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.