Have I understood correctly that when approaching the event horizon,

Math doesn't work in event horizon.

>>8577269
yes it do. 2 + 2 = 4 here and same in event horizon.

>>8577269
>hurr durr its magic!

>>8577271
No it's not, you can't prove me wrong because everything breaks in log horizon, so math has to break as well.

gravity is so strong that math cant escape from there

>>8577269
This is sort of right.

For instance if you look at the Schwarzschild metric, [math]{\operatorname{ds} ^2} = - \left( {1 - \frac{{{r_s}}}{r}} \right){\operatorname{dt} ^2} + \frac{1}{{\left( {1 - \frac{{{r_s}}}{r}} \right)}}\operatorname{d} {r^2} + {r^2}\operatorname{d} {\Omega ^1}[/math], it blows up at r_s = r.

i.e. It blows up at the event horizon, however you can play with the math to get rid of this singularity.

>>8577269
How can math be real if numbers aren't real?

>>8577566
This.

>>8577257
Pretty sure you'd get time dilated until you died of something before you got close enough to the event horizon to be incinerated. If it were a smaller black hole, you would be spahgettified.

>>8577880
Do black holes not exist, because the matter time dilates infinitely and nevery actually reaches the singularity?

>>8577916
We don't know. Predictions made by GR are unreliable at the "center" of a black hole because there would be quantum effects.

>>8577880
Why even give such an answer? Obviously I'm not asking about a person approaching the event horizon but a hypothetical invincible observer.

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>>8577296
>everything breaks in log horizon

except the DATABASE

>>8577916
not how time dilation works m8

Please answer question.

Pleasse the quetion anser .

>>8577257
It's hard to say what one would perceive as they passed through the event horizon. However, it is understood that (at least for a large enough black hole) as you fall through, you don't feel anything change. The amount of the universe which you could actually receive information from changes dramatically, but a gyroscope or something wouldn't alter its behavior dramatically.

I would imagine the plane you describe actually begins to envelop the space behind you as well, eventually closing in to a point as you approach the singularity. This is because the geodesics inside the event horizon don't allow for light to travel "across" the black hole. They go straight for the singularity, in a sense. So depending on your own trajectory, the light from outside that may cross your path comes from a smaller and smaller region of space as you fall closer.

I hope that makes sense. Technically you could take >>8577757 and compute the geodesics with relative ease and see for yourself.

>>8577269
This is surprisingly a correct answer.

>>8580782
>This is surprisingly a correct answer.
No it isn't. The singularity at the event horizon can be removed by a coordinate change, which is just adjusting the Schwarzschild metric's coordinates. The reason it appeared in the first place is because the chart originally chosen didn't actually contain the entire geodesic of an in-falling object. Hence motion inside and outside the event horizon weren't connected. This is remedied by maximally extending the chart and "pulling" the connected point of the geodesics from outside to inside the new chart. In the new coordinates, the geodesic of an in-falling object is continuous and uninteresting as it crosses the event horizon.

>>8577757
It's as right as "math doesn't work at the north pole".

>>8577257
>when approaching the event horizon, the horizon appears
Taking "appears" literally, you cannot perceive the event horizon until you have passed it.

>Does an observer at exactly the event horizon perceive it as planar instead of curved?
Yes. Definitely for a Schwarzschild black hole (static, noncharged, nonrotating), the event horizon does not have extrinsic curvature. Don't trust me, work it out yourself. It's also moving outward at the speed of light. I think the flat thing is generally true for any static black hole; otherwise the event horizon's surface area would be increasing.

The event horizon does have intrinsic curvature, though, due to gravity, and so it doesn't extend infinitely, it wraps around.

>>8580977
>Taking "appears" literally, you cannot perceive the event horizon until you have passed it.
This is not true.

>Definitely for a Schwarzschild black hole (static, noncharged, nonrotating), the event horizon does not have extrinsic curvature. Don't trust me, work it out yourself. It's also moving outward at the speed of light. I think the flat thing is generally true for any static black hole; otherwise the event horizon's surface area would be increasing.
What the fuck are you talking about? What is "extrinsic" curvature? Why would you say it's moving outward at the speed of light?

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>>8577880
>>8577916
The event horizon has infinite time dilation in the same sense that the north pole has infinite longitude dilation. The only thing that goes to infinity there are human-made coordinates.

For any observer accelerating at constant acceleration a (as measured in his own rest frame), there is a personal event horizon at distance c^2/a from him from which no light will ever catch up with him. It's called the "Rindler horizon". You can introduce space and time coordinates that treat anything that accelerates so as to avoid the same horizon as static; these are called Rindler coordinates, and in these coordinates, there is infinite time dilation as you approach the horizon. But of course nobody who falls through the Rindler horizon notices anything.

>>8580996
>>Taking "appears" literally, you cannot perceive the event horizon until you have passed it.
>This is not true.
It is true. To get from inside the event horizon to outside it, you have to travel faster than light.

>What is "extrinsic" curvature?
Extrinsic curvature is what you think of as curvature.
Intrinsic curvature means that the angles of triangles don't add up to 180 degrees. It's the kind of "curvature" we're talking about when we say "spacetime is curved". It's called "curvature" because surfaces with extrinsic curvature often have intrinsic curvature. Intrinsic curvature is what you can measure from inside the surface without considering any higher dimensions the surface is bent in.

>Why would you say it's moving outward at the speed of light?
This is from the perspective of any observer falling through the event horizon. Maybe you would want to say that it's really the observers moving, not the event horizon, but that's not how physics works; whether something is moving is always relative to who's measuring it; there's no one true rest frame. Also, the event horizon, since it's moving at the speed of light, doesn't have a rest frame. The reason the event horizon doesn't escape the black hole is because the black hole's intense gravity distorts the space above it.

>>8581025
You said you cannot perceive the event horizon until you have passed it. This is false, and it has nothing to do with escaping a black hole. If you are approaching one, you can clearly perceive the horizon, since it is a region which no light will be coming from.

>you have to travel faster than light
No such thing. There are no outward-going geodesics inside the event horizon.

>Extrinsic curvature is what you think of as curvature.
[math]\textbf{d}\omega + \omega\wedge\omega [/math] is what I think of as curvature.
I looked up "extrinsic curvature" and it seems to have something to do with embedding. This is not relevant to the study of general relativity.

>the rest of your post
I don't know what kind of education you have, but it reads to me like you can put together sentences that are syntactically correct but mostly meaningless.

except the 'singularity' isnt a tiny dot at the center of the swirly that youre trying to figure out what happens when u approach it, its thousands of miles equidistant like a sphere that you touch and evaporate through

>>8581054
Complete unfiltered nonsense.

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>>8577296
>math has to break as well.
Stop being retarded. 2 + 2 = 4 no matter where you are. It's not math that breaks down, but the math associated with the physics we currently accept to be true outside of black holes

>>8577269
See this >>8581068

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>>8577269
So at an event horizon numbers that are prime everywhere else will no longer be prime there? At an event horizon Fermat's Last Theorem doesn't hold true? What in the goddamn FUCK are you talking about?

>>8581046
>You said you cannot perceive the event horizon until you have passed it. This is false, and it has nothing to do with escaping a black hole. If you are approaching one, you can clearly perceive the horizon, since it is a region which no light will be coming from.
If you take a point in the dark part of your field of vision, and trace back where a hypothetical light ray arriving at that angle would have had to come from, it's certainly not going to be from the event horizon.

>>you have to travel faster than light
>No such thing. There are no outward-going geodesics inside the event horizon.
Assuming by "outward-going" you mean "increasing Schwarzschild radius", this is false. The correct statement would be that there are no timelike outward-going geodesics inside the event horizon. There are spacelike ones, which is what I just said.

>I looked up "extrinsic curvature" and it seems to have something to do with embedding. This is not relevant to the study of general relativity.
The event horizon is a 2+1-dimensional surface embedded in a 3+1-dimensional spacetime.

>>8581025
>Also, the event horizon, since it's moving at the speed of light, doesn't have a rest frame.
To be clear, there are certainly coordinate systems in which the event horizon's spatial coordinates are not changing. What there is not is a rest frame for a point on the event horizon in the special relativity sense. (Special relativity still works as a local approximation; mathematically the local SR approximations are the tangent spaces of the manifold.)

>>8581077
Yes exactly

Have you heard of Horizon Primes (HjPs)?

>>8577757
not really sort of. not by the technical definition of radius of curvature. you have to switch to kruskal coordinates to see that, and even then, it is not infinite in both directions for all directions.

>>8581094
>If you take a point in the dark part of your field of vision, and trace back where a hypothetical light ray arriving at that angle would have had to come from, it's certainly not going to be from the event horizon.
I think mathematically these end up coming from the white hole event horizon, which is nonphysical. So for a real black hole, you'd be looking at an extremely time-dilated version of the star's collapse.

As you approach a black hole, the horizon always retreats from you.

Anything "ahead" of you is invisible, as all possible paths for the light to travel from them only lead closer to the singularity. Anything "behind" you is visible, for the sane reason.

So assuming you were capable of observation, the universe would appear split in half, the half closer to the singularity than you(apparently empty space), and the half farther away than you.

So yes, approaching the even horizon would appear like closing in on an infinitely black ball until the surface of the ball appears as a flat plain. At which point you are beyond it, and it will always appear so, unchanging.

>>8581289
>this

congrats m8