Re: Bending of light not well authenticated

Gene McGraw wrote:
> On 22 May 2005 "Ken S. Tucker" wrote:
> >IMO there's a problem. You should get b=R-2*M.
>
> It isn't a matter of opinion.

In this limited forum, communication is challenging,
so misinterpretation is possible.

> At the perihelion r = R we have dr/dt =
> 0. Insert this into the equation of motion

> (dr/dt)^2 / c^2 = (1 - 2 U)^2 (1 - (1 - 2 U) b^2 / r^2)

Does that come from Schwarz...?
If so give a hint how.

> with r = R, and solve for b. The result is b = R / sqrt(1 - 2U). This
> is just simple grade-school algebra.

Simple for you maybe:-).

> >That's from ds=0, C(radial)=c*g_00 = c*(1-2U).
>
> Non-sensical. The fact that the paths of light have ds = 0 has
already
> been accounted for in the derivation of the equations of motion. The
> radial speed of light (which is what I assume you mean by C(radial))

yes

> is irrelevant at the perihelion because the light is not moving
> radially. That's the meaning of dr/dt = 0 ar r = R.

There's nothing inherently wrong with a simplified
approach as long as you understand it, and IMO I
think your approach is unique and interesting.

Consider when M=0 then the spacetime field is Cartesian
and the Cartesian Radius R^2 = X^2 + Y^2 + Z^2 holds,
and there is a *direction* where dR/dt=0 at the closest
approach to the M=0 point. That direction from "M" could
be aligned with a distant star.
When M>0 the deflection in that aligned *direction* is
now

dr/dt= -2M/R

where "r" is in terms of a new CS that *contracts* the
Cartesian "R", thus defining the meaning of dr/dt.
From Schwarzschild we'll find ds=0 yields the coordinate
radial velocity of light to be,

dr/dt = g_00 = 1-2U ,c=1 and of course dR/dt = 1,

with the diff being -2U = -2M/R along the trajectory,
IOW's,

dr/dt = dR/dt - 2M/R .

Now at the perihelion, (the alignment I used above),
dR=0, and the above integrates, (using R=ct=t),

r/t = R/t-2M/R => r=R-2M .

At this point you can sub the so-called impact parameter
"b" to get,

b=R-2M,

which is a constant from the integration above, because
dR=0.

That's a fast and cheap way to go from the Schwarz...to
deflection 2M/R and integrated contraction r=R-2M.
Regards
Ken S. Tucker
kxsxt

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