# The Smallest blackhole possible?

1. Nov 9, 2004

### Alkatran

What is the minimum size of the event horizon of a blackhole? Obviously you need some certain amount of mass for a blackhole, because I don't get sucked into my chair, to overcome the other forces.

I just had an image of a tiny black orb hovering in front of me, and wondered what would happen to the baseball bat I hit it with.

2. Nov 9, 2004

### pervect

Staff Emeritus
I seem to recall reading that the smallest black hole is expected to be around a planck mass. I'm not sure if this is correct, or how it was determined without a full theory of quantum gravity.

3. Nov 9, 2004

### mathman

Try to calculate the circumference of a black hole with the mass of your chair. The formula is 18.5 kilometers x (mass of object)/(mass of sun).

4. Nov 9, 2004

### Wave's_Hand_Particle

Event Horizons around a Blackhole are 'OBSERVER-DEPENDANT', so let me ask you, where are you 'looking' from?..and what is the Area of your theoretical BH?

I ask this because the 'AREA' of a blackhole dictates the 'OBSERVABLE' horizon

P.S.
The situation of your posting is thus:BH = orb..will not be hovering in front of you. its actually the reversed..YOU will be hovering in in an unmovable position with regard to anyone situated 'further away'.

You cannot mainifest the energy needed to move your had to swing the bat to 'hit' the orb..you just cannot do this theoretiaclly or physically!

Last edited: Nov 9, 2004
5. Nov 9, 2004

### Chronos

A Planck mass black hole has a Swarzschild of 1 planck length. Interesting. I'm trying to figure out why I initially found that surprising and then didn't. Talk about confusing yourself.

6. Nov 9, 2004

### Alkatran

I didn't mean I was inside the event horizon. I was just wondering what would happen to a 'rigid' bat, since the part which would hit the black hole would begin travelling through time slower and slower....

7. Nov 9, 2004

### DivineNathicana

Yeah, the minimum size of a black hole is supposedly around the Planck mass... According to string theory. But don't forget guys, there's an enormous amount of tension on a string, and it's a lot heavier than you'd imagine! In fact, it's the weight of an average dust particle. Not much? But think of the size! Around the Planck length, was it? Think of the ratio. So one can see how that is a perfectly plausible candidate for a black hole.

- Alisa

8. Nov 9, 2004

### franznietzsche

If mass is quantized, then it could not be smaller than the Planck amss, it would have to be the planck mass.

9. Nov 10, 2004

### Chronos

I think the limit is more fundamentally tied to the planck length of 1.6E-35 meters, or about 1E-20 times the size of a proton. Mass is quantized in much smaller units than the planck mass: about 2.176E-8 kg. The mass of a hydrogen atom, for example, is about 1.673E-27 kilograms.

10. Nov 10, 2004

### DivineNathicana

Yes, I know that it could not be less than the Planck mass, but I was unsure of whether that was the minimum mass, or whether the latter was actually a bit heavier. But I'm pretty sure that's correct.

What do you think, would the aforementioned supposed black hole be a stable black hole, or would it be more along the lines of a type of quantum fluctuation?

- Alisa

11. Nov 10, 2004

### pervect

Staff Emeritus
Planck units share the characteristic with geometric units that G=c=1, so I'd expect the Schwarzschild _radius_ 2GM/c^2 to be twice the mass. Unless I'm blowing it badly.

12. Nov 11, 2004

### Chronos

Hi pervect! I just did a quick and dirty calculation using the same formula you cited and solved for M, assuming a radius of 1 planck unit. It looked to be about a planck mass. On the other hand, I may have botched the calculation. Not like it would be the first time that ever happened.

13. Nov 11, 2004

### yogi

I will take issue with the Planck mass/size limitations. The whole idea of fundamental units al la Planck is nothing but cosmological numerology. There are other factors that can be combined to yield different values for a unit of mass such as the electron charge and the like. Why attach any significance to the planck mass - why not use the electon mass as a fundamental entity if you must come up with non scientific deductions - in which case the black hole radius would be on the order of 10-57 meters

14. Nov 11, 2004

### Chronos

It's more than numerology. The Planck length is a fundamental unit in nature. Why? Because it is impossible to quantify anything smaller, and this plays a vital role in quantum physics. Consider this:

"There is only one truly fundamental length in nature a length free of all reference to the dimensions and rate of revolution of the planet on which we happen to live, free of any appeal to the complex properties of any solid or gas: free of every reference to the mysterious properties of any elementary particle: what we call today the Planck length,
L=(hG/C^3)1/2= 1.6X10^-33 cm
And what we identify with the characteristic scale of quantum fluctuations in the geometry of space".
- John A Wheeler "At Home in the Universe" p169

And here is a paper that further explains things:

http://xxx.lanl.gov/abs/gr-qc/0201030
Uncertainty in Measurements of Distance
Authors: John C. Baez, S. Jay Olson

And, as suspected, a Planck mass black hole occupies a Planck volume
http://zebu.uoregon.edu/~js/glossary/planck_time.html [Broken]
"Contained within a Planck volume is a Planck mass (hc/G)1/2, roughly 10-5 g. An object of such mass would be a quantum black hole, with an event horizon close to both its own Compton length (distance over which a particle is quantum mechanically "fuzzy") and the size of the cosmic horizon at the Planck time."
- Encyclopedia Britannica

Last edited by a moderator: May 1, 2017
15. Nov 12, 2004

### hellfire

If I take a look to the vacuum Schwarzschild solution, in which a black hole appears, it seams to me that black holes are only due to singularities which arise in case of point masses with infinite density and not in case of any mass distribution with finite density. Is this incorrect?

16. Nov 12, 2004

### da_willem

Take another good look; I don't believe this is correct. You can find a certain finite mass/radius ratio for wich the schwarzschil metric explodes. The (r,r) term is:

$$-\frac{1}{1-\frac{2MG}{rc^2}}$$

The condition for a black hole to form is that the radius of the object is smaller than $2MG/c^2$. This yields a finite density! The singularities do not follow from general relativity but are predicted by other theories. But no theory of gravity and QM has been found so I would not put your trust in these preditions....

17. Nov 12, 2004

### Chronos

Not at all incorrect and an excellent question, hellfire. I like the planck density as a fundamental limit. Unfortunately, many theorists approaches seem to discount planck and compton lengths as fundamental to anything. It sort of bothers me, but I will get over it.

18. Nov 12, 2004

### hellfire

What you are talking about is the coordinate singularity at r = 2GM in Schwarzschild coordinates. This can be removed chosing proper coordinates to describe the Schwarzschild spacetime.

As far as I know, the condition to have a coordinate independent singularity is an infinite value in the Riemann curvature tensor or any of its contractions. Following S. Carroll (Lecture Notes on General Relativity), in case of the Schwarschild metric:

$$R_{\mu \nu \rho \sigma} R^{\mu \nu \rho \sigma} = \frac {12 G^2 M^2} {r^6}$$

If one considers the interior Schwarschild solution, which looks like the vacuum Schwarzschild solution but with m(r) instead of M, it seams that for a homogeneous mass distribution (m(r) ~ r3), the singularitiy in the mentioned contraction disappears.

Now, suddenly I am aware ( sorry for the confusion) that there are indeed mass distributions which lead to a singularity (m(r) < r3). But I do not know how realistic these distributions are.

Another question is then whether event horizons (and therefore black holes) may exist without singularities (https://www.physicsforums.com/showthread.php?t=52060)

19. Nov 12, 2004

### yogi

Chronos - I am well aware that almost every writer has jumped on the Planck "units" bandwagon - to assert such sweeping generalities based upon nothing that has ever been proven is sheer folly. There is no guarantee that any of the 3 "so called constants" h, c and G that make up planck units are temporally invariant. Show me one single thing that has been proven experimentally that gives support to the notion of their fundamental significance - all you can find is one author quoting another - its another example of modern physics having gone copycat. Why pick Planck's constant h as being more significant than the electron charge e - which leads in combination with c and G to a different (so called) fundamental length, mass and time. There are more interesting cosmic coincidences that result from the electron mass and size that those that occur between the Planck mass and its volume (if you are of a notion that there must be some minimum value that rates a designation as fundamental). Interestingly, John Wheeler held to another view all his life, that all matter was made of electrons (even though he dutifully acknowledged quarks as the constituents of heavier particles).

20. Nov 12, 2004

### da_willem

I agree. But I mentioned this element because it is probably no conicident that exactly at the schwarzschild radius the metric (in certain natural coordinates...) diverges. But the condition for a black hole still stands. The radius of the object has to be smaller than $2MG/c^2$ this yields a minimum density of $3c^6/32M^2G^3$, not infinite! So could you please elaborate on how it follows from the Schwarzschild solution that only point masses with an infinite density lead to black holes.

21. Nov 12, 2004

### hellfire

In my previous post when I wrote the contraction of the Riemann tensor I realized that a point mass is not the only case which leads to black holes. Thank you for your comment.

22. Nov 12, 2004

### hellfire

But, wait a moment... according to the formula I wrote in post #18, it seams that the existence of a singularity (and thus the formation of a black hole) depends on the profile of the mass distribution m(r), rather than on the density. For the density you mention, if its an homogeneous density within a sphere (m(r) ~ r3, the formula for the contraction of the Riemann tensor leads not to an infinite value. So according to this criterium (I got from S. Carroll's lectures) no black hole is formed.

Could you explain why do you claim that a black hole is formed if the condition: mass M is confined within a radius 2GM/c2 holds?

23. Nov 12, 2004

### da_willem

The Schwarzschild solution is found from Einsteins field equations by assuming spherical symmetry, static field and asymptotic flatness, it is also an external solution so the metric only applies to empty spacetime. From this metric follows an infinite redshift for photons at r=2GM/c2. Combined with the fact that the Schwarzschild solution is an external solution it follows that if you reach r=2GM/c2 before you reach the object's surface the surface is a black hole. No light can escape from within r=2GM/c2 to infinity.

Ofcourse the Schwarzschild solution is actually for a point mass M. But I always heard it also applies to a spherical symmetric mass distribution. The solution then is ofcourse only valid for values of the radial coordinate larger than value for the surface of the object; it is still an external solution. I dont' know is this is really true but it sounds not unreasonable to me. Especially because in the classical analogon this also applies. The gravitational field outside a spherical mass distribution is the same as if the total mass sits at the center of the distribution; a point mass. Ofcourse the Poisson equation from wich this follows (Gauss's law) is only an approximation and the idea does not hold in the general relativistic case. Any ideas?

24. Nov 12, 2004

### Orion1

Relativity Relation...

Radial solution for a spherically symmetric BH sphere horizon is:
$$r_1 = \sqrt[3]{ \frac{3 M_s}{4 \pi \rho_s}}$$

Schwarzschild radial solution for a spherically symmetric gravitational BH event horizon is:
$$r_s = \frac{2 G M_s}{c^2}$$

Quantum Mechanics + classical General Relativity quantum shutdown:
$$r_1 = r_s$$

$$\sqrt[3]{ \frac{3 M_s}{4 \pi \rho_s}} = \frac{2 G M_s}{c^2}$$

Schwarzschild Density solution for spherically symmetric gravitational BH:
$$\boxed{\rho_s = \left( \frac{3c^6}{32 \pi G^3 M_s^2} \right)}$$

Note that the Schwarzschild solution is only a solution for Schwarzschild BHs with zero angular momentum $$L = 0$$, this is a highly improbable state.

Neutron star spin increases with increased density, therefore an object generating in the core of a neutron star or supernova without spin is...impossible. Only BHs with angular momentum $$L > 0$$ can exist in the Universe, a rotating Kerr BH.

There is a solution for a 'point-like' Schwarzschild Singularity, however there is a philosophical consequence, a 'point-like' object must exist in at least one dimension. Although such an object is implied to exist in only one spatial dimension, it actually must exist in a minimum of two. (1 space + 1 time)

Classical Schwarzschild Singularity Dimension Number:
$$n = 1$$ - dimension #
$$dV = 2r_p$$ - volume
$$L = 0$$ - angular momentum

$$r_p = \sqrt{ \frac{\hbar G}{c^3}}$$

Solution for 'non-rotating' Classical Schwarzschild Singularity Density for one dimensional 'point-like' object:
$$\boxed{\rho_s = \frac{M_s}{2} \sqrt{ \frac{c^3}{\hbar G}}}$$

'non-rotating' Classical Universe-Schwarzschild Singularity Density solution for a one-dimesional 'point-like' object:
$$\boxed{\rho_u = \frac{M_u}{2} \sqrt{ \frac{c^3}{\hbar G}}}$$

$$\boxed{\rho_u = 1.263*10^{85} kg*m^{-1}}$$

The density values for these types of objects are a large values, however they are not infinite.

Note that no form of energy or object has ever been demonstrated to exist in less than four dimensions (3 space + 1 time). Therefore, only four dimensional BHs can exist in the Universe.

BH Singularity Density infinities do NOT exist.

Last edited: Nov 12, 2004
25. Nov 13, 2004

### Chronos

Well said Orion. Singularities are mathematical artifacts, not reality.