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"Androcles" wrote in message ...
"Jim Greenfield" wrote in message om... | Paul Lawler wrote in message .125.202... | "Androcles" wrote in news:TmuVc.3953 | : | | | Risking showing my naivitie, what is REALLY being discussed- a "one | of" change in gravitational field strength (pulse/wave), or a SERIES | of equal strength waves emanating from a static body? | | Jim G | c'=c+v Jim, this discussion is not about the existence or non-existence of gravity waves, but about their amplitude being great enough to be detectable. Simply spinning the Earth in the lunar gravity produces tides and when we include solar gravity we have neap and spring tides. If the lunar orbit were highly elliptical we'd have higher tides at perigee than at apogee. Thus we would have a detectable gravity 'wave'; they do exist, and can be detected. Yep. As I told Old Man, I weigh less with moon above than 12 hours later. So the moon produces a gravity wave of frequency 1/24hour, right? LIGO, however, is about detecting a gravitational field from a supernova at a distance of a kiloparsec = 3260 light years, where some quantity of matter is completely converted to energy (E= mc^2) and the resultant gravity field is reduced. That would be a step pulse. Or it could be the field from a pulsar in orbit about a neighbour that is periodically approaching and receding from us, and that would be a sinusoidal wave. So the answer to your question is : both. However, the supernova (which may produce a pulsar as a remnant) is the greater. If a star explodes, the "center of gravity" of that star remains in the same place afterward. As I intuitively feel that the particles which comprise EMR DO exert gravitational force themselves, therefore no pulse/wave, as the star still "acts" the same after exploding. However, as the EMR dissipates, opening up the angle from us (from a point to an expanding cloud), there should be a gradual decline in field strength towards/from that center of gravity. Undetectable change until the outer ring of the burst is over a significant arc to us = no wave (detectable = Ligo wont work for Sn If you want to express the problem mathematically: let delta be the smallest amplitude detectable by the instrument used. Let a pulse (or wave) of amplitude A be emitted at 0 and the amplitude at r where the instrument is placed be A/r^2 = delta. Then the amplitude at A/(r+epsilon)^2 (epsilon 0) is less than delta and is not detectable. LIGO has a real delta, so from that estimate the greatest imaginable A and calculate A/r^2 = delta r^2/A = 1/delta r = sqrt(A/delta) Androcles. Don't math me :-( Jim G c'=c+v |
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"Jim Greenfield" wrote in message m... | "Androcles" wrote in message ... | "Jim Greenfield" wrote in message | om... | | Paul Lawler wrote in message | .125.202... | | "Androcles" wrote in news:TmuVc.3953 | | : | | | | | | Risking showing my naivitie, what is REALLY being discussed- a "one | | of" change in gravitational field strength (pulse/wave), or a SERIES | | of equal strength waves emanating from a static body? | | | | Jim G | | c'=c+v | | Jim, this discussion is not about the existence or non-existence of gravity | waves, but about their amplitude being great enough to be detectable. Simply | spinning the Earth in the lunar gravity produces tides and when we include | solar gravity we have neap and spring tides. If the lunar orbit were highly | elliptical we'd have higher tides at perigee than at apogee. Thus we would | have a detectable gravity 'wave'; they do exist, and can be detected. | | Yep. As I told Old Man, I weigh less with moon above than 12 hours | later. So the moon produces a gravity wave of frequency 1/24hour, | right? Relatively speaking, yes. Of course it is the spin of the Earth that actually changes your position and produces the wave, you are simply moving in a fixed field, the strength of which is a function of distance. 1/24hour may be misconstrued, though. 1 cycle every 24 hours is not the same as 1 cycle in 2 minutes and 30 seconds, which is 1/24th of an hour. To be a little more precise, the moon orbits the Earth 13 times a year (but not exactly) and the Earth orbits the sun. 24 hours is the time from noon to noo n (sun at zenith) but a distant star moves about 4 minutes a day from midnight to midnight. Which star is overhead at midnight depends on your longitude. But rougly speaking, the gravity wave has a frequency of one cycle per day. | | LIGO, however, is about detecting a gravitational field from a supernova at | a distance of a kiloparsec = 3260 light years, where some quantity of matter | is | completely converted to energy (E= mc^2) and the resultant gravity field is | reduced. That would be a step pulse. | Or it could be the field from a pulsar in orbit about a neighbour that is | periodically approaching and receding from us, and that would be a | sinusoidal wave. So the answer to your question is : both. However, the | supernova (which may produce a pulsar as a remnant) is the greater. | | If a star explodes, the "center of gravity" of that star remains in | the same place afterward. Yes. There may be a shell of matter that leaves the star, but momentum is conserved. The same is true for a rocket. We see the rocket accelerate, but the exhaust is flying away in the opposite direction and the centre of mass of the combined exhaust and rocket only moves with its original velocity before the engine was fired. The combined momentum of the entire Universe is zero. | As I intuitively feel that the particles | which comprise EMR DO exert gravitational force themselves, therefore | no pulse/wave, as the star still "acts" the same after exploding. | However, as the EMR dissipates, opening up the angle from us (from a | point to an expanding cloud), there should be a gradual decline in | field strength towards/from that center of gravity. Undetectable | change until the outer ring of the burst is over a significant arc to | us = no wave (detectable = Ligo wont work for Sn Intuition is a dangerous tool. I don't recommend it. Better to prove a theorem mathematically and then see if intuition agrees. Thunder and lightning arrive at different times, and a child's intuition is that they are seperate events. An adult sees it differently. Until Copernicus, intuition told us the Earth is at the centre if the universe. After all, we see the sun cross the sky daily, it MUST be going around us. With greater knowledge we revise our view that we are turning toward and away from the sun. Never trust intuition, it is bane of science and the boon of religion. | | If you want to express the problem mathematically: let delta be the smallest | amplitude detectable by the instrument used. | Let a pulse (or wave) of amplitude A be emitted at 0 and the amplitude at r | where the instrument is placed be A/r^2 = delta. | Then the amplitude at A/(r+epsilon)^2 (epsilon 0) is less than delta and | is not detectable. | LIGO has a real delta, so from that estimate the greatest imaginable A and | calculate | A/r^2 = delta | r^2/A = 1/delta | r = sqrt(A/delta) | | Androcles. | | Don't math me :-( | | Jim G | c'=c+v -- don't math me. :-) Androcles |
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"Androcles" wrote in message ...
"Jim Greenfield" wrote in message m... | "Androcles" wrote in message ... | "Jim Greenfield" wrote in message | om... | | Paul Lawler wrote in message .125.202... | | "Androcles" wrote in news:TmuVc.3953 | | : | | | | | | Risking showing my naivitie, what is REALLY being discussed- a "one | | of" change in gravitational field strength (pulse/wave), or a SERIES | | of equal strength waves emanating from a static body? | | | | Jim G | | c'=c+v | | Jim, this discussion is not about the existence or non-existence of gravity | waves, but about their amplitude being great enough to be detectable. Simply | spinning the Earth in the lunar gravity produces tides and when we include | solar gravity we have neap and spring tides. If the lunar orbit were highly | elliptical we'd have higher tides at perigee than at apogee. Thus we would | have a detectable gravity 'wave'; they do exist, and can be detected. | | Yep. As I told Old Man, I weigh less with moon above than 12 hours | later. So the moon produces a gravity wave of frequency 1/24hour, | right? Relatively speaking, yes. Of course it is the spin of the Earth that actually changes your position and produces the wave, you are simply moving in a fixed field, the strength of which is a function of distance. 1/24hour may be misconstrued, though. 1 cycle every 24 hours is not the same as 1 cycle in 2 minutes and 30 seconds, which is 1/24th of an hour. To be a little more precise, the moon orbits the Earth 13 times a year (but not exactly) and the Earth orbits the sun. 24 hours is the time from noon to noo n (sun at zenith) but a distant star moves about 4 minutes a day from midnight to midnight. Which star is overhead at midnight depends on your longitude. But rougly speaking, the gravity wave has a frequency of one cycle per day. More precise !!!!,how for goodness sake did you lot manage to get control of astronomy I will never know but the trail leads back to Newton. Poor Isaac,gave you a system that competes with the Tychonian quasi-geocentric system based on Flamsteed.With concentration now turning to Copernicus and the attempt by Brahe to return to a quasi-geocentric view,it is only a matter of time before Newton's alternative version rears its ugly head. "PH?NOMENON IV. That the fixed stars being at rest, the periodic times of the five primary planets, and (whether of the sun about the earth, or) of the earth about the sun, are in the sesquiplicate proportion of their mean distances from the sun. http://members.tripod.com/~gravitee/phaenomena.htm This is how that statement works out graphically - http://www.absolutebeginnersastronomy.com/sidereal.gif Even if you ever get round to reading 'The Book Nobody Read' you are unlikely to pick up on the idea of heliocentric motion taken from the center of the Earth's orbit rather than the Sun's center,poor Isaac in the best tradition of mathematicians fudged what he could not understand. Pity you can't read astronomical language and pity it took me a long time to understand that,the ability is a gift rather than acquired.The recent recovery of a more accurate history of investigative techniques in geology, astronomy,clockmaking and multiple other avenue paves the way for setting off the awful mistake Newton made in adopting his particular nasty quasi-geocentric outlook. You live your life living with Newton's personal revenge but defiance until death is sure a lousy reward for anyone,aetherist,relativist and bottom line.Simply stated,Newton got it wrong and in a very fundamental way via Flamsteed. http://hypertextbook.com/facts/1999/JennyChen.shtml That is how close your entire discipline is to oblivion. | | LIGO, however, is about detecting a gravitational field from a supernova at | a distance of a kiloparsec = 3260 light years, where some quantity of matter | is | completely converted to energy (E= mc^2) and the resultant gravity field is | reduced. That would be a step pulse. | Or it could be the field from a pulsar in orbit about a neighbour that is | periodically approaching and receding from us, and that would be a | sinusoidal wave. So the answer to your question is : both. However, the | supernova (which may produce a pulsar as a remnant) is the greater. | | If a star explodes, the "center of gravity" of that star remains in | the same place afterward. Yes. There may be a shell of matter that leaves the star, but momentum is conserved. The same is true for a rocket. We see the rocket accelerate, but the exhaust is flying away in the opposite direction and the centre of mass of the combined exhaust and rocket only moves with its original velocity before the engine was fired. The combined momentum of the entire Universe is zero. | As I intuitively feel that the particles | which comprise EMR DO exert gravitational force themselves, therefore | no pulse/wave, as the star still "acts" the same after exploding. | However, as the EMR dissipates, opening up the angle from us (from a | point to an expanding cloud), there should be a gradual decline in | field strength towards/from that center of gravity. Undetectable | change until the outer ring of the burst is over a significant arc to | us = no wave (detectable = Ligo wont work for Sn Intuition is a dangerous tool. I don't recommend it. Better to prove a theorem mathematically and then see if intuition agrees. Thunder and lightning arrive at different times, and a child's intuition is that they are seperate events. An adult sees it differently. Until Copernicus, intuition told us the Earth is at the centre if the universe. After all, we see the sun cross the sky daily, it MUST be going around us. With greater knowledge we revise our view that we are turning toward and away from the sun. Never trust intuition, it is bane of science and the boon of religion. | | If you want to express the problem mathematically: let delta be the smallest | amplitude detectable by the instrument used. | Let a pulse (or wave) of amplitude A be emitted at 0 and the amplitude at r | where the instrument is placed be A/r^2 = delta. | Then the amplitude at A/(r+epsilon)^2 (epsilon 0) is less than delta and | is not detectable. | LIGO has a real delta, so from that estimate the greatest imaginable A and | calculate | A/r^2 = delta | r^2/A = 1/delta | r = sqrt(A/delta) | | Androcles. | | Don't math me :-( | | Jim G | c'=c+v -- don't math me. :-) Androcles |
#67
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On Tue, 24 Aug 2004 05:46:06 GMT, (Eric Flesch) wrote:
On 23 Aug 2004 17:36:06 -0700, (Jim Greenfield) wrote: ..... But...but... shouldn't we find out what `gravity' really is before we go too much farther. Newton - a force, Einstein - an alteration in spacetime around a mass. But why does one mass move toward another mass against its own inertia. If it is `falling into another masses spacetime `hole' - why? Don't say it's a force because then we are back to Newton, who admitted he didn't know why the observation was valid. Just a simple answer if possible: what is gravity? We have all observed the action... now - why? what is a `graviton'? what is a `gravitational wave'? what is quantum gravity? |
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* dkomo:
I found an old PBS documentary on VHS from 1991 called _The Astronomers_ at the local public library. One of the programs in the series was "Waves of the Future" about gravitational waves. In the program Kip Thorne was shown making a bet with one of his collaborators on gravity wave theory that these waves would positively be detected by 2000. I found this both humorous and a touch sad. The program described some of the early planning for LIGO (Laser Interferometer Gravitational Wave Observatory). Curious, I went to the LIGO web site to see what was going on. I found nothing of substance there -- just a lot of slick PR. So my question is, what are the prospects that gravity waves will be detected anytime soon? Is LIGO still having technical problems or what? It is now 2004, after all. Other detection labs are being built around the world. Are these labs going to have any better luck? From the chronology availble at the web-site, url: http://www.ligo.caltech.edu/LIGO_web/PR/scripts/chrono.html, it seems nothing has happened after August 2002 when the first "scientific operation of all three interferometers in S1" was run (note: without me knowing anything about it I'd say 3 interferometers are the least number needed to establish any kind of direction for a short-duration signal, but then it would still be open which of the two halves of the sky it came from). I remember SciAm had an article essentially saying between the lines that the noise levels are too high to expect any useful (non-ambigous) result. So a _detection_ would probably be a major event, contradicting theory... Also, what are people's opinions about gravity waves? Is it possible that these are a scientific dead end like the decay of the proton turned out to be? Anything is possible, but gravity waves are associated with some extremely precise predictions of increasing rotation rate of pulsars. Hulse and Taylor got the Nobel Prize for that work in 1993. So it seems gravity waves are real, or else the universe is playing us Yet Another magician's joke! If gravity waves are never detected, what are the implications for the general theory of relativity? Serious. -- A: Because it messes up the order in which people normally read text. Q: Why is it such a bad thing? A: Top-posting. Q: What is the most annoying thing on usenet and in e-mail? |
#69
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Intuition is a dangerous tool. I don't recommend it. Better to prove a theorem mathematically and then see if intuition agrees. Thunder and lightning arrive at different times, and a child's intuition is that they are seperate events. An adult sees it differently. Until Copernicus, intuition told us the Earth is at the centre if the universe. After all, we see the sun cross the sky daily, it MUST be going around us. With greater knowledge we revise our view that we are turning toward and away from the sun. Never trust intuition, it is bane of science and the boon of religion. But aren't you using intuition to discard relativistic addition of velocities in your c'=c+v (or is that c=c'+v?). daveL |
#70
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"Dave" wrote in message ... | | Intuition is a dangerous tool. I don't recommend it. Better to prove a | theorem mathematically and then see if intuition agrees. Thunder and | lightning arrive at different times, and a child's intuition is that | they are seperate events. An adult sees it differently. Until | Copernicus, intuition told us the Earth is at the centre if the | universe. After all, we see the sun cross the sky daily, it MUST be | going around us. With greater knowledge we revise our view that we | are turning toward and away from the sun. Never trust intuition, it | is bane of science and the boon of religion. | | But aren't you using intuition to discard relativistic addition of | velocities in your c'=c+v (or is that c=c'+v?). | | | daveL You'd really need to ask Jim that. You have my response to him (above) confused with his statement x' = c+v. However, I will state that the vector addition of velocities, c+v, is used by Einstein in his equation ½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v)) following which he clearly states "But the ray moves relatively to the initial point of k, when measured in the stationary system, with the velocity c-v, so that x'/(c-v) = t." Reference : http://www.fourmilab.ch/etexts/einstein/specrel/www/ So you'd need to ask Einstein where he gets his "velocity c-v" from and why he is not using (c+w)/(1+w/c) = c in that equation. For myself, I'm wondering why the idiot couldn't make up his tiny ****ing mind. Maybe 'w' and 'v' are too confusing for the moron. Oops, sorry... didn't mean to offend your tin god... Oh, wait! Yes I did. Androcles. |
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