Watching something fall into a black hole from far away

Question

I am observing (theoretically) an object falling into a black hole from a safe distance away. My understanding is that from far away it appears as if the body will asymptotically approach the event horizon, but I will never see it cross it. Although the object will flatten itself and encompass the entire event horizon its center of mass will slow down.

I deduce that there is a radially repulsive force to counter act gravity from my point of view. Is there a different way to do $\sum F=m \ddot{r}$ where I do not have this fictitious force, and if not what do we call this force? This is equivalent to a centripetal force needed to keep something in orbit, although in this case it is the opposite.

Update

From the responses I see the free-fall EOM is

$$\sum F=ma=\frac{GMm}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}}$$

So the local gravitational acceleration is $$g = \frac{GM}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}}$$

and any motion besides a free fall is $F_{ext} = m (\ddot{r}-g)$ or

$$ \ddot{r} = \frac{r_S c^2}{2 r^\frac{3}{2} \sqrt{r-r_S}} $$ with $r_S = \frac{2 G M}{c^2}$.

So how can this produce $\ddot{r} \approx 0$ near the event horizon?

Answer 1

In Newtonian mechanics the equation you use to describe the trajectory of the falling particle is the usual $F = ma$. In General Relativity you shouldn't be surprised that you use a more sophisticated equation that gives the Newtonian equation as the low curvature limit. This equation is called the geodesic equation:

$$ \frac{d^2x^\gamma}{dq^2} + \Gamma^\gamma_{\mu\nu} \frac{dx^\mu}{dq} \frac{dx^\nu}{dq} = 0 $$

The variable $q$ parametrises the trajectory - it's proportional to the proper time i.e. the time measured by the falling particle. The equation gives you $t(q)$, $r(q)$, $\theta(q)$ and $\phi(q)$ in whatever coordinate system you wish to work in.

You can calculate the EOM of the falling particle then invent a fictional force that is a function of distance to make Newton's equation work, but this would be an intellectual exercise only and wouldn't have any physical significance.

If you're interested, Phil's answer to What is the weight equation through general relativity? analyses a similar problem.

Later:

Re your comment to Stan's answer, there's nothing physically opposing the motion of the falling body. What you're seeing is analogous to the length contraction and time dilation you get in special relativity. The radial coordinate $r$ is defined as the circumference of a circle drawn round the black hole divided by $2\pi$. It is not the same as the length you'd measure if you let down a tape measure towards the event horizon. So you shouldn't be surprised that the speed you get by dividing $dr$ by $dt$ turns out a bit odd.

To see this look at this diagram:

The $dr$ you're using to calculate the speed is the distance on the flat plane, and it's immediately obvious that this isn't the same as the proper distance, $ds$, you get by integrating $dr$ along the trajectory. You can calculate the proper distance simply by holding the other parameters constant and integrating the metric:

$$ \int ds = \int_{r_1}^{r_2} \frac{dr}{\sqrt{1 - r_s/r}} $$

As $r_2 \rightarrow r_s$ this integral goes to infinity. So your result is that the falling body takes an infinite time to move an infinite distance.

But this doesn't mean that the distance to the event horizon is infinite any more than it means time stops there. All it means is that $r$ and $t$ are not the simple quantities that we dwellers in low curvature spacetime think they are.

Answer 2

I will assume a Schwarzschild black hole with mass $M$ and Schwarzschild coordinates, $-+++$ sign convention, and units of $G = c = 1$.

Because the metric is independent of $t$, $\partial_t$ is a Killing vector field, and hence for any geodesic with tangent vector $u$, there is a constant of motion $$\epsilon = -\partial_t\cdot u = \left(1-\frac{2M}{r}\right)\frac{dt}{d\tau}\text{.}$$ Similarly, $\partial_\phi$ is another Killing vector corresponding to angular momentum conservation, but for radial freefall we don't care about that. Also, I assumed the geodesic is timelike, and hence the affine parameter can be taken to be the proper time $\tau$ along it.

Together with the timelike condition $u\cdot u = -1$, the constant of motion above implies that $$\frac{1}{2}\dot{r}^2 - \frac{M}{r} = \frac{\epsilon^2-1}{2} = e$$ is constant, where $e$ is the relativistic analogue of specific mechanical energy of the orbit, and overdot represents differentiation with respect to $\tau$. Differentiating this, we get: $$\ddot{r} = -\frac{M}{r^2}\text{.}$$ Since we actually want to get the equation of motion with respect to Schwarzschild coordinate time, $$\frac{dr}{dt} = \dot{r}\frac{d\tau}{dt}\text{,}$$ $$\frac{d^2r}{dt^2} = \frac{d\tau}{dt}\frac{d}{d\tau}\left[\dot{r}\frac{d\tau}{dt}\right]\text{,}$$ and all that remains is plugging and chugging.

But it is obvious that these quantities go to $0$ as $r\to 2M$: $\epsilon>0$ if the particle starts anywhere outside the horizon, and since $\dot{r}$ is finite is well-behaved there while $\frac{d\tau}{dt} = \left(1-\frac{2M}{r}\right)\epsilon^{-1}\to 0$, the claim follows.

Answer 3

As other answers pointed out, you can interpret it as either dilation of time, increase in distance, increase in inertial mass, depending on the system of coordinates and terminology you use.

I also can point out that you can interpret it as frame-dragging, a second-order gravitational force.