sci.physics.particle

On Mar 4, 12:32 am, Koobee Wublee wrote:
> On Mar 3, 4:53 pm, carlip-nos...@ 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.