Praying?!? Sorry pal, God's busy with other worlds for the next few centuries, can I take a message?
POSTED BY: chrisisall
UPDATED: Friday, July 10, 2009 17:49
VIEWED: 9477
PAGE 6 of 15
Quote:
Originally posted by Bytemite:
Here's something that's LIKE what you ask for...Quote:
Yang-Mills theory is an extension of Maxwell theory that describes interactions in two other forces called the weak and strong nuclear forces. However, ground state fluctuations have a much more serious effect in a quantum theory of gravity. Again, each wavelength would have a ground state energy. Since there is no limit to how short the wavelengths of the Maxwell field can be, there are an infinite number of different wavelengths in any region of spacetime and an infinite amount of ground state energy. Because energy density is, like matter, a source of gravity, this infinite energy density ought to mean there is enough gravitational attraction in the universe to curl spacetime into a single point, which obviously hasn’t happened.
~The Universe in a Nutshell, page 46.
I'll have to keep looking for other (better) examples.
I'm not sure what that has to do with singularities. It merely says that Yang-Mills would seem to suggest that the universe should have collapsed into a singularity, which since that hasn't happened indicates there's something missing from Yang-Mills, not that Singularities don't exist.
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It's small, but it's not zero. What you have observed, however, is part of a noted paradox in Hawkings radiation.
http://en.wikipedia.org/wiki/Hawking_radiation#Trans-Planckian_problem
Quote:
The trans-Planckian problem is the observation that Hawking's original calculation requires talking about quantum particles in which the wavelength becomes shorter than the Planck length near the black hole horizon. It is due to the peculiar behavior near a gravitational horizon where time stops as measured from far away. A particle emitted from a black hole with a finite frequency, if traced back to the horizon, must have had an infinite frequency there and a trans-Planckian wavelength.
Here's a wikipedia entry on Stephen Hawking's no singularity Big Bang model.
http://en.wikipedia.org/wiki/Big_Bang#Speculative_physics_beyond_Big_B
ang_theory
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The Hartle-Hawking no-boundary condition, in which the whole of space-time is finite; the Big Bang does represent the limit of time, but without the need for a singularity.
http://en.wikipedia.org/wiki/Hartle-Hawking_state
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Quote:
Originally posted by citizen:
I'd be surprised if he did say such a thing. Hawking once lost a bet that Singularities 'exist' only within Black Holes (his position), when computer models showed naked singularities were theoretically possible. Hell, it was Steven Hawking, George Ellis, and Roger Penrose who came up with the Big Bang from a singularity theory. So in fact it would be Hawking saying his entire life's work is bullshit. I suspect such a statement would be big news.
Anyway, I don't see any reason why QM effects, even if they had the effect you say would negate a singularity. In fact the singularity of a rotating black hole is already known to not be a single point but a 'disc'.
Amazingly enough, that's EXACTLY what I'm saying. He refuted his own work that he did in the 1960s on the Big Bang model, because he thought a no singularity big bang model better explained his observations.
Now THERE'S an example of a non-biased scientist! I love that he was willing and able to say he was wrong in the quest for greater understanding.
And prior to reading his work on the subject, I would have said anyone who argued against a singularity based big bang was a total nutjob. But he makes very convincing, easy to follow arguments all based in quantum mechanics, so, like I said, I like the idea. I think it has potential. He even proposed ways in which it could be proved or disproved, that I believe the LHC is going to be looking into... If they ever get it online, that is. ._.
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The Singularities wavelength being shorter than the planck length means it's effect on the 'probability cloud' is meaningless to the singularity. It means it's as good as Zero.
Quote:
Originally posted by Bytemite:
Amazingly enough, that's EXACTLY what I'm saying. He refuted his own work that he did in the 1960s on the Big Bang model, because he thought a no singularity big bang model better explained his observations.
Now THERE'S an example of a non-biased scientist! I love that he was willing and able to say he was wrong in the quest for greater understanding.
And prior to reading his work on the subject, I would have said anyone who argued against a singularity based big bang was a total nutjob. But he makes very convincing, easy to follow arguments all based in quantum mechanics, so, like I said, I like the idea. I think it has potential. He even proposed ways in which it could be proved or disproved, that I believe the LHC is going to be looking into... If they ever get it online, that is. ._.
And I'm saying I find it unlikely until I read it for myself
.As for the big bang, the only updated thinking I've seen from Hawking was that the singularity appeared spontaneously and then began to expand like a bubble.
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Quote:
Originally posted by citizen:
Anyway, I don't see any reason why QM effects, even if they had the effect you say would negate a singularity. In fact the singularity of a rotating black hole is already known to not be a single point but a 'disc'.
And I think that the idea of a two dimensional disk violates the concept of a singularity all by itself, without quantum effects.
Even if it doesn't... Hey, look! Quantum effects! You can smear a disk just like you can smear a point.
The only thing we seem to disagree on is whether that smearing does in fact happen, even though you appear to acknowledge wavelengths of particles in a singularity, and Hawking radiation.
Looks like I'm going to be doing a refresher in quantum gravity right now, since your argument seems to be that the gravity well at a black hole singularity locks particles into place and renders their probability cloud non-existent.
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Quote:
Originally posted by Bytemite:
And I think that the idea of a two dimensional disk violates the concept of a singularity all by itself, without quantum effects.
Well it doesn't, the singularity in a rotating black hole is still called a singularity.
Quote:
Originally posted by Bytemite:
The only thing we seem to disagree on is whether that smearing does in fact happen, even though you appear to acknowledge wavelengths of particles in a singularity, and Hawking radiation.
We disagree on whether it's relevant. Everything has a wavelength, but with anything more massive than an electron it becomes increasingly irrelevant. And Black holes are much more massive than an electron. Even so, QM making a singularities position 'fuzzy' doesn't actually mean it's not a singularity any more than it stops an electron being an electron.
Quote:
Originally posted by Bytemite:
Looks like I'm going to be doing a refresher in quantum gravity right now, since your argument seems to be that the gravity well at a black hole singularity locks particles into place and renders their probability cloud non-existent.
Is there a formalised theory of Quantum Gravity? There's a lot of hopefull's, but it's still very much the holy grail.
Anyway, no, my argument is that the singularity has enough mass to make it's wavelength irrelevant. I'm also saying that the probability cloud is a way of us getting to grips with the fact that a particle could be anywhere within that cloud. But the particle isn't actually a cloud, it is a specific point, it's just we can't be sure where it is without measuring it, and when you do that you change it's state, and you also can't know where it's going to be.
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Quote:
Originally posted by citizen:
There's a lot of hopeful's, but it's still very much the holy grail.
Excuseshe me?
The laughing Chrisisall
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Of physics, bitch. Stick Sean Connery in a Wheel chair and you're on to something.
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Here's some stuff I found on the wikipedia page for Schrodinger's equation. Has to do with particles in very small or one dimensional (point) spaces.
Quote:
For potentials V(x) that are bounded below and are not infinite in such a way that will divide space into regions which are inaccessible by quantum tunneling, there is a ground state which minimizes the integral above. The lowest energy wavefunction is real and nondegenerate and has the same sign everywhere.
To prove this, let the ground state wavefunction be ψ. The real and imaginary parts are separately ground states, so it is no loss of generality to assume the ψ is real. Suppose now, for contradiction, that ψ changes sign. Define η(x) to be the absolute value of ψ.
η = | ψ |
The potential and kinetic energy integral for η is equal to psi, except that η has a kink wherever ψ changes sign. The integrated-by-parts expression for the kinetic energy is the sum of the squared magnitude of the gradient, and it is always possible to round out the kink in such a way that the gradient gets smaller at every point, so that the kinetic energy is reduced.
This also proves that the ground state is nondegenerate. If there were two ground states ψ1(x) and ψ2(x) not proportional to each other and both everywhere nonnegative then a linear combination of the two is still a ground state, but it can be made to have a sign change.
For one-dimensional potentials, every eigenstate is nondegenerate, because the number of sign changes is equal to the level number.
Quote:
Relativity is incompatible with a single particle picture. A relativistic particle cannot be localized to a small region without the particle number becoming indefinite. When a particle is localized in a box of length L, the momentum is uncertain by an amount roughly proportional to h/L by the uncertainty principle. This leads to an energy uncertainty of hc/L, when |p| is large enough so that the mass of the particle can be neglected. This uncertainty in energy is equal to the mass-energy of the particle when
L = {\hbar \over mc} \,
and this is called the Compton wavelength. Below this length, it is impossible to localize a particle and be sure that it stays a single particle, since the energy uncertainty is large enough to produce more particles from the vacuum by the same mechanism that localizes the original particle.
I'm not entirely sure what this MEANS. But that last paragraph digests tasty in regards to how those virtual particles could be emitted from black holes.
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