Quantized redshift anomaly

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Michiel</i>
<br />In the Meta Model the light carrying medium consists of elysons. What would happen if one of these elysons ceases to exist at the moment a photon is passing by? I think the photon would lose some of its energy. On the other hand if an elyson came into being while a photon is passing, the medium would get diluted, again resulting in energy loss.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Think of elysons as like water molecules in a wave on the surface of the ocean, which is the analog for a "photon" (a singlet wave). If we subtract water molecules from a wave, it carries less energy; and if we add molecules, it carries more energy and does more damage if it strikes something.

However, "energy" in this sense is related to wave amplitude, not wave frequency or wavelength. So when the analogy is applied back to photons, adding elysons increases the photon's amplitude and therefore its intensity, but does not change its "energy". -|Tom|-

Tom:"...if we add molecules, it carries more energy and does more damage if it strikes something."

I would say it carries the same amount of energy, but divided over more molecules. Unless the molecules are created with the extra wave-energy included, then there really is more energy.

Tom:"So when the analogy is applied back to photons, adding elysons increases the photon's amplitude and therefore its intensity, but does not change its 'energy'."

Don't all photons of a certain wavelength have the same energy/intensity?

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Michiel</i>
<br />I would say it carries the same amount of energy, but divided over more molecules. Unless the molecules are created with the extra wave-energy included, then there really is more energy.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">That is one way to look at it.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Don't all photons of a certain wavelength have the same energy/intensity?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Same energy, yes: E = hf. However, there is no quantum for intensity. The amplitude of a single lightwave can be anything.

Nonetheless, amplitude does carry another form of energy because lightwaves can apply a significant pushing force to low-density objects such as balloon satellites. So formulas for the momentum carried by light are different than the energy formula. No inconsistency arises because there is no specific mass associated with any photon. -|Tom|-

If the amplitude of a single photon could be altered then surely its energy would change accordingly. The intensity of EM-radiation is associated with the combined energy of a number of photons in transit. So for a single photon the intensity is exactly one time the energy. Or am I missing a crucial point from the Meta Model?

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Michiel</i>
<br />If the amplitude of a single photon could be altered then surely its energy would change accordingly. The intensity of EM-radiation is associated with the combined energy of a number of photons in transit. So for a single photon the intensity is exactly one time the energy. Or am I missing a crucial point from the Meta Model?
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">These specific points have nothing to do with MM. They are standard physics. The energy that lightwaves carry is absorbed by matter and heats the matter. The momentum that lightwaves carry exerts a pushing force on matter, accelerating it. The standard formulas are:

Photon energy: E = hf (f = frequency)
Photon momentum: p = hf/c = h/l (l = wavelength)

Both of these are consistent with classical formulas such as E = mc^2 and p = mc if the lightwave is envisioned as having a frequency-dependent mass equivalent to hf/c^2.

Power is the rate of flow of energy, and intensity is power per unit area. Intensity for any wave, including light, is proportional to the square of wave amplitude. So it is more practical to compute the pressure of light in terms of intensity than any of these frequency-dependent quantities, especially since we are usually dealing with white light (which is a mix of all frequencies). -|Tom|-

Ah, so with amplitude you mean 'the amplitude of the local disturbance in the medium'. Of course.
Sorry, I must have been tuned in on a different wavelength for a while. %)

Still.. In the double slit experiment we see a wave which is clearly spread out (hence the interference), but when it hits the screen the energy is transferred to a small dot, with always the same intensity.
I can't think of a way to explain that in terms of oceanic waves.