## gravitational lensing derivation using equivalence principle

 Quote by grav-universe I have been trying to work this out for the last couple of weeks, but I just keep getting the Newtonian deviation in angle for a path of a photon travelling from x=-∞ to x=∞. [..] Locally I am applying the equivalence principle, using the relativistic acceleration formulas to find the change in radial speed for a freefaller initially falling at the same coordinate speed vr1 as the photon, radially only with no tangent speed. [..] But this hasn't been working out so far, giving only the Newtonian change in angle instead of twice that value which GR predicts. Could someone please show me how to derive the GR value using the equivalence principle in this way? [..]
I hope to find time this evening to check it in a GR book* at home, which, I think, describes how to do it. Including length contraction and time dilation should suffice to obtain the light bending of GR directly from SR together with the equivalence principle - but I never actually calculated it myself (lazy me - I always postpone doing what you now are doing!).

*Adler et al, Introduction to General Relativity

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 Quote by Bill_K A null geodesic in Schwarzschild is given by 0 = (1 - 2m/r)-1(dr/ds)2 + r2(df/ds)2 - (1 - 2m/r)(dt/ds)2
 Quote by JDoolin Is this a null geodesic? I think this is a more general case, here. The null geodesic is a much simpler problem; you can set $c d\tau=0$. Here, you are letting the ds vary, which means its applicable to any path.
On further reflection, I realized that because of the zero on the left-hand-side, this does have to be a null-geodesic after all.

You guys don't know how much time I spent trying to work through the metric using the isotropic coordinates and then integrating it, only to get a bunch of complex results and elliptical integrals, before realizing I could just transform the radiuses in the original program lol. I've been working on some more stuff and it's starting to get involved, so for future reference, I want to go ahead and post what I've found so far. A while back in another forum, someone asked what the coordinate speed of light would be that a distant observer would measure depending upon the angle to the gravitating body, so I posted this:

 Well, let's see. Let's say the x axis lies along the radial direction and a hovering observer emits a pulse of light locally at c along the angle θ from the x axis toward the mass. The pulse travels a very short distance so that we can ignore the gravity gradient and our measurements remain local. According to the local observer, then, the pulse travels a distance d in a time t, ending up at coordinates x = (cos θ) d = (cos θ) c t, y = (sin θ) d = (sin θ) c t, y being revolved to include the z axis. To the distant observer using regular Schwarzschild coordinates, the distance the pulse travels along y is the same, the distance travelled along x is contracted to x' = sqrt(1 - r_s / r) x, the time that passes is t' = t / sqrt(1 - r_s / r), and there is no simultaneity difference, where r is the distance of the hovering observer from the mass according to the distant observer. So the distant observer measures a speed for the light pulse at r in the gravitational field using the locally measured angle θ of c' = d' / t' = sqrt[(1 - r_s / r) x^2 + y^2] / [t / sqrt(1 - r_s / r)] = sqrt[(1 - r_s / r) (cos θ)^2 c t + (sin θ)^2 c t] / [t / sqrt(1 - r_s / r)] = c sqrt[(1 - r_s / r) (cos θ)^2 + (sin θ)^2] sqrt(1 - r_s / r) = c sqrt[(1 - r_s / r) (cos θ)^2 + 1 - (cos θ)^2] sqrt(1 - r_s / r) = c sqrt[1 - (cos θ)^2 (r_s / r)] sqrt(1 - r_s / r) The angle that the pulse travels according to the distant observer is cos θ' = x' / d' = sqrt(1 - r_s / r) (cos θ) / sqrt[(1 - r_s / r) (cos θ)^2 + (sin θ)^2] = sqrt(1 - r_s / r) (cos θ) / sqrt[1 - (cos θ)^2 (r_s / r)] Solving for θ, we get [1 - (cos θ)^2 (r_s / r)] (cos θ')^2 = (1 - r_s / r) (cos θ)^2 (cos θ)^2 [1 - r_s / r + (cos θ')^2 (r_s / r)] = (cos θ')^2 (cos θ) = (cos θ') / sqrt[1 - r_s / r + (cos θ')^2 (r_s / r)] And substituting that into the equation to gain the coordinate speed c' in terms of the angle θ' the distant observer measures, we have c' = c sqrt(1 - (cos θ')^2 (r_s / r) / [1 - r_s / r + (cos θ')^2 (r_s / r)]) sqrt(1 - r_s / r) = c sqrt([1 - r_s / r + (cos θ')^2 (r_s / r)] - (cos θ')^2 (r_s / r)) sqrt(1 - r_s / r) / sqrt[1 - r_s / r + (cos θ')^2 (r_s / r)] = c (1 - r_s / r) / sqrt[1 - r_s / r + (cos θ')^2 (r_s / r)] = c (1 - r_s / r) / sqrt[1 - (sin θ')^2 (r_s / r)]
I'm glad I already had that worked out. :) Okay, so below is an image of the path a photon will travel in relation to the angle and radius. Over infinitesimal amounts of time and distance, the path can be considered straight for the derivation, but of course it will curve over a larger finite time and distance. I already used θ (now used for what the distant observer measures) for the angle of travel of the photon from the radial direction, so I will use dφ for the infinitesimal change in angle of the radius (although greatly exaggerated in the image). Following the image, one can see that that

((r1 + dy)^2 + dx^2 = r2^2

(r1 + (cos θ) c' dt)^2 + ((sin θ) c' dt)^2 = r2^2

r1^2 + 2 (cos θ) c' dt r1 + (cos θ)^2 c'^2 dt^2 + (sin θ)^2 c'^2 dt^2 = r2^2

where (cos θ)^2 + (sin θ)^2 = 1 so

r2 = r1 sqrt(1 + 2 (cos θ) c' dt / r1 + c'^2 dt^2 / r1^2)

and expanding the square root and taking only first order infinitesimals,

r2 = r1 (1 + (cos θ) c' dt / r1)

dr = r2 - r1 = (cos θ) c' dt

(dr / dt) = (cos θ) c'

So we now have dr / dt. We can also find

(sin dφ) r2 = dx = (sin θ) c' dt

where (sin dφ) = dφ for first order infinitesimal dφ, so

dφ r2 = (sin θ) c' dt

(dφ / dt) = (sin θ) c' / r

using r since we are not finding for the difference between r1 and r2 and the difference is infinitesimal, so r = r1 = r2 here. We now have dφ / dt also, but both values we have found for are expressed in terms of θ and c', so we will need to get rid of those.

From those two values, we have

(dr / dt)^2 = (cos θ)^2 c'^2

(dr / dt)^2 = c'^2 - (sin θ)^2 c'^2

(dr / dt)^2 = c'^2 - (dφ / dt)^2 r^2

c'^2 = (dr / dt)^2 + (dφ / dt)^2 r^2

From the quote, we get

c'^2 = c^2 (1 - 2 m / r)^2 / (1 - (sin θ)^2 (2 m / r))

c'^2 - (sin θ)^2 (2 m / r) c'^2 = c^2 (1 - 2 m / r)^2

and substituting, we get

[(dr / dt)^2 + (dφ / dt)^2 r^2] - [(dφ / dt)^2 r^2] (2 m / r) = c^2 (1 - 2 m / r)^2

(dr / dt)^2 + (1 - 2 m / r) (dφ / dt)^2 r^2 = c^2 (1 - 2 m / r)^2

c^2 dt^2 (1 - 2 m / r) - dr^2 / (1 - 2 m / r) - dφ^2 r^2 = 0

which gives us the metric.

Here's something more. If we replace the speed of light with the speed of a massive particle, the coordinate speed v' will be found in exactly the same way as c' was, just replacing c with v_loc. The same thing goes for the rest of the post. But of course c and v_loc are the locally measured scalar speeds. So the metric then becomes

v_loc^2 dt^2 (1 - 2 m / r) - dr^2 / (1 - 2 m / r) - dφ^2 r^2 = 0

Used in this way, the metric is always null, even for speeds less than c, at least according to the distant observer. If we substitute the function for v_loc into the metric, then we can solve the metric using that. Does anybody know what that would be, if v_loc is given for a particular radius? Does it depend upon the direction of travel also?

I'm trying to find another way to solve the metric. I still don't understand what BillK posted about s approaching t as r approaches infinity making E = 1, which if that is meant to be ds approaching c dt, then since ds = 0 always, not even infinitesimal, I don't see how it could approach c dt or how we could divide the other terms by it in the first place, but if ds were non-zero, then that would also make the metric equal unity instead of zero when divided by ds as far as I can tell. Anyway, I'll keep working on it.
Attached Thumbnails

 After all of that, I visited JDoolin's site and from what he had on a page about black holes, realized that since it was already given that the coordinate speed of light that the distant observer measures is just v_r' = v_r (1 - 2 m / r) and v_t' = v_r sqrt(1 - 2 m / r) by simply comparing the temporal and distance transforms in those directions, one can derive vt^2 + vr^2 = c^2 locally (vt' / sqrt(1 - 2 m / r))^2 + (vr' / (1 - 2 m / r))^2 = c^2 ((sin dφ) r / dt)^2 / (1 - 2 m / r) + (dr / dt)^2 / (1 - 2 m / r)^2 = c^2 c^2 dt^2 (1 - 2 m / r) - dr^2 / (1 - 2 m / r) - dφ^2 r^2 = 0 although I didn't see that it could be worked out so directly until having worked through it, um, more rigorously. Hah. :)

 Quote by harrylin I hope to find time this evening to check it in a GR book* at home, which, I think, describes how to do it. Including length contraction and time dilation should suffice to obtain the light bending of GR directly from SR together with the equivalence principle - but I never actually calculated it myself (lazy me - I always postpone doing what you now are doing!). *Adler et al, Introduction to General Relativity
OK, it's not clear if you already found the solution but indeed, in a nutshell (sorry not enough time now):

- from SR + equivalence principle:
time dilation and vertical length contraction, approx. correction factor 1 +/- gh/c2 = 1 +/- Δψ/c2
They obtained from that and the equation for a Lorentz line element an approximate Schwartzschild equation.

Then apply the Huygens principle just like Einstein did in 1916: p.198, 199 of the English version of
- http://web.archive.org/web/200608290...ry/gtext3.html

However, those few lines correspond to about a page of derivation in the textbook...
Interestingly, the time dilation and the length contraction contribute equally.
 Thanks harrylin. I found what I needed for the most part, but now I'm backtracking and trying to find a way to solve the metric that makes sense to me, solving for dθ/dr, and then find what I was looking for more precisely from that. I also still need to know the relation for a massive particle between the local speed v
 Blog Entries: 4 Recognitions: Gold Member It appears that others have noted the same difficulty that I have: From Wikipedia: http://en.wikipedia.org/wiki/Schwarz...bital_equation "For lightlike (or null) orbits (which are traveled by massless particles such as the photon), the proper time is zero and, strictly speaking, cannot be used as the variable q. Nevertheless, lightlike orbits can be derived as the ultrarelativistic limit of timelike orbits, that is, the limit as the particle mass m goes to zero while holding its total energy fixed." Now, when I tried it (last summer), I took no such "limit as the particle mass m goes to zero" but instead, just plugged a zero into the left-hand-side of the Schwarzschild metric directly. (I believe this may be what grav-universe was referring to in post #21) However, at the time, I was unable to make progress on what seemed to be a much more difficult problem of a non-null geodesic. Attached Thumbnails
 The last few days I have been trying to transform between the coordinate systems of the local observer and the distant observer. If I figure that many local observers are at a distant r on the surface of a planet, then the distant observer will measure the planet as spherical, and the local observers, all observing the same thing from any position and since the contraction is radial only, might determine their planet as spherical also. At what radius though, I'm not sure, since the inferred radial distance is greater but the circumference is the same as the distant observer measures. If the tube is perfectly straight in free space and is brought down to the planet, local observers should still measure the tube as straight, right? If they then lay the tube level on the ground, each end will rise a slight distance from the ground. After finding that distance that the local observers measure in the radial direction for the end of the tube from the ground of a spherical planet, I then contracted that radial distance for what the distant observer measures between the ends of the tube and the ground, bringing the ends closer to the ground, while the center of the tube remained on the ground at the same place on the surface. The distant observer, then, will measure a slight bend in the tube this way that is the same as what was found earlier that it would need to be, but only half the necessary value, so there's still something missing. If I simply extend a single local observer's local coordinate system globally, so that if the local observer is located at some point upon the y axis in the radial direction, say, then all distances in the y direction are 1 / sqrt(1 - r_s / r) greater than the distant observer measures and all distances along the x axis are the same. This would make the planet look like an ellipse to that local observer with him at the peak. If I then place the straight tube with the center at the peak, then transform in the same way back to the distant observer's coordinate system, then there is no bend. But then, if we're dealing with single local observers at this point, then there should be another at the end of the tube to measure that distance, but would then observe the planet and tube differently than the first local observer that was at the center of the tube, so might not say the tube is straight, hard to say. My latest attempts were to start with the distant observer's coordinate system, already knowing what the bend should be, then trying to transform that back to what the local observer would measure that would make the tube appear straight, but the results are so far indefinite. So I'm stuck, help please.
 Blog Entries: 4 Recognitions: Gold Member I've also been trying to work this problem out (again); I was working on the same problem about a year ago. Grav-universe, your approach seems unfamiliar; I don't know whether it might be ultimately equivalent, but I can show you the direction someone else took, and whose footsteps I tried to follow in. Kevin Brown appears to have the entire derivation online here... http://www.mathpages.com/rr/s8-09/8-09.htm I was working through the derivation on physics-forums here, but I did not get all the way through it. (In fact, I could not even follow the derivation to Kevin Brown's equation (8.9.2), at the time) http://www.physicsforums.com/showthr...=510985&page=2 I spent some time trying to remember what I was doing, retracing my steps from last year, and recording it with screen-recording software. Maybe this might give you some ideas. http://www.spoonfedrelativity.com/mi...ild-Metric.swf http://www.spoonfedrelativity.com/mi...warzschild.swf http://www.spoonfedrelativity.com/mi...Derivation.swf http://www.spoonfedrelativity.com/mi...warzschild.swf
 Yay, I finally got it. The tangent speed from which the metric was derived in an earlier post was $$v_t = (sin \ d\theta) r / dt$$ The numerator is the tangent distance which the distant observer and local observer will measure the same, so $$(sin \ d\theta') r' = (sin \ d\theta)r$$ (primed for the local observer's measurements) where $$r' = r / \sqrt{1 - r_s/r}$$, so $$sin \ d\theta' = (sin \ d\theta) \sqrt{1 - r_s/r}$$ The angle we're finding for is that between the radial direction and along the line from the center of the planet to the end of the tube, with some very small x'/r' and x/r, which the local observer will measure as $$sin \ \theta' = x' / \sqrt{r'^2 + x'^2}$$ with the radial distance r' and the length of the straight tube x' in the tangent direction. For the distant observer, we have $$sin \ \theta = x / \sqrt{(r - y)^2 + x^2}$$ where the radial distance to the end of the tube is r-y and the tangent distance x, where the tube bends some distance y in the radial direction. Since both agree upon the tangent distance, then x' = x, and we have $$x' / \sqrt{r'^2 + x'^2} = (x \sqrt{1 - r_s / r}) / \sqrt{(r-y)^2 + x^2}$$ $$\sqrt{(r-y)^2 + x^2} = \sqrt{r'^2 + x^2} \sqrt{1 - r_s/r}$$ $$r^2 - 2 r y + y^2 + x^2 = r^2 + x^2 (1 - r_s/r)$$ $$y^2 - 2 r y + x^2 (r_s / r) = 0$$ Applying the quadratic formula, we get $$y = \left(2 r _-^+ \sqrt{4 r^2 - 4 x^2 (r_s/r)}\right)/2$$ and taking the negative value since the positive would give us 2 r, we have $$y = r - r \sqrt{1 - (x / r)^2 (r_s / r)}$$ which to first order gives us $$y = r \left(1 - \left(1 - \left(\frac{x}{r}\right)^2 \left(\frac{r_s}{r}\right) / 2\right)\right)$$ $$y = \left(\frac{G M}{c^2}\right) \left(\frac{x}{r}\right)^2$$ precisely as it would need to be for the equivalence principle to work out. Cool. $$\ddot{\smile}$$

 Quote by Bill_K I have no idea. I do know that no matter how you do it, by dimensional analysis the deflection angle must come out N M/b, where N is some numerical constant. The only trick is getting N right! Would you like to see the correct derivation? I think it's considerably simpler. A null geodesic in Schwarzschild is given by 0 = (1 - 2m/r)-1(dr/ds)2 + r2(dφ/ds)2 - (1 - 2m/r)(dt/ds)2 where s is an affine parameter. There are two immediate first integrals, L ≡ r2(dφ/ds) E ≡ (1 - 2m/r)(dt/ds) We choose to scale s so that s → t as r → ∞. This determines the value of E = 1. The orbit equation then reduces to (dr/ds)2 = 1 - (L2/r2)(1 - 2m/r) (dr/dφ)2 = (dr/ds)2/(dφ/ds)2 = r4/L2 - r2(1 - 2m/r) Let r = b be the perihelion, the place where dr/ds = 0. This determines L: L2 = b2/(1 - 2m/b) and the final form of the orbit equation: (dr/dφ)2 = (r4/b2)(1 - 2m/b) - r2(1 - 2m/r) ≡ f(r) This is exact. The only thing that remains is to approximate m/b << 1 and integrate: Δφ = ∫dr/√f(r) ≈ ∫dr/(r√(r2/b2 - 1)) + m ∫dr(r3/b3 - 1)/r2(r2/b2 - 1)3/2) The first integral is easy, while the second one takes an integral table or computer. But the result is the correct one: Δφ = π + 4m/b
Here also is a Wiki link to some other methods for solving the metric. I don't understand half of them and the other half don't make sense to me. For instance, as I stated earlier in the thread, I don't see that it is mathematically rigorous to divide by ds or dτ when either is zero for a photon. Even if we were to consider a particle that travels just under c, we are not actually using the particle as the origin as we would for SR, where the distance of the particle from the particle's own origin would be zero, leaving just ds^2 = c^2 dτ^2 in terms of the proper time of the particle, but rather we are using the distance of the particle from the gravitating body, so even in the particle's own frame, this distance would be non-zero and so should be included in the metric. Anyway, I think I have found a very simple way to solve the metric. If L is the locally measured angular momentum, and if this quantity is conserved, then we have

L = m v_t' r' = constant

Since m is also a constant, we can drop that, leaving

L = v_t' r' = (v_t / sqrt(1 - 2m/r)) (r / sqrt(1 - 2m/r)) = v_t r / (1 - 2m/r)

where the primed values are locally measured and unprimed is measured by a distant observer. Since L is constant, then at the point of closest approach b, where the photon travels perfectly tangent to the gravitating body, we have

L = v_t' b' = c (b / sqrt(1 - 2m/b))

L_b = L_r, c b / sqrt(1 - 2m/b) = v_t r / (1 - 2m/r)

c b / sqrt(1 - 2m/b) = (dφ r / dt) r / (1 - 2m/r)

dt = dφ r^2 sqrt(1 - 2m/b) / (c b (1 - 2m/r))

So applying that to the metric, we get

v_r'^2 + v_t'^2 = c^2

(v_r / (1 - 2m/r))^2 + (v_t / sqrt(1 - 2m/r))^2 = c^2

where v_r = dr / dt and v_t = (sin dφ) r / dt = dφ r / dt

(dr / dt)^2 / (1 - 2m/r)^2 + (dφ r / dt)^2 / (1 - 2m/r) = c^2

dr^2 / (1 - 2m/r) + dφ^2 r^2 = c^2 dt^2 (1 - 2m/r)

dr^2 / (1 - 2m/r) + dφ^2 r^2 = c^2 [dφ r^2 sqrt(1 - 2m/b) / (c b (1 - 2m/r))]^2 (1 - 2m/r)

dr^2 / (1 - 2m/r) = dφ^2 [r^4 (1 - 2m/b) / (b^2 (1 - 2m/r)) - r^2]

dr^2 / dφ^2 = r^4 (1 - 2m/b) / b^2 - r^2 (1 - 2m/r)

This seems much simpler and somewhat more mathematically rigorous than any of the other methods, don't you think? Now I'm only left with trying to find a way to prove that the angular momentum is conserved as measured locally. Does anybody know a way to show that?
 Here's something very interesting. If we just start with the already known solution to the metric and reverse engineer it, we'll end up with (dr / dθ)^2 = r^4 (1 - 2m/b) / b^2 - r^2 (1 - 2m/r) dr^2 = dθ^2 r^4 (1 - 2m/b) / b^2 - dθ^2 r^2 (1 - 2m/r) dr^2 + dθ^2 r^2 (1 - 2m/r) = dθ^2 r^4 (1 - 2m/b) / b^2 dr^2 / (1 - 2m/r) + dθ^2 r^2 = dθ^2 r^4 (1 - 2m/b) / (b^2 (1 - 2m/r)) and substituting according to the metric, c^2 dt^2 (1 - 2m/r) = dθ^2 r^4 (1 - 2m/b) / (b^2 (1 - 2m/r)) and re-arranging gives c^2 b^2 / (1 - 2m/b) = (dθ r / dt)^2 r^2 / (1 - 2m/r)^2 c^2 b^2 / (1 - 2m/b) = v_t(r)^2 r^2 / (1 - 2m/r)^2 c b / sqrt(1 - 2m/b) = v_t(r) r / (1 - 2m/r) and transforming c to the tangent speed measured by the distant observer so that all values will be as measured by the distant observer, we have c_dist b / (1 - 2m/b) = v_t(r) r / (1 - 2m/r) so according to the metric and its solution, this relation must be conserved regardless of whether we consider it angular momentum or whatever, but that's definitely what I would call it. That's not the interesting part though. The angular momentum was found at the point where the photon travels tangently at a distance b. For a photon that approaches the body and then escapes, that's fine. But in the case of a photon orbitting the body, there will be two tangent points, at the aphelion and perihelion, with the photon travelling tangent to the body with a locally measured speed of c at both points. If the angular momentum is conserved at each of those points, then we have c_a a / (1 - 2m/a) = c_p p / (1 - 2m/p) and transforming to the locally measured speeds, (c_a' sqrt(1 - 2m/a)) a / (1 - 2m/a) = (c_p' sqrt(1 - 2m/p) p / (1 - 2m/p) c_a' a / sqrt(1 - 2m/a) = c_p' p / sqrt(1 - 2m/p) But the locally measured tangent speeds at each of these points is just c_a' = c_p' = c, giving c a / sqrt(1 - 2m/a) = c p / sqrt(1 - 2m/p) a = p And so it would appear that a photon cannot have an elliptical orbit. A circular orbit is still allowed, but only if it is perfectly circular, so it looks like photons do not orbit a body at all, only approach and escape or fall in.

 Quote by grav-universe [..] And so it would appear that a photon cannot have an elliptical orbit. A circular orbit is still allowed, but only if it is perfectly circular, so it looks like photons do not orbit a body at all, only approach and escape or fall in.
That's interesting! I never thought of that, but it makes perfect sense to me as a photon cannot store potential energy. Thanks.
 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 travelling 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 travelling 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 travelling radially will not accelerate at all, always measured at c locally in the radial direction, while a photon travelling 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 travelling 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.

 $$\frac{(1 - (v_p'/c)^2)}{(1 - 2m/p)} = \frac{(1 - (v_q'/c)^2)}{(1 - 2m/q)} = K (constant)$$
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
 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 $$\tau \approx \left(\frac{b}{c}\right) \sqrt{\frac{b}{2 m}} \left(\frac{\pi}{2}\right)$$
 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