On Feb 25, 4:41 am, "harry"
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?

Einstein explained it as follows:

The light nearer to the sun (or the earth) propagates slower than the light
farther away

------ light ray velocity vectors
-----

X sun

Using the Huygens construction,http://en.wikipedia.org/wiki/Huygens...snel_principle
we can find the wave fronts as follows:

¦------/
¦-----/
¦ /
wave fronts

X

The light propagates perpendicular to the wave fronts; consequently, the
light must bend.

Interesting. It contradicts my toy notions of gravity as being a
series of spatial "wells" around which I would have naively expected
the shorter distance (and time) to be along a path nearer the massive
object. Maybe I should be thinking of space-near-mass as a higher
density of "ether' with a refractive index? Maybe those gravitons are
rest stops?

--
David Winsemius

On Mar 4, 12:32 am, Koobee Wublee wrote:
On Mar 3, 4:53 pm, wrote:

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.

I have seemed misinterpreted the mathematics. At low speeds, the
particle would be deflected right into the sun. At high speeds, the
deflection should be a lot more with twice the Newtonian deflection at
the speed of light. Light at the speed of '1' represents the lowest
deflected angle. Lightman's result is wrong.

In sci.physics Koobee Wublee wrote:
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.

[...]
2 O ~= pi + 2 u (1 + 1 / B^2)

This is correct.

If (B^2 = 1), we have a deflection of (4 u) which is twice the
Newtonian result.

That's right.

However, at the speed just below 1, the deflection
angle is higher than twice the Newtonian result.

No. Now calculate the *Newtonian* prediction. You will find that,
in your notation

2 O ~= pi + 2u/B^2

(If you get stuck on this, look at chapter 3 of Goldstein, _Classical
Mechanics_.) The GR prediction is thus, as I said, (1+B^2) times the
Newtonian one. For speeds just below c, the GR deflection is just
below twice the Newtonian deflection; as v becomes small, the GR
deflection approaches the Newtonian value.

Steve Carlip

On Mar 2, 8:12 am, Koobee Wublee wrote:
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.

Why do you reference me to textbooks you disagree with? Be consistent
in your whining.

On Mar 4, 9:59 am, wrote:
In sci.physics Koobee Wublee wrote:

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

This is correct.

But according to you, Lightman gave a result below.

2 O = pi + 2 u (1 + B^2)

If (B^2 = 1), we have a deflection of (4 u) which is twice the
Newtonian result.

That's right.

However, at the speed just below 1, the deflection
angle is higher than twice the Newtonian result.

No. Now calculate the *Newtonian* prediction. You will find that,
in your notation

2 O ~= pi + 2u/B^2

At (B = 0.99), (1 + 1 / B^2 = 2.02) which results

2 O = pi + 4.04 u

The deflected amount at lower speed of (B = 0.99) is higher than
light, and this makes sense. Deflected angle of light represents the
lowest angle of deflection.

(If you get stuck on this, look at chapter 3 of Goldstein, _Classical
Mechanics_.) The GR prediction is thus, as I said, (1+B^2) times the
Newtonian one. For speeds just below c, the GR deflection is just
below twice the Newtonian deflection; as v becomes small, the GR
deflection approaches the Newtonian value.

If Lightman is wrong (not your typo), how about Goldstein and others?

In sci.physics Koobee Wublee wrote:
On Mar 4, 9:59 am, wrote:
In sci.physics Koobee Wublee wrote:

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

This is correct.

But according to you, Lightman gave a result below.

2 O = pi + 2 u (1 + B^2)

I did not say that. I said that the result was proportional to
1+B^2. It is -- the GR prediction is 2u/B^2 (1+B^2), that is,
1+B^2 times the Newtonian prediction for a given speed and impact
parameter.

I suggest that you try for a graceful retreat.

Steve Carlip

You wrote:
“ Anything which has inertial mass has energy
and anything which has energy has energy has inertial mass. �.

Very beautifully put, thanks.

On another topic... Do you think the following is true ? why ?

This “ Inverse Gaussian Probability density function �
( describing the distribution of
“ the time a Brownian Motion with positive drift takes
to reach a fixed positive level � ):
http://upload.wikimedia.org/wikipedi...F_invGauss.png

Looks a lot like this “ Black body spectrum �:
http://upload.wikimedia.org/wikipedi...ns_law.svg.png

On Mar 4, 6:03 pm, wrote:
In sci.physics Koobee Wublee wrote:

On Mar 4, 9:59 am, wrote:
In sci.physics Koobee Wublee wrote:
2 O ~= pi + 2 u (1 + 1 / B^2)

This is correct.
But according to you, Lightman gave a result below.
2 O = pi + 2 u (1 + B^2)

I did not say that. I said that the result was proportional to
1+B^2. It is -- the GR prediction is 2u/B^2 (1+B^2), that is,
1+B^2 times the Newtonian prediction for a given speed and impact
parameter.

You were very misleading then. You left out a very important factor,
1 / B^2.

I suggest that you try for a graceful retreat.

I did say the result was not what I initially expected. What else do
you want me to do?

However, I need no graceful retreat from the comment on the absurdity
of the geodesics following the path with the least accumulated
spacetime. When Christoffel derived the geodesic equations, there was
nothing wrong with his assumption that the shortest distance between
two points in 3-dimensional space is the actual distance through each
local point even if that space is curved. It is not necessarily a
straight line as an observer observes it using his choice of
coordinate system. However, when the Goettingen group of
mathematicians including Hilbert, Klein, Schwarzschild, and Minkowski
worked on this problem, they simply extended to the 4-dimensional
spacetime. In doing so, no one really thought out except 100 years
later by yours truly that this model does not allow photons to
propagate with a coherent path. Are you still standing by your
conjecture that this model of geodesic is a valid one?

Koobee Wublee wrote on Tue, 04 Mar 2008 22:32:43 -0800:

On Mar 4, 6:03 pm, wrote:

However, I need no graceful retreat from the comment on the absurdity of
the geodesics following the path with the least accumulated spacetime.
When Christoffel derived the geodesic equations, there was nothing wrong
with his assumption that the shortest distance between two points in
3-dimensional space is the actual distance through each local point even
if that space is curved. It is not necessarily a straight line as an
observer observes it using his choice of coordinate system. However,
when the Goettingen group of mathematicians including Hilbert, Klein,
Schwarzschild, and Minkowski worked on this problem, they simply
extended to the 4-dimensional spacetime. In doing so, no one really
thought out except 100 years later by yours truly that this model does
not allow photons to propagate with a coherent path. Are you still
standing by your conjecture that this model of geodesic is a valid one?

There four main complaints about geodesic models:

i)
It have been not empirically verified, really. Same 'tests' of GR are
verified by non-geometrical theories where gravity is a force

In those 'tests', the beatiful artistic works

http://bp2.blogger.com/_I-n4UWp0ZqM/...U/EdXg-C5_-Fo/
s1600-h/curvedspacetime.gif

are empirically indistinguisable from alternative models

http://bp3.blogger.com/_I-n4UWp0ZqM/...c/joqIPrW5O-k/
s1600-h/flatspacetime.gif

ii)
Geodesic model is limited in several ways: e.g. geodesic motion has not
correct Newtonian limit (GR literature no Newtonian limits is completely
incorrect at this point); no complete N-body solution is geodesic, no
quantization...

iii)
Geodesic motion introduces further difficulties: conservation, unphysical
boundaries, systems of reference...

iv)
It breaks unification with rest of forces.


--
I apply http://canonicalscience.org/en/misce...guidelines.txt

On Mar 5, 3:17 am, "Juan R." González-Álvarez
wrote:
Koobee Wublee wrote on Tue, 04 Mar 2008 22:32:43 -0800:

On Mar 4, 6:03 pm, wrote:
However, I need no graceful retreat from the comment on the absurdity of
the geodesics following the path with the least accumulated spacetime.
When Christoffel derived the geodesic equations, there was nothing wrong
with his assumption that the shortest distance between two points in
3-dimensional space is the actual distance through each local point even
if that space is curved. It is not necessarily a straight line as an
observer observes it using his choice of coordinate system. However,
when the Goettingen group of mathematicians including Hilbert, Klein,
Schwarzschild, and Minkowski worked on this problem, they simply
extended to the 4-dimensional spacetime. In doing so, no one really
thought out except 100 years later by yours truly that this model does
not allow photons to propagate with a coherent path. Are you still
standing by your conjecture that this model of geodesic is a valid one?

There four main complaints about geodesic models:

i)
It have been not empirically verified, really. Same 'tests' of GR are
verified by non-geometrical theories where gravity is a force

In those 'tests', the beatiful artistic works

http://bp2.blogger.com/_I-n4UWp0ZqM/...U/EdXg-C5_-Fo/
s1600-h/curvedspacetime.gif

are empirically indistinguisable from alternative models

http://bp3.blogger.com/_I-n4UWp0ZqM/...c/joqIPrW5O-k/
s1600-h/flatspacetime.gif

Name one competing alternative model to GR that does not include GR as
a subset which survives all experimental observation. In your
response, try to use something other than argument-by-picture.


ii)
Geodesic model is limited in several ways: e.g. geodesic motion has not
correct Newtonian limit (GR literature no Newtonian limits is completely
incorrect at this point); no complete N-body solution is geodesic, no
quantization...

Liar. You know this to be wrong, yet you say it anyway.


iii)
Geodesic motion introduces further difficulties: conservation, unphysical
boundaries, systems of reference...

Yes, more things to confuse poor Juan R.


iv)
It breaks unification with rest of forces.

Not even close to relevant.


--
I applyhttp://canonicalscience.org/en/miscellaneouszone/guidelines.txt