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![]() "Henri Wilson" HW@.... wrote in message ... On Wed, 25 Apr 2007 09:16:22 -0600, Art Deco wrote: Androcles wrote: "bz" wrote in message 9.198.139... At any speed below c, the grating will 'see' longer wavelengths as it goes away from the source. Photons do not have wavelength anymore than cars so. Roads have wavelengths, cars have frequency. hahahahaahahahahahahahahahahhahahahahahaha Good one, thanks. Leave Andro alone. He's not as stupid as some others here. There was an obvious typo in my last message. 'c' should have been 'v'. This is what I was asking: The equations for gratings include 'wavelength' and not light speed or 'frequency'. If a grating is used to inspect light coming from a star moving at v towards us, then the diffracted angles are indicative of the relative speed between the star and the grating. If the grating is now moved away at 'v', why should those angles change? Certainly the movement of the grating has not altered the light's wavelength in any way. I smell a flaw in a theory somewhere. The flaw is that you haven't applied ballistic theory yet, you are assuming the conventional analysis. If the light arrives at c and you move the grating at v then the relative speed is c-v. The wavelength is the same so the frequency drops. If the reflected signal is emitted at c relative to the grating, then the reflected wavelength is increased to c/(c-v) times the incident wavelength. If on the other hand the reflected wave moves at c-v relative to the grating then the wavelength stays the same as for the static grating, but of course the speed is different. In both cases you need to then work out the angle where the reflections are in phase. George |
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