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  1. 13 Aug '08 11:32
    Preon stars are stars with a density higher than that of a neutron star, but less than that of a black hole. It is thought to consist of preon matter, a kind of matter bilieved to be some strange sub-quark particles and leptons.

    Experiment has been made at the large particle accellerators but no preons has been found (yet).

    But - here is the interesting part! - the existance of preon stars can explain some observations of gravitational lensing of gamma rays.

    If a preon star has the same mass as the Earth, it would have the size of a tennis ball, but still be bigger than a black hole.

    I don't know much of preon matters and preon stars, perhaps others will fill in...?
  2. Standard member sonhouse
    Fast and Curious
    13 Aug '08 18:25 / 1 edit
    Originally posted by FabianFnas
    Preon stars are stars with a density higher than that of a neutron star, but less than that of a black hole. It is thought to consist of preon matter, a kind of matter bilieved to be some strange sub-quark particles and leptons.

    Experiment has been made at the large particle accellerators but no preons has been found (yet).

    But - here is the interes ...[text shortened]... ck hole.

    I don't know much of preon matters and preon stars, perhaps others will fill in...?
    I don't know why there would be anything special about gravitational lensing around prion stars v neutron stars, it wouldn't just bend gamma rays, it bends EVERYTHING, like X-rays, Neutrino's, gravity waves, protons, etc. Gamma rays are just a shorter wavelength version of radio or TV or light waves or Terahertz radiation. They all get bent to the same degree, with one exception (You are hearing this first from me, this is my work):
    Gravitational lensing is frequency dependent to a degree, it's a geometrical effect. Suppose you chart the gravitational lens 'strength' around an ordinary star like our sun. You probably are fully aware of the 1.75 arc seconds of bending a light path takes when it goes by the surface of the sun, right? Ok, now if you back off from the sun you find our sun is like a poor lens with an F stop of around 117,000:1. If you do the trig, it works out that the first focal point for the sun happens about 52 billion miles following the path of the radiation skimming the surface of the sun. Like if you had two parallel laser beams well focused so it doesn't spread out much and one skimmed the surface on the left side of the sun and the other skimmed the surface on the right side of the sun, the beams would be bent by that 1.75 seconds of arc and would meet some 83 billion Km out in space, like ten times the distance to pluto. But if those same laser beams were to be moved so the beams passed by the sun at 1 solar radii out they would meet at 4 times the distance, or about 320 billion Km out in space. So there is a continuous focal line between those two points and beyond of course.
    I mention this to illustrate my own little discovery in gravitational lensing: As the beams skim by the sun at further and further distances, say 2 solar radii, 3 solar radii, and so forth, there comes a point where the focus point would be a million light years out into space, in otherwords, a point of diminishing returns. Now think about a single electromagnetic wave. If it is the size of light, say half a micron, then the two laser beams gets pretty well focused. But look at what happens if the wavelength increases, lower frequencies. There comes a point at which some VERY low frequency, (one hertz would be a wavelength of 300,000 Km and one tenth of a hertz would have a wavelength of 3,000,000 Km)
    Think about that for a minute. A one micron size EM pulse would be bent at that famous 1.75 seconds of arc but a very low wavelength wave would go by the sun with no focus simply because it would straddle the whole sun and the gravitational lens effect may twist the wave in a phase change but the wave is simply too large to be focused. So that means that for every lens. be it a star, a black hole, a Prion star, neutron star, planet, whatever the mass concentration is, there will be a frequency that will not bend. That means if you had a variable wavelength generator and antenna system, you could in theory make an independent estimate of the mass of a star or whatever, by charting the frequencies that no longer lens. Get that? An independent way to measure mass. The idea would be, you have a generator and antenna on one side of a star, say 100 billion Km out in space and a receiver on the opposite side of the star at maybe the same distance, and vary the frequency starting at some very long wavelength and at the other side of the sun the receiver would be charting the strength of the radiation. So say starting at a frequency of 1/100 hertz, which is a wavelength of 30,000,000 km, you would note when the signal gets to the receiver, the strength would be so and so, the same as if the star was not there, but when you start raising the frequency, 1/10th hertz, 1 hertz, 10 hertz, etc., now at 10 hertz, 30,000 Km wavelength, the signal would show an increase in strength beyond that of what you would expect with no interloping mass, and that frequency would be unique for every mass. I have been in the process of writing up this idea for a few years but am too busy at work and with music and kids, etc. to finish it up. Do you get the main idea here?
  3. Standard member flexmore
    Quack Quack Quack !
    14 Aug '08 13:40 / 2 edits
    Originally posted by sonhouse
    I don't know why there would be anything special about gravitational lensing around prion stars v neutron stars, it wouldn't just bend gamma rays, it bends EVERYTHING, like X-rays, Neutrino's, gravity waves, protons, etc. Gamma rays are just a shorter wavelength version of radio or TV or light waves or Terahertz radiation. They all get bent to the same degr usy at work and with music and kids, etc. to finish it up. Do you get the main idea here?
    are you suggesting using the strength of the lensing effect to measure the mass of the lense? you seem to be saying more - but i don't quite get it.(the lensing effect is proportional to the mass of the lense and inversely proportional to the distance from the centre of the lense)
  4. 14 Aug '08 20:40
    Originally posted by sonhouse
    I don't know why there would be anything special about gravitational lensing around prion stars v neutron stars, it wouldn't just bend gamma rays, it bends EVERYTHING, like X-rays, Neutrino's, gravity waves, protons, etc. Gamma rays are just a shorter wavelength version of radio or TV or light waves or Terahertz radiation. They all get bent to the same degr ...[text shortened]... usy at work and with music and kids, etc. to finish it up. Do you get the main idea here?
    I think there is a dormant proposal of a space observatory that would be located in the area of the Neptunus orbit that could use the sun as a gravitational lens an thus enhance observations.
  5. Standard member sonhouse
    Fast and Curious
    15 Aug '08 05:35 / 1 edit
    Originally posted by FabianFnas
    I think there is a dormant proposal of a space observatory that would be located in the area of the Neptunus orbit that could use the sun as a gravitational lens an thus enhance observations.
    there won't be any gravitational lensing from the sun anywhere in the solar system proper. Do the math: Take a right triangle, long skinny one, base 704,000 Km, (radius of the sun) and the angle of 1.75 seconds of arc for the angle, the 90 degree one is from the center of the sun as locus and that line extends to infinity. How far out does the angled line go to meet the center of sun line? That is the first focal point. I wish I could draw it out but you should get the picture. Do the trig!
    One addendum: the angle is actually 90 degrees MINUS 1.75 seconds of arc. That is the inside angle of the triangle where the 90 degree angle is at the center of the sun.
  6. 15 Aug '08 07:56
    Originally posted by sonhouse
    there won't be any gravitational lensing from the sun anywhere in the solar system proper. Do the math: Take a right triangle, long skinny one, base 704,000 Km, (radius of the sun) and the angle of 1.75 seconds of arc for the angle, the 90 degree one is from the center of the sun as locus and that line extends to infinity. How far out does the angled line g ...[text shortened]... That is the inside angle of the triangle where the 90 degree angle is at the center of the sun.
    What distance from the sun is needed to be able to use the sun as a gravitational lens?
  7. Standard member sonhouse
    Fast and Curious
    15 Aug '08 09:27 / 1 edit
    Originally posted by FabianFnas
    What distance from the sun is needed to be able to use the sun as a gravitational lens?
    Just solve that triangle! I'll give you a hint: Its 117,000 times 704,000Km.
    That said, that is only the first focal line. The lens is not a very good one in terms of glass lenses. That is because the further from the center of the sun you skim your light beam, the less it gets bent, so if the bend at one solar radii (just skimming the surface) is bent at 1.75arc seconds, at 2 solar radii, the angle is now 1.75/2 or 0.875 arc seconds. So that means if you have two laser beams on the opposite sides of the sun skimming at one radii out, 704,000 Km above the surface, the distance to where those beams meet will be four times the distance that two beams skimming just above the surface will meet.
  8. Standard member sonhouse
    Fast and Curious
    15 Aug '08 09:36
    Originally posted by flexmore
    are you suggesting using the strength of the lensing effect to measure the mass of the lense? you seem to be saying more - but i don't quite get it.(the lensing effect is proportional to the mass of the lense and inversely proportional to the distance from the centre of the lense)
    Yes but I am saying something new here: Radiation of very low frequencies will not be bent by the sun sized lens, simply because the wave is just too large to get focused. If the wavelength of a wave is 100,000,000 Km, a very low frequency indeed, it will pass by the sun pretty much unscathed by the lens. Only as the wavelength gets closer in size to the sun does the lens start focusing that energy. So if the wavelength is now say, 1000 Km, then you can see the whole wave can fit above the surface of the sun and get bent as a whole.
    So that means if you have a very low frequency RF generator on one side of a mass and a reciever on the other side, way apart in distance, say a 1000AU or so, then as you sweep up in frequency you will find a frequency that starts to get gain from the lens, That pegs the mass.
  9. 15 Aug '08 10:12 / 1 edit
    Originally posted by sonhouse
    Yes but I am saying something new here: Radiation of very low frequencies will not be bent by the sun sized lens, simply because the wave is just too large to get focused. If the wavelength of a wave is 100,000,000 Km, a very low frequency indeed, it will pass by the sun pretty much unscathed by the lens. Only as the wavelength gets closer in size to the su ...[text shortened]... n frequency you will find a frequency that starts to get gain from the lens, That pegs the mass.
    …If the wavelength of a wave is 100,000,000 Km, a very low frequency indeed, it will pass by the sun pretty much unscathed by the lens. .…

    Are you talking here about ‘diffraction‘ here? (http://en.wikipedia.org/wiki/Diffraction)
    If so, I am no expert on this but I know diffraction isn’t dependent on gravity and I don’t think gravitational lensing works by causing diffraction so I think you could be confusing these two different kinds of phenomena up? -if so, then, I could be wrong but; gravitational lensing should work for all frequencies of electromagnetic radiation even if the wavelength of radiation is much greater than the diameter of the object doing the lensing? Can a real expert here please either confirm or refute this because I admit I am not certain of this.
  10. Standard member sonhouse
    Fast and Curious
    15 Aug '08 10:52
    Originally posted by Andrew Hamilton
    [b]…If the wavelength of a wave is 100,000,000 Km, a very low frequency indeed, it will pass by the sun pretty much unscathed by the lens. .…

    Are you talking here about ‘diffraction‘ here? (http://en.wikipedia.org/wiki/Diffraction)
    If so, I am no expert on this but I know diffraction isn’t dependent on gravity and I don’t think gravitational ...[text shortened]... real expert here please either confirm or refute this because I admit I am not certain of this.[/b]
    This is exactly my conjecture, that wavelengths way longer than the size of the lensing body will pass by without being focused. Look at like this: suppose you have an RF wave whose wavelength is the size of the solar system, how much could it possibly be bent by the sun?
  11. Standard member flexmore
    Quack Quack Quack !
    16 Aug '08 11:36 / 4 edits
    Originally posted by sonhouse
    This is exactly my conjecture, that wavelengths way longer than the size of the lensing body will pass by without being focused. Look at like this: suppose you have an RF wave whose wavelength is the size of the solar system, how much could it possibly be bent by the sun?
    This sounds like an experiment to be performed ... find something that gives long wavelengths and try to measure how much it is bent ... you do know of course that this would be a major rewriting of physics as i know it ... gravity affects EVERYTHING as far as most of us know.

    You are proposing that a long wavelength wave will be unaffected ... worth a shot i guess.
  12. 16 Aug '08 11:55
    Originally posted by flexmore
    This sounds like an experiment to be performed ... find something that gives long wavelengths and try to measure how much it is bent ... you do know of course that this would be a major rewriting of physics as i know it ... gravity affects EVERYTHING as far as most of us know.

    You are proposing that a long wavelength wave will be unaffected ... worth a shot i guess.
    If gravitation is bent space, and not an attractive force, then it's wavelength cannot be affected, can it?
  13. Standard member flexmore
    Quack Quack Quack !
    16 Aug '08 12:02 / 2 edits
    Originally posted by Andrew Hamilton
    Are you talking here about ‘diffraction‘ here? (http://en.wikipedia.org/wiki/Diffraction)
    If so, I am no expert on this but I know diffraction isn’t dependent on gravity and I don’t think gravitational real expert here please either confirm or refute this because I admit I am not certain of this.[/b]
    I think the optical analogy is more like dispersion
    http://en.wikipedia.org/wiki/Dispersion_%28optics%29
    the longer wavelengths (red) are turned less as they move through a lense.

    sonhouse suggests that different frequencies might be affected differently as they move through a differentially adjusted thickness of space ...
  14. Standard member sonhouse
    Fast and Curious
    16 Aug '08 22:27
    Originally posted by flexmore
    I think the optical analogy is more like dispersion
    http://en.wikipedia.org/wiki/Dispersion_%28optics%29
    the longer wavelengths (red) are turned less as they move through a lense.

    sonhouse suggests that different frequencies might be affected differently as they move through a differentially adjusted thickness of space ...
    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 picture a very low frequency EM, a LOT bigger a wave packet than the solar system and you can see as it propagates through space there is only a small distortion of the wavefront which could be described in terms of phase shift of a portion of the wavefront but will not effect the direction of travel of the main wave simply because the the volume of space of the wave is so much larger than the volume of bent space around the mass, star, black hole, whatever.
    In black holes, for instance, the mass is incredible but the size is small and the volume of space effected is the same as if it were a star with that much mass but there still will be a frequency below which that wavefront will not bend considering the wave as a whole. There will be a bending of the EM lines of force and maybe the generation of higher frequency components, frequencies that may show up on examination of the wavefront as it passes by the gravitational mass, and those sub-wavefronts may diverge from the path of the main wavefront.
  15. Standard member sonhouse
    Fast and Curious
    16 Aug '08 23:47
    That leads me to another line of reasoning:
    If "harmonics" of the main wave are generated, there may be yet another way to measure the mass of a lens, by detecting how much energy is lost of the main wave. The energy to make the harmonic has to come from the energy of the main wave so if you can very accurately measure the strength of the main wave it may show a loss, that loss going into the generation of higher frequency components of the main wave. They wouldn't be actual numerical harmonics, like doubling or quadrupling the original wavelength, but a new frequency completely unrelated in a harmonic way to the original wavefront frequency. At least that's what my present line of thinking is leading to.