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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
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AW: Physik jenseits Einstein: neue Energie-Masse Äquivalenz, Anti-Schwerkraft, Äther
Zitat:
Starke Leistung! |
#12
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AW: Physik jenseits Einstein: neue Energie-Masse Äquivalenz, Anti-Schwerkraft, Äther
Zitat:
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. |
#13
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AW: Physik jenseits Einstein: neue Energie-Masse Äquivalenz, Anti-Schwerkraft, Äther
Thanks for your comment, really appreciated!
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#14
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AW: Physik jenseits Einstein: neue Energie-Masse Äquivalenz, Anti-Schwerkraft, Äther
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
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AW: Physik jenseits Einstein: neue Energie-Masse Äquivalenz, Anti-Schwerkraft, Äther
Zitat:
The formulations are very simple and clear. |
#16
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AW: Physik jenseits Einstein: neue Energie-Masse Äquivalenz, Anti-Schwerkraft, Äther
Oh I see - so we consider this equality of [Xyd13/eq(3)
Zitat:
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 |
#17
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AW: Physik jenseits Einstein: neue Energie-Masse Äquivalenz, Anti-Schwerkraft, Äther
Zitat:
I do not have any time and interest to continue such kind of conversations since they are pointless. |
#18
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AW: Physik jenseits Einstein: neue Energie-Masse Äquivalenz, Anti-Schwerkraft, Äther
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. ☐ |
#19
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AW: Physik jenseits Einstein: neue Energie-Masse Äquivalenz, Anti-Schwerkraft, Äther
Zitat:
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? |
#20
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AW: Physik jenseits Einstein: neue Energie-Masse Äquivalenz, Anti-Schwerkraft, Äther
Zitat:
Your ansatz is wrong in any way. That's all.
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Gruß, Johann ------------------------------------------------------------ Eine korrekt gestellte Frage beinhaltet zu 2/3 die Antwort. ------------------------------------------------------------ E0 = mc² |
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