1. Standard memberflexmore
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    18 Aug '08 04:291 edit
    Originally posted by sonhouse
    Visualize gravity from a mass as a dimple on a flat sheet of rubber, you all have seen that analogy, right? Well the bending of space does go to infinity but the area which translates to a volume in real space has definite limits as far as how much the bend of space can effect adjacent masses.
    That also applies to the frequency of EM radiation. Just pictur ...[text shortened]... he gravitational mass, and those sub-wavefronts may diverge from the path of the main wavefront.
    I am still struggling to catch what you say:

    If i imagine a very wide wavefront going past the sun then yes .... most of the wave would go around the sun undisturbed ... and just a short/insignificant section will be held back by the sun's gravity.

    But if it is narrow wavefront and must go near the sun, then regardless of the wavelength it seems to me that the geodesic(shortest path) will be out further from the sun ... the relativistic "bending" of space means there is simply a larger distance to be crossed if the laser goes in the traditional "straight line" near the sun than if it goes around a little further out. If it can go near the sun's surface and keep up with the other rays on slightly wider paths then it needs to be going faster than the speed of light ... are you suggesting that!
  2. Subscribersonhouse
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    18 Aug '08 05:111 edit
    Originally posted by flexmore
    I am still struggling to catch what you say:

    If i imagine a very wide wavefront going past the sun then yes .... most of the wave would go around the sun undisturbed ... and just a short/insignificant section will be held back by the sun's gravity.

    But if it is narrow wavefront and must go near the sun, then regardless of the wavelength it seems to me aths then it needs to be going faster than the speed of light ... are you suggesting that!
    Not at all, not suggesting anything material goes faster than C.
    But the bending of a light's path in response to a large mass is a geometric thing. I went to Big Al's original equations and found, to my suprise, the bending of space around a star is not like gravity which follows the inverse square law but a simple inversion, that is to say, the bending of light just grazing the surface of the sun, defined as X, that distance is one solar radii from the center. So the bending of light at 2 solar radii is not 1/4 as I had assumed but simply 1/2 so the bend at 3 solar radii (2,112,0000 Km from the surface) is 1/3 that of a beam skimming the surface and so forth.
    That means there is a diminishing effect of the bending as you go away from the surface of the sun and so it's obvious if a very long wavelength EM wave goes by the sun the most that can happen is the portion of it skimming by the surface or within a few million Km will be kind of torn from the original wave, and I say, thus, generating a new waveform going off at a slight angle to the main wave. It's not like an EM wave going by, say, a high intensity magnetic field where the polarization only is effected, both pieces of the EM wave, the electric and the magnetic component are effected simultaneously, thus making a new wave of much smaller wavelength and bent by the average bending angle at the average of the altitudes above the sun.
    So when you look at shorter and shorter waves, say, one meter wavelength, 300Mhz, that is so short compared to the size of the sun and so it will be bent as a whole, and if it was a wavefront that was as big as the solar system, say, from being generated at some interstellar distance, then the entire wavefront will bend in around the sun and meet at some distance out, thus being amplified. That distance for say, a one micron laser beam skimming the surface, and another on the other side of the sun 180 degrees away from the first laser beam, those beams will meet at around 84 billion Km from the sun and those at one solar radii out, 700,000 km above the surface, will meet at four times that distance, or around 330 billion Km from the sun (following the path of said laser beams as it leaves the vacinity of the sun). That said, I think there is some shearing of the wave when it comes together because of the gravity differential from the side of the beam nearer the sun vs the side of the beam away from the sun. That effect is more pronounced the lower the frequency, it is reacting to gravitational tidal effects, differences in gravity from one altitude to another, small as it would be at one meter of wavelength for instance. I am still trying to wrap my head around that smearing effect, what it means if you were an EM wave going by the sun. I follow it in my head and it seems the front part of the wave gets together first and the wave away from the sun meeting later, so it is smeared sideways, ( the sun, at the surface, is equivalent to a glass lens of F-stop of 117,000:1) so at one meter, the bottom of the wave would focus 117,000 meters before the top of the wave. Not sure what that means in terms of what you would actually measure could you be there to accurately track a one meter beam as it skims by the sun.

    One other effect I just noticed: Time flows slower in a gravity well so it takes a longer time for a wave to go past the sun, that is to say, its geodesic will be a longer path than one passing further from the sun. So maybe that effect cancels out the differential gravity smearing I was looking at so maybe there is no smearing.
  3. Standard memberflexmore
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    18 Aug '08 07:261 edit
    Originally posted by sonhouse
    ... One other effect I just noticed: Time flows slower in a gravity well so it takes a longer time for a wave to go past the sun, that is to say, its geodesic will be a longer path than one passing further from the sun. So maybe that effect cancels out the differential gravity smearing I was looking at so maybe there is no smearing.
    Yes this is the crux of it ... time flows slower down near a heavy planet (equivalent to saying space is thicker, which is simpler conceptually i think) .... the lensing is not a feature of the wave passing near the massive lensing object - it is a simple geometrical result from the nature of the space itself: the wave moves at a constant speed c and will move along the quickest possible path as it propagates, this quickest path is one that bends slightly around the gravity well.
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    18 Aug '08 09:141 edit
    Originally posted by flexmore
    Yes this is the crux of it ... time flows slower down near a heavy planet (equivalent to saying space is thicker, which is simpler conceptually i think) .... the lensing is not a feature of the wave passing near the massive lensing object - it is a simple geometrical result from the nature of the space itself: the wave moves at a constant speed c and will m ...[text shortened]... le path as it propagates, this quickest path is one that bends slightly around the gravity well.
    I would love to find out if a wavefront that starts out in sync, like a laser beam, but one coming from so far away the wavefront is a lot bigger than the sun, and then to go to the first set of focal points and see if the resultant wave is still in sync. It will be for sure amplified but will it still be coherent?
    I think it may not be, I bet there will be differences because the sun is slightly flattened by it rotation so I expect a beam going over the poles will have a slightly different bend than the beams going over the equator of the sun. They would at least be in different altitudes compared to the average radius.
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    18 Aug '08 10:54
    Originally posted by sonhouse
    I would love to find out if a wavefront that starts out in sync, like a laser beam, but one coming from so far away the wavefront is a lot bigger than the sun, and then to go to the first set of focal points and see if the resultant wave is still in sync. It will be for sure amplified but will it still be coherent?
    I think it may not be, I bet there will b ...[text shortened]... quator of the sun. They would at least be in different altitudes compared to the average radius.
    Is the sun flattened? Nut much, is it?

    The sun has an average period of revolution of 25 days, give or take. This slow rotation doesn't flatter the sun very much, does it?
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    18 Aug '08 12:40
    Originally posted by FabianFnas
    Is the sun flattened? Nut much, is it?

    The sun has an average period of revolution of 25 days, give or take. This slow rotation doesn't flatter the sun very much, does it?
    It's an oblate spheroid, somewhat skinnier through the poles than the equator. I will have to find a link later, I am still at work and we have a company meeting in a few minutes.
  7. Standard memberflexmore
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    18 Aug '08 23:471 edit
    Originally posted by sonhouse
    It's an oblate spheroid, somewhat skinnier through the poles than the equator. I will have to find a link later, I am still at work and we have a company meeting in a few minutes.
    from wikipedia: Sun .... Flattening ... equalling the relative difference between its equatorial radius, , and its polar radius,

    0.000009
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    19 Aug '08 02:081 edit
    Originally posted by flexmore
    from wikipedia: Sun .... Flattening ... equalling the relative difference between its equatorial radius, , and its polar radius,

    0.000009
    That seems to be about a 6,300 Km difference between north and south pole vs east and west pole🙂
    The next step is to figure out how much gravitational difference there is between poles and equator. 9 parts in a million geometrically is what percent of solar gravity?
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    19 Aug '08 04:38
    Squash that, my calculator slipped a couple of decades🙂
    Make that 6.3 Km difference between poles and equator. Funny what a decimal point or 3 can do🙂
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    23 Aug '08 21:29
    Just a couple thoughts. I'd think it would be harder to see stuff behind the sun. ;o

    Also if a electromagnetic wave has a higher frequency the energy is higher and thus it's proportionate "mass" should be higher. And by that i mean, higher energy waves should be affected more by gravitation.

    Also, how do you know you are looking though a lens if you don't know it's there? You'd need a second observation or at least a reliable indication that what you see is distorted or sumsuch.
  11. Subscribersonhouse
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    28 Aug '08 13:343 edits
    Originally posted by Tera
    Just a couple thoughts. I'd think it would be harder to see stuff behind the sun. ;o

    Also if a electromagnetic wave has a higher frequency the energy is higher and thus it's proportionate "mass" should be higher. And by that i mean, higher energy waves should be affected more by gravitation.

    Also, how do you know you are looking though a lens if you don't observation or at least a reliable indication that what you see is distorted or sumsuch.
    Well that is the point of the gravitational lens, if you go to the first of an infinite # of focal points you will begin to see a wavefront that passes by the surface of the sun, the stuff that actually hits the sun would of course not be lensed. So imagine a wavefront of energy coming at the sun and you are in line with the direction of travel of that energy but on the opposite side of the sun and about 60 billion miles away from the sun on that same line, that going back to the sun, would intersect the sun dead center, right? So there you are, 60 billion miles, 100 billion Km or so, at the first focal point of the focal line, and a portion of that waveform that passes by the surface of the sun gets all squished into one small volume which I have used the unit of one cubic wavelength for my analysis, which by definition, makes the sun a lens (not a perfect lens because the focal point of a regular lens is not like the sun, the sun has a focal line, as the wave front from progressively higher up from the surface focuses further and further from the sun.
    My point is, at a sufficiently long wavelength, there comes a wavelength that is not going to focus simply because it's too big, say if the wavelength is 10 million miles or 16 million Km long, an RF wave that takes about one minute to go from crest to crest, about 0.02 Hz (!)
    then only about ten percent of it can skim the surface of the sun so the wave in total cannot be focused. But suppose we look at a wave of 60 Hz, the standard AC power frequency in a large portion of the world, which is a wavelength of about 5000 Km or 50 Hz, a wavelength of 6000 Km, those waves would be focused maybe down to one cubic wavelength at around 60 billion miles from the sun. And of course higher frequencies, shorter wavelengths will have that many more cubic wavelengths coming together so there is more energy focused at higher energy photons.
    If anyone is interested I can go into more detail of what I figured out about this phenomena, it helps me to put it together in a paper if I can get it out to people in a digestible form and maybe people can find flaws in my arguments to keep me from embarrassing myself in print🙂
    One thing about a real article about this is I can include drawings, illustrations and such which are hard to put into words, a picture being worth at LEAST a thousand words! I do have a LOT more to say about this stuff including some mindblowing images which at the present are in my head, I need the help of a graphics artist to get it down to a PDF or powerpoint image or whatever, JPEG...
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