## Maths behind the ‘Black hole in your pocket’ Video by kurzgesagt – Part 1

- Par AstroNomad
- août 2, 2015
- Aucun commentaire

I guess the tittle explains it all. I assume you arrived at this page after having watched the recent video by kurzgesagt on youtube. If you’ve not watched it already, I invite you to have a look at the video below.

Now that you’ve watched the video I assume that you’ve gathered some background in order to read further. What would be the aftermath if you had a black hole in your pocket? Kurzgesagt suggests 2 scenarios to tackle the explanation.

a) Black hole with the mass equivalent to that of a coin

b) Black hole with the size equivalent to that a coin

In this article I shall try to give you some idea how the Maths works out as seen in the video. In order to get the gist of it you don’t need to know general relativity or quantum mechanics, although some know-how might help those with right amount of enthusiasm to dig in further and explore more. Therefore I assume you possess some knowledge on general physics and gravity. High-school level can already be a solid base.

a) Black hole with the mass equivalent to that of a coin

Let’s assume a US nickel whose diameter is 21.21 mm. So this has a radius of 10.605 mm. Or this is 1.0605 cm. Let’s consider it’s 1 cm for the simplicity of the calculations. Of course you can use the exact value if you prefer.

Now we’re going to use an equation that relates to the gravitational potential energy and kinetic energy of an object present in a gravitational field. (Forget the black hole for a second and I’m going to explain here how to derive this equation)

Let’s assume that the mass dominating the gravitational field is of spherical mass M and the object we consider has a mass m. Let’s consider that our object is at r distance away from the center of mass of M and that the universal gravitational constant is G. Now we could write the gravitational potential energy of m due to M, being E_{p} and the kinetic energy of m, being E_{k}. We also assume that the object m has an instantaneous velocity of v.

E_{p} = – GMm/r

E_{k} = mv^{2}/2

In fact we the total energy this object has, at distance r away from the center of M at this instant is the sum of E_{p} and E_{k}. Then we would consider the same object m at infinity distance away from the center of M. If we were to write the same 2 expressions modified according to the latest position of m, the total sum would be simply 0 for it could be assumed that gravitational potential field of M no longer holds at the distance of infinity. Since the object m is the same in 2 situations we could assume that its total energy is conserved and hence write the following relationship.

E_{p} + E_{k} = 0

– GMm/r + mv^{2}/2 = 0

Then,

GMm/r = mv^{2}/2 , we could cancel the ‘m’ from both sides out:

v^{2} = 2GM/r

If we isolate v in this relationship we get now the equation for the escape velocity for a given mass, M. With the concept of escape velocity we have to take into account the fact that the object willing to escape rests on the surface of the large body of mass M, and hence r becomes R which is the radius of the aforementioned spherical mass. You can notice that the escape velocity is independent of the own mass of the object willing to escape. No matter what size the object is, it will escape the gravity of the massive body if it achieves the escape velocity of the massive body. Just to clarify, it only depends on the mass and the radius of the massive body. For example if you want a ball to escape Earth’s gravity and shoot into space, you have to give it a velocity which is at least 11.2 km/s. You may obtain this value by substituting in the following equation, M, R and G by mass of the Earth, radius of the Earth and the value of universal gravitational constant respectively.

v= √(2GM/R)

Now let’s come back to the black holes. It’s known that the escape velocity of a black hole is C, the speed of light in a vacuum. Therefore our relation now becomes as follows.

C= √(2GM/R)

So now we have to calculate the radius of a black hole with a mass of 5g which is what is assumed for the nickel coin by kurzgesagt. So you have to plug in the values of C, G and M to come up with R, which is the radius of this black hole. The values of the constants are as follows.

C = 3 x 10^{8} ms^{-1}, G =6.67 x 10^{-11} Nm^{2}kg^{-2 }, M = 5 x 10^{-3 }kg

Then, R = 7.41 x 10^{-30 }m.

So I hope it’s clear how this number, or rather its magnitude shows up in the video. This radius is incredibly small and they try to give some analogy comparing this size to that of a Hydrogen atom whose size is about 10^{-11} m. Not bad.

Next you have to deal with the concept of Hawking radiation, resulting in the evaporation of a black hole. All the black holes evaporate……at certain point they finish and will be no more. YES ! What matters is the mass of the black hole. More massive it is, the longer it lives, or rather longer it takes it to evaporate away. Let me just brief a bit without throwing you directly at some digits with flabbergasting magnitude differences.

In fact one would think that the space is empty. However when it comes down to the sub-atomic level it isn’t necessarily the case. It is known that a pair of sub-atomic particles spontaneously is created in the void out of the nothing. One of these particles has positive mass while the other has negative mass. These 2 particles exist for a very tiny amount of time and annihilate each other. This is happening everywhere in the space all the time. Stephen Hawking did the thought experiment asking himself what would happen if this pair of particles was created at the event horizon (edge) of a black hole. He arrived at the conclusion that the particle with the positive mass would have just enough energy to escape from the black hole while the other particle with the negative mass would fall into the black hole. Then he figured out that consequently, the particles falling inside black hole would decrease the mass of the black hole due to their effective negative mass. The particles with positive mass that escape the black hole give rise to a faint glow around the black hole, which is known as the Hawking radiation, behaving just like heat. In a nutshell aka kurzgesagt, this Hawking’s result implies that black holes always emit radiation, instead of being black. Thus eventually the black hole gets smaller and smaller as its mass decays. However its temperature keeps increasing as it’s being dwarfed. When the remaining mass becomes critical, it explodes loosing all of its energy and this is the evaporation of the black hole. However this process takes place in cosmological time frames for all the black holes with significant amount of mass (in the order of solar masses).

So here is the equation that relates the mass M, of the black hole with its time for evaporation, t_{years}. M_{s} is the solar mass. Check here if you want to know how to derive this.

t_{years} = 2.0903 x 10^{67} x (M/M_{s})^{3}

We know already that the mass of this black hole would be equal to that of the original coin we had, which is, 5g. For the record, one solar mass is 1.989 x 10^{30 }kg.

So you just need to substitute the above numerical values in the previous equation and convert the years to seconds. Then you should have the following value for the evaporation time for the black hole.

t_{seconds} = 1.0477 x 10^{-23 }

and this is of the same magnitude as the value shown in the video. However for real black holes this value is very large, so large that some black holes easily have a value greater that the current age of the universe. As said before, it all depends on mass.

Then in the video, they give an estimate of the energy liberated during this evaporation / exploration of the video. This is rather easy and we need to think about the famous Einstein’s relationship that relates mass, M to the energy, E.

E = MC^{2}

Again, we have to plug in here the values of the mass of the coin (5g) and the speed of the light in a vacuum. Then one should obtain the following results.

E = 450 x 10^{12} J

It is also important to keep in mind that this amount of energy would be shared by the different wavelengths across the spectrum, i.e visible, ultra-violet, infra-red radiation. There will also be a contribution of the energy for the sound created during the explosion. Whatever happens, this is still a lot of energy and could be catastrophic.

Well, this concludes the first part of the explanation for this video, relating to the black hole with mass equivalent to that of a coin. It has taken longer than I expected and I will be publishing the second part of the explanation in a different blog post although I intended to combine the 2 parts of the explanation in a single post. Just to clarify, the next and last part will be dedicated to the scenario of a black hole whose size is equal to that of a coin.

…and voilà, this is the raison d’être for this blog post. I hope now you have more of an insight into the appearance of the numbers in this video. Looking forward to your comments and questions if any. Of course kudos to kurzgesagt for the awesome video !!

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