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Greg Egan
I am a science fiction writer and computer programmer. Web site: [mirrored at … ]
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Greg Egan proslijedio/la je tweet
Prof. Peter Doherty 14 h
Jim Molan tells us on that, as an LNP politician, he is constantly barraged with denialist nonsense that comes across his desk every day and is, as a consequence, confused. Why doesn't he take some time out to talk to real climate scientists? There are plenty at ANU.
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Greg Egan 20 h
Odgovor korisniku/ci @agolian
I want to conduct an experiment on 10,000 mathematicians to see how much space it takes between the 2 and the a in: 1/2a before they stop seeing it as 1/(2a) and switch to a/2.
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Greg Egan proslijedio/la je tweet
Craig Foster 2. velj
Not seen their family Not tasted freedom Not lived without security guards at their door Under Australia’s care ‘Three and a half years ago I got my refugee status..for what crime have we been imprisoned for 7 years?
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Greg Egan 22 h
Odgovor korisniku/ci @ctrlcreep
It’s not God you have to beat at chess.
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Greg Egan 2. velj
Odgovor korisniku/ci @AnActualWizard
Smashwords will distribute everywhere, for a cut, but since it’s possible to deal directly with Amazon and iTunes I do that, and use Smashwords just for their own site, Barnes & Noble, and Kobo.
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Greg Egan 2. velj
Odgovor korisniku/ci @silvascientist @johncarlosbaez
I *think* it has to expel material during the collapse.
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Greg Egan 2. velj
Odgovor korisniku/ci @Quasilocal
This is curious! Clearly the real universe is full of subsystems that violate the limit, sometimes by enormous factors. I don’t know if there’s any global, cosmological limit on the specific angular momentum. Maybe if L is too high you get something weird, like Gödel’s cosmos.
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Greg Egan 2. velj
Odgovor korisniku/ci @johncarlosbaez
MTW Section 33.8 explains why you can’t. If I understand correctly, it turns out that at the extremal point, any mass that could actually be *captured* by the hole would raise the mass of the hole, as well as its angular momentum, by at least enough to maintain the limit.
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Greg Egan 2. velj
Odgovor korisniku/ci @johncarlosbaez
Yeah, it’s not all that surprising in retrospect, but given that stars that actually *do* collapse into black holes can also be close to the limit, it gives some sense that extremal Kerr holes aren’t really all that exotic. Having that much specific angular momentum isn’t rare.
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Greg Egan 2. velj
Odgovor korisniku/ci @gregeganSF
J = (G/c) M_E ≈ 7.9×10^{30} should be: J = (G/c) M_E^2 ≈ 7.9×10^{30}
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Greg Egan 2. velj
Odgovor korisniku/ci @AnActualWizard @AndrewCStuart
Idealised rotating black holes have ring singularities: But nobody really knows the exact final state of the interior geometry when a black hole forms from a collapsing star.
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Greg Egan 2. velj
Odgovor korisniku/ci @gregeganSF
One reference claims the Sun also exceeds its Kerr limit, but that’s based on a calculation that assumes a uniform density and rate of rotation. The accepted value of the Sun’s angular momentum is: J_S = 1.92×10^{41} [] which is 20% of the Kerr limit.
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Greg Egan 2. velj
The Earth has more angular momentum than any Earth-mass black hole could have! m_E≈6×10^24 kg r_E≈6.4×10^6 m ω_E≈72.7×10^{-6} s^{-1} J_E ≈ (2/5) m_E r_E^2 ω ≈ 7.1×10^{33} if uniform density, but realistically J_E > 3×10^{32} The Kerr limit is J = (G/c) M_E ≈ 7.9×10^{30}
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Greg Egan 2. velj
Odgovor korisniku/ci @atomicthumbs
The greatest angular momentum a black hole can have is (in geometric units where G=c=1): L=M^2 or in conventional units: L = (G/c) M^2
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Greg Egan 2. velj
Odgovor korisniku/ci @gregeganSF
I inadvertently used a mix of conventional units (ω, g and c) and geometric units (R and M). In geometric units, G=c=1, and mass and distance have the same units, so R/M is dimensionless. I should have written: ω < g / [c (2R/R_s–1)] R_s = 2GM/c^2
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Greg Egan 1. velj
Odgovor korisniku/ci @thenicknelson
Nobody really knows the fate of the matter that collapses while forming a black hole, but it’s unlikely that anything becomes literally infinitesimal or literally infinite.
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Greg Egan 1. velj
Odgovor korisniku/ci @war_mindz
Yes, this is essentially the same thing as the Lense–Thirring effect. And I point out at the end of the thread that a perceptible deflection can’t actually coexist with a mere 1 gee’s acceleration.
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Greg Egan 1. velj
Odgovor korisniku/ci @gregeganSF
[4/4] Fine print: alas, *no* choice of black hole mass M, black hole rotation, and distance R from the hole could make the distance the ball swerved visible to the naked eye, if weight is 1 gee. Frame dragging per se can be made arbitrarily large, but: ω < g / [c (R/M–1)]
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Greg Egan 1. velj
Odgovor korisniku/ci @gregeganSF
[3/4] that the habitat is hovering above a rotating black hole — not orbiting the hole, but using its engines to stay put. It’s not rotating with respect to infalling starlight — but it *is* rotating relative to the definition of “non-rotating” built into the local spacetime.
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Greg Egan 1. velj
Odgovor korisniku/ci @gregeganSF
[2/4] You toss a ball straight up. As it rises, it veers sideways, and it lands over to your left — as if the habitat were spinning along an axis that passed through the floor, and the Coriolis force sent the ball askew. But … the stars aren’t moving! The only explanation is
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