 We've seen that as viewed by the bookkeeper, an object, let's say a rock, falling from rest into a black hole first accelerates. Then, as it approaches the event horizon, it decelerates and ultimately comes to rest at the horizon. At the same time, any radiation emitted by it is ever more redshifted, so this coming to rest is unobservable. Yet, according to the rock's clock, it continues to accelerate right down to the horizon, where our equations of motion break down. So for the case simulated, an observer falling with the rock, the red curve, measures 7.41 seconds traveling to the horizon. To the remote bookkeeper, the same sequence of events, the blue curve, takes an infinite amount of time. He perceives the Schwarzschild Singularity, the event horizon, as an absolute limit of motion. In his reckoning, nothing ever falls through the horizon. The rock's equations of motion fail at the horizon, but there doesn't seem to be anything particularly notable about its motion there, according to an observer falling with it. Let's look at the details. To simplify the expressions, we'll assume the black hole mass is one-half, so that the horizon radius 2m is one. And we'll assume that the rock has dropped from r equals h. For motion in the r direction, the metric has only two terms, giving the elapsed time ds on a clock that moves between events separated in bookkeeper time by dt and r-coordinate by dr. In an appendix video, we derive the equation of motion shown here. The problem is due to the highlighted terms in brackets. At the horizon, r equals 1, this expression has a singularity and can't be directly evaluated. Now, if we divide the metric by ds squared, we get 1 on the left. And on the right, we'll have the square of dt over ds, which we call ut, the velocity of the time coordinate. And we'll have the square of dr over ds, ur, the velocity of the r-coordinate. Multiplying through by minus 1 and rearranging, the right side becomes identical to the highlighted expression, which we can therefore replace by minus 1. And thus we end up with an equation of motion that is well behaved at r equals 1. We can solve it to get the velocity ur at any value of r from h right through the horizon all the way down to r equals 0. The Schwarzschild singularity at r equals 1 is actually not a singularity for the rock. In its own reference frame, nothing particularly notable happens there. On the other hand, the central singularity at r equals 0 cannot be gotten rid of by coordinate manipulation. It's a true physical singularity where the curvature of spacetime becomes infinite, and as the rock approaches it, the r-coordinate changes infinitely fast. So relativity contains a description of how the rock, by its own reckoning, moves through the event horizon into its ultimate fate at r equals 0. Even though these events inside the horizon are forever hidden from outside observers. But this seems to create a paradox. According to the bookkeeper, the rock and all other objects that fell into the black hole at t equals 0 are frozen at the event horizon. Now, what happens if one million years later, according to the bookkeeper's clock, light falls into the black hole? Does the bookkeeper predict that the rock will be illuminated by this light? After all, the light is headed for the horizon and the rock by the bookkeeper's reckoning is still there. Yet the rock only experienced a few seconds of events before being consumed by the central singularity. To get a handle on this, let's solve the equations of motion for a photon and a rock released at r equals 3 and t equals 0. The light pulses on the left and red. The rock is on the right and blue. Light always travels at the speed of light while the rock falls from rest and initially accelerates. According to the bookkeeper, the speed of light decreases to 0 at the horizon, and given enough time, the photon and the rock both approach arbitrarily close to it. Plotting the trajectories in bookkeeper coordinates, the photon at left and red and the rock at right and blue, we see at the top of the graph that the motion is initially very different. The photon's trajectory is steeply sloped due to its high speed while the rock trajectory is gently sloped due to its falling from rest. At the bottom of the graph, however, the trajectories have similar shapes indicating that the objects are moving with similar velocity, the speed of light, which itself, according to the bookkeeper, is approaching 0. If at a given r coordinate, 2.8 in the case shown here, we draw a horizontal line between the curves. Its length represents the delay of the rock relative to the photon in reaching that coordinate. As we move to smaller r values, this delay grows. But not without limit. Arbitrarily close to the horizon, the rock is never more than a given length of time behind the photon. For a photon in a rock released at a given r coordinate, we can calculate this ultimate time difference, and the result is shown here. For the r equals 3 case we've been considering, we see that this is slightly more than 7 seconds. Now let's see the significance of this by considering a slightly different scenario. We'll again drop a rock at t equals 0, then at t equals 4, 5, 6, 7, 8, and 9, we'll shoot photons after it. Looking at the trajectories, we see that the t equals 4, 5, and 6 photons pass the rock. However, the t equals 7 photon just barely catches up to the rock. And none of the photons launched later ever catch the rock, no matter how long we wait. So the bookkeeper simultaneously predicts that, one, the rock is frozen forever on the horizon, and two, that it will not see light that falls into the black hole more than a few seconds after it fell in. We're left with a fascinating picture of the event horizon. Particles that have fallen into the black hole at different times can be thought of as forming separate layers. Over time, these layers collapse toward the horizon. Ultimately, from the bookkeeper's perspective, the entire history of everything that has ever fallen into the black hole is frozen on the horizon in invisible, infinitesimally thin, yet not overlapping layers. And yet, in each particle's perspective, it is zoomed through the horizon, encountering nothing but empty space, quickly continuing on to oblivion at the central singularity. This is such a mind-bending prediction that it probably should come as no surprise that the event horizon and central singularity of a black hole play a significant role in the quest for a theory of everything, merging general relativity with quantum mechanics. Now we'll turn attention to phenomena outside the horizon, where relativity makes clear and potentially observable predictions. In particular, let's see what its predictions are for circular orbits. The event horizon of a black hole of mass m is located at r equals 2m. In an appendix video, we look at the equations of motion and show that the velocity, as measured by a local observer, required for a circular orbit at a given r-coordinate, is square root of m over r minus 2m. At r equals 3m, this becomes 1, the speed of light in our units, which defines the photon sphere, the r-coordinate at which light can orbit the black hole. Circular orbits cannot exist at smaller values of r, because the orbiting particle would have to travel faster than the speed of light, as measured by a local observer. Indeed, at the horizon, it would have to be traveling infinitely fast. Another important r-value is twice this, r equals 6m, where the locally observed velocity would be half the speed of light. Circular orbits inside this are unstable, while those outside are stable. Here we'll look at circular orbits about a black hole of mass m equals one-half. The boundary of the stable region is then r equals 3. We'll place a particle at r equals 3.1, just inside the stable region, and launch it with the numerically exact velocity required for a circular orbit. Solving the equations of motion, we indeed obtain a circular orbit. We'll launch the particle from r equals 2.9, just inside the unstable region. It too travels in a circular orbit. Circular orbits are possible in both regions. To look at stability, we'll repeat these two numerical experiments with the initial velocity decreased by a tiny amount, just one-hundredth of one percent. When the particle is launched inside the stable region, it travels in a very nearly circular orbit. There's no dramatic change from before. The same tiny perturbation to an orbit in the unstable region soon produces drastic changes in the particle spirals into the black hole.