How is light influenced really close to the event horizon?

[MisterMuoN]

I was just wondering as it seems pretty counter intuitive that there is a really defined horizon where light can't escape from a black hole. It would make more sense to me if light gradually curved into one. Or does it do this? Please enlighten me ;)!

 

[phinds]

I was just wondering as it seems pretty counter intuitive that there is a really defined horizon where light can't escape from a black hole. It would make more sense to me if light gradually curved into one. Or does it do this? Please enlighten me ;)!

What do you mean by "defined horizon"? If you are asking if there is some actual physical boundary, then no, but light near the event horizon has one of three possibilities. Outside, it can escape to infinity, at the EH it just sits there, and inside in cannot escape and heads for the singularity. If the singularity, whatever it is, is point-like or spherical then the EH is a perfect sphere (i.e. "defined" mathematically). Your concept of "gradually" doen't seem reasonable to me. It's binary (well, ternary actually; outside, at, and inside)

 

[Ibix]

It would make more sense to me if light gradually curved into one.

Are you thinking of the "black star" concept from Newtonian gravity? A body so dense that the escape velocity exceeds the speed of light? Under Newton, one could imagine throwing an object up at the speed of light and having it come back down like a thrown ball. Unfortunately, Newtonian physics is a low-speed, low-energy approximation to relativity. It is a very bad guide to how light and gravity interact, especially in strong gravitational fields like a black hole.

A better description is that in curved spacetime near a mass, fewer of the possible paths a particle could follow lead away from the mass than lead towards it. The event horizon is the place where all paths always lead inwards except for one "radially outward" path that doesn't lead anywhere at all and is only accessible by light ("it just sits there", as phinds says). Inside the horizon the curvature is such that all paths lead inwards, full stop.

 

[MisterMuoN]

Okay awesome thankyou both. By defined horizon I just meant the event horizon. Which according to my knowledge is at a certain (defined) distance to the singularity. Where outside of it Things have a chance of following a path that's not leading into the black hole. I've studied SRT last semester but we didnt do anything involving gravity and light. I believe its general relativity that describes this right? My question comes from the assumption light gets attracted like a mass, of which I now see is wrong. Thinking of it like a mass is why i found it weird how (the bending of)light would behave so binary.
Now that you gave me new brainfood. Could one of you explain more about light that just 'sits' at the EH?

Thanks a lot again
OP

 

[phinds]

Now that you gave me new brainfood. Could one of you explain more about light that just 'sits' at the EH?

The escape velocity outside the BH is less than c, so things can escape. The escape velocity inside the BH is > c so nothing can escape [actually, things get weird inside the EH so I'm not even sure if "escape velocity" is a meaningful concept in that region]. The escape velocity AT the EH is exactly c and light travels at c, so it just sits there moving at c (locally) but not overcoming the escape velocity.

 

[jbriggs444]

Which according to my knowledge is at a certain (defined) distance to the singularity.

A black hole is not a spherical volume with a singularity in the center. Its "radius" is not a distance to the singularity. The so-called radius of a black hole is a number computed by dividing the surface area of the event horizon and solving $a=4 \pi r^2$ for r. But it's not a distance.

Relative to any locally inertial reference frame covering a portion of the event horizon, the horizon is a surface that moves at the speed of light.

 

[pervect]

I was just wondering as it seems pretty counter intuitive that there is a really defined horizon where light can't escape from a black hole. It would make more sense to me if light gradually curved into one. Or does it do this? Please enlighten me ;)!

Call a velocity that escapes "positive", and a velocity that falls in "negative". And of course a velocity that doesn't go anywhere is zero. Then if the velocity is a continuous function, if it is positive at some point (i.e. where light escapes), and negative at some other point (where it falls in) must have a value where it's zero, as long as it is continuous. Which in this case, it is.

 

[PeterDonis]

the event horizon. Which according to my knowledge is at a certain (defined) distance to the singularity.

Not really; it's more correct to think of it as a defined time to the singularity. The reason for that is that the singularity is not a place in space; it's a moment of time; and this moment of time is to the future of all events inside the hole's horizon.

I believe its general relativity that describes this right?

Yes. More specifically, it's a particular solution of the Einstein Field Equation of GR, which is called the Schwarzschild solution.

My question comes from the assumption light gets attracted like a mass, of which I now see is wrong.

No, it isn't. The paths of light rays moving tangentially past a massive object get bent by gravity just like the paths of ordinary objects. Light bending by the Sun is one of the classic predictions of GR.

However, the light we are talking about light that is moving radially. What defines the horizon is that radially outgoing light stays at the same radial coordinate. Sometimes this is phrased as the light "staying in the same place", but that's not really a good description because the horizon is not a "place". One way of seeing why is to observe that, as @jbriggs444 pointed out, in any local inertial frame that contains the horizon, the horizon is moving outward at the speed of light; and nothing that moves at the speed of light can ever be at rest in any local inertial frame, which is what a "place" would have to do.

i found it weird how (the bending of)light would behave so binary.

It isn't binary. Consider three rays of light, all directed radially outward, and all emitted at the same time in some local inertial frame that contains the horizon. One is emitted just inside the horizon, one at the horizon, and one just outside. As far as the local inertial frame is concerned, these three rays of light are all moving at the same speed--the speed of light--and the same direction. So within the local inertial frame, there is no discontinuity between them.

If we consider how the light rays behave beyond the local inertial frame where they were emitted, there is still no discontinuity between them. The ray at the horizon stays at the horizon: that is, its radial coordinate never changes. The ray just inside the horizon is falling towards the singularity, but at first its radial coordinate decreases very slowly, and the closer to the horizon it was emitted, the more slowly its radial coordinate decreases. Similarly, the ray just outside the horizon is escaping towards infinity, but at first its radial coordinate increases very slowly; and the closer to the horizon it was emitted, the more slowly its radial coordinate increases. So there is a continuous series of light rays, in terms of how their radial coordinates change; there's no discontinuity or binary division between them.

 

[PeterDonis]

@MisterMuoN I have written a four-part series of PF Insights articles discussing the Schwarzschild solution. The first one is here:

https://www.physicsforums.com/insights/schwarzschild-geometry-part-1/

You might find them of interest.

 

[MisterMuoN]

It isn't binary. Consider three rays of light, all directed radially outward, and all emitted at the same time in some local inertial frame that contains the horizon. One is emitted just inside the horizon, one at the horizon, and one just outside. As far as the local inertial frame is concerned, these three rays of light are all moving at the same speed--the speed of light--and the same direction. So within the local inertial frame, there is no discontinuity between them.

If we consider how the light rays behave beyond the local inertial frame where they were emitted, there is still no discontinuity between them. The ray at the horizon stays at the horizon: that is, its radial coordinate never changes. The ray just inside the horizon is falling towards the singularity, but at first its radial coordinate decreases very slowly, and the closer to the horizon it was emitted, the more slowly its radial coordinate decreases. Similarly, the ray just outside the horizon is escaping towards infinity, but at first its radial coordinate increases very slowly; and the closer to the horizon it was emitted, the more slowly its radial coordinate increases. So there is a continuous series of light rays, in terms of how their radial coordinates change; there's no discontinuity or binary division between them.

Well that's interesting.. What is the exact definition of a radial coordinate? And when somethins radial coördinaten increases slowly does that mean its radial speed is slower? It probabiliteit won't because c=c but what do you mean then?

Also thanks for the four part series, I will look into them!

 

[PeterDonis]

What is the exact definition of a radial coordinate?

The radial coordinate of an event is defined as $r = \sqrt{A /4 \pi}$, where $A$ is the area of the 2-sphere on which the event lies.

when somethins radial coördinaten increases slowly does that mean its radial speed is slower?

Yes. In a curved spacetime, the coordinate speed of light does not have to be $c$.

 

[Ibix]

What is the exact definition of a radial coordinate? And when somethins radial coördinaten increases slowly does that mean its radial speed is slower? It probabiliteit won't because c=c but what do you mean then?

A typical way of defining coordinates on the surface of the Earth is latitude and longitude. Imagine walking along the equator at 1m/s, a slow walking pace. It'll take around 31 hours to change your longitude by one degree, so your coordinate velocity is about $9\times 10^{-6}$degrees per second.

Now imagine walking in a circle of radius 1m around the north pole. This time you need barely a sixtieth of a second to change your longitude by one degree. Your coordinate speed is around 57 degrees per second.

The point is that the same change of coordinates doesn't have the same physical meaning when you are in different places. This is a general property of coordinate systems on curved spaces. In Schwarzschild coordinates, the decision is made to make spatial coordinates on any spherical surface work as normal (analogous to distances north-south on a Mercator projection). But this means that the coordinates don't match up in either the time or radial directions - both have a variable relation to what clocks and rulers actually measure (analogous to east-west directions on a Mercator projection). So the coordinate speed of light varies, but that's because the radial coordinate isn't a fair measure of distance.