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Originally Posted by Richard J. Hanak
In the same way that a mass object moving in space does not have to be pushed (or push itself) to maintain its motion, no action is required on the part of the photon to keep itself moving. That accords with the equivalence of energy and mass.
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It appears you saying a photon has mass.
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Originally Posted by Richard J. Hanak
My model of a photon is not an a priori conception. It is a synthesis of prior knowledge and experimental observations. I hope these few additional comments will give all of you a more complete understanding of my interpretation.
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It helped, but I have a few questions.
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Originally Posted by Richard J. Hanak
A photon has three intrinsic properties: its energy, the periodic transverse extension of its EM field, and its forward/propagation velocity. Therefore, we should expect to observe three distinct kinds of effects from a photon.
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It also has spin, but since it doesn't affect your arguments below, we can ignore it for now.
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Originally Posted by Richard J. Hanak
The second effect is dependent on the periodic transverse extension of the field of the photon. We observe the transverse extension of a photon by the striking efficacy of a half-wave dipole antenna ...
snip.
The third effect is dependent on both the periodic transverse extension and the forward velocity of the photon. The relationship of the periodic transverse field extension and the distance traveled is what we call the wave nature of the photon.
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You mention half-wave antenna in the second and then state the third is the wave nature of the photon. I'm not quite clear on this. Wouldn't you need to have the wave nature to affect a half-wave antenna?
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Originally Posted by Richard J. Hanak
Tensor:Since we can't know a photon's energy until it is measured (and that energy is dependent on the observers frame), how do you propose to determine what that independent energy is?
You claim that we can't know the energy of a photon until we measure it. In this thread we have not been considering an isolated photon in space.
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I was just responding to this:
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Originally Posted by Richard J. Hanak
Assume for a moment that two observers in different reference frames could measure the energy of the same photon. Which of the two energies would the photon have?
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It appeared to me that you were talking of a single photon here.
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Originally Posted by Richard J. Hanak
We are measuring the wavelength. And if we can identify the quantum state responsible for the photon when it was emitted, doen't that tell us the energy of the photon without having to directly measure its energy?
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The frequency (f) of a photon is equal to c divided by the wavelength (L) and the energy (E) of a photon is its frequency times Planks constant (h). Mathematically f = c/L and E = fh or through sustitution E = ch/L. Since c and h are constants, when you measure a photon's wavelength, you
are measuring it's energy. Also, don't you need to take a mesurement to determine the quantum state?
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Originally Posted by Richard J. Hanak
I happily stand corrected about the spin interpretation. Thanks.
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You're welcome.