User talk:WillOakley

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Hello, WillOakley, and welcome to Wikipedia! Thank you for your contributions. I hope you like the place and decide to stay. Here are a few links to pages you might find helpful:
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Please remember to sign your messages on talk pages by typing four tildes ( ~ ); this will automatically insert your username and the date. If you need help, check out Questions, ask me on my talk page, or, and a volunteer should respond shortly. Again, welcome! Vsmith (talk) 00:13, 26 August 2017 (UTC)

I've undone you recent edit to Gravitational constant as it appeared to be personal commentary with an attempt at referencing that appears to be your work based on your choice of username. I would suggest that you address your concerns on the article talk page. We aren't here to promote our own stuff ... sorry 'bout that. Vsmith (talk) 00:13, 26 August 2017 (UTC)

Again ... please read WP:SELFCITE. Vsmith (talk) 02:52, 26 August 2017 (UTC)


 * To extend this: Wikipedia is not the right place for the nonsense you published in some predatory open-access journal.
 * "where c is the numerical value of c in cgs units" - do you really think the cgs units have a fundamental role in physics? Does the Moon suddenly change its orbit if we use inches instead of centimeters in calculations? --mfb (talk) 01:08, 29 August 2017 (UTC)

Basics of editing on a user talk page
Hello, Will Oakley – I saw your recent  to Vsmith's talk page. There are two things you neglected to do:

1) When you are posting a comment to a talk page that is on a new topic – that is, there is no existing section on that topic to which you want to add – you need to begin a new section. The easiest way to do that is to type in a heading – any heading, preferably not too long, that makes sense – in the "Subject/headline" space at the top of your edit window. That will create the new section heading after you save your edit. If you forget to do that, you can go back into edit mode on the page and type pairs of equals signs on either side of a heading just above your comment, then save your edit. That, too, will create a section heading on the talk page. Two equals signs on either side of the heading will create the right size font.

2) When you have finished typing your comment, you need to sign it with four tildes: ~ . That will automatically enter your user name, the date, and the time. If you forget, a bot will add it for you, but it looks better if you sign with four tildes. Best regards, – Corinne (talk) 21:58, 29 August 2017 (UTC)

Welcome to The Wikipedia Adventure!
-- 22:30, Tuesday, August 29, 2017 (UTC)

Regarding #3, EM waves propagating rectilinearly in free space do not exhibit charge only possessing an E field. But an EM wave in a quantum loop can have an E field directed outward and thus appear as a charge, So an EM wave localized in 3D, (e.g. as a toroid), could be appear to be a localized charge. An EM wave can only exist as a quantum loop if propagating rectilinearly near an event horizon of closed geometry at the particle scale. To evidence mass the EM wave has to be relativistic. GRT as presently formulated only applies to cosmic phenomena, but the same equations apply to space-time curvature at the particle scale. Gravity (and thereby space-time curvature), must originate with curvature at the particle scale. The question is how does this translate to observer domain gravity.

Consider an EM wave propagating near an event horizon in a relativistic state is orthogonal to radii from the event horizon in two dimensions. If an EM wave is of the electron energy, and considered to be relativistic by alpha^-1 over all three dimensions, the relativistic energy in the two dimensions orthogonal to the radii is alpha^-2/3, about 26.58. EM energy E propagating in a curved metric orthogonal to a radius will appear as an effect reduced by c^2, i.e. E/c^2 = m. Consider, E = hbar.c/r and the strong force (F)= E/r and acts in the same direction as the energy. So F = hbar.c/r^2 is the circumferential force in a quantum loop. Inserting E^2 twice, gives F = (hbar.c E^2)/E^2.r^2, in the rotating frame. Radially F/c^4 = hbar.c.m^2/E^2r^2. = (hbar.c/E^2)m^2/r^2, i.e. Gm^2/r^2, where G = hbar.c/E^2 = hbar.c/(alpha^-2/3me, c^2) for the electron. This differs from the classical dimension of G by (alpha-2/3.c^2)^2. As F/gravity = 5.71 x10^44. Using (alpha-2/3.c^2)^2 = 5.71 x10^44, please compute c. You will get c = 2.998 x 10^10. same as cgs units. By assuming gravity force in observer space between masses I the same space a numerical factor of C^4 in those units was hidden in G, as was the relativistic factor. This connects QED,te strong force, and GRT in a simple electron model as in my published article. The implication of this for physics are huge!!WillOakley (talk) 19:19, 3 September 2017 (UTC)


 * Now you are just making strings of buzzwords. And you repeat the unit errors over and over again. --mfb (talk) 23:31, 3 September 2017 (UTC)

There are no unit errors, sorry, you do not understand, let me try again. Gravity does not act between masses in observer space, it acts between energies rotating within particles at relativistic velocities. So the m in the dimensions of G, i.e. (hbar.c/m^2), gets replaced by (alpha^-2/3.me.c^2)^2. This value was essentially measured by Cavendish when he measured G. But he thought gravity acted between masses so he got the dimensions of G wrong by c^4. Further, by assuming G a constant he indirectly and unknowingly assumed (alpha^-2/3.me.c^2)^2 was a constant in the units he was using, cgs units. Thus the numerical value of (alpha^-2/3.me.c^2)^2 was inadvertently included in G as a fixed number i.e. 5.71x10^44. This has no effect anywhere else in physics, only in G. This is the same number as classically obtained for the ratio (Strong Force)/gravity. Gravity is not force in the observer domain but force in the rotating energy space in each particle, c^2 remote from mass space,(i.e. observer space), So the ratio depends on the units of c used to measure G. The classical "Force ratio" for electrons, hbar.c/Gme^2 = 5.71x 10^44. [the mystery Large Number in physics]. The fact that (alpha^-2/3.c^2)^2 = the same number with c in cgs units, 5.71x 10^44 is PROOF that the above is valid. The electron EM energy is two spin half quantum loops, and loop crosstalk adds a small adjustment. So G = hbar.c/(alpha^-2/3.me.c*^2)^2, where only c* is the numerical value of c in cgs units, gives G = 6.674273033 x 10^-11 in SI units. But again G is just a convenient scaling factor for a force that acts elsewhere, i.e. c^2 from observer space for each particle. So the energies of two electrons are c^4 apart, which reduces the gravitational force from the strong force level by c^4. As this concept describes an EM wave in an effective relativistic state propagating close to an event horizon, and it produces G, it directly connects G, general relativity, the strong force and QED in the same particle model. Basically, Cavendish measured gravity in a domain where it does not actually work, and inadvertently hid the problem by including it in the dimensions of G. WillOakley (talk) 03:55, 4 September 2017 (UTC)

Please stop
If you keep adding nonsense to articles, your account here will be blocked from making further edits. You are not helping anyone. Get a book, learn the actual science, then you'll understand the misconceptions you currently have. --mfb (talk) 11:40, 2 September 2017 (UTC)

Newton's Gravitational constant.

So, Mfb, I guess you did not read my input, or you know little physics. Which of my statements did you find nonsense? 1. That the Moon orbits in curved space time and is not under a central force as described by Newton? That is Einstein's general relativity, 1917. Not in question. 2. Our space is very flat as shown by observations using the Hubble telescope. Obviously not curved at the lunar radius. Again not in question. 3. Electrons and other particles are only localized electromagnetic energy. Dirac, 1920. Again not in question. 4. Newton's G was introduced as part of his gravity equation, and no other independent place in physics. 5. Newton's equation assumes gravity force in observer space acting on masses in observer space where we are the observer. Once again, not in question.

The above #5 is incompatible with #1 thru #4, so Newton's equation must be wrong, unless you wish to contradict Einstein, Dirac, etc. Please do not respond, clearly Wiki editors are not up to the task. I'm abandoning wiki as not viable in the sciences.

Will Oakley

Newton's gravitational equation is clearly wrong as simple logic now shows. It describes a force in the observer domain acting on masses in the same domain. General relativity has replaced Newtonian gravity and shows gravity acts via curved space-time. The Moon orbits Earth by following curved space and is not held in place by a central force as described by Newton. But looking near the Moon we see the star background is not distorted, meaning our space is flat, and therefore gravity does not act in our space-time domain. Newton introduced G as a scaling factor for his equation, but the equation is clearly wrong so G cannot be a natural fundamental constant. Further, science has known since about 1920 that particles, electrons etc., are just localized electromagnetic energy, so gravity must act via energy, not mass.


 * You are asking questions and writing "please do not respond". I'm not sure if you would like answers or not. If not, just ignore my reply.
 * 2. Hubble, Planck and so on are only sensitive to large-scale curvature of space, their measurements wouldn't change even if you would replace the Earth by a black hole of the same mass, which certainly does have a large impact on spacetime directly around it. The light deflection by the Moon is incredibly tiny. Gaia is able to measure the light deflection by planets, that is currently the limit of the experimental sensitivity.
 * 3. is wrong. An electromagnetic field cannot have an electric charge, for example. And some particles don't even interact via the electromagnetic interaction.
 * 4. is just an artifact of our unit system. You can introduce units where G=1, or 8 pi G = 1 as it is often done in cosmology.
 * 5. What?
 * Newtonian gravity is only an approximation. The source term in general relativity is the stress-energy tensor. Mass contributes to it (in the T00 component), but it is not the only contribution. That has been known for more than 100 years. What is your point?
 * Concerning your worry about my physics knowledge: Thanks, but it is not necessary. It is my job knowing these things. --mfb (talk) 16:47, 3 September 2017 (UTC)