Exploring the Risks of the Large Hadron Collider

[humanino]

In other words, I subscribe to the general statement that present day micro black holes are unstable. Period.

If you think that I question that, you must have read very few pages of this discussion.

Sure, but I'm not sure how to do that.

My question is : how is the geometrical cross section related to the Schwarzschild radius ? It's a simple and elementary question. Which classical radius should I use to set up an absolute upper bound on any possible quantum cross section ? The even horizon, the acoustic horizon, the apparent horizon (...?) ?

 

[gendou2]

If you think that I question that, you must have read very few pages of this discussion.

Please understand, I did not mean to question you at all.

How is the geometrical cross section related to the Schwarzschild radius? It's a simple and elementary question. Which classical radius should I use to set up an absolute upper bound on any possible quantum cross section? The even horizon, the acoustic horizon, the apparent horizon?

Oh, sorry I misunderstood your question. As I understand, R_s is the radius of the event horizon, which is the surface of the black hole. So, the quantum cross section should have a radius of R_s.

It is reasonable to consider the semiclassical cross sections with form
factors greater than unity as loose upper bounds on the black hole cross sections, which may increase by a factor of a few as the trapped-surface cross sections increase.

- From http://arxiv.org/abs/hep-ph/0609055

 

[humanino]

I found this article: http://arxiv.org/abs/hep-ph/0609055
Might be helpful.

Thanks for the link. After reading the beginning, I realize that my question is moot. If one is after a cross section upper bound estimate for safety purposes, a factor 3, 4 or even 10 is irrelevant. The upper bound must provide safety with plenty orders of magnitude anyway. Still, the article discusses this, and other aspects as well.

 

[gendou2]

Yeah, that trapped surface cross section stuff is a little over my head. Sorry for the initial confusion.

 

[malawi_glenn]

How much time does it takes to the protons to reach 99.999% of c in the LHC?

that is just 220GeV.. not so long time. It depends on what you mean-> starting a single proton bunch at 0 to 99.999c or when LHC will start running protons at that energy?

 

[Nick666]

LHC live cam.

http://www.cyriak.co.uk/lhc/lhc-webcams.html

 

[Almanzo]

Humanino: Assuming that the initial mass of the micro black hole would be equal to ten thousand proton masses, I calculate a Schwarzschild Radius of 1.2 * 10^-50 meter. The impact parameter would not be appreciably larger, because the force of gravity is 10⁴⁰ times weaker than the electromagnetic force, so the hole would not be able to draw anything in by its gravity. Neither would it be able to polarize or damage anything by its tidal force.

R= GM/c², where G = 6.67 * 10^-11 m³/kg*s², c = 3.00 * 10⁸ m/s, and M = 1.67 * 10^-23 kg.

 

[Vanadium 50]

that is just 220GeV.. not so long time. It depends on what you mean-> starting a single proton bunch at 0 to 99.999c or when LHC will start running protons at that energy?

220 GeV is below injection energy for the LHC.

 

[Hans Dorn]

Since the postulated minature black holes are so incredibly small, wouldn't they just pass through protons or neutrons unaffected?

Hadrons are composite particles after all.

 

[Orion1]

(4+n)-dimensional black hole...


The Schwarzschild radius R_s of an (4+n)-dimensional black hole: (ref. 2)
$$R_s = \frac{1}{\sqrt{\pi} M_p} \left[ \frac{M_{BH}}{M_p} \left( \frac{8 \Gamma\left(\frac{n+3}{2} \right)}{n+2} \right) \right] ^{\frac{1}{n+1}}$$

The Schwarzschild radius R_s of an (4+n)-dimensional black hole: (ref. 3)
$$R_s = \frac{1}{M_p} \left[ \frac{M_{BH}}{M_p} \left( 2^3 \sqrt{\pi}^{(-n-1)} \frac{\Gamma \left(\frac{n+3}{2} \right)}{n+2} \right) \right]^{\frac{1}{n+1}}$$

Why does the second equation solution dimensionally act upon $$\sqrt{\pi}$$, however the first equation solution does not?

Theorists have thus suggested that there could be extra spatial dimensions below this scale, curled up into tiny loops. Gravity could then be much stronger than we have measured, but with most of it being absorbed into the "invisible" extra dimensions. Stronger gravity corresponds to a smaller Planck mass, lowering the predicted Higgs mass and thus solving the hierarchy problem.

It is plausible that extra dimensional micro-black holes can quantum-gravitationally interact with normal matter equivalent to the strong nuclear force via Strong Gravitation.

Strong Gravitation: (1 Tev)
(Quantum BH strong nuclear reaction with a proton)
$$\boxed{t_p = \frac{4 E_b^2}{3} \sqrt{\frac{m_p r_p^7}{2 (\hbar c)^5}}}$$
$$\boxed{t_p = 3.362 \cdot 10^{-16} \; \text{s}}$$

Reference:
http://www.youtube.com/watch?v=kVsZdgz5oFM"
http://arxiv.org/PS_cache/hep-ph/pdf/0106/0106295v1.pdf"
http://arxiv.org/PS_cache/hep-ph/pdf/0609/0609055v2.pdf"
http://www.wissensnavigator.ch/documents/OTTOROESSLERMINIBLACKHOLE.pdf"
http://physicsworld.com/cws/article/print/26016"

Leave, leave Geneva every last one of you,
Saturn will be converted from gold to iron,
RAYPOZ will exterminate all who oppose him,
Before the coming the sky will show signs.

 

[malawi_glenn]

220 GeV is below injection energy for the LHC.

yes that is true for sure :-)

 

[Almanzo]

Humanino: Assuming that the initial mass of the micro black hole would be equal to ten thousand proton masses, I calculate a Schwarzschild Radius of 1.2 * 10^-50 meter. The impact parameter would not be appreciably larger, because the force of gravity is 10⁴⁰ times weaker than the electromagnetic force, so the hole would not be able to draw anything in by its gravity. Neither would it be able to polarize or damage anything by its tidal force.

R= GM/c², where G = 6.67 * 10^-11 m³/kg*s², c = 3.00 * 10⁸ m/s, and M = 1.67 * 10^-23 kg.

The above post contains a mistake, which I realized last night. The event horizon would indeed have an apparent radius of 10^-50 meters, but the impact parameter might be quite a bit larger.

Consider an electron in the outer shell of a carbon atom. It is 0.77 * 10^-10 m from the center of the atom, and feels, on average, the attraction of just one elementary charge, because the other charges are compensated for by the other electrons. It would feel a force F = Q²/4*pi*epsilon*R², which comes to 2 * 10^-8 Newton. Dividing this by the electron's mass gives an acceleration of 10²² m/s².

To feel the same acceleration from the gravity of a 10.000 proton mass (whether as a black hole or in any other form), one would have to approach it to a distance equal to the square root of that mass times G, divided by 10²² m/s². This comes to 3 * 10^-28 meter. Still a very tiny distance, but rather larger than the Schwarzschild Radius.

I wonder, however, whether an electron might actually come within this distance of the hole, and what would happen if it did. If the electron is actually a point sized particle, as suggested by the Wikipedia article, it might be simple. The point would be drawn in, and as soon as it crosses the event horizon inwards, it can never cross it again outwards. But if the electron has a "classical" radius of 7 * 10^-15 meter, it would be huge compared to the impact region. Only a tiny part could enter this region, and the force felt by the electron as a whole would be correspondingly tiny. In that case we may be back to an effective size of 10^-50 m for the black hole, although this wouldn't matter very much, because the size of the electron now dominates its probability of encountering the hole.

But there is something bothersome about this disparity in size. What if such a huge electron encounters such a tiny hole? How can more than a tiny part of it cross the horizon? I wouldn't like to envision the hole digging a tunnel through the electron, or taking just a nibble of it. Electrons are, after all, elementary particles; they may be annihilated by positrons, absorbed into nucleons or swallowed by (large) black holes, but they probably cannot be disassembled. I also would not like to envision the electron being stuck to the black hole, like a huge but nearly empty balloon being glued with superglue to a grain of sand. Because it would be stuck by virtue of being partially inside. No information can however leave the black hole, so the confined part of the electron would have no way to communicate its location to the rest of it.

One might, of course, envision the electron being deformed. To deform it, the black hole's influence would have to overcome the Poincare stresses keeping it in shape, and these would probalby be 100.000 squared, or ten billion times stronger than the force keeping it in orbit around the nucleus. This could only happen in a radius of 10^-33 meter around the hole, which would therefore not be able to deform the entire electron.

Anyway, if the entire electron were to be deformed, one might envision two ways of doing so. It might assume an oblate "pancake" shape, which would (seen from afar) seem to wrap itself around the event horizon. But it would have to envelop the horizon kazillions of times, self-intersecting and becoming as thin as gossamer. The tidal force (the inhomogeneity of the gravity field) would in any event suggest an opposite effect; the electron assuming a prolate "cigar" shape. But a cigar slender enough to pass as a whole through the event horizon would have to be slender indeed. The word "needle" would be more appropriate, and one would have to think of a needle which, if it were scaled up to stretch across the observable universe, would still have a waist of subatomary size.

Both deformations would greatly increase the mass of the classical electron, by intensifying the electric field near its surface. Considering this, I realized that if a 10^-50 meter size black hole were to acquire an electron charge, the electric field around it would have a mass of several kilograms! Which would cause a new event horizon to form, surrounding the best part of the field, at a radius of 10^-22 m around the original hole. And even if the charge were later to be neutralized, the hole would retain this mass. A frightening idea; if this were actually to happen, there would be no "dormant" stage at all.

But I don't expect this to happen, because the hole could not suddenly acquire a charge. What might happen, is that it starts to grow on approaching a charge, and eventually attains the minimum size which a black hole with an elementary charge must have. This minimum size I calculate to be roughly 10^-36 m, corresponding to a mass of a few micrograms.

This leaves however the question where the mass (and hence the energy) came from. And I can see only four possibilities. (1) Energy might not be conserved. (2) Charge might not be conserved. (3) Micro black holes beneath microgram mass can exist, but cannot acquire charge. (4) Micro black holes beneath microgram mass cannot exist.

Personally, I don't believe (1) or (2) to be true. (But it is only a belief, nothing more. After all, people are talking about recreating the conditions of the Big Bang.) I would greatly prefer (4) to be true, but I can see no reason why. If (3) is true, which would be my guess, 10.000 proton mass black holes could be relatively harmless, because even neutrons are assembled from charged particles. They could only grow by encountering each other, and even if the LHC were to create trillions of them, the Sun would be in its grave before they did.

 

[JustinLevy]

LHC: What is definition of micro-blackhole?

Are the following correct?

• The higgs is its own anti-particle
• The higgs is a scalar (spin 0) particle

If so, it seems like every quantum number should be zero for the Higgs. zero spin, zero electric charge, zero color, zero "weak charge", zero lepton number, zero baryon number ... zero everything I can think of.


Does this mean the higgs is the lightest possible black hole? If not, then how are they defining a "blackhole", when they are looking for black holes at the LHC to help probe quantum gravity?

(For reference, no other fundamental particle in the standard model can be considered a black hole because its spin or charge is too large compared to its mass... it has an "undefined" event horizon if you tried to naively treat it as a point particle with GR, this is not so for the Higgs boson.)

 

[Vanadium 50]

Are the following correct?

• The higgs is its own anti-particle
• The higgs is a scalar (spin 0) particle

If so, it seems like every quantum number should be zero for the Higgs. zero spin, zero electric charge, zero color, zero "weak charge", zero lepton number, zero baryon number ... zero everything I can think of.

Yes.

Does this mean the higgs is the lightest possible black hole? If not, then how are they defining a "blackhole", when they are looking for black holes at the LHC to help probe quantum gravity?

No, the Higgs has nothing to do with black holes. They are different beasts - as an example of one difference, every Higgs has the same mass, and black holes have a range of masses.

 

[JustinLevy]

No, the Higgs has nothing to do with black holes. They are different beasts - as an example of one difference, every Higgs has the same mass, and black holes have a range of masses.

Black holes have a discrete spectrum of allowed masses. It seems like the Higgs would be the smallest allowed one.

Shouldn't the smallest allowable black hole also be expected to have quantum numbers of zero for everything?


EDIT: Why was my original post moved into this thread? This thread seems to be discussing the decay rates and formation rates of black holes. I feel my question will pull this discussion off track.

Using the same math as my other post, I calculate the evaporation time of a black hole this size to be 1.38 * 10^-40 seconds.

But that is assuming Hawking radiation which is thermal. The whole point of hoping for micro-blackholes is that they will give some information about quantum gravity, and in these situations the radiation must NOT be thermally distributed (otherwise quantum mechanics is wrong).


Almanzo,
Yes, your number 3 is probably correct. According to GR the range of allowed electrically neutral black hole masses extends below that of the allowed range of charged black holes.

 

[Vanadium 50]

Black holes have a discrete spectrum of allowed masses. It seems like the Higgs would be the smallest allowed one.

Not in conventional BH theory. Your "seeming" sounds like you have your own ideas for a theory that differs from the conventional one. Can you express it mathematically?

 

[JustinLevy]

Black holes have a discrete spectrum of allowed masses. It seems like the Higgs would be the smallest allowed one.

Not in conventional BH theory. Your "seeming" sounds like you have your own ideas for a theory that differs from the conventional one. Can you express it mathematically?

I'm not sure which statement (or both?) you are objecting to.

- Blackholes have a discrete spectrum of allowed masses.

I assume we agree on that one. It appears to be required by quantum mechanics, as both LQG and String Theory, and most if not all other candidate theories, predict this.


- It seems like the Higgs would be the smallest allowed one.

This starts to get into the definition of a black hole. What I'm using as the definition is just a point particle which can classically have an event horizon ("classically", solely because it is not clear yet without a full quantum theory of gravity how to answer this quantum-mechanically). I aksed how people here are defining it, so please do feel free to share your working definition if you disagree.

Using that definition, and assuming (as the standard model does) that the lightest mass particles with zero for all quantum numbers is the Higgs, then yes... it seems like the Higgs would be the smallest allowed black hole.

 

[Vanadium 50]

I'm not sure which statement (or both?) you are objecting to.

- Blackholes have a discrete spectrum of allowed masses.

I assume we agree on that one. It appears to be required by quantum mechanics, as both LQG and String Theory, and most if not all other candidate theories, predict this.

I don't necessarily agree. Why would that be the case?

It seems like the Higgs would be the smallest allowed one.

You seem to be defining a black hole in a way totally independent of anything to do with gravity. Also, why is a pi0 not a black hole in this case? It also is chargeless, etc. This line of reasoning sounds circular to me - you define a Higgs to be a BH and voila! The Higgs is a BH.

 

[JustinLevy]

- Blackholes have a discrete spectrum of allowed masses.

I assume we agree on that one. It appears to be required by quantum mechanics, as both LQG and String Theory, and most if not all other candidate theories, predict this.

I don't necessarily agree. Why would that be the case?

Please be more specific. Is the problem with:
That LQC and String Theory, and most if not all other candidate theories, predict this?
Or "it appears to be required by quantum mechanics"?

If it is the first, please cite reasons for your disagreement otherwise it is hard to answer your Why? question. As for the later, I feel the first gives sufficient evidence for this discussion.

The general idea though is this: in quantum mechanics the energy of bound states are discrete. Therefore quantum-gravity objects should have discrete energy levels as well. Even in the classical limit (quantum applied in a fixed spacetime), Hawking found the spectrum of black hole event horizon areas is discrete (and thus its mass).

I hope that helps some.

You seem to be defining a black hole in a way totally independent of anything to do with gravity.

Please reread what I wrote:
"What I'm using as the definition is just a point particle which can classically have an event horizon ("classically", solely because it is not clear yet without a full quantum theory of gravity how to answer this quantum-mechanically)."

My definition depends directly on gravity (GR to be exact).

Also, why is a pi0 not a black hole in this case? It also is chargeless, etc.

pi0 is not a point particle. It is composite, and neither of its components are black holes either.

Further more, if you wanted to treat pi0 as an "effective" point particle, you will find that the actual size of this composite particle is larger than the event horizon if all the mass was at a point. Therefore treated it as an "effective" point particle for these purposes is not justified.

This line of reasoning sounds circular to me - you define a Higgs to be a BH and voila! The Higgs is a BH.

Where are you getting this from?
I stated what I was using as a definition of a BH. The Higgs fits this definition, but it is in no way circular.

Again, my first post asked for the definition of black hole being used here. I gave mine. Please, if you are going to disagree, give me your definition. You keep disagreeing without providing any discussion, so if I am wrong I can't learn from this.

 

[Rascalking]

So, the overall scientific consensus is that this machine is safe? I've yet to see a rebuttal to Orion1's stable black hole given a (4+n) dimension theory on the previous page. Can anyone comment on that?

Sorry everyone, just a concerned parent over here.

 

[JustinLevy]

So, the overall scientific consensus is that this machine is safe?

Yes. Nature herself throws much much higher energy particles at the Earth than we could ever possibly create ourselves. If a black hole hasn't already been created to gobble up the planets in the 4 billion years they have existed, then we're not going to be able to accidentally do it.

I've yet to see a rebuttal to Orion1's stable black hole given a (4+n) dimension theory on the previous page. Can anyone comment on that?

No one suggested a stable black hole. In Orion1's post a decay time was listed for the black holes even in this very hypothetical scenario.

 

[Rascalking]

Yes. Nature herself throws much much higher energy particles at the Earth than we could ever possibly create ourselves. If a black hole hasn't already been created to gobble up the planets in the 4 billion years they have existed, then we're not going to be able to accidentally do it.


No one suggested a stable black hole. In Orion1's post a decay time was listed for the black holes even in this very hypothetical scenario.

I appreciate the response, however, I wasn't able to find a decay time for Orion1's post. And if his hypothetical scenario is also safe, why does he have references from those who are clearly against the LHC? Then of course followed by a Nostradamus quote.

 

[JustinLevy]

I appreciate the response, however, I wasn't able to find a decay time for Orion1's post.

Eq 3., in the paper in the second link shows the lifetimes of these hypothetical "4+n" dimensional black holes. None of these things are stable.

Remember, nature performs much much more violent "experiments"/collisions with particles hitting the Earth at much higher energies than we could ever obtain in a laboratory. Theorists are hoping for micro-blackholes for it will give us information about gravity that we couldn't measure otherwise. But even these "out-there" (in the good sense) theories don't predict the Earth disappearring.

Some people will always worry. Some people will always assume the worse. And some people will always predict the end of days. Make up your own mind based on the facts... and if you don't feel you have a good enough background to understand the facts, please realize that thousands of scientists from many countries worked on this. If you really don't trust all those brilliant minds, I don't see why you'd bother asking for advice on the internet ... :)

 

[humanino]

Remember, nature performs much much more violent "experiments"/collisions with particles hitting the Earth at much higher energies than we could ever obtain in a laboratory.

I disagree with "than we could ever obtain". The GZK limit is not so far. Do not confuse the energy of a cosmic ray and the energy in the center of mass : LHC's 14 TeV is roughly one thousandth of the GZK limit.

 

[vanesch]

I disagree with "than we could ever obtain". The GZK limit is not so far. Do not confuse the energy of a cosmic ray and the energy in the center of mass : LHC's 14 TeV is roughly one thousandth of the GZK limit.

Ok, so if we make for a factor of 7 every two decades, then we will be there at the end of the 21st century. However, that's as stupid a prediction as using Moore's law to say that 15 years from now, we should have 4 THz processors, and 30 years from now, 4 PHz processors. This exponential growth only occurs during the maturing of a technology. After that, one hits physical limits.

 

[Almanzo]

Calculation of a minimum size for a black hole carrying electric charge.

(It might be more appropriate to make this a new thread, but as threads about tiny black holes seem to be merged with this one, I will give the calculation here.)

The Schwarzschild Radius R of a black hole is related to its mass M by R = GM/c².
Here G is the gravitation constant, and c the speed of light.

If the hole carries a charge Q, the electric field E outside the event horizon will be have a strength |E| = Q/(4pi*epsilon*r²).
Here pi is the circumference of a circle with diameter 1, epsilon is the dielectric constant, and r the (apparent) distance to the hole's center of mass.

The electric field contains energy, and the energy density (dU/dV) is equal to epsilon*[E|²/2.
This is equal to Q²/(32pi²*epsilon*r⁴).
Integrating dU over a spherical shell with r between r₁ and r₂, theta between 0 and pi, and phi between zero and 2pi, we may find the total energy within this shell.
The volume element dV is equal to dr*(r*dtheta)*r*sin(theta)*dphi = r²sin(theta)*dr*dtheta*dphi.

This energy is Q²/(32pi²*epsilon)*(1/r₁ - 1/r₂)*{cos(0)-cos(pi)}*(2pi-0) = Q²/(8pi*epsilon)*(1/r₁ - 1/r₂).
The energy, however, has mass, E = mc² or m = E/c².
The mass in the spherical shell becomes equal to Q²/(8pi*epsilon*c²) * (1/r₁ - 1/r₂).

Now consider the field in a shell between r₁=R and r₂=R+dR, just outside the horizon. This part of the field has a mass dM. Adding this mass to the mass of the hole itself, we obtain a mass M+dM within an (apparent) radius R+dR of the hole's center. The Schwarzschild Radius of such a mass is G(M+dM)/c². If this radius proves to be as large as R+dR, or larger, the event horizon must envelop a part of the field outside the event horizon, which is a contradiction.

It must therefore be ensured that G(M+dM)/c² < R+dR. But GM/c² = R, so GdM/c² must be < dR.

dM, however, is equal to Q²/(8pi*epsilon*c²) * {1/R - 1/(R+dR)}.
1/R - 1/(R+dR) = (R+dR)/{R(R+dR)} - R/{R(R+dR)} = dR/{R(R+dR)}, which, in the limit for very small dR, becomes equal to dR/R².
So, dM = Q²/(8pi*epsilon*c²) * (dR/R²) and GdM/c² = GQ²/(8pi*epsilon*c⁴) * (dR/R²)

This must be smaller than dR, so GQ²/(8pi*epsilon*c⁴) * (1/R²) must be smaller than 1.
R² must be larger than GQ²/(8pi*epsilon*c⁴)

G= 6.67 * 10-11 m³/kgs² or nearly (2/3) * 10^-10 m³/kgs²
By a curious coincidence, 36pi*epsilon = 10^-9 C²/Nm², so that 1/(8pi*epsilon) = (2/9)*10⁹ Nm²/C⁴.
c = 3.00 * 10⁸ m/s, so 1/c² is nearly (1/9) * 10^-16 s²/m².
If one takes Q to be the elementary charge, 1.60 * 10^-19 C, Q² becomes 2.56 * 10^-38 C².

Calculating first Q²/(4*pi*epsilon*c²), this comes to roughly 5 * 10^-46 kg*m.
Calculating G/c² this comes to roughly 7 * 10^-28 m/kg.
Multiplying these gives the square of the minimum radius, some 4 * 10^-74 m².
Dividing them gives the square of the minimum mass, some 7 * 10^-19 kg².

Curiously, it follows that a black hole with even one elementary charge must have a microgram-sized mass.

 

[vanesch]

Calculation of a minimum size for a black hole carrying electric charge.

I think the fundamental error you make is that you do this in Euclidean space. You integrate with an Euclidean space element and you calculate distances using an Euclidean metric. But near the Schwarzschild radius, the metric is far from Euclidean. So your energy calculations from the E-field and so on are all off.

 

[JustinLevy]

You can get the minimum size for a black hole containing one electric charge from the following:
http://en.wikipedia.org/wiki/Reissner-Nordström_metric

I calculated it once before and remember it to be around that of a Planck mass. Since we don't even know how gravity works on that small of a scale, I don't know how useful the exact number GR gives will be. The important point was merely that uncharged black holes can be smaller.


Can someone answer my question on the previous page regarding the definition of black hole being used here? For it still seems to me that the Higgs would be a small black hole.

(Is it acceptable to break out some of the individual threads in this massive intertwining thread? I feel things are getting buried, but I don't want to upset any moderators.)

 

[vanesch]

(Is it acceptable to break out some of the individual threads in this massive intertwining thread? I feel things are getting buried, but I don't want to upset any moderators.)

You can of course start a discussion on a scientific topic concerning black holes ; however, we try to group all the LHC-will-create-a-black-hole and will it or not destroy the Earth stuff in one single thread (this one). The reason for this is that we wanted to avoid a "pollution" of the particle physics forum, with the same questions, fears, and answers discussed over and over.

If you want to discuss a specific scientific topic, unrelated to what will happen in the LHC, you are of course free to do so in a separate thread.

However, no, the Higgs (at least, the standard Higgs in the standard model) is not a BH: a BH is a concept from GR, while in the standard model, there isn't even any gravity present. Now, as to whether a kind of elementary black hole could play the role of the Higgs in one or other quantum gravity theory, I'm out of my depth.

 

[JustinLevy]

You can of course start a discussion on a scientific topic concerning black holes ; however, we try to group all the LHC-will-create-a-black-hole and will it or not destroy the Earth stuff in one single thread (this one).

I did start a thread for discusison on a scientific topic concerning black holes, and no it did not involve anything with "destroying the earth" or any of that non-sense. Yet my question was moved here. That I why I was confused and thought I should ask.

However, no, the Higgs (at least, the standard Higgs in the standard model) is not a BH: a BH is a concept from GR, while in the standard model, there isn't even any gravity present. Now, as to whether a kind of elementary black hole could play the role of the Higgs in one or other quantum gravity theory, I'm out of my depth.

What I'm using as the definition is just a point particle which can classically have an event horizon ("classically", solely because it is not clear yet without a full quantum theory of gravity how to answer this quantum-mechanically). I aksed how people here are defining it, so please do feel free to share your working definition if you disagree.

Using that definition, and assuming (as the standard model does) that the lightest mass particles with zero for all quantum numbers is the Higgs, then yes... it seems like the Higgs would be the smallest allowed black hole.

(For reference, no other fundamental particle in the standard model can be considered a black hole because its spin or charge is too large compared to its mass... it has an "undefined" event horizon if you tried to naively treat it as a point particle with GR, this is not so for the Higgs boson.)

A couple people have said I'm wrong so far (and I'm completely willing to accept that), but they never give a replacement definition for black hole ... so I never learn anything. Please, if you disagree, give your definition of a black hole.

 

[tiny-tim]

Higgs boson

A couple people have said I'm wrong so far (and I'm completely willing to accept that), but they never give a replacement definition for black hole ... so I never learn anything. Please, if you disagree, give your definition of a black hole.

Hi JustinLevy!

I'd prefer to define an event horizon …

An event horizon is a surface boundary between ordinary (+,-,-,-) space (or the (+,+,-,-) space inside an ergosphere) and (-,+,-,-) space.

The Higgs boson is an ordinary electroweak-theory particle, and has nothing to do with event horizons.

 

[JustinLevy]

I'd prefer to define an event horizon …

An event horizon is a surface boundary between ordinary (+,-,-,-) space (or the (+,+,-,-) space inside an ergosphere) and (-,+,-,-) space.

The Higgs boson is an ordinary electroweak-theory particle, and has nothing to do with event horizons.

Please reread my post. I did refer to an event horizon, and that is because the Higgs as a fundamental point particle would have an event horizon (while none of the other particles in the standard model will).

So if your definition of a BH is also that is it a mass which would have an event horizon ... then you too are also calling the Higgs a black hole.

 

[ZapperZ]

Please reread my post. I did refer to an event horizon, and that is because the Higgs as a fundamental point particle would have an event horizon (while none of the other particles in the standard model will).

Where exactly in the Standard Model is this stated or formulated?

Zz.

 

[JustinLevy]

Where exactly in the Standard Model is this stated or formulated?

Zz.

Did you read my previous post?

I defined a black hole as such:
A point particle which can classically have an event horizon ("classically", solely because it is not clear yet without a full quantum theory of gravity how to answer this quantum-mechanically).

If you do not like that definition, then please do give me your definition.

Apply my suggested definition to the currently observed particles in the standard model and you will find that none have an event horizon due to their charge or angular momentum being too large. The Higgs boson, if observed, would be the only particle in the standard model which would have enough mass, and small enough charge (zero) and angular momentum (zero) to classically have an event horizon.


I'm sorry if I am not clear, but since people's questions seem already answered to me, I am also not sure what the confusion is, so am unsure how to fix it. Please, please, if you disagree can you provide your own definition of a black hole so that discussion can move forward.

 

[Vanadium 50]

The general idea though is this: in quantum mechanics the energy of bound states are discrete. Therefore quantum-gravity objects should have discrete energy levels as well. Even in the classical limit (quantum applied in a fixed spacetime), Hawking found the spectrum of black hole event horizon areas is discrete (and thus its mass).

Three counter-arguments. Pick your favorite:

(1) I can take a black hole of mass $$M_1$$ and turn it into a black hole of mass $$M_2$$ by firing a photon of the correct energy at it. I believe that energy is $$E = \frac{M^2_2 - M^2_1}{2M_1}$$.

(2) While a quantum mechanical potential gives rise to discrete states, the energies of the discrete states are functions of continuous parameters of the potential. So while the energy of any given example is discrete, the range of possible examples is continuous.

For example, the energy of a simple harmonic oscillator is quantized in units of its natural frequency, but I can have a SHO with any frequency I like.

(3) The quantization of energy states comes from matching wavefunctions of the internal structure of the system. Black holes don't have an internal structure.

Now, you mentioned the Hawking argument. We have to consider the entropy of a black hole. It's true that A ~ S, and it's also true that $$S = k \log(\Omega)$$ where $$\Omega$$ is the number of microstates per macrostate. While it's true that $$\Omega$$ is discrete, if you look at the minimum it can be (2), you get a minimum black hole size of about the Planck Mass. (Ignoring factors of 4, pi and log(2)). Apart from the general problem of statistical mechanics not being truly applicable for small $$\Omega$$, i.e. small system size, if you blindly plow ahead you end up with a minimum mass much, much larger than the Higgs mass.

Finally, if you sweep all that aside, the Higgs still can't be a black hole because the couplings are all wrong. For example, a black hole certainly interacts with photons (hence the name "black") but the Higgs does not.