# B Flashlights and Guns inside an Event Horizon

1. Feb 2, 2014

### Glurth

Lets say I fall into a super-massive black hole's event horizon. Facing outwards towards the event horizon (with the singularity directly at my back), turn on my flashlight.

Will those photons emitted by my flashlight actually be able get any further away from the singularity than I was, when I turned it on?

Does the answer change if I ask if they get any closer to the event horizon?

What if I'm shooting bullets from a gun, rather than photons from a flashlight?

What I'm trying to understand is this:
I would expect that as one gets even closer to the singularity, the warping of spacetime that prevents escape from the event horizon, would become even more pronounced. Does this imply "onion-like layers" of event-horizons all the way down?

2. Feb 2, 2014

### Staff: Mentor

Think of the singularity as a moment in time rather than a position in space. If you are inside the event horizon then the singularity is a time in your future.

If you do that, then you realize that your question is similar to the following: "Will those photons emitted by my flashlight actually be able to get any further away from Friday than I was, when I turned it on?""

3. Feb 2, 2014

### Glurth

Are you saying they won't reach the singularity after me?
Wait, there must be more to it that just a time component, how can I even orient myself to face "friday"?

4. Feb 2, 2014

### Bill_K

You can't!

You can't do this either. As DaleSpam said, the singularity is in your future. It hasn't happened yet. All that's in "front" of you and in "back" of you are other victims!

5. Feb 2, 2014

### Glurth

>> you Can't

Ok, I need a bit more explaination.
I thought I wasn't going to feel anything when I crossed the event horizon (feet first). I'm still waving my arms up and down, surely they aren't going back and forth between tuesday and wednesday.
I know I passed through the event horizon feet first; Am I not facing the singularity when I look down, and the event horizon when I look up? If not, what AM I facing in those directions?

6. Feb 2, 2014

### Staff: Mentor

No, what I was saying is that this question:
"Will those photons emitted by my flashlight actually be able to get any further away from Friday than I was, when I turned it on?"

Has the obvious answer: "No."

If you emit the photons on Tuesday then they will never get any further away from Friday than Tuesday. Similarly, inside the event horizon the photons that you emit will never be further away from the singularity than when you emitted them.

Here the analogous question would be: "Are you saying they won't reach Friday after me?"

This question doesn't have such an obvious answer. I could see it being answered any of the following ways:
"No, they will all reach Friday at the same time, on Friday"
"The question doesn't make sense, how could one object reach a moment in time before or after another"
"The events of different objects reaching Friday are spacelike separated, so there is no frame invariant temporal ordering."

I prefer the last one. So the analogous answer would be "The events of different objects reaching the singularity are spacelike separated, so there is no frame invariant temporal ordering."

You cannot, that is the point of the analogy. You cannot orient yourself towards or away from the singularity any more than you can orient yourself towards or away from Friday. Whichever direction you face, you will inevitable reach Friday.

However, suppose that Adam, Bob, and Chuck all cross the event horizon in order, Adam first, Bob second, and Chuck last. Then Bob can turn to face Adam or Chuck. But he cannot face towards or away from the singularity any more than you can face towards or away from Friday.

7. Feb 2, 2014

### Staff: Mentor

No, you can only go forward in time. Suppose Bob is surrounded by friends on all sides, and suppose that all of these friends free fall with Bob through the horizon. Bob will still be able to wave his arms back and forth nearer or closer to any of his friends, but not towards or away from the singularity. The singularity lies in the future, not in any direction. If it were in some specific spatial direction then you could go away from it, but you cannot go away from the future.

8. Feb 2, 2014

### PAllen

On the other hand, you can measured distance from the horizon using a family of spacelike slices. You can rephrase the OP's questions in terms of distance from the horizon (e.g. using Kruskal foliation).

9. Feb 2, 2014

### Glurth

>>"The events of different objects reaching Friday are spacelike separated, so there is no frame invariant temporal ordering."

Ok, I THINK I almost got this: the bullets from my gun will indeed move away from me, but regardless of the direction I shoot, both me and the bullets will reach the singularity at the same time (along with everything else that ever fell in). Or, does the "no frame invariant" part mean that the bullet and me will have different perceptions on who reaches the singularity first?

>>On the other hand, you can measured distance from the horizon ...
That was indeed part of my original post. How would the answer change when considering distance to the event horizon?

10. Feb 2, 2014

### Glurth

>>The singularity lies in the future, not in any direction
I guess I dont see why it doesn't lie BOTH in the future AND a direction.
Before I pass through the event horizon, my feet are pointed straight down. Are they not pointing at the singularity at this point?
If so, do they stop pointing at it as I approach the event horizon? The very moment I pass through?
Shouldn't there be some kind of gradual change going on here...

11. Feb 2, 2014

### PAllen

At the risk of confusing the OP, I will try to say something about the topology of the interior (of the idealized, non-spinning BH, that we discuss for simplicity).

Outside of the horizon, if you want a continuous family of non-intersecting 2-spheres of the same surface area in 4-d space-time, the only way to achieve this is connect them in a timelike direction. That is, they can be interpreted as the history of a 2-sphere.

Inside the horizon, the only way to have a continuous family of non-intersecting 2-spheres of the same surface area is to connect them spatially. This is a 3-d analog of a cylinder. Instead of a circle moved along an axis, you have a 2-sphere moved along a spatial axis. You can't picture this topology embedded 3-space, but it is readily possible to embed in 4-dimensions. Easiest is to just picture a cylinder, but imagine each circle is 2-sphere (of the same surface area). [While there are many other ways to take spatial slices of the interior spacetime, with different topologies - e.g. each Lemaitre slice is a space filling series of concentric 2-spheres of different areas, which is much more 'normal'; the invariant fact remains that the only continuous family of non-intersecting same area 2-spheres must be connected spatially rather than via time.]

Then, any 'history' of an infaller involves following this 3-cylinder as the 2-spheres shrink in size, and the axis stretches without bound (that is, the proper distance between two infall trajectories measured in the axial direction increases without bound). The singularity is when the 2-spheres have zero surface area and the the axial distance between any two infallers has become infinite. This is another way of describing spaghettification.

If, as the horizon passes you during infall, you are watching someone following you in, then, inside, they are separated from you in the axial direction of this 3-cylinder (rather than being separated from you radially as was the case when you were both outside the horizon). In the other axial direction are things you were watching falling in before you.

Despite all this, your very local space looks normal. You can shine a light toward someone you watched following in. They will soon see your flashlight turned on. However, in terms of terms of these 3-cylinders I described, your flashlight shined toward the later infaller is moment to moment on a smaller area two sphere; so is the later infaller. When it reaches their axial position, both the light and and the later infaller will be on a smaller 2-sphere than when you turned on the flashlight (as will you).

Hopefully, after thinking about this, you can make at least partial sense of it. It is hard grasp.

Last edited: Feb 2, 2014
12. Feb 2, 2014

### Staff: Mentor

Yes. The "no frame invariant" part just refers to the relativity of simultaneity. Different coordinate systems disagree with which events are simultaneous and which are not.

13. Feb 2, 2014

### JesseM

If you're familiar with the basics of how spacetime diagrams work in special relativity (light cones and such), you might try reading this conceptual description of the Kruskal Szekeres coordinate system, these coordinates made things much clearer for me when I first learned about them (for now you can ignore the stuff about the white hole and the other universe, these ideas would only apply to an idealized black hole that had existed for an infinite time in the past). If you use a coordinate system like this where light rays always travel at the same coordinate speed, you find that in this coordinate system the event horizon is expanding outward at the speed of light, so that gives an intuitive picture of why you can't get back out of it. And in this coordinate system the singularity is a spacelike surface in the future light cone of everyone inside the event horizon, it doesn't lie in any particular spatial direction.

The more traditional Schwarzschild coordinates for describing a black hole are more confusing, even though they do allow the event horizon to have a fixed position. For one thing, light doesn't travel at a constant speed in these coordinates, instead it slows to a halt at the horizon. But more confusingly, the Schwarzschild time coordinate is only timelike outside the horizon (meaning that a line generated by varying the time coordinate while keeping the other coordinates constant would represent a timelike worldline)--inside it becomes spacelike! Similarly, the radial coordinate in Schwarzschild coordinates is only spacelike outside the horizon, inside it becomes timelike. With Kruskal-Szekeres coordinate you don't have such switching of the meaning of coordinates inside and outside the horizon.

14. Feb 2, 2014

### pervect

Staff Emeritus
Some belated info that might help:

The easiest way to understand the geometry of a black hole is a penrose diagram. I assume we are talking about the idealized Schwarzschild interior geometry, there is some debate over what the actual geoemtry of a real gravitational collapse will be.

A penrose diagram of the black hole part of the Schwarzschild geometry can be found at http://casa.colorado.edu/~ajsh/kitp06/penrose_Schw.html

For complete disclosure, I'll mention that I'm not going to talk about the white hole part of the geometry because I don't think it's relevant to the OP's question. If Im wrong, the OP can ask about that I guess, but I think there is enough to cover just dealing with the back hole part.

As the labelling shows, time goes consistently up the page, and light cones travel at 45 angles. Recall from special relativity (hopefully!) that light moves at a constant velocity relative to all observers, so that what is lightlike to one observer is lightlike to all.

Note that the event horizon on this diagram is not a place, it is lightlike. This is probably the most confusing aspect of black holes, it is almost irresitible to think of the event horizon as a place. But the event horizon is intrinsically light like, and thinking of it that way makes it much easier to understand the geometry.

In particular, when the event horizon passes you at "c", because of special relativity, you have no way of ever catching up to it - whether you fire a light beam, or whether you fire a bullet.

You can also see from the diagram that any light or bullet in the future light cone of an observer past the event horizon must strike the jagged line at the top, which represents the singularity at r=0.

Confusingly, the singularity is actually a space-like surface, as others have mentioned.

Bullets and lightbeams will strike the singularity at r=0, the jagged line at the top of the diagram, at different places than the infalling observer (I won't sketch them out, I'll assume, perhaps incorrectly, that the OP can figure out the penrose diagram enough to sketch their trajectories for himself). But they will all strike the singularity, there is no avoiding it.

It turns out that there is a way to maximize the amount of time it takes to reach the singularity, this is mentioned in:

http://arxiv.org/abs/0705.1029

"No Way Back"

which is an interesting but perhaps advanced read. Basically, if you are on the ideal trajectory (getting on the ideal trajectory typically involves falling through the black hole and then breaking with rockets), you have the longest possible "time to live" according to your wristwatch.

If you fire a bullet, different observers will give in general different orderings of when the bullet reaches the event horizon, because of SR effects called the "relativity of simutaneity". But due to relativistic time dilation effects, etc, the time reading of a clock carried on such a bullet (it's proper time to use the technical term) will always be shorter than the time reading on a clock you carry with you (your proper time) when you reach the horizon when you are on this ideal trajectory which maximizes your time to live.

Last edited: Feb 2, 2014
15. Feb 2, 2014

### Glurth

You guys are AWESOME!

*goes back to reading (and re-reading) posts.

16. Feb 4, 2014

### Glurth

Again, you guys really are great, thank you so much!

The Penrose diagram makes some things clear, but is messing me up on a couple of points.
-If I start free fall 5 light years away from a tiny black hole, it looks like I will hit the event horizon in less than 5 years!
-Why is the singularity line drawn jagged? (as opposed to curved, angled or straight)

So, outside the horizon we have a 2-sphere forming a 3-d volume of space, good-old, X,Y,Z (or just as good & old: R,latitude,longitude). The "continuous family of non-intersecting" comes from the existence/persistence of that volume of space through time, T.
I'll write that as (X,Y,Z),T or (R,Lat,Long), T

Does this imply that T would become part of the expression of the 2 sphere's surface area? Like, (X,Y,T), Z or (Lat,Long,T),R

That graph is really great, and does indeed make it easy to see stuff. It looks to me like the two sides of a light cone emitted inside the event horizon will require different distances to reach the singularity! Doesn't this imply direction DOES matter?

I'm afraid I still don't quite understand what this means I will perceive (other than the very well explained tidal forces). In particular I'm still thrown off by the fact that hovering outside the event horizon, I can point with my plumb-bob and say, in that direction is (or will be at some point in time) the singularity location in space [I could even triangulate with Bob and Chuck]. If I letup on the rockets enough to drop through the event horizon, what does the change of geometry actually mean to my plum-bob setup? How will it ever cease pointing in the direction of where the singularity will be?
(Am I throwing everything into insanity because I'm not in free fall anymore?)

Which direction you shoot your rockets?

AH, ok. so allow me to rephrase my initial bullet question in terms of proper time.
If am in free fall, (not the ideal trajectory Jesse mentioned), would my bullets get to experience a longer proper time than me, before hitting the singularity? And as a follow-up to what I've been hearing, how is the bullet's "proper time to splat" is effected by which direction the gun is fired, if at all?

This very clearly answers my initial question regarding the flashlight, and about the "onion-layers" (smaller and smaller 2-spheres).

17. Oct 29, 2016

### Stephanus

So, if I carry a gun and a flashlight, when I'm inside EH and fired a bullet and turn on my flashlight, both of them OUTWARD the singularity, they (the bullet and the photon) will reach the singularity at the same time. But because you say that singularity is an event not a time or a position.
And can I rephrase that question.
Two objects enter the singularity not simultaneously, they each will always have differents event of singularity?
And what about this.
Adam (and his gun and his flashlight) crosses the EH, and right after that Bob crosses EH perhaps in 1 billionth time interval, or perhaps Adam and Bob distance is 30 cm). So, if Adam and his bullet (fired from his gun) will reach the singularity event simultaneously (perhaps the word simultaneously is more appropriate than "at the same time") Adam can't never shot Bob? And consequently Adam can't see Bob?

Thank you very much.

18. Oct 29, 2016

### Staff: Mentor

Due to the relativity of simultaneity there is no frame invariant answer to that question. However, with that caveat I would say yes.

I don't think that I ever said that the singularity is an event. If I did it was a mistake. The singularity is a spacelike surface.

Provided that the black hole is massive enough then Adam and Bob will have enough time to see and shoot each other as normally.

19. Oct 29, 2016

### Stephanus

Thank you very much Dale.
I will respon to this first, before I study your answer.
I'm sorry if I make a false statement here.
I think it's my interpretation of your statement.
This I hardly can understand. Get any further away from Friday.
Not that English is not my first language, but this relativity thing is very confusing for me
Get any something is a sentence to denote space.
Further away as opposed to farther away. Further is the comparative form of far, but it denotes time.
So I think if this singularity has the adjective of a place and time I think it's an event not a location or a time. But, again, it's me who can't understand what you say
[Add: And interpreting the answers above, now I think singularity is a time not a space?]

And one more thing, while I study your answer.
Can I ask a question here?
If a clock crosses EH for a black hole around 1 million solar mass which has Schwarzschild radius exactly 3 billion KM, so just before the clock get destroyed by spaghettification, will the clock shows 10 thousands second?
As opposed when a very [Edit: FAST] far object reaches a rest point A and travel to another rest point B 3 billion KM away, its clock will show less than 10 thousand second

Thank you very much.

20. Oct 29, 2016

### Staff: Mentor

No. There is only one event horizon. Consider the definition of the event horizon: it is the boundary between the region of spacetime (outside the horizon) that can send light signals to infinity, and the region (at or inside the horizon) that can't. That is a binary property--can/can't. There are no "layers" of it.

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