Re: Bending of light not well authenticated


"Dirk Van de moortel" <dirkvandemoortel@xxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote
in message news:CBdbe.73679$zJ3.4824160@xxxxxxxxxxxxxxxxxxxxxxxx
>
> I'll just copy the relevant equations from project D
> of Taylor and Wheeler's "Exploring Black Holes",
> http://www.eftaylor.com/general.html
>
> Schwarzschild metric
> Coordinates t, r,f (f=phi) - theta is taken constant
> b = impact parameter
> M = solar mass
> R = sun radius
> space = multiplication
>
> From the radial and angular components of motion of light:
> dr/dt = +/- (1-2 M/r) sqrt( 1-(1-2 M/r) b^2/r^2 )
> r df/dt = +/- b/r (1-2 M/r)
> on page 5-8 of chapter 5
> you get
> [1] (df/dt)^2 = b^2/r^4 (1-2 M/r)^2
> [2] (dr/dt)^2 = (1-2 M/r)^2 - (1-2 M/r)^3 b^2/r^2
>
> from [1] and [2]:
> [4] df = dr / ( r^2 sqrt( 1/b^2 - 1/r^2 (1-2 M/r) ) )
>
> with u = R/r this gives
> [5] df = -du / sqrt( R^2/b2 - u^2 + 2 M/R u^3 )
> and since r goes from R to infinity, u goes from 1 to 0.
>
> Eliminating the impact parameter (such that the light just
> grazes the sun) with
> [7] R^2/b^2 = 1-2 M/R
> gives

Chapter 5 is not available for download. However, this impact factor (b)
sounds like the actual distance of the photon to the center of the sun at is
closest approach to the sun. It is nothing more than an integration
constant which we have to determine by matching the boundary conditions
carefully. That is very tricky to apply in this case. If this is the case
which is very likely, we have

b^2 = R^2 / (1 - 2 M / R)

Which indicates a push away from the sun by the approximate amount of (G M
/ c^2). This is a deflection in the other direction which is not what is
observed. Thus, the correct 'impact equation' should be

R^2 = (R - dR)^2 / (1 - 2 M / (R - dR))

The deflected distance is just (solve for dR)

dR ~= G M / c^2, towards the sun (not away)

> [8] df = -(1-u^2)^(-1/2) du / sqrt( 1 - 2 M/R (1-u^3)/(1-u^2) )
>> Approximate for small M/R:
> [10] df ~= -du / sqrt(1-u^2)
> - M/R du / (1-u^2)^(3/2)
> + M/R u^3 du / (1-u^2)^(3/2)

[8] should then be very simply expressed as

[8xx] (du/df)^2 = 1 - u^2 + 2 M/R u^3

This is not an easy differential equation to solve because the range for u
is not from 0 to 1 as indicated by [12] below but 0 to (1 - dR/R). Although
(1
>> dR/R), the amount of deflection is in the range of dR/R. dR/R is not
negligible in [8] and [8xx]. Therefore, both [8] and [8xx] cannot be
approximated. Despite all that, [8xx] indicates a symmetry between the
inbound trip and the outbound trip of the photon. The overall deflection at
(M / r = 0) is that no deflection takes place at all. The closest
deflection is when the photon is deflected by (G M / c^2) at its closest
approach towards the sun (not away as [7] indicates) and then corrects its
deflection (un-deflect itself) during the outward bound.

This derivation has a very different result from Ciufolini/Wheeler which Mr.
Hobba is very proudly posted.

http://www.pupress.princeton.edu/sample_chapters/ciufolini/chapter3.pdf

However, [1] and [2] can be derived through the spacetime equation with (ds
= 0).

Thus, we have the spacetime equation for a photon described as the following
equations [A] and [B].

{A] 0 = c^2 (1 - 2 U) dt^2 - dr^2 / (1 - 2 U)

Where

** U = G M / c^2 / r

[B] ds^2 = 0 = c^2 dtau^2 - drho^2

Where

** dtau = local (or proper) time
** drho = local (or proper) displacement to the center of the sun

Applying the principle of minimal action to equation [A] with dt as the
parameter to be minimized, we get equations [1] and [2] with the appropriate
integration constant of R and not that mumbo-jumbo Ciufolini/Wheeler result.

Since the elapsed time represents the very minimal action of an event,
applying this principle to equation [B] by minimizing dtau, we have a local
(or proper) spacetime without any distortion. The boundary conditions for
the result of no distortion (or deflection) at U = 0 (infinite distance away
from the sun) in the local spacetime must be in the result of equation [A]
because [A] and [B] are derived through the same spacetime equation.
Therefore, the photon, after being deflected by an amount at its closest
approach to the sun, it straightens itself out as it leaves the sun. This
is a case of a perfect gravitational lens as it should be. Does that mean
GR is wrong because the prediction does not explain what is observed? In my
opinion, it is no because the observed data is heavily favoring the
erroneous predictions of GR (especially that 1919 Eddington expedition).
And that is not science but politics at work. The parameter that affect the
deflection of the photon is in the 1st order effect of the metric.

sqrt(1 - 2 U) ~= 1 - U - U^2 / 2

Where the 1st order is (-U) which any post Newtonian gravitational theories
besides GR must conform to (-U) to be compatible with Newtonian physics.
GPS relies on the 1st order effect. Thus, any theories should explain GPS
not just GR. The real battle for the test of all these post-Newtonian
gravitational theories lies in the prediction of Mercury's orbital anomaly
because it relies on the 2nd order effect. In GR's case, the 2nd order
effect is (-U^2 / 2) which does not give any anomaly at all with the
conservation of angular momentum also identified from Noether's Theorem.
So, does anyone want to discuss that?

> Integrate [10] with u between 1 and 0
> [12] f_total = Int{ u=1 to 0 ; df }
> = pi + Delta(f)
> where a tricky integration gives
> [13] Delta(f) = ... = 4 M/R




.

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