# Thought Experiment on the Dual Nature of EM Radiation

• Jim Lundquist
In summary, the thought experiment is based on the dual nature (particle/wave) of electromagnetic radiation.

#### Jim Lundquist

The following question/thought experiment is based on the dual nature (particle/wave) of electromagnetic radiation. Consider the emisson of a single photon on a 3D grid along the x-axis with its origin at (0,0,0). The only matter in this experiment consists of “electrons” found only in alignment along the x-axis. To eliminate the repulsive nature of like-charged particles, consider these particles to be charge-less, so that we can align them as close to each other as possible. Question 1: Is the associated electromagnetic wave propagated in 3 dimensions around the x-axis and with the probability of both the electric component and the magnetic component “existing” in any dimension at any particular moment? Question 2: If you number the x-axis 1, 2, 3, 4…, and establish the wavelength of the wave to be “1”, is it possible for the photon to interact with an “electron” at any point on the grid other than at (1,0,0), (2,0,0), (3,0,0), etc.? Or should I say, is it more probable that the photon will interact with an “electron” at co-ordinates (1,0,0), (2,0,0), (3,0,0),etc. than at any other point? I know I could get butchered on terminology and construction, but it is a thought experiment after all.

A photon does not have a position. You cannot release it at some point.
If you have a plane wave (this is not a photon!) traveling 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.

mfb said:
A photon does not have a position. You cannot release it at some point.
If you have a plane wave (this is not a photon!) traveling 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 can occur at proton 1, it seems that the probability would decrease from the 2nd to the nth proton. Am 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?

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

Jim Lundquist said:
Am 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?
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

Last edited by a moderator:
Nugatory said:
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
Thank you Nugatory and mfb. At 69, that "mathematical price of admission" is a bit steep, so I have ordered the book.

Last edited by a moderator:
Jim Lundquist said:
so I have ordered the book.
Until it arrives: view these to get started

PF sees double ?

BvU said:
Until it arrives: view these to get started
Thank you, BvU. Now I am excited! The question is, "Will I emit a photon?"

You do it all the time !

Thought experiments with single photons are never simple.
Jim Lundquist said:
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 can occur at proton 1, it seems that the probability would decrease from the 2nd to the nth proton.
The probability would go down, but mainly due to the inverse square law for realistic setups.

Jim Lundquist said:
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 can occur 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.

Jim Lundquist said:
Thank you, BvU. Now I am excited! 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

I'm going to wait to read Feynman's book, but the point I was actually getting at was that if a photon is propagated along the x-axis as depicted in the diagram as an electromagnetic wave, the most "logical" point of interaction with matter and where it would again be expressed as a photon/particle would be where the electric component and magnetic component cross the x-axis and thus be predictable if the wavelength and/or frequency is known. But I'm not a physicist and have not yet read Feynman's book.

The picture is one moment in time. The next moment, the whole field arrangement will be shifted a bit to the right.
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.

mfb said:
The picture is one moment in time. The next moment, the whole field arrangement will be shifted a bit to the right.
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.
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.

Jim Lundquist said:
It seems like all of these graphic depictions of physical phenomena are more misleading than helpful to the lay enthusiast.
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.