Creating a Black Hole: Possible?

[SpaceTiger]

Certainly 10^25cm is small, but even many OOM larger, the gravitational field of a 1kg BH must be greater than the Earth's gravity at the surface (to lazy to do calculations)

I think this may be the primary point of confusion here. The gravity of a 1 kg black hole is exactly the same as the gravity of a 1 kg baseball at radii comparable to that of the baseball. You only notice it's a black hole once you get much closer. The "effective radius" that you're talking about (in other words, the distance from the black hole an object has to get to be pulled in by the gravity) depends on the velocity of the incoming object. It can be approximated by saying that an object will fall in when the potential energy is comparable to the kinetic energy. Thus:

$$\frac{1}{2}mv^2=\frac{GMm}{r}$$
$$r \sim \frac{2GM}{v^2}$$

The typical velocity of particles on the surface of the Earth can be approximated via

$$kT=\frac{1}{2}m_pv^2$$
$$v \sim \sqrt{\frac{2kT}{m_p}}$$

where $$m_p$$ is the mass of the proton. For $$T \sim 300 K$$, this gives $$\sim 2 km/s$$. Plugging this into the original equation, along with $$M=1 kg$$, we get $$\sim 3 \times 10^{-15} cm$$. This is still seven orders of magnitude smaller than the Bohr radius (approximate radius of an atom). Remember that atoms are almost entirely empty space, so the chance of a proton being captured by the black hole is very small (many orders of magnitude smaller for an electron).

Furthermore, this "capture" radius isn't even really that, it's just the radius at which the black hole's gravity has a noticable effect. I suspect that you would need a three-body interaction to bind the nucleon to the black hole. Otherwise, its path would just be bent by the gravitational field and it would pass right on by.

Note 1: Above calculation is very rough (good to about an order of magnitude)
Note 2: I used the Newtonian approximation because the capture radius is well beyond the event horizon of the black hole.

 

[Billy T]

Very nice work SpaceTiger, but I still have a few questions, comments below.

...The "effective radius" that you're talking about (in other words, the distance from the black hole an object has to get to be pulled in by the gravity) depends on the velocity of the incoming object. It can be approximated by saying that an object will fall in when the potential energy is comparable to the kinetic energy. Thus:

$$\frac{1}{2}mv^2=\frac{GMm}{r}$$
$$r \sim \frac{2GM}{v^2}$$

The typical velocity of particles on the surface of the Earth can be approximated via

$$kT=\frac{1}{2}m_pv^2$$
$$v \sim \sqrt{\frac{2kT}{m_p}}$$

where $$m_p$$ is the mass of the proton. For $$T \sim 300 K$$, this gives $$\sim 2 km/s$$. Plugging this into the original equation, along with $$M=1 kg$$, we get $$\sim 3 \times 10^{-15} cm$$. This is still seven orders of magnitude smaller than the Bohr radius (approximate radius of an atom). ...Furthermore, this "capture" radius isn't even really that, it's just the radius at which the black hole's gravity has a noticable effect. I suspect that you would need a three-body interaction to bind the nucleon to the black hole. Otherwise, its path would just be bent by the gravitational field and it would pass right on by...

If I understand you correctly, you have evaluated the distance from the BH where the depth of its potential well (infinity being zero potential) is equal to the KE of the typical room temperature proton. Proceeding on the assumption this is correct:
1)First I note that "equal partition of energy" should apply and that the typical oxygen molecule (A= 32proton masses) will be moving almost 6 times slower than your proton.
2)Even if the velocity of the proton were directly away from the BH, since by assumption it is equal to the depth of the potential well, it would just be able to separate from the BH. But of course in the typical case the velocity will not be directly away form the BH. A more reasonable case is that the proton/molecule is in a bound orbit about the black hole, although this does not make much sense either as it is much bigger than your 10^-15cm.
I don't know exactly how to think about this, but bet that the gradient rips molecules apart, and probably even the outer shell electrons of the oxygen atom nearer the BH also are stripped away. Thus in addition to gravity, we are going to have strong electrical forces acting on the +ion oxygen nucleus and a "cloud" of electrons orbiting the BH. My poor classsical mind is not up to doing this right, but you get the idea, which I will try to sumarize as follows:

Because the gravity gradient rips up atoms, makes electrons clouds orbiting the BH, etc. we have opportunity for disapative events (even some EM radiation? - probably very harse UV continium, not lines). Even if some strange "BH / O+ ion" molecule forms which is stable against these disapative processes, (except for line radiation), it is not likely that it will remain stable when others join this "soup of orbiting charges"

3) I forget my statistics, but think the Maxwellian distribution's "hot tail" contributes a relative large amount to the average energy compared to the number of particles in it. That is, unless I am wrong, the median velocity is significantly less than you have calculated. In any case, there are lots of relatively slow moving oxygen molecules for the HB to rip appart and eat.

Again, you probably are right, but I am still not willing to throw in the towel. I think several OOM still more distant from the BH than your now relaxed 10^-15cm there are molecules that just happen to be headed roughtly towards the BH and that although they will initially not feel much of its gravity - little effect on their trajectory, but as they come closer (on their own momentum, not falling in yet) they will get close enough to get ripped up by the gradient, disapatively interact with the electons they once owned and feed that hungry BH.I beet you do not really deny this is a posibility, but may think it takes too long. If this is the case, we need to know how the presence of an atmosphere modifies the rate of "vacuum polarization," if it does. I still would like to look into this alittle and need to know where you 10^-16 second life time comes from.

I am prety confident that you error by setting the KE equal to the potential well depth to evaluate the effective "capture cross section" of BH in Browian motion air, but you certainly are correct to note that on these small scales, most of what you have arround is "nothing but empty space."

What do you think? Should I throw in the towel, or do I still have a point?

 

[SpaceTiger]

1)First I note that "equal partition of energy" should apply and that the typical oxygen molecule (A= 32proton masses) will be moving almost 6 times slower than your proton.

Absolutely right, that was one of the reasons put in the disclaimer. On the other hand, the length I used for comparison (the Bohr radius), was actually extremely conservative (or liberal depending on your point of view). In reality, I should be using the mean distance between molecules:

$$l \sim \frac{1}{n^{1/3}}$$

2)Even if the velocity of the proton were directly away from the BH, since by assumption it is equal to the depth of the potential well, it would just be able to separate from the BH. But of course in the typical case the velocity will not be directly away form the BH. A more reasonable case is that the proton/molecule is in a bound orbit about the black hole, although this does not make much sense either as it is much bigger than your 10^-15cm.

For a particle entering from infinity (in any direction) to enter a bound orbit, it needs to lose energy. A simple two-body interaction is not enough to do that (unless there are tidal forces, but I think that's negligible here). This makes it harder to capture into the black hole.

I don't know exactly how to think about this, but bet that the gradient rips molecules apart, and probably even the outer shell electrons of the oxygen atom nearer the BH also are stripped away.

You're talking about tidal forces. Remember that gravity is $$\frac{1}{r^2}$$ (tidal forces $$\frac{1}{r^3}$$), just like the electrostatic force, so unless the black hole's gravity is comparable in magnitude to the attraction/repulsion from neighboring charges, these effects will be negligible. My previous calculation shows this not to be the case.

Because the gravity gradient rips up atoms, makes electrons clouds orbiting the BH, etc. we have opportunity for disapative events (even some EM radiation? - probably very harse UV continium, not lines). Even if some strange "BH / O+ ion" molecule forms which is stable against these disapative processes, (except for line radiation), it is not likely that it will remain stable when others join this "soup of orbiting charges"

These effects will be negligible for black holes of 1 kg, but you've piqued my curiosity about the possibilities for more massive ones. If I get the chance, I'll look into it.

3) I forget my statistics, but think the Maxwellian distribution's "hot tail" contributes a relative large amount to the average energy compared to the number of particles in it. That is, unless I am wrong, the median velocity is significantly less than you have calculated. In any case, there are lots of relatively slow moving oxygen molecules for the HB to rip appart and eat.

Again, we're not talking more than an order of magnitude.

Again, you probably are right, but I am still not willing to throw in the towel.

Right or not, this has been a very interesting discussion. I appreciate your open-mindedness.

I am prety confident that you error by setting the KE equal to the potential well depth to evaluate the effective "capture cross section" of BH in Browian motion air

Are you not used to order of magnitude calculations? I'm not erring by more than a factor of a few, and there's no point in more sophistication with the results being so many orders of magnitude off from the desired result.

 

[Sariaht]

If we make a one gram black hole, charged it and kept it in a box, we would eventually find out if they blow when they evaporate. How long time would it take?

 

[SpaceTiger]

If we make a one gram black hole, charged it and kept it in a box, we would eventually find out if they blow when they evaporate. How long time would it take?

If the usual evaporation equations apply, $$10^{-25} seconds$$.

 

[Billy T]

SpaceTiger: Ref your post 38:

Can you give estimate of mean free path for air at STP (I forget so much. Volume of Mole of Oxygen, 22.4 L? Advogradra’s No. is 6x10^23? Etc. - That stuff was 45 years ago for me, and I am too lazy now too work as hard as I did then. You seem like you are in the stage of life where it is still good for you to do so.)

Also would be interesting to assume half your vacuum life time (5x10^-17s) and see how far typical O2 molecule can travel in that time. Then inside sphere of this radius, R, how many of them are there? Throw away the half that are going away from BH, and what remains?

I’m well aware that BH will not capture passing molecule unless (1) QM effects such as tunneling and/or lack of definite position come into play or (2) have third body to transfer energy/ momentum to. This is why I talked about stripping off outer shell electrons by gravity gradient etc. “Co-valent bonding” and Van De Wall’s forces don’t seem (intuitively to me) too much to compete with for the BH.
You seem to be saying that 1kg BH won’t rip apart molecules, but surely it is just a question of how close they need to get, not whether or not it can. Because each O2 (or O atom for that mater) comes with bunch of electrons and they can be stolen away with a few ev of effort, my intuition tells me that there will be a “charge cloud” around the BH. , harse UV, etc -I.e. no problem with lack of “third body” for capture. All this under very questionable assumption that the BH does not evaporate before electrons get stripped of any molecules /atoms etc.

All I was doing by being critical of your KE=PE radius for capture was to note tht it should be bigger. How many OOM bigger is my “R” of second paragraph? Or the MFP of paragraph 1?

I agree we’re still “give or take a few OOM“ and also find this interesting to think about, even though my classical mind is not prepared for the task.

 

[SpaceTiger]

I started going about the process of answering some of your questions more rigorously and, after a bit, realized that I had something that might be publishable. If this is the case (and even if it isn't), I'll put something together and post it for you once I'm done.

 

[Billy T]

I started going about the process of answering some of your questions more rigorously and, after a bit, realized that I had something that might be publishable. If this is the case (and even if it isn't), I'll put something together and post it for you once I'm done.

Very glad to hear this. Perhaps we have helped each other. Please send me a private msg when you post again b/c I will stop looking here for a while.

I suspect you know it well, but since you told me the better known r^-2 and r^-3 laws and did not mention it, I will just note that the gradient of a 1gm BH is greater than that of a 1kg one (at the event horizon). Thus, it really is a question of how close the atom must get before it is ripped up and ionized.

PS if you want to know more about me and my major concerns now visit www.DarkVisitor.com

 

[SpaceTiger]

Very glad to hear this. Perhaps we have helped each other. Please send me a private msg when you post again b/c I will stop looking here for a while.

I should note that the part that I think was interesting was more of a tangent off of your original idea. I can still assure you that a 1 kg black hole is not going do much. However, you might find this interesting:

Hawking Paper

I suspect you know it well, but since you told me the better known r^-2 and r^-3 laws and did not mention it, I will just note that the gradient of a 1gm BH is greater than that of a 1kg one (at the event horizon).

You need to get close to the event horizon for that to be the case and I had already determined that to be extremely unlikely to happen, so I was checking the tidal forces at large radii.

 

[Walter L. Wagne]

BillyT:

Nice to see you posting about mini black holes.

However, I do want your readers to know that magnetic monopoles are almost certainly real. It is to be noted that in 1975 I discovered the tracks of a novel particle on a balloon-borne cosmic ray detector. "Evidence for Detection of a Moving Magnetic Monopole", Price et al., Physical Review Letters, August 25, 1975, Volume 35, Number 8. A magnetic monopole was first theorized in 1931 by Paul A.M. Dirac, Proceedings of the Royal Society (London), Series A 133, 60 (1931), and again in Physics Review 74, 817 (1948). While some pundits claimed that the tracks represented a doubly-fragmenting normal nucleus, the data was so far removed from that possibility that it would have been only a one-in-one-billion chance, compared to a novel particle of unknown type. The data fit perfectly with a Dirac monopole.

I first theorized that colliders might create miniature black holes [that quickly evaporated by Hawking radiation] way back in 1987. However, by 1999 I was uncertain as to whether or not Hawking radiation actually works the way it's supposed to, and so wrote that letter to Scientific American [which I believe is the first written science reference to collider-produced black holes] suggesting that colliders might create mini black holes. They tapped Frank Wilczek to write a response, and he seriously lampooned the idea, erroneously believing that one needed a huge gravitational force to create one. Once created [by whatever means], their singularity nature allows them to remain in existence. Only if Hawking's theory is right, would they "evaporate". If they don't evaporate, we could be in some deep doo-doo. Ones made by nature [high-E cosmic rays striking a stationary planet] would zip right on through, neutrino-like, traversing our moon in 0.1 seconds. Non-evaporative ones made at a collider would hang around indefinitely, orbit our Earth at slow speed, gloming on to quarks and growing, slowly at first, but faster as their Schwartzschild radius grew. Google on "LHC black hole" for a wealth of science articles showing Frank was wrong and I was right.

Anyway, food for thought. No weirder than magnetic monopoles.

Regards,


Walter L. Wagner (Dr.)

 

[Orion1]

Hawking-Planck Temperature

$\text{Quantum Shutdown:}$
$\lim_{r \rightarrow r_p} F_n = F_p \; \; \; \frac{G M_p^2}{r_p^2} = \frac{c^4}{G}$

$\text{Hawking-Planck temperature:}$
$T_b = \frac{\hbar c^3}{8 \pi G M_p K_b}$

$\text{Planck mass:}$
$M_p = \sqrt{\frac{\hbar c}{G}}$

$\text{integration by substitution:}$
$T_b = \frac{\hbar c^3}{8 \pi G K_b} \sqrt{\frac{G}{\hbar c}} = \frac{}{8 \pi K_b} \sqrt{\frac{\hbar c^5}{G}}$

$\text{Hawking-Planck temperature:}$
$\boxed{T_b = \frac{}{8 \pi K_b} \sqrt{\frac{\hbar c^5}{G}}}$


I compiled these equations based upon this thread discussion, what do you agree and disagree about these equations and my solution?

Reference:
http://en.wikipedia.org/wiki/Hawking_radiation
http://en.wikipedia.org/wiki/Planck_mass