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- Thread starter Jim Lundquist
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mfb

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If you have a plane wave (this is not a photon!) travelling in the x direction, it can interact with electrons at every point on the x axis, independent of their position. Standing waves can be different.

Chargeless particles do not interact with photons, but there are other ways to keep electrons on fixed places.

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If you have a plane wave (this is not a photon!) travelling in the x direction, it can interact with electrons at every point on the x axis, independent of their position. Standing waves can be different.

Chargeless particles do not interact with photons, but there are other ways to keep electrons on fixed places.

Thank you for your reply, but I feel like the question is harder for me to ask than it is for you to answer. I am not a physicist, and I realize that a lot of the science is expressed in probabilistic terms, so let me try to put this in more conceptual terms. Please remember that this is a “simple” thought experiment, and certain concepts may not have direct equivalency in the world of physics, and a lot of the variables normally found in the real world are eliminated. Let’s replace the electrons with protons with a diameter of 1 fm. I know that in spontaneous photon emission that the direction the photon travels is expressed as a spherical wavefront, but if by measuring the recoil momentum and direction of the source atom, you can determine the direction of the photon, and if you establish this direction as the x-axis, and on this x-axis are lined protons side-by-side, if given the frequency and wavelength of the photon, can it be hypothetically predicted to some finite degree which proton that photon would interact with? If it can only truly interact with one proton, there must be some form of probability gradient here, because if the interaction

This will be my final attempt at this problem. I thank the forum for your patience.

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Nugatory

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It's better than some of the other non-technical descriptions floating around, but it's not correct either. There's really no substitute for learning the underlying theory (although the mathematical price of admission is very steep), but Richard Feynman has written a layman-friendly book you might want to try: https://www.amazon.com/dp/0691024170/?tag=pfamazon01-20Am I incorrect in thinking that a photon is emitted by a source and is propagated in the form of a wave and will continue under the guise of a wave until it interacts again with matter, and this interaction is expressed in the form of a photon?

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Thank you Nugatory and mfb. At 69, that "mathematical price of admission" is a bit steep, so I have ordered the book.It's better than some of the other non-technical descriptions floating around, but it's not correct either. There's really no substitute for learning the underlying theory (although the mathematical price of admission is very steep), but Richard Feynman has written a layman-friendly book you might want to try: https://www.amazon.com/dp/0691024170/?tag=pfamazon01-20

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BvU

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[edit]PF sees double ?

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BvU

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You do it all the time !

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mfb

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The probability would go down, but mainly due to the inverse square law for realistic setups.can it be hypothetically predicted to some finite degree which proton that photon would interact with? If it can only truly interact with one proton, there must be some form of probability gradient here, because if the interactioncanoccur at proton 1, it seems that the probability would decrease from the 2nd to the nth proton.

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Let’s replace the electrons with protons with a diameter of 1 fm. I know that in spontaneous photon emission that the direction the photon travels is expressed as a spherical wavefront, but if by measuring the recoil momentum and direction of the source atom, you can determine the direction of the photon, and if you establish this direction as the x-axis, and on this x-axis are lined protons side-by-side, if given the frequency and wavelength of the photon, can it be hypothetically predicted to some finite degree which proton that photon would interact with? If it can only truly interact with one proton, there must be some form of probability gradient here, because if the interactioncanoccur at proton 1, it seems that the probability would decrease from the 2nd to the nth proton.

Yes. The surfaces of objects receive more sunlight than the centers of objects.

And that's because the molecules between the sun and the center of an object cause a reduction of the probability of the absorbtion of a photon by a molecule at the center of the object.

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davenn

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Thank you, BvU. Now Iamexcited! The question is, "Will I emit a photon?"

you are emitting zillions of photons every second of ever day in the form of infrared radiation ( radiative heat)

Dave

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mfb

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In addition, it shows the fields at the x axis only. For a planar wave, the fields are the same for all points with the same x-value. There are no preferred points in space.

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Thanks, mfb, I was going to leave this alone until I have read Feynman's book, but I read other replies and got sucked back in, thinking that it all may be due to my lack of adequate description. So I take it that photon/particle expression has nothing to do with the intersection of the two fields. It seems like all of these graphic depictions of physical phenomena are more misleading than helpful to the lay enthusiast.

In addition, it shows the fields at the x axis only. For a planar wave, the fields are the same for all points with the same x-value. There are no preferred points in space.

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mfb

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Oh, it is not that bad if you want to describe the electromagnetic fields of a wave. Just don't try to add particles to that picture.It seems like all of these graphic depictions of physical phenomena are more misleading than helpful to the lay enthusiast.

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