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by Miles Mathis

In other papers I have reminded the reader that planets and moons appear to have strong exclusionary fields, even when they do not have strong magnetic fields. How can a planet like Venus exclude the Solar Wind without a strong magnetosphere?

The current upper limit on the size of the magnetosphere of Venus is 10,000 times smaller than that of the Earth. In addition, “There are no belts of trapped radiation such as Earth's Van Allen belts, and there is no 'magnetotail' composed of fields of planetary origin.”1

The current explanation of this is not really an explanation, it is only a word: an “induced” magnetosphere. The ionosphere of Venus creates a boundary called an ionopause, and this creates the same bowshock as would a magnetosphere.

That is it. That is the whole explanation. Current scientists never attempt to answer the gigantic begged questions, such as,

1) How do these ions repel the Solar Wind, if they are not strongly magnetic?

2) If an ionosphere is enough to repel the Solar Wind, why do we not assign the Solar Wind exclusion by the Earth to its ionosphere, by this “induced” field? If the ionosphere is capable of exclusion, why give the Earth’s bowshock to the magnetosphere?

3) Why does Venus create non-magnetic ions, while the Earth creates magnetic ones?

4) Why are non-magnetic ions just as good as magnetic ones, in excluding the Solar Wind.

And so on. Obviously, the standard model doesn’t try to answer this, because they have no way to do so. They can begin to answer question number 3 by proposing that the Earth’s interior is different than Venus’, but they have no proof of that one way or the other. On the other questions, the standard model mostly admits ignorance, which is refreshing.

The answer to number 2 is that we give the bowshock to the magnetosphere when we are looking at the Earth, due to the fact that we do see “a magnetotail of planetary origin.” It is clear that the magnetosphere causes the exclusion, and not the ionosphere.

But for the rest, it is a mystery. The current gravitational and E/M theories don’t have the tools to solve it. Current theory has no mechanical explanation of the E/M field, but we require a mechanical explanation in order to solve this problem. Obviously, it is not a problem we can solve with virtual photons or messenger photons.

As another example, the Encyclopedia of the Solar System2 explains this phenomenon this way: "If conducting paths exist across the planet's interior or ionosphere, then electric currents flow through the body and into the solar wind, where they create forces that slow and divert the incident flow." Then the authors tell us that Venus and Mars have these paths, but the Moon doesn't. But we are never told how the body creates these paths, or why Venus should have them but the Moon should not. We have a theory with no mechanics. We are shown a diagram where the wind is diverted, but we are given no reason for the diversion. Proof by diagram. At the very least we could be given a bald theory for why the Moon blocks such a path, but we get nothing. Nor do we get an answer to the begged question: "Given this conducting path through the body of Venus, why are unmagnetized ions conducted but not magnetized ions? A through field like this should create a magnetic difference or potential, not just an electrical difference. Why no magnetic field?"

I can solve this problem where the standard model cannot because I have assigned real particles and real motions to the field. I have shown that the E/M field is an outcome of the charge field, and the charge field is a straight bombarding field. That is, charge is an emission of real photons. Charge is always positive, never negative, and the field has mass equivalence. The linear momentum of this emitted field is the electrical part of the field, and the summed spin of the photons in the field is the magnetic part of the field. If we look at one photon only, the linear energy of that photon creates the electric field, and the spin of that photon creates the magnetic field.

Now, each photon can be spinning either CW or CCW, so even the charge has charge, in a way. This is how we get particles and anti-particles, for a start. Anti-particles are simply upside down compared to particles, with reversed z-spins. But if we look at the field of photons, we can also get a field or an anti-field. Some photons are CW, some are CCW. We sum the angular momenta of all the photons in the field to create the magnetic strength of the field, and each CW cancels a CCW. So if we have an equal amount of CW and CCW photons, we will have an electric field without a magnetic field. The photons cancel each other’s spins, but they do not thereby cancel each other’s linear momenta.

Normally this doesn’t happen, because normally we are measuring small or limited fields that are created by homogeneous substances or objects. These objects tend to emit their E/M fields in a pretty consistent way, so that most of the photons are CW, for example. This tends to match the electrical field strength to the magnetic field strength. But given a very heterogeneous makeup, or a large body, the magnetic field can sum to a low number, due to the amount of CCW photons present.

Nor is it only anti-protons or anti-matter that can give a large object this heterogeneity. I have shown that positrons and electrons also emit the E/M field (although to a smaller extent). Yes, electrons and positrons both have a positive charge, in this respect. Therefore, a large number of positrons could provide the CCW photons necessary to flatten out a magnetic field.

I predict this is what we see with Venus. For some reason, Venus has a greater production of CCW photons than the Earth. It could be that, being closer to the Sun, it has absorbed more positron emission, which it then uses to emit its up-side down photon field. Or it could be that Venus has a higher concentration of anti-matter than the Earth. Or it could be that the Solar Wind turns Venus’ free electrons into positrons directly, by bombardment, randomizing Venus’ overall charge-spin. Without more information, it is difficult to say what the cause is, but the mechanics must be that Venus has an emission field that is much more balanced, in terms of CW and CCW, than the Earth.

This explains the power of Venus’ exclusion, because, as I said, CW and CCW photons only cancel angular momenta, they do not cancel linear momenta. Due to the fact that Venus is about the same size and density as the Earth, it must have about the same size total charge emission. It is this total charge emission that creates any ionosphere or magnetosphere, since the ions are created by these photons colliding with matter in the atmosphere. In other words, Venus keeps the full E component of the field, even if it loses the entire M component. Because the summed E component of the field has a vector that is radial out from the surface (that is how it is emitted from a sphere), this electrical component is perfectly capable of excluding external fields.

One final clarification. As I have said many times, the E component of the E/M field only creates the electrical field: it is not the field itself. In other words, these emitted photons create the E/M field, but they are not the electrical field directly. The photons are the charge field, or the foundational E/M field. Given a material field, these photons will drive free electrons and positrons and ions, and it is the motion of these larger particles that we call the electrical field and electricity and so on. So you could have a very strong charge field of photons, and a very weak electrical field. The charge field requires ions that can be driven by the stream, and if you have a poor source of ions, you will have a weak electrical field, no matter how rich your photon field is.

This also means that you cannot measure a charge field at the macro-level simply by trying to measure a local electrical field. Venus could have a weak electrical field at any or all points on its surface or lower atmosphere, but have a strong field at higher altitudes, if it had few free particles close to the surface and lots of free particles higher up.

This also applies to the Moon. We have been told that the Moon has weak magnetic and electrical fields on the surface, and this is used to counter my claim that the Moon must have a large charge field. But this is simply because the Moon has no atmosphere, and little ion content near the surface. This means that the charge field cannot be expressed, not that it doesn’t exist. My theory says that the Moon must have a very large charge field, and that we know this because it finally finds a place to express itself here on the Earth, with tides and other phenomena.


1VENUS: MAGNETIC FIELD AND MAGNETOSPHERE, J. G. LUHMANN AND C. T. RUSSELL, Originally published in: Encyclopedia of Planetary Sciences, edited by J. H. Shirley and R. W. Fainbridge, 905-907, Chapman and Hall, New York, 1997.
2Academic Press, 1999, p. 481. Edited by JPL.

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