The Photon

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Offline Ron Hughes

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« on: 10/02/2010 23:30:57 »
An accelerated charged particle will emit radiation perpendicular to the direction of acceleration.  http://www.cv.nrao.edu/course/astr534/LarmorRad.html   It seems reasonable to think that accelerating a charged particle changes the position of it's field with respect to it's original position (observer) and that change propagates at C. I propose that change is what we call a photon and have shown that it can propagate without the need of virtual particles as the medium. The medium is the charged particle's own field. One should ask the question, " If that is true why doesn't the particle emit radiation just moving through space? ". The particle must be accelerated with respect to any observer to emit radiation. If the observer is accelerated at the same rate as the particle then the field lines of the observer change at the same rate as the particles and therefore does not see the particle emitting radiation.
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Offline Ron Hughes

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« Reply #1 on: 11/02/2010 16:26:23 »
No comments.

Proposing ideas outside the standard model is like trying to dig your way through a concrete wall using only your finger nails. Changing mainstream  physics is as difficult as changing the orbit of the moon.
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Offline BenV

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« Reply #2 on: 11/02/2010 16:35:30 »
No comments.

Proposing ideas outside the standard model is like trying to dig your way through a concrete wall using only your finger nails. Changing mainstream  physics is as difficult as changing the orbit of the moon.
Well, to be fair it hasn't been here very long.

But yes, changing mainstream physics should be very, very difficult.  There's a vast body of work to go up against.  It's only right that new ideas should not be immediately accepted, but thoroughly tested, poked and prodded first.

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Offline Ron Hughes

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« Reply #3 on: 11/02/2010 17:33:07 »
Your right Ben. Right now QM is mainstream physics in that it has been so successful at predicting the probability of events. That success has led to the standard model which has been around for forty or fifty years. Someone outside the mainstream can invent a hypothetical mechanism to make their theory work and the mainstream people will dump all over him/her. They on the other hand can invent the Higgs, graviton and the virtual particles without which the standard model goes down the toilet. I am using nothing but actual observed phenomena to justify the idea.
« Last Edit: 12/02/2010 03:27:03 by Ron Hughes »
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Offline Geezer

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« Reply #4 on: 12/02/2010 00:26:54 »
Ron: Can EM radiation be accelerated? I would have thought it would travel between source and destination at C.
There ain'ta no sanity clause, and there ain'ta no centrifugal force Šther.

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Offline Ron Hughes

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« Reply #5 on: 12/02/2010 03:25:06 »
As you already know EM waves propagate at C in a vacuum so it cannot be accelerated. It does travel from the source (the accelerated charged particle) to the observer at C.
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Offline Geezer

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« Reply #6 on: 12/02/2010 07:43:29 »
As you already know EM waves propagate at C in a vacuum so it cannot be accelerated. It does travel from the source (the accelerated charged particle) to the observer at C.

Ron: Thanks! I understand your point about acceleration now.
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Offline Ron Hughes

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« Reply #7 on: 15/02/2010 18:48:12 »
It should be understood that a charged particle has an electric field extending from it three hundred and sixty degrees in all directions. A portion of that field will be represented by a line in the direction of the field in the diagram below.


[diagram=559_0]

The dashed line above the particle on the right represents a field line of the stationary charged particle. The horizontal line between the vertical dashed and solid line represents the changing field as the stationary particle is accelerated. The field of the stationary position of the particle would disappear at C. The new field position of the accelerated particle would expand at C. To an observer the horizontal line would appear as the magnetic component of the photon and the vertical portion of the accelerated particle as the electric component of the photon. A representation of how it would look after acceleration below.

[diagram=560_0]

The vertical dashed line represents the field of the stationary particle disappearing at C. The horizontal is the magnetic portion of the new photon and the vertical is the electric portion of the new field that expands at C. It should be understood here that a three dimensional picture of the photon would resemble an expanding torus. Using this picture of the photon I will explain the photon double slit experiment and the electron double slit experiment in my next post.
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Offline Ron Hughes

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« Reply #8 on: 15/02/2010 19:27:42 »
In the photon double slit the experimenter designed the emitter device to emit a single photon, but if my view of the photon from my last post is correct it can be seen that the experimenter was releasing a complete wave that would always pass through a single or double slit producing the corresponding interference pattern.

In the case of the electron double slit The details of exactly how the devise was designed is relatively inadequate. The emitter releases an electron that is accelerated toward a positive plate that has the double slit. Again from the post above we know the electron will produce a wave that will pass through a single or double slit producing the corresponding interference pattern.
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Offline Farsight

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« Reply #9 on: 16/02/2010 01:27:07 »
Ron, I'm confused. An electron is a charged particle, the only acceleration you can perform on a photon is to change its direction, eg via Compton scattering.

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Offline Ron Hughes

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« Reply #10 on: 16/02/2010 04:02:25 »
The thread is about charged particles and how they produce the photon.
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Offline Ron Hughes

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« Reply #11 on: 16/02/2010 22:33:17 »
A good example of the first premise. Suppose that the only thing created by the BB  was a single proton. The proton's field would start expanding creating space time. Now suppose some force suddenly shifted the proton. It's field would start expanding again and the difference between the old still expanding field and the new expanding field would be that Universe's first photon
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Offline yor_on

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« Reply #12 on: 16/02/2010 23:59:56 »
Okay Ron, take a look at this movie and see if you feel that you can cover all of it. there are several experiments described, one after another. http://www.youtube.com/v/DfPeprQ7oGc and then try to describe how your EM-wave/photon is thought to cover them all. And try to give a extensive explanation, so you can be sure of that we understand all your thoughts on it.

But observe the last experiment and how it changes the outcome, although you really will need to cover all of them. And you will also have to explain how 'black body radiation' comes to be, the photoelectric effect, the discrete emission of electrons from metal surfaces illuminated by light. . Those are the first crucial experiments I can think of.
==
You might need to let it load a while first, but it's worth it :)
« Last Edit: 17/02/2010 00:13:45 by yor_on »
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Offline Ron Hughes

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« Reply #13 on: 17/02/2010 17:20:01 »
   I'm sure the part of the double slit experiment you want me to explain is where the electron acts like a wave in one instant but a particle in the next. In the electron double slit they heat a tungsten element until it starts emitting electrons at high speed and they can set it to emit one electron at a time. The electrons pass through one slit or the other and are detected by a device called an electron multiplier. I won't go into a complete description of how it works, suffice it to say that if a single electron hits the emissive material it creates an avalanche of electrons that can be detected. All of this is done in a vacuum. There is one problem with this type of detector. One photon with the right energy can also trigger this avalanche. The designers of this experiment had obviously ruled out the possibility of any photons in the experiment and as we know that is not the case. Now if we put a detector in front of one slit in order to find out which slit the electron/photon goes through we get one maxima which implies that it is acting like a particle but guess what, if you put a detector in front of one slit in a photon double slit device you get exactly the same result. So the interference pattern is coming from the photons created by the accelerated electrons.

There may be something we can add to the blackbody-radiation idea. Let me study it for a couple of hours and I will get back to it.
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Offline Ron Hughes

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« Reply #14 on: 17/02/2010 20:12:45 »
Yor, the number of different problems in blackbody-radiation is enormously interesting. I will work on explanations and post as I go along. Thank you for pointing me to this.
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Offline Ron Hughes

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« Reply #15 on: 17/02/2010 20:39:56 »
This, http://galileo.phys.virginia.edu/classes/252/black_body_radiation.html , appears to be a very good description of blackbody-radiation. I will try to post explanations using my description of the photon.Bear with me.
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Offline yor_on

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« Reply #16 on: 17/02/2010 20:53:44 »
Nice link Ron :)
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Offline Ron Hughes

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« Reply #17 on: 18/02/2010 16:10:18 »
We know that electrons emit and absorb radiation equal to the energy of their orbitals. That energy is still created by the acceleration of a charged particle. The acceleration must be exactly right to produce a photon with the correct energy to be absorb and re-radiated by that electron. Any acceleration that  produces radiation outside the required energy level will not be absorbed. Blackbody-radiation is created by the charged particles that it is made of vibrating (accelerating) to create the photons that are emitted. I do not see anything in this paper that contradicts my theory. If you see something that needs clarification Yor-on please point it out to me.
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Offline yor_on

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« Reply #18 on: 19/02/2010 07:14:27 »
Ron I'm still trying to get to grips with your idea. Take the sun for an example. Assuming that 'light' travels in a vacuum, where do you see the electrons to be? In your proposition you say that "a charged particle changes the position of it's field with respect to it's original position (observer) and that change propagates at C. I propose that change is what we call a photon" Now, either you are suggesting that this field can stretch from the sun to here? Or that there is electrons following the photons? Or is there a third way you look at it.

How far away do you assume this field to stretch. Or are you thinking that the photon is a 'emission' freed from what particle that 'fired' it. To me a 'field' imply a continuous pattern, and in this case still connected to its originator?

How do you see it?
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Offline Ron Hughes

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« Reply #19 on: 19/02/2010 16:28:52 »
Yor, a charged particle, from the instant it was created after the BB, had a field that started propagating away from it at C. That field is still propagating away from the point of it's creation at C . That first field will expand forever and every time the particle was accelerated new fields start propagating again at C. When a charge is accelerated it's field line in one direction is like water spraying out of a hose. When you shut the water off the line of water that was coming out of the hose dose not just disappear but keeps on traveling until it hits the ground. If you are holding the hose with the water shooting out and move the water hose to the left the old stream of water keeps traveling away and a new stream is formed. That difference in position is the photon and how quickly you accelerate the hose (charged particle) is the energy of the photon.
« Last Edit: 19/02/2010 16:31:35 by Ron Hughes »
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Offline yor_on

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« Reply #20 on: 19/02/2010 21:22:00 »
Okay, I see your proposition better now. Then we're discussing some kind self contained 'fields' created by a charged particle, right? Like an electron accelerated by interacting in a EM field. So what would then differ the field accelerating the electron from those 'photons' emitted by our electron. They must be different to keep their continuity as entities in that larger field, don't they?
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Offline Ron Hughes

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« Reply #21 on: 19/02/2010 23:21:28 »
Yor, I have a much better drawing and explanation here.  http://hypography.com/forums/strange-claims-forum/22596-photon-creation.html#post292773 Click on the field drawingpdf.pdf about midway down in the first post.
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Offline yor_on

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« Reply #22 on: 20/02/2010 03:54:44 »
Yes, the drawing was nice, and you seem to have some knowledgeable opponents discussing it there too. I'll follow it with interest, as for how you suggest that the electron will let lose photons due to accelerating?

Are you then suggesting that it will do the same falling (accelerating) into a gravity well (Black Hole) too? Or do you see it as a consequence of interaction with a EM field accelerating it?
« Last Edit: 20/02/2010 03:56:17 by yor_on »
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Offline Ron Hughes

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« Reply #23 on: 20/02/2010 04:16:00 »
To an observer stationary with respect to the black the electron will emit radiation.
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Offline yor_on

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« Reply #24 on: 20/02/2010 06:07:55 »
Then you either will have to endow gravity with EM, or you will have to define a field that change due to interacting with itself, it seems to me Ron?

Although, we will see :)
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Offline Ron Hughes

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« Reply #25 on: 20/02/2010 16:16:50 »
Yor, the release of energy is also relative to the observer. The fact that the observer sees the electron emit radiation is not because the electron is falling but because the observer is accelerating. He/she must be blasting their rockets in order to remain stationary with respect to the black hole. Note: The electron sees the observer as emitting radiation.
« Last Edit: 20/02/2010 16:19:59 by Ron Hughes »
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Offline yor_on

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« Reply #26 on: 20/02/2010 18:03:13 »
Ahh, thats the question here Ron :)
In the BH scenario, isn't he from his own frame seen 'free falling'?
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Offline Ron Hughes

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« Reply #27 on: 20/02/2010 18:23:54 »
If the observer is free falling with the electron then he does not see the electron emitting radiation.  Here is a site that may help me to determine how the photon is turned into matter. Farsight, you should be very interested.  http://www.commonsensescience.org/pdf/articles/nature_of_the_physical_world_P2_FoS_V7N2.pdf
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Offline yor_on

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« Reply #28 on: 20/02/2010 22:00:34 »
Ron, my point is, where is the 'acceleration'?
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Offline Ron Hughes

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« Reply #29 on: 20/02/2010 22:26:48 »
With respect to what?
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Offline yor_on

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« Reply #30 on: 20/02/2010 23:02:00 »
My thought there was that acceleration normally depends on 'energy spent' but when free falling you don't expend any energy at all. And that also goes for in falling objects towards a black hole. So, if you like, it may all fall back to how you define acceleration. Any thoughts on that?
==

Look at it this way, while accelerating due to a EM field will produce a radiation, 'accelerating' due to 'free falling' seems different to me. There is no 'interaction' in the usual way of expression. Not if you don't consider gravity to be an 'energy'. If you chose not to do so then the charge/electron will have to interact with its field to somehow produce that radiation, well, as I see it.

To me it's about 'interactions' and in this case, how to define gravity/acceleration.

1.Gravity is 'energy' and therefore it 'works'
2. 'acceleration' will produce this effect, no matter how it is produced. Then you have a new definition of acceleration, as I see it, as it now without being an energy itself and neither is expending energy still can produce an radiative change.

And maybe there are more variables here that I haven't considered?
« Last Edit: 21/02/2010 02:02:28 by yor_on »
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Offline Ron Hughes

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« Reply #31 on: 21/02/2010 02:33:30 »
It doesn't make any difference whether it is gravity or force it's still acceleration and will produce the required radiation with respect to a stationary observer. Emitting radiation obeys general relativity. It is always relative to an observer.
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Offline yor_on

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« Reply #32 on: 21/02/2010 04:07:13 »
Wish I knew this one for sure Ron. Looking at the two scenarios I think of, it seems as a free fall without any energy spent, as compared to a acceleration with energy spent is different, although I'm not sure how exactly. But to me it seems to come back to 'potential energy' if it would be shown as equivalent. Either that, or that I should look at gravity as an 'energy'.

If I look at it as 'potential energy' I will have the fact that 'real energy spent' as in that rocket accelerating then will have an exact equivalent counterpart in 'potential energy' as seen from our particle falling down that Black Hole. And as you never will be able to measure 'potential energy' in that 'accelerating' free fall towards the Black Hole. How does our electron/charge know that it should radiate? And why does it separate the frames of observation? 'Near and far observer'

Then it seems easier to look at it as if 'gravity' was an 'energy'? But if I do that where did Einsteins Geodesics go? instead of geometrical 'deformations' of space due to mass we now will have to define it as an 'energy'? That's no longer General relativity is it? Now we're talking Quantum Mechanics instead. When we go down to that level everything seems 'foggy'. No 'matter' is defined and everything will seem as 'probabilities' at least as I see it.

Well that's my dilemma Ron, and maybe your's too?
==

There is another way to illustrate why I feel that they're not equivalent.

When you accelerate your rocket with the electron/charge on, the 'gravity well' created will act from behind your accelerating rocket:
G<---Rocket - - direction - > with our rocket 'dragging' the gravity well behind it.

But in our free falling electron/charge the gravity acting upon it will act from the front;
Electron-> direction--> Black Hole (Gravity)

This is a simplification, but I hope you see my point here.

Another point being that we're not talking about a 'constant uniform acceleration' even though it is an acceleration as seen from the observer at rest with the black hole. And to be equivalent the acceleration from the charge carried by the rocket will have to act the same.
« Last Edit: 21/02/2010 04:34:30 by yor_on »
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Offline Farsight

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« Reply #33 on: 21/02/2010 12:29:00 »
Here is a site that may help me to determine how the photon is turned into matter. Farsight, you should be very interested. http://www.commonsensescience.org/pdf/articles/nature_of_the_physical_world_P2_FoS_V7N2.pdf
I read the paper Ron. I'd say it's not wholly wrong, but there are better papers around.

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Offline Ron Hughes

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« Reply #34 on: 21/02/2010 18:14:32 »
I am interested. If you have some sites please tell me. I would like to look at them.

Yor, how about this as a way to explain it. A particle accelerating in a gravity well experiences time dilation. A particle being accelerated by some other force also experiences time dilation. In either case an outside observer sees this dilation but if the observer is moving in the same frame of reference of one or the other particles he does not see time dilation for the particle in his frame of reference.
« Last Edit: 21/02/2010 18:25:10 by Ron Hughes »
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Offline yor_on

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« Reply #35 on: 21/02/2010 22:30:07 »
It's an idea Ron. But if you consider that the far observer (at rest with the Black Hole) will see our particle 'hang' at the 'EV' why would he then also see radiation, to observe it we must assume that they have an frequency right? And if we let our electron shine a flashlight up the gravity well the far observer would describe that light as redshifted and somewhere near the EV I expect that redshift to be so 'large' that the waves wouldn't be measurable for our observer. Now, put that in view with the idea of the electrons charge radiating as it falls down to the EV. Shouldn't that radiation then 'keep on coming' no matter where that electron was? And how would it do that?
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Offline Ron Hughes

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« Reply #36 on: 22/02/2010 03:21:17 »
You answered your own question. At the EV the electron disappears from the Universe. The case of a black hole applies to my theory only in that it is a gravity well and well cause the electron to emit radiation. The fact that it is red shifted is as you say a consequence of the enormous strength of that gravity well. The photo electric effect and possibly Compton scattering needs explaining and to do that I have to figure out exactly what causes charge.
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Offline yor_on

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« Reply #37 on: 22/02/2010 08:31:22 »
Yes, you're right, at the EV all oscillations should disappear for the 'far observer' at rest relative the BH. But it seems to me as if a gravity field acts like a 'dampener' of all oscillations/waves seen by him even before, which then will add yet another difficulty to the concept. Otherwise I agree that the free falling frame must, from inside that frame, be experienced the 'same as always' measuring inside 'the black box'.
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Offline Ron Hughes

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« Reply #38 on: 22/02/2010 15:57:35 »
Please revisit drawing here,  http://hypography.com/forums/strange-claims-forum/22596-photon-creation.html#post292585 ,  It is a much better picture of the action.
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Offline Liesch

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« Reply #39 on: 24/02/2010 21:43:45 »
There is an interesting video on You Tube (from BBC) about the theme, not specific but related. i will post it once I find it again.

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Offline Vern

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« Reply #40 on: 27/02/2010 02:41:28 »
The way you make a photon is to change the amplitude of a magnetic field or an electric field. Either one will do it.

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Offline Ron Hughes

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« Reply #41 on: 27/02/2010 15:53:37 »
Here is a neat question. The electron has a field, how does it make that field?
From a drop of water a logician could infer the possibility of an Atlantic or a Niagara without having seen or heard of one or the other. Sherlock Holmes.

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Offline Ron Hughes

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« Reply #42 on: 27/02/2010 16:02:53 »
Vern, and the amplitude of an electromagnetic wave is ,what? I forget.
From a drop of water a logician could infer the possibility of an Atlantic or a Niagara without having seen or heard of one or the other. Sherlock Holmes.

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Offline Ron Hughes

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« Reply #43 on: 04/03/2010 15:33:35 »
Has anyone considered the possibility that this idea suggests inertia is also relative?
From a drop of water a logician could infer the possibility of an Atlantic or a Niagara without having seen or heard of one or the other. Sherlock Holmes.

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Offline frenrg

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« Reply #44 on: 13/07/2011 03:05:50 »
Regarding large diameter motor coupled to small diameter generator.
If the motor was twice the diameter of the generator, this would at first appear to give the motor a two to one advantage as far as leverage goes.
However, by doubling the motor diameter, this in effect halves the available finite power for the magnetic field generation, so any gain by leverage is negated here.