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Theorien jenseits der Standardphysik Sie haben Ihre eigene physikalische Theorie entwickelt? Oder Sie kritisieren bestehende Standardtheorien? Dann sind Sie hier richtig.

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  #11  
Alt 03.04.13, 12:58
Slash Slash ist offline
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Zitat:
Zitat von Solkar Beitrag anzeigen
...and again:
What would be my reward for reading this paper and guessing what that kind of fantasy formalism is meant to mean?
Die ständige, fast schon nötigende Wiederholung erinnert mich an Samuel L. Jackson alias Jules Winnfield von Pulp Fiction.

Starke Leistung!
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  #12  
Alt 03.04.13, 13:13
Ioannis Ioannis ist offline
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Zitat:
Zitat von Solkar Beitrag anzeigen
So even after I pointed you twice to the fact that you use a partial symbol "∂" instead of a latin "d" in that integrand, you don't get that.

Very well.
It's quite a motivating to know that the author of a paper targeting the most complicated physical topics is not fluent in basic calculus.

...and again:
What would be my reward for reading this paper and guessing what that kind of fantasy formalism is meant to mean?
OK, Solkar, you are right about it. Just ignore this error since it will not harm the rest of the work. If you continue read on the same page you will see that it is derived the same equation without the use of the integral. The integrals that follow, they do not have this small mistake.

If you are not motivated to read my work, just do not read it and do not be provocative. I am an Electronic Engineer in profession and not a physicist as also errors are human. Because I want to be honest, high mathematics are not my strength (I left them after finishing the Fachhochschule, 15 years ago).
There are more than 100 equations on my paper and I corrected a lot of mistakes in the past. Well, I missed that what you mentioned.

The rewards come to those who have the intuition, insight and will beyond prejudices to explore new ideas. I believe that you do not belong on this category because you started having as your primary guide, the prejudices.
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  #13  
Alt 03.04.13, 13:15
Ioannis Ioannis ist offline
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Zitat:
Zitat von Slash Beitrag anzeigen
Dear Ioannis,

I agree with you and your impression of the partially very disrespectful way to discuss.

I can not make any statement about the scientific content oft your paper.

Best regards

Slash
Thanks for your comment, really appreciated!
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  #14  
Alt 03.04.13, 13:41
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Solkar Solkar ist offline
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Zitat:
Zitat von Ioannis Beitrag anzeigen
If you continue read on the same page you will see that it is derived the same equation without the use of the integral. The integrals that follow, they do not have this small mistake.
OK, let's talk about significance of mistakes - is it also insignificant for the validity of your theory that photons are uncharged?
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  #15  
Alt 03.04.13, 13:50
Ioannis Ioannis ist offline
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Zitat:
Zitat von Solkar Beitrag anzeigen
OK, let's talk about significance of mistakes - is it also insignificant for the validity of your theory that photons are uncharged?
If I understood correctly, you mean that photons are uncharged. Of course they are uncharged and they additionally do not have mass. I do not write something different in regard to this. Eq. (3) and Eq.(7) resulted from the interaction of photon radiation with the Electric field of the stationary charge.
The formulations are very simple and clear.
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  #16  
Alt 03.04.13, 14:39
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Oh I see - so we consider this equality of [Xyd13/eq(3)
Zitat:
∆E_γ = ∆E_Field
insignificant as well, or do we?

Ioannis, why don't you simply emphasize (eg. by boldface), which parts of [Xyd13] are meant to be taken serious and which are to be disregarded?

[Xyd13]Ioannis Xydous. The secret of the Electron-Positron pair. v7.0 13.03.2013 20:31,
http://www.ioannisxydous.gr/SEPPv7.pdf
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  #17  
Alt 03.04.13, 15:15
Ioannis Ioannis ist offline
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Zitat:
Zitat von Solkar Beitrag anzeigen
Oh I see - so we consider this equality of [Xyd13/eq(3)

insignificant as well, or do we?

Ioannis, why don't you simply emphasize (eg. by boldface), which parts of [Xyd13] are meant to be taken serious and which are to be disregarded?

[Xyd13]Ioannis Xydous. The secret of the Electron-Positron pair. v7.0 13.03.2013 20:31,
http://www.ioannisxydous.gr/SEPPv7.pdf
This applies only for you: discard the entire content of the paper because it does not worth to be read. Are you happy now?

I do not have any time and interest to continue such kind of conversations since they are pointless.
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  #18  
Alt 03.04.13, 15:27
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Put it this way:

Because you wrongfully assume neutral photons interact with Coulomb fields like charged particles in your ansatz for [Xyg13], your whole ansatz is plainly wrong, thus your paper [Xyg13] is null and void.

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  #19  
Alt 03.04.13, 15:49
Ioannis Ioannis ist offline
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Zitat:
Zitat von Solkar Beitrag anzeigen
Put it this way:

Because you wrongfully assume neutral photons interact with Coulomb fields like charged particles in your ansatz for [Xyg13], your whole ansatz is plainly wrong, thus your paper [Xyg13] is null and void.

We will see about who is void and null. In the meantime, could you please provide your opinion about the below:

We have a propagating photon with an exact threshold energy 1.022MeV which travels near a heavy nucleus (acting as momentum absorber). At a specific moment the photon energy is transformed to an Electron-Positron pair. I would like to have your opinion (all those who are interested) in regards to what kind of interaction we have on this particular situation:

i) Does the photon fall upon the nucleus? Obviously NO due to three reasons. First the photon Energy is very small and cannot trigger a photo-fission process. Secondly, the photo-fission process has nothing to do with the pair creation phenomenon. Third, the only outcome in the pair production process is the Electron-Positron pair. (see Eq. (1))

ii) Today's Physics supports that the heavy nucleus due to its mass is used as momentum absorber (which actually this is equal to the definition of a blocked nucleus deflection/scattering) that will actually will enable the photon to decay to an Electron-Positron pair by conserving the Energy and momentum, simultaneously. If the statement (i) is valid then the mass of the nucleus (only the mass property) is not involved in the process and it does not make sense the definition of nucleus mass (only the mass property) as momentum absorber because actually the photon does not fall upon nucleus.

iii) Then, if the above two statements are valid, what kind of interaction takes place?
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  #20  
Alt 03.04.13, 19:26
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Zitat:
Zitat von Ioannis Beitrag anzeigen
We have a propagating photon with an exact threshold energy 1.022MeV which travels near a heavy nucleus (acting as momentum absorber). At a specific moment the photon energy is transformed to an Electron-Positron pair. I would like to have your opinion (all those who are interested) in regards to what kind of interaction we have on this particular situation:
How does that negate the point, Solkar spoke about?
Your ansatz is wrong in any way. That's all.
__________________
Gruß, Johann
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Eine korrekt gestellte Frage beinhaltet zu 2/3 die Antwort.
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E0 = mc²
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