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The
Solution to Tides Part 2
by
Miles Mathis
I had
intended to save the explanation of the spring and neap tides
until the end of this series on tides, but since my recent
publication of Part 1, my readers have already raised a small
outcry. They cannot imagine that I have a mechanical solution to
spring and neap tides that does not include the Sun.
Well,
I never said it didn't include the Sun. What I said is that this
variation was not caused by the Sun's gravitational or E/M
fields. I proved this already regarding the E/M field, showing it
is too small at the distance of the Earth to cause the size of
the variation we see. And few will care about that anyway, since
I imagine that few had any hopes for the E/M field to begin with.
Most, I imagine, are more concerned with the loss of the
Sun’s gravitational field, which I likewise dismissed. They may
be concerned with the loss, but I can’t imagine they disagree
with my analysis. Using the current explanation, there is no way
to logically explain spring or neap tides, and this should be
clear to anyone with a speck of honesty. Even if we accept that
gravity is a pulling force, or the equivalent of a pulling force,
there is no way to get two spring tides every month of the size
we see.
Let me repeat the argument here, before I show
the real cause. Spring tides are now explained by an alignment of
the Moon and Sun. There are roughly two alignments each month, of
course, one in conjunction and one in opposition. If we treat
gravity as a direct pulling force, neither the vectors nor the
numbers work out. Any fool can see that, since the Sun cannot add
itself to the Moon's force in the same way in opposition and
conjunction. To answer this obvious problem, the tidal field is
now treated as a differential field. This creates tides both
front and back, from both Sun and Moon. At first glance this
appears to work. But a closer analysis shows that the numbers
don't add up. In order to get the necessary 46% solar tide, the
standard model fudges the math, applying the same equation to
solar and lunar tide. Unfortunately, the Earth is not orbiting
the Moon, so the equations can't be the same. The standard model
is desperate to hide this fact, which is why they are playing cat
and mouse at Wiki, changing the page and deleting all the math.
They are also desperate to hide the fact that the barycenter tide
would swamp both solar and lunar tides, since this would swamp
all spring and neap variation. They could hide the
barycenter just by giving up on the idea, but the barycenter is a
necessary outcome of current gravitational theory, and the
standard model is very attached to the idea. Amazingly, they
continue to mention it in tidal theory, even though it is a
conspicuous Achilles heel.
All this means that neither
the differential field nor gravity as a direct pulling force can
explain tides. But since no one can come up with anything better,
they stick with what they have and guard it jealously, as if it
is another "crown jewel of physics".*
What
really causes the spring and neap tide variation is the Solar
Wind. The basic electromagnetic field of the Sun is 15 times
smaller than the Lunar field at the distance of the Earth, but
the Sun has many electromagnetic effects that are caused not by
the standard radiation of all matter, but as byproducts of
fusion. This is already wellknown. The Solar Wind is powerful
and mostly unidirectional (radial out from the Sun). This makes
it the perfect candidate for spring and neap tide variation. But
to even consider it as a candidate, you first have to be
explaining tides as an electromagnetic phenomenon. Since no one
else is doing that, no one else is anywhere near the true
solution.
The quickest way to show this is at neap tide.
At neap tide the Sun and Moon are orthogonal. That means the
Solar Wind will be blowing at a 90^{o} angle to either of
the two opposite lunar positions. In Part 1, I have already shown
that the tidal mechanism is both electric and magnetic, and in
the two neap positions, the Solar Wind is at the perfect angle to
obliterate the magnetic component of the tide. The magnetic
component is orthogonal to the electrical force, and in this
position, the Solar Wind is also orthogonal. The Solar Wind
therefore has little or no effect on the electrical component of
the neap tidal force, but it interferes with the magnetic
component to a very large degree.
In the spring tide
positions, we have a different encounter with the Solar Wind.
When Moon and Sun are on the same side, the Solar Wind has little
effect on the magnetic component, but it would be expected to
augment the electric component. When the Sun and Moon are in
opposition, the magnetic component is once again little affected,
but the electric component would be expected to be diminished.
That is, it would be expected to be diminished, except that
the E/M field is shielded from the Solar Wind in that position.
We would therefore expect a small diminishment only. Most of the
effect of the Solar Wind will be blocked by the Earth.
You
will say that the Solar Wind must be blocked by the Moon, when
Moon and Sun are on the same side. And to a small degree, this is
true. But because the Moon is much smaller than the Earth, the
blocking is much greater when the roles are reversed. When the
Moon and Sun are in conjunction, relatively little blocking takes
place.**
Now you will say that the magnetosphere of the
Earth blocks the Solar Wind at all times, since if it didn’t we
would be fried by Solar radiation. Once again, this is true to a
degree. The Earth’s magnetosphere does block a part of the
Solar Wind. But it does not block it all, or we would not have
the Aurora Borealis and all the other known effects. There is a
significant yearround effect even with the partial blocking, and
this is important since we need the effect to be consistent in
all positions.
In fact, even if the magnetosphere were
proved to block all direct Solar Wind effects from
reaching the surface of the Earth, we could still use it as part
of my tidal theory. The reason for this is that we do not need a
huge number of ions actually hitting the oceans' surfaces in
order to use the Solar Wind in the equation. The very fact that
the magnetosphere blocks the Solar Wind is proof that the E/M
field of the Earth is greatly affected by it. The energy that the
field expends resisting the Solar Wind is energy it can't use in
that position to create tides. The field at any dt is a limited
field of energy, and if field strength has to be used in the
upper atmosphere, say, to deflect ions from the Sun, then that
much field strength must be subtracted from the field as a whole.
So the tidal effect is the same whether the ions are deflected in
the upper atmosphere or whether they make it to the ocean’s
surface and deflect some electron there.
The next problem
concerns the ions themselves. It will be said that the Solar Wind
is composed of both positive and negative ions. Won’t this skew
my theory? No, it won’t. I have already stated that the main
effect is on the magnetic part of the field, when the Moon is
orthogonal to the Sun. In those positions, the Solar Wind acts
like cross traffic in the field. As I have shown in another
paper, cross traffic impedes the motion of the field, no matter
which direction the cross traffic is. Which is as much to say
that both positive and negative ions will have the effect of
impedance on the total tidal force in those positions. This is
even easier to understand and accept if you follow the reasoning
of the previous paragraph. If the E/M field of the Earth/Moon is
being diminished in those positions by its role in having to
block the Solar Wind, then that diminishment is absolute: it does
not depend on the charge of the Wind at specific points. In other
words, the total E/M field of the Earth/Moon does not lose when
it blocks positive ions and gain when it blocks negative ions. It
loses either way, since the total field strength is a matter of
total potential. The Solar Wind will sap field potential either
by interacting with positive or negative parts of the field.
One more way to show this is to point out that although
the Earth's magnetosphere may block the Solar Wind, and although
the Moon’s magnetosphere may do likewise, the E/M field of the
Earth/Moon—which only together causes tides—spreads across
the entire distance from Earth to Moon. The greatest effects of
the Solar Wind will be at neap tide, as I said, and at neap tide
the Moon is orthogonal to the Earth, compared to the Sun. In this
position, the E/M field of the Earth/Moon is highly exposed. It
is literally out in the breeze. The magnetospheres of the Earth
and Moon cannot protect the midmost points of the combined field,
and it is here that much of the brunt of the Solar Wind will be
felt.
Now let us
look at the force of the Solar Wind. The speed of the wind
averages about 500 km/s near the Earth. The mass is 10^{9}
kg/s. But we need a density and a flux at 1AU: D/s = 10^{9}
kg//s4π(1AU)^{3}/3 = 7 x 10^{17} kg/skm^{3} Φ
= 3.5 x 10^{14} kg/s^{2}km^{2}
If
we express mass as m^{3}/s^{2}, we obtain Φ =
3.5 x 10^{20} m/s^{4} a = √Φ = 1.87 x
10^{10} m/s^{2}
Now we compare this to
the force from the Moon, calculated in Part 1, which was a = 4.7
x 10^{10} m/s^{2}. We have found that the force
from the Solar Wind is about 40% of the force from the Moon. The
standard model finds it needs the effect from the Sun to be 46%
of the effect from the Moon, so we are close to that already,
with only the first rough equations completed.
Just to be
clear, I treated mass as a 3D acceleration above, which allowed
me to use the square root of the flux for the "field
acceleration." I could take this pretty little mathematical
shortcut since we are comparing an E/M radiation field to an ion
radiation field. What I mean is that the Earth/Moon field is a
field composed of E/M photons, which are treated as massless. But
the Solar Wind is a field of ions, which are far from massless
compared to the photons. Therefore the ion field acceleration has
to include the acceleration due to mass on the ions. It is
automatically included any time the kilogram variable is used, so
all I have to do is remember that, and use it in my equation.
Seen this way, the flux is then an acceleration of an
acceleration, which is why we get the s^{4} in the
denominator.
You will say, "Even by your theory, we
should have to actually find the acceleration due to the average
mass of the ions and 'subtract' it from the flux. How can we just
take a square root?" We can take the square root because the
mass of the ions determines all the other variables in this
field. Look at the variables: flux is determined by density and
velocity, and both velocity and density are determined by the
mass of the ions. If they weighed more, the field would
accelerate them less, and the velocity would drop. You can see
that the question above could be stated, "why should the
acceleration due to mass equal the acceleration due to the
field?" Because, I have shown that they are the same thing.
Gravity and mass are the same thing, stated in different ways.
The "field" that is accelerating the ions is a field
defined by the mass of the Sun and the masses of the ions. The
mass of the ions determines both the acceleration due to mass,
which is the gravitational expansion of the ions, and the motion
in the greater field. Therefore the flux in this case can just be
written as the expansion of the ion squared.
So the short
answer is, the ions are moving as they do because of a
gravitational field. A gravitational field is exactly the same
thing as a mass field, except that one is expressed as motion and
one is expressed as force in collision. If the motion in question
is caused only by gravity, then a total potential force in
collision is just the mass squared, expressed as acceleration.
You can see that this gives us an easy way to calculate force
even though we are only given a velocity.
So, we have
40%, but we need 46%. Fortunately, we can now bring in the E/M
acceleration from the Sun, calculated in part 1, and add it to
the Solar Wind. It provides precisely the extra 6%.
1.87
x 10^{10}
m/s^{2}
+ 3.13 x 10^{11}
m/s^{2}
= 2.18 x 10^{10}
m/s^{2} 2.18
x 10^{10}
m/s^{2}/4.7
x 10^{10}
m/s^{2}
= 46.4%
The
acceleration from the Sun is two phenomena stacked: the Sun's
charge field and the Solar Wind. Since the two phenomena are
caused separately, we can stack them in effect. The Solar Wind is
an ionic field, and the ions are caused by nuclear reactions. The
charge field, however, is a fundamental field, independent of the
ionic field, being simply a field of charge photons. Jupiter, for
instance, has a charge field, but Jupiter does not have a
planetary wind like the Sun's wind. For this reason, we can add
the two types of E/M fields from the Sun.
I will continue
the math of tidal variations later. But in Part 3, we will return
to the main effects. I will show the cause of M2, the lunar
semidiurnal, as well as some of the primary variations of that
lunar tide.
Until then you can read my newest paper on
tides, More Trouble
with Tides, where I return to the beginning of modern tidal
theory and critique Newton's Proposition 66 from The
Principia.
*Watch the standard
model—every time they call something a "crown jewel"
or "the highest achievement of physics", it means they
are covering a very big hole in the floor. Every time they tell
you to look up, in a spirit of awe, look down instead: you are
probably standing on slender ferns, covering a yawning pit.
**The greatest effect would be in total eclipse, of course,
but the Moon doesn't hit that position very often, and it never
hits it for the entire Earth. For this reason alone, the Moon
cannot be expected to be a good blocker of the Solar Wind.
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