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Black holes have such a strong gravity that escape velocity from it is more than speed of light which basically means nothing could escape it.

Everything in the universe have escape velocity. For Earth it is 11.2 km/s which means if any body has speed of 11.2 km/s from earth's surface then it could reach infinite distance away and would never come back to earth.

If something has half the speed of escape velocity say 5.6 km/s then the body would return back to earth after travelling some distance.

Suppose a black hole have escape velocity double the speed of light the it would mean photons can't escape it.

But as in our previous example of earth where body with 5.6 km/s speed travelled some distance before coming back to earth in the same way any photon emitted from that black hole would travel some distance before coming back to black hole.

It would mean if any person is standing outside the range of that photon he wouldn't see black hole but he could see it if he go close in range of the photon.

A very similar question could be that if any photon is emitted from the surface of earth then would its speed decrease or energy decrease as it move away like blue light photon turn less intense .

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  • $\begingroup$ To avoid confusion it may be worth noting that black holes themselves do not emit light or any other EM radiation, but the accretion disk around a black hole can emit a very great deal of energy. $\endgroup$
    – gandalf61
    Oct 2, 2023 at 9:36

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Your intuition is that of Newtonian mechanics, I can say this because you said "Black holes have such a strong gravity that escape velocity from it is more than speed of light which basically means nothing could escape it". In fact, gravity is much more complicated than that. In the frame of general relativity, it is not surprising to have a Black-Hole whose surface gravity is less than Earth's.

As your question seems to focus only on the horizon's gravity, I will not consider the internal structure of a static black hole solution in General Relativity. But I do think that it is noteworthy to consider it too in order to have a good understanding of a black hole. Furthermore, I will say "infinite amount of" because the maths blow up at the coordinate singularity (the event horizon) in the coordinates I chose. But it is to be interpreted as "when you approach the horizon from the exterior, things tend to either $0$ or $\infty$".

Firstly, what is an event horizon in General Relativity, in simple words? To keep things simple, I will only consider Schwarzschild's black holes, which do not rotate. An event horizon is a surface in space such that the first component of the metric tensor is precisely $0$. To understand this statement, let us focus on the Schwarzschild line element, which we can interpret as a deviation of the geometry of space-time from the Minkowski picture: \begin{equation} ds^2=\underbrace{\left(1-\frac{r_s}{r} \right)}_{\text{radial time dilation}}\times\overbrace{dt^2}^{\begin{array} \\ \text{infinitesimal}\\ \text{time parameter}\end{array}}-\underbrace{\left( 1-\frac{r_s}{r} \right)^{-1}}_{\text{radial length contraction}}\times\overbrace{dr^2}^{\begin{array} \\ \text{infinitesimal} \\ \text{radial parameter} \end{array}}-\underbrace{r^2d\Omega_2^2}_{\text{angular part}} \end{equation} Concretely, $ds^2$ is the "square" of an infinitesimal length in space-time and is called the "space-time interval" (among other appellations). Note that when $r=r_s$, the "radial time dilation" becomes 0, this means that the event horizon is located at a radius $r_s$ from the origin center. Of course, the "radial length contraction" factor blows up at this radius, but this is a feature of Schwazrchild black holes. In the Kerr geometry, which describes a rotating black hole, this is no longer the case.

But what does it mean to have a time dilation factor of $0$? It means that, for the faraway observer, times will appear to stop at the horizon. So if you are far away from the horizon and drop an apple towards it, the apple will gradually stop moving, and ultimately be completely stopped right at the horizon. Of course from the point of view of the apple, time does not stop. But if the apple has a camera stuck to it that films you, it will see you in accelerated motion. Ultimately, when the apple is at the horizon, it would film the death of the universe (of course, if the camera has an infinite lifetime, and if the black hole does not evaporate).

Now, I have focused on the falling of the apple only, but not on its appearance. As you may know, you see electromagnetic radiations from ~400nm to ~800nm. But, as the apple falls towards the horizon, the light it scatters in your direction will be redshifted. So light still does travel at $c$, but the more it is emitted near the horizon (and directed outwards), the more its wavelength is dilated. Ultimately if the light, which has a non-zero wavelength, is emitted toward the faraway observer from the horizon, then from the point of view of this very observer${}^{\ast}$, light has an infinite wavelength, and so a frequency of $0$, and so does not exist anymore. Thus, black holes are classically... well, black.


I still must clarify some things about your question. The phenomenon you describe well, the fact that objects having a speed below the escape velocity fall back to the Earth, happens for light too, but when you are inside of the black hole!

Also, what I described is purely in terms of the faraway coordinates $(t,r,\theta,\varphi)$. In other coordinate systems, you have other conclusions. Among those is, for example, the fact that you can fall in a finite time into the black hole. But in this case, the time parameter is not the cosmological time $t$ anymore. See this Wikipedia link for the other representations of the line element I wrote above.

${}^{\ast}$Finally, I said that you will no longer see the emitted light, but this is if you stay still for an infinite amount of time! Indeed, due to the "radial length contraction" term, which blows up at the horizon, light has to travel an infinite distance to reach you, and so it takes an infinite amount of time.

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Let us conduct a following experiment:

We send a photon to a mirror hovering near the event horizon of a black hole, and we observe what happens when we move the mirror closer to the event horizon.

We will note that the time it takes for the photon to return approaches infinity, as the mirror approaches the event horizon.

And we will notice that the mirror has to be made to have the ability to reflect photons whose energy approaches infinity, as the mirror approaches the even horizon.

So now we conclude that at the event horizon photon takes an infinite time to move a short distance and the energy of a photon changes an infinite amount when it moves a short distance.

Now is this is a complete description of what happens to photon? Well yes. Photon is a simple thing, and so is a black hole. So it makes sense that the description is simple.

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  • $\begingroup$ the energy of a photon changes an infinite amount when it moves a short distance” - The energy of a photon doesn’t change in the Schwarzschild spacetime when properly measured in the same coordinates. The effect you are describing is that the same constant energy appears increasing when measured in different coordinates closer and closer to the horizon. “photon takes an infinite time to move a short distance” - It takes about a second when the mirror is one Planck length off the horizon. For longer times the mirror must be at a sub-Planck distance that has an unclear physical meaning. $\endgroup$
    – safesphere
    Oct 2, 2023 at 16:25
  • $\begingroup$ @safesphere I know that. p of E=pc approaches infinity, c approaches zero, when photon approaches event horizon. But I will not 'correct' my answer. The simplicity would be spoiled. $\endgroup$
    – stuffu
    Oct 5, 2023 at 12:08
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No. Light does not escape up to any point. It isn't the case that light goes to some distance away from, say, an $r$ slice if we emitted from a constant $r$ surface inside a Schwarzschild black hole before falling back, like the example of a ball thrown up, which first escapes to a certain distance. This is why we say that the expansion of null congruences for surfaces inside the event horizon is negative, or that the null orthogonal expansion is negative along all directions (future inward and outward orthogonally) for trapped surfaces and (future inward/outward, and zero along the other orthogonal direction) marginally trapped surfaces for any $r$ slice in a black hole. So no -- light does not reach any point inside the black hole, but instead has a negative expansion, and hits the singularity inevitably.

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If a luminous object falls into a black hole, that object will continue to emit light as it does so (for some finite time measured on its own clock), even if it is inside the event horizon.

That light cannot emerge from the event horizon or travel to any observer that is outside the horizon, even if they wait an arbitrarily long time for the signal to arrive.

Using Newtonian physics to describe black holes is a bad idea. The concept of escape velocity is not a valid description of why nothing can escape from inside the black hole event horizon. A better description can be found in numerous duplicate questions (e.g. Why can't light escape from inside event horizon of Black Holes?), but I am fond of a "first principles" description I wrote here.

As another example: It may be the case that, at the event horizon, there is a numerical coincidence between the Newtonian escape velocity and the speed of light, but light could only escape from just above the event horizon if it is directed radially outwards. A Newtonian escape velocity is actually an escape speed and it would not matter in which direction an escaping object is launched. However, if light is directed at any other angle than radially in the scenario described above, it will actually not escape and curve back towards the black hole - non-Newtonian behaviour.

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