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As you come in closer to a black hole, how do you see the event horizon? Is it always like a clear-cut surface? Or it only looks clear-cut from a distance, but as you come closer to the black hole, you start seeing it's a blurry layer, and everything around you gets gradually darker as you fall through?

I know that the Schwarzschild radius would define an exact (clear-cut) sphere around the singularity (the surface of no return), but what actually happens to light itself? Anyway, my question might be lacking in imagination, since light gets bent badly already outside the event horizon.

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Actually in most cases you don't see the event horizon, but instead the photon sphere. For example if you are looking from some distance, if light emitted from some star goes inside the photon sphere (where light can travel theoretically in circular orbit, though the orbit is unstable), which is located outside the event horizon, it is more or less doomed to go inside the event horizon also, unless it scatters from anything. This is because a photon inside the sphere can bend only more towards the BH, which means that if it's direction was inside the BH while crossing the photon sphere, it can never start pointing outwards without interaction.

Some time ago I simulated what would it look like if you were nearby a black hole.

Picture of a black hole.

In the picture, the edge of the black area marks the photon sphere. However, if you were inside photon sphere, then of course my argument of photon sphere does not hold, as light coming to you is anyways pointing inside the hole. I don't really know what is the edge that you see if you are already inside, but as John Rennie noted, this edge is should be always sharp.

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  • $\begingroup$ I didn't know about the photon sphere, thank you. It really aids the imagination. If I cross into the photon sphere, but not yet the event horizon, then I will be "blinded" by starlight coming from all directions, right? All photons are spiralling into the event horizon, and they're coming from all sides of the photon sphere. I can't imagine I could distinguish the image of any star though, due to chaotic distorsions. $\endgroup$ – CamilB Jul 23 '15 at 19:31
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    $\begingroup$ Well, the intensity probably does increase, as you fall in, but it does so gradually. You can see individual stars, though the picture is of course highly distorted. Also you can see several images of the same star, because a ray can either come almost straight to the viewer or make a full loop around the BH and then hit the observer or make even more loops. Every configuration will give a different image. However, the images resulting from rays that have made several loops are extremely distorted and almost impossible to see. $\endgroup$ – kristjan Jul 24 '15 at 15:24
  • $\begingroup$ @CamilB The viewer can't see anything very special while he is falling through the photon sphere or event horizon. The videos in John Rennie's link are actually very helpful. $\endgroup$ – kristjan Jul 24 '15 at 15:29
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Calculating what you would see as you fell into a black hole is straightforward but tedious. Fortunately there are lots of sites that have done this for you. Actually, if you've been to the cinema recently the film Interstellar does a pretty good job of it.

Less spectacularly, have a look at this site that has videos of what the journey would look like. There is a sharp cutoff between the light and dark areas, though the bending of the light means the cutoff isn't simply the edge of the black hole.

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The event horizon or 'point of no return' is the point where the escape velocity of an object falling into the black hole has to be greater than the speed of light (about 299,792,458 m/s). That concludes that light can't escape it, therefore there is a sharp, clear distinction between the event horizon itself and the background. If you moved towards the black hole and you stopped just before the event horizon and turned your head to either left or right, the light which was reflected from the back of your head would then be bent by the curvature of space-time around the black hole, and you would be able to see the back of your head. Light itself, once passed through the event horizon, will be trapped in the black hole, adding to its mass (light doesn't have mass but it has energy hence E=mc^2). You will also be trapped inside, however the length of your journey would depend on the size of the black hole's Schwarzchild radius, until the gravitational force difference exerted on your feet and head is so enormous that you would experience spaghettification, which is a violent way of gravity ripping you apart.

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