# B What do I see while falling into a black hole?

1. Apr 4, 2017

### snoopies622

If I fall into a Schwarzschild black hole and, as I'm falling, look in the direction away from it, what do I see?

I once heard someone say the rest of the universe would appear to move faster and faster, and as I cross the event horizon it would appear to move infinitely fast.

I'm not sure this make sense to me because it would imply that I see the end of the universe, yet from the outsider's point of view my existence ends in a finite time. (Or does it?)

2. Apr 4, 2017

### Orodruin

Staff Emeritus
You need to define "away from it" better.

"I once heard someone say" is not a valid reference. Please state where you got this.

The event horizon is a null surface. To a local observer, it moves at the speed of light.

This statement makes no sense. You need to define what "finite time" means in terms of the outsider's point of view.

3. Apr 4, 2017

### George Jones

Staff Emeritus
An earlier post of mine:

4. Apr 4, 2017

### Khashishi

People don't know how close to a black hole general relativity still remains valid. But if we assume it stays valid up to the singularity, then you don't notice anything catastrophic as you pass the event horizon. This site explains it all. http://jila.colorado.edu/~ajsh/insidebh/schw.html

Some people believe that quantum gravity effects take over. A theory is that you will hit a firewall which instantly burns you to a plasma as you fall into a black hole.

5. Apr 4, 2017

### haael

There are countless movies on Youtube that show the simulation of the observer falling into a black hole.

The exact image depends on the type of black hole and the theory assumed for calculations.

6. Apr 4, 2017

### snoopies622

Thanks all! For some reason I didn't get any email notifications of these replies after the first one.

This implies that if I am at location B, I'll see a clock at location A running very fast, no? And another (stationary) clock even farther away than location A running even faster? As long as we all stay at a constant position with respect to the black hole. This is what makes this interesting to me - the thought that the closer I get to the event horizon, the faster events in the rest of the universe appear to go.

7. Apr 4, 2017

### pervect

Staff Emeritus
It's unclear to me when you ask what you "see" when you fall into a black hole, if you actually mean that literally (which is what George answered, it's what you would photograph), or something else, something involving a mental model in your head.

What we can answer most accurately is what George answered - what you literally photograph.

To answer what sort of mental model you use turns out to be difficult, due to the fact that it's not clear what simultaneity even means in such cases. The mental model differs from the photograph in the fact that the photograph includes effects like light-speed delays, and typically the mental model involves some concept of "now". The problem is that the concept of "now" is fundamentally ambiguous even in special relativity, and even worse (if that's possible) in General relativity.

So lets answer the easy question - what you'd directly observe. A stationary observer would observe blue-shifted light, and speeded up motions.

Red shift and blue shift not only change the frequency of the light, but the rate at which information carried by the light is received. If it took 1 million carrier cycles to encode a complete frame of video, the information encoded on that video wouldn't change when the signal was blue-shifted. The duration of 1 million cycles would be less for a blue-shifted signal than the original signal. So the blue-shifted signal would play the information back faster, as well as having an effect on the carrier frequency.

An observer falling in from a large distance would have such a high velocity that the doppler shift would more than overcome the gravitational blueshift, so they'd directly observe a net red-shift. The doppler shift involves light-speed delay effects, and is what's actually observed- it's not a mental model.

Note that if you just want to observe the far future history of the universe, it turns out to be just as eas to do it without a black hole than with. All one needs to do is to use the thrust that one would use to hover near the event horizion of a black hole to accelerate to relativistic speeds. Relativistic time dilation would do the rest.

8. Apr 4, 2017

### snoopies622

Yes, that's what I was wondering about. And since the dt term in the Schwarzschild metric approaches zero as r approaches the event horizon, it would seem that — with sufficient technology — one could park oneself arbitrarily close to the event horizon and watch the rest of the universe fly by with any desired speed.

9. Apr 4, 2017

### Khashishi

Yeah, but you would have to accelerate very rapidly to keep yourself from falling in. If you had that kind of thruster, then it's probably safer to just fly back and forth in deep space at some speed near c.

10. Apr 5, 2017

### timmdeeg

Means that hovering close the event horizon might create a pancake-like shape out of your body.

11. Apr 5, 2017

### newjerseyrunner

PBS Spacetime did an episode of that a few days ago. The host is an astrophysicist so the science is usually good.

Apparently, if you travelled towards the black hole, you'd see everything falling into it happen in reverse because you're encountering light that left in the past. If you try to push out of the black hole, you'll encounter light still coming into it so you'll see the events after you went in unfold.

12. Apr 5, 2017

### jbriggs444

Since you cannot exceed the speed of light locally, you are not going to catch up to light emitted in the past to see things happen in reverse.
Since you cannot push out of the black hole, the part about what you would see if you did is not physically meaningful.

13. Apr 5, 2017

### Staff: Mentor

Not if it said the things you say in the rest of your post. See below.

This is not correct. The light that "left in the past" from events outside the hole will have fallen into the hole and reached the singularity before you do. You will never see it. Any light signals you see coming from outside the hole after you fall in will be from events that happened outside after you fell in.

This is also true if you don't try to "push out" of the black hole. The only difference the "pushing out" (which I take to mean accelerating in a radially outward direction--which won't actually stop you from falling further and further in) makes is in the redshift/blueshift of the light you see coming in from events that happened after you fell in.

14. Apr 5, 2017

### pervect

Staff Emeritus
That's true. Perhaps the OP and the PBS special is thinking of viewing light from things that have fallen into the hole before the observer, for instance light being emitted from the falling object "outwards", away from the black hole, just as the falling object is at the event horizon. Then the observer falls into the black hole a little bit later, and sees the light from the previously falling object just as they pass through the event horizon.

15. Apr 5, 2017

### PeroK

I don't think that's what he said at all! On the first point, he was referring to light from the collapsing star that was emitted at the "outer" extreme (*) of the light cone, just "below" or "after" the event horizon. That worldline can be intersected by a particle falling into the black hole later.

(*) Outside the event horizon this would be radially outwards, but inside the horizon it's still towards to singularity.

On the second point, he indicated that your maximal proper time after the event horizon is achieved by just letting yourself fall and any attempt to struggle will shorten the proper time until the singularity.

16. Apr 5, 2017

### pervect

Staff Emeritus
Let's try to work out the formula for the amount of crush you'd experience if you attempted to hover over a Schwarzschild black hole to get some time dilation factor $\gamma$.

First we need the formula for the amount of crush, which is the proper acceleration. I believe the formula is:

$$a = \frac{GM}{r^2\sqrt{1-\frac{r_s}{r}}}$$

for instance https://en.wikipedia.org/wiki/Komar_mass which assumes that G=1, or https://www.quora.com/What-would-the-acceleration-due-to-gravity-be-on-the-surface-of-a-black-hole. I recall seeing this formula in Wald's book "General Relativity", but not which page I saw it on, if I had more time I'd look it up to verify my fallible memory.

Gravitational time dilation is given by

$$\gamma = \frac{1}{\sqrt{1-\frac{r_s}{r}}}$$

Let r be some multiple $\eta$ of $r_s$, so that $r = \eta \, r_s$.

Then we can write:

$$a = \frac{GM}{\eta^2 r_s^2\sqrt{1-\frac{1}{\eta}}}$$

and

$$\gamma = \frac{1}{\sqrt{1-\frac{1}{\eta}}}$$

Making use of the defintion of $r_s$

$$r_s = \frac{2GM}{c^2}$$

and solving for $\eta$ as a function of $\gamma$

$$\eta = \frac{\gamma^2}{\gamma^2 - 1}$$

we can re-write the expression for a as:

$$a =\frac{GM}{\eta^2 r_s^2\sqrt{1-\frac{1}{\eta}}} = \frac{GM\gamma}{\eta^2 r_s^2} = \frac{GM}{r_s} \frac{\gamma}{{\eta^2 r_s}}=\frac{c^2}{2} \frac{\gamma}{\eta^2\,r_s} = \frac{c^2}{2\,r_s} \gamma \left( \frac{\gamma^2-1}{\gamma^2} \right)^2$$

The term to focus on is $c^2/2r_s$. For $\gamma=2$, the term on the right side turns out to be 18/16, so it's close to 1. For higher values of $\gamma$, the term becomes approximately equal to $\gamma$.

A large r_s makes $c^2 / 2\,r_s$ smaller, to minimze crush we want it as small ass possible. So we want a large black hole. Suppose we have a billion solar mass black hole. This is larger than the one at the center of our galaxy, which is about 4 million solar masses. The Scharzschild radius for one solar mass is about 1.5 km <<wiki link>>, so for our billion solar mass black hole, we'd have $r_s$ = 1.5 billion km = $1.5\,10^{12}$ meters.

Plugging this in, we get $\frac{c^2}{3\,10^{12} meters}$ = 30,000 meters/second^2. Ouch. Assuming I didn't make any mistakes, something I'm not as good as avoiding as I used to be :(.

This is why I say it's time for plan B when trying to get large time dilation factors. Hovering squishes you flat, trying to orbit the black hole runs into two problems. The first problem is that orbits don't exist inside the photon sphere, which is r=3m. One can get high time dilation due to velocity as one approaches the photon sphere, but one can get the same time dilation with high velocity outside the photon sphere. Furthermore, around a Schwarzschild black hole, the orbits at the photon sphere are unstable, so you'd need active correction to go this route. Not only does the gravity of the black hole not help much to achieve time dilation, it makes surviving the experience more difficult.

An interesting and non-obvious take of this is from Kip Thorne, the science consultant for "Interstellar". Tasked with finding a stable orbit with a high time dilation, he envisioned a very rapdily rotating black hole at the center, an came up with a possible (though not very likely) scenario that had some basis in physics that would work for the movie. But this isn't a Schwarzschild black hole at all, it's an extremal Kerr black hole. I rather suspect that in Thorne's scenario, the gravity also turns out to be much less important than the velocity, but I haven't actually computed anything.

Last edited: Apr 5, 2017
17. Apr 5, 2017

### Staff: Mentor

Btw, a very counterintuitive property of what firing your rockets to accelerate does once you're inside a black hole's event horizon is illustrated by the following thought experiment: consider two observers, Alice and Bob, who fall into a black hole together--i.e., they are right next to each other, in free fall, as they cross the horizon. But just after they cross, Bob fires his rockets and accelerates radially outward. Bob sees Alice fall away below him, and Alice sees Bob rocket away above her, just as if they were both in a large region of empty, flat spacetime. And Alice will reach the singularity "before" Bob does, if we order events by light signals--in other words, Bob will be able to see light signals emitted by Alice all the way down to the singularity, but Alice will stop seeing light signals from Bob (because she hits the singularity) while Bob is still well away from the singularity. However, if we compare the elapsed times on Bob's and Alice's clocks, from the event where they separate (i.e., where Bob starts firing his rockets), until the events where they, respectively, hit the singularity, we will find that Bob's elapsed time to the singularity is less than that of Alice! In fact, we can make Bob's elapsed time as small as we like, in principle, by increasing the proper acceleration he feels due to his rockets.

18. Apr 6, 2017

### Arkalius

Yeah, as explained in the video, time and space kind of reverse roles within the event horizon. Instead of being stuck moving forward in time but being free to move in space, you're stuck moving in space toward the singularity, while having the freedom to kind of move forward and back in "time" (within the event horizon). Since all paths in space lead to the singularity in the black hole, any attempt to accelerate will only reduce the amount of proper time you experience before you reach it.

19. Apr 6, 2017

### Staff: Mentor

Which is apparently wrong. The video is not a valid source if it is saying what you are saying.

This is not correct. You are always stuck moving forward in time; you can't move backward in time. The reason you are forced to move toward the singularity inside the event horizon is that in that region, the singularity is in your future; its direction from you is a timelike direction, not a spacelike direction.

Also, you can still move in all directions in space inside the event horizon.

This is not correct. There are spacelike paths that start inside the hole and do not lead to the singularity. Some of them remain inside the hole; others lead outside the hole. There are no timelike or null paths from inside the hole to outside the hole.

This is true, but the reason it's true is not what you claim. See above.

20. Apr 6, 2017

### Arkalius

I think that's the point. Time-like paths inside the event horizon are effectively unidirectional (they move only toward the singularity, there is no time-like path that moves you further from it) as opposed to allowing freedom to move toward or away from any particular location in space when outside the event horizon.

But they all lead closer to the singularity at different rates, just like how outside the event horizon, while you can alter your "velocity" through time, you can only move forward in it.

I'd actually be interested in seeing what you have to say after actually watching the video instead of responding to people's descriptions of it. We may be iterating the ideas presented within it imprecisely.

Last edited: Apr 6, 2017
21. Apr 6, 2017

### Staff: Mentor

You're still misunderstanding the physics. Timelike paths inside the horizon only move you towards the singularity because the singularity is in the future. "Towards the singularity" is a timelike direction, not a spacelike direction. So your "freedom to move" is the same as it is outside the horizon.

In other words, there is no difference between the inside and the outside of the horizon in this respect. Yes, that's correct.

I don't know that I'll have time to watch the whole thing any time soon. Particularly when it already seems apparent to me that it's not a good source, so rather than try to explain all the things it says that are wrong, I feel like it's better for me just to recommend that people not use it as a source, period.

22. Apr 6, 2017

### PAllen

The singularity is simply not a location in space at all. There are still three spatial directions you can move in, inside the horizon, and none of them is the direction of the singularity. But just like outside the horizon, you always move forward in time, which is toward the singularity. There is no inversion of space and time except for the artifact that for two different historical coordinate patches (used for the interior and exterior, and undefined on the horizon), the coordinate called t is a time in the exterior patch, and it is a spatial direction inside. But this is nothing but a historical naming oddity of one coordinate system.

[edit: in these interior coordinates it would be less confusing if t was called z since it functions as the axial direction of S2xR hypercylinder surfaces. Meanwhile, r should have been called t.]

Last edited: Apr 6, 2017
23. Apr 6, 2017

### Arkalius

The point the video makes has to do with the Schwarzschild metric applied to the spacetime interval (for a body moving directly into the black hole) given as $$s^2 = \frac {\Delta r^2} {\left(1 - \frac {r_s} r \right)} - \left(1 - \frac {r_s} r \right) c^2 \Delta t^2$$

Here, in order to have a time-like spacetime interval (ie, negative), you must have $\Delta r^2 \gt c^2 \Delta t^2$. In addition, $\Delta r$ cannot be 0 and $\Delta t$ can be 0. Effectively, time and distance swap roles in this equation for time-like intervals. Even the language I've seen you guys use points to this, saying that the singularity isn't a location, but a future time. It's not literally the case that time and space are swapped, but more that their roles in causally valid motion are.

The meaty part of the video is only 10 minutes long. If you have time to spend writing posts here, you have time to view that video at some point. Anyway, I suspect the video isn't necessarily a bad source, but it's trying to teach a complex idea from general relativity to a well-informed but lay audience. There is possibly something lost in translation, so I'm eager to know how you would react to the way the content in the video is presented, not so much to tell us what's wrong with the video, but more to tell us what's wrong with our understanding of it.

24. Apr 6, 2017

### PAllen

No, despite the metric looking the same, there are two independent coordinate patches represented by the schwarzschild metric, interior and exterior. Neither covers the horizon. Inside, the time coordinate is called r, outside it is called t. Physics doesn't care about the letters we use. As I said before, when used inside the horizon it would be less confusing naming conventions to call t z, and r as t. Note, in other coordinates for the same spacetime, there is no such confusion. There are coordinates e.g. Lemaitre or Kruskal that cover interior and exterior and horizon in one patch. In these coordinates, there is no switching evident. The time coordinate remains timelike throughout the spacetime. The switching in swarzschild coordinates is an accident of labeling interior coordinates in a confusing way, inherited from history.

Last edited: Apr 6, 2017
25. Apr 6, 2017