🚀 go-pugleaf

On Feb 6, 2:00 pm, Albertito <albertito1...@gmail.com> wrote:
> On 6 feb, 18:54, Randy Poe <poespam-t...@yahoo.com> wrote:
>
>
>
> > On Feb 6, 12:39 pm, Albertito <albertito1...@gmail.com> wrote:
>
> > > On 6 feb, 16:48, "Androcles" <Headmas...@Hogwarts.physics> wrote:
>
> > > > <srp2...@gmail.com> wrote in message
>
> > > >news:4695358e-db97-4485-94b4-06f417a451d4@i7g2000prf.googlegroups.com...
> > > > On 6 fév, 02:34, Pentcho Valev <pva...@yahoo.com> wrote:
>
> > > > >http://philipball.blogspot.com/2007/09/arthur-eddington-was-innocent-...
> > > > > "With the technology then available, measuring the bending of
> > > > > starlight was very challenging. And contrary to popular belief,
> > > > > Newtonian physics did not predict that light would remain undeflected
> > > > > - Einstein himself pointed out in 1911 that Newtonian gravity should
> > > > > cause some deviation too. So the matter was not that of an all-or-
> > > > > nothing shift in stars' positions, but hinged on the exact numbers."
>
> > > > | Right.
>
> > > > | Classical Newtonian mechanics predicts that light will bend with
> > > > | twice the angle observed.
>
> > > > How?
>
> > > I'll tell you how. In order to see how newtonian gravity can
> > > bend the trajectory of a photon, relativists fake it. They
> > > magically tranform a photon with energy E into a particle
> > > with mass m = E/2c^2, it is saying  half the mass in E = mc^2.
> > > Then, relativists assume that particle passes by the massive
> > > body from infinity travelling locally at c, and then they can
> > > apply newtonian gravity to see how its trajectory is deflected
> > > into a hyperbolic one. As a result there is a deflection angle
> > > twice the observed one. A particle with mass m = E/c^2,
> > > travelling locally at c from infinity, would be deflected in
> > > the correct angle, under newtonian gravity.
>
> > Why don't you do the Newtonian calculation before
> > making this claim?
>
> >           - Randy
>
> The calculations have been already done by you relativists,http://www.theory.caltech.edu/people/patricia/obsertop.html

Correct.

> I'm only witness of that, and I see that if we double the mass
> assumed for a photon we attain the correct answer.

There are a great many assumptions in those
equations which you are not aware of. You are trying
to change the meaning of the symbols midstream.

Here's what those orbital mechanics equations look like for
planetary orbits:
http://www.go.ednet.ns.ca/~larry/orbits/kepler.html
Note that E is the sum of kinetic energy and potential
energy. So what are you going to use for this E? Do
you really want to use mc^2 and claim it's a Newtonian
model?

If you were doing Newtonian gravitation the way Newton
did it to calculate the planetary orbits, you'd say
the kinetic energy is (1/2)*mc^2, and the potential
energy is -GMm/r. Then you'd find out that the mass m
drops out, that is the parameter E/m = [c^2/2 - GM/r]
does not depend on m.

Let's start from the basics: What is the initial velocity,
what is the acceleration, what is the result?

Start out with a velocity c, mass m, distance R
as it passes near mass M.

The acceleration experienced by the mass m is
GM/R^2. That is what you would apply to the velocity
vector to get the change in velocity. It does not
depend on m.

Trajectories of small bodies through solar systems do
not depend on mass.

            - Randy