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Pastimes : The New Qualcomm - write what you like thread. -- Ignore unavailable to you. Want to Upgrade?

To: Maurice Winn who wrote (4186)	2/12/2002 3:30:58 AM
From: John Walliker	Read Replies (1) \| Respond to of 12232

Maurice, Okay John, so exactly where and why does a 2GHz photon get absorbed please? ... You do believe in asking difficult questions... In the case of an antenna, the photon does not get absorbed there. The antenna acts as a scoop, diverting the photon down the antenna feed cable to the receiver where it is absorbed ... somewhere in the resistance (electric friction) of the electrical circuits. The antenna (aerial) does not need to be the right length, its just more efficient when it is resonant. In the case of photons passing through water, brains and the like the one thing which it is not possible to answer is "precicely" where a photon will be absorbed. But you knew that anyway, or you wouldn't have asked:-) As a photon passes through water its alternating electric field will interact with the electrically polarised water molecules causing them to vibrate a bit as they try to line up with it and hence absorb energy. This vibration of a polarised molecule then in turn re-radiates the photon and it carries on its journey. This process slows down the passage of the photon. (Of course if it was just one photon and one molecule then it would not necessarily get re-radiated in the same direction. There is a need for the photon to spread itself over many water molecules any one of which might have absorbed it to get the right effect.) If the water is in the form of ice the re-radiation is not much disturbed because all the molecules are locked down in a crystal lattice. If the water is in the liquid state however, the vibrating molecule may bump into another water molecule and share some of its 2GHz vibration with its neighbours. This dissipates the energy as heat (random vibrations) and means that it is no longer available to be re-radiated. It is impossible to predict which water molecule will do this so your question cannot be precisely answered. I made all this up as I went along and it is probably mostly wrong... John