Re: Bending of light not well authenticated


"Randy M. Dumse" <rmd@xxxxxxxxxxxxx> wrote in message
news:9zRhe.128$Jq.1249@xxxxxxxxxxxxxxxxxxxx
>
> Proper spacetime implies proper time of a local observer. In a local
> non-accelerated frame, the speed of light is always c. However, if the
> frame is accelerated, light appears bent/slowed.

This point does not have major impact on our core discussion. However, you
seem to claim to be able to observe a light beam being bent as it travels
near an accelerated frame. When you receive or observe a photon directly,
there is no way to find out how many times or how this photon has been
deflected. The deflection can only be determined through indirect means,
and you can never be able to observe the bent path of the photon.

> And actually, one can construct an explanation of the bending of light by
> setting up a series of local frames, and following the net effect of all
> those frames across the path of the photon. Will does that very nicely in
> "Was Einstein Right?".

Yes, I agree. Therefore, according to the coordinates of the series of
proper frames, there is no deviation from flat space because for a photon,

ds^2 = c^2 dtau^2 - drho^2 - rho^2 dtheta^2 = 0

After integrating the angle to get the angle of deflection, one should
easily arrive at the conclusion that there is no distortion of spacetime in
the proper frames. At proper frames too far away from the gravitational
source, they should also exhibit no distortion. And these proper frames
must equate with the observed frame where Schwarzschild metric applies. So,
no distortion of proper frames at infinity must equate with no distortion of
observed frame at infinity. That means the photon at the ends of travel
must not have any deflection. However, in between, there is a deflection.
The spacetime equation should tell us how this photon is deflected and then
after perihelion undo its deflection. I get the following two equations in
accordance with most derivations except Ciufolini/Wheeler's.

** (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 from the conservation of angular momentum which
Taylor calls it the impact factor

I disagree with Taylor's (moortel posted his derviation earlier in this
thread where you should have no problems finding it) conclusion on the value
of b. However, b does not affect the overall trajectory of the photon.
Rather, it affects only how close the photon is going to get to the sun. I
have (b = R) where R is the perihelion of the photon to the sun with mass of
the sun = 0. This is worth another session of discussion if you are
interested. Thus, we can calculate how close the photon is deflected at
perihelion through the following equation.

R = (R - dR) / sqrt(1 - 2 G M / c^2 / (R - dR))

Where

** dR = deflected distance at perihelion

Note: If dR is very large due to very strong gravity, the photon would get
absorbed by the sun.

So, during the inbound trip,

** dr/dt < 0
** r = infinity to (R - dR)

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). Although (R >> dR),
the angle of deflection is proportional to (dR / R). Thus, I cannot agree
with Einstein's integration limit. Also, at strong gravity, Einstein's
limit does not allow the photon to be captureed by the sun. I am,
therefore, very certain that he is wrong in this case.

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), 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.

>> Clearly, the photon should behave with no anomaly in its orbit in the
>> reference frame of spacetime. At infinite distance away where the
>> gravitational effect of the sun is asyptotically approaching zero,
>
> Correct, but, there is no "orbit" to have an anamoly in flat spacetime. No
> curvature, not orbit, no deflection, no problem.

All GR derivations of the deflection of the photon treats the photon as in
orbit around the sun. This is by definition what it is if using polar
coordinate. You cannot get out of this one.

> Well, when you model the spacetime around the sun as deflecting it toward
> the sun, then at the other side, deflecting away from the sun, if I took
> your meaning, then it is not worth bothering with the analogy, because
> instead of a prism, you are using a flat plate. So I'm not interested in
> going there.

No, I am using a prism shaped like a ball where the index of refraction
increases towards the center of the ball. Your prism analog is excellent.
It agrees with my calculations. However, it does not agree with Einstein's
derivation. Your intuition is telling you that Einstein's derviation is
bogus.

>> In equal distance away from the sun, the degree of curvature in spacetime
>> is the same. So, your excellent analogy of the gravitational lens
>> behaves like a prism firmly applies.
>
> Huh..., and I thought you didn't like it.

Again, I am very fond of your analogy of using the gravitational lens as a
prism. However, I don't understand how you are confused with your own
analogy. Perhaps, I can help.

> Light does not approach the gravitating body like it was a plane of glass,
> first deflecting down on contact with the slower indexed material. Like an
> accelerated frame (equivalence principle) the light appears to bend

Again, you have to show me how a photon during its travel can emit another
photon to be observed by you for you to determine the original photon is
bending its path.

> Having placed himself among the others predicting Mercury's orbital
> anamoly doesn't seem to bad a thing, since Einstein got it right to high
> accuracy and without having to bother with any special assumptions of
> tweaking of constants or variables, etc., but from his original principles
> unaltered.

You have to explain with no handwaving that how Einstein is able to fudge a
2nd order effect out of only 1st order knowledge. I would love to discuss
Mercury's orbital anomaly with you step by step without any handwaving.
Please understand this point. My invitation is not a bluff but a very
sincere invitation. If your intention is honorable and not of a troll, I am
very certain that I can convince you through simple mathematics that the
current derivation under the concept of GR is totally BS.

> Again, Will is an interesting author explaining how Brans Dicke gained
> prominence, for a time, because that theory had a slightly different
> smaller effect. However, with better accuracies, GR was found to be the
> more correct of the two.

You keep quoting away from references out of reach by me. We live in the
dawn of the golden age of internet. You must be able to quote something
from the internet. Is there anything you have problems with
Ciufolini/Wheeler?

http://www.pupress.princeton.edu/sample_chapters/ciufolini/chapter2.pdf

> No. The impact parameter is not the distance from r=0 to r="the closest
> point of approach" of the photon. Or said another way, which I believe now
> you might be saying, the impact parameter b is not the radius of the
> peirhelion.

I never said the impact factor, b, is the closent point of approach to the
sun. Thus, you are correct that the impact factor is not the distance of
perihelion. I have mentioned at least 3 times already that this impact
factor is nothing other than an integration constant.

>> The authors seem to try to make it as confusing as possible.
>
> I'd say instead, the authors try to match the level of understanding of
> their intended audience, and also the level the dilligent student should
> have attained upon reaching that point in the book.

A better explanation is that the authors have already established that
Einstein's calculation must be correct. Some of them must have discovered
something wrong. So, they try to make GR as complex and confusing as
possible. In the field of science, nothing works better at confusing anyone
by using fancy vocabulary. For example, the word 'geodesic' is nothing more
than the path of an event following the principle of minimal actions. One
sub-example is how the Snell's Law comes about. Only Fermat's derivation
follows this sounding principle while Huygens' derivation was nothing more
than fudging of geometry. Guess whose derivation Feyman endorses.

> The problem with understanding section 25.6 in MTW Orbit of a photon...,
> is it requires understanding 25.5 Orbits of particles, which ties back to
> the Schwarzschild geometry, etc.

See what I mean. If you go back to 25.1 or whatever the 1st equation is,
you will find it is nothing more than the application of the principle of
minimal action. And minimal action means minimal amount of elapsed time and
not necessarily spacetime. Ciufolini took the minimal action to be an
application of spacetime. In doing so, his derivation of bending photon is
very much different what the main camp. After studying this, maybe you have
a comment or two about this.

> Yes, actually it does require sophisticated mathematical skills. From the
> Schwarzschild geometry, you can get differential equations that describe
> the motion of particles. MTW, D'INV, etc., start with those equations (See
> Binet's Eg. http://scienceworld.wolfram.com/physics/BinetsEquation.html)
> in the relativistic sense, and move on to massless versions, then extract
> the necessary calculation. If you can't follow those mathematical
> gymnastics, which they assume their readers are already equiped to do,
> then pronouncing them bogus is perhaps premature.

Oh, yes. Would you like to get into more details of combining the equations
of

** (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)

Into

(dr/dH)^2 = ...

I have done it, and I have done the substitution of (u = 1 / r) trick. Have
you?


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