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From: Albertito on 16 Sep 2009 06:21
On Sep 16, 5:22 am, eric gisse <jowr.pi.nos...(a)gmail.com> wrote:
> Albertito wrote:
>
> [...]
>
> > div(\psi) = \phi
>
> > Does \psi have any physical meaning?
> > What name should we give to \psi?
>
> > Thanks in advance
>
> For someone who has been posting so much about Newton and Maxwell's
> equations, the obvious seems to have eluded you.

And the obvious is ...?

If gravitational potential is

\phi= -GM/|r|

where |r| is the norm of the radial vector
distance r, and GM is gravitational parameter

then \psi should be

\psi = -GM ln(r/r_0)

where r_0 is a scalar, constant of integration

therefore, \psi has dimensions of a gravitational parameter
mesured in km^3 s^(-2). But, when we look inside the
natural logarithm we see the vector distance r in it.
A vector r divided by scalar r_0 is still a vector.
How is the natural logarithm of a vector defined?
if the components of that vector r are (r_1, r_2, r_3),
is it another vector as
ln(r/r_0) = [ln(r_1/r_0), ln(r_2/r_0), ln(r_3/r_0)] ?


The interesting issue arise when you solve for r/r_0
in the above \psi equation,

r/r_0 = Exp[-\psi/GM]

Wow, an hyperbolic function ready to be attached to anything!
:-D

Actually, that's all gravitation is about. If you add a small
perturbation to the exponent, -\psi/GM, you attain a very accurate
model of the behavior of gravitation. That perturbation actually is
a ratio between two speeds of propagations, multiplied by a factor
of 2,

2c/c_g,

where c is speed of light and c_g is
speed of gravity

So, the radial distance r od an orbiting
test body in a 2-body problem actually is

r/r_0 = Exp[-\psi/GM + 2c/c_g]

But, from the dispersion relation

v = c^2/c_g

where v is radial velocity of the test body, we have

r/r_0 = Exp[-\psi/GM + 2v/c]

Then, the gravitomagnetic field, B_g, multplied by v in
a cross product, arises as the gradient of the divergence
of that small perturbation, as this


v x B_g = grad(div(2v/c))


Of course, 2v/c is a vector, because c is a scalar and v
is a radial velocity.




From: funkenstein on 16 Sep 2009 08:23
On Sep 16, 12:21 pm, Albertito <albertito1...(a)gmail.com> wrote:
> On Sep 16, 5:22 am, eric gisse <jowr.pi.nos...(a)gmail.com> wrote:
>
> > Albertito wrote:
>
> > [...]
>
> > > div(\psi) =  \phi
>
> > > Does \psi have any physical meaning?
> > > What name should we give to \psi?
>
> > > Thanks in advance

I'm not sure I follow your motivation. When in physics do we consider
that a potential is the divergence of some other function?

>
>
> And the obvious is ...?
>
> If gravitational potential is
>
>         \phi= -GM/|r|
>
>         where |r| is the norm of the radial vector
>         distance r, and GM is gravitational parameter
>
> then \psi should be
>
>         \psi = -GM ln(r/r_0)
>
>         where r_0 is a scalar, constant of integration
>


You are looking for a solution to
div \psi = -GM/|r|

the function you suggest doesn't solve that differential equation.

if you assume \psi is only a function of |r|, (i.e. in spherical
coordinates no a function of the two angles) then you can rewrite your
differential equation as:

d/dr (r^2 * \psi) = GMr

A solution doesn't jump out at me, what is my motivation to work more
on it? :)

Cheers

From: eric gisse on 16 Sep 2009 08:50
Albertito wrote:

> On Sep 16, 5:22 am, eric gisse <jowr.pi.nos...(a)gmail.com> wrote:
>> Albertito wrote:
>>
>> [...]
>>
>> > div(\psi) = \phi
>>
>> > Does \psi have any physical meaning?
>> > What name should we give to \psi?
>>
>> > Thanks in advance
>>
>> For someone who has been posting so much about Newton and Maxwell's
>> equations, the obvious seems to have eluded you.
>
> And the obvious is ...?

I'm sorry, I meant "obvious for people who have actually studied the
subject".

>
> If gravitational potential is
>
> \phi= -GM/|r|
>
> where |r| is the norm of the radial vector
> distance r, and GM is gravitational parameter
>
> then \psi should be
>
> \psi = -GM ln(r/r_0)
>
> where r_0 is a scalar, constant of integration

All aboard the field theory failboat!

Let's see where you pilot this trainwreck.

>
> therefore, \psi has dimensions of a gravitational parameter
> mesured in km^3 s^(-2). But, when we look inside the
> natural logarithm we see the vector distance r in it.

r isn't a vector. Christ almighty.

> A vector r divided by scalar r_0 is still a vector.
> How is the natural logarithm of a vector defined?

It isn't.

> if the components of that vector r are (r_1, r_2, r_3),
> is it another vector as
> ln(r/r_0) = [ln(r_1/r_0), ln(r_2/r_0), ln(r_3/r_0)] ?

Normally when a student does such a thing, the teacher patiently explains
why the student is incorrect and then refers he/she/it to the relevant
textbook or explains a little bit/a lot on the board.

Thank god its' easy for me to suppress that instinct.

>
>
> The interesting issue arise when you solve for r/r_0
> in the above \psi equation,
>
> r/r_0 = Exp[-\psi/GM]
>
> Wow, an hyperbolic function ready to be attached to anything!
> :-D
>
> Actually, that's all gravitation is about. If you add a small
> perturbation to the exponent, -\psi/GM, you attain a very accurate
> model of the behavior of gravitation.

You say this even though you have no way of quantifying "very accurate".

> That perturbation actually is
> a ratio between two speeds of propagations, multiplied by a factor
> of 2,
>
> 2c/c_g,
>
> where c is speed of light and c_g is
> speed of gravity

Newtonian gravitation has an infinite speed of light. Its' like you are
allergic to learning.

>
> So, the radial distance r od an orbiting
> test body in a 2-body problem actually is
>
> r/r_0 = Exp[-\psi/GM + 2c/c_g]

Sure. Why not? I could explain why its' wrong and stupid but I don't see the
point.

>
> But, from the dispersion relation
>
> v = c^2/c_g

Nice to see you continue to call your idiocy a 'dispersion relation' even
though you have no idea what a dispersion relation is.

>
> where v is radial velocity of the test body, we have
>
> r/r_0 = Exp[-\psi/GM + 2v/c]
>
> Then, the gravitomagnetic field, B_g, multplied by v in
> a cross product, arises as the gradient of the divergence
> of that small perturbation, as this
>
>
> v x B_g = grad(div(2v/c))
>
>
> Of course, 2v/c is a vector, because c is a scalar and v
> is a radial velocity.

Newton doesn't have gravitomagnetism, and your only reference for the
subject is a poorly written Wikipedia article that attempts to explain a
subtle concept from general relativity.

You can (and will) keep prattling about your inane bullshit but I am rather
confident that you and your ideas will remain irrelevant until entropy
consumes the stars.

From: eric gisse on 16 Sep 2009 08:53
funkenstein wrote:
[...]

> You are looking for a solution to
> div \psi = -GM/|r|
>
> the function you suggest doesn't solve that differential equation.

Shhhh...

[...]
From: Albertito on 18 Sep 2009 07:11
On Sep 16, 1:50 pm, eric gisse <jowr.pi.nos...(a)gmail.com> wrote:
> Albertito wrote:

[snip already read text]

> Newtonian gravitation has an infinite speed of light. Its' like you are
> allergic to learning.

You're partly wrong. What is infinite in Newtonian
gravitation is NOT the speed of light, c, but the
speed of gravity, c_g. So in the equation I proposed,

r/r_0 = Exp[-\psi/GM + 2c/c_g]

if c_g -> oo, then that limit yields

r/r_0 = Exp[-\psi/GM]

which is the Newtonian limit. You can attain the
Newtonian gravitational potential \phi, as

\psi = -GM ln(r/r_0),

\phi = div(\psi) = - GM/|r|

In the generic case, you can also attain a
gravitational potential \phi, as

\psi = -GM [ln(r/r_0) - 2c/c_g],

\phi = div(\psi) = - GM/|r| + 2GM div(c/c_g),

IOW, since the radial velocity, v, of the test body is
a vector, we have

v/c = c/c_g

is also a vector in the direction of v. So, the above
equation for \phi yields

\phi = div(\psi) = - GM/|r| + 2GM div(v/c),

And the force of gravitation, for the test body of mass m,
is deduced as the gradient of \phi, as

F = GMm r/|r|^3 + 2GMm grad[div(v/c)],

Therefore, the gravitomagnetic part obviously is

B_g = 2GM grad[div(v/c)].


I said that you are partly wrong, because what in Newtonian gravity
is regarded as infinite is not c, but the speed of gravity, c_g.
Through the dispersion relation, we have v/c = c/c_g, thus you are
partly right too, as far as the radial velocity of the test body
remains meaningfully less than c. So, in the limit c -> oo, we get
a null gravitomagnetic field, B_g = 0, which is the Newtonian limit.


Have you found this article interesting? yes? No? Reason the answer
if you are smart, or leave it in blank if you are stupid.
My ideas and derivations aren't yet in textbooks, but I'm
pretty sure you can appreciate my genialities :-)


>
>
>
> > So, the radial distance r od an orbiting
> > test body in a 2-body problem actually is
>
> > r/r_0 = Exp[-\psi/GM + 2c/c_g]
>
> Sure. Why not? I could explain why its' wrong and stupid but I don't see the
> point.
>
>
>
> > But, from the dispersion relation
>
> > v = c^2/c_g
>
> Nice to see you continue to call your idiocy a 'dispersion relation' even
> though you have no idea what a dispersion relation is.
>
>
>
> > where v is radial velocity of the test body, we have
>
> > r/r_0 = Exp[-\psi/GM + 2v/c]
>
> > Then, the gravitomagnetic field, B_g, multplied by v in
> > a cross product, arises as the gradient of the divergence
> > of that small perturbation, as this
>
> > v x B_g = grad(div(2v/c))
>
> > Of course, 2v/c is a vector, because c is a scalar and v
> > is a radial velocity.
>
> Newton doesn't have gravitomagnetism, and your only reference for the
> subject is a poorly written Wikipedia article that attempts to explain a
> subtle concept from general relativity.

I think I have showed above the reason why there is not
gravitomagnetism
in Newtonian gravitation.
>
> You can (and will) keep prattling about your inane bullshit but I am rather
> confident that you and your ideas will remain irrelevant until entropy
> consumes the stars.

Don't be so confident. Maybe, my ideas aren't so irrelevant after all.

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