Discussion:
Interesting. ChatGPT fails defending starlight deflection when photons graze Sun's surface
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rhertz
2025-01-17 18:05:00 UTC
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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.
- Poor's critics were more than justified, but relativists joined forces
to bury him down in history. No other significant scientist DARED to
confront the IGNORANCE WALL that relativists built since 1919.

ChatGPT is trained TO OBEY, DISTORT, OPENLY LIE AND DECEIVE when
defending Einstein. It's a Jewish site, after all.
LaurenceClarkCrossen
2025-01-17 19:39:59 UTC
Reply
Permalink
Post by rhertz
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.
The deflection angle for light passing close to a massive object is
Δθ = 4GM/(Rc^2)
For a photon grazing Sun's surface, Δθ ≈ 1.75arcseconds
Please, give me the equation of the hyperbola that verify such
deflection.
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
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.
*******************************************************************
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.
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.
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
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
- 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.
******************************************************************
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
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.
- Einstein refused, for the rest of his life, to show his calculations.
- Poor's critics were more than justified, but relativists joined forces
to bury him down in history. No other significant scientist DARED to
confront the IGNORANCE WALL that relativists built since 1919.
ChatGPT is trained TO OBEY, DISTORT, OPENLY LIE AND DECEIVE when
defending Einstein. It's a Jewish site, after all.
According to Einstein's 1915 explanation, doubling involves more
deflection before and after the closest approach of the wavefront to the
Sun instead of at its closest approach as with Newtonian gravity. -
https://www.mathpages.com/rr/s8-09/8-09.htm = "Doubling the Deflection"
LaurenceClarkCrossen
2025-01-17 21:31:48 UTC
Reply
Permalink
That is: "Gravitational Deflection of Light: A Heuristic Derivation at
the
Undergraduate Level"
rhertz
2025-01-17 21:46:00 UTC
Reply
Permalink
On Fri, 17 Jan 2025 19:39:59 +0000, LaurenceClarkCrossen wrote:

<snip>
Post by LaurenceClarkCrossen
According to Einstein's 1915 explanation, doubling involves more
deflection before and after the closest approach of the wavefront to the
Sun instead of at its closest approach as with Newtonian gravity. -
https://www.mathpages.com/rr/s8-09/8-09.htm = "Doubling the Deflection"
This article EXACTLY follows what I've been posting for years, based on
the equation 7c of his 1915 paper on Mercury's phenomena:

THAT "HIS CALCULATIONS" for doubling the 1911 value for starlight
deflection, when razing the Sun's surface IMPLIES A VARIABLE SPEED OF
LIGHT:

1) Light speed DECREASES FROM its nominal value c when approaching the
Sun.
2) Light speed INCREASES TO its nominal value c when going away from the
Sun.
3) Light HAS MASS.

EXCERPT:
----------------------------------------------------------------

"Therefore, the speed of light (along this path) is actually given, to
the first order in the small quantities m/r, by the expression

c(t) = 1 - m/r (1 + x^2/r^2)
----------------------------------------------------------------

The above equation is in geometrical units: c = G = 1.

Now compare with equation 7c:

Ί(r) = -α/2 (1 + B^2/r^2) ; Equation 7c from Mercury's paper. α =
2GM/(rc^2) in physical units



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

B = mr^2ω, the constant angular momentum, under Newton's Law of
Gravitation.


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

At r = R_Sun, c = R_Sun.ω, which is THE LOWER VALUE OF c on the
trajectory of starlight. Then, c decreases approaching the Sun, to later
increase to the nominal value c_0.

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


Huygens or not involved here, IT PROVES THAT GR IS BASED ON A VARIABLE
SPEED OF LIGHT when it passes by gravitational masses of celestial
bodies.


DISCLAIMER: I don't believe that the above is true. I'm just analyzing
the CRAP that general relativity contains.

For me, c IS CONSTANT, and the phenomena of deflection is due to
REFRACTION.

No gravitational lensing exist.
LaurenceClarkCrossen
2025-01-17 22:00:32 UTC
Reply
Permalink
Post by rhertz
<snip>
Post by LaurenceClarkCrossen
According to Einstein's 1915 explanation, doubling involves more
deflection before and after the closest approach of the wavefront to the
Sun instead of at its closest approach as with Newtonian gravity. -
https://www.mathpages.com/rr/s8-09/8-09.htm = "Doubling the Deflection"
This article EXACTLY follows what I've been posting for years, based on
THAT "HIS CALCULATIONS" for doubling the 1911 value for starlight
deflection, when razing the Sun's surface IMPLIES A VARIABLE SPEED OF
1) Light speed DECREASES FROM its nominal value c when approaching the
Sun.
2) Light speed INCREASES TO its nominal value c when going away from the
Sun.
3) Light HAS MASS.
----------------------------------------------------------------
"Therefore, the speed of light (along this path) is actually given, to
the first order in the small quantities m/r, by the expression
c(t) = 1 - m/r (1 + x^2/r^2)
----------------------------------------------------------------
The above equation is in geometrical units: c = G = 1.
Φ(r) = -α/2 (1 + B^2/r^2) ; Equation 7c from Mercury's paper. α =
2GM/(rc^2) in physical units
In physical units, Eq 7c is Φ(r) = - GMm/r [1 + B^2/(mcr)^2]
B = mr^2ω, the constant angular momentum, under Newton's Law of
Gravitation.
Making r.ω = c at the perigee of the trajectory, at which the test
particle has maximum speed, gives (1 + r_Sun^2 ω^2/c^2) = (1 + 1) = 2,
which doubles the 1911 value.
At r = R_Sun, c = R_Sun.ω, which is THE LOWER VALUE OF c on the
trajectory of starlight. Then, c decreases approaching the Sun, to later
increase to the nominal value c_0.
It implies a variable speed of light, besides ACCEPTING that "photons"
have mass and suffer gravitational attraction.
Huygens or not involved here, IT PROVES THAT GR IS BASED ON A VARIABLE
SPEED OF LIGHT when it passes by gravitational masses of celestial
bodies.
DISCLAIMER: I don't believe that the above is true. I'm just analyzing
the CRAP that general relativity contains.
For me, c IS CONSTANT, and the phenomena of deflection is due to
REFRACTION.
No gravitational lensing exist.
From the same article's conclusion: "The preceding discussion explains
the meaning of Einstein’s oft-quoted comment in Appendix 3 of his 1916
book “Relativity”, where he states that “half of this deflection is
produced by the Newtonian field of attraction, and the other half by the
geometrical modification (‘curvature’) of space caused by the sun”. As
explained in Section 8.5, Newton’s theory can essentially be expressed
as a metrical theory with a line element given by (1), according to
which the spatial geometry is flat, but curved in time. This leads to
the Newtonian prediction 2m/R. The line element given by (2) includes
the effects of spatial curvature, as indicated by the variable
coefficients of the space-space components. As
we’ve seen, when the effect of this spatial curvature is taken into
account, the total predicted deflection is 4m/R."

This quote shows that the Newtonian curvature is doubled by curving
space itself as if it were a surface. That involves the elementary error
of reification fallacy.
ProkaryoticCaspaseHomolog
2025-01-18 00:57:06 UTC
Reply
Permalink
Post by rhertz
For me, c IS CONSTANT, and the phenomena of deflection is due to
REFRACTION.
======================================================================

In the Earth's ionosphere, radio waves are strongly refracted by free
electrons. This refraction is strongly dependent on frequency:

n^2 = 1 - ω_p^2 / [ω(ω + iν)]

where ω_p is the plasma frequency. Propagation of optical frequency
light is virtually unaffected by free electrons in the ionosphere.

Both radio frequencies and optical frequencies are refracted by BOUND
electrons in atoms and molecules. In the Earth's atmosphere, water
molecules are extremely important in the refraction of radio waves,
but of rather less importance in the refraction of optical frequencies.

======================================================================

In the solar corona, the situation is quite different. Radio waves
are, of course, refracted by free electrons in the solar corona, so
VLBI measurements of the gravitational deflection of radio waves must
be corrected for refraction. This is done by performing measurements
at multiple wavelengths and using the known relationship between
wavelength and refraction by free electrons to determine the
un-refracted path of the radio waves.

It is unnecessary to correct for refraction by BOUND electrons because
of their almost complete absence in the solar corona. Above a
transition zone a few thousand kilometers above the surface, the
coronal gases are heated to temperatures greater than 1,000,000°C. At
such temperatures, all of the lighter elements (hydrogen, helium,
carbon, nitrogen, and oxygen) are stripped down to bare nuclei.
Spectral lines visible in the corona comes from heavier trace elements
like iron and calcium which are able to retain a few of their
electrons.

(Do not be confused by the F corona spectral lines, which are due to
scattering of light by dust particles.)

Visible light passing by the Sun is not refracted by FREE electrons in
the solar corona, and the complete stripping of electrons from all
elements up through oxygen means that there are virtually no BOUND
electrons in polarizable atomic species capable of refracting visible
light either.

In other words, not only is the solar atmosphere far too tenuous to
contribute significantly to the measured deflection of visible light,
its composition is entirely wrong.

======================================================================
LaurenceClarkCrossen
2025-01-18 01:34:39 UTC
Reply
Permalink
Post by ProkaryoticCaspaseHomolog
Post by rhertz
For me, c IS CONSTANT, and the phenomena of deflection is due to
REFRACTION.
======================================================================
In the Earth's ionosphere, radio waves are strongly refracted by free
n^2 = 1 - ω_p^2 / [ω(ω + iν)]
where ω_p is the plasma frequency. Propagation of optical frequency
light is virtually unaffected by free electrons in the ionosphere.
Both radio frequencies and optical frequencies are refracted by BOUND
electrons in atoms and molecules. In the Earth's atmosphere, water
molecules are extremely important in the refraction of radio waves,
but of rather less importance in the refraction of optical frequencies.
======================================================================
In the solar corona, the situation is quite different. Radio waves
are, of course, refracted by free electrons in the solar corona, so
VLBI measurements of the gravitational deflection of radio waves must
be corrected for refraction. This is done by performing measurements
at multiple wavelengths and using the known relationship between
wavelength and refraction by free electrons to determine the
un-refracted path of the radio waves.
It is unnecessary to correct for refraction by BOUND electrons because
of their almost complete absence in the solar corona. Above a
transition zone a few thousand kilometers above the surface, the
coronal gases are heated to temperatures greater than 1,000,000°C. At
such temperatures, all of the lighter elements (hydrogen, helium,
carbon, nitrogen, and oxygen) are stripped down to bare nuclei.
Spectral lines visible in the corona comes from heavier trace elements
like iron and calcium which are able to retain a few of their
electrons.
(Do not be confused by the F corona spectral lines, which are due to
scattering of light by dust particles.)
Visible light passing by the Sun is not refracted by FREE electrons in
the solar corona, and the complete stripping of electrons from all
elements up through oxygen means that there are virtually no BOUND
electrons in polarizable atomic species capable of refracting visible
light either.
In other words, not only is the solar atmosphere far too tenuous to
contribute significantly to the measured deflection of visible light,
its composition is entirely wrong.
======================================================================
Alexander Unzicker differs with you on that in his book, "The Liquid
Sun."

LaurenceClarkCrossen
2025-01-17 21:30:25 UTC
Reply
Permalink
Post by rhertz
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.
The deflection angle for light passing close to a massive object is
Δθ = 4GM/(Rc^2)
For a photon grazing Sun's surface, Δθ ≈ 1.75arcseconds
Please, give me the equation of the hyperbola that verify such
deflection.
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
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.
*******************************************************************
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.
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.
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
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
- 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.
******************************************************************
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
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.
- Einstein refused, for the rest of his life, to show his calculations.
- Poor's critics were more than justified, but relativists joined forces
to bury him down in history. No other significant scientist DARED to
confront the IGNORANCE WALL that relativists built since 1919.
ChatGPT is trained TO OBEY, DISTORT, OPENLY LIE AND DECEIVE when
defending Einstein. It's a Jewish site, after all.
Einstein never gave it, but his followers attempted it. Here's another
article purporting to explain the derivation, resorting to non-euclidean
geometry as usual.
rhertz
2025-01-17 22:23:09 UTC
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I forgot. Here is a link to the original 1915 Mercury paper (in German).

Equation 7c is in Page 30.

Note HOW CROOKED/COOKED IS THE PAPER that Eq. 7c was added AFTER
equations 6c, 7, 7a, 8, 8a, 9, 7b AND 10.

Such paper, presented to the Prussian Academy of Science (who??) on
Nov.18, 1915. It CLEARLY SHOWS how the fucker struggled to get his final
COOKED result in the previous WEEK, as he had promised to deliver HIS
Field Equation in that date.

But, as HE FAILED TO DO SO, he delivered this paper (that also announces
the doubling of the 1911 deflection), in order TO GAIN MORE TIME to get
HILBERT SOLUTION. He knew that Hilbert was going to deliver a lecture on
his (the ONLY solution that ever existed) by Nov. 20, 1915.

So he got together all the CRAP that he accumulated since 1913, with
Besso and, with the EVIDENT HELP OF TWO ASTRONOMERS (Freundlich and
Schwarzschild), he presented this paper AS A DIVERSION OF ATTENTION to
his promise about the field equations.

The CROOK presented his paper with the field equation in Nov. 25, 1915.
Just one week after this crappy paper, and 5 days after Hilbert's
lecture.


The Mercury paper DIDN'T NEED the field equations, but he and Besso had
been struggling for years with a solution including EVERY FUCKING
PLANET, until an astronomer (Freundlich) taught him to do it in the
Newtonian way:

1) Forget every planet. Use Le Verrier/Newcomb calculations on the
Newtonian influence of every planet, except Mercury.

2) Focus ONLY on Mercury and the Sun, as a differential.

3) Use Gerber's 1898 formula, and FUCKING REVERSE ENGINEER IT!


In that way, he forged the fucking paper.
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