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I heard that a black hole is not black because it's escape velocity is greater than or equals to the speed of light. But instead it is black because the light that gets emitted from a black hole gets red shifted due to the curvature of space-time near that black hole.

  1. Is it true or not?

  2. If it is true then tell me how does light even escape from a black hole to be red-shifted in the first place?

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marked as duplicate by Rob Jeffries, user36790, Gert, ACuriousMind, Sebastian Riese Feb 19 '16 at 2:14

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No, this is a poor description of what happens at the event horizon. However the concept of an escape velocity is also a poor description of what happens, since that implies the light tries to escape but falls back to its doom.

There isn't any simple way to explain what happens because general relativity involves concepts that are profoundly alien to everyday experience. We think we know what we mean by a length, and we know what we mean by a time interval, so a velocity must surely just be a length divided by a time interval. The trouble is that the simple ideas of length and time break down at the event horizon. Technically our length and time coordinates become singular there, and there is no longer a simple definition of a velocity.

There are various ways round this. On method is to use a different coordinate system that isn't singular, but then the coordinates don't have an intuitive meaning. I take this approach in my answer to Why is a black hole black? where I use the Gullstrand–Painlevé coordinates to calculate the speed of the light. With this approach you can show that the velocity of an outgoing ray of light is zero at the horizon, so if you send a light ray outwards right at the horizon the light is frozen there unable to move. However this velocity is not what you and I mean by a velocity when we are far from the black hole.

Probably the most intuitive way is to consider the velocity of the light as we approach the event horizon. I do this in my answer to Speed of light in a gravitational field?. That answer includes a graph showing that the speed of light falls towards zero as we approach the horizon, though just to illustrate how strange the behaviour is, the speed of an ingoing light rays to zero as well as the speed of an outgoing light ray.

Light does red shift as it leaves the vicinity of a black hole, and the red shift tends to infinity as the source of the light approaches the horizon, however this isn't usually associated with the change in velocity (though I suppose it could be). I would describe the red shift as a result of time dilation. Time runs more slowly near the black hole, so for the same light there are more cycles per (time dilated) second close to the black hole than far away from it i.e. an observer near the horizon observes the light to be bluer than an observer far from the horizon.

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If a black hole were black because the light gets red-shifted, then it wouldn't be a black hole, since we could see it. Any light that is "emitted" comes from accretion disks around the black hole, where reactions that emit photons occur. Inside the event horizon, the escape velocity is above that of the speed of light, and since light has mass (from its energy), it cannot escape.

I apologize if I misunderstood the question or if I just have false knowledge, and I hope I got the terminology right.

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Can light escape a black hole?

No. If it could, it wouldn't be a black hole.

I heard that a black hole is not black because its escape velocity is greater than or equals to the speed of light. But instead it is black because the light that gets emitted from a black hole gets red shifted due to the curvature of space-time near that black hole. 1. Is it true or not?

No. It's black because the light can't get out. The light doesn't get emitted at all. And note that even if light did get emitted, it doesn't change frequency. Gravitational redshift is where the light is emitted at a lower energy. See what Einstein said: "An atom absorbs or emits light at a frequency which is dependent on the potential of the gravitational field in which it is situated". Also note that spacetime curvature relates to the second derivative of potential, and the emission frequency depends on the depth of potential.

  1. If it is true then tell me how does light even escape from a black hole to be red-shifted in the first place?

It isn't true. Light doesn't escape from the black hole. Because at the event horizon the "coordinate" speed of light is zero. Like John Rennie said, the velocity of an outgoing ray of light is zero at the horizon, and the speed of an ingoing light rays is zero too. The thing that people don't tell you about is that the speed of light varies in a gravitational field. See Einstein saying so in the second paragraph here. A black hole is where this is taken to extremis. Note that the "waterfall analogy", where a gravitational field is described as space falling inwards, is popscience nonsense.

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