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Author Topic:   Tired Light
Melchior
Inactive Member


Message 26 of 309 (191893)
03-16-2005 9:28 AM
Reply to: Message 25 by Percy
03-16-2005 9:06 AM


Re: Millisecond Pulsars
Yes, via something called the Compton effect. It's commonly used as a theoretical basis for experiments with Xrays, so you might have encountered it if you attended basic physics in university.
Essentially, photons can 'collide' with electrons. When this happens, some of the photons energy transfers into kinetic energy for the electron.
The effects are that the path of the photon is changed (you can check up the formulas for the change in angle on the net) and it's frequency is lowered.
So there is still the problem with shattering, although energetic light DO become tired by interacting with free electrons.

This message is a reply to:
 Message 25 by Percy, posted 03-16-2005 9:06 AM Percy has replied

Replies to this message:
 Message 27 by lyndonashmore, posted 03-16-2005 9:55 AM Melchior has not replied
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Melchior
Inactive Member


Message 38 of 309 (191988)
03-16-2005 6:02 PM
Reply to: Message 36 by Percy
03-16-2005 10:51 AM


Re: Mossbauer effect?
quote:
quote:
If you cool the stuff then the nucleus cannot recoil so all the energy is given to the photon. A second identical nucleus can absorb it. The photon is not redshifted - analogous to electrons and glass.
Sorry, unable to figure this part out.
He's saying that if you cool an atom to low enough temperatures, it can not be moved. And if it can't be moved, it can't recieve kinetic energy. And if it can't recieve kinetic energy, it can not take away any from photons.
I have no idea why he thinks that cooling atoms makes them behave this way, though. It is clear enough why they behave like this in a crystal lattice (they can't move because they are held into place) but I can't see any way to make this apply to free atoms. Please clarify what you mean by this, lyndonashmore.
Also, if an atom first absorbs and then emitts a photon, why would it have to recoil in the first place?
This message has been edited by Melchior, 03-16-2005 06:11 PM

This message is a reply to:
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Replies to this message:
 Message 39 by NosyNed, posted 03-16-2005 6:48 PM Melchior has replied

Melchior
Inactive Member


Message 41 of 309 (192011)
03-16-2005 8:37 PM
Reply to: Message 39 by NosyNed
03-16-2005 6:48 PM


Re: Recoil
Yes, but the momentum the atom gains when it absorbs the photon would be exactly as great but opposite in direction as the momentum it loses when it reemitts an identical photon, wouldn't it?
This is under the assumption that there is no shattering going on.

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Melchior
Inactive Member


Message 46 of 309 (192120)
03-17-2005 11:12 AM
Reply to: Message 45 by lyndonashmore
03-17-2005 10:41 AM


I guess what we want to know is; how can an electron tied to an atom absorb only a part of the photons energy. Basic theories of quantum mechanics taught in universities claims that the electron must absorb all of it and that is why atoms can only absorb specific wavelenghts of light.

This message is a reply to:
 Message 45 by lyndonashmore, posted 03-17-2005 10:41 AM lyndonashmore has replied

Replies to this message:
 Message 48 by NosyNed, posted 03-17-2005 1:29 PM Melchior has replied
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Melchior
Inactive Member


Message 50 of 309 (192156)
03-17-2005 3:12 PM
Reply to: Message 48 by NosyNed
03-17-2005 1:29 PM


Re: "partial" absorbtion
I think this is something in general that puzzles me, and not exactly something lyndon said, but I still find it rather odd if atoms can absorb a range of frequencies instead of just set ones.
I can understand that, due to scatter, the emitted photon can have a different energy because the atom has a different kinetic energy, but I can't see how it can work from the other end.
What is the actual mechanism that allows the atomic nucleus to pick up and convert photons into kinetic energy?

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 Message 48 by NosyNed, posted 03-17-2005 1:29 PM NosyNed has not replied

Melchior
Inactive Member


Message 196 of 309 (193045)
03-21-2005 12:28 PM
Reply to: Message 194 by Eta_Carinae
03-21-2005 12:00 PM


Re: My God man....
If you check his website you will see that he is just expressing himself in an odd way and that you are missunderstanding his actual point.
When he says per cubic metre he actually means 'n', which means number of free electrons. This 'n' should use the same lenght unit as used elsewhere in the formula, so if you use m^3, you use m^-3, if you use ft^3, you use ft^-3.
I'm not sure if there is a previously established number for 'n', but Lyndon sets this as exactly 1 per cubic metre.
On his webpage, he says that as a formula this should be expressed as H = 2nhr/m. Note that the 2 in there is probably a typo on his part, but I copied and pasted the formula directly from his webpage.
This message has been edited by Melchior, 03-21-2005 12:34 PM

This message is a reply to:
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Replies to this message:
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