Suppose you are in a spacecraft near a black hole, hovering at a fixed distance thanks to your engines. A fellow astronaut conducting repairs to the exterior of the spacecraft has a mishap ... and plummets towards the hole.

As you watch them fall, you never actually see them pass through the event horizon. If they’re carrying a flashing light, you will see that light grow dimmer and redder, and to you it will appear to be flashing ever more slowly as it approaches the horizon. You might even imagine that they’ve become frozen in time (as you measure it), granting them a miraculous reprieve by giving you as long as you need to rescue them.

**But that’s not the reality at all!** Although the light emitted by your unlucky companion as they approach the horizon will take ever longer to reach
you, there is a *finite time* from the moment they fell until the moment when it becomes *physically impossible* for you to swoop down and snatch
them back to safety before they cross the event horizon. If you wait any longer than that before you act, even a beam of light you sent after them
wouldn’t catch up with them before they were inside the black hole — and your spacecraft certainly can’t do better.

How can we compute this time interval? A convenient set of coordinates to use for this problem are the Eddington-Finkelstein
coordinates^{[1]}. The space-time metric for a non-rotating black hole of mass *M* (using geometric units in
which the gravitational constant and the speed of light are both equal to 1) is given in these coordinates by:

g= –(1 – 2M/r)dV^{2}+ 2dVdr+r^{2}(dθ^{2}+ sin^{2}θ dφ^{2})

Here *r* is the Schwarzschild *r* coordinate, defined so that the area of a sphere of constant *r* has area 4π*r*^{2},
θ is an angular coordinate measured from some chosen axis, and φ is an angular coordinate measured around the axis.

*V*
is a time coordinate, chosen for the very useful property that for a ray of light falling straight into the hole, the value of *V* is constant.
If we can derive the trajectory of our infalling astronaut as a function *V*(*r*), then since the event horizon is at *r*=2*M*,
the value *V*(2*M*) will be the *V* coordinate of the last infalling ray of light that can reach the astronaut before they cross
the horizon. It’s then easy to use the metric to compute the amount of time that passes for the spacecraft, at *r*=*r*_{0},
as *V* advances from *V*(*r*_{0}) to *V*(2*M*).

The easiest way to get a differential equation that we can solve to obtain the trajectory is by making use of the fact that
the vector field ∂_{V} is an isometry: if you shift every point of spacetime by increasing its *V* coordinate
by the same amount, this has no effect on the geometry. This is obvious when you look at the metric, because none of its components are functions of *V*.
As a consequence, the dot product between a unit-length tangent along a geodesic and ∂_{V} will be constant along the whole
geodesic.^{[2]}

If we work with variables *v* = *V*/*M* and ρ=*r*/*M*, the constant *M* disappears from the equations, so
we can solve them in all generality without needing to worry about black holes of different masses. The differential equation for
*v*(ρ) that we obtain by requiring the unit-length tangent to the trajectory to have a constant dot product with ∂_{V} is:

ρ^{2}ρ_{0}+ 4 ρ (ρ_{0}– ρ)v'(ρ) – 2 (ρ – 2) (ρ_{0}– ρ)v'(ρ)^{2}= 0

Here ρ_{0} = *r*_{0}/*M* is the value of ρ from which the astronaut begins their fall.
This equation can be solved in closed form:

v(ρ) = ρ – ρ_{0}+ √[ρ (ρ_{0}– 2) (ρ_{0}– ρ)]/√2

+ (ρ_{0}+ 4) ArcCos[2 ρ/ρ_{0}–1] √(ρ_{0}– 2)/(2 √2)

– 2 Log[ρ_{0}(ρ_{0}– 2)] + 2 Log[2 √(2 ρ(ρ_{0}– 2) (ρ_{0}– ρ)) + ρ (ρ_{0}– 4) + 2 ρ_{0}]

The constant of integration has been chosen so that *v*(ρ_{0})=0. We can then compute:

v(2) = (ρ_{0}+ 4) ArcCos[4/ρ_{0}–1] √(ρ_{0}– 2)/(2 √2) + Log[64/ρ_{0}^{2}]

Finally, we multiply this by a factor of √(1 – 2/ρ_{0}) in order to give the elapsed proper time back at the spacecraft,
divided by the mass: this is the quantity τ/*M* plotted on the graph.

As a check, the corresponding classical result is plotted as well. This assumes Newtonian gravity, but still measures the time until
a ray of light could catch up with the falling astronaut before they reached the same point, *r*=2*M*.
The Newtonian calculation is based on conservation of energy (in fact so is the relativistic one, with the dot product
between the falling body’s relativistic momentum and
∂_{V} acting as a kind of conserved relativistic energy). Equating the kinetic energy to the change in potential
energy gives us:

½ (dr/dt)^{2}=M/r–M/r_{0}

½M^{2}(dρ/dt)^{2}= 1/ρ – 1/ρ_{0}

Mdρ/dt= –√[2 (1/ρ – 1/ρ_{0})]

(1/M)dt/dρ = –√[½ ρ ρ_{0}/ (ρ_{0}– ρ)]

This is solved by:

t(ρ)/M= (ρ_{0}^{3/2}/√2) (√[ρ (ρ_{0}– ρ)]/ρ_{0}+ ArcCos[√(ρ/ρ_{0})])

The dashed curve in the plot at the top of the page is *t*(2)/*M* – (ρ_{0}–2), i.e. the time of arrival of the falling astronaut at
*r*=2*M*, minus the time it would take for a message at lightspeed to reach *r*=2*M* from *r*=*r*_{0}, all divided by *M*.

Somewhat surprisingly, the full, **general-relativistic** analysis
yields exactly the same differential equation for the proper time τ experienced by the falling astronaut as we just obtained for the
Newtonian time, and hence the same result:

τ(ρ)/M= (ρ_{0}^{3/2}/√2) (√[ρ (ρ_{0}– ρ)]/ρ_{0}+ ArcCos[√(ρ/ρ_{0})])

So the time measured by the falling astronaut until they pass through the horizon looks very similar to the dashed curve for the Newtonian time-for-rescue,
only differing by the linear term ρ_{0}–2 that no longer needs to be subtracted.

Suppose you’re too late to rescue your friend before they pass through the event horizon. How much time do you have
to send a parting message to them at lightspeed that will reach them before they hit the black hole’s singularity
at *r*=0? Of course in most
black holes they’d be killed by tidal forces long before that happened, but in a giant black hole they might get quite close to the singularity.

We can plug *r*=0, or equivalently ρ=0, into our formula for *v*, to obtain:

v(0) = π (ρ_{0}+ 4) √(ρ_{0}– 2)/(2 √2) – ρ_{0}– 2 Log[(ρ_{0}– 2)/2]

As before, we need to multiply this by a factor of √(1 – 2/ρ_{0}) in order to give the elapsed proper time back at the spacecraft.
It’s clear from the plot of trajectories of falling astronauts that
these times are finite, and proportionately not much greater than the times to the horizon.

The time experienced by the falling astronaut before they hit the singularity is easily found from the last formula of the previous section:

τ(0)/M= π ρ_{0}^{3/2}/(2 √2)

On a more optimistic note, suppose you *do* act quickly enough to reach your falling friend.
If we care about the time that passes at the spacecraft’s original location, *r*=*r*_{0}
(as opposed to the time you personally experience during your rescue mission)
it’s not hard to show that this can increase without bound. Your spacecraft can’t do any better for the round trip than
two beams of light: one that travels in and catches up with the falling astronaut, and another that immediately
comes back in the other direction.

The ingoing null geodesics in Eddington-Finkelstein coordinates take the simple form *V*=constant.
From the metric, we can check that *outgoing* null geodesics will satisfy the equation:

dV/dr= 2r/(r– 2M)

Switching to the simpler variables where we divide everything by the mass of the black hole,
*v* = *V*/*M* and ρ=*r*/*M*, we get:

dv/dρ = 2ρ/(ρ – 2) = 2 + 4/(ρ – 2)

This equation is solved by:

v_{ON}(ρ) = 2ρ + 4 Log[ρ – 2] +C

where we’re writing *v*_{ON}(ρ) for the outgoing null geodesic, to avoid confusion with
our previous function *v*(ρ) for the infalling astronaut.

If you take a certain time τ_{R} to respond to the fact that your companion has fallen, the soonest
you could possibly reach them is at the event where an ingoing light ray would reach them, for which the *v*
coordinate is given by:

v_{R}= (τ_{R}/M) / √(1 – 2/ρ_{0})

The value of ρ at which the ingoing light ray would reach the infalling astronaut would be the
ρ_{R} that solves the equation:

v(ρ_{R}) =v_{R}

We can’t get an explicit formula for ρ_{R}, but if we find the value numerically we could then plug it
into our equation for the outgoing geodesic to solve for the constant *C* that identifies the
particular light ray we’re interested in:

v_{ON}(ρ_{R}) = 2ρ_{R}+ 4 Log[ρ_{R}– 2] +C=v_{R}

C=v_{R}– 2ρ_{R}– 4 Log[ρ_{R}– 2]

Now we can find the *v* coordinate for the outgoing light ray when it arrives at *r*=*r*_{0}:

v_{ON}(ρ_{0}) =v_{R}+ 2 (ρ_{0}– ρ_{R}) + 4 Log[(ρ_{0}– 2)/(ρ_{R}– 2)]

The longer you wait before responding, the closer the falling astronaut will be to the event horizon when
you reach them – or in the best-case scenario, when an ingoing light ray would reach them. And as
ρ_{R} gets closer to 2, the last term above increases without bound.

So the time back at the
spacecraft’s original position (as measured, say, by the crew of a second craft that didn’t take part in the
rescue) increases without bound, as a consequence of a *finite* delay in your response. In other words,
once *v*_{R}=*v*(2) the rescue is literally impossible, but even as *v*_{R} gets close to
*v*(2), the rescue can entail arriving back where you started after an arbitrarily long time has passed
for the astronauts who stayed put.

Of course, for *you* the time experienced during the rescue can be greatly reduced by relativistic time dilation,
but that doesn’t change the fact that there’s an unmissable deadline at the start of the process.

Our formula for outgoing light rays:

v_{ON}(ρ) = 2ρ + 4 Log[ρ – 2] +C

makes it clear that there is no bound on the time it can take, back at the spacecraft, to receive light sent by the infalling astronaut just before they cross the horizon. This is one source of the tempting misconception that the falling astronaut is somehow “frozen” at the horizon.

If light from the falling astronaut can be misleading, what if we ignore that
and instead think about the way a stationary astronaut might try to
extend their local notion of “space, at this moment”?
A simple calculation allows us to compute the family of spacelike geodesics that are orthogonal to the world lines of anyone with a
fixed *r* coordinate. In this case, the dot product with ∂_{V} along the geodesic has a constant value of 0, and the
differential equation we need to solve is:

v_{SP}'(ρ) = ρ/(ρ – 2) = 1 + 2/(ρ – 2)

The solutions are:

v_{SP}(ρ) = ρ + 2 Log[ρ – 2] +C

These are plotted in the diagram on the right. The geodesics come arbitrarily close to the event horizon but never cross it.

If we consider the sequence of geodesics that intersect the world line of the stationary astronaut at regular clock ticks, the infalling astronaut will cross an infinite number of them before crossing the horizon — illustrating another sense in which the fall might be considered “infinitely prolonged” according to observers who remain at a fixed distance from the hole. But the time limit on any kind of rescue — or even sending a message at lightspeed to the falling astronaut — makes it clear nonetheless that in every physically meaningful sense the fall is finite.

[1] *Gravitation* by Charles Misner, Kip Thorne and John Wheeler, W.H. Freeman, New York, 1973. Section 31.4.

[2] *op. cit.* Section 25.2.