Should Not Photon Spheres Be INSIDE Event Horizons?

[amigocabal]

&;I have read articles claiming that a photon sphere is a distance outside an object, such as a neutron star or black hole, where the circular orbital velocity is the speed of light (c), and that this distance is 1.5 Schwarzschild radii. (a Schwarzschild radius is the radius where the escape velocity is c.)

There seems to be a flaw in this claim. The formula for escape velocity is ve=sqrt(2GM/r), where v is in meters per second, G is the gravitational constant, M is the mass in kilos, and r is the radius in meters. The formula for the circular orbit velocity is vcir=sqrt(GM/r) Setting r to Schwarzschild, the ratio of the circular velocity at 3r/2 to the speed of light should be sqrt(2GM/3r)/sqrt(2GM/r)=sqrt(1/3). This means that the circular velocity at this claimed radius is less than the speed of light. (To be more specific, at 1.5r, vcir should be c*sqrt(1/3), about 568,000,000 ft/sec)

Plugging these equations, the circular orbital velocity at the event horizon itself should be about 0.707c (about 695,000,000 ft/sec). The photon sphere should be inside the event horizon at one half Schwarzschild radius. (This implies objects can avoid hitting the center of a black hole if it enters at a high enough speed and at a large enough angle to be captured in an orbit greater than one half Schwarzschild radius, instead of plunging straight down.)

Am I missing something?

[Kuroneko]

There's a bunch of ways to think of this. But the most immediate one is simply that your formula for circular orbital velocity is wrong. You might have expected it to be correct because the Newtonian formula for the escape velocity happens to be right, but that's peculiarity of the niceness of the Schwarzschild solution and the particular choice of the Schwarzschild radial coordinate. It doesn't mean that other Newtonian formulae can be safely transplanted into general relativity, not even for something as simple as Schwarzschild spacetime.

It turns out that for this radial coordinate and using the proper time, radial freefall in Schwarzschild spacetime has exactly Newtonian form. It's not really Newtonian, of course, since the Schwarzschild radial coordinate r is not the radial distance (it is defined by the area being 4πr², but because the spatial geometry is non-Euclidean, it can't be the distance from the origin in any naive sense), while the proper time τ measured by the free-falling test particle is not the Newtonian absolute time. Still, in those coordinates, the form is exactly the same: the acceleration is r̈ = -GM/r², it has the same cycloid solution for free-falling from rest at some apoapsis, etc.

The simplest way to resolve your question is to just look at the Schwarzschild metric (in units of G = c = 1):
[0] ds² = -(1-2M/r)dt² + dr²/(1-2M/r) + r²(dθ² + sin²θ dφ²).
If we're interested in a light ray orbit, then the spacetime interval is null, ds² = 0. If the orbit is circular, then dr = 0. If we also suppose it happens in the equatorial plane, then θ = π/2 and dθ = 0. Therefore:
[1] (1-2M/r)dt² = r²dφ².
Since physically we should expect that in this spherically symmetric spacetime, a circular orbit would have constant dφ/dt, we are in fact looking for critical points of W(r) = (1/r²)(1-2M/r), which is a simple calculus problem with a unique solution of r = 3M, i.e. 3/2 the Schwarzschild radius.

That's really it. One can say more things about the correspondence between the Newtonian and Schwarzschild cases, so... Spoiler

In Newtonian theory with gravitational potential -GM/r, there are six conserved constants of motion. Three of which constitute the specific angular momentum vector, the orientation of which defines the orbital plane. Thus, once we fix the orbital plane, we only have to care about the magnitude of the specific angular momentum; let's call this l = r²(dφ/dt). The remaining three can be understood as specific energy of the orbit, E/m, and the direction of the inter-focal vector. So we really only have to care about specific angular momentum and specific energy. Just by decomposing the velocity into its radial and angular components, we can find the effective potential:
[2] E/m = (1/2)(dr/dt)² - GM/r + (1/2)(l/r)²,
where the last term can be understood equivalently as the centrifugal potential. A circular orbit occurs at the extremum of the effective potential (last two terms), which is possible for any value of l.

In Schwarzschild spacetime, things are similar but different. Let's say we take just the equatorial plane (θ = π/2, dθ = 0). Because the metric is independent of t and φ, there are two Killing vector fields that generate conserved quantities:
[3] ε = (1-2GM/r)(dt/dτ), l = r²(dφ/dτ)².
For a massive particle, the spacetime interval should be timelike, ds² = -1 = -dτ², and if substitute this into the metric, we can get the Schwarzschild effective potential:
[4] (1/2)(ε²-1) = (1/2)(dr/dτ)² - GM/r + (1/2)(l/r)² - GMl²/r³,
so in addition to the Newtonian version, we have a quadrupole-like extra term. You can already see here, that for radial orbits, things are going to look Newtonian because l = 0 matches the Newtonian effective potential, but for any non-radial orbits, much less circular ones, we simply can't count on any Newtonian formula to be correct because of that additional term.

However, that doesn't give us light orbits because light rays travel along null geodesics instead, ds² = 0. Doing the same procedure as above but with that change would give us the effective potential for light ray orbits, but it turns out that the shape of the orbit is not sensitive to ε and l individually, but rather only to the ratio b = |l/ε|. So:
[5] 1/b² = (1/l²)(dr/dλ)² + (1/r²)(1-2M/r),
where I'm only using λ instead of τ because we can't interpret it as proper time. The quantity b is sometimes called the impact parameter, and we can analyze light deflection as a scattering process. In this case, the effective potential (last term) has a local extremum at r = 3M, as we already saw previously.