitational redshift which involves both length and time dilation. My point has been that there can't be any detectable length contraction b/c space itself is contracting, so anything occupying that space is contracted precisely the same amount -- and is thus undetectable. There is no contraction of either arm. Isn't that a pretty clear point?
dimreepr Posted June 5, 2017 Posted June 5, 2017 My point has been that there can't be any detectable length contraction b/c space itself is contracting, so anything occupying that space is contracted precisely the same amount -- and is thus undetectable. There is no contraction of either arm. Isn't that a pretty clear point? Yes but no, you do realise it has been detected?
aramis720 Posted June 5, 2017 Author Posted June 5, 2017 Yes but no, you do realise it has been detected? The point of my OP was that the premises underlying this experiment seem questionable. False positives happen pretty regularly in science.
dimreepr Posted June 5, 2017 Posted June 5, 2017 False positives happen pretty regularly in science. Yes but rarely when predictions matches observations.
swansont Posted June 5, 2017 Posted June 5, 2017 I could have stated more clearly my summary of your previous statements, I'll agree on that, but my point remains. You had agreed that length and wavelength change to the same degree exactly as the wave, and thus are undetectable, that was my point. You now clarify that because the speed of light remains constant that the interferometer will register a difference. But think again about what is going on with the measurement apparatus. Light does not exist in some other realm than the space we exist in. So if space itself is being distorted so is any light occupying that space. You write:"Because if light takes a different amount of time to travel one arm than the other, it will have a different phase when it gets back." But why would it take a different amount of time to travel one arm? We know, of course, that light is affected by gravity in the same way as mass -- this was the basis for the famous 1919 Eddington experiment looking at the curving of light around the sun during a full eclipse. I had this same question some time ago. Strange's link in post #3 is IMO a good explanation. You have to think of the light as a clock rather than a ruler. Yes, the wavelength changes, but you are not detecting a change in wavelength, or counting the number of waves. You are detecting a phase difference from a difference in travel time between the two arms. Since c is an invariant, if the length changes (regardless of the wavelength of the light) the travel time will vary.
Lord Antares Posted June 5, 2017 Posted June 5, 2017 The point of my OP was that the premises underlying this experiment seem questionable. False positives happen pretty regularly in science. But what did they detect then? How do you think they detected it with the precision that they did if it was undetectable? How did they get the mathematics of the prediction to match the experiment? If if was undetectable because the instruments contracted by the same degree as the space around it, as you said, how could they have gotten a false positive? They would have gotten a null result. Don't you see a contradiction there?
aramis720 Posted June 5, 2017 Author Posted June 5, 2017 I had this same question some time ago. Strange's link in post #3 is IMO a good explanation. You have to think of the light as a clock rather than a ruler. Yes, the wavelength changes, but you are not detecting a change in wavelength, or counting the number of waves. You are detecting a phase difference from a difference in travel time between the two arms. Since c is an invariant, if the length changes (regardless of the wavelength of the light) the travel time will vary. But the length of the arms isn't changing. That's been my point all along. But what did they detect then? How do you think they detected it with the precision that they did if it was undetectable? How did they get the mathematics of the prediction to match the experiment? If if was undetectable because the instruments contracted by the same degree as the space around it, as you said, how could they have gotten a false positive? They would have gotten a null result. Don't you see a contradiction there? There's no contradiction. The wave forms reported in the experiment are modeled and then data is matched to the modeled wave forms. Since both detectors got the same signals something was detected, yes, but if my objections here are correct that detection event wasn't from grav waves. We certainly shouldn't put the cart before the horse and argue that the premises must be correct b/c they found what they were looking for. If the premises aren't correct the experiment can't work, period. We then should look to what was detected in the apparent false positive. Again, I'm almost certainly wrong about my objections here, since I'm not even a physicist. But I do follow this stuff pretty closely and I still haven't seen any responses to my concerns that satisfactorily explain how the alleged detection event is taking place.
dimreepr Posted June 5, 2017 Posted June 5, 2017 (edited) But the length of the arms isn't changing. That's been my point all along. Since C can't change something has to, why not length? Edited June 5, 2017 by dimreepr
Lord Antares Posted June 5, 2017 Posted June 5, 2017 Since both detectors got the same signals something was detected, yes, but if my objections here are correct that detection event wasn't from grav waves. . You do realize that the experiment matched the prediction with those exact mathematics? So what did they detect which behaved exactly like gravitational waves were supposed to without being grav. waves themselves? Also, swansont is a professional physicst; I suspect he understood your point.
swansont Posted June 5, 2017 Posted June 5, 2017 But the length of the arms isn't changing. That's been my point all along. That's not what you were saying in your initial posts. I read the arguments to be that whatever "ruler" you might use also stretches and contracts, so you can't use a ruler to make the measurement. e.g. "I guess I'm still not seeing how the interferometer arm is detecting anything at all b/c if a gravitational wave is defined (as it is) as a ripple in spacetime itself, then anything in that ripple of spacetime will be distorted exactly the same degree to which spacetime is distorted." and "But again there is no detectable change in any dimension because the waves are literally waves of space, so anything occupying that space (whether it's falling masses or interferometer arms or a space gerbil) will be distorted by EXACTLY the same degree as space itself."
aramis720 Posted June 5, 2017 Author Posted June 5, 2017 Since C can't change something has to, why not length? Why would the invariance of c in this case lead to any length change? Again, the grav wave is waving space itself. The speed of light would be as affected by this wave as anything else in that space, as we know from various experiments finding the bending of light from gravitational masses like in 1919.
swansont Posted June 5, 2017 Posted June 5, 2017 Why would the invariance of c in this case lead to any length change? Again, the grav wave is waving space itself. The speed of light would be as affected by this wave as anything else in that space, as we know from various experiments finding the bending of light from gravitational masses like in 1919. The bending of light did not change the local speed
aramis720 Posted June 5, 2017 Author Posted June 5, 2017 That's not what you were saying in your initial posts. I read the arguments to be that whatever "ruler" you might use also stretches and contracts, so you can't use a ruler to make the measurement. e.g. "I guess I'm still not seeing how the interferometer arm is detecting anything at all b/c if a gravitational wave is defined (as it is) as a ripple in spacetime itself, then anything in that ripple of spacetime will be distorted exactly the same degree to which spacetime is distorted." and "But again there is no detectable change in any dimension because the waves are literally waves of space, so anything occupying that space (whether it's falling masses or interferometer arms or a space gerbil) will be distorted by EXACTLY the same degree as space itself." "Change" implies detectable change. So if the arms are waving to exactly the same degree as the space they occupy (as they must under the definitions of these terms in this context), then it's entirely undetectable. So the lengths of the arms are not changing. Which brings us back to my initial question: what is being measured? The bending of light did not change the local speed So how in this case is the speed of light being changed by grav waves, and in such a way that any change in speed would be detectable by an interferometer?
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