Finite Relativism & Special Relativity Disproof

Phil:

Incredibly easy. This is just three successive integrations, of a linearly
independent quadratic, a trig function, and a constant. First year calculus
stuff.

The inner definite integral is:

Integral [p=0, p=r1] p^2/r^2 dp

This is just an integral in the form k * x^2, its solution is:

(1/3)(r1^3)/r^2 - (1/3)(0^3)/r^2

= (r1)^3/3r^2

The next level out of the integral is:

Integral [theta = 0, theta = pi] sin(theta) * (r1)^3/3r^2 d(theta)
= (r1)^3/3r^2 Integral[theta = 0, theta = pi] sin(theta) d(theta)

= (r1)^3/3r^2 (-1* cos(0) - -1*cos(pi))

= -(2/3r^2) (r1)^3

The next level out of the definite integral is:

Integral [phi =0, phi =2 * pi] (-(2/3r^2) (r1)^3) d(phi)
= -(2/3r^2) (r1)^3 * Integral [phi =0, phi =2 * pi] d(phi)
= -(2/3r^2) (r1)^3 * 2*pi
= - 4*pi/3r^2 * (r1)^3

You don't actually say what this is supposed to be - what "f" is supposed to
represent, calling it the "respective factor" doesn't help.

If it is a force, as you imply, then it increases with the cube of the
distance, and not the inverse square, which we know to be an excellent
approximation. If this is the what your formula is supposed to represent,
then it is wrong.

More generally, if you can't solve simple freshman Calculus 101 integrals,
then you don't have anywhere near the mathematical machinery to understand
GR, let alone propose a replacement.

You really need to study basic mathematics and then tackle problems
appropriate to your knowledge and skill sets. GR is currently completely
beyond you, as is any attempt to replace it.

Should you wish to learn some basic mathematics, I am happy to point you at
resources appropriate to your knowledge, and which are appropriate to your
interests (eg vector calculus, functions of a complex variable, and
eventually transformation groups, tensors etc).

I will not provide any further help on FR until you have learned these
basics, as it is a waste of time.

Peter Webb


"Phil Bouchard" wrote in message
...
Peter Webb wrote:

Well, he has produced zero so far.

Hopefully he will post his equations and their derivations. If he is
having problems solving his triple integrals, I am sure somebody here can
help.

It obvious I have the solution if I rendered a graph.

We might be able to:

* Provide a full, analytic solution using techniques unknown to Phil
(nobody know every single trick for solving integrals)

* Provide a full, analytic solution for some special cases, such as
highly symmetric configurations of masses (eg spherically symmetry as
Schwartzschild did for GR, or 2D solutions only)

Even if we can't do this, we/I can definitely provide a numeric solution
for specific test configurations that have already been experimentally
tested.

Its then simply a matter of looking at a few experimental results, and
seeing how well FR agrees with what is actually observed, and we can see
if FR is complete crap or if Bouchard is a genius.

I await his formulas, and an opportunity to practice my calculus skills.

The following integration should be very easy:
http://www.fornux.com/personal/phili.../fr-test-1.pdf

But I do have an interesting question that follows.

doug wrote:

More ignorance from phil. You really have no idea of how relativity
was developed at all.

Publishing unreferenced blunders from unknown students isn't acceptable.

Pointing out your mistakes is helping you. I have no idea why
you consider it intimidationn. Maybe you want to go cry to your
mother about it.

You're not intimidating me, but other physicists.

With the corrections to the Schwartzchild radius, my measurement unit is
now kg^2/m^4. This makes the radius more important.

Like I said, it is undoubtedly flawless now. The equivalence principle
is very misleading but is in fact a misunderstanding from Einstein.

You have not understood what was said in the article. It is clear you
do not understand statistics either.

Well I wrote my own calculator 5 years ago featuring non-linear
regressions you cannot find on your handheld calculator.

Peter Webb wrote:
Phil:

Incredibly easy. This is just three successive integrations, of a
linearly independent quadratic, a trig function, and a constant. First
year calculus stuff.

Thanks for your help Peter, but this is a wrong assumption that may
needed to be mentioned. You cannot split the integral into 3 separate
ones. It is literally a triple integral.

[...]

You don't actually say what this is supposed to be - what "f" is
supposed to represent, calling it the "respective factor" doesn't help.

Well it is a function and should be called f(x2, y2, z2).

If it is a force, as you imply, then it increases with the cube of the
distance, and not the inverse square, which we know to be an excellent
approximation. If this is the what your formula is supposed to
represent, then it is wrong.

No it is really the inverse square law.

[...]

Phil Bouchard wrote:
doug wrote:


More ignorance from phil. You really have no idea of how relativity
was developed at all.


Publishing unreferenced blunders from unknown students isn't acceptable.

You are the one making the blunders and you are trying to publish
them. I know you do not believe me but you should at least read
something about relativity so you will not look so stupid.

Pointing out your mistakes is helping you. I have no idea why
you consider it intimidationn. Maybe you want to go cry to your
mother about it.


You're not intimidating me, but other physicists.

I am trying to help you but you are too stupid to want help.

With the corrections to the Schwartzchild radius, my measurement unit is
now kg^2/m^4. This makes the radius more important.

Like I said, it is undoubtedly flawless now. The equivalence principle
is very misleading but is in fact a misunderstanding from Einstein.

You have not understood what was said in the article. It is clear you
do not understand statistics either.


Well I wrote my own calculator 5 years ago featuring non-linear
regressions you cannot find on your handheld calculator.

Wonderful. So why do you not understand the article you posted?
Have you figured out why your inside the sphere calculation is
totally wrong?

Phil Bouchard wrote:
Peter Webb wrote:

Phil:

Incredibly easy. This is just three successive integrations, of a
linearly independent quadratic, a trig function, and a constant. First
year calculus stuff.


Thanks for your help Peter, but this is a wrong assumption that may
needed to be mentioned. You cannot split the integral into 3 separate
ones. It is literally a triple integral.

Not for someone who knows math. It is a simple single integral.
But, you have even set it up wrong.

[...]

You don't actually say what this is supposed to be - what "f" is
supposed to represent, calling it the "respective factor" doesn't help.


Well it is a function and should be called f(x2, y2, z2).

That is a meaningless statement. But we are not surprised.

If it is a force, as you imply, then it increases with the cube of the
distance, and not the inverse square, which we know to be an excellent
approximation. If this is the what your formula is supposed to
represent, then it is wrong.


No it is really the inverse square law.

Well, then we know for absolute sure you are wrong. Why do you refuse
to look at any textbox to see how easy the answer is? Do you enjoy
looking stupid?


[...]

"Phil Bouchard" wrote in message
...
Peter Webb wrote:
Phil:

Incredibly easy. This is just three successive integrations, of a
linearly independent quadratic, a trig function, and a constant. First
year calculus stuff.

Thanks for your help Peter, but this is a wrong assumption that may needed
to be mentioned. You cannot split the integral into 3 separate ones. It
is literally a triple integral.


I solved it for you.

If you think it is more complicated than the very simple solution I gave
you, point out the error in my working.


[...]

You don't actually say what this is supposed to be - what "f" is supposed
to represent, calling it the "respective factor" doesn't help.

Well it is a function and should be called f(x2, y2, z2).


Well, calling it a function doesn't help. That means little more than it is
an equation.

You need to specify a physical interpretation of "f " if this is to have any
physical meaning. Usually f is force, but that is just a guess. What is "f"
supposed to be, yes I know its a function, but what does it represent.

BTW, in your definition of f , x2, y2, and z2 don't actually appear, so it
aint a function of them. In your definition, f is a function of r1.


If it is a force, as you imply, then it increases with the cube of the
distance, and not the inverse square, which we know to be an excellent
approximation. If this is the what your formula is supposed to represent,
then it is wrong.

No it is really the inverse square law.

Now that I have solved your integral, are you going to put the solution into
the book? Do I get credit in the references?

"Sam Wormley" wrote in message
news:PSNGl.66942$DP1.37515@attbi_s22...
Phil Bouchard wrote:
You cannot split the integral into 3 separate ones.

http://en.wikipedia.org/wiki/Integral
http://mathworld.wolfram.com/topics/...dAnalysis.html

Of course you can, even if you don't know how. I did it for you. Yours is an
incredibly simple integral, the sort of thing that is used as the first
example of multiple integrals in a Calculus 101 course. If you have any
questions about my solution, just ask.

I note that this is the first equation in the first chapter of the book. I
assume it is supposed to be important. As it has an incredibly simple
solution, this is obviously also important.

I am very happy for you to provide both the analytic solution and my
derivation of it in your book, it should be right under the derivation of
the integral; my closed form solution is far easier to use than the original
integral, but exactly equal to it.

This would be a huge improvement for your book, your key equation (a triple
integral) turns out to have an incredibly simple analytic solution. A
significant breakthrough, I would have thought.

doug wrote:

You are the one making the blunders and you are trying to publish
them. I know you do not believe me but you should at least read
something about relativity so you will not look so stupid.

Are you saying "Relativity - The special and general theory" by Albert
Einstein is wrong?

I am trying to help you but you are too stupid to want help.

Not following what I am told to do helps me too.

Wonderful. So why do you not understand the article you posted?
Have you figured out why your inside the sphere calculation is
totally wrong?

Because the innermost integral should be p^2/r^4 and not p^2/r^2. But
this is irrelevant because I have proved the single and therefore the
simplest version can handle it by juxtaposing the plots.

doug wrote:

[...]

Well, then we know for absolute sure you are wrong. Why do you refuse
to look at any textbox to see how easy the answer is? Do you enjoy
looking stupid?

Well the point of having 2 versions is to confirm the first was right.
The second version turns out to be a single integral also.

BTW that Schwartzchild radius really was the fudge factor of GR.

Sam Wormley wrote:
Phil Bouchard wrote:
You cannot split the integral into 3 separate ones.

http://en.wikipedia.org/wiki/Integral
http://mathworld.wolfram.com/topics/...dAnalysis.html

http://books.google.com/books?id=knn...plit#PPA325,M1