Is this where I'm supposed to use the phrase "self-appointed defender of the orthodoxy"?
March 30, 2011 11:52 PM Subscribe
The strong nuclear force is ≈10^39 times as strong as gravity. It takes ≈10^39 protons to make a black hole with a Schwarzschild radius about the size of a proton. That's not just a coincidence, right? Why?
So something got me thinking about the LHC and all the OMG! Quantum Black Holes! crap that was on the web a while back and I found myself thinking about how pathetic the force of gravity was relative to the strong nuclear force and wondering, "How massive does a black hole have to be to have an event horizon the diameter of a single proton?" It kind of surprised me that the difference in masses was almost exactly the difference between the strength of gravity and the strong nuclear.
Part of me is feeling like I've just stumbled on a fundamental secret of the cosmos and part of me realizes that the diameter of black holes and protons are surely defined by some property of spacetime and the strength of their respective fundamental force and of course they're going to be different in direct proportion to their relative strength!
Physicists probably have a word or two that explains this and Planck's constant is almost certainly involved. Probably. Can anyone elucidate?
So something got me thinking about the LHC and all the OMG! Quantum Black Holes! crap that was on the web a while back and I found myself thinking about how pathetic the force of gravity was relative to the strong nuclear force and wondering, "How massive does a black hole have to be to have an event horizon the diameter of a single proton?" It kind of surprised me that the difference in masses was almost exactly the difference between the strength of gravity and the strong nuclear.
Part of me is feeling like I've just stumbled on a fundamental secret of the cosmos and part of me realizes that the diameter of black holes and protons are surely defined by some property of spacetime and the strength of their respective fundamental force and of course they're going to be different in direct proportion to their relative strength!
Physicists probably have a word or two that explains this and Planck's constant is almost certainly involved. Probably. Can anyone elucidate?
Best answer: Since we don't actually have any idea what determines the strength of the gravitational force (we have no quantum theory of gravity for a start) I don't think there's an explanation for this co-incidence.
The strong nuclear force increases with distance (up to the point where the force required to separate particles any further requires sufficient energy to create new particles at which point you start all over again...) whereas gravity decreases, so the ratio chosen is approximate anyway, since it depends on how close you place your colour carrying particles.
Who knows, maybe a unified theory will explain this as being a reflection of some underlying unity between the forces? We'll have to wait and see.
posted by pharm at 1:53 AM on March 31, 2011
The strong nuclear force increases with distance (up to the point where the force required to separate particles any further requires sufficient energy to create new particles at which point you start all over again...) whereas gravity decreases, so the ratio chosen is approximate anyway, since it depends on how close you place your colour carrying particles.
Who knows, maybe a unified theory will explain this as being a reflection of some underlying unity between the forces? We'll have to wait and see.
posted by pharm at 1:53 AM on March 31, 2011
Best answer: I'd bet you'd be interested in Dirac's Large Numbers Hypothesis.
posted by Mapes at 3:02 AM on March 31, 2011
posted by Mapes at 3:02 AM on March 31, 2011
Best answer: My husband, a physicist, says that it's a coincidence, as far as we know.
And then he says, "There's a more involved answer, to do with dimensional analysis, but it all still boils down to it basically being a coincidence." And he mentioned Dirac too, but then I had to go and eat some cheese, so I might have missed that bit.
posted by lollusc at 3:18 AM on March 31, 2011 [17 favorites]
And then he says, "There's a more involved answer, to do with dimensional analysis, but it all still boils down to it basically being a coincidence." And he mentioned Dirac too, but then I had to go and eat some cheese, so I might have missed that bit.
posted by lollusc at 3:18 AM on March 31, 2011 [17 favorites]
To be fair, there are a lot of numbers between 10^38 and 10^40, many of which could be rounded to 10^39 with enough inpercision. Rounding with exponents and assuming equal magnitude might be a very bad plan.
Also, on my way to work today I passed 10 school busses and 10 lights. I'm pretty sure there was minimal correleation asides from a population density needing both safety and service.
posted by Nanukthedog at 5:25 AM on March 31, 2011 [4 favorites]
Also, on my way to work today I passed 10 school busses and 10 lights. I'm pretty sure there was minimal correleation asides from a population density needing both safety and service.
posted by Nanukthedog at 5:25 AM on March 31, 2011 [4 favorites]
To support what Nanukthedog says above I will demonstrate "real" dimensional analysis via the funniest thing that ever happened to me in a physics class:
One of the perks of studying undergrad physics at MIT was taking third-semester quantum mechanics from someone who had an honest-to-god Nobel Prize. He (who shall remain nameless) was doing a test prep session with the class one night and at one point got to an expression that looked like this:
posted by range at 7:01 AM on March 31, 2011 [180 favorites]
One of the perks of studying undergrad physics at MIT was taking third-semester quantum mechanics from someone who had an honest-to-god Nobel Prize. He (who shall remain nameless) was doing a test prep session with the class one night and at one point got to an expression that looked like this:
3 √α ---- 2 π... at which point he stares hard at the board, then looks at us (~50 senior physics majors). Then at the board. Then us. Then back to the board, where he (a little sheepishly) reduces it to
When we all got done laughing he retaliated with: "Look. Experimentally, we don't know the value of this number [points at alpha] better than within 2 orders of magnitude, and nobody can think of a way to measure it any better. The difference between pi and 3 is 5%. The simpler expression is going to hold true enough for some time between 50 years and forever. So shut up."3√α ---- 2π
posted by range at 7:01 AM on March 31, 2011 [180 favorites]
Response by poster: I'm not saying that there must be a fundamental reason for this - hell, I'm absolutely sure there is more going on in the domain of a proton than 1 significant figure of strong nuclear force. I'm equally certain there's room for one or two PhD theses in what I don't know about black holes (for very large values of one or two).
I just wanted to make sure I wasn't dividing, oh, let's say h bar squared by the force of gravity and then multiplying by the strong nuclear force and being amazed that my result was the ratio of the strength of the strong nuclear force to gravity. If not, I'm pretty much going to put it in the nebulously cool things about the universe file and move on.
If you ever meet me in a bus station and I'm ranting about how I've united quantum physics and general relativity using only Microsoft Excel, Wikipedia and basic arithmetic, do the right thing and smother me with a pillow.
Rounding pi to 3 doesn't make me flinch, but:
∙ I've had to deduct ≈10^39 partial points for sig figure issues when I was a lab assistant in college.
∙ I do woodworking where the second sig figure of a measurement can change with the weather.
∙ I'm learning machine work, where the third and fourth significant figure can make the difference between a cool project and hot shrapnel.
∙ I've had to explain to a colleague, who was all handswavingfreakoutish out about the high RSD of her low standard, that the RSD of her blank was infinity.
posted by Kid Charlemagne at 8:06 AM on March 31, 2011 [6 favorites]
I just wanted to make sure I wasn't dividing, oh, let's say h bar squared by the force of gravity and then multiplying by the strong nuclear force and being amazed that my result was the ratio of the strength of the strong nuclear force to gravity. If not, I'm pretty much going to put it in the nebulously cool things about the universe file and move on.
If you ever meet me in a bus station and I'm ranting about how I've united quantum physics and general relativity using only Microsoft Excel, Wikipedia and basic arithmetic, do the right thing and smother me with a pillow.
Rounding pi to 3 doesn't make me flinch, but:
∙ I've had to deduct ≈10^39 partial points for sig figure issues when I was a lab assistant in college.
∙ I do woodworking where the second sig figure of a measurement can change with the weather.
∙ I'm learning machine work, where the third and fourth significant figure can make the difference between a cool project and hot shrapnel.
∙ I've had to explain to a colleague, who was all handswavingfreakoutish out about the high RSD of her low standard, that the RSD of her blank was infinity.
posted by Kid Charlemagne at 8:06 AM on March 31, 2011 [6 favorites]
When we all got done laughing he retaliated with: "Look. Experimentally, we don't know the value of this number [points at alpha] better than within 2 orders of magnitude, and nobody can think of a way to measure it any better. The difference between pi and 3 is 5%. The simpler expression is going to hold true enough for some time between 50 years and forever. So shut up."
As a person who majored in physics, let me be the first to say that shit like this is the reason why I threw my hands up and left the field.
The MIT guy is right, but there's no way you could come up with that on your own, especially on an exam or problem set. Approximations annoy the bejeezus out of me. (There are also approximations made in every single calculation made in the engineering world, although those at least follow a very specific set of rules)
posted by schmod at 10:59 AM on April 1, 2011
As a person who majored in physics, let me be the first to say that shit like this is the reason why I threw my hands up and left the field.
The MIT guy is right, but there's no way you could come up with that on your own, especially on an exam or problem set. Approximations annoy the bejeezus out of me. (There are also approximations made in every single calculation made in the engineering world, although those at least follow a very specific set of rules)
posted by schmod at 10:59 AM on April 1, 2011
As a person who majored in physics, let me be the first to say that shit like this is the reason why I threw my hands up and left the field.
I've been watching the Susskind Physics videos on itunesU and he does this alot. He also forgets i's all the time, mixes pluses and minuses and seems to just randomly add hbars and c's all over the place and then drops them later (only sometimes mentioning that 'oh, btw, i just set hbar = 1')
You kind of get the idea of what he was trying to do anyway, but it definitely makes things unnecessarily confusing, because I don't know when I'm just not following what he's doing, and when he's just making mistakes.
posted by empath at 11:24 AM on April 1, 2011
I've been watching the Susskind Physics videos on itunesU and he does this alot. He also forgets i's all the time, mixes pluses and minuses and seems to just randomly add hbars and c's all over the place and then drops them later (only sometimes mentioning that 'oh, btw, i just set hbar = 1')
You kind of get the idea of what he was trying to do anyway, but it definitely makes things unnecessarily confusing, because I don't know when I'm just not following what he's doing, and when he's just making mistakes.
posted by empath at 11:24 AM on April 1, 2011
Best answer: Mrs. sebastienbailard sez: The Schwarzschild radius is
R = 2 G_N m / c^2,
where G_N is Newton's constant and m is the mass of the black hole. The radius of a proton is about a femtometer; a photon (say) with this wavelength would have energy corresponding to the rest-mass of a proton. Setting hbar = c = 1 (sorry schmod) to express things in energy units, we just say the radius of the proton is "equal" to 1/m_proton. In the same units, Newton's constant is G_N = 1/M_Planck^2. Solving for the mass of the black hole, you then get
m = (M_Planck^2 / m_proton^2) * m_proton.
M_Planck^2 / m_proton^2 is about 10^{39} -- the "relative strengths" number comes from comparing the strength of gravity and the strength of the nuclear force between two protons separated by a proton diameter, which basically boils down to this.
posted by sebastienbailard at 8:36 PM on April 1, 2011 [3 favorites]
R = 2 G_N m / c^2,
where G_N is Newton's constant and m is the mass of the black hole. The radius of a proton is about a femtometer; a photon (say) with this wavelength would have energy corresponding to the rest-mass of a proton. Setting hbar = c = 1 (sorry schmod) to express things in energy units, we just say the radius of the proton is "equal" to 1/m_proton. In the same units, Newton's constant is G_N = 1/M_Planck^2. Solving for the mass of the black hole, you then get
m = (M_Planck^2 / m_proton^2) * m_proton.
M_Planck^2 / m_proton^2 is about 10^{39} -- the "relative strengths" number comes from comparing the strength of gravity and the strength of the nuclear force between two protons separated by a proton diameter, which basically boils down to this.
posted by sebastienbailard at 8:36 PM on April 1, 2011 [3 favorites]
To support what Nanukthedog says above I will demonstrate "real" dimensional analysis via the funniest thing that ever happened to me in a physics class:
Pedant alert: that's just approximation, not dimensional analysis.
Dimensional analysis is when you use the dimensions (or 'units') of a quantity (e.g. 'length' or 'time' or 'length divided by time squared') to check relations between several quantities.
For example, if a car's been travelling for 3 hours at a steady 60 mph, how much distance has it covered? Now, it's obvious that to get the answer you simply multiple 3 by 60, but you can use dimensional analysis (and a bit of algebra) to check that hunch.
On the other hand, if you remembered your speed relationship wrong, and thought the answer could be got by dividing the speed by the time taken, dimensional analysis would show that that is definitely wrong:
Using a similar technique, dimensional analysis is the reason why E=mc4 and E=mc3 are definitely wrong; the only valid equation of that form has to be E=mc2.
In other words, it allows you to narrow down the dimensional properties of some otherwise completely unknown element of a relationship. I think this is what lollusc's husband was getting at above.
posted by chrismear at 2:34 AM on April 4, 2011 [2 favorites]
Pedant alert: that's just approximation, not dimensional analysis.
Dimensional analysis is when you use the dimensions (or 'units') of a quantity (e.g. 'length' or 'time' or 'length divided by time squared') to check relations between several quantities.
For example, if a car's been travelling for 3 hours at a steady 60 mph, how much distance has it covered? Now, it's obvious that to get the answer you simply multiple 3 by 60, but you can use dimensional analysis (and a bit of algebra) to check that hunch.
miles
3 hours x 60 -------
hours
3 x hours x 60 x miles
= ------------------------
hours
3 x 60 x miles x hours
= ------------------------
hours
hours
= 3 x 60 x miles x -------
hours
= 180 x miles
And, as expected, you've got an answer in miles (length). So you've got some confidence that that was the correct relationship/calculation to use.On the other hand, if you remembered your speed relationship wrong, and thought the answer could be got by dividing the speed by the time taken, dimensional analysis would show that that is definitely wrong:
60 x miles / hours
--------------------
3 hours
60 x miles
= -----------------
3 hours x hours
= 20 (miles / hours2)
which is an answer in terms of miles per hour per hour (which is a unit of acceleration), not an answer in terms of miles as you were expecting, so you know you've gone wrong.Using a similar technique, dimensional analysis is the reason why E=mc4 and E=mc3 are definitely wrong; the only valid equation of that form has to be E=mc2.
In other words, it allows you to narrow down the dimensional properties of some otherwise completely unknown element of a relationship. I think this is what lollusc's husband was getting at above.
posted by chrismear at 2:34 AM on April 4, 2011 [2 favorites]
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I am not a particle physicist, but this seems to make plenty of intuitive sense to me. The strong force holds stuff like protons together. If gravity is stronger than it, the protons will collapse inwards. To get enough mass to overwhelm a force X times as strong, you need X times as much matter smashed into region on atomic scales.
posted by aubilenon at 1:52 AM on March 31, 2011