sci.physics.particle

On Mar 3, 4:53 pm, carlip-nos...@physics.ucdavis.edu 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.