Willem F Esterhuyse

If we build a time machine, anyone may use it and cause the past or future to change. Even if we establish an institution for using it that checks for altering the past, we still have the problem of sensing the natural past from an altered past, because we can't know whether some John or Jane were born - if they aren't famous. So unintended changes may cause someone not to be born, but was born in the natural past.

Say you got a formal proof: Line # Statement Reason 1 A Premise 2 ... ... n B m, Some reason n+1 A -> B n, Deduction Theorem. Now we have proven: A -> A -> B, so we proved: true -> B = true (by truth table of "therefore"). So we proved A -> true (by deduction theorem). Now A could be false, then we shouldn't be able to derive "true".

This has become an escape clause - used too lightly. "Something" is defined as occupying space.

I have circuits in my mind that can be triggered regardless of the laws of physics.

Particles being points is in conflict with our common sense notion that they are "something".

Because in my model neutrinos come from pi-mesons. I meant all massive particles. No they have diagrams of electron, muon and tau neutrinos with various amounts of mass 1,2,3.

They call it mass 1,2,3.

The following ref. says neutrinos have 3 types of mass (see timestamp 31:08): Then all particles must have 3 types of mass and theory does not predict this.

They don't decay in one step to Pb for example.

Right. No, I mean with daughter isotopes the isotopes when only one reaction happened. " I wonder what you mean by daughter isotope ? or are you meaning that the 'daughter' must be an isotope of the same element as the starter ? That is the only way I can make sense of this absolute claim Because clearly all decay is to an isotope of some element or another." I mean they don't decay in one step to any isotope.

I mean the direct daughter isotopes (one decay away). According to the diagram U decays directly just to Io with 90 protons and an isotope with p=91 though I don't know why this isotope is not graphed halfway between two horizontal lines.

I don't read it as a farce. It is expected to contain buzzwords. Anyway, they peer reviewed my article and accepted it for publishing, therefore it has at least some prestige.

They don't decay to any isotope whatsoever.

Abstract We first prove that nuclear structure is not randomly arranged nucleons. We then develop a geometric model for nuclear structure using JP data from the Internet. The model shows why Be(p = 4, n = 4) is unstable while Be(4, 5) is stable. It predicts correctly the mode of decay of unstable isotopes and it predicts the daughter isotope. It also predicts why Tc(43, n = 54 or 53) is unstable and why Tc(43, 52) is not allowed and why all other isotopes of Tc are unstable. There is also a clear indication of why nuclei of p > 26 require energy to form. Knowing the relative position of the particles may make field computations easier. The model also explains why an isotope can have zero Orbital Angular Momentum. Also shown is why the magic number 8 yields a threshold of minimal binding energy. The model predicts that 16 should also be a magic number for stability. It also predicts that for nuclei with A = 22 or more has the formula for radius as follows: R = R0 (2*Z)^(1/3) and not R = R0 (A)^(1/3). Thus the model predicts smaller radiuses for heavy nuclei. It is predicted that for F(9,9) decaying to O(8,10) there would be released 4 photons per nucleus. The transition energies i.e. frequencies for this decay is computed. It is shown how to calculate transition energies. It is predicted that Na(11, 11) with JP = 3+, will behave anomalously w.r.t. the Electromagnetic Interaction (i.e. it will not have the same chemistry as other isotopes of Na), as will the stable Ge(32, 41) at JP = 9/2+, Kr(36, 47) at JP = 9/2+, Rh(45, 58) at JP = 1/2- , Pd(46, 59) at JP = 5/2+ and Ag(47, 60) at JP = 1/2-. There are a few other properties one would not see without the model. See: https://ijasr.org/2021 volume 4 issue 2 March - April.html and scroll to article number 3.

The acceleration vector due to the Suns gravity is in the opposite direction at the sunny side than on the dark side. Centripetal force is towards Earth on the sunny side (and if I'm correct cancels the force from the Sun exactly), while centripetal force is away from Earth on the dark side. Thus the object would be lighter on the dark side. Yes. The Moons gravity was neglected.

I disregarded centripetal force.

We have renormalization and the collapse of the wavefunction: both operations imposed by our knowledge (experimental evidence), hence non-mathematical.

The Earth is accelerating and the acceleration vector points in the direction of the Sun. Is something lighter on the side facing the Sun than the same thing on the side facing away from the Sun? If not, why not. F = m(g-a) so total a is smaller on the side facing the sun.

Then they must not state that the telescope orbits this point.

The law states: F = GMSm/r2 + GMEm/(r+d)2 and this must equal zero for a neutral gravity point. It doesn't.

We have: P(A|B) = P(B|A)P(A)/P(B). Is P(B) = P(B|nothing) or is P(B)=P(B|everything)?

The picture shows L4 as having repulsive gravity and L1, L2 and L3 to have attractive gravity from one direction and repulsive gravity for another direction. How does the gravity of Earth add together with the gravity of the sun to cause L3 to have neutral gravity? It isn't a subtractive operation. I don't believe L2 and L3 have neutral gravity.

To me it looks like at the Lagrange points space would not be curved at all (for small enough distances). The space grid would be just cubes of equal size.

I've read again that a space telescope is to orbit a point of equal gravity on a line connecting Earth and Sun. How is this possible, is there curvature causing attraction throwards this point?

No, you learn it from the truth table for "therefore". There is a reason for it though, more of a convention.