# I Looking out from a event horizon

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1. May 28, 2016

### Vidar Martinsen

Lets look at the known story of Alice and Bob outside a Black hole. Alice are falling inn to the hole, and Bob are outside.

What would Alice observe from the event horizon of a Black hole. Looking outward.

We know Bob who is outside, se Allices clock standing still, but Allices clock are not standing sill according to her, so would she se the clock outside running incredible fast. Actual infinite fast, if her clock was standing still.

Or is there another explanation here?.

2. May 28, 2016

### Staff: Mentor

Not if she is actually falling (= no thrust to hover close to the black hole), because you have to take the Doppler effect into account.
There are literally thousands of websites and multiple forum threads here describing all the different views in detail.

3. May 28, 2016

### PAllen

Also note that it is exactly as impossible to hover at the event horizon as it is for body with mass to travel at c. For any observer falling through the event horizon, they would not be able to locally detect this (per classical physics), nor would they see any change in what they observed looking out as they passed the event horizon. They would see essentially the same blueshift just below the horizon as they saw just above it. The amount would depend on initial conditions and thrust applied (it could even be redshift).

4. May 28, 2016

### pervect

Staff Emeritus
One of the websites that has some views of what you "see" when you fall into a black hole is http://jila.colorado.edu/~ajsh/insidebh/schw.html

Note that they interpret "see" literally. It shows you what a camera would photograph. Also, while they show the view from a couple of different trajectories, I don't think they have one for "falling straight in".

As far as seeing goes - often people want a mental model, rather than a camera photograph. This is harder to communicate, and I'm not sure if it's possible without getting heavy into math.

5. May 28, 2016

### rootone

Would she see anything at all?
She will be accelerating towards whatever the singularity is at a faster rate than any photons (or anything else) which fell in after her.

6. May 28, 2016

### PAllen

Wrong. She will see the world outside virtually unchanged from just before passing the horizon to just after. Inward directed photons would overtake a free-faller inside the horizon just as easily as outside.

7. May 29, 2016

### timmdeeg

Alice would see Bob redshifted if Bob is stationary and presumably still but less redshifted if Bob is in free fall.

8. May 30, 2016

### Vidar Martinsen

I se that nobody has claimed my assumption to be false.
My point is :

If what I am claiming is true ( Allice experience time outside the black hole run at infinite speed).
She will never enter the black hole at all, and never cross the event horizon because the black hole will have disappeared and evaporated due to Hawking radiation long before she reaches inside.

Do anyone have any comments on that.

9. May 30, 2016

### Staff: Mentor

She does not, as I mentioned in post 2 already. Unless she hovers above the event horizon with diverging thrust to weight ratios, which she cannot do for a long time.

10. May 30, 2016

### Vidar Martinsen

Ok , Thanks.
So what you say is if she somewhat manages to stay put for a short time at the event horizon due to a sort of infinite rocket power trust outward, she would experience time outside go at infinite speed.

Of course it is a theoretical situation, but I just wonder about the behavior of time and space on the outside as seen from the event horizon.

11. May 30, 2016

### pervect

Staff Emeritus
No. What we are saying is that something that moves at c (light for instance) can "stay put" at the event horizon. But anything else cannot, not even for an instant.

The relative velocity between any material object whatsoever, and the event horizon, will be c, no matter how fast the object moves. This is because the speed of light is equal to "c" , for all observers, no matter how fast they are moving.

The FAQ "rest frame" of a photon spells this out in more detail. https://www.physicsforums.com/threads/rest-frame-of-a-photon.511170/

It's very short, so I'll quote the current version
THe idea that "something with a point of view" can stay at rest at the event horizon is equivalent to the idea that "something can move at the speed of light". It's just not possible. Not even for an instant.

A corollary of this is that light does not have a point of view.

The point of view of an object falling through the event horizon is a concept that DOES makes sense, and we've attempted to answer that question.

12. May 30, 2016

### timmdeeg

What you are claiming isn't true. Alice would experience the time outside at infinite speed only if she was hovering at the event horizon, which is not possible however as explained in post #3. Hovering a bit outside the event horizon she would perceive the time far outside running extremely fast while her wristwatch ticks just normal. In contrast, Bob hovering far outside would see Alice's wristwatch time running very slow compared to his clock in case she is hovering slightly outside the event horizon and would see it slowing down to zero in case she is falling freely. Google coordinate time (the time Bob experiences Alice's wristwatch which is far away) vs. proper time (the time Alice reads on her wristwatch).
So Alice will never enter the black hole from Bob's perspective, but will enter it in free fall from her own perspective. As mentioned being in free fall she will see things outside redshifted which means she will perceive clocks outside ticking slower than her wristwatch.

13. May 30, 2016

### Vidar Martinsen

Another question is of course:

How fast speed will you experience at the event horizon if you are falling in from a significant height?.

If it is close to light speed, you will of course experience a significant time dilation and length contraction looking outwards.

14. May 30, 2016

### Vidar Martinsen

I was to fast here, and did not read your answer first. If I understand it correctly all matter will have the speed of light at the event horizon.

So the answer is. Since she has the speed of light, time will stand still and there will be a total length contraction looking outwards.

Last edited: May 30, 2016
15. May 30, 2016

### Staff: Mentor

Speed relative to what? You never experience any speed relative to you.
Relative to the horizon, in some coordinate systems.
That's not how it works.
You cannot apply special relativity here any more, you need general relativity. If you fall in freely, you don't notice anything special while crossing the event horizon.

16. May 30, 2016

### jartsa

Yeah, in that case all black holes above evaporate at infinite rate according to Alice. And the black hole below evaporates at the same rate as similar black holes very far above.

A micro black hole at same altitude with Alice evaporates at normal rate according to Alice, that black hole is the slowest evaporating black hole in the universe according to everyone.

Last edited: May 30, 2016
17. May 30, 2016

### pervect

Staff Emeritus
Here's a good way of looking at it. You have a spaceship with a flashbulb on the bow, and a person in the stern.

When the bow of the ship reaches the event horizon, the flash bulb goes off.

The person in the rear of the ship watches the bulb go off, and doesn't see anything unusual. They don't even know they've fallen though the event horizon, there's nothing special there to mark it. The flash from the bulb reaches them normally, they see it normally, and if they measure it's speed, they measure the speed of the flash as "c", the speed of light.

Additionally, the person on the ship computes the fact that the outward travelling light stays stationary with respect to the event horizon. So they conclude that measuring the speed of light is equivalent to measuring the speed at which the event horizon approaches them.

It's common to ask "at what speed does the ship approach the event horizon". This is unfortunately phrased, and the phrasing of the question leads to confusion. From the point of view of the ship, ship is not moving. What's moving from the point of view of the ship is the event horizon. And the event horizon is moving towards the ship at "c", the same speed as the flash of light is moving.

Now, consider what happens in Schwarzschild coordinates.

Some of the flash, that part of it that was headed directly outward, "stays put" at the event horizon. And the infalling observer catches up with it.

And that's all there is to it. As long as one avoids postulate things that don't exist (such as the point of view of light), things are simple. Postulating things that don't exist leads to confusion, such as trying to imagine light that stays still while at the same time also travelling constantly at "c".

18. May 31, 2016

### Vidar Martinsen

Thank you all again for giving me a greater understanding of these things :)
Regards
Vidar

19. Jun 5, 2016

### Staff: Mentor

No, we don't. We know that Bob, who is outside, can't see Alice's clock at all if Alice is just at the event horizon. Light emitted outward at the horizon stays at the horizon.

No. Actually, if Bob is very far away from the hole and is at rest relative to it, Alice will see ingoing light from Bob redshifted by a factor of 2. One way to understand this heuristically is that there are two effects involved: Bob's light blueshifts as it falls towards the hole, but Alice is falling inward as well so there is a Doppler redshift when the light from Bob reaches her. If you work through the math, the redshift effect outweighs the blueshift effect for an overall redshift factor of 2, as above.

Note that this redshift factor is not necessarily the same as the rate Bob's clock runs relative to Alice's. The latter requires a definition of simultaneity, and if Alice is at or inside the horizon, there is no natural definition of simultaneity between her and Bob, so there is no natural way to define their relative clock rates.