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Author Topic: The Photon  (Read 15084 times)

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 »
 

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'?
 

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
 

Offline yor_on

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

Offline Ron Hughes

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

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 »
 

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.
 

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 »
 

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.
 

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 »
 

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?
 

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.
 

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'.
 

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.
 

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.
 

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.
 

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?
 

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.
 

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?
 

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Offline frenrg

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« Reply #44 on: 13/07/2011 03:05:50 »
Shrunk
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.
 

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« Reply #44 on: 13/07/2011 03:05:50 »

 

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