Pioneer Anomaly Anomalous No More.

Max Keon wrote:
"George Dishman" wrote in message
...
Max Keon wrote:
George Dishman wrote:
Maths is a tool independent of the physics. Your
equation is supposed to tell me what effect your
anisotropy has and I have just applied that.

Don't be ridiculous. Did the maths invent the big bang universe,
or did that universe determine how the maths would be applied?
If theory predicts a CMBR, the maths is applied accordingly. If
a gravity anisotropy is predicted, the maths is applied
accordingly. ...

The maths must be designed
to incorporate such a thing.

And I assume that is what your equation does.

It certainly does in the zero origin universe.

Then when I use your equation, it will tell me what will
happen in the zero origin universe. That's what I do and
the result is a consequence you didn't anticipate and don't
like, but that's what it is.

You are still in the wrong universe. My equation doesn't design
the zero origin universe, the universe designs the equation.

Sure, and then the equation tells us what will happen to
the planets in your universe - they spiral into the Sun.
Theories often have consequences that weren't anticipated
by their originators. Planck could never have known the
consequences of his original lecture on quantisation as a
solution to black body radiation.

And that sent you all off on another wild goose chase. _Blackbody
radiation has never had anything to do with quantitization_. The
value for the photon has just been plucked out of the air. It
could have been anything at all. But you must surely know that?

Take this part of the formula for example: (2*pi*h*c^2).
Since every component is a constant, the whole thing can be
treated as a simple multiplier. That's all it is, isn't it! So
why isn't it stated as 3.747e-16. But 3.747e-16 what? It's
nothing more than a nominated surface area per unit radius from
the blackbody source on which to establish an energy quantity
according to wavelength. If the unit radius is 1 meter, the
designated surface area is 3.747e-16 square meters.

According to Planck, the energy received from a 4000 K radiator
at 7.25e-7 meter wavelength is
(2 * pi * h * c ^ 2) / (w ^ 5 * ((EXP((h * f) / (k * t))) - 1))
= 1.315e+13 (# units of energy, whatever they are)

Here's an equation that gives the emissive power per wavelength
received at unit radius from the same blackbody source for a 1
square meter surface area. w is wavelength, m is surface area.
m^2 / w^5 / (1.0145^(m / (w * t)) - 1) = 3.509e+28 (# units)

3.509e+28 * 3.747e-16 = 1.315e+13, as the Planck "photon" would
specify it for the tiny surface area.

Try any temperature-wavelength combination you like and you will
always get the right answer.

Planck's photon is entirely dependent on "e" which has no claim
to fame other than it never repeats. That is crap George.

Nice rant Max, but it doesn't even touch on the reason
why quantisation is required, the shape of the spectrum.


How hard would it be
to detect the anomalous acceleration on the outward and inward
legs? How hard would it be to notice that one is pointing in the
opposite direction to the other, relative to the Sun? Whatever is
the cause, there should be some sort of anisotropy evident in the
result

Sure, but that gets you no farther forward in identifying
the _cause_ of the effect.

It would certainly prove that the effect is real though. Then we
can move on.

We already know the anomaly is real without a doubt.

But GR's amazing feat is quite irrelevant anyway because
instantaneous action at a distance was never a requirement in
anybody's gravity.

It was in Newton's.

Only because he didn't understand gravity at all. The fact is
that something was missing in the way he analyzed planetary
motion. And that is clearly still the case today.

No, GR resolved that problem 90 years ago. Science
has moved on a long way since then.

---

No Max, you are off into fantasy world again. The
focal point of elliptical orbits is the barycentre. The
slight effect of a delay would move them from there
so you get a diagram like this:

0-


+ . +


-0

I've fixed the diagram for you.

No, you have screwed it up again so I restored what
I wrote. Go and learn vectors, then read up on the
two-body problem

The following accompanies an animation at this address.
http://www.optusnet.com.au/~maxkeon/binstar.html

The animation depicts a binary star pair (blue) in a concentric
orbit about each other. The star masses are identical. This an
impossible scenario of course because the orbit velocity of each
star is faster than light speed.

I assume you mean your illustration shows impossible
speeds.

The geometry can obviously be
altered to overcome that problem, but my point is still made.

The geometry is not a problem. There are many such
binary systems in reality but obviously the orbital speed
is much slower. Something around 300 km/s is the top
speed for a "grazing" binary IIRC but binaries of nearly
equal mass are not unusual.

The action of gravity at one point around the orbit only is
shown, and that is obviously common throughout the entire orbit
cycle. The nearest red dot in the direction of motion is the
retarded position of each companion star from the viewpoint of
each star.

The '\' line travels the straight line distance to the intersect
point with its companion at the speed of light. The action of
gravity on each star is necessarily shifted to point toward the
retarded image of its companion, which is 90 degrees offset from
its true position around the orbit path. The apparent gravity
link with each companion star can obviously never be generated
where the companion currently resides.

Sure, but only because you have assumed there is abberation
which isn't the case in Newtonian gravity or GR, you are only
describing a strawman so I don't see the point.

The focal point of the orbit radius is perpendicular to the
natural tangent,

No, an ellipse has two focii and for the simpler case
of a circle, they coincide at the centre.

which would cause the stars to fly apart. But
the distance from each star to where its companion appears to be
is .707 of the instantaneous distance. According to 1/r^2, the
pull of gravity toward each other is 1 / .707^2 which is double
that required to hold the stars at the instantaneous orbit
radius. The stars would be pulled inward to point in the
direction of the instantaneous orbit tangent.

Even if that's not enough to do the trick, the universe has no
rules which dictate that a binary star pair of specific mass can
only reside at a specific radius for a specific orbit velocity.
A stable orbit exists for every circumstance. Denying that is
absurd.

Crap. For any radius, if the speed is too fast for a circular
orbit the object will move farther away and if it is too slow
it moves closer. Both give stable elliptical orbits because
energy and momentum are conserved, but as sonn as you
start to remove energy the orbit decays. That's what happens
in nature so that's what our maths has to copy.

There is no reason whatever why the universe should comply with
theory. Nature sets the rules, not you, or me.

Yeah, Max. You see the thing is that we know nature is just
going to do its own thing and ignore us, so the rule for science
is that the maths we use is required to mimic nature

The equation representing an upward moving mass relative to a
gravity source is ((c+v)^2/c^2)^.5*G*M/r^2-(G*M/r^2), while
((c-v)^2/c^2)^.5*G*M/r^2-(G*M/r^2) represents a downward moving
mass.

In which case the anisotropic force pushes the body
in the same direction regardless of the direction of
motion, the opposite of what you told me.

Even an idiot could understand my meaning.

Probably, but your equation says the opposite of your meaning.

That wasn't the case at all. I'll try to explain each step of the
process in some detail (for my own benefit).

The conclusion was that no energy (or momentum) can be
immediately absorbed by the matter of the universe,

That's fine, but energy is being removed by the anisotropy
immediately, it is up to you to say where it goes.

so the force
remains like a spring ..

No it isn't. A spring always pulls in the same direction
like gravity but your anisotropy changes direction. A
spring removes and stores energy while being stretched
and returns it while being relaxed because the force is
in the same direction as the speed.

The Mercury-universe relationship is no different to a mass
cycling around at the center of a stretched elastic sheet.
http://www.optusnet.com.au/~maxkeon/merc-un.gif.jpg

Wrong, you have no anisotropy in that.

which is applying a constant restraining
force on Mercury's orbital motion. That action will of course
initially slow Mercury, which in turn will begin a slow
acceleration toward the Sun. As you say, that reaction cannot be
elastic, and momentum is lost.

As is energy.

But still no energy has been
transferred away from what can only be a locally closed system.

You could hypothesise that the energy and momentum
are stored somewhere locally but that is for you to
work out.

The momentum loss immediately converts to potential energy, which
in turn slowly converts to kinetic energy.

Wrong, momentum and energy are not interchangeable, they
are separate quantities, both of which must be separately
conserved (as must the three spatial components of momentum).

In an inelastic, head on collision between two equal masses
the two masses are brought to a halt relative to each other.
Heat energy is obviously generated during the process. So where
does that come from if not from lost momentum?

From lost energy. There is no momentum lost in the crash.

e=pc^2 . That's what I think.

It is nonsense. Until you learn what a vector is, you won't
even understand this answer. Momentum is a set of three
numbers, one for each of the three spatial directions. Energy
is a single number, it doesn't have a direction so you cannot
equate them that way at all.

According to your maths, that process continues until Mercury
hits the Sun in a million years or so.

No, according to _your_ equation, that's what happens.
I just applied it for you without making your mistake
of confusing energy and momentum.

The eccentricity would not be affected by the universe generated
gravity anisotropy at all when Mercury arrives at a stable orbit.

You won't be able to work out what it does until you
learn how to handle vectors, differential equations and
the difference between energy and momentum.

The need for instantaneous action at a distance to overcome the
problem of the planets spiraling into the Sun should have
sounded alarm bells.

It did and people were trying to resolve it for 200
years until Einstein found the solution.

You know as well as I do that the "solution" fails the Pound and
Rebka test.

What are you talking about Max? Pound-Rebka is one of
the classic tests which _confirm_ GR ! There are no tests
whatsoever that GR fails, only known limitations in merging
it with QM, which could as easily indicate a failing of QM.

I suggest you start looking for another answer.

I suggest you learn how to handle negative numbers and
what a vector is because we cannot continue this
discussion beyond primary school level until you have
those basic maths tools at your disposal.

George