Gravitational lensing derivation using equivalence principle

[grav-universe]

Here is the conservation of energy. The locally measured energy of a photon just depends upon the local time t, since the observed frequency of a photon depends only upon the local gravitational time dilation. If a hovering observer emits a photon, then another photon a time t later, the first photon will follow some path, curved or otherwise, to some other point in the gravitational field, and the second will follow exactly the same path and arrive at the same point a time t later, so the frequency that the photons pass any point remains the same to the observer, but a observer at the point where they arrived will measure a different frequency that depends only upon his own rate of time. So we'll have

\frac{f_q}{f_p} = \frac{dt_p}{dt_q} = \frac{\sqrt{1 - 2m/p}}{\sqrt{1-2m/q}}

for a photon traveling from a radial distance p to a radial distance q in the field, although the path can be curved and non-radial. So there is also a dependence upon r. Since f_p and f_q are directly proportional to the locally measured energy of the photon, we have

\frac{E_q}{E_p} = \frac{\sqrt{1 - 2m/p}}{\sqrt{1 - 2m/q}}

Since this ratio of energy holds for light, we can try it for massive particles as well and see how that holds up. If we just take this ratio of energies directly for massive particles, we get

\frac{E_q}{E_p} = \frac{\sqrt{1 - 2m/p}}{\sqrt{1 - 2m/q}}

\frac{(m \ c^2 / \sqrt{1 - (v_q'/c)^2})}{(m \ c^2 / \sqrt{1 - (v_p'/c)^2})} = \frac{\sqrt{1 - 2m/p}}{\sqrt{1 - 2m/q}}

\frac{(1 - (v_p'/c)^2)}{(1 - 2m/p)} = \frac{(1 - (v_q'/c)^2)}{(1 - 2m/q)} = K (constant)

K = 0 for a photon. Of course, that must also reduce to Newtonian gravity, so let's find that for a massive body traveling from p to q = p - dp. The scalar speeds for the energy are locally measured while the distances are measured by a distance observer, whereas dp = p - q and dp' = (p - q) / \sqrt{1 - 2m/p}.

(1-(v_q'/c)^2) = \frac{(1-2m/q) (1-(v_p'/c)^2)}{(1-2m/p)}

(v_q'/c)^2 = 1 - \frac{(1-2m/q) (1-(v_p'/c)^2)}{(1-2m/p)}

(v_q'/c)^2 - (v_p'/c)^2 = 1 - (v_p'/c)^2 - \frac{(1-2m/q) (1-(v_p'/c)^2)}{(1-2m/p)}

(v_q'/c)^2 - (v_p'/c)^2 = \frac{(1 - (v_p'/c)^2)}{(1-2m/p)} [(1 - 2m/p) - (1 - 2m/q)]

(v_q'/c)^2 - (v_p'/c)^2 = \frac{(1 - (v_p'/c)^2)}{(1-2m/p)} \left[\frac{2 m (p - q)}{p \ q}\right]

a' = \frac{d(v'^2)}{2 dp'} = \frac{[(v_q'/c)^2 - (v_p'/c)^2] \sqrt{1 - 2m/p}}{2 \ (p-q)} = \frac{m \ (1 - (v_p'/c)^2)}{p \ q \ \sqrt{1 - 2 m/p}}

and dropping infinitesimals at this point, the locally measured acceleration at r = p with instantaneous speed v' is

a' = \frac{(1 - (v'/c)^2) (G M / r^2)}{\sqrt{1 - 2m/r}}

We can see that for a body falling from rest at r with v' = 0, the acceleration reduces to Newtonian to first order. Okay, but this is an acceleration meant for a scalar speed. For instance, according to the equation, the acceleration of a photon with v' = c is zero, so only its direction changes, but not its scalar speed. So we'll want to divide that up into its radial and tangent components. We can use the equivalence principle for that. If a rod freefalls at r with original radial speed v_r', matching the radial component of a particle that also freefalls with some tangent speed, then in order for the equivalence principle to hold, the radial acceleration of both must match over an infinitesimal time of freefall. With no tangent component, the radial speed of the rod is its scalar speed, so for the rod we have just

a'_{rod} = \frac{(1 - (v'/c)^2) (G M / r^2)}{\sqrt{1 - 2m/r}} = \frac{(1 - (v_r'/c)^2) (G M / r^2)}{\sqrt{1 - 2m/r}}

The radial component of the particle must match that, so we have

a'_r = \frac{(1 - (v_r'/c)^2) (G M / r^2)}{\sqrt{1 - 2m/r}}

There should be no tangent acceleration, so the apparent tangent component of the acceleration comes from the tidal gradient I think. For instance, for a circular orbit, as a particle passes the x-axis at a vertical distance y, its tangent speed begins to drop while its radial speed increases, although the scalar speed stays the same. Technically, then, we would really be extending the accelerations over some distances in the x and y directions instead of radially and tangent, but the apparent acceleration along infinitesimal y should still match the instantaneous radial acceleration if gravity only acts in that direction. So then, from that equation, a photon traveling radially will not accelerate at all, always measured at c locally in the radial direction, while a photon traveling in a circular orbit will have zero radial speed, so reduces to just

a'_r = \frac{(G M / r^2)}{\sqrt{1 - 2m/r}} = \frac{c^2}{r'}

\frac{(m / r^2)}{\sqrt{1 - 2 m / r}} = \frac{\sqrt{1 - 2 m / r}}{r}

(m / r) = 1 - 2 m / r

3 m / r = 1

r = 3 m

giving r = 3m for a photon traveling in a perfectly circular orbit. As far as photons go, as a photon falls toward a body, it spirals inward. If it doesn't pass a tangent point, it will fall all the way in. If it passes a tangent point, however, then it will spiral back out and eventually escape. It cannot turn around to spiral back in again because that would require that it passes another tangent point at a greater radius than the first, so angular momentum wouldn't be conserved. So other than a perfectly circular orbit at r = 3m, which would require that it be "born" there, and so precise with no disturbances as to make it virtually impossible, these are really the only two options for photons.

 

[grav-universe]

\frac{(1 - (v_p'/c)^2)}{(1 - 2m/p)} = \frac{(1 - (v_q'/c)^2)}{(1 - 2m/q)} = K (constant)

I introduced the constant K for energy conservation because I want to use it to show that ds = c dτ for the proper time of a particle or freefaller, as I wasn't sure about that earlier in the thread. For a freefaller passing a static observer at r with a scalar speed of v' in any direction, the locally measured time dilation of the freefaller's clock is sqrt(1 - (v'/c)^2). Since there are no simultaneity differences between static observers, only a gravitational time dilation between them, the time dilation that the distant observer measures of the freefaller's clock becomes dτ / dt' = sqrt(1 - (v'/c)^2) slower than for the local static observer that coincides in the same place, with the static observer's clock ticking dt' / dt = sqrt(1 - 2 m / r) slower than the distant observer's own clock, so the distant observer measure's a time dilation for the freefaller of (dτ / dt) = (dτ / dt') (dt' / dt) = sqrt(1 - (v'/c)^2) sqrt(1 - 2 m / r). Okay, so we have

K = (1 - (v'/c)^2) / (1 - 2 m / r)

1 - (v'/c)^2 = K (1 - 2 m / r)

v'^2 = c^2 [1 - K (1 - 2 m / r)]

and the metric becomes

v'^2 = v_r'^2 + v_t'^2

v'^2 = (dr' / dt')^2 + (dθ' r'/ dt')^2

v'^2 dt'^2 = dr'^2 + (dθ' r')^2

v'^2 dt^2 (1 - 2 m / r) - dr^2 / (1 - 2 m / r) - dθ^2 r^2 = 0

and substituting for v'^2,

c^2 [1 - K (1 - 2 m / r)] dt^2 (1 - 2 m / r) - dr^2 / (1 - 2 m / r) - dθ^2 r^2 = 0

c^2 dt^2 (1 - 2 m / r) - dr^2 / (1 - 2 m / r) - dθ^2 r^2 = c^2 dt^2 K (1 - 2 m / r)^2

ds^2 = c^2 dt^2 K (1 - 2 m / r)^2

ds^2 = c^2 dt^2 [(1 - (v'/c)^2) / (1 - 2 m / r)] (1 - 2 m / r)^2

ds^2 = c^2 dt^2 (1 - (v'/c)^2) (1 - 2 m / r)

ds = c [dt sqrt(1 - (v'/c)^2 sqrt(1 - 2 m / r)]

ds = c dτ

 

[grav-universe]

Okay, now I want to work out what the proper time of the freefaller will be upon reaching a radius h when falling from rest from a radius b with an original clock reading of zero. So we have

v'^2 = v_r'^2 + v_t'^2

where v'^2 = c^2 [1 - K (1 - 2 m / r)] and v_t' = 0 for a particle falling radially, giving

c^2 [1 - K (1 - 2 m / r)] = (dr' / dt')^2

c^2 [1 - K (1 - 2 m / r)] \left(\frac{d\tau}{\sqrt{1 - (v'/c)^2}}\right)^2 = \left(\frac{dr}{\sqrt{1 - 2 m / r}}\right)^2

c^2 [1 - K (1 - 2 m / r)] \ d\tau^2 = K \ dr^2

d\tau^2 = \frac{dr^2}{c^2 [\frac{1}{K} - (1 - 2 m / r)]}

where

K = \frac{(1 - (v_b/c)^2)}{(1 - 2 m / b)} for a particle falling from rest at b, where also v_b = 0 in that case, so

K = \frac{1}{(1 - 2 m / b)}

d\tau^2 = \frac{dr^2}{c^2 [(1 - 2 m / b) - (1 - 2 m / r)]}

\tau = \ _h\int^b \frac{dr \sqrt{r}}{c \ \sqrt{2 m} \sqrt{1 - \frac{r}{b}}}

\tau = \left(\frac{b}{c}\right)\left[\sqrt{\frac{h}{2 m}} \sqrt{1 - h / b} + \sqrt{\frac{b}{2 m}} \tan^{-1} (\sqrt{b / h - 1}) \right]

When falling to h = 2 m where b >> 2m, the atan approaches π / 2 and the first term in brackets helps to make up the difference, so we can approximate (the same as using h = 0) with

\tau \approx \left(\frac{b}{c}\right) \sqrt{\frac{b}{2 m}} \left(\frac{\pi}{2}\right)

 

[grav-universe]

Geesh. All I really needed to show that ds = c dτ, without any reference to energy, is

v'^2 = v_r'^2 + v_r'^2

for the locally measured radial and tangent speed of a freefaller with scalar speed v', which transforms to

v'^2 = (dr' / dt')^2 + (dθ' r' / dt')^2

v'^2 dt'^2 = dr'^2 + dθ'^2 r'^2

c^2 dt'^2 - [v'^2 dt'^2] = c^2 dt'^2 - [dr'^2 + dθ'^2 r'^2]

c^2 (1 - (v'/c)^2) dt'^2 = c^2 dt'^2 - dr'^2 - dθ'^2 r'^2

where dτ = sqrt(1 - (v'/c)^2) dt' as the clock of the freefaller is observed locally, so

c^2 dτ^2 = c^2 dt'^2 - dr'^2 - dθ'^2 r'^2

ds^2 = c^2 dτ^2 = c^2 dt'^2 - dr'^2 - dθ'^2 r'^2 = c^2 dt^2 (1 - 2 m / r) - dr^2 / (1 - 2 m / r) - dθ^2 r^2