“aumic” theory and the propagation of light.

1. Aug 12, 2004

McQueen

I am starting a new thread on this subject because I feel that the subject under discussion is of the greatest importance. The “aumic” theory on the nature of light brings to our attention , for the first time , several peculiar and interesting aspects on the nature of light and indeed of all electromagnetic radiation:

(1) In the course of their investigation into light physicists had discovered something very strange , namely that the speed of light disregards the classical laws of Galilean Transformations . This seems to be impossible , but experiment after experiment proved just the opposite. Thus in the normal sense if two cars are approaching each other at 80 km/hr on the highway , then we say that their speeds relative to each other is 160 km/hr. In the same way if two cars are heading in the same direction , one going at 80km/hr and the other at 100 km/hr , when the car moving at 100 km/hr passes the car moving at 80km/hr , the driver of the slower car would note that the faster car was moving away from him at a relative speed of 20 km/hr. Yet the speed of light is the same for all frames of reference , thus if you are standing still or are moving at any velocity , the speed of light would remain constant. For instance if you are moving towards a distant light source at a speed of 100,000 km/sec , then it is assumed that the speed of light relative to you would be 400,000 km/sec , this is not the case , the light would have a speed relative to you of 300,000 km/s . Similarly if you are moving away from the source of light at 100,000 km/s , then it should be assumed that the speed of light would be 200,000 km/s. (i.e., the speed of light minus your speed.) Yet this is not the case the speed of light remains constant at 300,000 km/s. This was the paradox which worried physicists and which was finally solved when it was stated that the speed of light is a fundamental constant. Notice , no explanation is given for why the speed of light is constant in all frames of reference , it is merely stated as an axiom that the speed of light is a fundamental constant. Now for the first time “aumic” theory explains that the speed of light is due to the manner in which light ( and all electromagnetic radiation ) propagate. Thus , the front of the propagating wave is constantly renewing itself , so that at any point in time it can be said to have a constant velocity of 300,000 km/s. The front of the propagating wave as it advances and spreads over a wider area , has gaps in it which are occupied by virtual photons , which are promoted to real photons by a transfer of energy from real photons at the back of the propagating wave , the photons at the back of the wave which give up their energy are transformed into virtual photons , thus the number of photons and the total energy involved is always constant. However when a virtual photon is promoted to a real photon it naturally , results in a new line of force being formed , thus light can , at any moment of time be thought of as having a constant velocity , or as starting anew at each point. The other great question concerning light was the problem of how light could be represented as a traveling wave and as individual photons , because light manifests both properties. This problem is also solved by the “aumic” theory explanation for the propagation of light. This is a very significant contribution. The “aumic” theory for the propagation of light explains every aspect of light , including the superposition of waves which lead to fourier wave forms and the constancy of the eigen values of the photons comprising the waves.

(2) The other point of interest arises from the statement in “aumic” theory that the number of photons reaching a surface 1 metre from the source or 100,000,000 kms from the source remains constant. Thus if you have a 1 m sq surface and place it t a distance of 1 m. from the source , the number of photons reaching the surface would remain the same even if you shifted the 1 m sq. surface to a distance of 100,000,000 kms from the source , always supposing that the wave can propagate so far. “aumic” theory states that the number of photons remain the same and it is only the intensity of the photons which changes. Note , not the eigen energy of the photons , which also remains constant , only the intensity of the photons undergoes change. This goes completely against all accepted views held by QM which states in effect that the number of photons reduces. Even in these forums terms , such as intensity is reduced because one in a hundred photons reaches its destination.

(3) Intensity of a source of electromagnetic radiation , depends according to “aumic” theory , on the number of excited electrons per area which are emitting photons of a given eigen value and the duration for which the photons are emitted. Thus when dealing with light , the intensity of the source depends upon the number of photons of a given eigen value in each line of force (i.e., aligned lines of virtual photons.) When dealing with electromagnetic radiation of radio wave-lengths , intensity depends solely upon the power flowing in the conductor and the resultant number of lines of force this power gives rise to .

(4) Lastly , is the question of why light travels at 300, 000 km/s ( in a vacuum)? ‘aumic” theory explains this as being due to the fact that this is the fastest speed at which light can travel , to go faster than this would mean ending up at it’s destination before leaving. So in this sense the speed of light is a manifestation of the fine structure constant 1/137 and deserves to be investigated with more enthusiasm and vigour. For instance what is the criteria which governs this particular speed , is it the time taken for light to travel the distance represented by the radius of an atom ? How does it connect to the fact that we live in a causal Universe?

2. Aug 12, 2004

Staff: Mentor

That's all well and good, but you're changing the definition of a "photon" there (something I'm pretty sure would go against how we observe photons to behave). You'll need to specify what you mean, because as we know them, photons only have one energy and all photons of the same frequency are identical. We know this from observation - we can observe and test individual photons.

3. Aug 12, 2004

AWolf

This implies that the speed of light is dependant on time - (distance/time).

Is it not possible that time is determined by the duration taken for light to travel its minimal distance. The distance being directly related to the size of a photon and a virtual photon - somewhere in the region of Planck's Constant.

If this were the case, then it gives a very straight forward explanation for why light does suffer from time dilation when dealing with relativity.

Light (or the transmission of energy) determines time and subsequently is not dependant on it.

4. Aug 12, 2004

McQueen

That's all well and good, but you're changing the definition of a "photon" there (something I'm pretty sure would go against how we observe photons to behave). You'll need to specify what you mean, because as we know them, photons only have one energy and all photons of the same frequency are identical. We know this from observation - we can observe and test individual photons.

My mistake , I should have specified that the light I referred to was monochromatic light , which I had chosen because it poses the most uncomplicated case. In the case of white light there would be electrons emitting photons of different eigen values , at different frequencies. However the point you had raised does serve to illustrate , that light of different frequencies and wave-lengths can interfere to give a new wave-length (white light ) while maintaining their own individual wave-lengths and energy values as can be seen when sun light passes through a prism.

5. Aug 12, 2004

McQueen

This implies that the speed of light is dependant on time - (distance/time).

Is it not possible that time is determined by the duration taken for light to travel its minimal distance. The distance being directly related to the size of a photon and a virtual photon - somewhere in the region of Planck's Constant.

The speed of light is very fast and seems to have a fundamental role to play in the working of the Universe. Suppose and the operative word here is suppose , that a plane existed which could fly at the speed of light. It starts out at the airport goes round the world and returns before it has taken off. So it’s really a very basic thing without the barrier represented by the speed of light our ordered Universe would become chaotic. What it means is that if the speed of light could be exceeded time would cease to exist.

6. Aug 12, 2004

Staff: Mentor

If one photon is fired at a time, it is, necessarily, monochromatic. But that isn't the issue. The issue is that it can and has been shown that that one photon has an energy that doesn't change with distance. Your "intensity" issue is not observed. It is false.

It has also been shown that an individual photon is capable of interfering with itself.

7. Aug 12, 2004

what_are_electrons

8. Aug 13, 2004

Chronos

Last edited: Aug 13, 2004
9. Aug 13, 2004

McQueen

The issue is that it can and has been shown that that one photon has an energy that doesn't change with distance. Your "intensity" issue is not observed. It is false.

The energy of a photon refers to its eigen value , that is the eigen value of an individual photon , the intensity depends on how many photons are available in a single line of force or to put it another way how many successive photons of the same eigen value deliver their energy at a single location ( electron ) in a second. The physical structure of the photon in "aumic" theory and the method of propagation both support this statement , which in fact is exactly what is observed. OK , take the Voyager spacecraft , sending back signals from the vicinity of Saturn 1.5 billion kms away , how would you account for the inensity in the received signal , would you for instance say that the intensity of the signal is reduced with distance because there are fewer photons per receiving area ? Or is there some other explanation ?

Last edited: Aug 13, 2004
10. Aug 13, 2004

NEOclassic

Hi McQueen,
I have noticed that all but three of your thread starts are in this venue and I find that your reason for being here is quite similar to mine in that I believe that all light is corpuscular - i.e., Faraday/Maxwell equations are classic only and light, perse, is niether electric nor magnetic. Because this venue is my favorite I feel comfortable submitting you as "BUDDY". I anticipate your interest in sharing so many of our apparant common interests in expanding Quantum-Mechantics into the much broader theory which I call Quantum-Realm that includes the two varieties of positronium quantum orbitals. Feel free to contact me personally - click on my moniker, NEOclassic, to view my internal private message address. Cheers, Jim Osborn

11. Aug 13, 2004

Staff: Mentor

Actually, I was referring to the fact that the double-slit experiment can be done with single photons fired one at a time and you still get an interference pattern.

edit: To elaborate a little more, if you fire one photon at a time, it "should" only go through one slit and not even know the other slit is there. You should not get an interference pattern.

You can also get a beam splitter prism to split a single photon and recombine it. But if you try to detect either split photon, you get a whole photon. The photon "knows" if you're going to try to detect it before it gets recombined.
http://www.uni-konstanz.de/quantum-optics/quantech/nonlocality.html
Thanks (and no prob) - I don't always give the gory details partly because I can respond to more posts that way and partly because I'm an engineer, not a physicist and there are a lot of people who know the specifics a lot better than me.

Last edited by a moderator: Apr 21, 2017
12. Aug 13, 2004

Staff: Mentor

Ok, good - to be more correct, though, its photons per unit area. But you're contradicting yourself (and observations) now. You said both that the number of photons remains the same while the intensity varies and also that the number of photons varies:
For that to be possible, the photons have to multiply. Same number of photons per unit area with more area means more total photons. From earlier:
Ok, here you are saying that while you see the same amount of photons per unit area, the "intensity" of the individual photons varies. But that directly contradicts what you said in the first quote.

Science doesn't work via making stuff up as you go along.

13. Aug 14, 2004

McQueen

Russ Waters
Actually, I was referring to the fact that the double-slit experiment can be done with single photons fired one at a time and you still get an interference pattern.

edit: To elaborate a little more, if you fire one photon at a time, it "should" only go through one slit and not even know the other slit is there. You should not get an
interference pattern.

This is a classical statement proving that the virtual photon field exists . It is almost impossible to get individual photons for the experiment but the point is electrons , protons and neutrons , which can be found individually , behave in exactly the same way. If there was an all pervasive virtual photon field when both slits are open the particle would follow wherever the field was most dense . Surely this is a better explanation than trying to state that particles can become disassociated b’cos of their wave-particle properties and pass through both slits at once. Or that they can intuitively sense when both slits are open ?

Ok, here you are saying that while you see the same amount of photons per unit area, the "intensity" of the individual photons varies. But that directly contradicts what you said in the first quote.

Science doesn't work via making stuff up as you go along.

This is just a question of semantics , what I was saying is that intensity is due to the number of photons present in a given line of force , in other words the amount of energy that could be given up by a succession of photons all arriving at the same location , represents intensity. The individual photons involved maintain their eigen values. If we look at the photoelectric effect , it is surely odd that if it is the eigen value that is responsible for the ejection of electrons from the metal surface , why are no electrons ejected at all ( taking a 1W ultraviolet LED as the source ) at a distance of 10m ? At 10 m. you can perform the calculations yourself , there would still be a sufficient number of photons 10 7 photons striking every millimeter of the metal surface, according to the present theory . Surely at least one of these photons should strike an electron and eject it from the metal surface ? Normally what happens is , suppose an electron absorbs a photon of a given eigen value , it almost immediately re-emits a photon of the same eigen value. To actually eject an electron from the metal’s surface a succession of photons of a given energy are needed to strike the electron in rapid succession. If electrons can emit photons of a given eigen value at the same rapid frequency ( for instance in order to give rise to light of a certain colour ) it should also be possible for electrons to absorb photons at the same frequency , in rapid succession , resulting in so much gain of energy that they are ejected from the surface of the metal. So the photoelectric effect depends upon the intensity of the photons striking the electron , in rapid succession and on their combined energy and not on the energy of a single photon as has been previously believed. An option exists here , w can either have a pointless argument or we can conduct the experiment perform the calculations and find out which opinion holds good. If you remember this has been the accepted practice , when somthing has been proved indisputably wrong by observation and/or experiment , it has to be accepted and things have to be re-worked or re-thought. Bohr was the first to accept ,when the objection was raised ,that his model did not explain how the electron stayed in orbit around the nucleus or of how it made quantum jumps from orbit to orbit. It is this ability to accept the situation when observation and theory are in conflict that leads to good science.
Russ waters
Ok, good - to be more correct, though, its photons per unit area. But you're contradicting yourself (and observations) now. You said both that the number of photons remains the same while the intensity varies and also that the number of photons varies:
I had stated that both the number of photons and the total energy remains the same , the photons are just re-distributed into a thinner and thinner line. Thus at the source the lines of force are heavily populated with photons , as the wave propagates the photons from these lines of force move into gaps created in the propagating front as it advances. Once the propagating front reaches a stage where it is only a single photon deep the propagating wave breaks up and the photons comprising it lose their energy and become “virtual” photons. This is what I had stated at my site : http://www.geocities.com/natureoflight/pgindex.com

Last edited: Aug 14, 2004
14. Aug 14, 2004

McQueen

Neoclassic

Thanks buddy !! I was thinking of constructing a time – line against which we can check how quantum mechanics evolved and where it has made assumptions that might turn out to be wrong. Here goes :

1) 1803 - Thomas Young conducts the first “Double Slit Experiment “
proving once and for all the wave nature of light , diffraction ,
reflection and interference. Only waves can experience
interference.
2) 1900 – Max Planck puts forward his quantum hypothesis and introduces the Planck constant

3) 1905 - Albert Einstein explains the photoelectric effect , apparently
proving indisputably the particle nature of light , which he called
quanta or photons.

(Now physicists were in a quandary , Young had proved that light was a wave and Einstein , equally indisputably had proved that light had a particulate nature )
4) 1923 Neils Bohr puts forward his theory of the specific orbit model of the atom. But wasn’t able to explain either quantum jumps or the classical paradox of why the electron did not spiral into the nucleus.
5) 1924 Prince Louis de Broglie , introduces the concept of matter waves in order to resolve the differences between Young and Einsteins observations.
6) 1926 Erwin Schrodinger , his interest sparked by the de Broglie hypotheses , introduces his concept of the standing wave model of the electron , thereby explaining the classical paradox of the electron but offering no solution for the phenomenon of quantum jumps.
7) 1927 The Copenhagen interpretation of quantum mechanics is formulated .

It is easy to see , looking at this time line that many interpretations had been made which in the course of time turned out to be wrong. The most important of these being the inclusion of wave-particle duality.

Last edited: Aug 14, 2004
15. Aug 14, 2004

Staff: Mentor

McQueen, what is this "line of force?" Line? Force? That doesn't make any sense. Are you talking about photons traveling one behind the other on the same line? Intensity is photons travleing next to each other - photons per unit area.

16. Aug 14, 2004

McQueen

McQueen, what is this "line of force?" Line? Force? That doesn't make any sense. Are you talking about photons traveling one behind the other on the same line? Intensity is photons travleing next to each other - photons per unit area.
A line of force pertains to the virtual photon field and denotes how the field lines up. This is how the energy of the photon is transmitted according to "aumic" theory. If we take your theory then , taking the example of a 40w light bulb shining on a 1 m sq surface at a distance of 1m you would have 10 10 photons per hundreth of a square millimetre ? (presuming the radius of action of the photon is about 105m. Your distribution still doesn't make sense , b'cos each electron would have to absorb more than (one !) photon. Which is not in agreement of your definition of intensity as 1 photon , one eigen value per electron.

Last edited: Aug 14, 2004
17. Aug 14, 2004

Staff: Mentor

Gibberish, McQueen, all of it. The geometry of light intensity - number of photons per unit area - is junior high simple. Your "theory" just plain doesn't work. Photons don't multiply in transit.

18. Aug 15, 2004

McQueen

Russ Waters
Gibberish, McQueen, all of it. The geometry of light intensity - number of photons per unit area - is junior high simple. Your "theory" just plain doesn't work. Photons don't multiply in transit.

I did not say that photons multiply in transit. I quote from one of my own posts in this thread both the number of photons and the total energy remains constant …………… Well Russ all I can say is , let it be gibberish , p’raps that’s a more sensible viewpoint for you to take , than a disinterested look into what should be an interesting observation namely that the present conception of treating the intensity of a source of electromagnetic radiation and the eigen value of the photons comprising that radiation as one and the
same thing !

19. Aug 15, 2004

Chronos

Try applying the inverse square law.

20. Aug 16, 2004

McQueen

Try applying the inverse square law.
I think it is time for a more in-depth explanation of my previous comments. In order to understand the arguments I had previously made it is first necessary to make some assumptions :
1) The first assumption is that all photons , whether belonging to x-rays , radio waves or light , have a radius of action of about 10-7 m. This is not unreasonable given the fact that the classical radius of the hydrogen atom is 10 -8 m. which makes it reasonable that the radius of larger more complicated atoms might be in the region of 10 -7 m. to 10-8 m.
(N.B: in order to understand how all photons can have the same radius of action , when it is well known that wave-lengths vary from 10 -12 m to 106 m it should be understood that “aumic” theory states that EM radiation with a wave-length longer than 8 x 10-7 m
are composite waves , consisting of linked together chains of conduction photons the wave-length of which is 8 x 10-7 m.)
2) If we make the assumption that the radius of action of all photons is approx: 10 -7 m
then this means that when photons propagate they are limited by this distance , i.e ., the propagating wave of photons will be split into discrete 10 -7 m sq blocks. This in turn means that the amount of radiation reaching a destination from a source is limited by the same distribution(i.e., photons are distributed evenly across the receiving surface in blocks of 10-7 m sq. ) A corollary to this is that the number of photons striking a given area does not vary with distance , thus whether the receiving surface is 1m distant from the source or 1,000,000 km from the surface does not make a difference , the number of photons remains the same . Thus a 1m sq surface at a distance of 1m from the source receives 10 14 photons and a 1m sq surface at a distance of 1,000,000 km receives the same number of photons.( 1014 photons ) What does this mean ?
3) Each photon emitted by a radiating source , results in a line of force , (i.e., an alignment of the Virtual photons of the “virtual photon field” . Close to the source these lines of force are heavily populated with real photons. As the distance from the source increases , the depth of photons in these “lines of force “ decreases , while the area over which the photons spread increases , this increase in the area over which the photons spread out is made possible by the structure of the photon and is inversely proportional to the square of the distance from the source. The result is that as the depth of photons in the “lines of force “ decreases so does the intensity . At the same time the photons retain their original eigen values and are present in their original number. Thus intensity is not due to the eigen value of individual photons but to the number of photons with that particular eigen value which are absorbed in rapid succession by an electron at the receiving surface.
4) Example: Take a 4w ultra-violet diode with a wave length of 300 nm. A photon of this wave-length has an eigen value of 3.1 eV which translates into 3.1 x 10 -19 J. Thus the number of photons emitted in one sec. is equal to 4 / 3.1 x 10 –19 J = 1.2 x 10 19 photons thus at 10 cms from the source ( we consider that the light from the diode is directed ) the number of photons spread across a 10 cm sq surface would be 106 x 106 photons , or 10 12 photons. Since the total number of photons emitted is 1.2 x 10 19 photons each line of force at this point would have an intensity of 1.2 x 10 19 / 10 12 photons
= 107 photons per line of force . This results in what we know as the photo-electric effect , wherein electrons are ejected from the metal’s surface. In other words each ejected electron , has absorbed in rapid succession , a number of photons giving it the necessary energy to be ejected from the metal surface.
Now if we increase the distance from the source to 10m the number of photons striking this surface would be 108 x 10 8 or 10 16 photons . Thus at this point each “line of force” would have 1.2 x 10 19 / 10 16 photons or 103 photons . Note the number of photons striking the surface remains the same per area of surface only they are striking with less intensity , thus instead of 107 photons per line of force we have 103 photons per line of force. The result is that no photo-electric effect is observed at this distance. A corollary to this is that high energy photons travel shorter distances than low energy photons.

Last edited: Aug 16, 2004
21. Aug 16, 2004

Staff: Mentor

But you also say the number of photons per unit area stays the same. That's a contradiction: either the photons must multiply in transit or you don't understand the inverse square law.

And your assumptions are unreasonable. There is nothing about photons that limits the number that can occupy a given space.

22. Aug 16, 2004

AWolf

This just says that 'the lines of force are proportional to the distance and inversely proportional to the intentisity'.

There is nothing here about high energy photons and the distance they can travel, only that your lines of force effectively sub-divide. The photons emitted have not been altered and if all the photons are the same wavelength, then they will have the same energy.

23. Aug 17, 2004

McQueen

The photons emitted have not been altered and if all the photons are the same wavelength, then they will have the same energy.
True. All photons of the same wave-length have the same energy. I did not state that all photons have the same wave-length but only that they had the same radius of action. Thus even photons of different energies share the same radius of action regardless of their wave-length. The longest permissible wave-length for any photon is 8 x 10 -7 m. All wave lengths longer than this are composite wave lengths. You know that visible colours exist in light, how do these colours exist. It is not that your eye detects single photons of this particular energy , but that the electron emits a whole string of photons in rapid succession of that particular eigen value (energy). Thus for blue light the electron would emit , vary rapidly and in succession photons of 550nm wave length at a frequency of 1014 photons per second. This is how blue light becomes visible to us. It is not that several different electrons are emitting single photons of 550nm at the same time , it is the frequency with which they are emitted which gives us our impression of blue light. This is a common place phenomenon , thus intensity is represented by the number of photons absorbed and not only by the eigen energy value of the photon which is absorbed. A photon with a wave length of 550nm and an eigen energy value of 3 x 10 -1 eV would have all the characteristics of blue light but its intensity would be negligible. Light follows the same principles of intensity as radio-waves.

Last edited: Aug 17, 2004
24. Aug 17, 2004

Staff: Mentor

More gibberish, McQueen. What is this, out with the old gibberish, in with the new?

First off, this "radius of action" thing doesn't exist.

Second, maximum wavelength...? This morning on the car ride to work, I was listening to the radio at 94.1mhz. What's that wavelength? How, precisely does a radio tower produce this wavelength?

25. Aug 17, 2004

McQueen

Second, maximum wavelength...? This morning on the car ride to work, I was listening to the radio at 94.1mhz. What's that wavelength? How, precisely does a radio tower produce this wavelength?
Russ , you can't do this for yourself ? I'll give you a tip 94.1 MHz is so many KHz.