"Max Keon" wrote in message
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"George Dishman" wrote in message
ups.com...
Repeating post lost by ISP:
"Max Keon" wrote in message
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"George Dishman" wrote in message
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"Max Keon" wrote in message
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I'll try to abide by your new set of rules and use a + - switch
to identify velocity direction. I should still get the same
answer though.
You will if you stick with just one of your equations,
the changed sign inherent in using the standard
convention means you don't need to swap between c+v
and c-v.
I've updated the web page accordingly.
http://www.optusnet.com.au/~maxkeon/pionomor.html
You still have the incorrect equations show, they should
be removed.
"According to the conventional method of identifying
gravity force direction, and the conventional method
of identifying velocity direction relative to a gravity
source, just the one equation is all that's required.
But what it attempts to describe is not as clear.
((c+v)^2/c^2)^.5*(-G*M/r^2)+(G*M/r^2)"
To anyone familiar with maths at high school level or
beyond, your versions are both incorrect and very confusing.
The Lorentz equations include ((c^2+v^2)/c^2)^.5 and it is
very easy to misread your version as that, I did for some
time before noticing what you were really saying. By far
the clearest way is to write this way:
a = (-G*M/r^2) * (1 + v/c)
Well why didn't Newton do that?
He didn't have your (1+v/c) term because it doesn't
exist. Other than that, we now take calculus for
granted as something we all learnt at school while
he had to invent it, so in explaining the physics
he had to write in a much more detailed way. I can
write "a" above and you know I mean acceleration
which is the second derivative.
And why do you think that a = (1 + v/c) * (-G*M/r^2) + (G*M/r^2)
is not as confusing as a = ((c+v)^2/c^2)^.5*(-G*M/r^2)+(G*M/r^2)
It would be, why have you added an extra term? The
total acceleration in your theory is
a = (-G*M/r^2) * (1 + v/c)
of which the Newtonian part is
a = (-G*M/r^2)
and the anisotropy is
a = (-G*M/r^2) * (v/c)
Add the two together to get the total.
Your simplified equation is not describing the cause of the
gravity anisotropy at all. It's very confusing in my opinion.
I think you just need to learn these basic laws of
arithmetic well enough to be able to read equations:
http://www.mathsisfun.com/associativ...tributive.html
In the original format, as two separate equations, each equation
described very clearly how an object moving toward or away from a
gravity source will be affected. In order to achieve that goal,
there was nothing wrong with the way they were written in this
extract from the web page.
Except that the have the acceleration pointing in the same
direction whether the body is moving towards or away from
the Sun.
--
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. Even matter in a fixed position relative to a gravity
source is traveling outward through dimension because dimension
is traveling inward through it, hence the action of gravity.
--
It's obvious that velocity is added to or subtracted from the
speed of light. And that should be obvious to anyone at all.
Sure, but if you add a positive number or subtract a
negative number, the result has the same sign. Your
equations may look pretty to you but they give the
wrong answer.
It is then obvious that your are taking the conventional
Newtonian equation (-G*M/r^2) which everyone recognises
and modifying it by the (1 + v/c) term. It also makes it
clear that the result is first order where anyone looking
at your versions would see the speed appearing only as
"(c+v)^2" and assume it is second order.
and that is why your version is confusing.
While I was there, I made another attempt at finding the words
to better describe how Mercury's apparent loss of orbit momentum
is conserved when it finally comes to rest in a stable orbit that
counteracts the influence from the universe.
It is still wrong, it does not conserve the momentum. Have
you learned stuff about vector addition yet? Momentum is a
vector so you _cannot_ conserve it without dealing with the
direction issue.
snip stuff on vectors, it is too advanced at the moment
Just as a reminder of where your theory is at the moment,
applying the equation above, either your form (the one with
the correct sign) or mine, with a value of "the mass of the
rest of the universe" taken from the Pioneer anomaly produces
a simple decay:
http://www.georgedishman.f2s.com/max/Mercury.png
That is to be expected of course because the math was never
required to incorporate a velocity related gravity anisotropy.
That graph uses _your_ math of course.
You later claim that momentum must be immediately conserved, but
that is clearly an impossibility. There is always a time delay
involved.
Sorry, a delay is impossible, the word "conserved" means
that the total is unchanging. If Pioneer loses some momentum
at some time and the Sun gets it after a delay, then for the
duation of the delay, the total is lower so the value isn't
conserved, it changes.
In a collision between two objects, momentum is redistributed
between them which can be considered local.
And a time delay is involved in the process if there is any
component separation at all, even within the colliding
components themselves. It takes time to redistribute momentum.
Nope, if that were true, it would not be conserved by
definition.
The effect of
gravity is not local. Right now Pioneer is being slowed by
the Sun and losing momentum. If the total is to be conserved
then that has to be matched by some other equal and opposite
change at the same time.
The Sun-Pioneer relationship is fairly constant, so even though
they are both being drawn toward each other in a way the appears
to immediately conserve momentum, the ten hour signal delay is
still present. Neither the Sun or Pioneer could react instantly
if the other suddenly ceased to exist.
That is true in GR but not in the Newtonian gravity
that you are modifying.
The gravitational effect on the Sun
does that even though the craft is more than ten light hours
away. If you abandon GR for Newton then you have what he
called "instantaneous action at a distance" and whether the
distance is ten light hours or ten light years makes no
difference. Your equation is based on -GM/r^2 which applies
instantaneously, there is no delay term in the equation.
That has always been a totally absurd statement. Of course
there's no delay term built into the equation. Why should there
be? Nature provides the delay, not mathematics.
Don't be stupid Max, you know that in order to work,
the maths must be a model of nature. If nature has a
delay, that must be reflected in the maths.
That statement
implies that math dictates how nature must behave.
No, it says the maths must be written to reflect nature.
You can't
simply gesture hypnotically and brush the obvious truth aside,
that a time delay in the transfer of momentum in the physical
world is very clearly a part of nature.
But this has nothing whatever to do with Newton anyway. The zero
origin universe has its own very specific rules, which most
certainly don't include instantaneous action at a distance. Light
speed is the absolute limit, **for a very good reason**. Such a
time delay is certainly expected in that universe.
Then change your maths to show that delay. What you
will find is something Newton knew, that it will
produce aberration of the gravitational force and
again cause the planets to spiral into the Sun. He
didn't like the instantaneous nature of forces but
he knew he had no choice if his maths was to work.
Energy can be almost immediately removed from interacting
charges and stored in space in the form of E/M radiation. But
gravity is equivalent to only a single wave that extends to
infinity. So there is obviously nowhere to store the energy
equivalent of Pioneer's momentum loss due to its motion
relative to the mass of the universe. Momentum is by no means
immediately conserved, but it is conserved in time.
The word "conserved" means it has the same value
AT ALL TIMES.
The fact that the Pioneer anomaly exists demands that the
question be resolved, one way or another, and a dedicated mission
seems to be the only way to do that. Such a mission can tell us
**much** about the universe, and is absolutely vital to the
progress of physics.
Afterall, truth is the ultimate goal, whatever the cost. So why
waste time and money chasing rainbows if that experiment has the
potential to turn physics upside down? It needs to be resolved
before we go searching for dark matter, surely?
Many people would like to see a mission perhaps as an adjunct
to an existing plan, but spending money to chase a gas leak
or whatever when major plans are being postponed or cancelled
to fund Bush's publicity stunts is unjustifiable.
Gas leaks or not, until the Pioneer anomaly has been properly
resolved, there is no point whatever searching for the pot of
dark matter at the end of the rainbow. Which is what it all boils
down to in the end, isn't it?
Dark matter does not explain the Pioneer anomaly, and
your idea doesn't explain either that or galactic
rotation curves. However, you skills in maths are a
long way short of being able to manipulate the equations
to the point where you can follow the derivations. Either
you start learning stuff like this or you will be stuck
with taking my word for it (or that of others). It is
basic arithmetic that you should have learnt at least in
your first years of senior school so I don't know how you
could have missed out:
http://www.mathsisfun.com/associativ...tributive.html
George