Gravity waves finally detected, Einstein proven right
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Hm, quick thought. If effect scales with 1/r....
[affected length(4 thousandths of proton)]*[distance/distance to potentially-nearby star]
4*10^-3*0,855*10^-15m * (1 300 000 000 ly/4 ly) = 1,1115*10^-9 m
About one millionth of a millimetre over 22 meters. That's what, the scale of electron microscope? Meanwhile, I'd guess at that distance 3 solar masses would vaporize planets. Maybe ignite fusion in gas giants.
Not only is gravity is damn weak, but hoo boy.
Detecting something on that tiny scale from another galaxy?
Terrifyingly impressive.
[affected length(4 thousandths of proton)]*[distance/distance to potentially-nearby star]
4*10^-3*0,855*10^-15m * (1 300 000 000 ly/4 ly) = 1,1115*10^-9 m
About one millionth of a millimetre over 22 meters. That's what, the scale of electron microscope? Meanwhile, I'd guess at that distance 3 solar masses would vaporize planets. Maybe ignite fusion in gas giants.
Not only is gravity is damn weak, but hoo boy.
Detecting something on that tiny scale from another galaxy?
Terrifyingly impressive.
Vorpal
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I've been going through the LIGO paper and its references, and apparently they did not use squeezed light in the aLIGO detectors responsible for this news, despite such schemes having already having been tested for years. Apparently, squeezed light schemes are implemented in LIGO's GEO600 detector in Germany, but not the aLIGO detectors at this time. Instead, what the aLIGO detectors are doing to deal with shot noise basically boils down to putting a lot of circulating laser power (100 kW) into each arm via several schemes for increasing resonant buildup, as well as signal recycling.
@'Lement: The four-thousandths of a proton bit comes from the peak strain being 1.0E-21 over a 4 km interferometer arm, so 0.004 fm. Also, like many distances in astronomy, the distance is quite imprecise: (410+160-180) Mpc. But anyway, yes, these 3±0.5 solar masses worth of gravitational waves could go off at an Earth-Moon distance away from you, and you still wouldn't notice any effect on your body (if this all happens in a complete vacuum, that is**).
(Also: I made a dumb statement about ringdown earlier, when the combined black hole radiates away imperfections in its horizon. Since we can think of the black holes as spiralling into each other by radiating away their orbital energy, the most violent phase would be right before ringdown. Oops.)
**Obviously, matter very close to the black holes might be excited enough to ruin your day. That may have happened here, as there were gamma rays detected 0.4s after LIGO's gravitational wave signal.
That's something like a tiny gamma ray burst. But if you're that close, it being 'tiny' by GBR standards would not help you at all.
@'Lement: The four-thousandths of a proton bit comes from the peak strain being 1.0E-21 over a 4 km interferometer arm, so 0.004 fm. Also, like many distances in astronomy, the distance is quite imprecise: (410+160-180) Mpc. But anyway, yes, these 3±0.5 solar masses worth of gravitational waves could go off at an Earth-Moon distance away from you, and you still wouldn't notice any effect on your body (if this all happens in a complete vacuum, that is**).
(Also: I made a dumb statement about ringdown earlier, when the combined black hole radiates away imperfections in its horizon. Since we can think of the black holes as spiralling into each other by radiating away their orbital energy, the most violent phase would be right before ringdown. Oops.)
**Obviously, matter very close to the black holes might be excited enough to ruin your day. That may have happened here, as there were gamma rays detected 0.4s after LIGO's gravitational wave signal.
Article: At a distance of 410+160−180 Mpc implied by the GW observations (Abbott et al. 2016), we obtain a source luminosity between 1 keV and 10 MeV of 1.8+1.5−1.0 × 1049 erg s-1. The uncertainties reflect the range of possible distances to the progenitor, uncertainties in the spectral fit parameters and the range of arrival directions along the arc.
That's something like a tiny gamma ray burst. But if you're that close, it being 'tiny' by GBR standards would not help you at all.
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Is that 0.4s after the start of a detectable gravitational wave signal, or 0.4s after the gravitational signal peaks? I would've expected them to arrive at the same time.
Vorpal
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Hmm...
LIGO: 09:50:45.39, ~0.45 s duration
Fermi: 09:50:45.8, ~1 s duration
So it looks like the gravitational waves are faster than light Fermi: 09:50:45.8, ~1 s duration
.Gamma rays have very high frequencies, so an explanation in terms of refractive index (though we need only n-1 ~ 10-17) due to non-perfect vacuum of the intervening space is probably far too unworkable by itself.
Ok, LIGO says they observed a gravitational wave sweeping from 35 to 250 Hz, with a peak strain of 1.0E-21 around f = 128 Hz or so (estimated from one of their figures), originating from roughly 410 MPc away. Because this is going to be purely back-of-the-envelope, let's scale that to 0.4 light-seconds away (strain A = 1E-4), call that an angular frequency of ω = 2πf = 800 Hz, and assume the wave is plus-polarised in the z-direction. The strain is small enough for linearised gravity, so in some small patch, we can approximate the spacetime metric as
ds² = -dt² + (1 + A sin(ω(t-z)))²dx² + (1 - A sin(ω(t-z)))²dy² + dz².
The gravitoelectric tensor simplifies to just Eαβ = -(1/2)hαβ,00, so the only nonzero components are given by the double time derivatives,Exx = +(Aω²/2) sin(...), Eyy = -(Aω²/2) sin(...).
Since A≪1, we can ignore the distinction between distances in coordinates and proper orthonormal frames, so the gravitational tidal force across a small distance d will be bounded by Aω²d/2 in magnitude, or about 3 Earth-gravities across one meter. A meter is very small compared to the inverse frequency times the speed of light, so that's valid.Ow! How naive looking at just the strain scaling was! At Earth-Moon distance (1.3 light-seconds), the strain may be bounded by a tiny, negligible-looking 3E-5, but the frequency-squared factor makes makes this a whopping 1.0 gee across 1 meter. Definitely noticeable... might be highly unhealthy, actually, as you're basically alternatively squeezed and stretched at high frequency with forces comparable to your own body.
Still, now we can start addressing your concern: if the black hole system is surrounded by matter, it's completely believable that introducing such strong varying tidal forces into it will produce violence through friction. This matter doesn't have to be right at the black hole merging event, and its heating would itself introduce further delays. The Fermi gamma-ray event was still a half-dozen orders of magnitude less than the gravitational waves, so transferring a millionth of the energy when close by isn't unbelievable.
Iandude0
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Well, once again Einstein was proven right. Of course, the dude was obviously a time traveler who accidentally went back to 1900 Europe after getting past the speed of light somehow.
Thus why he's so adamant that no one should ever breach it or even try.
Clearly I have the truth in my grasp here.
But yes, it's great to see that science is continuing to attain a greater understanding of everything that makes the universe tick. I'm actually kind of giddy that we can figure out how to find them, even if it took colliding black holes. Because now that we know they're there, we will find them.
And we will kill them.
Thus why he's so adamant that no one should ever breach it or even try.
Clearly I have the truth in my grasp here.
But yes, it's great to see that science is continuing to attain a greater understanding of everything that makes the universe tick. I'm actually kind of giddy that we can figure out how to find them, even if it took colliding black holes. Because now that we know they're there, we will find them.
And we will kill them.
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Quoting for posterity! Vorpal says FTL is possible!
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On that subject, if FTL=time travel, doesn't quantum entanglement allow for time travel? You just send a dude to a moon with one halves of entanglement A and entanglement B. Tell him to, if entanglement B has not been broken, to break entanglement A upon landing. When Houston with the other halves of the entanglements see that entanglement A is broken (which would be 1.3 seconds in moon-dude's past), they break entanglement B (which moon-dude sees 1.3 seconds in Houston's past). Meaning that moon-dude gets the signal to not break entanglement A roughly 2.6 seconds before landing... which means he doesn't break A... which means Houston doesn't break B, which means moon-dude doesn't get the signal... etc etc ad infinitum.
If I understand correctly, the reason why the above doesn't work is because moon dude and Houston have no way to check if entanglement has been broken without breaking entanglement, thus prohibiting transfer of information to the past. Which seems like a bit of a cop-out, it seems like arranging for a if/then device that produces different results depending on a quantum entanglement shouldn't be impossible... or is it?
If I understand correctly, the reason why the above doesn't work is because moon dude and Houston have no way to check if entanglement has been broken without breaking entanglement, thus prohibiting transfer of information to the past. Which seems like a bit of a cop-out, it seems like arranging for a if/then device that produces different results depending on a quantum entanglement shouldn't be impossible... or is it?
Hotdog Vendor
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This is the best response I've seen on this topic. I wish I had room in my sig for part of it.Well, once again Einstein was proven right. Of course, the dude was obviously a time traveler who accidentally went back to 1900 Europe after getting past the speed of light somehow.
Thus why he's so adamant that no one should ever breach it or even try.
Clearly I have the truth in my grasp here.
But yes, it's great to see that science is continuing to attain a greater understanding of everything that makes the universe tick. I'm actually kind of giddy that we can figure out how to find them, even if it took colliding black holes. Because now that we know they're there, we will find them.
And we will kill them.
I work in plasma physics, specifically in plasma-surface interaction, which is like a weird mix of chemistry, material science, plasma physics and, sometimes, nuclear physics. @Vorpal is much more qualified to answer these questions than me.
I do remember that quantum entanglement does not allow for FTL signal transmission.
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Really?
Because what I was getting out of this was, "Holy shit, my high school advanced maths classes were apparently a lot longer ago than I thought!"
Vanathor
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This shit went waaaaaay over my head. The concepts when Vorpal dumbed them down are easy enough, short of that though, strikes me as something that would be indistinguishable from Treknobabble.
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Vorpal
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I'm completely and utterly sure I'm not that. And I hope you're kidding, because basic familiarity with some technical topics relevant to this thread shouldn't be taken as expertise even in them, much less over-generalised so broadly.
There's actually several different points of failure in this scheme. Say for simplicity, we're talking about entanglement between two qubits (e.g., down/up spins of two electrons). A state like |01〉 = |0〉⊗|1〉, which says that the first qubit is 0 and the second is 1, is a product state, so it's by definition not entangled. Neither are |10〉 or even (|0〉+|1〉)⊗((|0〉+|1〉) in which the both qubits is in superposition. Note that for the last state, the two qubits can each be measured to be either 0 or 1 with equal probability, but knowing the outcome of one doesn't tell you anything at all about the probabilities of the other.
A (pure) state is entangled whenever it isn't a product state. It is essentially equivalent to classical random variables being not-indepedent whenever their joint probability distribution is not factorable into a product of individual distributions. Here's an ur-example of an entangled state:
|ψ〉 = |01〉 - |10〉.
If we're talking about electron spins, this would be the case where each individual electron has equal probability of spin-down or spin-up, but the total spin is zero, so knowing the outcome of one completely determines the other. More generally, you could have weaker correlations than that, too.Because entanglement is simply a quantum-mechanical version of subsystems of being correlated, one should immediately realise that measuring entangled systems doesn't involve transfer of information. Correlation is not causation, brah. The issue really does boil down to something that simple.
Given some hypothetical device that tells the dude whether the entanglement exists or is broken, this device only has access to half of the entangled pair, so it can't verify what happened or didn't happen to the other half. So, at minimum, it would need access to the other half of the entangled pair as well. Unless this happens no faster than light, you're basically presupposing an FTL scheme to make another FTL scheme.
However, it turns out that it still wouldn't work as advertised because there can be no entanglement observable for the device to measure, not even if it has access to the entire system! Since all observables in quantum mechanics are linear operators, what we want is a linear operator  that's with eigenvalue 0 ('no') for |01〉 and |10〉, but 1 ('yes') for the singlet state |ψ〉, or in any case something different from zero. But that's not possible because linearity means that Â|ψ〉 = Â|01〉 - Â|10〉 = 0 - 0 = 0.
In other words, there is no possible measurement that can determine whether or not a system is an entangled state in the first place.
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Wait never mind I see how you'd do it. If you have Alice and Bob both have the entangled electrons, then you have Alice measure first, then Bob measure second. If Bob always gets the opposite of Alice's, that's 100% correlation, which is statistically improbable if its chances of being one spin or the other is 50-50. Moreover Alice would be able to correctly intuit the spin of Bob's half, post-measurement of her electron, even if Bob was light years away. Well assuming they entangled it properly and it didn't collapse early. Is that correct?
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Vorpal
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Yes. There is no measurement of one pair you could make that's equivalent to 'are they entangled, y/n?', but what you could do is make a device that prepares them in an entangled state, and statistically verify that it works correctly.
You could have an observable that's equivalent to 'are they in such-and-such specific entangled state, y/n?', but you'll still back to square one because you'll need access to both particles: this can only be FTL if you already have FTL. Also, you'll still need statistics over many repeated instances because probabilistically for many other states have a chance to the answer 'yes' to this measurement (the specific state you're testing for will have probability 1 for this, though).
BTW, if you're testing for a specific state, you'll can't just keep measuring the spin along one axis (that would underdetermine the entangled state). And it's this freedom to switch axes, or more precisely that any superposition of spin-down and spin-up along one axis is a definite spin-up along some other axis, that allows quantum mechanics to have stronger correlations than those allowed by any classical system. Some people consider this 'spooky', but it's usually misattributed to QM's treatment of entanglement (which is actually identical to CM) rather than QM's treatment of superpositions (which is genuinely different).
magic9mushroom
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Now that I think about it, could gravitational radiation itself be the cosmic censor? Calling @Vorpal.
You mean the guy who posted before you?
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Hm, yeah - a spinning black hole would radiate gravity waves, wouldn't it?
Doesn't feel like it'd be a meaningful amount in a eyeblink, but over million years, at spin sufficient to be naked.....
Wouldn't work for the extremely charged ones, but kinda harder to make those happen.
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magic9mushroom
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A fully-formed spinning black hole doesn't radiate, to my understanding, because it's a perfectly symmetrical ring rotating about its axis (thus, the distribution of mass remains identical as it rotates). A similar effect is responsible for monatomic gases having no rotational EM spectra (hence why air doesn't absorb microwaves).
However, a progenitor of a strongly-spinning black hole, such as a neutron star with an active accretion disc or a pair of inspiraling neutron stars or black holes, will radiate very strongly (inspiral was both the first indirect confirmation of gravitational radiation and now directly confirmed as a powerful radiator, and indeed the limiting factor in inspiral is how fast the system can dump angular momentum). And a forming naked singularity wouldn't be subject to the No Hair Theorem and possibly could also radiate.
But frankly I'm out of my depth here; I've done second-year astrophys but we didn't do quantitative GR.
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Vorpal
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An isolated rotating black hole is stationary (but not static), and you can think of the spacetime geometry as axisymmetrically rotating into itself, so there can't be radiation losses. Saying it's a ring is inaccurate because the black hole is defined by the event horizon, not by the singularity. The Kerr interior is unstable, so not only are its specific predictions about the singularity of dubious physical significance, but the event horizon means that it doesn't actually matter whether something inside radiates, gravitationally or otherwise.
I don't get the diatomic gas analogy (I gather you meant diatomic rather than monatomic, because air is overwhelmingly diatomic). Diatomic molecules can have a rotational spectrum, but those relevant to air (N2, O2) have an additional mirror symmetry that prevents them from having an electric dipole moment, which is enabled by there being two opposing signs of electric charge.
Anyway, a black hole merger is very far from a plausible scenario for a naked singularity, because the orbiting black holes must radiate away angular momentum in order to merge, but you would need more angular momentum for a naked singularity in Kerr spacetime (J > GM²/c). From the point of view of black hole thermodynamics, an extremal Kerr black hole has absolute zero temperature (and does not Hawking-radiate), so if one takes the area～entropy and surface gravity～temperature correspondence seriously, black hole mergers producing a naked singularity is prohibited by the laws of thermodynamics.
Specifically, Hawking proved in 1972 that no classical process can decrease the area of a black hole horizon (assuming the weak energy condition, which is trivial in an black-holes-only spacetime), and in particular the a black hole merger results in a black hole with an area no smaller than the sum of the horizon areas of the initial black holes. It's the key part of why area～entropy identification is taken seriously in the first place. (Hawking radiation decreases the horizon area, but is not a classical process, and quantum gravity effects on black holes of this size are ludicrously small, as mentioned previously in this thread.)
I don't think there is anything in general preventing naked singularities from forming. It's certainly at least very hard—a fast-spinning body would repel particles that would add angular momentum, and one provably can't make an overextremal Kerr spacetime by dumping particles one at a time into a black hole. However, in complete generality, the 'cosmic censorship' hypothesis seems to be more driven by hope rather than clear physical criteria. It's not a big deal if there are singularities in GTR for physically reasonable situations, because GTR is an effective field theory that has to be wrong for extreme enough conditions anyway.
Whether gravitational radiation is the 'cosmic censor'... uh, I'm not sure how to interpret that question. It's certainly true that in the weak-field limit, the spacetime geometry can be written as Minkowski + retarded radiation from sources + radiation from infinity, completely analogously to the electromagnetic case. But it's extending it to the strong-field case that we actually want is not well-defined at all. On the flip-side, tolerating this ambiguity to make an analogy is probably OK too, because it's basically how it works in perturbative string theory: even though the formalism assumes some background, essentially any choice of background can be written as any other plus some coherent state of gravitons and other excitations.
So I'm not sure I'm interpreting what you want to ask in the way you meant, but I would answer 'is gravitational radiation the cosmic censor?' as somewhere between 'no, that doesn't really make sense in GTR' and 'only in as much as any spacetime geometry whatsoever is produced by gravitational waves/gravitons, so geometries with horizons would be included by default'.
I don't get the diatomic gas analogy (I gather you meant diatomic rather than monatomic, because air is overwhelmingly diatomic). Diatomic molecules can have a rotational spectrum, but those relevant to air (N2, O2) have an additional mirror symmetry that prevents them from having an electric dipole moment, which is enabled by there being two opposing signs of electric charge.
Anyway, a black hole merger is very far from a plausible scenario for a naked singularity, because the orbiting black holes must radiate away angular momentum in order to merge, but you would need more angular momentum for a naked singularity in Kerr spacetime (J > GM²/c). From the point of view of black hole thermodynamics, an extremal Kerr black hole has absolute zero temperature (and does not Hawking-radiate), so if one takes the area～entropy and surface gravity～temperature correspondence seriously, black hole mergers producing a naked singularity is prohibited by the laws of thermodynamics.
Specifically, Hawking proved in 1972 that no classical process can decrease the area of a black hole horizon (assuming the weak energy condition, which is trivial in an black-holes-only spacetime), and in particular the a black hole merger results in a black hole with an area no smaller than the sum of the horizon areas of the initial black holes. It's the key part of why area～entropy identification is taken seriously in the first place. (Hawking radiation decreases the horizon area, but is not a classical process, and quantum gravity effects on black holes of this size are ludicrously small, as mentioned previously in this thread.)
I don't think there is anything in general preventing naked singularities from forming. It's certainly at least very hard—a fast-spinning body would repel particles that would add angular momentum, and one provably can't make an overextremal Kerr spacetime by dumping particles one at a time into a black hole. However, in complete generality, the 'cosmic censorship' hypothesis seems to be more driven by hope rather than clear physical criteria. It's not a big deal if there are singularities in GTR for physically reasonable situations, because GTR is an effective field theory that has to be wrong for extreme enough conditions anyway.
Whether gravitational radiation is the 'cosmic censor'... uh, I'm not sure how to interpret that question. It's certainly true that in the weak-field limit, the spacetime geometry can be written as Minkowski + retarded radiation from sources + radiation from infinity, completely analogously to the electromagnetic case. But it's extending it to the strong-field case that we actually want is not well-defined at all. On the flip-side, tolerating this ambiguity to make an analogy is probably OK too, because it's basically how it works in perturbative string theory: even though the formalism assumes some background, essentially any choice of background can be written as any other plus some coherent state of gravitons and other excitations.
So I'm not sure I'm interpreting what you want to ask in the way you meant, but I would answer 'is gravitational radiation the cosmic censor?' as somewhere between 'no, that doesn't really make sense in GTR' and 'only in as much as any spacetime geometry whatsoever is produced by gravitational waves/gravitons, so geometries with horizons would be included by default'.
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magic9mushroom
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I actually did mean monatomic; the brain-fart was mentioning air.
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Hum.
(36^3+29^3)^0,333333≃41,4 min solar masses.
This is 62.
I'm ...bit flummoxed on what circumstances would result in bleeding off more of their rest mass.
If you flung two black holes almost at each other, then they'd bleed off at most all angular momentum or less imparted such when merging.
As for the non-spinning naked singularities and cosmic censorship goes, I'm reminded of the xkcd what-if of turning moon into electrons.
Now, obviously, the electrons need not be all the bundled together as near each other as they are there when black hole forms, so that can massively drop potential energy1. On the other hand, blacks holes are pretty small compared to their mass, and I'd imagine if electron was on the border of black hole with greater electric potential than gravitational potential, it'd fly away (a proton for negatively-charged hole would, of course, have far greater minimum inescapable distance).
1For a sphere of electrons with electrons gathered on rQ, this can probably be relatively trivially calculated (I'd guess n*(q*n*q/r) off the top of my head).
Then, since rs = 2 G(M0+ (Q^2/rQ)/c^2)/c^2 and rQ = (G/piε0)0,5*Q/2c^2, when looking at specific case of rs=2rQ, it becomes G(M0+ 2Q*(piε0/G)^0,5)/c^2 = (G/piε0)0,5*Q/2c^2 →[crat=(G/piε[sub]0[/sub])0,5]→ GM0+2*Q/crat=Q*crat/2 → M0 = Q*(crat/2+2/crat).
Now, brief look into WolframAlpha reveals that a) The resultant absolute value for crat suggests mass slightly over 2 kg for 1 Coulomb, which would be over million^2 times excess of minimum mass for electrons ( and hundred million for protons) and b) the units are incompatible, which probably indicates I made a mistake somewhere. Still, I'd imagine that much electrons or positrons wouldn't take too long to fluctuate into being near the 'hole'.....Maybe. Well, it seems like what happens to vacuum between dual event horizons of charged black hole might be interesting topic.
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You can't find most of them on your keyboard, you are supposed to use a character map or their alt key numeric core for them.
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