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What happens to a photon shot straight into a black hole? Does it gain infinite momentum before it crosses the horizon? If it has a finite momentum going in, then it would seem that a photon of the same momentum could escape the BH by time reversal.

Does the photon enter the BH at the speed of light as we perceive it from the outside? Or does time dilation take place as the photon gains energy?

Does the photon add to the mass of the black hole an amount of mass m = e/c^2, where e is photon energy? If so, is it the energy the photon had at the beginning, or the energy the blue shifted photon has as it crosses the horizon?

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2 Answers 2

When you're asking a question about general relativity you need to state what coordinates you want to use. This isn't just a mathematical nicety - as you'll see shortly, the different coordinate systems attached to different observers will describe very different behviours.

The obvious interpretation of your question is to ask what happens when an observer well outside the event horizon shines a torch at the black hole. The coordinates used by this observer are called Schwarzschild coordinates. For the Schwarzschild observer the time coordinate $t$ is just what the observer measures on their clock. The distance coordinate $r$ is more subtle. We can't measure the radial distance to the centre of the black hole because there's an event horizon in the way. Instead we say that since the circumference of a circle is $2\pi r$, we define the value of the radial distance $r$ by measuring the circumference of a circle centred on the black hole, then divide the circumference by $2\pi$ to get our $r$ coordinate. So we infer the radial distance: we don't measure it directly.

Anyhow, in the Schwarzschild coordinates the metric describing a stationary black hole is:

$$ ds^2 = -\left(1-\frac{2M}{r}\right)dt^2 + \left(1-\frac{2M}{r}\right)^{-1}dr^2 + r^2 d\Omega^2$$

If our light ray is radial then $d\Omega$ is zero. We can also exploit the fact that for light (or any massless particle) the proper time $ds$ is zero, and our equation simplifies to:

$$ 0 = -\left(1-\frac{2M}{r}\right)dt^2 + \left(1-\frac{2M}{r}\right)^{-1}dr^2 $$

And we can rearrange this to get the radial velocity $dr/dt$:

$$ \frac{dr}{dt} = \pm \left(1-\frac{2M}{r}\right) $$

The $+$ sign is for a light ray heading outwards i.e. $r$ increasing with time, and the $-$ is for a light ray heading inwards i.e. $r$ decreasing with time.

So now we can put $r$ equal to the event horizon radius $r = 2M$ to see what happens to light at the event horizon and we get:

$$ \frac{dr}{dt} = \pm \left(1-\frac{2M}{2M}\right) = 0 $$

We find that the velocity of light at the event horizon is zero. Indeed, if we integrate $dr/dt$ to get the equation of motion of the light ray we find it would take an infinite time to reach the event horizon, or conversely light starting at the event horizon would take an infite time to escape.

This isn't some mathematical trick. If you or I throw any object, whether it's a light ray or a massive object, into a black hole and time its motion then we would see it slow at the approach to the event horizon and we would have to wait forever for it to even reach the horizon let alone cross it.

A light ray doesn't have a rest frame, so you can't ask what the photon sees as it approaches the black hole. However you can ask what you or I would see if we jumped into a black hole. The answer is that our coordinates are locally flat Minkowski spacetime and we wouldn't see any horizon. An infalling observer sees the event horizon retreat before them and they never cross it. However they do hit the singularity in a finite (usually very short!) time as measured by their wristwatch.

There is a calculation of the trajectory of a light ray crossing the event horizon in Why is a black hole black?, but note that the coordinate system used for this calculation does not correspond to the experience of any observer. There is a related discussion in Does someone falling into a black hole see the end of the universe?, but again this note that this also uses an (even more) abstract coordinate system.

But to finish by returning to your question, we can ask what someone hovering just above the event horizon sees - we call this observer a shell observer. The answer is that they do see the light blue shifted, and as they hover closer and closer to the horizon the blue shift tends to infinity. There is no way to hover at the horizon because that would require an infinitely powerful rocket motor, but the limit of the blue shift is infinite as you approach the horizon. However the red shift for outgoing light is also infinite, so the light still can't escape.

Although the shell observer sees the light blue shifted as it reaches them, the shell observer still sees the light slow to a stop as it passes them and approaches the event horizon.

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"An infalling observer sees the event horizon retreat before them and they never cross it." Can you link to a derivation of this? I'm not doubting you, I'd just really like to see it. –  user27578 Mar 3 at 20:42
    
Well, the Physics FAQ seems to implicitly disagree with you: the first paragraph implies that an infalling observer does recognize the event horizon when they cross it (by optical effects, not by the local space-time geometry which is Minkowski) rather than seeing it retreat the whole way down. –  user27578 Mar 3 at 20:59
    
@dgh: Suppose, as you fall, you hold your arm out in front of you, and in your hand you hold a torch shining at your face. If you cross a trapped surface there must be a point at which you can't see the light from the torch because it can't escape from the trapped surface. However this never happens. A freely falling observer can see the torch right up until they hit the singularity. The freely falling observer can certainly calculate where the event horizon is, but they will never cross anything that appears to them like an event horizon. –  John Rennie Mar 4 at 8:10

Does it gain infinite momentum before it crosses the horizon?

Momentum is frame dependent so, when asking for the momentum, one must specify according to whom?

Since the Schwarzschild metric is independent of time, the time component of the four-momentum of freely falling particle is constant.

$$p_0 = -E$$

Now, imagine that the particle is at some radius $r > 2M$ and that there is a freely falling observer that is momentarily at rest at $r$. What is the energy of the particle as measured by that observer?

Using geometrized units, the metric in Schwarzschild coordinates is (suppressing the angular components)

$$ds^2 = -(1-\frac{2M}{r})dt^2 + \frac{1}{1-\frac{2M}{r}}dr^2$$

so, for the momentarily at rest observer at $r$, the time component of the four-velocity is

$$U^0 = \frac{dt}{d\tau} = \frac{1}{\sqrt{1-\frac{2M}{r}}}$$

It's straightforward to show that the energy of this particle relative to this observer is

$$-U^0p_0 = \frac{E}{\sqrt{1-\frac{2M}{r}}}$$

Now, for a photon, the energy and momentum (magnitude) are proportional so we see that, as $r \rightarrow 2M$, the momentum of a photon, as measured by freely falling observer momentarily at rest there, goes to infinity.

Also, note that $E$ is the measured energy for $r \rightarrow \infty$

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