Re: Bending of light not well authenticated

On Sat, 21 May 2005 "Koobee Wublee" wrote:
> I get the following two equations in accordance with most derivations
> (r dH/dt)^2 / c^2 = (1 - 2 U)^2 b^2 / r^2
> (dr/dt)^2 / c^2 = (1 - 2 U)^2 (1 - (1 - 2 U) b^2 / r^2)
>Where H = angle of the photon where at perihelion H = 0
> U = G M / c^2 / r, b = integration constant

Those are indeed the equations of motion of a pulse of light (making
allowances for your ambiguous notation of two divisions in a row).
Hopefully you understand that those two equations are derived from the
metric by the variational technique (corresponding to Fermat's
Principle of Least Time), and they represent the set of light-like
extremal paths. So the hard work is already finished by this point,
and all that's left is to examine the path with the appropriate values
of M and b to see how much deflection there is.

If you really want to derive the deflection of light for general
relativity, from scratch, you have to start with the field equations,
derive the metric, and solve for the light-like paths to give the
equations of motion that you cited above. From this point (which you
seem to be taking as your starting point) it's fairly trivial to
determine the deflection.

> I have (b = R) where R is the perihelion of the photon to the sun
> with mass of the sun = 0.

The mass of the sun is not zero, it is M, and the value of the
constant of integration b is not exactly equal to M, as can be seen
immediately from your second equation of motion above. Remember, the
perihelion distance R is, by definition, the distance from the path to
the sun at the point on the path closest to the sun, and at this
minimum point the derivative dr/dt is zero. Plug this into your second
equation of motion above (with r = R to signify the perihelion
distance) and solve for b. This gives b = R / sqrt(1 - 2U). Do you
understand this?

> The objection I have is with the integraton limit. Einstein has it at
> (r = infinity to R), and I have (r = infinity to R - dR).

There is no "dR" involved at this point. The perihelion distance R is
not being varied. The variational work has already been done, and the
equations of motion for light-like geodesics have already been
found... they are the two equations you started with. If you are
interested in understanding the calculus of variations, and how
Fermat's principle leads to those equations of motion, you can learn
about that on the web as well, but from this point, where you have
already stipulated the equations of motion, the appropriate value of R
is simply the radius of the sun (when examining starlight that just
grazes the disk of the sun). There is no "dR" involved here. Do you
understand this?

> Although (R >> dR), the angle of deflection is proportional to (dR / R).

No. The parameter R signifies the closest distance from the Sun of the
path of light. We know the starlight we are going to examine is just
grazing the disk of the sun, so we need to know how much deflection to
expect for a light-like geodesic path whose closest approach to the
sun is essentially the radius of the sun. We already have the
equations of motion for all the light-like geodesic paths, so we just
need to insert the values of M and b = R / sqrt(1 - 2U) to determine
the deflection.

> By integrating from (r = infinity to R), we get only the angle of the
> photon at (r = R) which is (- 4 G M / c^2 / R)...

You're badly mistaken (again). The integration from r = infinity to R
gives pi/2 + 2GM/(Rc^2), as does the integration from r = R to
infinity, so the total deflection, accounting for both branches of
the hyperbolic orbit, is 4GM/(Rc^2).

Do you even know what it is you are integrating? Your result is so
grossly wrong that I have to wonder.

> according to choice of the impact factor (b) during the inbound trip,
> and during the outbound trip, we have the angle of the photon at
> (4 G M / c^2 / R) where (dr/dt > 0). The correct value of deflection
> is at perihelion which is (r = R - dR) where you will find the angle of
> deflection to be zero.

No. The deflection on the inbound trip is 2GM/(Rc^2) and the
deflection on the outbound trip is the same (the two legs of the
hyperbola are symmetrical about the perihelion). This gives a total
deflection of 4GM/(Rc^2).
.

Relevant Pages

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    ... > of M and b to see how much deflection there is. ... > equations of motion that you cited above. ... >> with mass of the sun = 0. ... > perihelion distance R is, by definition, the distance from the path to ...
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  • Re: Bending of light not well authenticated
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