Re: Pioneer Anomoly


"John C. Polasek" <jpolasek@xxxxxxxxxx> wrote in message
news:l4gbb19gq4jg7rvp9uk4bi458jrfsb0vua@xxxxxxxxxx
> On Sun, 19 Jun 2005 18:54:29 +0100, "George Dishman"
> <george@xxxxxxxxxxxxxxxxx> wrote:
>
>>
>>"John C. Polasek" <jpolasek@xxxxxxxxxx> wrote in message
>>news:fhvab11tui0d443qm3p2lbnmqvelqbasae@xxxxxxxxxx
>>> On Sun, 19 Jun 2005 09:10:15 +0100, "George Dishman"
>>> <george@xxxxxxxxxxxxxxxxx> wrote:
>>>
>>>>
>>>>"John C. Polasek" <jpolasek@xxxxxxxxxx> wrote in message
>>>>news:8dd8b1dt8rrcade7ph486esrjivbrq29ih@xxxxxxxxxx
>>>>> On Sat, 18 Jun 2005 15:29:28 +0100, "George Dishman"
>>>>> <george@xxxxxxxxxxxxxxxxx> wrote:
>>>>...
>>>>>>> On Sat, 18 Jun 2005 14:08:02 +0100, "George Dishman"
>>>>>>> <george@xxxxxxxxxxxxxxxxx> wrote:
>>>>...
>>>>>>>>OK, I have it now. There are some obvious problems
>>>>>>>>without getting into the detail yet. The first you
>>>>>>>>point out yourself: the effect of a large distant
>>>>>>>>mass would apply to all the bodies in the solar
>>>>>>>>system hence we would not detect anything. In fact
>>>>>>>>this is currently happening both because the Milky
>>>>>>>>Way galaxy is bound in the Local Group. You might
>>>>>>>>also find it interesting to look up the "Great
>>>>>>>>Attractor":
>>> Any point in our universe of particles is located at x y z cT. There
>>> is image in Dual Space of antiparticles located at x y z 0. (Espace
>>> is the dual to our vaccum). The range I used for the Newton
>>> acceleration is R = cT. It is not a distance xyz w/re to the Sun e.g.
>>
>>That comment seems entirely unrelated to what
>>you quote above.
> Some perspective. Forget the xyzicT of Einstein.

x, y, z and t are all you can measure. Whatever your
theory, you have to relate them back to those values.

You are still not saying anything relevant to the
comment above. Your suggestion of a remote black
hole or other large mass cannot explain A_p for
the reason you correctly stated yourself.

<snip stuff>
> There's no use trying to encapsulate a whole new theory in a half
> dozen paragraphs.

Then stop trying to do so and respond to the
objection I raised instead.

>>>> All of relativity is in Fig. 1
>>> paper#2
>>
>>You have used normal trig instead of hyperbolic
>>so the craft clock appears to run fast instead
>>of slow but it is analogous,
>
> In DS time-arrow diagram Fig.1 #2,

Above you said fig 1 was relativity, not DS. That
conflict is what I asked you to clarify.

> the horizontal line is along the
> 4th dimension time axis and every xyz velocity in Uspace is drawn
> normal to the time line.

Since velocity is the rate of change of the spatial
vlalues, those should be distances, not velocities
or the slope of the line becomes acceleration.

> Our velocity is at c on the 4th dimension
> ever since our launching at creation.
>
> DS does not use time dilation. It's all reduction in c. Look again at
> Fig. 1b. The relative velocity V of the craft causes its time arrow to
> be rotated by angle a. The craft's time arrow remains the same length
> because it's totally equivalent, just a different direction in
> space-time:

Relativity also rotates the arrow and keeps it the
same length.

> Craft's clock on the hypotenuse will read slow in our frame being
> reduced by cosine a or Lorentz.
> Similarly for force F in 1b, which is intended to push the craft,
> has a diminishing effect Feff = F cosine a, just like Lorentz. This
> differs with relativity by avoiding the use of the odious gamma to
> "jack up" m so as to resist the force. This avoids all the
> relativistic mass debates.

"Relativistic mass" was a poor attempt to shoehorn
relativity into Newtonian equations that causes a
lot of misunderstanding. Mass is really invariant
in relativity.

> The little 4th side delta c in Fig. 1c divided by c is z, the
> redshift.
>
>>I know what you
>>mean. I guess you did the same later too.
> No, I did not waver. This same diagram solves general relativity
> problems too. We need a velocity V to represent strength of gravity
> and it is naturally the terminal velocity or escape velocity from
> infinity to that point.
> Again,cx/c is the actual slowdown ratio of a clock in the gravity
> well, and we can just as well use the 4th side, delta c/c to compute
> any deficits. delta c = 0.5(V/c)^2
>
> This is not just geometry. No, a mass at infinity gets "free energy"
> by accumulating velocity of 11,180m/s at earth. But this energy comes
> from a decrement in c of 0.208m/s.
>
> The immediate application is seen in the 4 corrections in the GPS
> system.
> This is clearly and numerically illustrated in Fig. 2 and Cols. 3 & 4
> in table 1 in #2 paper. Each clock correction comes off as dc/c x
> 86,400 sec/day. The results are plain and clear, contrasted with the
> relativity version in Eq. 6 & 7.
>
>>
>>> with total energy never deviating from 1/2mc^2.
>>> A hyperbolic 4space is an abomination as you can see from Fig. 4 #2,
>>
>>The energy seems to be in fig 3 in document 2,
>>not fig 4. Even then it is not a style I've
>>seen before. Where did you get the diagram?
>
> Yes fig.3. The diagram is a natural embodiment of relativitys total
> energy equation 10, TE being the sum of the squares of (already
> quadratic) terms. As you can see, in Fig. 3b, we have cx on the
> horizontal and real V normal to it. DS uses that same philosophy,
> except that in ours, horizontal c is our speed thru the Espace and in
> relativity its undefined I guess.

In relativity, we plot the mometum on the
horizontal and energy on the vertical as shown
below. Mass is the hypotenuse and the length is
unchanged by rotation though it looks as though
it changes because the Euclidean geometry of
our monitors.

>>This is a simple version I did some time ago
>>but it needs more annotation:
>>
>> http://www.georgedishman.f2s.com/relativity/momenergy.gif
>>
>>The mass is an invariant scalar and as usual
>>the rotation in spacetime known as velocity
>>produces the increase we call kinetic energy.
>>
>>As for your figure 4, the one on the left is
>>the relativistic version (if you used the
>>hyperbolic trig as above) with 'v' being the
>>velocity and alpha being the corresponding
>>rapidity. The one on the right is Newtonian
>>where R1 and R2 are the velocities. Alpha in
>>that diagram has no equivalent in Newtonian
>>theory. Did you get the labels the wrong way
>>round?
>
> Not at all. Look again. The effect of a real velocity V is to rotate
> the original time-arrow c as shown in fig. 4a so that after each
> rotation its value remains c.

That's also true of SR.


> The right hand one is relativistic, c
> becoming a larger gamma c.

No, c is a fundamental constant that does not
change in relativity and the axis rotates as
you show in 4a except that it rotates the
other way from what you have shown.

>>> the relativity total energy with its abominable gamma mc^2, resulting
>>> from hyperbolic 4space.
>>
>><snip>
>>>>Then unless you have a way to turn that
>>>>general result into a direction and show
>>>>that it will be within 1.5 degrees of a
>>>>line towards the Sun, you don't have an
>>>>explanation, sorry.
>>>>
>>>>George
>>> It's new physics, there's 12 chapters ahead of chapter 13, and you
>>> seem to insist on my solving it with what we already know, and so far,
>>> it can't be done.
>>
>>I'm not insisting on anything, you said you
>>had done it without any prompting from me:
>
> There's far too much to cover in order to prove my P10 explanation. I
> again got trapped into trying to teletype it all to you in a few
> breathless paragraphs, but it's plainly not working.

What is not working is that you keep trying
to tell me how you worked out your theory
instead of how you applied it to Pioneer.
Look back at what you posted, there isn't
a word about the spacecraft in it. Just
stick to the point and you might get
somewhere.

> New physics you need, new physics you haven't got, new physics you
> have to study.
> You seem to have objected to enough stuff in paper #2 to merit a
> re-reading based on my responses.

Your responses show only some significant
misapprehensions about relativity. You
might try breaking out the textbooks, I have
only given hints of where the errors lie so
you'll need to work out the details yourself.

George


.

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