# Would time go by infinitely fast when crossing the event horizon of a black hole?

If you were to fall into a black hole, my understanding is that from your reference point, time would speed up (looking out to the rest of the universe), approaching infinity when approaching the event horizon. If this is correct, would you see the whole universe's future "life" flash before your eyes as you fall in, assuming you could somehow withstand the tremendous forces, and assuming black holes don't evaporate? If it is correct that black holes evaporate due to Hawking radiation, would you be "transported" forward in time to where the black hole fully evaporates?

This is considering the "alternate" frame of reference from my question: Does matter accumulate just outside the event horizon of a black hole? In that question, I thought about what happens to matter falling into a black hole from the perspective of someone observing from the outside (e.g. as seen from earth). Here I am considering the perspective of the thing falling into a black hole.

This also takes into account ideas discussed in: Why does time get slow near a black hole?

Note: This answer to another question provides some insight here too (refer to the last part of the answer): http://astronomy.stackexchange.com/a/3713/1386

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(I will assume a Schwarzschild black hole for simplicity, but much of the following is morally the same for other black holes.)

If you were to fall into a black hole, my understanding is that from your reference point, time would speed up (looking out to the rest of the universe), approaching infinity when approaching the event horizon.

In Schwarzschild coordinates, $$\mathrm{d}\tau^2 = \left(1-\frac{2m}{r}\right)\mathrm{d}t^2 - \left(1-\frac{2m}{r}\right)^{-1}\mathrm{d}r^2 - r^2\,\mathrm{d}\Omega^2\text{,}$$ the gravitational redshift $\sqrt{1-\frac{2m}{r}}$ describes the time dilation of a stationary observer at a given Schwarzschild radial coordinate $r$, compared to a stationary observer at infinity. You can check this easily: plug in $\mathrm{d}r = \mathrm{d}\Omega = 0$, the condition that neither the radial nor the angular coordinates are changing (i.e. stationary observer), and solve for $\mathrm{d}\tau/\mathrm{d}t$.

The conclusion is that if you have the rocket power to hover arbitrarily close to the horizon, you will be able to see arbitrarily far into the history of the universe over your lifetime. However, that doesn't actually cover what happens to an observer that crosses the horizon. In that case, $\mathrm{d}r\not=0$, and the coefficient of $\mathrm{d}r^2$ above becomes undefined at the horizon: as in the other question, the Schwarzschild coordinate chart simply fails to cover the horizon and so is ill-suited for talking about situations cross the horizon.

But that's a fault of the coordinate chart, not of spacetime. There are other coordinate charts that are better adapted to questions like that. For example, the two Eddington-Finkelstein charts are best suited for incoming and outgoing light rays, respectively, and the Gullstrand-Painlevé chart is adapted to a freely falling observer starting from rest at infinity.

If this is correct, would you see the whole universe's future "life" flash before your eyes as you fall in, assuming you could somehow withstand the tremendous forces, and assuming black holes don't evaporate?

No. I think this is best seen from the Penrose diagram of Schwarzschild spacetime:

Light rays run diagonally. In blue is an example infalling trajectory, not necessarily freely falling. Note the two events where it crosses the horizon and where it reaches the singularity. Shown in red are inward light rays that intersect those events. Thus, the events that the infalling observer can see of the external universe consist of the region between those light rays and the horizon. The events occurring after that won't be seen because the the observer will have already reached the singularity by then.

Now suppose the observer tries a different trajectory after crossing the horizon, accelerating outward as much as possible in order to see more of the future history of the external universe. This will only work up to a point: the best the observer can do is hug the outgoing light ray (diagonally from lower-left to upper-right) as much as possible... but since the observer is not actually allowed to go at the speed of light, seeing all of the future of history will be impossible. The best the observer can do is to meet the singularity a bit more on the right of the diagram.

By the way, since the light ray worldlines have zero proper time, trying to do that will actually shorten the the observer's lifespan. If you're in a Schwarzschild black hole, you would live longer if you don't struggle to get out.

The above is for an eternal, non-evaporating black hole, as that's what you're asking about here. (The 'antihorizon' is there because the full Schwarzschild spacetime is actually an eternal black hole and its mirror image, a white hole in a mirror 'anti-verse', which not shown on this diagram. That's unphysical, but not relevant to the situation we're considering here.)

If it is correct that black holes evaporate due to Hawking radiation, would you be "transported" forward in time to where the black hole fully evaporates?

An evaporating black hole is morally the same as above: only an ideal light ray can reach the point when the black hole fully evaporates; everyone else gets the singularity. (Since this ideal light ray right along the horizon would be infinitely redshifted, arguably not even that.) You can repeat the above reasoning on its Penrose diagram yourself: