Inside the Photon sphere of a black hole

Question

A photon sphere is defined as the innermost orbit in which the gravitational force of the compact object causes light to orbit it. The event horizon is the boundary at which the escape velocity is the speed of light. So, why is it that any photon crossing the photon sphere, at 3/2 Schwarzschild radius spirals into the black hole? Why do we see a black hole shadow?

Answer 1

Imagine pointing a laser pointer in the direction of a black hole. Two things can happen:

• If you point close enough to the black hole the laser beam will disappear into the black hole.
• If you point further away from the black hole, it will mass the black hole, but the light beam will be deflected by the black hole.

Which of these two things happens depends solely on how far away from the black hole you point the laser (the impact parameter $b$).

Consequently, there must be a critical value of the impact parameter $b_{crit}$ such that when $b < b_{crit}$ the beam is absorbed and when $b > b_{crit}$ the beam if deflected.

This leaves the question of what happens when $b = b_{crit}$. This is of course an unrealistically fine-tuned case, but we can see what happens mathematically. The answer is that the beam must asymptotically approach an unstable circular orbit. Since there is only one lightlike circular orbit in Schwarzschild spacetime, it has the the light-ring at $r=r_{LR}$.

So the critical case of $b = b_{crit}$ is exactly the case where the pericenter passage of the beam is $r_{peri}=r_{LR}$. When $b > b_{crit}$ we will also have that $r_{peri}>r_{LR}$. Moreover, any beam that passes closer to the black hole than $r_{LR}$, must have $b < b_{crit}$ and will be absorbed.

When looking at a black hole you get this same picture, but in time reverse. Instead of following a laser beam travelling forward in time towards the black hole, you now follow the beams travelling towards your eye backwards in time. Since the math is exactly the same, we conclude that you will only see light beams coming from the stars at infinity approach with an impact parameter $b>b_{crit}$. In particular $b_{crit}$ is the size of the shadow cast by the black hole.

Answer 2

Just as one constructs effective potentials for orbiting bodies in Newtonian gravity, we can do this in the Schwarzschild geometry (the spacetime of a black hole) for both massive objects and light beams.

Hence, the plunge you describe can be understood by just the same reasoning.

Mechanics of freely falling objects and light in general relativity are given by the geodesic equation; really a set of 4 ODEs for motion through each dimension of the spacetime.

In the Schwarzschild geometry, two of the equations reveal two conserved quantities, specific angular momentum and specific energy. One of the equations can be gotten rid of by assuming motion is in the equatorial plane of the black hole (this is okay because spherical symmetry allows us to rotate the coordinate system to make this true). The final equation describes radial motion.

Combining this information, we can derive the following equation describing the dynamics of light beams near the black hole: $$\frac{1}{2}\mathcal{E}^2 = \frac{1}{2}\left(\frac{dr}{d\lambda} \right)^2 + \frac{\mathcal{L}^2}{2r^2} - \frac{R_s}{2} \frac{ \mathcal{L}^2}{r^3} = K.E. + V_{eff}(r)$$ where $R_s$ is the Schwarzschild radius, $\mathcal{L}$ is the aforementioned angular momentum and $\mathcal{E}$ the energy; we have written the total energy of the light beam as the sum of a term resembling classical kinetic energy resulting from radial motion and a term resembling an effective potential.

The photon sphere is discovered by requiring the effective potential is constant - this is the condition of circular orbit. You can carry out the calculation by solving $$\frac{d}{dr} V_{eff}(r) = 0$$.

So then passing closer to the black hole than the photon sphere causes light to plunge for the following reason:

The yellow curve is the effective potential as a function of r and the black horizontal line is the total energy. The purple vertical line is the photon sphere's radial value.

As this image is drawn now, a photon would orbit in the photon sphere: its total energy equals its potential energy and hence it exhibits no radial motion.

But a little push to smaller $r$ would mean an increase in kinetic energy; the distance between the black line and yellow curve is the difference in total energy and potential energy - the kinetic energy.

As a little bit of KE has been added, we move closer to the black hole and the effect snowballs with the stronger pull of the central mass (which we can't call a force; remember GR says gravity is the curvature of spacetime); the state of the previously orbiting light beam plunges down the yellow curve, and so $r\to 0$.

Answer 3

Why do we see a black hole shadow?

The image shows what you call "shadow", but The word "shadow" is really inappropriate. Note that the background is not lighted either. In this link there is a discussion of the so called "shadow", from the abstract:

This signature is a sharp-edged dip in brightness that is coincident with the black-hole shadow, which is the projection of the black hole's unstable-photon region on the observer's sky. We highlight two key mechanisms responsible for producing the sharp-edged dip: i) the reduction of intensity observed in rays that intersect the unstable-photon region, and thus the perfectly absorbing event horizon, versus rays that do not (blocking), and ii) the increase of intensity observed in rays that travel along extended, horizon-circling paths near the boundary of the unstable-photon region (path-lengthening)