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Perhaps there is another way to calculate the escape velocity, or perhaps light always has the same kinetic energy regardless of the energy of the light, but shouldn't the event horizon of a black hole be different distances from the singularity for different colors (and thus energies) of light?

The amount of potential energy due to gravity is calculated via

$U_{grav} = -\frac{Gm_1m_2}{r}$

Escape velocity is the velocity at which the kinetic energy is equal to the gravitational potential energy.

However, with light, the energy depends on the color or frequency. If that energy is greater than the potential energy due to gravity, then the photon should be able to escape the gravitational force (unless, of course, the energy I'm think about for light is different than the kinetic energy that is necessary to overcome the gravitational potential energy). Couldn't, then, low energy light be captured by something close to a black hole while high energy light is able to escape? Could you, then, theoretically, though not practically, have a purple hole?

I also realize that things become more complex than that light can't reach escape velocity to escape a black hole (or else an indestructible human could build an indestructible tower to get out). So, (1) does my hole argument (pun intended) fall apart with that simply having one form of energy doesn't mean that you can escape gravity, but it has to be kinetic energy or something? (2) If not, does other aspects of a black hole either forbid high energy light from escaping even when it has more energy than (the absolute value of) gravitational potential energy or allow low energy light to escape even when it has less energy than (the absolute value of) gravitational potential energy?

CLARIFICATION

I know that the escape velocity at the event horizon is greater than the speed of light and that no colors of light can go faster than the speed of light. However, I'm saying that "escape velocity" is only a special case of "escape energy." That is, objects don't need some speed in order to escape gravity, they need a certain kinetic energy that is greater than or equal to the absolute value of gravitational potential energy. The only reason that "escape velocity" works for most objects is because the kinetic energy is calculated as half of its velocity squared times its mass. Therefore, an object's kinetic energy is determined by its mass and velocity. Because the gravitational potential energy is also determined by its mass, additional mass of an object does not give it enough energy to escape. Therefore, for most cases, only velocity contributes to an object's ability to escape gravity. Nevertheless, if some other property increased the object's kinetic energy, then it would be able to escape gravity at the same speed as something else that didn't have enough kinetic energy to escape.

But light's kinetic energy (as far as I know) is not determined only by its velocity. For example, different colors of light have different energies while having the same speed (unless that energy doesn't contribute to light's ability to escape gravity). In order to determine whether light can escape a gravitational field, you would check whether its kinetic energy is greater than its gravitational potential energy rather than whether its speed is greater than this "escape velocity." It's not the velocity that matters but the energy. Usually, velocity is the only factor that contributes to the energy (other than mass which is canceled out by the gravitational potential energy's dependence upon it). But with light, the energy isn't only determined by its speed. It's also determined by its frequency. Does that mean that high energy light has enough energy to overcome gravity, or does the energy due to frequency of light not contribute to its ability to overcome gravity (i.e. not kinetic energy)?

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    $\begingroup$ Do you understand why light objects and massive objects have the same escape velocity even though they have different amounts of kinetic energy? $\endgroup$
    – BowlOfRed
    Feb 20, 2019 at 17:51
  • $\begingroup$ @BowlOfRed Because they also have different gravitational potential energies. $\endgroup$ Feb 20, 2019 at 19:24
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    $\begingroup$ BTW, there's generally no point in using Newtonian physics on black holes, although it can sometimes coincidentally give correct results, eg calculating the Schwarzschild radius from the mass. $\endgroup$
    – PM 2Ring
    Feb 20, 2019 at 20:08
  • $\begingroup$ @PM2Ring Exactly. Once you accept that the energy of a photon is dependent on its frequency which can take indefinitely large values, Newtonian physics fails to give you any true black hole. OP seems to be taken in mainly by this very confusion. $\endgroup$
    – user87745
    Feb 22, 2019 at 10:14
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    $\begingroup$ @BillAlsept With all due respect, if you are making claims that are based on your own research and are not generally accepted in the scientific community, you should mention it explicitly. $\endgroup$
    – user87745
    Feb 22, 2019 at 16:50

3 Answers 3

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All wavelengths of light have exactly the same velocity in the vacuum surrounding a black hole. This means that a "purple hole" cannot form in the manner you propose.

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  • $\begingroup$ But do different colors of light have the same kinetic energy? I know that $\frac{1}{2}mv^2$ doesn't work because light is massless. I also know that different colors have different energies, but would those energies contribute toward overcoming the gravitational potential energy? I suppose a better question to ask is if "escape velocity" should be "escape energy" because the escape velocity is derived by solving for what velocity gives the object enough energy to overcome gravitational potential energy. $\endgroup$ Feb 20, 2019 at 19:28
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    $\begingroup$ @Elliot Gravitational lensing demonstrates that light rays of all frequencies experience identical gravitational deflection. $\endgroup$
    – PM 2Ring
    Feb 20, 2019 at 20:08
  • $\begingroup$ @PM2Ring I suppose that makes sense considering General Relativity. I guess I'm just realizing one example of how Newtonian Physics is wrong. $\endgroup$ Feb 21, 2019 at 15:34
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Although the existing answers are perfectly correct, I think they don't directly address the reason as to why OP thinks it is possible to have different effective horizon radii corresponding to different colors of light--despite the fact that OP seems to (somewhat contradictorily) understand that there would be a true horizon radius beyond which no light can escape. So, I will give it a try.

The proposal by OP seems to suggest that there would be an ordinary true horizon radius for all the massive particles because the only thing that determines their kinetic energy is their speed and thus if their speed needs to be greater than the speed of light then they simply can't escape. And if you go close enough to a black hole, your speed would be required to be greater than the speed of light in order for you to escape and thus, you cannot escape. And this radius beyond which this starts happening is the ordinary horizon radius. Now, OP suggests that the story is a bit different for light because it is not the speed that determines the energy of the light but the frequency. Thus, for different colors of light, there would be different horizon radii beyond which they cannot escape. Now, this is untrue for a multitude of reasons I will try to explain the most basic flaw in this argument.

First of all, this seems to suggest that there would be no true black hole. Given enough frequency, the light will always escape any gravitational pull and thus, there would not be a true black hole which prevents absolutely everything from escaping its pull beyond a certain distance. The proposal by OP seems to be deeply embedded in Newtonian thinking. You see, in a Newtonian black hole, the interior of a black hole is a region in space--so you can think of "starting at a point" inside the black hole and of approaching towards the horizon. If you have enough energy, you will escape just like a rocket with enough energy escapes the earth--if not, too bad--you will eventually reach a zero velocity and then you will be pulled towards the center again. This seems to motivate OP to think that if only a photon had high enough frequency (which it can have--there is no limit on how high the frequency of a photon can be as long as quantum gravity effects don't take over) then it can just escape any pull of gravity. Now, all of this is simply untrue because the black holes we have in nature are not Newtonian black holes. They are general relativistic black holes. The interior of a black hole is not a region in space but rather a chunk of time. As is often said, going towards the center of a black hole inside the event horizon is not a journey towards a point in space but an inevitable flow towards the future that nobody can avoid. Therefore, no matter how high the energy of a photon is, it will simply head towards the center of the black hole if it is inside the event horizon.

The proposal by OP can perhaps be seen to mean something more benign. Let's say we all know and understand why there is only one true horizon radius and no light can escape the black hole if it is once inside this radius. Now, you can think of massive particles with low enough velocities outside the horizon who will inevitably fall into the black hole simply because they don't have the required (and, in principle, achievable) velocity to escape. The proposal by OP can be seen to mean an analog of this phenomenon for light which simply doesn't exist as I will try to elaborate. In particular, OP might suggest that the light can have low enough frequency outside the event horizon so that it will nonetheless fall into the black hole. If so, then we can resurrect the idea of an orange hole suggested by OP which is to say, okay, if you had a high enough frequency, you could escape this potential since you are outside the event horizon, but since you actually have an orange frequency which doesn't have as high an energy as required (as dictated by how close you are to the horizon), you will not really escape. This is not true the moment you start to think about the effect that the pull of the black hole exerts on a light outside the horizon. You see, for massive particles, decreasing energy would mean decreasing velocity and thus, eventually, they will reach a zero velocity (and will return then towards the black hole) if they didn't start with the required amount of energy based on their distance from the horizon. But this is not true of light because decreasing energy simply means decreasing frequency for light. The effect of the pull of gravity will never be to decrease the speed of light as it can't. Thus, the light will keep becoming reddish and reddish but it will inevitably escape the black hole given it started outside the horizon--no matter how low the energy of the light actually was. (This is the famous gravitational redshift one talks about regarding the light observed coming from a black hole.)


Main Points to Keep in Mind (IMHO)

  • Even if we accept the notion of the speed limit, Newtonian gravity doesn't give us true black holes provided that we also accept that the energy of light depends on its frequency which can have as high a value as one wishes.
  • Nature is more relativistic than Newtonian, and thus, luckily, we do have true black holes for which there is a true horizon beyond which nothing can escape including all the variety of light.
  • Massive particles outside the horizon will need to have certain velocities depending on how far they are from the horizon to actually escape the black hole, but the light will always escape the black hole once it is outside the horizon--no matter how low its energy is.

Note: I have assumed that all motion is radial, if you have, for example, purely orbital motion, the light will have to be even farther than the horizon to stay outside the horizon forever. Although interesting, that is not much relevant to the question that OP asked here.

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  • $\begingroup$ A provocative comment: Be cautious whenever you start to think that something fundamental depends on the color of light ;-) $\endgroup$
    – user87745
    Feb 22, 2019 at 9:43
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The speed of light is the speed of light. But the escape velocity required to overcome the gravitational pull of a black hole is greater than the speed of Light. And as we know that nothing travels faster than light, so light, no matter how energetic or what wave-length cannot escape.

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  • $\begingroup$ Your statement is only true at the event horizon and within. Outside the event horizon it can escape even though it’s still in the gravitational pull of the black hole. $\endgroup$ Feb 22, 2019 at 22:48
  • $\begingroup$ I was basing this answer on the OP's original post. $\endgroup$
    – Rick
    Feb 23, 2019 at 2:05

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