- #26

- 136

- 0

- Thread starter kcodon
- Start date

- #26

- 136

- 0

- #27

- 40

- 0

While a photon may have 0 rest mass - and it's meaningless to talk of a photon at rest as it doesn't exist, it does have momentum and velocity (c) so it has mass when moving (or perhaps more accurately, there is a mass equivalent of energy contained within it).

There is also conservation of momentum with the excited atome that emitted it as well. It will recoil. If there were such a thing as the photon travelling in all directions until it 'decided' to be really at one location, say to be absorbed (and detected) by the 3rd element from the top and fourth element from the right of a ccd sensor in a camera on the telescope, then there would be no recoil for the atom emitting it until that 'decision' was made.

Seems like I recall something of a 'pilot wave' - at least a thought experiment - trying to convey this concept of all directions. In terms of the universe, such things lose out and rather appear indicative of serious problems in interpretation of the concepts(at least to me).

The question is if there is nothing between a single atom and a spherical shell 10 miles away with one small telescope peering through a port, why would there be more photons reaching it than reaching any other point on the sphere?

- #28

- 303

- 0

The r=ct sphere (in my opinion) represents the leading edge of the propogating sinusoid. However, by my interpretation, the amplitude of the sinusoid snakes its' way through space along a poynting vector (direction of propogation). This places the trailing edge of the sinusoid in what amounts to "imaginary space" - physically inside the r=ct sphere, and yet to pass through (not "visible" at time t, per se). In order to observe the sinusoid as such, you would have to freeze time as the sinusoid passes, and vary position along the poynting vector. The sinusoid is exactly one wavelength long within this time-frozen 3-space.So theres no such thing as an EM "wave" so to speak? This was kind of my original question. Does that mean that any "wave" is actual an increasing sphere so to speak? Then my question was how does one get a sinusoidal wave from a sphere or vice a versa? And then a photon I believe is not spherical...but I think you explained this later on, I just didn't understand.However, you bring up a common misconception: EM waves are not planar - they are sperical. Always. They exist on a sphere of constant radius/time.

Plane waves are merely a convenient way of approximating spherical waves at very large radii, and are only valid were the included spherical surface area has negligible curvature (phase error).

The infamous r=ct sphere about whatever we have constructed in order to radiate a single photon represents the past of the item at the sphere's center. When some portion of that radiated energy reaches another item constructed to detect it at the sphere's surface, that energy arrives in that second item's present.This I also found interesting...

Do you mind elaborating?I'm of the opinion that the propogation of EM waves are the causal equivalent of the past of one reference frame finding its' way into the past/present/future of another.

The emission effectively takes place in the future of the detector, arrives (later) in the present of the detector, from a point in the past of the emitter.

Follow?

Given a Euclidian sense of 4-space, time would be orthogonal to 3-space. Just as z=3 can be true anywhere on the x-y plane, t=0 can be true for any x-y-z. By the same token, any t can be true for any x,y,z - beit the present of whatever is at that point, or the past/future of another item that either has passed, or will pass through it (in some way). "Local time" is relative within the framework of absoloute time - because of stipulations regarding "c", and some arbitrary "points of reference" affected by time's passage.

Since a photon allegedly does not experience time (by virtue of its' velocity), a transaction occurring over tens of millions of years is effectively instantaneous within the photon's frame. Hence the notion of the past (energy decrease) finding its' way into the future (energy increase). Energy is still conserved ... over time.

Regards,

Bill

- #29

- 81

- 0

I can see some people are getting into this thread, and thats cool; its just a little over my head lol. So I was wondering if someone could give me simple answers on the following please?

#1 OK cbacba raises a really good, logical point

So this implies that there is no wave nature of light at all in the case of a single photon being emitted, or spherical nature, and that light must be a particle, does it not?If there were such a thing as the photon travelling in all directions until it 'decided' to be really at one location ... then there would be no recoil for the atom emitting it until that 'decision' was made.

#2 The sphere in this case would be of r=ct and would not really physically represent anything...except that at one point on its surface the photon would be located - at least this is what I think??. Mathematically, would the surface area represent the probability of receiving the photon, as jtbell mentioned?

#3 So sort of combining the above two, can one then think that the "wave" nature of a photon, is actually just some form of mathematical representation of the probability of finding the photon at a given point in space and time? I assume this can be related to what Bill said in the last post about sinusoidal wave forms etc? So really light is "corpuscles" as Newton put it?

And finally, thanks Bill,

I think I see what you're saying...I've never thought of that before.The emission effectively takes place in the future of the detector, arrives (later) in the present of the detector, from a point in the past of the emitter.

Follow?

Anywho, thanks for any replies on these 3 points, and hopefully some light can be shone on my questions,

Kcodon

- #30

- 303

- 0

While that seems to be a reasonable interpretation, I don't think that it would be accurate. Consider the source of the emission: change in energy level of an *orbiting* electron.#1 OK cbacba raises a really good, logical point

So this implies that there is no wave nature of light at all in the case of a single photon being emitted, or spherical nature, and that light must be a particle, does it not?If there were such a thing as the photon travelling in all directions until it 'decided' to be really at one location ... then there would be no recoil for the atom emitting it until that 'decision' was made.

For the sake of a description, let us assume that an electron orbits an atom in a plane with period T. Sensors at constant radius in this plane might sense the changing electric field emanating from the electron as having a maxima when the electron is closest, and minima when the electron is furthest - vis: a sinusoid of period T.

If this electron were to drop to a lower energy level, all sensors at constant radius will record the same change in amplitude of the radiated field. The detection of this change (photon?) becomes fleeting as r increases due to the rate at which the event propogates around the r=ct arc (or shell, for the 3-D version).

I doubt there would be any "recoil" to speak of - aside from, perhaps, a very breif oscillation with no net change in position/direction.

I'd offer that the r=ct sphere physically represents a region of "constant time" with respect to the emitter, and the probability that the photon exists on that sphere at time t is 1. However, given the description I provided above, it may well be true that at any given radius, the photon might only present itself at a particular point on that sphere - as it moves to a slightly different point on a slightly larger/smaller sphere (r=c(t+dt)) while following the angular trajetory of the affected electron. Curiously, this interpretation satisfies both the particle (one point on a sphere of radius r=ct) and wave properties (all spherical angles) of photons simultaneously.#2 The sphere in this case would be of r=ct and would not really physically represent anything...except that at one point on its surface the photon would be located - at least this is what I think??. Mathematically, would the surface area represent the probability of receiving the photon, as jtbell mentioned?

I think it would be interesting to arrange an array of detectors over a narrow angular range, and broad radial range to see if it might be possible to record the same photon wave in multiple locations/times. I'd imagine that a steep cone arrangement would suffice.

The above strikes me as having a little too much emphasis on the particle properties, and too little on the wave. Both aspects should carry equal weight since either could be observed.#3 So sort of combining the above two, can one then think that the "wave" nature of a photon, is actually just some form of mathematical representation of the probability of finding the photon at a given point in space and time? I assume this can be related to what Bill said in the last post about sinusoidal wave forms etc? So really light is "corpuscles" as Newton put it?

Regards,

Bill

- #31

- 303

- 0

In order for your pilot wave to work, you would have to concentrate the energy radiated in one general direction. The sperical distribution of the wave energy still holds (thanks to such phenomena as diffraction), but the energy density on the shell can be modified to favor a particular direction above all others.Seems like I recall something of a 'pilot wave' - at least a thought experiment - trying to convey this concept of all directions. In terms of the universe, such things lose out and rather appear indicative of serious problems in interpretation of the concepts(at least to me).

Imagine an omnidirectional emitter at the focus of a parabolic bowl that creates a hemispherical "shadow" (focus and rim in plane with the emitter). The size of the aperture in terms of wavelengths relates directly to the configuration's directivity along the focal axis. A simple raytrace will show that any ray from the emitter/focus, to the parabola, and then to the aperture plane (plane of rim) will follow a path of equal length. Thus, the energy emitted in a spherical fashion becomes "columnated" along the focal axis as a disk of constant phase energy (rather than a hemisphere of constant phase).

While the modes of propogation are defined over the entire sphere, there are any number of non-zero order modes (limited by the aperture's size in wavelengths) that can be used to characterize the resultant spherical energy distribution.

Perhaps because that is where the emitter was "pointed" at that point in time.The question is if there is nothing between a single atom and a spherical shell 10 miles away with one small telescope peering through a port, why would there be more photons reaching it than reaching any other point on the sphere?

Regards,

Bill

- #32

- 40

- 0

I think the real point of qm is that you cannot detect the condition without altering the condition. That and it's a probability thing.

When it comes to the emission of a single photon, there are two types. These are spontaneous and stimulated. There are probability factors for each. Spontaneous means it's going to go off on its own time schedule and in its own direction - ostensibly, completely random within the context of what is known about the nature of the atom or molecule and the state it is in. Stimulated is a forced emission caused by the nearby passing of a photon of the same energy and direction and phase of the one that is being emitted (laser stuff).

BTW, it's not my 'pilot wave'. I don't believe it and was attempting to debunk it. I don't know whose it is but I can't help remembering seeing feynman in movie clip explaining it somewhere.

When it comes to the emission of a single photon, there are two types. These are spontaneous and stimulated. There are probability factors for each. Spontaneous means it's going to go off on its own time schedule and in its own direction - ostensibly, completely random within the context of what is known about the nature of the atom or molecule and the state it is in. Stimulated is a forced emission caused by the nearby passing of a photon of the same energy and direction and phase of the one that is being emitted (laser stuff).

BTW, it's not my 'pilot wave'. I don't believe it and was attempting to debunk it. I don't know whose it is but I can't help remembering seeing feynman in movie clip explaining it somewhere.

Last edited:

- #33

- 81

- 0

Hmmm I think I disagree...photons have momentum and there must be some change in momentum of the atom emitting it - instantly - to conserve momentum. This would of course be much to small to detect etc etc, but still occurs. However:I doubt there would be any "recoil" to speak of - aside from, perhaps, a very brief oscillation with no net change in position/direction.

cbacba is right here I think...and this is what so much annoys me about QM, and although I'm know very little of it, I'm guessing the QM interpretation would indicate something like the molecule would be in a superposition of possible states until the photon emitted was measured, so it nicely avoids this problem.I think the real point of qm is that you cannot detect the condition without altering the condition. That and it's a probability thing.

Its just the whole probability thing again then. The small telescope would maybe only detect a photon one in every gazillion emissions, but it would eventually. Asking then why it hit that certain telescope is kind of like asking when you toss a coin and get heads,Perhaps because that is where the emitter was "pointed" at that point in time.The question is if there is nothing between a single atom and a spherical shell 10 miles away with one small telescope peering through a port, why would there be more photons reaching it than reaching any other point on the sphere?

I also think in a round about way we agree on #2, however for #3...

What is the wave properties of the emitted photon in this exact case, without passing it through a double slit or anything? We've all been saying - I think - that all the energy from the emission is concentrated in one place (particle nature), however there is a possibility of it being anywhere in the sphere r=ct (wave nature to some degree).The above strikes me as having a little too much emphasis on the particle properties, and too little on the wave. Both aspects should carry equal weight since either could be observed.

Kcodon

- #34

- 303

- 0

The non-qm version (as I understand it) would require that if some amount of energy is received at radius r, then there is that much less energy available on the r+dr sphere. How the conservation of energy might be probablistic is something that I don't quite follow.I think the real point of qm is that you cannot detect the condition without altering the condition. That and it's a probability thing.

Does a photon cease to propogate further (in all other directions) once detected? Or, is the condition of the photon wave simply altered to account for the loss of energy?

I find it hard to rationalize how both must necessarily be true if a single isolated detector records a "hit".

I find it curious that "stimulated emissions" presuppose the existense of photons already travelling in the direction you desire. Where would such a stimuli originate?When it comes to the emission of a single photon, there are two types. These are spontaneous and stimulated. There are probability factors for each. Spontaneous means it's going to go off on its own time schedule and in its own direction - ostensibly, completely random within the context of what is known about the nature of the atom or molecule and the state it is in. Stimulated is a forced emission caused by the nearby passing of a photon of the same energy and direction and phase of the one that is being emitted (laser stuff).

Sorry, I didn't know it came with baggage - just that you brought it up.BTW, it's not my 'pilot wave'. I don't believe it and was attempting to debunk it. I don't know whose it is but I can't help remembering seeing feynman in movie clip explaining it somewhere.

If photon emission is unidirectional (and I'm not saying I think it is), you might be able to place a "spontaneous emitter" at the center of a perfectly reflective sphere and wait to see if it moves - once, upon emitting the photon, and accelerated once again by absorbing the reflection (since the reflection would pass back through the center of the sphere in the direction of the recoil).

Another possibility that would allow for detection/confirmation of the emission would be to place the emitter at one focus of a perfectly reflective ellipsoid, and an omnidirectional detector at the other. While the detector might see a photon travelling directly from one focus to the other, it should certainly notice a spherical wave converging on it from all directions.

I doubt the emitter would have moved in either case, but that's just my opinion.

Regards,

Bill

- #35

- 303

- 0

Is that so?photons have momentum and there must be some change in momentum of the atom emitting it - instantly - to conserve momentum.

Consider that in the omni-directional case I attempted to describe (poorly I guess), for any photon travelling in one direction, there may well be another (the same?) travelling in the opposite direction (and 180deg out of phase). The momentum of each cancels the other, leaving the atom without change in momentum - just change in energy.

I've read over and over again how it takes an infinite amount of energy to accelerate a finite mass (however small) to the speed of light - so what better way to counter that infinite force than with another of equal magnitude and opposite direction?

An atom that emits a photon does still lose mass in the process; so how else could that instantaneous change in mass convert itself to a photon (wave, or otherwise)?

Just because it does?

Please re-read the 2-D scenario I described, and see if you can imagine how the detection points of a given photon might traverse a spiral of 360degs in phi for every one wavelength change in r.

Well - the way I see it is that the total energy of the wave exists in different directions at different times due to the nature of the source (an orbiting electron). The suggestion that one could capture all of the wave energy at a single point in space strikes me as absurd.What is the wave properties of the emitted photon in this exact case, without passing it through a double slit or anything?The above strikes me as having a little too much emphasis on the particle properties, and too little on the wave. Both aspects should carry equal weight since either could be observed.

The other possibility is that the wave exists everywhere on the r=ct sphere, but is only detectible where the leading edge of the propogating wave form is not imaginary (essentially equivalent to not "existing" there yet). Refer back to my orbiting electron for the rationale for phase to vary at constant radius.We've all been saying - I think - that all the energy from the emission is concentrated in one place (particle nature), however there is a possibility of it being anywhere in the sphere r=ct (wave nature to some degree).

As an aside, has anyone noticed that if you increase the radius of a circle by one wavelength ( [itex] 2\pi [/itex] radians), the circumference grows by the wavelength squared ( [itex] 4\pi^2 [/itex] radians)?

Regards,

Bill

- #36

- 81

- 0

Is this the 2D scenario you were referring to? I actually just reread the whole thread and came across it and looked at it under new light. However I don't come close to fully understanding what you are saying...note that I have pretty much no knowledge of relativity or the mathematics I believe you are speaking of. If there is an easier way to put it, I'd appreciate it, otherwise don't worry about it.Consider, if you will, an electron orbiting an atom. That electron exists on a shell of constant energy about the atom. If this electron were to move to a lower energy shell, a photon is released to conserve energy. This photon propogates outward equally in all directions at the speed of light - compressing the spatial dimension representing the shell's diameter to zero (radially) by Lorentz contraction. At any given point in space, that pseudo electron shell (representing the change in energy) can then be pojected on a sphere of constant time with respect to the emitting atom. The phase of this pseudo-electron orbit does not vary with time at our given point, but it will vary with radial distance from the emitter. The entire orbit of this pseudo-electron can then be construed to exist at a point (or "patch") on a spere - only by varying radius can you "observe" the sinusoidal amplitude of the wave as its' complex amplitude varies through 360 degrees (one wavelength).

I am talking about one singular photon being emitted. So there may not be another photon emitted in the opposite direction. Imagine then a H atom with an electron in the first excited state; it can only emit one photon - to conserve energy - and thus the H atom must now move to conserve momentum.there may well be another (the same?) travelling in the opposite direction (and 180deg out of phase)

I am not sure if when a photon is emitted the atom loses mass...? The photon has no mass, but has momentum. The electron loses energy, so a photon is given off...i.e. energy.An atom that emits a photon does still lose mass in the process; so how else could that instantaneous change in mass convert itself to a photon (wave, or otherwise)?

Firstly we are talking about a singular photon, so its energy can only be in one place. There is a probability of it being anywhere on the surface, but it's energy is only at one point. This is I think where we disagree the most, and maybe we need some others' opinions?Well - the way I see it is that the total energy of the wave exists in different directions at different times due to the nature of the source (an orbiting electron). The suggestion that one could capture all of the wave energy at a single point in space strikes me as absurd.

Kcodon

- #37

- 81

- 0

I didn't understand this before but re-reading this again, I have a few questions. Firstly the wording of the first sentence confuses me a little...do you mean to say the sine wave represents say the electric/magnetic field in the source, with respect to time? And then does this mean that there is one "dent" per photon. So if the electron in the radio transmitter oscillated once, this would be one dent, representing one photon, and one wavelength? So then the apparent oscillations at the fixed locations are in fact just measures of the amplitude of these dents so to speak...i.e. the strength of the electric field at that point in space, with respect to time? So again somewhat like a sound wave...if you plotted the amplitudes with respect to time you would get a sine wave?One thinks of the sine waves as that is going one when creating the wave - such as a radio transmitter and the sine wave represents a pure or single frequency. For the wave propagation, special relativity demands that time is stopped when moving at the speed of light so there is no oscillation happening in the wave itself. Rather the oscillations appear to happen at a fixed location as the wave ripples past due to the fact it is traveling at a velocity (c) and it has a wavelength which is a separation of these 'dents' which are the presence of an electric field perpendicular to a magnetic field.

Thanks anyways,

Kcodon

- #38

- 136

- 0

As far as I understand Quantum Mechanics, if we know the path of a photon, then we observe particles (not waves) but if we have no information on the path (as in Young's slits) then the photon 'is' a wave and exhibits interference to prove it.

In the case of a single photon starting out in free space and being 'oberved' by one telescope 10 miles away, then if we observe it, we know its path exactly so it must be a particle. But, if we don't observe it? Does it mean its a particle that shot off in the other direction to our telescope, or the fact that we did not observe it means that we could not know its path, (we didn't observe it) so it must be a wave. If its a wave, we did not observe it because of the radiation pattern from a small dipole denies us its observation. OR, we did'nt observe it because its a particle photon and we would then knows its path exactly, and that path is impossible because the photon went the other way.

- #39

- 40

- 0

Conservation of momentum demands that the atom recoil since a photon has energy and momentum - proveable by the radiometer toy - if sufficient vacuum it turns away from the white sides dues to 2x momentum from reflected light, if not enough vacuum it moves away from the black sides as there is heat and interaction with the vacuum gases overwhelming the 2x momentum coming from the white side. Look at a whole batch in store selling them sometime.

One too must remember that according to modern physics, EM waves are all the same stuff from different viewpoints. X and gamma rays don't show much wave behavior. What's more, even traveling huge distances with regard to the size of the universe, they red shift a bit (expanding U. and all that) but they don't dissipate out in inverse sqr and neither do other forms of Em. It's an all or nothing proposition. The difference between a hard gamma ray and a 200meter AM broadcast station 'photon', according to special relativity, is the speed at which you are approaching the broadcast station's antenna.

Note too that photons/EM waves are properties of space very much like waves on the ocean. They are not properties of the atoms/molecules or antennas emitting them.

For an antenna, I've read that EM waves actually form outside at the point where the E and M fields are perpendicular to each other (and in a particular phase relation?). This may be perhaps one or more wavelengths out. One may also be taught conceptually that the propagation is happening because each changing field is propelling the other forward - along the poynting vector. These changes are ovservable at a fixed point in space where the transmitter/antenna is located. There no time variation in the frame of reference in which the wave is traveling. Whether a photon has 1 full cycle, 5 or 5000 cycles of waveform in it, it's a fixed pattern moving though space at the appropriate c velocity. It only appears to oscillate for a stationary observer (or at least one moving at less than c).

There are other examples of what seem like similar circumstances going on in physics. Temperature is a bulk property of a huge bunch of molecules / atoms/ electrons / photons etc. Matter, like the skillet on your stove or the air in your room tend to have a measurable temperature and that means they are are following the Boltzman statistical distribution of kinetic energy. It's like the blackbody curve. A particular atom will have the liklihood of having a particular energy according to the Boltzman distribution but that particular atom will not be measureable without modifying it (bouncing some energy off it or letting it change state somehow and detecting it). That atom will not have a unique temperature as it is the statistics of the energies of all of the atoms that create what we call temperature.

Perhaps that is too close to qm and how things are being interepreted. We ascribe probabilities and perhaps even more to single objects for which we have no way of knowing the specific details and then expect the bulk properties of a whole batch of them to be directly applicable to a single hypothetical example. If an atom is an isotropic radiator and we let it sit there radiating its little heart out for a long enough time, we will get a uniform distribution for a radiation pattern. If we put a double slit in there and shine coherent light at it, we'll see a diffraction pattern. If we shut down the intensity even to 1 photon per second, we'll measure the same pattern eventually as it builds up. This leads to the qm conceptual assumptions of traveling through both slits at the same time and a photon interfering with itself. And, supposedly it's been tested quite properly that closing one slit at a time while building up the pattern via single photon emissions results in the destruction of the waveform - implying that photon does travel through both or is somehow 'aware' of both slits.

So far, I've yet to find a satisfactory conceptual answer which is even close to being correct. And, it would seem that things get even worse when considering things when dealing with astronomical distances.

- #40

- 40

- 0

I use the term 'dent' for lack of a better understanding of what an electric field is. It's the problem of everything is defined in terms of the fundamental undefinables. The general relativity guys like to conceptualize gravity by having matter bend their euclidean space table down so I thought I'd do likewise. Think of it as a deformation in the 'electric' direction as opposed to the 'up' direction.To cbacba,

I didn't understand this before but re-reading this again, I have a few questions. Firstly the wording of the first sentence confuses me a little...do you mean to say the sine wave represents say the electric/magnetic field in the source, with respect to time? And then does this mean that there is one "dent" per photon. So if the electron in the radio transmitter oscillated once, this would be one dent, representing one photon, and one wavelength? So then the apparent oscillations at the fixed locations are in fact just measures of the amplitude of these dents so to speak...i.e. the strength of the electric field at that point in space, with respect to time? So again somewhat like a sound wave...if you plotted the amplitudes with respect to time you would get a sine wave?

Thanks anyways,

Kcodon

Generally, the sine wave refers to the electric field or the displacement of molecules in air for a sound wave or the displacement up and down for the height of the water in an ocean wave. It is generally used for waves of a uniform single frequency and other more complicated waveforms can be created from combinations of multiple sine waves.

The photon is thought of a packet, statistically having a certain size or length. Even a sine wave of non infinite unchanging length tends to be represented by multiple sine waves - implying certain conditions and perhaps limitations on how pure a single frequency there is.

I don't know for certain hardly anything about a photon. However, they should be essentially of a single frequency and the implication being that it is a multiple of wavelength. The energy and particle nature of an AM broadcast station transmission is masked totally by the fact the photons are of such long wavelength and low energy. Again, it's the properties of space that define how it works.

In general at a fixed point you should get an electric field behaving with a sine wave magnitude - antenna input to oscilloscope display sort of thing.

- #41

- 81

- 0

I found this bit the most helpful:

And I think now with your explanation and others, I think I know whats going on with EM radiation to a much greater degree. I'm going to let this new found knowledge settle, however I'm much happier with my understanding and for now I don't really mind whether its right or wrong!In general at a fixed point you should get an electric field behaving with a sine wave magnitude - antenna input to oscilloscope display sort of thing.

Thanks again to all,

Kcodon

- #42

- 303

- 0

Don't sell yourself short - "understanding" is beleiving you are right, when others say you are wrong. Here's to making your "understanding" more robust (clink).Thanks so much cbacba,

I'm much happier with my understanding and for now I don't really mind whether its right or wrong!

Regards,

Bill

P.S. Treat your "understanding" objectively: correct a neccesary, and keep going.

- #43

- 81

- 0

And don't worry, I've got myself in the poop a few times with thinking I understand something that I soon realised I didn't - so I'll be objective. However its much nicer to have your understanding proven wrong with fact, than to have it verbally beaten into submission by those who think they understand - whether they are correct or not

Kcodon

- #44

- 136

- 0

we cannot know its position at all. Now a photon of an exact frequency e.g. from a

laser - the ferquency is known exactly, therefore on its travel from A to B we cannot

say anything about its position.

All I am saying is that the situation is not clear. There is confusion arising from speaking

about EM oscillations (antennae etc), to waves (dents in space-time, etc), to particles that cannot be defined (QM). And one fixed-frequency photon starting out from the center of a spherical volume seems to defy all the logical explanations at once.

highlight the problems.

- #45

- 40

- 0

One thing you can get rid of for comprehension is that there is no way you're going to exactly know anything, especially something like the position or exact energy of a photon. For one, it will have some energy because the atom/molecule will not be at rest because it will have kinetic energy due to temperature. Any measurement you make will be flawed, even a spectrometer will have a bandwidth or smear that means your result will only be so accurate. Note this doesn't mean it's the same thing as the qm limit, only that there are real limits. It's possible if not probable that a photon may not have just one frequency as it is of limited length (a packet). The same thing goes for position, it's not possible to determine exactly where it is. It's possible that being a packet, it may not have an infinitesimal size of a point or a definable and symmetric outer bound that permits the determination of a centroid for that point position.

- #46

- 303

- 0

The energy of the photon is real in the direction opposite the radial translation of the real electron, but imaginary in all other directions (this feature is new in my interpretation). The orbit of this imaginary electron grows radially at the speed of light, but continues with the same angular velocity of the parent electron in it's high-energy state (preserving it's wave nature).

If a ring of detectors were constructed in the plane of the electron's orbit, at constant radius from the orbit's center, one detector might receive the energy of the photon (where it's energy is real), but all should sense the change in periodic electric field variations due to the real orbiting electron (very nearly?) simultaneously.

Upon detection, (a portion of?) the real component of the wave energy is lost, but the imaginary component continues expanding outward - as if to signal the change in the real electron's state across space-time.

Regards,

Bill

- Last Post

- Replies
- 5

- Views
- 1K

- Replies
- 2

- Views
- 740

- Last Post

- Replies
- 15

- Views
- 4K

- Replies
- 10

- Views
- 2K

- Replies
- 0

- Views
- 2K

- Last Post

- Replies
- 18

- Views
- 1K

- Replies
- 2

- Views
- 3K

- Replies
- 17

- Views
- 9K