# Falling into a Black Hole

1. Jun 21, 2010

### dubsed

I am not trying to propose any personal theories I am just trying to understand currently accepted physics.

I have heard various descriptions of what would happen as you fall into a black hole. They usually end in spaghettification and eventually falling into the singularity itself.

I recently saw a show "Through the Wormhole with Morgan Freeman" in which two scenarios were discussed. The first scenario involves Alice and Bob where Alice fall towards the black hole Bob sees Alice fall in and gradually slow down and stop frozen/paused just before the event horizon. Alice Herself sees herself fall through and experiences no pain if she had here eyes closed could not tell she even crossed the event horizon.

The second scenario still involves Alice and Bob but this time Alice is in an air plane with a "String Theory Propeller". Bob sees Alice and the Plane fall towards the black hole he sees them slow down and as they slow down the "String Theory Propeller" gradually expands showing more and more detail till it covers the entire surface of the black hole. For Alice the experience isn't very different she never sees the propeller expand she only sees it's whirling hub.

My question is this. Why wouldn't Alice also see the "String Theory Propeller" expand to cover the surface of the black hole. After all the propeller is closer to the event horizon than Alice is. And unless I have greatly misunderstood the equation for gravitational time dilation or its application. Doesn't any observer no matter how close to the event horizon observe things even closer in as frozen/paused?

If this is not true then please explain what I am misunderstanding.

If this is true then please explain how it is possible to actually pass through the event horizon.

2. Jun 21, 2010

### nismaratwork

I don't know what a string theory propeller is, but there are a lot of threads here that cover the difference between frame of reference of an infalling "astronaut" and the distant observer.

3. Jun 21, 2010

### dubsed

My best description of a "String Theory Propeller" is a fractal that on the tips of the propeller are other propellers that spin even faster and at the tips of those propellers are more propellers that spin even faster and so on so that as you slow down time you can see more and more detail of the propeller till is spreads out apparently to infinity or that is the edge of the visible universe.

4. Jun 21, 2010

### Staff: Mentor

If you are stationary wrt the Schwarzschild coordinates then a signal from infinetessimally outside the event horizon is infinitely red-shifted. Why should that have any bearing on whether or not something can "actually pass through the event horizon".

5. Jun 22, 2010

### dubsed

It may be that I misunderstand, but, if there is an infinite red-shift at the event horizon doesn't that mean that time is literally stopped? If so then there can be no movement and the object stops getting closer eg, does not pass through.

6. Jun 22, 2010

### weaselman

Yes, that's why Bob never sees Alice fall through.
As for why Alice does not see her propeller stop, it is because she is falling down together with it. Her reference frame is inertial, so she should not be experiencing any "odd" effects, like stopping propellers.

7. Jun 22, 2010

### dubsed

Yes, but her propeller is in front of her, eg closer to the event horizon and thus experiencing a different time dilation. Or am I missing something?

8. Jun 22, 2010

### weaselman

Well, in theory, you are right - strictly speaking, Alice's and propeller's reference frames cannot be considered the same inertial frame, because the value of gravitational field is different between the two points.
I think, in this examples they just disregard the small distance between Alice and the propeller and consider the whole plane as a point-object.

Last edited: Jun 22, 2010
9. Jun 22, 2010

### Staff: Mentor

What do you mean by "time" here? If you mean "proper time" then no, the fact that a distant non-inertial observer sees an infinite redshift does not imply that an inertial observer's proper time has stopped. If you mean "coordinate time" then yes, in some coordinate systems time is stopped, but that is obviously a coordinate-dependent statement.

10. Jun 22, 2010

### Staff: Mentor

Hey weaselman, don't give up on your excellent point above so easily. An observer free-falling across an event horizon is an inertial observer. If it is an extended body then it will be subject to tidal forces which are finite at the event horizon and can be made arbitrarily small by having an arbitrarily massive black hole.

11. Jun 22, 2010

### dubsed

Dale, I really wasn't speaking about tidal forces, (at least that I know of). Actually, I considering the case you point out, with a very large black hole the gravitational gradient is still relatively flat at the event horizon. However, the time dilation gradient is very steep near the event horizon.

Are you saying that even though the propeller is closer to the event horizon than Alice that she never sees it slow down because they are moving at the same velocity?

12. Jun 22, 2010

### Staff: Mentor

Essentially yes, I would just modify it to "moving inertially at the same velocity".

13. Jun 22, 2010

### JesseM

Keep in mind that there's no coordinate-independent way to define the amount of time dilation for a clock at various distances from the horizon--what we're talking about is the rate a clock is ticking relative to coordinate time, so even if that rate approaches zero in Schwarzschild coordinates which are the most common ones to use for a nonrotating black hole, in a different coordinate system like Kruskal-Szekeres coordinates it wouldn't approach zero at the horizon at all.
What she sees should be coordinate-independent, it depends when light from various events on the worldline of the "propeller" reaches her eyes. And you can definitely still see something in front of you before, during, and after you cross the event horizon. In fact, thanks to the equivalence principle if you and the object in front of you are freefalling in a small region where tidal forces are negligible, your experiences of events in this region (like seeing the object) should not be noticeably different from what you'd experience if you were moving inertially in the flat spacetime of special relativity.

14. Jun 22, 2010

### dubsed

Thank you for your help, this gives me some more to learn about before I ask more.

15. Jun 22, 2010

### weaselman

Well, the problem is that all inertial frames are local in GR. An extended body near a black hole cannot all be in the same inertial frame. Alice's frame is inertial, propeller's frame is inertial, but those are two different frames. Just like Alice's and Bob's frames are both inertial but different.

To see this more clearly, imagine that Alice's ship is really really long. Let it be 10 light years long. Will then Alice see her propeller stop? I am sure, you'll agree, that she will. How about 5 light years? 1 light year? A kilometer? Ten meters?
Do you see where I am going with this? There is no fixed small length at which the propeller suddenly "jumps" into Alice's frame. With respect to her propeller, Alice is "distant observer" just like Bob is, the only difference is, because she is closer, the effects she sees will be smaller than those seen by Bob.
In fact, they may be small enough to be disregarded, as they always are in these examples, but they still do exist.

16. Jun 22, 2010

### JesseM

Incidentally, if you want to learn more about the intersection of black hole physics with string theory, including the "string theory propeller" and other topics, I recommend the book https://www.amazon.com/Black-Hole-War-Stephen-Mechanics/dp/0316016411 by Leonard Susskind.

Last edited by a moderator: May 4, 2017
17. Jun 22, 2010

### dubsed

Leonard Susskind was actually on the show I mentioned and was explaining the scenario. However he never said why Alice would not see the propeller slow down and show more detail. Hopefully learning more about the coordinate system you mentioned will shed some light.

18. Jun 22, 2010

### JesseM

There are actually a lot of different coordinate systems you can use for a black hole where it only takes a finite coordinate time to reach the horizon, see the bottom half of this page. Though as I said, the question of what anyone sees visually as they fall in doesn't depend on your choice of coordinate system (maybe some coordinates make it intuitively easier to track a series of light emissions from a falling object, and see where they catch up to an object falling right behind it, though...for example, Kruskal-Szekeres diagrams have the nice property that all light rays have diagonal worldlines, as discussed in this section of the wikipedia article)

19. Jun 22, 2010

### Staff: Mentor

What you are describing here is tidal effects, and again, they are finite and can be arbitrarily small given an arbitrarily massive black hole, which is exactly what the OP is considering.

No, not if the black hole is sufficiently large. In fact, (this is just an educated guess) if the black hole is large enough that the ship stays intact then I suspect that it is guaranteed that Alice will not see the propeller stop.

The main difference between Alice and Bob is not the distance, but rather the fact that Alice is inertial and Bob is non-inertial.

20. Jun 22, 2010

### nismaratwork

What I am getting from this, is that the Discovery channel has a lot to answer for. :grumpy: