On Feb 25, 7:47 am, Tom Roberts wrote:

The best model we have for the propagation of light near a massive
object like the sun is GR, in which the curvature of spacetime is the
important aspect in determining the path light follows. And it agrees
with measurements to part-per-million accuracy over an enormous range.

First, derive a set of geodesic equations a massed particle traveling
at high speed near the sun. Then, gradually reducing the mass to zero
and increasing the speed to c, do you see a discontinuity at mass = 0
and speed = c?

As you know, the geodesic equations are independent of mass. What
does that tell you when the model predicts a 1x deflection traveling
at speed just a hair below c and suddenly jumps to 2x deflection at
speed = c?

F*cked up.

On Mar 1, 8:53 pm, Koobee Wublee wrote:
On Feb 25, 7:47 am, Tom Roberts wrote:

The best model we have for the propagation of light near a massive
object like the sun is GR, in which the curvature of spacetime is the
important aspect in determining the path light follows. And it agrees
with measurements to part-per-million accuracy over an enormous range.

First, derive a set of geodesic equations a massed particle traveling
at high speed near the sun. Then, gradually reducing the mass to zero
and increasing the speed to c, do you see a discontinuity at mass = 0
and speed = c?

As you know, the geodesic equations are independent of mass. What
does that tell you when the model predicts a 1x deflection traveling
at speed just a hair below c and suddenly jumps to 2x deflection at
speed = c?

Show us the mathematics in which this "2x deflection" appears, along
with a consistent and clear definition of the terms.


F*cked up.

On Mar 2, 1:54*am, Eric Gisse wrote:
On Mar 1, 8:53 pm, Koobee Wublee wrote:





On Feb 25, 7:47 am, Tom Roberts wrote:

The best model we have for the propagation of light near a massive
object like the sun is GR, in which the curvature of spacetime is the
important aspect in determining the path light follows. And it agrees
with measurements to part-per-million accuracy over an enormous range.

First, derive a set of geodesic equations a massed particle traveling
at high speed near the sun. *Then, gradually reducing the mass to zero
and increasing the speed to c, do you see a discontinuity at mass = 0
and speed = c?

As you know, the geodesic equations are independent of mass. *What
does that tell you when the model predicts a 1x deflection traveling
at speed just a hair below c and suddenly jumps to 2x deflection at
speed = c?

Show us the mathematics in which this "2x deflection" appears, along
with a consistent and clear definition of the terms.

Just look in your books. Don't tell you collect them just to sit on.

On Feb 26, 3:28*pm, Cosmik de Bris
wrote:
pmbwrote:
On Feb 25, 2:16 am, Eric Gisse wrote:
On Feb 24, 9:28 pm, "Y.Porat" wrote:

On Feb 24, 11:25 pm, " wrote:
Ok...so I've been wondering how on earth does gravity bend light, I
mean I read all about how Einstein's General Relativity describes
space-time, and when there's a mass in space-time it creates a
curve(Bowling ball on a sheet or something), and that curve is
Gravity, and when light passes through the curve it bends....but why?
--------------------
just forget about curved spacetime
light bends next topthe sun
becuase of the most simple reason
LIGHT HAS MASS !!!
You are amazingly sure of this despite not having one shred of
evidence to support your claim. Then again, it is more important to be
original than be right, eh?

(even Einstein ddint know to read his momentous formula
E=mc^2
Show us the calculation of the deflection angle given the assumption
that photons have mass.

The reason that all text particles fall at the same rate given the
same inititial conditions is because the gravitational mass equals the
inertial mass and therefore the mass cancelts out in the equatuions of
motion. Einstein predicted that light is deflected in a gravitational
field because light has energy and energy has mass.

You are still propagating this myth. Energy and mass are equivalent that
doesn't mean that energy has mass.

I disagree. And its not a myth at all. The true myth is believing
otherwise. Anything which has inertial mass has energy and anything
which has energy has energy has inertial mass. As Feynman wrote in the
"Feynman Lectures" Vol -I page 7-11, Section entitled Gravitation and
Relativity
----------------------------------------------------------------------------
One feature of this new law is quite easy to understand is this: In
Einstein relativity theory, anything which has energy has mass -- mass
in the sense that it is attracted gravitationaly. Even light, which
has energy, has a "mass". When a light beam, which has energy in it,
comes past the sun there is attraction on it by the sun.
----------------------------------------------------------------------------

Or as Einstein wrote in "The Evolution of Physics," page 221. Einstein
is responding to the assertion regarding an observer in an
accelerating elevator "The inside observer, who believes in the
gravitational field acting on all objects in his elevator, would say:
there is no accelerated motion of the elevator, but only the action of
the gravitational field. A beam of light is weightless and, therefore,
will not be affected by the gravitational field."
----------------------------------------------------------------------------
But there is, unfortunately, a grave fault in the reasoning of the
inside observer, which saves our previous conclusion. He said : "A
beam of light is weightless and, therefore, it will not be affected by
the gravitational field." This cannot be right! A beam of light
carries energy and energy has mass. But every inertial mass is
attracted by the gravitational field as inertial masses and
gravitational masses are equivalent. A beam of light will bend in a
gravitational field exactly as a body would if thrown horizontally
with a velocity equal to that of light. If the inside observer has
reasoned correctly and had taken into account the bending of light
rays in a gravitational field, then his results would have been
exactly the same as those of an outside observer.
----------------------------------------------------------------------------

I understand that people may disagree with this since there is a
tendancy to confuse inertial mass (which is non-zero for light) and
inertial mass (which is zero for light). If you believe that you can
prove otherwise then please do. Thanks.

Best wishes

Pete

On Mar 2, 12:53*am, Koobee Wublee wrote:
On Feb 25, 7:47 am, Tom Roberts wrote:

The best model we have for the propagation of light near a massive
object like the sun is GR, in which the curvature of spacetime is the
important aspect in determining the path light follows. And it agrees
with measurements to part-per-million accuracy over an enormous range.

First, derive a set of geodesic equations a massed particle traveling
at high speed near the sun. *Then, gradually reducing the mass to zero
and increasing the speed to c, do you see a discontinuity at mass = 0
and speed = c?

As you know, the geodesic equations are independent of mass. *What
does that tell you when the model predicts a 1x deflection traveling
at speed just a hair below c and suddenly jumps to 2x deflection at
speed = c?


It tells us that they are completely exclusive cases. But I do
understand that YOU can't go there. Hell, you're still working on
transforming algebraic domains. By the way, how's that coming along?
Made any real progress lately? Have reached the level of an eighth
grader yet?

Apologies if this is a duplicate -- I'm having some news problems.

In sci.physics Koobee Wublee wrote:
On Feb 25, 7:47 am, Tom Roberts wrote:

The best model we have for the propagation of light near a massive
object like the sun is GR, in which the curvature of spacetime is the
important aspect in determining the path light follows. And it agrees
with measurements to part-per-million accuracy over an enormous range.

First, derive a set of geodesic equations a massed particle traveling
at high speed near the sun. Then, gradually reducing the mass to zero
and increasing the speed to c, do you see a discontinuity at mass = 0
and speed = c?

This is definitely a worthwhile exercise. I recommend that you do it.
If you get stuck, you can find the details in Lightman et al., _Problem
book in relativity and gravitation_, problem 15.9.

As you know, the geodesic equations are independent of mass. What
does that tell you when the model predicts a 1x deflection traveling
at speed just a hair below c and suddenly jumps to 2x deflection at
speed = c?

It doesn't. The model predicts a deflection proportional to 1+v^2/c^2,
which varies smoothly from the "Newtonian" value of 1 for small velocities
to 2 as v approaches c.

The moral is that before you decide that a model doesn't make sense,
you should check what the model actually predicts.

Steve Carlip

On Mar 3, 4:53 pm, wrote:

First, derive a set of geodesic equations a massed particle traveling
at high speed near the sun. Then, gradually reducing the mass to zero
and increasing the speed to c, do you see a discontinuity at mass = 0
and speed = c?

This is definitely a worthwhile exercise. I recommend that you do it.

Yes, indeed.

If you get stuck, you can find the details in Lightman et al., _Problem
book in relativity and gravitation_, problem 15.9.

There is no need to get angry. We just have to reason all these out.

As you know, the geodesic equations are independent of mass. What
does that tell you when the model predicts a 1x deflection traveling
at speed just a hair below c and suddenly jumps to 2x deflection at
speed = c?

It doesn't. The model predicts a deflection proportional to 1+v^2/c^2,
which varies smoothly from the "Newtonian" value of 1 for small velocities
to 2 as v approaches c.

The moral is that before you decide that a model doesn't make sense,
you should check what the model actually predicts.

Starting with a segment in spacetime, we have the following.

ds^2 = c^2 (1 - 2 U) dt^2 - dr^2 / (1 - 2 U) - r^2 dO^2

Where

** U = G M / c^2 / r
** R = Radius of the sun

The photon also propagates through space or spacetime with an
accumulation in spacetime of exactly zero. Thus the geodesic model
that dictates the path through spacetime is the one that accumulates
the least amount of spacetime is indeed absurd because it would never
allow a coherent trajectory for light. Thus, this derivation falls
back to the Fermat's principle where the geodesic path follows the
path with the least accumulated amount of time. Yes, this violates
the essence of SR which dictates relative simultaneity. Nevertheless,
this model of geodesics does allow a photon deflection with the photon
having a coherent path through out its course of being deflected by
the sun.

In doing so, we can easily find the Lagrangian that satisfy the
minimum elapsed time to be the following.

L c^2 = (ds/dt)^2 / (1 - 2 U) + (dr/dt)^2 / (1 - 2 U)^2 + r^2 (dO/
dt)^2 / (1 - 2 U) = c^2

Where

** L = 1, the Lagrangian
** T = Integral(t1, t2)[L dt], the action of accumulating time

We find the following Euler-Lagrange equations that indicate conserved
quantities.

** ds/dt = K (1 - 2 U)
** dO/dt = H R c (1 - 2 U) / r^2

Where

** K, H R c = Integration constants

So, the Lagrangian becomes the following.

K^2 (1 - 2 U) + (dr/dt)^2 / c^2 / (1 - 2 U)^2 + H^2 R^2 (1 - 2 U) /
r^2 = 1

Presenting the above equations and the equation describing the
conservation of angular momentum side by side, we have the following.

** r^4 (dO/dt)^2 / c^2 R^2 = H^2 (1 - 2 U)^2
** (dr/dt)^2 / c^2 = (1 - 2 U)^2 (1 - K^2 + 2 K^2 U - H^2 R^2 / r^2 +
2 H^2 R^2 U / r^2


Combining the above equations together, we have the follwing.

r^4 (dO/dr)^2 / r^2 = H^2 / (1 - K^2 + 2 K^2 U - H^2 R^2 / r^2 + 2 H^2
R^2 U / r^2)

Or

r^4 (dO/dr)^2 / r^2 = H^2 / (1 - K^2 + 2 u K^2 R / r - H^2 R^2 / r^2 +
2 u H^2 R^3 / r^3)

Where

** U = u R / r
** u = G M / c^2 / R

Applying the boundary condition, we find the following.

K^2 = 1 - B^2

Where

** B = Speed of the particle at (r = infinity)

Taking the sun away hypothetically, at the perihelion, the speed
remains B. Thus, we find the following.

H^2 = B^2

Then, the trajectory equation becomes the following.

r^4 (dO/dr)^2 / r^2 = B^2 / (B^2 + 2 u (1 - B^2) R / r - B^2 R^2 / r^2
+ 2 u B^2 R^3 / r^3)

At the perihelion, (dr/dt = 0), and we find the distance that is
deflected into the sun by the following amount.

dR / R ~= u / (B^2 + 4 u (1 - B^2)) ~= u / B^2

Where

** dR = Deflected distance into the sun
** B^2 u

Define the following.

p = R / r - u / B^2

The trajectory equation simplifies to the following.

(dO/dp)^2 ~= 1 / (1 - 2 u p - p^2 + 2 u p^3)

Or

dO/dp ~= 1 / sqrt((1 - 2 u p) (1 - p^2)) ~= (1 + u p) / sqrt(1 - p^2)

Twice the integral with proper integration boundaries, we have the
following.

2 O = 2 integral(-u / B^2, 1)[dp (1 + u p) / sqrt(1 - p^2)]

Or

2 O ~= 2 (sin^-1(1) - sin^-1(-u / B^2) + u sqrt(1 - u^2 / B^2))

Or
2 O ~= pi + 2 u (1 + 1 / B^2)

If (B^2 = 1), we have a deflection of (4 u) which is twice the
Newtonian result. However, at the speed just below 1, the deflection
angle is higher than twice the Newtonian result. I don't know where
the maximum deflection is as predicted by GR with the geodesics
obeying Fermat's principle. Perhaps, your students with lots of time
at hand are willing to sort through the second or third order effects
to find it.

So, after doing the mathematics as you have wisely recommended me to
do, I have to admit that I do not find this discontinuity I was
expecting. However, the absurdity in GR is in deflection more than
twice the Newtonian result at slightly lower speed than the speed of
light.

On Feb 29, 8:04*pm, Igor wrote:
On Feb 29, 2:35*am, "Y.Porat" wrote:





On Feb 28, 11:19*am, "Artful" wrote:

"Y.Porat" wrote in message

....

so thje formula shoul not be

E=gamma m * times c^2

but ....

E/Gamma = mc^2 !!

That is the same formula .. just written slightly differently.

AND m REMAINS * * COOOOOOOONSTANT * !! *(:-)

It is constant 9assuming you are talking about the one object) in both
formulas. *'m' means mass (rest mass or invariant mass) .. it doesn't
change. *rewriting the formula does not make any difference whatsoever ..
except that the formula you propose is less useful in that neither side
directly gives you a meaningful value (ie most useful formulas have a single
symbol on one side, and an expression on the other side that calculates the
value for that *formula.

----------------
thas exactly the point !!

*IT IS THE SAME FORMULA WITH A BASICALLY
DIFFERENT PHYSICAL UNDERSTANDING!!

And you're physical understanding is certainly different.- Hide quoted text -

- Show quoted text -

---------------
yess indeed
and some people already told me that
i am an orriginal thinker

now just wtite it before you
who told you that for the first time !!
and please dont forget .....

Y.Porat
------------------------------------

On Mar 3, 8:31*pm, Igor wrote:
On Mar 2, 12:53*am, Koobee Wublee wrote:





On Feb 25, 7:47 am, Tom Roberts wrote:

The best model we have for the propagation of light near a massive
object like the sun is GR, in which the curvature of spacetime is the
important aspect in determining the path light follows. And it agrees
with measurements to part-per-million accuracy over an enormous range.

First, derive a set of geodesic equations a massed particle traveling
at high speed near the sun. *Then, gradually reducing the mass to zero
and increasing the speed to c, do you see a discontinuity at mass = 0
and speed = c?

As you know, the geodesic equations are independent of mass. *What
does that tell you when the model predicts a 1x deflection traveling
at speed just a hair below c and suddenly jumps to 2x deflection at
speed = c?

It tells us that they are completely exclusive cases.
exclusive cases ?? (:-)
nice we are doing some progress!!

now if exclusive...
why not doing a little further step:

if the photon is an exclusive case
why should it not be the only mass that can move at c ??

(:-)

and you see as well that while masses are being s,aller
they can reach CLOSER AND CLOSER TO c !!
so
why not at a limit case not to reach exactly c ??

and i still didnt mension that
in E photon = hf(it id
6.6 x 10- ..... times 1 kilogram times 1 meter ^2/1 second !!
so if my eyesight is not wronn i see
even through my eyeglasses the KILOGRAM there

do you have any idea what is that **Kg* doing there?
do you have an idea how a physics formula is built
and used ???....

that h is is hiding something
for peoe who didnt use their own mind
even theoretically ........
now again
dont please forget who told you that the first time

ATB
Y.Porat
-------------------------


*But I do
understand that YOU can't go there. *Hell, you're still working on
transforming algebraic domains. *By the way, how's that coming along?
Made any real progress lately? *Have reached the level of an eighth
grader yet?- Hide quoted text -

- Show quoted text -

On Mar 4, 2:53*am, wrote:
Apologies if this is a duplicate -- I'm having some news problems.

In sci.physics Koobee Wublee wrote:

On Feb 25, 7:47 am, Tom Roberts wrote:
The best model we have for the propagation of light near a massive
object like the sun is GR, in which the curvature of spacetime is the
important aspect in determining the path light follows. And it agrees
with measurements to part-per-million accuracy over an enormous range.
First, derive a set of geodesic equations a massed particle traveling
at high speed near the sun. *Then, gradually reducing the mass to zero
and increasing the speed to c, do you see a discontinuity at mass = 0
and speed = c?

This is definitely a worthwhile exercise. *I recommend that you do it.
If you get stuck, you can find the details in Lightman et al., _Problem
book in relativity and gravitation_, problem 15.9.

As you know, the geodesic equations are independent of mass. *What
does that tell you when the model predicts a 1x deflection traveling
at speed just a hair below c and suddenly jumps to 2x deflection at
speed = c?

It doesn't. *The model predicts a deflection proportional to 1+v^2/c^2,
which varies smoothly from the "Newtonian" value of 1 for small velocities
to 2 as v approaches c.

The moral is that before you decide that a model doesn't make sense,
you should check what the model actually predicts.

Steve Carlip

------------------
lie!!
the model ddint predict
it was **fiddling** the data to the model

Y.Porat
-----------------------------