Re: Bending of light not well authenticated


"Koobee Wublee" <kublai@xxxxxxx> wrote in message news:2D_be.1197$Do2.105@xxxxxxxxxxxxx
>
> "Dan Winters" <dwinters@xxxxxxxxxx> wrote in message
> news:42705e70.535227546@xxxxxxxxxxxxxxx
> > On Mon, 25 Apr 2005 John C. Polasek <jpolasek@xxxxxxxxxx> wrote:
> >>I have not been able to find a really satisfactory derivation of
> >>general relativity's 1.75" bending of the sun's rays...does anyone
> >>know of a downloadable document that uses actual mathematics
> >>to prove the 1.75" deviation.
> >
> >
> > A quick search on google turns up this page, which includes a complete
> > derivation of the relativistic deflection of light:
> >
> > http://www.mathpages.com/rr/s6-03/6-03.htm
>
> At the perihelion, if (r = r0), then there is no deflection. To have a
> deflection, it should be that (r = r0 - dr) where (dr = deflected amount at
> perihelion). Granted
>
> r0 >> dr
>
> However, the angle of deflection is so small that including (r = r0 - dr)
> into the equating, it will null out the deflected angle at r = infinity.
> That means in the inward bound, the photon will be deflected closer to the
> sun, and at its perihelion, the deflected distance is (dr). However, as the
> photon leaves the sun, it will undo its deflection. It will end up with no
> deflection at all at (r = infinity).
>
> Since the derivation that leads to the Christoffel symbols is nothing but
> applying the principle of the least action. A simple Euler-Lagrange method
> with a few substitutions should leads to equations (2). So far, we are
> looking at how the photon is behaving in the observed spacetime whose
> curvature is influenced by the sun. The article also mentions (dtau = 0)
> which means we can also write an equation to describe the path of the photon
> in proper space using proper time. That is
>
> c^2 dtau^2 = drho^2 - rho^2 dtheta^2
>
> Applying the principle of least action (Euler-Lagange or Christoffel
> symbols), we have no distortion to the photon's travel. The points of
> interest are when the photon just stars out its journey and when the photon
> ends its journey. Since in proper spacetime there is no deflection, in
> distorted observed spacetime there must not be any deflection at all at (r =
> infinity). This agrees with the condition above where the perihelion occurs
> at (r = r0 - dr).
>
> GR does not predict a deflection of a photon near a gravitation mass in the
> end result. However, in between, the severity of the deflection is not a
> function of the deflected angle but its deflected position.

Gasp.
"An Aerospace Engineer's view on General Relativity":
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/AerospaceRelativity.html
Nice but futile trolling attempt.

Dirk Vdm


.

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