dirk van de moortel

Are you trying to get the title of
Aerospace Engineer Of The Month
again?

Dirk Vdm

joe fischer

: sits right at (0,R) where
: [snip]
: Does any one disagree with my reasoning?

Of course, first of all, the photon isn't
"deflected", it travels in a truly straight line
in space-time.

Does a cat feel lighter if one hair falls out?

Please don't think in Euclidean space, it
requires a force to act, when none is needed.

The very reason that light appears to be
deflected twice that of matter having mass, is
because light travels in a straight line (in space-time).

Einstein didn't think his elevator in space
out correctly in the 1907 and 1911 papers, but
did have it right in 1915.

You may get farly accurate results by thinking
Newtonian, but physics can be understood better if
the concepts of General Relativity are used.
Mass is important, but not by causing a
deflection, it simply defines the center of mass
of a system in a misleading way.
Objects don't orbit because of an attractive
"force", they orbit because of relative velocities
in the space-time.
General Relativity may not be as clear and
as understandable as Newtonian physics, but it
is a great improvement in understanding even
the most simple processes in the presence of gravity.

Joe Fischer

--
3

robert j. kolker

The photon has rest mass 0. Everything from here on fails.

Bob Kolker

australopithecus afarensis

Yeah, what do I get if I become the Aerospace Engineer of the Month. Who
won the previous month?

* * *

"Australopithecus Afarensis" <fossil.lucy@cox.net> wrote in message news:MTM5b.43997$cj1.19792@fed1read06...

Are you trying to get the title of
Aerospace Engineer Of The Month
again?

Dirk Vdm

australopithecus afarensis

I did not even introduce Newtonian mechanics. The very concept of the
conservation of energy and momentum still should preside under General
Relativity.

Towards the end of the article has data on several eclipses. They scatter
all over the place. The radio wave experiment is more consistent because of
the same initial energy of each photon.

http://www.mathpages.com/rr/s6-03/6-03.htm

----- Original Message -----
From: "Joe Fischer" <gravity1@shell1.iglou.com>
Newsgroups: sci.physics.relativity
Sent: Thursday, September 04, 2003 01:36 PM
Subject: Re: Deflection of a Photon Near the Sun

: Consider an (x,y) coordinate system with the center of the sun initially
: sits right at (0,R) where
: [snip]
: Does any one disagree with my reasoning?

Of course, first of all, the photon isn't
"deflected", it travels in a truly straight line
in space-time.

Does a cat feel lighter if one hair falls out?

Please don't think in Euclidean space, it
requires a force to act, when none is needed.

The very reason that light appears to be
deflected twice that of matter having mass, is
because light travels in a straight line (in space-time).

Einstein didn't think his elevator in space
out correctly in the 1907 and 1911 papers, but
did have it right in 1915.

You may get farly accurate results by thinking
Newtonian, but physics can be understood better if
the concepts of General Relativity are used.
Mass is important, but not by causing a
deflection, it simply defines the center of mass
of a system in a misleading way.
Objects don't orbit because of an attractive
"force", they orbit because of relative velocities
in the space-time.
General Relativity may not be as clear and
as understandable as Newtonian physics, but it
is a great improvement in understanding even
the most simple processes in the presence of gravity.

Joe Fischer

--
3

australopithecus afarensis

I sorry. There is a couple of typos. I am not very good at copying some
one else's work including my own.

sin($)^2 should be sin($ / 2)^2 which is corrected in the message bleow.

----- Original Message -----
From: "Australopithecus Afarensis" <fossil.lucy@cox.net>
Newsgroups: sci.physics.relativity
Sent: Thursday, September 04, 2003 01:02 PM
Subject: Deflection of a Photon Near the Sun


Consider an (x,y) coordinate system with the center of the sun initially
sits right at (0,R) where

R = radius of the sun, and

A photon starts its journey at (-oo,0) along the x-axis towards (oo,0).

The photon will then be constantly deflected out of the x-axis of travel.
In doing so, an energy exchange will take place. The sun will gain some of
the photon's energy in terms of the kinetic energy, and the photon will lose
some of its mass.

From conservation of momentum and energy, we can write down

(1 - %) m c' cos($) + M Vx = m c'
(1 - %) m c' sin($) + M Vy = 0
m c'^2 = (1 - %) m c'^2 + (M / 2) (Vx^2 + Vy^2), where

M = mass of the sun
m = initial mass of the photon
% = final percentage of mass lost
Vx = final speed of the sun along the x-axis
Vy = final speed of the sun along the y-axis
$ = final angle of deflection from the x-axis, then

Vx = (1 - (1 - %) cos($)) m c' / M
Vy = - (1 - %) sin($) m c' / M, thus

Vx^2 + Vy^2 = (2 (1 - %) (1 - cos($)) + %^2) m^2 c'^2 / M^2
Vx^2 + Vy^2 = (4 (1 - %) sin($ / 2)^2 + %^2) m^2 c'^2 / M^2, therefore

% = (4 (1 - %) sin($ / 2)^2 + %^2) m / (2 M), or

$ = 2 arcsin(sqrt(% (2 M / m - %) / (1 - %)) / 2), or since 1 >> %,

$ = sqrt(2 % M / m)

Without any mass loss by the photon, there would be no deflection. Gravity
and the sun's atmosphere are among the major contributors to such a photon's
mass loss. In addition, the equation above also shows that ? very
significantly depends on the initial mass of the photon. Therefore, if some
one tells us how much a photon will be deflected during a solar eclipse
without knowing the initial energy of the photon, this some one is not being
realistic.

Does any one disagree with my reasoning?

titan point

Actuslly you need to stare at any textbook of science which will clearly
tell you that photons have no mass, relativistic or not. They have
momentum, but no mass. If they had mass, they could not travel at the
speed of light because they would have infinite energy.

joe fischer

You introduce Euclidean space any time you
think an object should move in inertial motion in
and in a straight line in the absence of a force acting.

If only relative velocity and position is
considered, a more correct result is possible, and
more models can be treated.

: The very concept of the conservation of energy and momentum
: still should preside under General Relativity.

They do, but not in Euclidean space, and
not in exactly the same way as in Newtonian physics
and gravitation.

It would be silly to say an object in freefall
is in inertial motion, and then say that it has
kinetic energy or gravitational potential energy.

: Towards the end of the article has data on several
: eclipses. They scatter all over the place.

Do you think you are the only one who can read,
I knew that 50 years ago.
The fact is that all the observations together
provided enough information to say that the apparent
deflection was definitely greater than the 1911 math,
and definitely showed something other than Newtonian.

: The radio wave experiment is more consistent because of
: the same initial energy of each photon.
:
: http://www.mathpages.com/rr/s6-03/6-03.htm

If you knew any math at all you would know
that the ratio of the energy of a photon versus
the energy of the sun is probably greater than
10E+40.
That means you can't calculate any difference
in the energy between a radio wave photon and a
visible light photon passing the sun.

And of course you are protecting your identity
just in case you are making a fool of yourself??

You should have said

: The radio wave experiment is more consistent because of
experiments were

because they have been done many times, and
all have better accuracy because they can be compared
to the apparent deflection at all distances and angles,
an eclipse isn't needed.

Also, there is a very simple reason why objects
having mass do not appear to fall as fast as photons.
I don't think it has been discussed much, but
I will post it in a thread called Einstein's Elevator.

Joe Fischer

--
3

xxein

xxein: Me, of course!

All those mathematics!

See if this is what you get.

Ignore the supposed infinite mass of a pea traveling at c. It is
obvious that photons don't increase mass because of their speed.
Photons just travel at c.

If you accept that R=3M for least radius for stable circular orbit
around BH's, then you would be able to calculate that the orbit
velocity of such photon or pea is 173085256.327319 m/s and the escape
velocity at that radius is 244779516.944918 m/s. If you square and
then sum the velocities, you get c^2.

If the orbit velocity is not c, as expected, then neither is its mere
deflected velocity toward you to be c while it is in gravity.

First, figure out its path. Not so simple if its "path speed" is
diminished by a function of escape velocity. Next, you will find that
the path is incorrect because if its speed is diminished, it will
curve more.

When you finally get a result that will impact on the observer, you
will see that the curvature is 2x what you expected it to be.

For light there was no energy exchange and no consideration of
conservation of energy. But no energy exchange means that the
frequency of that light is merely a function of the source and the
reciever as if there was no deflection.

The hard part to understand is that the system of eclipse is moving.
If it were not moving, you would see the same unadultered frequency
all of the time from wherever it was sent and recieved. But because
it is moving, you see a doppler effect. It is not the standard
doppler you think of normally. It is a combination of two things.
Approching eclipse, the sent light you can recieve has to be the light
that is sent at a greater and greater angle away from you to
accomodate the intervening gravity. The increasingly longer path is
doppler effect. Eventually light cannot reach you.

But note that if the movement were to stop, it would normalise
(increase) to the standard frequency. But it doesn't stop. A
relatively non-moving seeable star in the same background position
does not suffer this frequency change.

But on coming out of eclipse, the reverse is true. The apparent
frequency rises until it is out of the "moving eclipse" effect.

You have a lot to think about.

dirk van de moortel

As you well know, some top-posting clumsy troll who used
to use the name Scholarly Fungi before he started using your
current wimp disguise:
http://groups.google.com/groups?&threadm=H9N0b.77035$F92.8566@afrodite.tele net-ops.be

Dirk Vdm