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energy of a photon

Hi,

Ok so this formula E = h*c/? gives the energy of a "photon". To simplify that formula, it could be thought of as E = amplitude / wavelength, with the h constant scaling the amplitude correctly, and the c constant scaling the wavelength constant correctly.

One interesting thing from this formula is it apparently shows that all "photons" aka electromagnetic waves resulting from electron orbital emissions, have the same amplitude! So this must mean that all electron orbital emissions and absorptions (which are frequency keyed) must all emit or absorb electromagnetic waves with the same fixed amplitude, regardless what the orbital number is. I find that to be quite amazing and hard to understand why that would occur.

cheers, Jamie

Reply to
Jamie
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Ahh I think I figured it out, this amplitude is proportional to the the electron's angular momentum (which is a constant I guess)

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cheers, Jamie

Reply to
Jamie

Hi,

New idea: Planck's constant describes the constant angular momentum transfer when an electron changes orbital states.

From this page:

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"an electron standing at rest has an angular momentum of hbar/2."

This shows that it is the electron's angular momentum that is the inherent source of Planck's constant, if the electron had inherently more "at rest" angular momentum, then the photoelectric emission graph of voltage over frequency would have a higher slope and Planck's constant would be a proportionately different number.

from this page:

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"The Planck constant has dimensions of physical action; these are the same as those of angular momentum"

So right there the Planck units even match those of the electron's angular momentum.

Einstein made up the photon to explain Planck's constant and the photoelectric effect when really he should have picked the electron!

cheers, Jamie

Reply to
Jamie

Nope, sorry. Electromagnetic fields add in instantaneous amplitude, just like they told you in Fields 101. Photons don't have well defined amplitude, because the total amplitude isn't a linear function of the photon number.

But as I have said till I'm blue in the face, thinking about E&M as photons rattling around will get you the wrong answer every time, guaranteed. Believe Maxwell for propagation, and think about quantization once you actually have photoelectrons to count.

Cheers

Phil Hobbs

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Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
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Reply to
Phil Hobbs

Hi,

I don't typically think of photons rattling around, as I don't believe in them either. However for the emission of electromagnetic waves from an atom, each emission must have a fixed amplitude, as seen from the photoelectric experiments. That amplitude is defined by Planck's constant, which I think is defined itself by the electron angular momentum, so that when an atom emits a "photon", the amplitude of the photon is directly related to the electrons angular momentum. That shows Planck's constant is derived from matter and not from quantized electromagnetic fields aka "photons".

cheers, Jamie

Reply to
Jamie

How about having the amplitude as square root of the power ? As just the energy is given, we need a time. Here comes the interesting relation. We'd naturally assign one wavelength giving a time of ?/c = 1/frequency. But then the bandwidth of the photon would be huge and the spectrum continuous. And a line spectrum requires the photon to be infinitely long... We can trade spatial uncertainty against frequency uncertainty and arrive at gaussian pulses, the eigenfunctions of the Fourier transform. Thus the length of a photon is (low natural number)/frequency.

For further reading search for "squeezed photons"

Rene

Reply to
Rene Tschaggelar

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