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The
Solution to Tides Part 1
by
Miles Mathis
In
a previous paper I did an extensive analysis of current tidal
theory, proving that the theory fails. I will admit that the
theory is ingenious in many ways, since it finds many clever
methods for forcing a solution in a field that cannot possibly
yield one. For instance, we are told that although the Sun's
field is stronger than the Moon's, its differential field changes
less quickly, and this differential field is the one that causes
tides. Ingenious, but false. The field that causes tides is
electromagnetic, not gravitational. Therefore all the math done
by the standard model is completely wrong from top to bottom. I
show this incontrovertibly in the previous paper, since even with
every trick they can throw at the problem, including the
barycenter trick, the numbers still don't work out. No tidal
theory has ever been successful at deriving the tides we see, and
current tidal theory exists only by subterfuge. Once I point out
all the fudges, it crumbles into a ugly pile of very dishonest
math.
It was interesting for me to return to Wikipedia
(which took the brunt of my critique) a year later. The entire
page had been rewritten, and all the math deleted. There had been
a link to a long page called "Tidal Theory", but that
is gone. If you will remember, I said that only Wikipedia was
foolish enough to leave its dirty laundry out in the open where
people could see it. Every other site on the web, including NASA,
was, and still is, publishing broad theories and fuzzy
mechanisms. Only Wiki let us in on the math. Now Wikipedia has
hired a rewrite from a pro, and we get lots of pretty artist's
renderings and macromedia presentations and (almost) fully edited
sentences. Unfortunately the content is still the same. Tidal
theory still depends on the barycenter fudge and the
Suncausesthespringtide fudge and all the other fudges. They
are just packaged much better now, with no math to confuse the
issue.
This interests me because it is clear that someone
at Wiki is reading my articles. I had spent a few days on the
discussion pages at Wiki when I was writing that first paper, and
I left some links for the Mandarins backstage. Clearly someone
followed them and was highly embarrassed. Possibly alarms went
off on campuses all over the country. Someone was questioning the
standard model. Man the walls and clean the crennelations!
In
fact, this is precisely why the Tidal Theory page had to be
scrapped. The discussion page had my critique all over it, and
Wiki has no policy for deleting or locking out discussion. The
only way to get rid of me was to delete the whole subject. Which
they did. No cost is too high in the flight from truth.
This
hiding of the math (and so censoring debate) is becoming more and
more common, and I expect that physics books at all levels,
beginning with high school, will soon become little more than
bound fourcolor computer models and glossy illustrations of
dogma, with no math or mechanics to get in the way.
I
came across a similar instance of "hide the math" on
the web when I did a recent search for Einstein's famous papers
on SR and GR. There is indeed a new website devoted to Einstein
and his papers, but none of the real stuff is there. We get his
correspondence and his papers after 1925, but the famous papers
are not to be found. They are quite difficult to find at Amazon
or Alibris or even big libraries. It is much easier to get a copy
of the Principia
than it is to get a copy of any of Einstein's real papers. There
are thousands of books on Einstein, but a websearch that seeks
his famous papers on Relativity turns up nothing. I don’t think
this is an accident. The standard model is in hiding.
It has taken
me many months to put together this new theory, since at first I
believed that Newton's and Einstein's equations could be made to
work. I believed this because I knew that both field equations
already contained the E/M field. That is to say, both Newton's
and Einstein's equations describe a compound or resultant field.
The problem is, tides are caused completely by the E/M part of
this field and not at all by the gravitational part.
Although
I have shown that gravity is just an acceleration, and that there
is no real force imparted by the field, many phenomena have
apparent motions caused by the gravitational field that are very
real. That is, they are not caused by a push or a pull, but they
are measurable phenomena. They register on our eyes and
instruments and therefore are just as physical as anything else.
Einstein showed that geometry is just as physical as mechanics,
and he was certainly correct. One such phenomenon is precession.
Another is the orbit itself, both its shape and its very
existence.
But tides are not geometric phenomena. They
are not caused, in the first instance, by curvature, expansion,
Relativity, the time differential or separation, or by any
pointofview considerations. They have nothing at all to do with
gravity. Therefore, to solve the problem of tides, we are
required to separate out the E/M field completely from the
combined fields of Newton and Einstein. Obviously this is not so
easy to do. But knowing it was required was perhaps the biggest
part of the solution. I
have recently achieved the separation, and I now have a
Unified Field Equation that is capable of expressing each field
individually. This will allow me to compare the E/M field
strengths of the Sun and Moon as measured here on Earth, and
finally solve the problem of tides.
Before I do the math
on the E/M field, proving that it fits the data, I will remind
the reader of a problem I pointed out in my paper The Trouble
with Tides. It is known that the Moon shows a negative tide in
front, where it should show a positive tide. The crust of the
Moon has been partially obliterated by some influence from the
Earth, but the standard model has no mechanism to explain this
negative tide. My use of the E/M field as a bombarding field
explains this all at once. If the Moon is being bombarded with
photons from the Earth, as the primary mechanism of tides, we
would expect the front of the Moon to be obliterated by them.
Since the photons are real and the bombardment is real, the Moon
must be damaged just as if the Earth were constantly firing
bullets at it. In this first part, I will be showing the math for
the Moon's influence on the Earth, but in later parts I will be
able to show mechanically how the Earth affects the Moon.
The
most astonishing thing I have discovered in my Unified Field is
that small objects have stronger E/M fields than larger ones.
Given two spherical objects of equal density and makeup, the
smaller of the two will have a stronger E/M field, not just
relatively, but absolutely. The Moon has a field that is 110
times stronger than the Earth's field. This fact seems
counterintuitive at first, but a little consideration shows that
is completely logical, given the things we already accept about
our universe and the way it works. All we have to do is postulate
that the E/M field is a real bombarding field or radiation field
made up of much smaller bodies.^{1}
This field is being emitted by every single atom in the field
(and even by the electrons in the atoms). Therefore, any field of
atoms of any size will create the same E/M field density,
provided that the density of the atoms remains constant. A small
area of atoms will create the same field as a large area of
atoms. If we look at areas that are cubic, then at the edge of
any size cube we will find the same E/M field density, given an
equal density of atoms in the cube. It does not matter if our
cube is one meter across or one light year across.
But if
we look at areas that are spheres, we find something strange. If
we postulate that all the radiation is emitted radially, as if
from the direction of the center, then smaller spheres will
create much denser E/M fields than larger ones. This is due to
the ratio of the surface area to the volume, of course. A smaller
sphere will have the same ratio of mass to volume as a larger
sphere, by the definition of density. But it will have a larger
ratio of density to surface area, which proves my point. This
is a bit confusing, so think of it this way: the smaller sphere
focuses the charge that is emitted, creating a greater charge
density. A
larger sphere will emit more total
charge, of course, but if we look at the amount of charge emitted
from a given radial line, the smaller sphere will actually emit
more. This is because there are fewer such radial lines with the
smaller sphere, so more density has to go with each line. Or,
just reverse this logic: as we move to larger spheres, the same
density has to emit through a greater surface area. Since the
surface area increases faster than the radius, the charge will
dissipate into space more with a larger sphere than with a
smaller sphere. This is important when we are looking at tidal
influences between celestial spheres like the Earth and Moon, and
we will see the importance later with other charge influences,
like those that cause axial
tilt and Bode's
law. When looking at linear influences between planets, the
charge density is more important than the total charge.
In another
paper, I have shown that, given a density of atoms in a sphere,
the E/M field density outside that sphere drops off by 1/R^{4}.
I was able to prove this by looking only at the known
gravitational numbers from the Moon and Earth, and subtracting
out the acceleration fields. The radius of the Moon is 1/3.67
that of the Earth and its density is 1/.6057, which gives a field
strength of 110 times that of the Earth. If the Moon had the
same density as the Earth, its E/M field would be 182 times that
of the Earth, as
a matter of charge density.
According to this theory, the Sun would appear to have no
possible effect on the Earth's tides. Its surface area overwhelms
its density in this regard, and then we add the distance, which
must also dissipate the field. Unless the Sun's density is very
much greater than we think it is (1408), its effect must be
negligible. Let's do the math. The Sun's density is about 1/4
that of the Earth. If the Sun were the same size as the Earth,
its E/M field would also be 1/4. But its radius is 109x that of
the Earth. So it’s E/M field at the surface would be,
(.009545)332,900/(11,990)3.92 = .0676 m/s^{2}
At a distance of 1AU, this field would
be (.0676)(696,000/1.5 x 10^{8})^{4}
= 3.13 x 10^{11}m/s^{2}
Because the Sun is a giant fusion reactor, we do get
electromagnetic effects of other sorts here on Earth. Solar
flares, solar wind, etc. These cause very positive E/M responses
on Earth, as everyone knows. But the standard E/M field of the
Sun is caused only by the density and type of atoms present, and
that field appears too weak at the distance of 1AU to cause
tides. Unless it is postulated that fusion creates a constant
boost in the E/M field, it would appear at first glance
impossible to link the Sun to terrestrial tides. The standard
model has not made this postulate (obviously, or they would
include it to answer the gaps in tidal theory) and I am not in
any position to make it here.
Some will say that I
have shown that the Sun causes a large precession due to
curvature on the Earth (3.36 arc sec), which would imply some
great force, but this precession due to curvature or expansion is
not a force at all. That is to say, it is not caused by a real
radiation field. It is caused by apparent motion due to
expansion. This is clear from my simple diagrams, where I show
that this bending—whether applied to starlight or to
precession—can be attributed to expansion alone. Whether you
think it is caused by curvature or expansion, it is an effect of
the competing acceleration vectors. It has nothing to do with
electromagnetism or any other emitted field.
Others will
laugh and say, "By your theory of increase (1/R^{4}),
a satellite 1km across with a density the same as the Moon will
have an acceleration at its surface of 10^{13}m/s^{2},
all due to electromagnetism. It would act like an inverse black
hole. And a proton with a density of the Moon would have an
acceleration of 10^{85}m/s^{2}."
Stated in that way, the point seems devastating. If anybody ever
got here before me, this would be another place to give it all up
as a bad job. But I don’t quit so easily. This point, like all
the other points, can be answered without that much effort.
I
have already said in many other papers that the electrical field
and gravitational field have to be taken in tandem now. This E/M
acceleration we are finding has to be applied to the mass that is
producing it, and mass is now a function of acceleration, too. As
you have seen, mass and gravity are two names for the same
fundamental acceleration, which I have shown is L^{3}/T^{2}
in my Unified Field
Paper. Therefore, we can’t just compare accelerations
directly; we have to look at velocities, too. To take a specific
example, I have shown that the Moon has an acceleration due to
mass/gravity of 2.67m/s^{2}.
So assuming v_{o}
= 0, after 1s it will have a velocity of 1.335m/s. That is the
velocity and distance in which we now must place the E/M
acceleration. But if we look at a marble with a radius of 1cm and
the same density as the Moon, it will have an acceleration due to
mass/gravity of 1.54 x 10^{8}
m/s^{2}
and a velocity of 7.68 x 10^{9}
m/s. If we apply the E/M acceleration to this velocity, we
obviously get a much smaller effect.
You see, with the
new theory of a compound field, and mass and gravity as
acceleration, all things are not in equilibrium regarding
velocity. Before, it was thought that things were in equilibrium
regarding a velocity due to expansion, since they weren’t
expanding. But now the equilibrium is given only to the relative
size. Things still remain the same size relative to eachother,
since we must keep a visually consistent universe. But velocities
are not equal, as I just showed. Since dv’s are not equal,
accelerations cannot be compared directly. The only time that
accelerations can be compared directly is when they are taken
from objects close in size. The radius difference between the
Moon and Earth is only a factor of 3, so the accelerations of
Earth and Moon can be compared pretty much directly. But when we
compare the computed acceleration of the Moon to a little marble,
we can’t do a direct comparison. The Moon is moving so much
faster during each dt.
It will be said that this offsets
only part of the huge numbers, and that is true. There are other
factors. Let us start with the proton problem and see what has to
be done. I have an acceleration of 10^{85}
to account for. Well, the standard model says that the electrical
field is 10^{40}
more powerful at the quantum level than gravity, so I only have
10^{45}
to account for now. I have shown that the number is a relative
number (E/M field relative to gravity field) not an absolute
number, so I am halfway there. In calculating the difference in
field strengths at the quantum level, the standard model assumes
the gravitational field decreases with the square of the
distance. I have shown that this is false. The inverse square law
applies to the E/M field part of the Unified Field Equation,
which means that gravity is some 10^{22}
stronger at the quantum level than they thought. That brings me
to within 10^{23}.
I have just shown that the velocity due to mass of a 1 cm marble
is 10^{9},
so a proton would have a velocity due to mass of 10^{22}.
That leaves me 10^{1}
out of 10^{85}.
Not a bad margin of error for such a quick analysis. And I didn’t
even have to discuss the density of the proton, which would of
course give me more room to play, since it is unknown. I
will firm up these numbers in an upcoming paper, but the
quick answer needed to be aired here, to deflect uneducated
criticism. Those who would scoff at my numbers need to look at
how they fit into my Unified Field. The solution to tides cannot
be achieved without freeing the E/M field from its historical
bounds, and that cannot be done without understanding how I have
redefined the Unified Field.
After all that, perhaps the
easiest way to see that the numbers must resolve in much the same
way they always did is to recognize that I simply give the 1/R^{2}
that we used to give to gravity to the E/M field instead. This
makes E/M 1/R^{4}
while gravity becomes directly proportional to radius.
Before:
gravity is 1/R^{2}
and E/M is 1/R^{2} After:
gravity is 1/R and E/M is 1/R^{4}
But let us
move on. If the Sun is not a cause of tides, then we must rule
out Venus and Mars as well. This leaves us with only the Moon.
How can the Moon cause all the variations we see?
Since
the mechanism was always assumed to be the gravitational field,
all the historical (and current) explanations have been
spectacularly flawed and simplistic. The basic mechanical
explanation, one that is still trumpeted by major websites and
institutions, is the heaping of waters nearest and farthest from
the Moon (with some delay). This mechanism is so naïve it is
pathetic to ponder. Even were the force caused by the Moon's
gravity, it seems the shape of the field might have been taken
into account by these magnificent theoreticians. A force, of
whatever kind, emitted spherically by the Moon, must arrive at
the Earth spherically. Assuming the field is traveling at or near
the speed of light, we can imagine no delay more than a second or
two. So the position of high tides relative to the Moon cannot be
pushed very far. They must be looked for in very specific places
at very specific times, and since this has never been done, we
can call it gloriously negligent.
To show the degree of
this negligence, let us return to the internet. Wiki presents us
with a new illustration to show the spring tide, which puts the
Moon and Sun on opposite sides of the Earth. The same thing
happens with the Moon and Sun on the same side, but this
illustration is better propaganda, since it appears to show the
tides being created by direct pulling action. The problem is, to
get the correct tidal mechanism working with the Sun on both
sides, you have to treat the field as a differential field. This
differential field creates tides both front and back, and it does
so with either Sun or Moon. But if gravity really creates such a
differential field, it must do so with the barycenter motion as
well. Amazingly, the standard model still trumpets the
barycenter, even in tidal theory, as you will see from looking at
the new Wiki page. This is a problem, since the differential
field of the barycenter would create tides that totally swamped
the Sun and Moon tides, making this spring and neap tide
illustration moot. No matter how you look at it, with either
gravity as a direct pulling force or with gravity as a
differential field, spring and neap tides cannot be explained.
This is why Wiki pulled the full mathematical analysis: the
differential explanation does not work. This illustration is a
fraud.
Before I continue, I must make one more comment on
the standard model. It is amazing to find that tidal theory still
rests on Newton, Maclaurin, Euler and Laplace. Wikipedia
redressed its earlier problems by falling back on more passing
mentions of these famous guys. All were great mathematicians in
their day, but their tidal theories were garbage even then. It
should be highly embarrassing to have nothing to add to this
after hundreds of years. Is no one capable of doing a little
independent thought? Must we continue to stupidly parrot Newton
and Laplace until the end of time?
Here is possibly the
meatiest paragraph on the whole new Wiki tidal page. It is the
closest the new author gets to doing any math:
The
depth of the oceans is much smaller than their horizontal extent;
thus, the response to tidal forcing can be modelled using the
Laplace tidal equations which incorporate the following features:
(1) the vertical (or radial) velocity is negligible, and there is
no vertical shear—this is a sheet flow. (2) The forcing is only
horizontal (tangential). (3) the Coriolis effect appears as a
fictitious lateral forcing proportional to velocity. (4) the rate
of change of the surface height is proportional to the negative
divergence of velocity multiplied by the depth. The last means
that as the horizontal velocity stretches or compresses the ocean
as a sheet, the volume thins or thickens, respectively. The
boundary conditions dictate no flow across the coastline, and
free slip at the bottom. The Coriolis effect steers waves to the
right in the northern hemisphere and to the left in the southern
allowing coastally trapped waves. Finally, a dissipation term can
be added which is an analog to viscoscity.
Wiki's hired pro still can't
spell or use the English language, but so what? More important
is that this entire paragraph, though trying to be rigorous, is
nothing but niaiserie. It is bushwa, babblement. Laplace's tidal
equations are all based on gravity as the mechanism, so they are
worthless. But even if Laplace had based them on the E/M field,
his postulates would still be illogical. Laplace loved
mathematics, but he was very thin on mechanics. His first
postulate cannot be true. As I will show, the vertical or radial
force must be significant and nonnegligible, and so must the
velocity created by it. Which also falsifies the second
postulate. Concerning the third, the Coriolis effect is a
tertiary or lower effect, which does not need to be addressed
until the major mechanisms are sorted through. It is a
terrestrial stirring of the tide, not a creator of the tide. It
is here only to give the author more words to capitalize and more
esoteric names and effects to drop. And the fourth postulate is
just absurd. Here we have been inundated with false mechanics in
order to avoid the real questions. Laplace's equations won't tell
us where the high and low tides are, and why, but they claim to
tell us that variable tangential velocity at different depths is
enough, by itself, to cause the rising and the falling. As if
tangential velocity, by being variable, becomes
antigravitational. The author must bury this in the most
impenetrable language ("negative divergence of velocity"),
otherwise the casual reader might recognize it for the flapdoodle
it really is. Then, to finish up, we are treated to trivial
"boundary conditions". As if we need to hear about
boundary conditions when a basic mechanism hasn’t even been
offered.
Before we
look for the position of high tides, let us calculate the force
from the Moon. How does the E/M force compare to the
gravitational force that was thought to apply? It is stronger or
weaker? Let us take some numbers from my previous papers. I found
that the acceleration due to the E/M field of the Moon is 1.051
m/s^{2*}, but this is at the surface of the Moon. This
acceleration will dissipate with distance, since the field
spreads spherically. In fact, it diminishes at R^{4}, so
we find, at the surface of the Earth,
a/1.050 =
(1,738/378,022)^{4} a = 4.7 x 10^{10}m/s^{2}
I said above that the charge from the Sun seemed to be
negligible, but now that we see the force from the Moon, we find
it is only 15 times the force from the Sun. Therefore we may be
able to bring in the charge from the Sun as one of the lesser
harmonics later (possibly what is now called the Solar diurnal).
But it is still not large enough to explain the 46% Solar input
we are trying to explain.
The acceleration from the Moon
looks too small to do anything, but we have a vector situation
here. The Moon is repulsing the Earth and the Earth is repulsing
the Moon, so the total repulsion will be a summation of the two.
I have shown that the total acceleration of the combined field is
.151m/s^{2}, but that isn't the number we need here
either. What we need is the Earth’s E/M field acceleration at
its surface, .009545m/s^{2}.^{(2)}
By
this way of looking at it, the Moon’s field is almost
negligible. All it does is give the Earth's field something to
bite on. Ironic, I think, that all the effects of foreign bodies
turn out to be negligible. First the Sun and now the Moon. The
Moon only supplies a force to be resisted—an acceleration that
awakens the Earth's own E/M field acceleration.
But why
is the "bite" at the level of the Earth's oceans, and
not at some level in space halfway between the Earth and Moon? It
is simply because the acceleration has to have something to
accelerate. The combined E/M field has to have a place to work
itself out, and the Earth's oceans are the nearest material
field. The fields can’t meet in space, since E/M fields don’t
work like that. A field cannot express itself upon another field,
since the particles in the field are too tiny. A field must
express itself in a material field, an atomic or molecular field.
We already know that from QED, but it is good to be reminded of
it in this place. Of course, in the ionosphere, we do have an
ionic field, and we do see lesser effects from this same cause
there. But this isn't the time to get into all that.
Now
that we have a new number, we can compare it to the old number.
What was the size of the acceleration due to gravity, which was
thought to cause tides? Well, the acceleration due to gravity at
the surface of the Moon was thought to be 1.62m/s^{2},
until I showed that acceleration was a compound number. The
acceleration was thought to diminish as R^{2}, so the
acceleration at the surface of the Earth due to the Moon would
have been a maximum of
3.34 x 10^{5} m/s^{2}
or .0000334 m/s^{2}. In my previous paper on the
problems with tidal theory, I showed that the standard model now
finds an acceleration from the Moon of Δa = 1.14 x 10^{6}
m/s^{2}, using the differential field. They try to add
50% more to that, but I show they cannot logically do so, since
the Earth is not orbiting the Moon. Either way, with the
differential field or the straight gravitational field, the
acceleration is much smaller than I have shown is attainable from
the E/M field.
Somewhat shocking, isn't it? We have found
286 times as much force using the E/M field as Newton found using
gravity. And we found it all coming from the Earth itself.
What
is going to shock you more is that we aren't half finished
finding the force. All the numbers above come from the E/M fields
that I have segregated out of the compound fields of Newton and
Einstein. But this force I have calculated of the Earth's E/M and
the Moon’s E/M coming together must still meet the Earth's
acceleration due to gravity. For at the same time that the two
E/M fields are meeting in the upper levels of the ocean, the
Earth is accelerating at 9.81m/s^{2}. You can give this
acceleration to the field or to the Earth, I am past caring; but
I have shown that as a vector, this acceleration must be in
opposition to the E/M field. [It is 9.81, not 9.8, since I have
proved that the solo gravitational field (without the E/M field)
is a fraction more than we thought. We have been measuring a
compound field all these centuries. But now I subtract out the
E/M field, and we have 9.81.] This means that at the center of
our circle, we must add 9.81 to .009545.
Good lord, you
say. That would mean that a person would weigh more than double
directly under the Moon! So let me put it another way. These
vectors are difficult to keep in line, I admit. When I say that
we must add 9.81 to .009545, that is the same as saying that we
must add .009545 to 9.81. Whereas, normally, we would subtract
.009545 from 9.81, to get our old friend 9.8. That is how the two
fields work when no other body is causing tides. The
gravitational force pulls us down, as an effect, and the E/M
field pushes us up, as an effect, so the result is mostly down,
to the tune of 9.8. But now I am saying that instead of
subtracting, we add. The Moon causes the vector situation to
switch. So now, directly under the Moon, we have about 9.82 m/s^{2}
as our resultant acceleration. And this makes the tidal
acceleration
.009545 x 2 = .0191 m/s^{2} And
that is 572 times the maximum tidal force from gravity.
So,
yes, you would weigh about .2% more directly under the Moon.**
But don’t be too surprised, since this was an effect of the old
tidal theory, too. Except that you would have weighed .00035%
less. I am changing the size and direction of the change, but the
idea is not all that different.
I will answer one final
question before moving on. Why does the Moon's presence appear to
switch the direction of the Earth’s E/M field vector? I said
that when the two E/M fields met, we added them together, which
made some sense. But then I said that we added this result to the
Earth’s acceleration. How does this make sense, when I created
an entire Unified Field Theory to show that the E/M field vector
was in opposition to the gravitational acceleration? You will
say, "If the vector is in opposition, it must point out.
That is why it repulses the Moon. The E/M field is a repulsive
field, according to all your novel theories. If so, the Moon
cannot reverse the direction of the Earth's E/M vector just by
sending in a negligible amount of force (.00000000047m/s^{2})."
Again, the problem is one of vectors. I have shown that
as effects, the vectors of gravity and E/M are in opposition. The
gravitational acceleration creates an apparent motion of
attraction (the bodies get nearer) and the E/M field creates a
real repulsion (the bodies get farther away). But as real
motions, the gravitational acceleration and the E/M field are in
the same direction. The surface of the Earth is moving outwards.
The Earth is also emitting radiation, and this radiation is
moving outward, bombarding all things. So in this case the
vectors are not in opposition.
This means that when you
are solving a problem, you have to be aware of which vectors you
are talking about. Are you talking about force vectors, as seen
by real eyes or instruments; or are you talking about
acceleration vectors that describe the actual motion of the
quanta? They aren't the same.
The Moon's presence seems
to reverse the E/M field vector of the Earth, because when the
Moon is present, we stop looking at the E/M field as an emission
of the Earth and start looking at it as a product of the
Earth/Moon interaction. We stop looking at it as an acceleration
and start looking at it as a force. As an acceleration, we
subtract it from the gravitational acceleration, since their
effects are in opposition. As a tidal force, we add it to the
gravitational acceleration, since they are a vector summation.
This can be stated more briefly in this way: when two
vectors are drawn in opposition, they can either be added or
subtracted. If the vectors are meant to represent a collision of
particles, they are added, since forces in collision add. If the
vectors are meant to represent velocities or accelerations that
do not collide, then they are subtracted.
Now,
where do we look for high tides, in the first instance? Let us
say that we have just turned on the E/M field, and we want to see
where the very first effects are. We let the field travel from
the Moon, spherically, and look where it draws a circle on the
Earth. There we should find a complete circle of high tides,
surrounding a very large low depression. On the far side of the
Earth, we would expect no primary effects, only secondary
effects, and we will look at those in Part 2. All we can be
certain of is that there will be no heaping or depressing over
there at first, since nothing is over there to mechanically cause
such a thing.
The two spheres of Earth and Moonfield
meet at an angle, as in the diagram.
r = 6,378km R = 384,400km S = 378,022km r^{2}
= x^{2} + (r/2)^{2} x = 5524km sinθ =
5524/6378 θ = 60^{o} 2θ = 120^{o}
So
the high tides will make a circle with a radius of 6,679km on the
curved surface of the Earth. The diameter of influence of the
field is initially 13,358km, which is exactly 1/3 of the
circumference of the Earth.
Is the number 3 a
coincidence? No, it is not. You may assume that any number that
precise and that small is not a coincidence, but in this case it
is quite easy to show its cause. The cause is the density and
radius of the Moon, compared to the Earth. The Earth’s density
is 1.65x the Moon’s, and its radius is 3.67x. 3.67 x 1.65 =
6
Not only the strength of the E/M field but its rate of
dissipation are determined by density and radius. In the diagram,
we are letting the field dissipate with time, in a spherical
fashion, as you see. And we are holding the Earth steady. This
last is why we find 3 instead of 6. Because we are holding the
Earth steady, we are calculating only half of the field changes.
We are calculating the Moon’s field moving toward the Earth,
but not also the Earth’s field moving toward the Moon. It is an
equal and opposite reaction, so it will double the compound
field. But we are only calculating the action, not the reaction,
so we get half, or 3.
If that was not clear, think of it
this way: the orbital distance of the Moon is not a coincidence.
I will show this in more detail in another paper; it is not the
subject of this one. But the orbital distance, which we are
calling R here, is a direct outcome of the two fields, E/M and
acceleration (gravity). These two fields cause the orbital
distance. The acceleration creates an apparent attraction, and
the E/M field keeps the Moon from being caught. The Moon's
"innate" velocity is also involved, of course, but the
two fields determine this as well, after any amount of time.^{3}
So R is completely determined by the size of the bodies and their
densities. The Moon must orbit at (or near) that radius where its
field intercepts 1/3 of the Earth's sphere. Study the diagram
closely. It is a lovely thing in many ways. It almost looks like
a piece of astrology, except that all the coincidences are now
explained with simple and clear mathematics and mechanics.
[Some have looked at my diagram and said that the number
1/3 is not even dependent on the Moon; it is dependent on the two
small circles I have drawn. But this is not true. If this is not
clear at first, consider the variable t that I have marked. That
is the differential that brings the Moon into the equation, and
it does so since that differential is not arbitrary or
accidental. You see, the question is, how far do you let the two
spheres of Earth and Moon interpenetrate in the diagram? Which is
the same as saying, what is the initial interval of influence? It
can’t be instantaneous, so what number do we use? We can’t
just draw the larger of the two big circles and do the math with
it, since it doesn’t correspond to the first touch at the
center. We have to let the first influence wrap around the Earth,
you see, and this is impossible to draw. The best way to diagram
the interval is therefore the way I have done. This of course
begs the question, "What is the value for t?” This
brings us to some more math that isn't crucial to the main
argument of this paper, so I will put it in an endnote.^{4}]
So we have
found the size of the acceleration and its initial shape, when it
first meets the Earth. What then? From the illustration we can
see that the first force will be nearest the Moon, but the other
forces in the great circle must come split seconds later.
Therefore the initial time lag is not terribly important. What is
more important is the direction. In the center of the circle the
force is radial. In other words, it comes straight down upon the
ocean. We cannot say that this force is unimportant, and that is
why I said that Laplace's first postulate was wrong. First of all
we have a force that is 572 times any force Laplace could have
been working with, most of it generated by the Earth itself.
Secondly, this is a force down instead of a force up. A force of
that magnitude over that area cannot be negligible.
You
can see that the initial force will change from radial to
tangential as we go out from the center of our circle. All around
the circumference of our circle of initial influence, the force
will be completely tangential. But unlike Laplace, I find the
tangential force less important than the radial. You will soon
see why.
Let's look at depth of penetration. At the
actual tangent, the depth of penetration will be almost nothing.
The force will be like that of the wind. It will be expected to
cause waves, not tides. Only as we move inward from the outermost
points of our circle of influence will the depth of penetration
become significant, and this depth will be caused by an
increasing radial component of the force. This contradicts
Laplace once again.
But let's be generous for a moment
and try to find an instance where Laplace was correct. If we move
inward on the circle a few kilometers, we can cause our force to
penetrate to some depth without losing much of its tangential
component. We have a nearly tangential force that causes a nearly
tangential velocity of the waters. This is the component that
Laplace took such interest in, and it is certainly important. I
will show that it must be outranked by one other component of the
tide, but it is a strong second.
This force is not
resisted in any large amount by the ocean, since the water is
relatively free to move tangentially. Ahead of this water is only
other water until the curve ends, and then there is atmosphere.
Both the water and the atmosphere are unconstrained to a large
extent. The water has viscosity and pressure and the atmosphere
has pressure, but other than that the force is free to act. The
mass of the Earth is not behind the water at all to a significant
depth, at this angle, so the freedom of the force to act is near
perfect (compared to the radial force at the center of the circle
of influence). The water can therefore develop a significant
tangential speed.
Now, if we look just beyond the
tangent—which is to say just beyond our circle of initial
influence—we find water that has not been touched by any force
at all. It is completely unaccelerated. As our accelerated water
meets this unaccelerated water, it will pile up behind it,
causing a swell. This is one of our high tides. In the initial
stages of our analysis, it must be a complete circle of high
tides, with a diameter on the curved surface of the Earth equal
to 1/3 the circumference of the Earth. It will travel at some
velocity around to the far side of the Earth, until blocked by a
land mass or resisted by a reverse tide.
But let us
return to our central force. I have shown that it is radial and
quite large, 572 times larger than any force we have yet had at
our disposal in the history of tidal theory. It hits the Earth
like a radial meteor, except that this meteor has a radius of
378,000km. It is like a meteor with a very low density. The main
difference between our force from the Moon and a real meteor is
that our force keeps arriving continuously. The Earth is hit by
this low density meteor every dt, and it keeps getting hit
without end. The waters are elastic, and they want to rebound
from this force, but because it is continuous, they can do so
only to a very limited degree. The freedom of motion is lateral,
and so the waters move sideways to the force. Also, this lateral
freedom of motion decreases with greater depth. As we go into
deeper water, the freedom to move decreases in all directions
simultaneously, which pushes the effect up and out on the surface
even more quickly. This means that although the force is radial,
the motion created is tangential. The water does not want to move
down, and at greater depths it does not want to move sideways,
either. So the result is motion sideways nearer the surface.
Another circular wave is created, traveling out from the center.
Initially this central wave is 60^{o}
behind the outer wave, and unless we show that it is moving
faster than the outer wave, it will stay 60^{o}
behind it.
But which wave is bigger, the central wave or
the outer wave? And what is the ratio? Already it is clear that
they cannot be equal, since the mechanism of creation is so
different, and this confirms what we know of tides. But to get a
usable number we are going to have to do a mountain of math and
mechanics yet.
Looking ahead, we see many things. One, we
have to include the idea that we have a continuous source of
waves from both the center and the outer rim. Two, we have to let
the waves travel to the far side of the Earth, where they will
interfere. Three, the interference waves have to travel back to
the near side, where they will interfere with the source. Four,
the Earth’s rotation has to be taken into account, making the
continuous sources travel in a westward direction (as a first
approximation).
Five, I have to explain spring and neap
tides in a logical way, without the Sun's gravitational field (it
is worth waiting for, I promise you).
But before we do
that in Part 2, we have to look at one other major factor: the
magnetic component of the field. Since with the central force we
are looking at lateral results, we must remember that the
magnetic field also works in this direction. So far all the
forces I have postulated and calculated have been electrical
forces. But the E/M field has two components. By the right hand
rule, if the electrical force is radial down, then the magnetic
force will be clockwise, looking down on the ocean. Toward the
center of our circle, this should have a magnifying effect on the
electrical force, giving it the effect of a screw instead of a
nail. This is true whether you imagine once big screw or billions
of tiny screws. If the screws were impacting a solid, this would
not be true regarding the tiny screws, but since we are dealing
with a liquid, we have freedom of motion in all lateral
directions, near the surface. The screws therefore cause a
spreading, which magnifies the lateral forces already in play
with the electrical field. The magnetic field and the electrical
field work in tandem to produce the central wave.
At the
periphery of our initial circle of influence, we find a different
mechanics at work, regarding both fields and their confluence. At
the tangent, we have the Earth's force meeting the Moon's, to
create a resultant acceleration. But the collision is not
headon. The two electrical forces, by themselves, are
orthogonal. This must cause a great reduction in the resultant
force, and also cause an angle in the motion of the resultant
vector. It will affect the magnetic field as well, since it must
suffer the same diminishment. In both cases we will be taking
sines or cosines of accelerations instead of the accelerations
themselves.
Go
to part 2 of the Solution to Tides.
^{1}The
standard model now does make this assumption, finally admitting
that the field must be physical and must be mediated by
photons. *See my paper on The
Unified Field Theory for a full derivation of this
number. **If this were the only factor, you would weigh .2%
more, but it isn't. So don't rush out with your gravimeter: I am
not making a prediction here, just doing some math. ^{3}The
tangential velocity of an old satellite is caused by the magnetic
field working at a right angle to the electrical field. The
electrical field causes the orbital radius and the magnetic field
causes the orbital velocity. Only very young satellites can be
retrograde, and therefore retain a large part of their "innate"
velocity. ^{4}Once
the E/M field reaches the ocean, it can no longer travel at speed
c, since it is then conducted by the water. So the time period is
determined, in part, by the speed of electricity in saltwater.
From the other end, the time period is determined by the
gravitational acceleration of the Earth. While the E/M field
disturbance from the Moon is racing through the water, the Earth
is accelerating outward at 9.81m/s^{2}.
So, in effect, the tangent of the Earth is changing while the
effect is moving from center to tangent. If the Earth was not
expanding, the tangent would be near the top of the Earth in the
diagram, and we would find nearly half the Earth hit by the
sphere arriving from the Moon. That is, x in our equations would
be nearly r. Knowing this, we can solve from the other end. We
don’t have to ask how long it takes electricity to travel 1/6
the circumference of the Earth, on the surface of oceans; it is
easier to calculate how long it takes the Earth to expand from
r=5524 to r=6378. Since both times are equal (the period t is
determined by when the two motions meet) we choose to do the
easier math. At 9.81m/s^{2},
it takes the Earth about 13.2s to expand 854km. Therefore, we may
assume it takes electricity about that long to travel the given
distance. So t=13.2s, the time for one complete "wave"
of disturbances to arrive from the Moon. Velocity of E/M field in
seawater=506km/s.
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