# A New look at the intensity of light.

1. May 22, 2005

### McQueen

Suppose that you are an observer in a room equipped with only one window measuring 1 m sq. The window is glazed with a glass pane 1 cm. Thick. The only source of illumination is a green surface outside the window. Hence the only light entering the room is the green light from this surface. Taking the following:-
1) Speed of light through glass = 2.5 x $$10^{5}$$km/sec
2) diameter of atom = $$10^{-10}$$m
3) thickness of glass = 3 mm.
Then we find that the light passes through the glass in approx. 1.2 x $$10^{-11}$$ secs. and that the time taken for a single photon to be absorbed and re-emitted by an electron as it passes through the glass in the process known as photon conduction is an incredible 4 x $$10^{-19}$$secs.
Having demonstrated the incredible speed with which a photon of light is absorbed and re-emitted by the electrons of the atoms belonging to the substance it is passing through , the glass pane in the window has served its purpose and can be removed. We turn next to the green surface outside the window whose reflected light illumines the interior of the room and examine the process by which this takes place. Making, for the moment , the assumption that the origin of the reflected light is an intense source of white light , we come to the following conclusions:- Some frequencies of the white light are absorbed by the atoms of the green surface , but are not re-emitted , their energies being absorbed instead by recoiling against the nucleus , this raises the energy of the atom resulting in the emission of heat , other frequencies of the white light find no corresponding energies in the electrons present in the atoms of the green surface and pass through the substance without any interactions taking place , lastly the energies and wave lengths corresponding to green light (i.e., those of 550Nm and an energy of 2.2 eV) are absorbed and re-emitted by the electrons of the atoms of the green surface at the rate of 5.4 x $$10^{14}$$Hz/sec}. We now deal with the question of the manner in which the electrons emit these photons , do they emit these photons at random in the manner of a spinning Catherine wheel or do they emit the photons in a single line? The manner in which photons are absorbed and emitted as they pass through a substance like glass would seem to indicate that the latter is true , however we will not draw any conclusions from this at present but depend on further observations to inform our views. To do this we must re-enter the room and join the observer. Let us assume that the observer is equipped with a particularly sensitive densiometer with which he can measure the intensity of the light coming through the window. At first he stands at a distance of 1m. from the window and notes down the intensity. He then shifts back from the window to a distance of 2m and again measures the intensity of the light , he finds that the intensity of the light is now 1/4 that of the intensity he had recorded at a distance of 1m from the window. He keeps repeating the readings at varying distances until he comes to the conclusion that the intensity of the light varies inversely with the square of the distance from the window and also that the area over which the light spreads varies directly with the square of the distance from the window. ( In actual fact of course the window is large and diffraction would make such a measurement difficult , remember this is an experiment.) He repeats his readings under identical circumstances over the course of several days and always gets the same results. What conclusions does he draw from his observations ? Firstly he notes that the intensity of the light varies inversely with the square of the distance from the window and secondly he notes that the area over which the light spreads varies directly with the square of the distance from the window. He concludes that originally the light was more intense because more photons were being absorbed per atom (i.e a number of photons were giving up their energy in succession) and also that due to the interaction of the light with the air or with some medium in the air , the number of photons being absorbed by each individual atom has been reduced and that this reduction has taken place uniformly , in other words the photons which originally were arranged in a single line had spread out into a number of lines while at the same time maintaining a uniform intensity (i.e., each of the new lines has an equal number of photons.) Thus it would seem that the intensity of light is not due to the energy of the photon alone but to its energy x frequency and that the number of photons due to this frequency varies inversely with the square of the distance from the source.. Thus at the source the intensity would be 2.2 x 5.4 x $$10^{14}$$ = 1.18 x $$10^ {14}$$eV per sec. or roughly 1 x $$10^{-5}$$ watts per atom per sec. And so on. Notice that if the duration is reduced the intensity is also reduced , thus the amount of energy reaching an electron in the destination point in $$10^{4}$$ secs would be 2.2eV . Further it must be assumed that as each electron absorbs a photon it immediately re-emits it at the same frequency.
Also light emitted by a photon bound to an atom will always obey the inverse square law because it is a point source while light from lasers also obey the inverse square law but since they are being emitted by a comparatively massive source of free electrons , travel longer in a coherent beam. At a distance of 1,609m , a helium-neon laser (which is a well collimated source ) will spread out over an area of 1.3m.

Last edited: May 22, 2005
2. May 22, 2005

### Claude Bile

Ouch, this is most definately not the case.

1) The inverse square law is only approriate for a perfectly isotropic source (i.e. not a laser).

2) Light from lasers are not emitted by free electrons, but by bound electrons (it is the nature of bound electrons, i.e. energy levels and so forth that allows things like population inversions and hence lasing to occur). The one exception is an electron laser, but these are hardly typical lasers.

Claude.

3. May 23, 2005

### Anzas

im not sure this is true, no laser sends exact parallel light there is always a little bit of scattering which at a distance R from the laser forms a part of a sphere with radius R so i think lasers do obey the inverse square law only instead of it being of the form a/r^2 its something like a/xr^2 with x being smaller with lasers which emit more parallel radiation.

4. May 23, 2005

### Jonny_trigonometry

" Thus it would seem that the intensity of light is not due to the energy of the photon alone but to its energy x frequency and that the number of photons due to this frequency varies inversely with the square of the distance from the source.. Thus at the source the intensity would be 1.18*10^14 eV per sec. or roughly 10^-5 watts per atom per sec. "

intensity = energy* frequency -> which frequency? the frequency of a photon, or the frequency of photons colliding with the detector at a distance r from the source?

Inverse square law only applies to a point source right? if light were emitted from an ideal plane doesn't the intensity decrease as 1/r rather than 1/r^2? or at least, until you are far enough away from the source that it appears to be a point source? If the plane had infinate boundries, intensity would decrease as 1/r.

shouldn't it be watts per atom, and not watts per atom per sec? since a watt = joule/sec

5. May 23, 2005

### Jonny_trigonometry

What are the differences in calculating light intensity between your way and cannonical physics?

6. May 23, 2005

### McQueen

Sorry Claude . The post should have read as follows :- Also light emitted by a electron bound to an atom will always obey the inverse square law because it is a point source while light from lasers also obeys the inverse square law but since they are the result of a comparatively massive collection of photons moving in unison , travel longer in a coherent beam. At a distance of 1,609m , a helium-neon laser (which is a well collimated source ) will spread out over an area of 1.3m.
Thank you Anzsas , this is the point I was trying to make .

7. May 23, 2005

### McQueen

Johnny : I will take my time over this , since I do not wish to make another gaff , as I did with the lasers.

8. May 24, 2005

### McQueen

Jonny Trigonometry wrote :

What are the differences in calculating light intensity between your way and cannonical physics?

Obviously one cannot explain the intensity of light in a new way without going into the whole question of what light actually is . The only available alternative is to examine one particular aspect of the subject. Take the photoelectric effect , according to the photoelectric effect , the emission of electrons from a metal surface when it is irradiated with a monochromatic light depends solely on the wave-length or the frequency of the monochromatic light and not on the intensity of the light, in other words the photoelectric effect does not depend on the amount of light but on the energy of the monochromatic light in question. (i.e., the type of light that the metal is being irradiated with) Now this according to the theory I have put forward , is a simple question of the use of AND and OR. The Laws of Photoelectric effect enunciated by Einstein , in 1908 and remember this was a century ago , states that the photoelectric effect depends on the wave-length OR the frequency of the light in question , while the theory I have put forward states that the photo-electric effect depends on the wave-length AND the frequency of the light. Thus the photoelectric effect as it is at present understood , attributes the property of frequency to a single photon , while in the classical sense the term frequency refers to a succession of events and therefore cannot apply to a single particle , especially when used in this sense (i.e., to state that it is the frequency of a single photon which is responsible for the photoelectric effect. ) This means that , according to my theory , the kinetic energy of electrons ejected from a metal surface should decrease with increasing distance from the source of illumination. This statement should be easily verifiable.
The reason that I make this statement is that at the time (i.e in 1908 ) when the photoelectric was first put into mathematical form , physicists could have had no idea at the speed with which an electron emits and absorbs photons , which as I had pointed out at the beginning of this thread can be as high as 4 x $$10^{-19}$$ secs per photon emission and absorption , when light is traveling through glass. In fact at the time photons had not even been discovered , it would take another twenty years or so before the term photon was coined. In view of this we have to examine the possibility , even given the calculations of the work function (i.e., the energy needed for an electron to leave the metal surface ) that the photoelectric effect is intimately connected with frequency when used in the conventional sense (i.e., as a succession of absorptions and emissions ) resulting in the raising of the energy of the electron. The problem does not end here because it then raises the question of how light is propagated , because we find that the intensity of light varies uniformly over a given area, which would seem to indicate that photons are emitted by the electron along straight lines. If the electron absorbed and emitted photons in the manner of a spinning Catherine wheel , this uniform lowering in intensity in a regulated manner varying with distance would be difficult to explain. So when light falls on a substance (which reflects that particular colour of light ) the electrons of the substance absorb and emit photons at the rate of approx . $$10^{14}$$ Hz in straight lines , obviously ( as we can observe ) different electrons would emit these photons in different directions.

Last edited: May 24, 2005
9. May 24, 2005

### Jonny_trigonometry

"The Laws of Photoelectric effect enunciated by Einstein , in 1908 and remember this was a century ago , states that the photoelectric effect depends on the wave-length OR the frequency of the light in question , while the theory I have put forward states that the photo-electric effect depends on the wave-length AND the frequency of the light."

the photoelectric effect uses the idea that the energy of a photon = it's frequency (or color) times plank's constant. It uses this idea because this idea is the only way of fully explaining the photoelectric effect. to say frequency OR wavelength is to refer to the same thing, because C=(wavelength)(frequency), so E=hf OR E=hC/(wavelength). That is all. It has nothing to do with the number of photons emmitted per second. The number of photons emmitted per second (from a monochromatic source) times the energy of the photons (since they all have the same energy because the source is monochromatic) divided by the cross-sectional area that they pass through (dependant on the distance from the source) is the intensity of the light. And the photoelectric effect demonstrates that the intensity is independant of the ammount of energy absorbed by bounded electrons in the metal surface.

"This means that , according to my theory , the kinetic energy of electrons ejected from a metal surface should decrease with increasing distance from the source of illumination. This statement should be easily verifiable."

In fact, the photoelectric effect verifies the exact opposite of your claim. You may be thinking that the frequency of the light steadily decreases as it propogates through space, but this is not the case, what really happens is the intensity decreases as the photons spread out since they aren't all traveling parallel, even in a lasar (although in a lasar they are very closely parallel). It is true that the NUMBER of electrons stripped from the metal surface per unit time increases when the distance from the source (assuming it is emitting photons with energy greater than the work function of that particular metal) decreases, but the kinetic energy of each ejected electron will not increase.

10. May 25, 2005

### Jonny_trigonometry

I think the confusion is with the use of the word frequency. There are two independant frequencies that we are considering here, the frequency of the photons (the color of the light) and the frequency of photons moving through a cross-sectional area. The first refers to the occilations per second of the electromagnetic disturbance (photon). The second refers to the number of photons (each with a specific frequency) moving through a cross-sectional area per second.

A photon doesn't output energy over time, only a photon source does. Photons have energy, but no power, a light source has power since it outputs energy (photons) over time. I shouldn't say a photon has no power at all times. It does output power when it collides with a particle, but it outputs all it's energy in roughly the time you suggested (4 x 10^-19 sec). Or at least, this is how (I think) they are thought to interact with particles.

11. May 25, 2005

### McQueen

Jonny Trigonometry

Much of what you write is true and as such is part of the established wisdom on the subject. For instance when you state that “I think the confusion is with the use of the word frequency. There are two independant frequencies that we are considering here, the frequency of the photons (the color of the light) and the frequency of photons moving through a cross-sectional area..” you are right in stating that much of the confusion arises from the fact that frequency is often treated as being synonymous with energy forgetting to add that it is also an inherent part of the properties of that particular electromagnetic radiation. You cannot talk of the number of photons moving through a cross sectional area per second without also taking into consideration that the frequency you refer to is an inherent property of that particular electromagnetic radiation and that it therefore follows that if there are a number of photons of a particular frequency moving through a cross sectional area , they would have to be in sequence , (i.e ., a number of photons of the same energy following the same path.). This leads to the second point of confusion which is that there did not exist in 1908 and that there still does not exist today in 2005 , apparatus which can determine whether an electron has absorbed a single photon or a number of photons in succession. To take one example , in the photoelectric effect , suppose the metal plate is 50 cms away from the light source then in the time taken for the light (using green light with a wave length of 555nM) to reach the metal plate in approx. 3.3 x 10^10 secs. an electron would have emitted 1.5 x 10^4 photons. Taking this into consideration is it possible to speak of single energies of photons giving rise to the photoelectric effect ? Although as I had earlier stated there exists no apparatus at present which can be used to determine whether an electron has absorbed a single photon or a number of photons in succession , observations and logical deduction can be used to establish whether in fact electrons emit single photons or whether they emit multiple photons in rapid succession in numbers corresponding to the frequency and wave length of the particular electromagnetic radiation under consideration. If we are to accept the first viewpoint put forward in your post , that frequency is a property of a single photon , then we must assume that photon emission with reference to visible light , consists of electrons emitting single photons at random , thus when a source is emitting green light we would have to assume (using your theory ) that an individual electron emits a single photon with a wave-length of 555nM and an energy of 2.2eV in a time of 5.4 x 10^14 secs and then becomes quiescent until it is again excited at some later time, in other words using your view of light , it must be assumed that photons are emitted at random and not in sequence . Yet our observations , particularly in the matter of light as it travels through glass , show that this is probably not true and that electrons are capable of emitting and absorbing photons at a phenomenal rate over long periods of time , which raises the conjecture that photons are emitted in sequence.

12. May 25, 2005

### Claude Bile

Okay, what Inverse square law are we talking about here, the inverse square law I am referring to is;

$$I=\frac{P}{4\pi r^2}$$

Where I is the Irradiance (W/m^2), P (W) is the emitted power and r (m) is the distance from the source.

There is a lack of understanding here about the propagation of light through a dielectric medium. In a dielectric, such as glass, with no free charges and currents, Maxwell's equations can be decoupled and formed into a wave equation. Thus light propagates in a dielectric medium much the same way it propagates through a vacuum (on a macroscopic level anyway).

The propagation of light through transparent media can also be discussed on this thread.

After reading your initial post, and subsequent posts, I'm still not quite sure what the point to your post was. What conclusions are you drawing, and why is this 'A new look at the intensity of light'. Do you mean intensity as in W/m^2 (more correctly called Irradiance), or W/sr?

Claude.

13. May 26, 2005

### Anzas

yes this is the inverse square law we are talking about

p is divided by 4*pi*r^2 because thats the surface of a sphere with radius r and the power is divded between the surface area.

no laser sends exact parallel radiation so at a distance r from the laser the radiation scatters as well and the power is divided between the surface area which is actually a part of a sphere with radius r. so lasers do obey the inverse square law.

14. May 26, 2005

### Claude Bile

Okay, I have a CO2 laser in my lab, which runs at 50W (which is smallish) at full pelt.

The inverse square law predicts that the Irradiance will be about 4 W/M^2 at a distance 1 m from the laser. What I actually get is about a 5mm diameter spot (a large estimate), so the irradiance is actually about 637 kW/M^2 a factor of about 160,000 off. In this instance, the inverse square law does not work. You cannot apply the inverse square law to lasers.

What I suspect you are claiming is the Irradiance goes down with increasing area, which is a statement of the bleeding obvious since Irradiance is in units W/m^2, so for a fixed power, this is exactly what you would expect.

Hardly a new look at Irradiance.

P.S. What do you mean the laser radiation scatters? Do you mean the laser radiation diffracts diffract?

Claude.

15. May 26, 2005

### McQueen

Claude Bile
Okay, what Inverse square law are we talking about here, the inverse square law I am referring to is; I=\frac{P}{4\pi r^2}
I have just read Anzas latest post which is self explanatory.Basicall I agree with what he has to say. My own theory is that the divergence of a laser is less because it results from a massive collection of photons all moving in unison rather than as the result of individual emissions by electrons , as a result of this the majority of photons in a laser beam are shielded from interactions as the beam propagates. Of course it is only a theory.
There is a lack of understanding here about the propagation of light through a dielectric medium. In a dielectric, such as glass, with no free charges and currents, Maxwell's equations can be decoupled and formed into a wave equation. Thus light propagates in a dielectric medium much the same way it propagates through a vacuum (on a macroscopic level anyway).
As regards your statements with regard to the propagation of light through glass , once again you are right and polarization has a lot to do with the manner which light propagates through glass , however “photon “ conduction also plays an important role. At higher temperatures, photon conductivity (radiation) becomes the predominant mechanism of energy transfer. This is a rapid sequence of absorptions and emissions of photons that travel at the speed of light. This mode of conduction is especially important in glass, transparent ceramics, and porous ceramics.
After reading your initial post, and subsequent posts, I'm still not quite sure what the point to your post was. What conclusions are you drawing, and why is this 'A new look at the intensity of light'. Do you mean intensity as in W/m^2 (more correctly called Irradiance), or W/sr?
Unfortunately , according to the rules of this forum I am not allowed to advertise or elaborate on my own theories and so am unable to post a link. That doesn’t matter .
However I can tell you is that my theory depends upon a new model of the photon which is a symbiosis of a particle and a wave i.e., it possesses the properties of a particle and a wave simultaneously , further this model of the photon allows for the storage and delivery of electrical energy while itself remaining electrically neutral , lastly this model of the photon also explains long wave length electromagnetic radiation such as radio waves and wave lengths of 5 x 10^6 m and more which present theories seemingly are unable to do in a satisfactory manner.

Last edited: May 26, 2005
16. May 26, 2005

### Claude Bile

First of all, what is the divergence of the laser beam less than? What interactions are the photons being shielded from? What does your theory predict that Gaussian optics does not? How can this theory be tested? You are being awfully vague with your terminology.

Is this 'photon conductivity' your own term you have coined? I have never seen this term used, and I have been involved with the field of photonics for quite a few years now (most people just use the word transmission). The explanantion you have given is the simplified version, oversimplified in my opinion as it contains many inconsistancies, for example why do the photons always emit in the same direction, when stimulated emission, by its nature is isotropic? (Also energy does not propagate at the speed of light in media).

Don't current theories model the photon as having both particle and wave properties? Don't they also predict electrical neutrality? I was not even aware that long wavelength radiation is unexplainable using current theory. What is unsatisfactory about current theories and what does your theory do to rectify it? Also, you would meet strong opposition if you tried to suggest that photons store and deliver electrical energy (I would give that role to electrons and other charged particles).

Again, I fail to see the relevance of all this to intensity.

Claude.

P.S. I just noticed an error on my previous post, obviously, the final 'diffract' should be omitted.

17. May 27, 2005

### Anzas

here i made a small pic in paint to show what i mean

the left figure is an antenna (from above) it sends radiation in all directions (even though in the image it looks 2d its actually a sphere) so its power is divided by the sphere surface which is
4*pi*R^2

the right figure is a laser (also from above) it sends radiation in only one direction but the radiation is not completely parallel (no laser sends exact parallel radiation) so at a distance R from the laser the power is divdided by a so called "part of a sphere" (im not sure what the exact term is in english) so the power is divided by 4*pi*R^2/x
x depends how parallel the radiation from the laser is.

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18. May 27, 2005

### McQueen

Claude Bile

First of all, what is the divergence of the laser beam less than? What interactions are the photons being shielded from? What does your theory predict that Gaussian optics does not? How can this theory be tested? You are being awfully vague with your terminology.

OK if a laser beam did not diverge , we would be using laser pulses for all deep space missions , which is not being done for the obvious reasons. Secondly my theory states firstly that photons are solenoidal points , this means that they are the fundamental unit of electromagnetic energy , and are therefore the source of electromagnetic radiation and not the symptom.

Finally , one of the fundamental hypotheses behind my theory is that electricity is transmitted in a wire by photons and not by electrons. Look at the facts , a current is established at close to the speed of c , whereas the drift velocity of electrons is about 10^-5 m/s . How do you account for this , it represents a discrepancy of about 10^15 , this is not a small discrepancy. Remember that electrons are separated by about 10^5 their diameter , for one electron to bump into another would be the equivalent of trying to hit one billiard ball with another at a distance of about 6 Kms. Again using extant theories , how is it possible for an ion about 10^-10 m in size to produce by vibrations , an electromagnetic wave of 5 x 10^6 m wave length ? Let’s face it , there a lot of facts to do with electromagnetic radiation which do not gel together. At present electromagnetic radiation in the range of visible light is supposed to have different origins than electromagnetic radiation of longer wave lengths , although both these phenomena exhibit identical properties . They preserve their energies , move with the speed of light , have no mass etc., Is this good science , attributing two different causative factors to two identical phenomena. The same applies to the magnetic field around a current flowing in a wire and the magnetic field around a permanent magnet. The first is thought to be due to the vibration of ions in the lattice structure of the wire while the other is thought to be due to the lining up of electron spin. Again the rationale behind the explanation for the double slit experiment is literally out of this world , for it states that particles possessing both dimensions and mass , such as neutrons are able to disassociate and be in two places at once , (i.e., pass through both slits at the same time!). What about the polarization of light ? All these explanations do not make sense , my theory , provides a simpler and more rational explanation for all these phenomena. My theory explains electricity , explains electromagnetism , explains magnetism , explains super conductivity and even gravity without deviating from a central hypotheses. Finally , there are several simple experiments which would prove my theory , it is also possible to prove the theory mathematically.
Don't current theories model the photon as having both particle and wave properties?
According to present theories light is either a wave or a particle , it can never possess both properties simultaneously.

Last edited: May 27, 2005
19. May 27, 2005

### McQueen

Claude Bile
Don't they also predict electrical neutrality? I was not even aware that long wavelength radiation is unexplainable using current theory. What is unsatisfactory about current theories and what does your theory do to rectify it?
Here is a very brief synopsis of my theory , without which , nothing I say will make sense. It is a widely accepted fact that electrons emit and absorb photons. Since electrons are charged particles I thought it not unlikely that what in fact the electrons were exchanging were pulses of electrical energy , when an electron had an excess of charge it emitted an electrical pulse and absorbed a pulse when it was deficient. Next I reasoned that because of the extremely small size of the electron , ( the classical radius of the electron is something like 10^-15m ) that these pulses would be released in short bursts . Thus the picture that emerges is of short bursts of electrical energy separated by a di-electric. The possibility then arose that the bursts of energy first released by the electron were more electrically charged than subsequent bursts of energy , therefore the system takes on polarity. Obviously a structure such as I have outlined would form around itself a solenoidal field and would therefore be electrically neutral. It would have no mass , because of its condensor like structure it would be able to store energy almost indefinitely and it would also deliver up all of that energy on contact. Thus such a structure would seem to answer to all the properties that a photon is said to possess and at the same time sounds more reasonable in terms of dimensions etc., for the electron to deal with. My theory also sets a limit on the size of the largest wave length photon that an electron can emit. This size I have worked out is about 8 x 10^-7m. This is only slightly larger than the longest visible wave-length. Thus all wave lengths greater in length than 8 x 10^7m are composite waves and are made up of joined together or linked together photons.

20. May 29, 2005

### Claude Bile

The "Part of a sphere" is termed a solid angle in english, whose units are steradians (which are dimensionless). All you are doing is manipulating the inverse square law so the power is only being spread out over a portion of the sphere and not the whole sphere right?

I'm not saying you're wrong, I an arguing that once this equation is modified, you are no longer using the inverse square law.

I asked what you were comparing the divergence of the laser to. In your earlier post you wrote;

Less than what? That was my original question.

What? even DC? Why then do metals make good conductors and dielectrics bad conductors?

Drift velocity and the group velocity of a disturbance in the electric field of a wire are two different quantities. Current theory does not claim that the electrons move at c, it is the disturbance (or, more correctly, the pertubation) that moves at c.

Easy, you just jiggle it up and down at around 50 Hz. The free electrons in a conductor are not bound to any nucleus, thus they do not require transitions of any kind to emit certain frequencies, they are free to emit all frequencies.

What do you mean they have different origins? Also, I would argue that electromagentic waves have properties that vary with wavelength. Try convincing someone that Gamma rays and visible light have identical properties.

This touches on Quantum Mechanics, one of the most outstandingly successful theories in modern physics. You need to have damn good reasons to question it's validity (i.e. much more than "it doesn't make sense").

So prove your theory then, (mathematical proofs don't count, they need to be backed up by observation). You'd probably get a nobel prize.

Light posesses wavelength (a wave property) and momentum (classically, a particle property), simultaneously.

What???? An electron with an excess of charge? Never mind the evidence that the charge on the electrons has been measured to 9 significant figures last time I looked.

Energy cannot posess charge, charge is a particle property.

What do you mean "for the electron to deal with". What about blackbody radiation?

So the longest wavelength photon is 800nm? What about all those lasers that emit wavelengths much longer than 800nm. I'm sorry to say that your theory falls apart right here.

Claude.