# Time dilation when falling into a black hole

PeterDonis
Mentor
2019 Award
What the infaller sees looking at earth - nothing bizarre, earth appears mostly normal
Yes--with the additional proviso that the last thing the infaller sees of Earth, just before he hits the singularity, will be not too far in Earth's future.

What earth sees looking at the infaller - nothing (redshifted out of visible spectrum)/not significant
Yes, with the qualifier that we are talking about when the infaller crosses the horizon--after that there is literally *nothing* getting back to Earth, not even redshifted out of the visible spectrum. Light can't escape from inside the horizon at all.

What is actually happening to the observer relative to earth - observer is frozen
I would not put it this way; I would say that, once the observer falls inside the horizon, Earth simply has no way of knowing "what is actually happening" to the observer. It "seems" to the people on Earth that the observer is frozen; but it is possible for the people on Earth to calculate that this appearance is due to "optics", which you said you are trying to ignore. In other words, the people on Earth can calculate that, even though the infalling observer appears to be "frozen", he actually does pass through the horizon. (They can do this by calculating the proper time the infalling observer will experience from when he starts to fall until he reaches the horizon, and finding that it is finite. They can also calculate a similar result for the proper time he will experience before reaching the singularity, and find that that is finite too.)

What is actually happening to earth relative to the observer - lots of time goes by
I don't see any interpretation of the physics that would make this true. Once the observer hits the singularity, he ceases to exist and can no longer receive signals from Earth. Since the last signal he receives before that is from not too far in Earth's future, the only interpretation of "what is actually happening to Earth relative to observer" would have not too much time passing on Earth.

I don't see any interpretation of the physics that would make this true. Once the observer hits the singularity, he ceases to exist and can no longer receive signals from Earth. Since the last signal he receives before that is from not too far in Earth's future, the only interpretation of "what is actually happening to Earth relative to observer" would have not too much time passing on Earth.
Okay, thanks a lot, that makes sense. I feel like I have a basic understanding now. Then, how would you fit this in?:

Nope. You would die in quite a small amount of your own time. It would take forever of earth time.

pervect
Staff Emeritus
Okay, thanks. I found this at http://casa.colorado.edu/~ajsh/schwp.html#slow:

which further supports that time on earth will move much faster.
You should also find

Answer to the quiz question 5: False. You do NOT see all the future history of the world played out. Once inside the horizon, you are doomed to hit the singularity in a finite time, and you witness only a finite (in practice rather short) time pass in the outside Universe.
Also check out the "redshift map" on the same page, which shows distant areas of the sky are redshifted, as per my description.

What has probably misled you into your incorrect conclusion is thinking in terms of "time passing faster. This is the wrong mental image. A more nearly correct mental image would tell you "it's wrong to try to compare time at different places".

Thanks for all the info. But, now I am sort of confused again. The bolded part contradicts what was previously established in this thread - that you would appear to freeze to someone on earth (if they knew where you were) as you neared the event horizon due to the extreme gravitational time dilation. I'm sure I'm missing something...
Consider the Special Relativistic twin paradox. As one twin speeds up towards the speed of light, the other twin's time appears to slow down, in the limit they'd freeze. But the situation is symmetrical, the other twin can say the same thing!

Hanging onto absolute time, but believing that this absolute time "slows down" and "speeds up" is not compatible with even special relativity, (much less general relativity).

So, it's prefetly possible that E (the Earth) thinks that R's (the rocket's) clocks are slow, while R thinks that E's clocks are slow. This happens in SR with the twin paradox - and something vaguely similar ,though different in detail, happens when R falls into a black hole.

There are very important differences - in the black hole case, R actually vanishes from E's sight, though E never vanishes from R's sight.

You should also find

Also check out the "redshift map" on the same page, which shows distant areas of the sky are redshifted, as per my description.

What has probably misled you into your incorrect conclusion is thinking in terms of "time passing faster. This is the wrong mental image. A more nearly correct mental image would tell you "it's wrong to try to compare time at different places".

Consider the Special Relativistic twin paradox. As one twin speeds up towards the speed of light, the other twin's time appears to slow down, in the limit they'd freeze. But the situation is symmetrical, the other twin can say the same thing!

Hanging onto absolute time, but believing that this absolute time "slows down" and "speeds up" is not compatible with even special relativity, (much less general relativity).

So, it's prefetly possible that E (the Earth) thinks that R's (the rocket's) clocks are slow, while R thinks that E's clocks are slow. This happens in SR with the twin paradox - and something vaguely similar ,though different in detail, happens when R falls into a black hole.

There are very important differences - in the black hole case, R actually vanishes from E's sight, though E never vanishes from R's sight.
Okay, thanks a lot, that makes sense too. I understand now that I have the wrong mental image.

I asked this to Peter but I'm sure you know the answer too. Where does this fit in?:

Nope. You would die in quite a small amount of your own time. It would take forever of earth time.
I know now it is wrong to compare times at different places, so how can they be compared here (small amount of own time = lots of earth time)?

PAllen
2019 Award
I know now it is wrong to compare times at different places, so how can they be compared here (small amount of own time = lots of earth time)?
A.T.'s quote is not quite accurate. Accurate is:

Earth would never see you reach the horizon, therefore would certainly never see you die. However, they could easily conclude that all they are seeing is light delay, and that you have actually died at the singularity. Everything about 'what your state is now' is a matter of convention, not physics, even in special relativity. There are conventions where earth concludes - that you never reach the horizon; and other conventions where they conclude you reach the singularity.

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PeterDonis
Mentor
2019 Award
Then, how would you fit this in?:
PAllen gave a good answer. The only thing I would add is that, using the convention that A.T. was using, it "takes forever" for you to reach the *horizon*--this convention simply can't comprehend *at all* anything that happens to you at or below the horizon. So assuming that you are still alive when you cross the horizon, it would take "more than forever", so to speak, for you to die from the viewpoint of the people on Earth. But as PAllen said, this is just a convention, and not the only possible one that the people on Earth could adopt.

PAllen
2019 Award
I thought it might be useful to make concrete the point about the conventionality of simultaneity in both SR and GR.

For non-inertial observers in SR, two common conventions are:

- Fermi-Normal coordinates (using the simultaneity of an instantaneously comoving inertial observer).
- Radar simultaneity - using round trip light signals to define simultaneity.

These become arbitrarily close to each other in a sufficiently small region of spacetime around the observer (as would be desired of any 'physically plausible' simultaneity). However, they are not only radically different 'far away', but they cover different regions of spacetime!. The Fermi-Normal convention typically covers only part of spacetime with a valid coordinate chart, while radar (in SR) will cover all of spacetime.

Similarly, I propose a physically plausible alternate simultaneity for SC geometry for a distant static observer (it closely matches SC coordinates over a small region of spacetime for the distant static observer). The convention is as follows:

- consider launching test bodies on radial geodesics at .9c from the distant static observer. For every SC (t,r) coordinate (both interior and exterior), there is an event on the the observe'rs world line that could reach that (t,r) with a test body probe. Compute the proper time along the test body world line, multiply by gamma(.9c), and declare this simultaneous with this amount past the launch event, on the observer's world line.

Having charted (t,r), the angular coordinates are the same as SC due spherical symmetry. This chart smoothly covers one exterior and one black hole interior section of the complete Kruskal geometry in physcially motivated way. You can now talk about 'when' any event inside the horizon occurs in relation to the distant observer, and even compute proper distances along these simultaneity surfaces to events inside the horizon.

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PeterDonis
Mentor
2019 Award
Similarly, I propose a physically plausible alternate simultaneity for SC geometry for a distant static observer (it closely matches SC coordinates over a small region of spacetime for the distant static observer).
I have an even easier proposal, which basically amounts to using Painleve coordinate time as the standard instead of Schwarzschild coordinate time. Since at very large r, the two are almost the same, the idea is this: just use proper time along any infalling geodesic as the standard of simultaneity. That is, if it takes me one year of my proper time to fall from a faraway space station to the horizon of a BH, then the event of my crossing the horizon is deemed to be simultaneous with an event on the station's worldline one year, by the station clock, after I leave the station.

One key thing I would want to look at, for both these proposals, is how they match up with light travel times. That is, when we combine the suggested simultaneity convention with the travel times of light, do we get something reasonable? For example, if the faraway space station sends a laser message after me, at what time by the station's clock would it have to be sent in order to reach me just as I cross the horizon?

In the case of your suggested convention, I would want to formulate a similar comparison: the "time" by the station clock at which I am deemed to cross the horizon is determined by the time at which a 0.9c test probe would need to be sent after me in order to reach me just as I cross the horizon, plus the elapsed proper time for the probe times gamma. At what time, by the station clock, would a laser message have to be sent in order to reach me just as the test probe does?

(One thing that bothers me a bit about your proposal is that the comparison I just described depends partly on gamma--i.e., on the velocity of the test probe--as well as on when I leave the station. In my version, that's not the case.)

PAllen
2019 Award
I have an even easier proposal, which basically amounts to using Painleve coordinate time as the standard instead of Schwarzschild coordinate time. Since at very large r, the two are almost the same, the idea is this: just use proper time along any infalling geodesic as the standard of simultaneity. That is, if it takes me one year of my proper time to fall from a faraway space station to the horizon of a BH, then the event of my crossing the horizon is deemed to be simultaneous with an event on the station's worldline one year, by the station clock, after I leave the station.

One key thing I would want to look at, for both these proposals, is how they match up with light travel times. That is, when we combine the suggested simultaneity convention with the travel times of light, do we get something reasonable? For example, if the faraway space station sends a laser message after me, at what time by the station's clock would it have to be sent in order to reach me just as I cross the horizon?

In the case of your suggested convention, I would want to formulate a similar comparison: the "time" by the station clock at which I am deemed to cross the horizon is determined by the time at which a 0.9c test probe would need to be sent after me in order to reach me just as I cross the horizon, plus the elapsed proper time for the probe times gamma. At what time, by the station clock, would a laser message have to be sent in order to reach me just as the test probe does?

(One thing that bothers me a bit about your proposal is that the comparison I just described depends partly on gamma--i.e., on the velocity of the test probe--as well as on when I leave the station. In my version, that's not the case.)
Yours is effectively the gamma=1 case of mine, I think. I am really proposing a family of possible coordinate charts, that depend on two parameters: reference r value, and gamma. I think you do need a specific reference r value since proper time on a radial geodesic from infinity is infinite. My motivation for using a significant gamma was simply to make it 'look' more like half of the radar simultaneity convention. It would be interesting to compute how gamma affects the chart, but I am not motivated to do that. I suspect the effect may be subtle, because multiplication by gamma removes a large part of the difference.

Also, I believe these charts lose the 'manifest' static representation of SC coordinates. Some metric components would seem to depend on the new t. I suspect this is an inevitable tradeoff.

[edit: one can even conceive of taking the gamma=infinity limit of these charts. Not sure how that would work out.]

[edit2: Maybe you are thinking of the family of free fall from infinity geodesics. In which case this becomes the (gamma,r)=(1,infinity) member of my family of coordinates. ]

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Thanks a lot guys, I think I have a good understanding of it now (or at least a good understanding of what I currently understand and what I don't :p). That previous post by A.T. was sort of what was throwing me off the whole time, because I was misunderstanding what was being said.