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Prikvačeni tweet
For readers in the US, nine of my books are available as DRM-free ebooks for $2.99 http://www.gregegan.net/BIBLIOGRAPHY/Ebooks.html …pic.twitter.com/P7WjOYAuxH
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Greg Egan proslijedio/la je Tweet
Jim Molan tells us on
#QandA 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.Hvala. Twitter će to iskoristiti za poboljšanje vaše vremenske crte. PoništiPoništi -
Greg Egan proslijedio/la je Tweet
#PeopleWhoHaveLost7Years 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?#GameOver http://GameOver.org.au pic.twitter.com/Q19ZBN85x1Hvala. Twitter će to iskoristiti za poboljšanje vaše vremenske crte. PoništiPoništi -
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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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} [https://arxiv.org/abs/1112.4168 ] which is 20% of the Kerr limit.
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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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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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[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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[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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[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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[1/4] “Frame dragging” by a rotating black hole is a very cool phenomenon — and less arcane than it’s often portrayed to be. Suppose you woke up in some kind of space habitat, with a dome that showed you a sky full of stars that remained perfectly still. You feel 1 gee’s weight.
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If you were inside an orbiting space station that rotated to keep one face towards the body it orbited, some free falling objects would execute elliptical motion around the centre of the station. https://www.gregegan.net/INCANDESCENCE/Orbits/Orbits.html …pic.twitter.com/rjPYErFBfe
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Greg Egan proslijedio/la je Tweet
[1/2] My new eBook, INSTANTIATION, has 11 stories: • “The Discrete Charm of the Turing Machine” • “Zero For Conduct” • “Uncanny Valley” • “Seventh Sight” • “The Nearest” • “Shadow Flock” • “Bit Players” • “Break My Fall” • “3-adica” • “The Slipway” • “Instantiation”pic.twitter.com/2uveQZgWto
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I like the way the yellow and grey links force the red rhombus to sit symmetrically, with K lying on the horizontal line through AD, and the brown and grey links do the same for the blue rhombus and point H.
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[2/2] So ∠DAF=2∠DAB, and because AD=FJ, AF=DJ: ∠HDJ=2∠DAB Put ∠DAB=θ, ∠HDJ=2θ. The red & blue rhombi have sides 1/2, 1/4, so: AK=cos θ DH=cos(2θ)/2 Since AD=1/2, the identity (1+cos(2θ))/2 = (cos θ)^2 ⇒ AD+DH = (AK)^2. Other links impose some symmetries we assumed.
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[1/2] If anyone’s still bamboozled, the angle-doubling arises from pairs of non-convex quadrilaterals with pairs of equal sides. ABCD has side lengths 1/4,1/2,1/4,1/2. AFGB has side lengths 1/8,1/4,1/8,1/4. Because ∠ABG = ∠ABC, these quads are similar, and ∠BAF = ∠DAB.pic.twitter.com/eXF1SQaDfX
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Greg Egan proslijedio/la je TweetHvala. Twitter će to iskoristiti za poboljšanje vaše vremenske crte. PoništiPoništi
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[2/2] … lies in the identity: (1+cos(2θ))/2 = (cos θ)^2 The squared distance is created by ensuring that an angle in the blue parallelogram is *double* that in the red parallelogram! The originals by
@ChocoLinkage are: https://twitter.com/ChocoLinkage/status/1212285259598794752 …https://twitter.com/ChocoLinkage/status/1222858247566061569 …
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[1/2] Now that the ingenious
@ChocoLinkage has posted linkages that compute both squares and cubes, I thought it would be fun to reverse-engineer the squaring one. If you want to work through this in detail yourself, note that all lengths are multiples of 1/16. The magic here…pic.twitter.com/K9xVNUuWSCPrikaži ovu nitHvala. Twitter će to iskoristiti za poboljšanje vaše vremenske crte. PoništiPoništi -
Greg Egan proslijedio/la je Tweet
This vision from the Dunmore brigade of the
@NSWRFS from early January is terrifying. It shows how quickly a fire can move
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36 years ago, I heard “Swampland” by The Scientists on 2JJJ. The next day, in my lunch hour, I walked to my favourite record shop in Darlinghurst, but they didn’t have it, so I bought a mini-album by the same band. Today, I finally remembered to buy the song I actually wanted.
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