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How
General Relativity Solves the Metonic Cycle
by
Miles Mathis
The
Metonic Cycle has been known since before the time of Homer.
Odysseus timed a secret meeting with Penelope at the precise
moment of Sun and Moon conjunction. Although the Chaldeans and
Babylonians both knew of the cycle, it gets it name from Meton, a
Greek astronomer who in 432BC calculated the interval of the
cycle to be 19 (Tropical) years with a remainder of 2 hours. The
Metonic Cycle describes the time it takes for the Moon and Sun to
return to the same positions, relative to the Earth.*
Although
astronomers currently accept the Metonic calculations (with some
tiny updates) they believe the cycle is just a coincidence. They
do not dismiss it as astrology, but they do dismiss it as having
no mechanical cause. It is real, but not due to mechanical
affects; or so they assume. Wikipedia puts it this way,
This
cycle is an approximation of reality. The period of the Moon's
orbit around the Earth and the Earth's orbit around the Sun are
independent and have no known physical resonance. Examples of a
real harmonic lock would be Mercury, with its 3:2 spinorbit
resonance or other orbital resonance.
I
will show this is false. The Metonic Cycle is caused by lunar
precession caused by General Relativity.** That is, the same
curvature that caused Einstein to add 43 seconds of arc per
century to the precession of Mercury causes the Moon to precess
one full month every 19 years, with a remainder of about 2 hours.
Since General Relativity is believed to describe a mechanical
cause (geometry is as good as mechanics, or better), the standard
model is wrong concerning the Metonic Cycle. It is not a
coincidence. It has a mechanical cause that can be derived with
simple mathematics. We are currently in possession of all the
math and numbers required to solve this problem, but no one has
yet applied them. I will do so now.
I
have shown elsewhere that the simplest way to solve any
problem of General Relativity is to invert the field. Rather than
give the curvature to the space, I give the curvature to the
size/time field. What allows me to do this is Einstein’s
postulate of equivalence. Einstein showed the mathematical
equivalence of acceleration and gravity: the only difference is a
vector reversal. Einstein then showed that equivalence required a
curved field, but this is not quite true. Mathematically, a
consistent field can also be achieved by reversing the vector "by
hand" and following the mechanics in that way. Einstein did
not want to pursue this method, since it seemed to imply that
physical objects were expanding. He preferred to curve the field.
However, mathematically, the two manipulations are equivalent.
Einstein’s postulate of equivalence states it outright, and
I have shown that the two maths get the same numbers. I prefer to
reverse the vector since it allows me to achieve the same numbers
in about 1/100th the time, with simple math. This mathematical
field reversal, used as a simplification of General Relativity,
need imply nothing about the actual motions. It is a valid
mathematical manipulation only, and it has no necessary physical
implications.
Some have not liked my method, preferring to
stick to the tensor calculus and the curved field. But these
folks are enmeshed in a mathematical contradiction, since they
have already embraced the field of Minkowski. The GR field as it
exists now contains the math and postulates of Minkowski,
including the fourvector field, in which time moves on an
imaginary axis that is at a right angle to the other three axes.
Minkowski admitted that this was just a clever mathematical
trick, one he used to express the field in a more elegant manner.
He never tried to defend the reality of time being on an
imaginary axis or the reality of time moving at a right angle to
x,y,z. The fourvector field has been used because it works, and
because it vastly simplifies the manipulations of GR. Well, in
the same way, my math vastly simplifies the field once again, by
a simple mathematical manipulation. And my manipulation is
underwritten by Einstein himself, since it is justified by the
postulate of equivalence. The postulate of equivalence not only
justifies my vector reversal, it all but begs it. In this way, I
beat Minkowski at his own game. I have achieved a greater
elegance, and I have done so without an ad
hoc manipulation like field
symmetry. My manipulation is a direct outcome of Einstein’s
own postulate.
As you will see, the vector reversal
immediately flattens out the field, allowing me to dispense with
the tensor calculus and all lengthy computations. I have already
used my method to solve
the perihelion precession of Mercury and the bending of
starlight. In both cases I achieved exactly the same number as
Einstein for field curvature. After these successes, I looked for
other problems to apply my new equations to, and the Metonic
Cycle is the first of many I will solve using them.
The
method is incredibly simple. We reverse all the acceleration
vectors in the problem and then look to the time differential
caused by light. Precession caused by curvature is due to this
time differential. That is precisely why it is relativistic.
Specifically, we let the Sun and Moon accelerate in all
directions while light is traveling from one to the other. Since
the light travels from Sun to Moon, the Sun’s acceleration
happens after the light leaves it, so we can ignore the Sun. All
we have to look at is the Moon’s acceleration while the
light is in transit.
The average distance of the Moon
from the Sun is 1AU, or 149,600,000km. The time of the Tropical
year is 365.242 days. The time of the Draconic year, which
measures the Moon relative to the Sun directly, is 346.620 days.
The sidereal month is 27.322 days. It takes light 499s to travel
1AU. The Moon’s gravity is 1.62m/s^{2}.
It’s mean orbital velocity is 1023m/s. It mean orbital
distance is 384,400km. That is all we will need to solve.
s
= (1.62m/s^{2})(499s)^{2}/2
= 199.2km The Moon accelerates that far (in each direction)
in 499s. tanθ=199,200m/1.496x10^{11}m
θ = .27465 seconds of arc
That is the angle of
curvature, given motion in one direction in the field. And
it is an angle to the Sun, since the light is coming from the
Sun. It is only half the total
curvature in one orbital plane, since the Moon can accelerate
both +x and –x. That is, we have only calculated the
distance traveled in one direction, so we only achieve the
curvature in that direction. If we want total curvature, we must
take the increase in diameter, not only the increase in radius,
and this will double our number above. But in the current
problem, we do not need total curvature, since we will be
comparing our acceleration to the orbital motion, and the orbital
motion is in one direction only. At any one moment, the Moon
orbits prograde, but not also retrograde, you see. So we don’t
care about the curvature in the reverse direction. It does not
affect the calculation.
To calculate precession from this
curvature, we must compare the velocity of the Moon due to this
acceleration to the velocity of the Moon due to orbiting the
Earth and Sun. To make this comparison, both velocities have to
be straightline velocities. Therefore, we must transform the
Moon’s orbital velocity, which is an angular velocity, into
a tangential velocity, which is a vector. That is done with this
equation:
v_{t}
= √(a^{2}
+ 2ar)
That is an equation I
got from Newton, used before the world forgot about the
difference between orbital and tangential velocity (but which is
still highly useful).
a = v^{2}/r
= (1,023)^{2}
/(3.844 x 10^{8})
= .002722m/s^{2}
v_{t}
= 1,446.7 m/s
Now we need a velocity due to acceleration.
Since we have taken only half the curvature, we only need half
the total acceleration.
v_{a}
= at/2 = 1.62/4 = .405m/s
Now we compare the two complete
rotations, given the two velocities. If it takes 27.322 days
going 1,446.7m/s, then it will take 97,597 days at .405m/s.
(4.7187 x 10^{6}
)(.27465 arc secs) = 97,597 days P_{M}
= 4850.1 arcsec/yr
We
have just found how much of the Moon’s orbit the Moon eats
up by accelerating into it. That is what precession is, by the
postulate of curvature. Now let us find a number for 19 years.
4850.1 arc sec/yr = 92,152 arcsec/19yr
But we
have a remainder of about 2 hours, remember, so we have to make a
small correction. That 2hour gap is found at the end of a period
of precession, and is the distance from precise conjunction.
Therefore the gap is not defined by precession, but by normal
orbital motion. The exact gap is 2.082 hours. At 1023m/s the Moon
travels (in a curve) 7,667,600m in 2.082 hours. During that same
time, the Earth has moved as well, pulling the Moon with it. At
29,780m/s the Earth travels 223,200,000m in 2.082hr. To solve, we
must know where in its orbit the Moon is at Metonic conjunction,
since we must know whether its relative motion is with the
Earth’s motion or against it. It could be either one, and
we must know in order to know whether to subtract or add at this
point. It turns out that at Metonic conjunction, the Moon is
nearer the Sun, so that we must subtract. [But this is only an
estimate for use in this paper. A precise calculation requires a
bit more rigor in this part than I feel required to do.]
So,
in two hours, relative to the Sun, the Moon moves about
223,200,000 – 7,667,600
= 215,530,000m = 297.2
arcsec
We take the difference of those two corrections
and subtract them from the 19year number to get
91,854
arcsec/19yr
Now if we divide one full rotation (1,296,000
arcsec/orbit) by that number, we obtain 14.109. In 19 years, the
Moon eats up about 1/14th of its orbit. But as we have seen, in
this case "its orbit" means its orbit around the Sun.
The Moon is precessing relative to the Sun, since the angle we
found initially is an angle to the Sun. This means that we must
divide the Tropical year by 14, to continue our calculations.
365.242/14.109 = 25.887 days
The Moon precesses
an amount of curvature equivalent to almost 26 days every 19
years. That amount of curvature is relative to the Sun, therefore
the curvature is (approximately) on the main orbital path of the
Earth. But the Moon does not travel on that path. It travels a
path about the Earth which curves much more quickly than the
Earth’s path. We could do a lot of math to compare the two
curves, but a quick analysis shows that 26 days of curvature on
one path must equal 26 days of curvature on the other, since Moon
takes the two paths simultaneously. Therefore, we apply the 26
days of curvature to the Moon's real path, and so appear to
already have a full month by some reckoning. But what reckoning
is it?
That is not one month by either the sidereal or
lunar reckoning, so we need one more step. What we need here is
the Draconic calendar, which is a direct comparison of the Moon
and Sun. It is also called the eclipse calendar, since it tells
us when the Moon and Sun conjoin. Obviously this is the calendar
we need, and the Draconic year is 346.62 days, as I said above.
If we divide that into the Tropical year, we get 1.054, and if we
divide 14.109 by 1.054, we get 13.386. Now, there are 13.368
Sidereal months in a Tropical year, so we have completed the
math. In 19 years, the Moon precesses an amount of curvature
equivalent to one full month, returning to its original position
relative to both Sun and Earth. [25.9 days x 13.368 = 1 Draconic
year, therefore 25.9 days as we found it above is a sort of
Draconic month whose background is the Tropical year. My final
step allows us to assign the 19year curvature to the Draconic
year directly, so that we obtain the correct number for the Moon
relative to the Sun (instead of the Moon relative to the Earth or
the Earth relative to the Sun)].
The small margin of
error is caused by two things. 1) Using mean values. 2) In the
correction of 2 hours, using the speed of the Moon relative to
the Earth in one curve to estimate its distance travelled on a
different curve (its orbit around the Sun). The two curves aren’t
equivalent, therefore the distances aren’t exactly equal.
We used the Draconic year as a correction, since we
needed numbers that compared the Moon and Sun directly. I could
have used Draconic numbers from the beginning, but the problem
had been defined by Sidereal and Tropical numbers, and those
numbers are what most people know. Therefore I have preferred to
make the correction at the end, with a single manipulation. Of
course I could have also used lunar or synodic numbers, and made
other transforms, but the method I used above is the fastest. In
conclusion, what we found is that the Moon precesses the
equivalent of one full month every 19 years, minus 2 hours, and
that this precession can be shown to be caused by the curvature
of General Relativity and the time separation supplied us by the
absolute speed of light. Since this precession causes one full
orbit, relative to both Earth and Sun, the Moon must return to
the same spot it occupied 19 years earlier. Although I have used
a variant math to simplify the calculations, my math is strictly
equivalent numerically and mechanically and axiomatically to that
of Einstein. However, I have shown that it is much easier to
apply, as well as being much more transparent as a matter of
kinematics and dynamics.
I might add here at the end that
this is the first time in history that anyone has shown the
mechanical cause of the Metonic Cycle. Since the time of
Einstein, the world has been in possession of the tools to do so,
but the difficulties of the tensor calculus made the solution of
such a complex problem too daunting. Even with my simplified
math, I needed several pages of equations and commentary. The
tensor calculus would have turned this paper into 50page
treatise, like other GR solutions, and made it completely
impervious to mechanical explanation or comprehension.
I know that
some will not understand how the Moon’s precession due to
curvature can be so much larger than Mercury’s, but these
readers simply haven’t yet understood Einstein’s
field. GR, due to its axioms, must take into account two related
but basically separate functions. Einstein represents both by
curvature. The first is Newton’s acceleration, which
Einstein expresses by curvature. The second is the time
differential, which comes out of SR, and Einstein also expresses
this relationship with curvature, since it is not linear once it
is imported into the curved field. But only the first curvature
diminishes with distance. Only in calculating this first
curvature would you expect the Moon to experience less curvature
relative to the main field. But using the time differential, you
must expect the reverse. In this differential field, curvature
increases with distance, for the simple reason that it has more
time to curve. Or, the angle of curvature is greater since light
bends all along the path. This is why the Moon’s curvature
is greater than Mercury’s, regarding precession. This
becomes very clear with my math above. This sort of precession is
not a phenomenon of the Newtonian field, which is why Newton
couldn’t calculate it. It is a phenomenon of the time
differential field, which increases with distance. That is why it
is relativistic, and why it is added as a vector to the curvature
of the Newtonian field. This second curvature is a differential
field compared to the first (a rate of change of the first), is
much smaller than the first, and increases with distance rather
than decreases.
Look at it this way. You could express
Newtonian gravity with curvature, ignoring relativity and the
time differential completely. Curvature is just a mathematical
way of expressing the acceleration, and it need have nothing to
do with relativity. Say you accepted Einstein’s postulate
of equivalence but not SR. Well, in that case you could still set
up a Gaussian or Riemannian field to express Newton’s
accelerations. That is what Einstein did, in fact. It was not SR
that forced him to go to the curved field, it was equivalence (he
thought). Only afterwards did he import the postulates of SR into
GR, bringing into the field the time separation I have used
above.
Once he did that, he had a sort of doubly curving
field, and it was only that doubly curving field that gave him
variations from Newton. Without the second curvature supplied by
the time differential, Einstein’s curved field would have
given him the same numbers as Newton. Everyone should know that,
since many of the tensors express only what we would call the
Newtonian curvature. It is only the full assortment of tensors
that allows one to calculate variations from Newton. Well, the
second curvature must increase with distance, as my inverted
field makes clear.
Now, admittedly this has not been
current wisdom. Most of the biggest names in physics don’t
currently comprehend how the field works, and once they get
beyond the problems Einstein solved himself, they start making
false and illogical assumptions. They take the great machine that
is the tensor calculus and just throw the main switch. They make
no effort to understand the mechanics, since the mechanics is all
hidden by the machine.
And this is precisely why none of
them has ever used GR to solve this problem of Meton. The machine
is so big they do not know how to drive it. I have simplified the
equations for this very reason. I knew that I must not only solve
the problem, but must do so in a way that was so transparent it
defied contradiction. Anyone who wants to dismiss my solution
must explain to the world how I got the right answer with such
simple math. If I am wrong about the Metonic Cycle, why did my
math work so well? If my math is wrong, then how could I get
Einstein’s numbers for Mercury’s perihelion
precession and the bending of starlight? If my math is wrong, how
is it possible that I have used it not only to match Einstein
precisely, but to discover things he could not?
*Obviously,
both the Moon and the Sun are in the sky at the same time during
this conjunction, which means that the Moon is nearer the Sun
than the Earth. This is important later in the calculations,
since in that quarter the Moon moves against the Earth’s
motion. **This is not the same as normal lunar precession,
which measures the Moon's ascending node around the ecliptic and
is equal to 18.613 years. The two are related but are not
equivalent, as I will show in a subsequent paper.
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