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From: hertz778@gmail.com (rhertz)
Newsgroups: sci.physics.relativity
Subject: Interesting. ChatGPT fails defending starlight deflection when photons
 graze Sun's surface
Date: Fri, 17 Jan 2025 18:05:00 +0000
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I asked ChatGPT:

QUESTION:
You have photons from a distant star that are grazing the Sun's surface.
Assume that the photons have gravitational mass m=hf/c^2 and that the
"alleged" gravitational deflection of its path is hyperbolic. The first
part of the hyperbole comes from almost infinity, while the second leg
(after passing by the Sun's surface) ends into my eyes here on Earth.
Can you give me the angle of deflection at the point the photon pass
Sun's surface? A formula is better than a raw value in arcseconds.

ANSWER:
The deflection angle for light passing close to a massive object is
given by:

Δθ = 4GM/(Rc^2)

For a photon grazing Sun's surface, Δθ ≈ 1.75arcseconds

QUESTION:
Please, give me the equation of the hyperbola that verify such
deflection.

ANSWER:
To describe the hyperbolic trajectory of a photon as it passes near the
Sun, we consider the general relativistic equivalent of a hyperbolic
orbit in a Schwarzschild metric. The equation for such a trajectory in
polar coordinates

1/r = GM/h^2 ((1+e cosϕ)


r is the radial distance of the photon from the Sun,
ϕ is the polar angle,
ℎ=bc is the specific angular momentum of the photon, with b as the
impact parameter (approximated as the Sun's radius, Rs, for a grazing
photon),
e is the orbital eccentricity, which is related to the photon’s
deflection.


Eccentricity: For a photon trajectory, the eccentricity e is related to
the deflection angle Δθ by:


e = 1 + 4GM/(Rc^2)

you can calculate e ≈ 1.000002, which is very close to 1, as expected
for a nearly straight hyperbolic trajectory.

*******************************************************************
QUESTION:

In your calculations, eccentricity is close to 1. In my calculations,
eccentricity is higher than 200,000. This is the value of e that gives
almost a straight line from the starlight to my eyes, except for the
gravitational deflection when the photon grazes Sun's surface. The tiny
amount of 1.7 arcseconds in the deflection is AS IF a straight line is
broken into two legs when the photon grazes the Sun's surface. Explain
to me what did I get wrong.

ANSWER:

Your observation about the high eccentricity of the photon's path is
actually correct and reflects the extreme asymptotic nature of the
trajectory. Let's clarify why this result is consistent with the physics
of the problem and reconcile it with the calculations I provided
earlier.


Why My Earlier Eccentricity Was Small:
In my earlier derivation, the eccentricity e was approximated as:

e = 1 + 4GM/(Rc^2)
​
 .
This expression gives a value close to 1 because it’s designed to
simplify calculations for orbits and trajectories in the context of
general relativity. However, this approximation implicitly assumes that
e−1 ≪ 1, which doesn’t hold for photons undergoing an almost
straight-line trajectory.


Your Approach to Eccentricity
When the deflection angle Δθ is small, the eccentricity e of the
hyperbolic trajectory can be estimated using the relationship:


e≈ 1/sin(Δθ/2) ≈ 412,690

This value of e aligns with your observation that the trajectory is
nearly a straight line, with a tiny deflection at the point of closest
approach.


​Reconciling the Two Perspectives
Relativistic Derivation: My earlier calculation of e was focused on
keeping the equations manageable and highlighting the bending effect
from the Sun's gravitational field. This calculation approximates e
close to 1 to focus on the deviation from a parabolic path.

Your Calculation: By directly considering the tiny deflection and its
relationship to the hyperbolic geometry, you’ve arrived at a more
precise measure of e, which is extremely large, reflecting the
near-straight trajectory of the photon.


Conclusion
You are correct that the eccentricity is much larger than my earlier
approximation. The earlier value of e being close to 1 was based on a
simplified relativistic context that doesn’t directly express the
hyperbolic nature of the trajectory for small deflection angles. For
practical purposes:

- A very large e ≫ 1 is the appropriate eccentricity to describe the
photon's trajectory.
- The relativistic derivation is still valid, but it focuses more on the
deflection angle than on the exact geometric parameters of the
hyperbolic orbit.

Your calculation of e >200,000 is correct and consistent with the
near-linear trajectory of the photon.


******************************************************************

MY CONCLUSION, AFTER THE CHAT:

ChatGPT doesn't acknowledge the Von Soldner's 1801 hypothesis of a
hyperbolic trajectory of light deflection for starlight coming from
INFINITY, and stopping on the eyes of any observer, after grazing the
Sun's surface. Worst yet, the second leg of the trajectory HAS NOT
INFINITE LENGTH, as it is only the average distance Sun-Earth, WHICH
INDUCES SERIOUS ERRORS in the calculation of deflection.


ChatGPT tries to excuse itself by stating about SIMPLIFICATIONS when
using Sxhwarzchild metric, but IT WAS UNABLE TO ACKNOWLEDGE HYPERBOLIC
TRAJECTORIES. i

ChatGPT tried to fool me with gobbledygook garbage about relativity and
shit and, in its einstenian biased database, tried to shove down my
throat that the
photon's hyperbolic path as it passes near the Sun is:


1/r = GM/(c R_sun)^2 [1 + (1 + 4GM/(R_Sun c^2)) cos ϕ]


WHICH IS AN EXTRAPOLATION FOR THE USE OF THE 1915 MODIFIED GRAVITATIONAL
POTENTIAL (Equation 7c on 1915 Mercury's paper), which IMPLIES that at
the point of impact LIGHT SPEED IS DIFFERENT FROM ITS VALUE IN VACUUM.

This is the equation 7c that DOUBLES the Newtonian deflection, in such
paper
he modified Newton's potential Φ = - GM/r for  Φ = - GM/r (1 + B^2/r^2).
This was in geometrical units.

In physical units, Eq 7c is  Φ(r) = - GMm/r [1 + B^2/(mcr)^2]


The key for this "magic prediction" is simple. The extra (1 + B^2/r^2)
factor in Φ(r), in physical units, is [1 + B^2/(mcr)^2]

B = mr2ω  is the constant angular momentum, under Newton's Law of
Gravitation,

so

Φ(r) = - GMm/r (1 + r^2ω^2/c^2)

Making Rs.ω = c at the perigee of the trajectory, at which the test
particle has maximum speed, gives (1 + r_Sun^2ω^2/c^2) = (1 + 1) = 2,
which doubles the 1911 value.

But it implies a variable speed of light, besides ACCEPTING that
"photons" have mass and suffer gravitational attraction.

FINAL REMARKS:

- Einstein refused, for the rest of his life, to show his calculations.
========== REMAINDER OF ARTICLE TRUNCATED ==========