OK, I've looked at your pdf in detail now.
Quote:
The important term here is the coefficient of r", which is negative for r < 2M. Thus we get that velocity outward leads to inward acceleration, and velocity inward also leads to inward acceleration.
I had to scratch my head a bit to figure out what the argument might be. Clearly there mere fact that both inward and outward velocities produce inward acceleration can't be the only thing that's going--that is, after all, how it is in Newtonian gravity as well. So it might be that the coefficient is not just negative, but diverges near the horizon, utterly dominating all other terms and washing out any attempt to have a positive initial dr/dτ.
That makes some sense, but it's hard to see whether in fact that would be the case--diverging functions can have finite integrals; perhaps the particle doesn't "spend enough time" near the divergence to actually beat an arbitrarily large outward initial velocity, especially because (dt/dτ) is
also divergent near the horizon, and that term is positive and opposes the argument.
Perhaps there is a simple insight that to sort this out, but nothing comes to mind; actually, a reason that there
wouldn't be one does: GTR is time-symmetric. The equation describing light orbits in Schwarzschild coordinates (more on how to derive it later) with θ = π/2 is:
[1] (dr/dt)² = [1-2M/r]²[1 + b²/r²]
for some constant b (externally, this is the the impact parameter). For the radial case, b = 0 and this becomes a separable ODE (and one you could read off directly from the metric, dθ = dφ = 0), with Schwarzschild time going as ±2Mlog|1-r/2M|, depending on the sign of the square root we take. That is what we want, because light cones in (t,r)-space should have two sides to them, and graphically they're reflections of each other: one that starts out asymptotic to the horizon at t = -∞, going up in Schwarzschild time to hit the singularity, while the other is a reflection of that, starting asymptotic to the horizon at t = +∞, going
down in Schwarzschild time to the singularity.
OK, but how did I know the directions of those curves? They're just relationships between t and r; there's nothing to indicate which way they go. One can resolve this problem like so. Draw two of those curves, one asymptotic in the Schwarzschild past and one in the Schwarzschild future, in such a way that they intersect at some 0
(1) Both from the horizon toward the singularity
(2) Both from the singularity toward the horizon.
(3) One toward and one away from the singularity--two ways.
Immediately, we can rule out (3) because the curves are the edges of light cones: according to (3), the light cone opens up forward and backward in Schwarzschild time, making the t direction timelike. And we know it isn't at r<2M from the Schwarzschild metric's signature.
Thus, either the null geodesics are both inward or both outward, which makes the Schwarzschild radius the timelike direction, as is appropriate for its signature in the metric. From continuity, we can also conclude that all of them have the same directionality, even for nonzero b, as the same half of a light cone has the time orientation.
But which is it? The key insight here is for a radial
lightlike path, (dr/dλ)² is constant (see below), where λ is the affine parameter of the path, so continuity means it can't change sign. A light ray that starts out going inward continues to do so. Thus, one of the curves we drew previously is toward the singularity. By the above reasoning, they
both do so.
Obviously, since the timelike paths stay within their local light cones, every material particle goes toward the singularity as well.
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Now for some math.
I'll drop the zenith direction, since orbit with initial position and velocity on the equatorial plane will clearly remain so (θ = π/2 and θ' = 0 imply θ" = 0), and by spherical symmetry every orbit whatsoever is planar (sans pertubative effects, of course). Then the φ-component of the geodesic equation is just the separable ODE (dφ')/φ' = -2dr/r, with solution
[2] φ' = l/r²,
which is the conservation of specific angular momentum. The primes are all derivatives with respect to the affine parameter.
Now, let's try massaging the t-component equation. Notice that it with A = 1-2M/r, A' = 2Mr'/r², and so the equation becomes t" = -(2M/r²)/[1-2M/r] t'r' = -(A'/A)t'. Another separable ODE, this time in t'! The solution is obviously
[3] t' = e/A,
for some constant e. This is the conservation of specific energy.
You can substitute the above into the radial component of the geodesic equation, and you'll probably get something interesting... but to avoid an integration, there's another way there as well. If u is a tangent vector of our orbit, then its norm is constant: g_{μν}u^μu^ν = ε. Therefore:
[4] At'² - (1/A)r'² - r²φ'² = ε
Substituting the above constants of motion here instead, we've
[5] e² - r'² - (1-2M/r)(l²/r²) = (1-2M/r)ε,
[6] (e²-ε)/2 = (1/2)r'² + V(r), V(r) = -εM/r + l²/(2r²) - Ml²/r³,
where V(r) is the effective gravitational potential. In ordinary units,
for a massive particle:
[7] V(r) = -GMε/r + l²/(2r²) - GMl²/(c²r³).
Qualitatively, the first term is the Newtonian potential present only for massive particles (ε = 1), the second the usual centrifugal force, and only on the last do we have a relativistic correction. However, the coordinates mean something a bit different compared to pre-relativistic mechanics. Interestingly, a ~1/r³ potential contribution is just what one would expect for an oblate spheroid in Newtonian gravitation, but attributing the precession of Mercury to this effect breaks the precession predictions of other planets.
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Side note: This is obtainable without the geodesic equation as well, since ∂/∂t and ∂/∂φ are Killing vectors on the spacetime. Roughly, geodesics are an extremization problem, while the the metric has a symmetry in the t and φ directions... extremization + symmetry = Noether charge. This particular case isn't as operationally complicated as it might sound at first blush... it just means we could have taken e = <∂_t,u> = (1-2M/r)(dt/dτ) by inspection simply because the metric is independent of t, etc. But the theoretical machinery justifying that might not be familiar, and in any case, once we got the geodesic equation, we might as well make use of it.
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