Time turners, Energy Conservation and General Relativity
post by Shmi (shminux) · 2013-04-16T07:23:13.411Z · LW · GW · Legacy · 169 commentsContents
169 comments
This post is a bit of entertainment for scientifically inclined Harry Potter fans.
Time turner from the Harry Potter series (and from the Eliezer Yudkowsky's venerable HPMoR fanfic) is a very useful device if you have some unfinished business in the recent past, like attending an extra class or saving a friend from a certain death. However, General Relativity has a few words to say about them, and they are not very flattering. I will only address one issue here: Energy conservation. TL;DR: if you use a time turner to vanish into the past, those around you will see you blown to tiny bits of Merlin-knows-what, quickly disappearing from view. When you appear in the past, this explosion appears in reverse.
Before we get to the time turners, however, let us consider an aside.
Let us start with a common question: if the Sun stop shining this instant, when would we notice? The common answer: it takes light 8.5 minutes to travel the distance of 150,000,000 km between the Sun and the Earth, so that's how long it will take. This glosses over the issue of what does "this instant" mean exactly at two different points in space, which is not so trivial given the relativity of simultaneity in Special Relativity. It is easily patched up, however, once we fix a global frame of reference. The Cosmic Microwave Background (CMB) is a natural one to use, and both the Earth and the Sun travel with a negligible fraction of the speed of light relative to the CMB. Anyway, the answer is still very close to 8.5 min.
Now another, deceptively similar question: if the Sun disappears this instant, how long before the Earth will stop orbiting the point where it used to be? The common answer: gravity travels with the speed of light, so also 8.5 min. This answer is obvious, simple and wrong. Yes, dead wrong. Why? because static gravity is not like light, it's more like electric field, only worse.
Let's first think of how you would make the Sun disappear. Maybe it turned into a black hole? Well, this would not really mean disappearance of gravity, the mass of the black hole will still be that of the Sun, and the Earth will happily (or unhappily, as the case may be) continue orbiting the Sun's corpse. So, in this case the answer is "it won't stop orbiting".
OK, so black hole was a bad example. How about a wormhole instead? You know, the evil Vogon-like aliens need to clear the room for a hyperspace bypass, and they build a wormhole from far away and suck all the matter in the Sun through it out of the way. What would happen then? There are a couple of hints: one is that from outside a wormhole is indistinguishable from a black hole, and the other is the Gauss Law. Both hints lead one to the same answer: just like with turning the Sun into a black hole, there is very little gravitational effect on the surrounding space. The rest of the now ex-Solar system will continue merrily on its way around the point where our Sun used to be.
An aside for those curious about the Gauss Law argument. The law in its integral form states that the flux of the gravitational field inward through any closed surface encompassing the Sun is proportional to the Sun's mass. To change the field, you need to remove some mass from inside this imaginary surface, by having it physically cross the surface. This last point may not be obvious, but it follows from General Relativity. Specifically, the Einstein's most misunderstood theory says that the spacetime curvature is determined by the (past and present) distribution of matter in spacetime. There are some exceptions, like the fixed-mass spherical objects, such as black holes and wormholes, which contains no matter, and gravitational radiation, which can carry away energy. But if you take a spherical object like the Sun and try to calculate what happens if you decrease its mass, General Relativity tells you that this mass has gone outward from the Sun in all directions in some form. It is not fussy about the form, as long as just the right amount of mass/energy has gone out.
Let me repeat for those who skipped the above paragraph: if you take the Sun and decrease its mass, the only way it can happen if this mass leaves the Sun outward and disappears into space. This happens all the time, of course, the Sun constantly loses its mass through radiation and solar wind, or in more drastic cases through Supernova explosions. Effects like this propagate no faster than light, of course. So they take forever to propagate all the way to infinity.
Now, back to the time turners. Hermione Granger might be but a small if incredibly studious girl, but she still has mass. If you were to peek at her using a time turner and disappear, her mass, small though it may be, still has to go some place, just like the disappearing Sun's mass had to go some place. The options are few: she can blow into tiny pieces flying past you, or disappear in a flash of brilliant light (and it takes a lot of light to carry away 50kg, what's with E=mc^2) . Basically, it will not be a pretty sight. What cannot happen is her simply vanishing, with no ill effects whatsoever. Well, it cannot happen if we are willing to keep Relativity around. Maybe we don't have to, what's with a certain deputy mistress turning into a cat and back, probably instantly changing her mass, with no ill effects on her or her surroundings. But if you give up on General Relativity, quite a few things will unravel, like all four Newton's laws.
Also don't forget the other side of the time turner action: Hermione appearing out of thin air just before walking into her extra class. The above process has to happen in reverse: an amount of matter equivalent to her mass has to travel inwards out of nowhere and coalesce into a person. Where did this matter come from? How did it form before collapsing into a person? How did it know that it would need to time its arrival into a certain point perfectly with whatever time turner will have been set to? That's some hard-core magic right there. Also, suck it, the Second law of Thermodynamics.
So, let me summarize: mass cannot just disappear, it has to spread out. mass cannot just appear, it has to coalesce. Thus time turners cannot be used inconspicuously, everyone around would be well aware of one's use, assuming they survive it. Actually, it probably cannot be used at all without breaking General Relativity and/or Thermodynamics. But hey, that's what magic is for.
EDIT: this post currently sits at -2 karma with 6 downvotes. I'd appreciate if any of the people who thought "I want less of this" explicate their logic to me, so I can do better next time.
EDIT2: OK, no one replied to my request... I'm guessing that some of you guys just quietly hate me :)
169 comments
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comment by ikrase · 2013-04-16T10:33:04.492Z · LW(p) · GW(p)
We have already seen that HPMOR!magic can break the conservation of mass (Transfiguration of objects of various mass such as a canteen of frozen water into a large rocket engine, Animagus), conservation of momentum ('Arresto Momento!, broomsticks (apparently), telekinetics) etc.
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-16T15:00:27.003Z · LW(p) · GW(p)
True, but those were local violations. It was not stated that General Relativity must fail for magic to work, at least I don't recall it. Also an occasional discussion of Quirrell communicating with one rather distant object would look differently if FTL is explicitly OK.
Replies from: DanielLCcomment by kpreid · 2013-04-16T15:42:43.682Z · LW(p) · GW(p)
By the way, anyone who likes to think about magic-and-physics should probably read Ra. Much like Harry, scientists find something which looks like magic, recognize it's a big deal, and proceed to apply science; but here the entire world is in on it, not a lone hero. Also, plenty of technical exposition (though less than if the author weren't trying to make it more character-focused).
Replies from: Nonecomment by wedrifid · 2013-04-16T10:55:55.712Z · LW(p) · GW(p)
Time turner from the Harry Potter series (and from the Eliezer Yudkowsky's venerable HPMoR fanfic) is a very useful device if you have some unfinished business in the recent past, like attending an extra class or saving a friend from a certain death. However, General Relativity has a few words to say about them, and they are not very flattering. I will only address one issue here: Energy conservation. TL;DR: if you use a time turner to vanish into the past, those around you will see you blown to tiny bits of Merlin-knows-what, quickly disappearing from view. When you appear in the past, this explosion appears in reverse.
Yes, the Time Turners as described violate conservation of energy. Something is happening that doesn't comply with Reality!Physics. Harry notices this the first time he encounters a witch. From Chapter 2:
"You turned into a cat! A SMALL cat! You violated Conservation of Energy! That's not just an arbitrary rule, it's implied by the form of the quantum Hamiltonian! Rejecting it destroys unitarity and then you get FTL signalling! And cats are COMPLICATED! A human mind can't just visualise a whole cat's anatomy and, and all the cat biochemistry, and what about the neurology? How can you go on thinking using a cat-sized brain?"
Professor McGonagall's lips were twitching harder now. "Magic."
"Magic isn't enough to do that! You'd have to be a god!"
Professor McGonagall blinked. "That's the first time I've ever been called that."
A blur was coming over Harry's vision, as his brain started to comprehend what had just broken. The whole idea of a unified universe with mathematically regular laws, that was what had been flushed down the toilet; the whole notion of physics. Three thousand years of resolving big complicated things into smaller pieces, discovering that the music of the planets was the same tune as a falling apple, finding that the true laws were perfectly universal and had no exceptions anywhere and took the form of simple maths governing the smallest parts, not to mention that the mind was the brain and the brain was made of neurons, a brain was what a person was -
And then a woman turned into a cat, so much for all that.
In both the case of Transmogrification and Time Turning either Conservation of Energy is part of a theory which just is not describing reality or there is some mass redistribution that Harry just hasn't detected. For example, conversion of excess mass into some kind of Dark Matter that exists in the MoR!Universe but neither wizards or physicists have studied yet.
Given the possibility of a Transmogrification mechanism and either any capability for sending information back in time or the presence of an excessively powerful predictor prior to the 'earliest' end of the time jump, Time Turning (or the appearance thereof to all observers) doesn't introduce any additional insanity.
Also don't forget the other side of the time turner action: Hermione appearing out of thin air just before walking into her extra class. The above process has to happen in reverse: an amount of matter equivalent to her mass has to travel inwards out of nowhere and coalesce into a person. Where did this matter come from?
Presumably the same place that the matter came from when MoR!McGonnagal!cat turned back into MoR!McGonnagal!human---either as yet undetected local conversion or from Magic.
How did it form before collapsing into a person?
Tricky question, which generalises to the general problem of determining what all of this 'magic' stuff reduces to or is based on. Simulation, advanced deception or extremely powerful technology from an overwhelmingly superior civilization or agent seem to be the obvious hypotheses. (That is, conditioning on the observations being accurate and not the result of insanity. Eliminating that as the most plausible hypothesis would take a heck of a lot of evidence.)
How did it know that it would need to time its arrival into a certain point perfectly with whatever time turner will have been set to?
Similar problem and (at least some) additional evidence for there being a powerful agent at work in one of the various ways that could occur.
That's some hard-core magic right there. Also, suck it, the Second law of Thermodynamics.
Indeed. In fact one could even go so far as to include that as part of the description of the genres of Science Fiction and Fantasy.
comment by cousin_it · 2013-04-17T12:06:16.855Z · LW(p) · GW(p)
Interestingly, the boxes from the movie Primer can be made to avoid that problem.
A short recap of how they work. You switch the box on, walk away from it to avoid running into past you, come back to the box several hours later, switch it off, climb inside, sit there for several hours, and climb out at the moment the box was switched on. One reason this model is cool is that it avoids one common problem with fictional time travel, the changing location of the Earth. You don't end up in interplanetary space because you travel back along the path of the box in spacetime.
So here's how you make the Primer boxes obey conservation of mass as well. The idea is that a box containing a time-reversed human should weigh less than an empty box. Let's say you weigh 70kg, and the box weighs 100kg when empty and switched off. When you switch the box on, a past version of you climbs out, and the box now weighs 30kg. Several hours later, a future you climbs in and the box now weighs 100kg, at which point the box switches off and sits there as empty as before.
At first I felt pretty smart for figuring this out because this whole issue never came up in the movie at all. And then I remembered the small detail that the boxes in the movie were an accidental invention, whose original purpose was to reduce the mass of objects. Wow.
That got me thinking about the other possible hole in the movie, namely all the abandoned timelines. Can this model of time travel be made to work correctly with not just spacetime paths and conservation of mass, but also causality and probabilistically branching timelines? For example, if you travel back in time and kill your past self, can that yield a unique consistent assignment of probabilities to timelines, where all time travelers "come from somewhere" and can't affect their "probability weight"? The result was this comment, for which I later found a proof of consistency which this margin is too small to contain ;-)
comment by Eliezer Yudkowsky (Eliezer_Yudkowsky) · 2013-04-17T17:39:21.922Z · LW(p) · GW(p)
This is nothing compared to the impossibility of Time-Turners given MWI, which is of course a given. I've been assuming that HPMOR runs on collapse QM.
Replies from: None, None, Luke_A_Somers, shminux↑ comment by [deleted] · 2013-04-19T18:18:21.849Z · LW(p) · GW(p)
What if a time turner functions like this:
The Time Turner, unknown to you and hours prior to any spinning, creates a near-duplicate of you (and itself) somewhere in the world
The Time Turner contrives for you to end up end up exactly like the duplicate it created, n hours later, and then compels you to turn itself (like ComedTea compels you to drink it).
On being turned, the TIme Turner destroys you and itself.
↑ comment by Luke_A_Somers · 2013-04-17T20:34:12.664Z · LW(p) · GW(p)
I think it fits best with a transactional form, actually. Time-turners just provide alternate routes for the transactions over their length, and you get a self-consistent outcome.
Hmm. Now I want to see what happens if you try to do a quantum interference experiment involving a time-turner. Good luck getting the phase to stay coherent over an hour, including the process of time-turning, though.
Replies from: ThrustVectoring↑ comment by ThrustVectoring · 2013-04-19T04:22:53.240Z · LW(p) · GW(p)
Now I want to see what happens if you try to do a quantum interference experiment involving a time-turner.
My prediction: "QB ABG ZRFF JVGU GVZR"
↑ comment by Shmi (shminux) · 2013-04-17T18:32:53.704Z · LW(p) · GW(p)
Hmm, I don't see how MWI would be interfering with time-turners. If anything, you can model a time-turner as spawning a new world at the destination point in the past, if you are willing to overlook the mass non-conservation issue. Unwinding a directed cyclic graph into a tree would also resolve the grandfather paradox and the "don't mess with time" message-in-a-bottle.
Replies from: roystgnr↑ comment by roystgnr · 2013-04-17T20:09:50.354Z · LW(p) · GW(p)
Isn't "resolving the grandfather paradox" the whole reason for the impossibility of Time-Turners (as opposed to generic time machines)? HP and HPMoR time travel seems to be of the Novikov variety, which isn't what you'd observe if trips into the past aren't constrained to reproduce previously-observed history. In HP you can't kill your grandfather because you didn't kill your grandfather (exact mechanism to be handwaved later). In MWI/spawning time travel you can kill your grandfather, and then that new history just won't produce another baby you.
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-17T20:49:56.155Z · LW(p) · GW(p)
Right. In general, spawning time travel is paradox-free, that's why I am not clear on why "the impossibility of Time-Turners given MWI". Presumably if you already spawn uncountable numbers of worlds all the time, it's not a big deal to spawn one more.
Replies from: pengvado↑ comment by pengvado · 2013-04-18T13:50:15.306Z · LW(p) · GW(p)
You can certainly postulate a physics that's both MWI and contains something sorta like Time-Turners except without the Novikov property. The problem with that isn't paradox, it just doesn't reproduce the fictional experimental evidence we're trying to explain. What's impossible is MWI with something exactly like Time-Turners including Novikov.
Replies from: Eliezer_Yudkowsky, MugaSofer↑ comment by Eliezer Yudkowsky (Eliezer_Yudkowsky) · 2013-04-18T18:11:35.708Z · LW(p) · GW(p)
(Nods.)
↑ comment by MugaSofer · 2013-04-19T10:49:44.235Z · LW(p) · GW(p)
What's impossible is MWI with something exactly like Time-Turners including Novikov.
I am ignorant on these topics, but isn't Novikov consistency predicated on QM? In that the "actual" paradox-free world is produced by a sum-over-histories? What about MWI prevents this?
Sorry if this is an incredibly stupid question.
Replies from: pengvado↑ comment by pengvado · 2013-04-22T00:05:07.504Z · LW(p) · GW(p)
Novikov consistency is synonymous with Stable Time Loop, where all time travelers observe the same events as they remember from their subjectively-previous iteration. This is as opposed to MWI-based time travel, where the no paradox rule merely requires that the overall distribution of time travelers arriving at t0 is equal to the overall distribution of people departing in time machines at t1.
Yes, Novikov talked about QM. He used the sum-over-histories formulation, restricted to the subset of histories that each singlehandedly form a classical stable time loop. This allows some form of multiple worlds, but not standard MWI: This forbids any Everett branching from happening during the time loop (if any event that affects the time traveler's state branched two ways, one of them would be inconsistent with your memory), and instead branches only on the question of what comes out of the time machine.
Replies from: MugaSofer↑ comment by MugaSofer · 2013-04-23T10:08:56.509Z · LW(p) · GW(p)
Hmm. So if, say, I committed quantum suicide, then traveled back, I wouldn't have any special information about the result of the RNG. Most of me would still end up in worlds where I died; God's dice get re-rolled every time round. No extra math to prevent paradoxes; although it still looks like Novikov for non-quantum events.
Whereas under standard Novikov Consistency, I'm restricted to the worlds where I survived, because otherwise I came from nowhere. In fact, the universe is restricted to those worlds; there are only worlds where I survived and came back and worlds where I died and didn't. Thus, no Everett branching. Right.
The degree to which the difference would be observable depends on the amount of quantum variance in your life, I guess.
comment by NancyLebovitz · 2013-04-16T13:21:49.139Z · LW(p) · GW(p)
From science fiction physics[1]: Could the conservation laws be expanded so all times are included?
[1] I don't remember where I saw the notion, but possibly Heinlein or Poul Anderson.
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-16T16:31:18.340Z · LW(p) · GW(p)
Could the conservation laws be expanded so all times are included?
This is indeed the standard approach in modern physics, yes. But then it gets even weirder.
Suppose that the person activating the time turner is turned into a blob of neutrinos which radiate in all directions undetected. Or just stayed where the person used to be, somehow. But they still have to assemble into this person some time in the past, so they cannot be ordinary neutrinos, they have to be the special traveling-back-in-time ones. Now, in the usual particle physics a particle traveling back in time is equivalent to an antiparticle traveling forward in time. Same mass, but (nearly) every other numerical quantum property reversed. In this case even their mass has to be negative.
So, here is a workaround. When Hermione (or Harry in HPMoR) appears in the past, an undetectable ghost of equal and opposite mass separates from her, leaving the total mass unchanged (zero). While she goes about her business, this invisible negative mass ghost-Hermione gets to the final destination in the future when and where she activated her time turner in order to go back, then annihilates the poor girl. Well, the "real" Hermione is now the other one, so that's OK. Voila, energy is conserved. If you don't mind negative-mass ghosts walking around for a time.
comment by Richard_Kennaway · 2013-04-16T11:02:59.667Z · LW(p) · GW(p)
Those who think they understand General Relativity might warm up to the task of retconning Time-Turners by solving the following conundrum, which takes place in the real world, and so must be solved under the constraint that "magic" is not an allowed answer.
The Earth orbits at 18 miles per second, and causal influences from the Sun travel no faster than light, which takes 8 minutes to reach the Earth from the Sun. Therefore the Earth "sees" the Sun 18x8x60 miles behind the straight line joining the two (using a Sun-centred frame of reference). This is an angle to the radial line of 18x8x60/93000000 = 0.0001 radians, or about 1/3 of a minute of angle. The gravitational attraction, being a causal influence transmitted at lightspeed, must be along that line also, thus retarding the Earth in its orbit with an acceleration of 0.0001 times the radial acceleration. This is enough to cancel out the Earth's orbital velocity in about 1000 years.
This has not happened. Where is the error?
Replies from: Viliam_Bur, Decius, kpreid, tgb, OrphanWilde↑ comment by Viliam_Bur · 2013-04-16T11:18:39.149Z · LW(p) · GW(p)
Teach the heliocentric controversy!
↑ comment by kpreid · 2013-04-16T15:39:35.090Z · LW(p) · GW(p)
I don't know the relativistic math, but: Gur Fha vf abg npgvat ba gur Rnegu; vg jbhyq npg vqragvpnyyl ba nal znff gung unccraf gb or naljurer nybat gung beovg. Gb hfr lbhe jbeqf, “Gur tenivgngvbany nggenpgvba, orvat n pnhfny vasyhrapr genafzvggrq ng yvtugfcrrq, zhfg or nybat gung yvar”, ohg vg vf ba rirel bgure yvar vagrefrpgvat gur Fha nf jryy. Gurersber yvtugfcrrq vf abg n pbafgenvag ba jung unccraf.
↑ comment by tgb · 2013-04-16T17:00:59.876Z · LW(p) · GW(p)
I'm pretty sure there's no general relativity in the error:
Qenj gur gevnatyr sebz gur rnegu gb gur fha gb gur cbvag jurer gur rnegu jvyy or jura vg vagreprcgf yvtug sebz gur fha gung vf orvat rzvggrq 'abj'. Gur shgher Rnegu gung vf frrvat gur yvtug sebz gur fha abj vf frrvat gur yvtug gung gur Rnegu rzvggrq nybat gur ulcbgrahfr, juvpu vf abg ng gur fnzr qverpgvba n gur Rnegu vf 'abj', ohg engure vf va gur qverpgvba sebz gur fha gb jurer gur Rnegu jvyy or jura vg vagreprcgf gur yvtug. Va bgure jbeqf, V qvfnterr gung gur Rnegu 'frrf' gur fha va gur jebat fcbg.
↑ comment by OrphanWilde · 2013-04-16T13:01:52.162Z · LW(p) · GW(p)
Jul ner lbh hfvat gur Fha nf gur senzr bs ersrerapr jura nfxvat jung gur Rnegu vf frrvat? (V'z abg fher vs guvf vf gur reebe lbh'er ybbxvat sbe be abg, ohg vg fgevxrf zr nf vaghvgviryl jebat.)
Replies from: Richard_Kennaway, shminux↑ comment by Richard_Kennaway · 2013-04-16T14:52:56.825Z · LW(p) · GW(p)
Because it's an inertial frame, to the accuracy required, but the Earth's frame isn't.
Mind you, I can't claim to know anything about GR, certainly not by this standard. My expectation is that when one does the mathematics, GR does not predict planets spiralling to destruction on thousand-year timescales, and that shminux has located the problematic step. But I do not know the mathematics. My expectation is also that verbal stories to justify why it does or does not happen are only helpful once one has worked out the real mathematical story, and cannot by themselves resolve the matter.
↑ comment by Shmi (shminux) · 2013-04-16T14:34:06.841Z · LW(p) · GW(p)
Actually the issue is the statement "Gur tenivgngvbany nggenpgvba, orvat n pnhfny vasyhrapr genafzvggrq ng yvtugfcrrq".
Replies from: Luke_A_Somers, Manfred↑ comment by Luke_A_Somers · 2013-04-18T23:55:50.219Z · LW(p) · GW(p)
Umm, what? I studied GR way less than you, but so far as I know, there's nothing even slightly problematic about that particular phrase. Perhaps you're referring to its failure to imply the phrase after it? Because the following phrase is definitely wrong.
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-19T01:26:59.966Z · LW(p) · GW(p)
"The gravitational attraction, being a causal influence transmitted at lightspeed" is wrong here because static fields do not propagate, they are just there. That's why black hole can retain electric charge and mass, even though "not even light can escape it". EM and gravitational waves propagate at the speed of light, but that's not related to gravitational attraction or the coulomb force between bodies. But you know all that. So it's a wrong premise which motivates a wrong conclusion, which is the next phrase.
Replies from: Luke_A_Somers↑ comment by Luke_A_Somers · 2013-04-19T02:03:10.197Z · LW(p) · GW(p)
I won't say anything about gravitational waves, but in EM, what you just said is unrelated, totally is related. If you just take the rules for electrostatics and the magnetic field from charges, and apply time-retardation, you recover the electromagnetic waves produced by that charge's motion. No corrections or adjustments are needed.
The electrical and magnetic fields at a given point is precisely the sum over all charged particles of their momentary electrical and magnetic fields at that point, causally offset... (... plus any background fields that the universe was shipped with, that did not arise from charges. I'm not saying that that's impossible)
But the point is, it's totally completely fair to talk about causal influences of a charged particle or mass propagating at lightspeed. If a particle's sitting there and then gets kicked and settles down, the electrostatic and gravitational fields update to their new values by the particle emitting waves.
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-19T02:48:58.446Z · LW(p) · GW(p)
I won't say anything about gravitational waves, but in EM, what you just said is unrelated, totally is related. If you just take the rules for electrostatics and the magnetic field from charges, and apply time-retardation, you recover the electromagnetic waves produced by that charge's motion. No corrections or adjustments are needed.
That's true, but that's waves, not static attraction. In essence, electric attraction is "magically" corrected for the straight-line motion, so the electric field from a moving but non-accelerating charge points exactly in the direction of the charge, not in the direction where the charge was after accounting for time-retardation. This breaks down once you add acceleration, hence EM waves. The original post made this (possibly deliberate) mistake: calculated retarded field for (nearly) uniform motion when calculating the direction of the attractive force. Unless I misunderstood it.
For gravity the corrections resulting in radiation appear even later, its "predictive power" is one order in time derivatives better than that of EM.
Replies from: Luke_A_Somers, Decius↑ comment by Luke_A_Somers · 2013-04-19T12:12:00.560Z · LW(p) · GW(p)
That's true, but that's waves, not static attraction. In essence, electric attraction is "magically" corrected for the straight-line motion
Wat. This is so severely counterintuitive I'm going to have to look it up and get a technical explanation. Is there a named effect for this?
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-19T19:41:10.276Z · LW(p) · GW(p)
I don't recall the name, but here is a neat java applet visualizing the situation. In essence, for uniform motion the field lines are always straight, pointing away from the charge, while the direction of light received from the charge points to the retarded position. There is a standard detailed calculation (like the one in Griffiths or Jackson) here, with the following conclusion:
Replies from: Luke_A_Somerswhich confirms that the electric field at R points along the direction from R to the present (not the retarded) position of the charge.
↑ comment by Luke_A_Somers · 2013-04-19T21:12:57.465Z · LW(p) · GW(p)
Very interesting...
But it still doesn't change the causal relationship: that alignment only applies if nothing happens to the other thing in the time since the center of that retarded cone. If no new information has been generated, then sure, you can use the new information instead of the old information. But if anything happens, you had better use the old information!
To get more formal about it: Consider or charge (equivalently, a mass) whose worldline coincides with (t, 0,0,0) for all t <= 0) in some reference frame
The field at the event (10, 10, 0, 0 ) occurs after (0,0,0,0) in every subluminal reference frame.
The field at (10, 10, 0, 0 ) is independent of whatever happens at (1, 0,0,0). If a laser comes in and knocks that charge aside, there's zero difference. None at all.
Suppose the charge was deflected so that it passes through (10, 1, 0, 0). You can't get the electrical field at event (10, 10, 0, 0) by looking at what the charge is doing at (10, 1, 0, 0) - you need to look at (0,0, 0,0).
This is what causality looks like. The causal influences propagate at lightspeed. Even electrostatic ones.
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-19T21:33:08.221Z · LW(p) · GW(p)
But it still doesn't change the causal relationship: that alignment only applies if nothing happens to the other thing in the time since the center of that retarded cone.
Absolutely. As I said
This breaks down once you add acceleration, hence EM waves.
However I am not sure I agree with the part in bold:
This is what causality looks like. The causal influences propagate at lightspeed. Even electrostatic ones.
When you say "Suppose the charge was deflected", you have broken the electrostatic assumptions, since the charge is now accelerating. Depending on the distance, you either get the near-field effects or the radiative effects, which do indeed propagate at lightspeed. Once the acceleration disappears and the light-speed transients died down, you are back in the electrostatic/magnetostatic mode with lag-free fields.
Replies from: Luke_A_Somers↑ comment by Luke_A_Somers · 2013-04-20T13:21:25.790Z · LW(p) · GW(p)
That was the point of the example - by time 10, the charge was no longer accelerating, but you know that you're not clear to use electrostatics being 'noncausal' yet because the light cone hasn't reached that far. Being lag free is a computational convenience that sometimes applies, and you need to know when by applying causality.
So, it is ALWAYS fair to say that fields are causal influences, whether they're static or dynamic. That was why I objected to your complaint in the first place.
↑ comment by Decius · 2013-04-19T03:41:15.695Z · LW(p) · GW(p)
Isn't "the direction of the charge" where you point a telescope to look at it, even if it is moving?
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-19T04:51:49.283Z · LW(p) · GW(p)
Not necessarily. This is somewhat counter-intuitive, but the light lags the direction of the Coulomb's law's attraction if a charge moves past you with a constant velocity (and has been doing so for some time). The attraction force points to the "true" direction of the charge, whereas light takes a bit to catch up.
Note that this apparent FTL effect cannot be used to transmit any information FTL, because, as soon as you try to wiggle the charge to telegraph something, this wiggling will only be sent as EM radiation (light), at light speed.
Replies from: Decius↑ comment by Decius · 2013-04-20T01:20:59.010Z · LW(p) · GW(p)
So, it's possible for the electrical force to act from a location that never had a charge? If the charge moves at a constant speed long enough and then makes a hard turn away, the charge will act on other objects (at least briefly) as if the charge had continued straight? Does the charge also act on the object as though it had not turned, even after it had, or is the force unilateral?
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-20T02:45:23.953Z · LW(p) · GW(p)
Yes to the second, not sure how the third is different. That's why, in part, Newton's second law does not in general hold for Electromagnetism, but momentum conservation does, if you account for the momentum of the electromagnetic field itself.
Replies from: Decius↑ comment by Decius · 2013-04-21T03:04:04.824Z · LW(p) · GW(p)
That behavior is contrary to naive expectations, right? If I run really fast towards a wall but turn before I reach it, I shouldn't hit it. It also shouldn't smash my face in after I make the turn.
The major force involved in billiard balls bouncing off of each other is electric in nature, right?
comment by Decius · 2013-04-17T00:57:11.187Z · LW(p) · GW(p)
Has it been established whether time turners can return someone to before the time turner was created? If they cannot, then it is simple to postulate that the time turner brings into existence a perfect copy of the user at the time and location of arrival, and then destroys the user at the time of departure. The time turner itself would have to contain enough energy to create the user to conserve energy.
That would also mean that a mass/energy conserving time turner has less stored magic during the periods of time when a duplicate exists.
Timeline, from the point of view of the two time turners, referencing the time on the clocks:
Time turner is created and becomes fully charged. (Long time ago)
Copy of user and new depleted time turner created, depleting charge on time turner. (1200 noon)
User and time turner destroyed by use, recharging new time turner. (1300)
From the point of view of the new time turner:
Magically created in a depleted state along with copy of user. (1200 noon)
Recharged by the destruction of the original time turner and user (1300)
This might not mean that one cannot acquire a time turner, trade it with one's future self's time turner, then use that time turner to travel back and trade with one's past self. The timelines would look like:
Manufactured
Created duplicate user and duplicate time turner.
(Exchanged)
Recharged by destruction of user and duplicate time turner.
and:
Created along with duplicate user
(Exchanged)
Destroyed along with original user to recharge original turner
A slightly more complex variation allows the time-tuner to contain multiple charges and behave in the same manner as !Harry uses it. An additional factor is required to explain the 8-hour limitation of multiple time turners.
Replies from: fractalman, shminux↑ comment by fractalman · 2013-06-03T05:50:07.333Z · LW(p) · GW(p)
It's 6 hours, i think, regardless of the chain of time-turners)
And given the 6-hour limitation, all you have to posit is a 6-hour ritual for creating the time turner before it's useable. presto, no going back to before the time-turner was created.
...but the BIG time machine in the ministry may or may not comply with the constraints of the smaller time turners. lastly, the 6-hour limitation itself... seems more like a way to premptively prevent a harry vs. quirremort TIME WAR than anything else.
Replies from: Decius↑ comment by Decius · 2013-06-03T08:13:58.698Z · LW(p) · GW(p)
Has it been established that, in addition to never being able to retreat more than 6 hours using any combination of !time turners, you cannot use more than one !time turner to travel back to the same hour repeatedly? I see many munchkin ways to abuse either possibility (one being to use multiple time-turners to be in an arbitrary number of places at the same time, and the other to take a tick back multiple times and then plant it on the opponent.
Replies from: fractalman↑ comment by fractalman · 2013-06-27T21:15:56.551Z · LW(p) · GW(p)
yes, it has. when harry was trying to stun Moody, moody asked harry if "are you going to give up, or do you think you're going to win?"
Harry replied that he was on his last copy because he had used up his remaining hours studying.
although...hang on, is it 6 or 7?
Anyways. Eliezer nerfed the time-turner precicely to prevent Harry from utterly totally munckinning it. "prevent or limit timetravel-based gamibt SPAM" and all that.
Replies from: Decius↑ comment by Decius · 2013-06-28T16:57:42.664Z · LW(p) · GW(p)
Is that an inherent limit, or simply the result of !Harry having access to only one time turner?
Replies from: fractalman↑ comment by fractalman · 2013-07-01T06:33:58.959Z · LW(p) · GW(p)
come to think of it, it might not be a full-blown limitation... But "time does not like being stretched". Even if it IS possible to squeeze one extra hour in by using a second time turner, it's not a good idea. (side effects may include insanity, extremely-rapid aging (more so than living 8 extra hours in a day would lead you to expect), and dizziness.)
Replies from: Decius↑ comment by Decius · 2013-07-01T18:03:37.637Z · LW(p) · GW(p)
If time is anthopic enough to like things in a literal sense, it can be bargained with. That might be the principle of operation of the time-turners.
If time can be bargained with, a rationalist who embraces a decision theory that one-boxes when the boxes are transparent has a lot of leverage.
Replies from: fractalman↑ comment by fractalman · 2013-07-01T21:05:33.870Z · LW(p) · GW(p)
Ok, I went and found Where the quote came from. chapter 17...
"Because it's pretty impressive if you're doing all that on just thirty hours a day." There was another slight pause, [snip] "I'm afraid Time doesn't like being stretched out too much"
I think this is more a case of Dumbledore using anthropic shorthand than actual anthropic reasoning. He seems to have...(ooh, here's a randomish theory: Dumbledore accidentally(?) performed a ritual that sacrificed a few pieces of his overall sanity in order to protect himself against dangerous time-travel related phenomena..)
Now where was I?...
Oh yeah. So that very much looks like extra time-clones MIGHT be possible with standard time-turners, but it is NOT a good idea.
Now, the Big Time Turner in the ministry (The one which had the ever hatching/un-hatching egg) probably COULD give you time-clone armies, laser-guided peggy-sues, and possibly even "change time" (whatever that means), but it is probably even more dangerous to use-you'd basically have to create your own reference frame, or else understand the real rules of Time.
But...maybe, just maybe: The BTT+philosophers stone+resurrection stone...
↑ comment by Shmi (shminux) · 2013-04-17T02:45:45.187Z · LW(p) · GW(p)
To create a user from nothing, the time turner would have to acquire negative mass equal to that of the transported user. To destroy the user at the moment of transport, the time turner would have to have had negative mass prior to the use. In short, user mass + time-turner mass = 0 at all times. Which would mean that a time turner is much more useful as a levitation aid.
In general, negative mass behaves weirdly. The Newton's laws still apply, so the gravitational force acting on the negative-mass time turner is mg (upwards if m < 0). However, if you let it go, it will fall down, not up (F=ma, so a=g, since m cancels). If you push on a negative-mass object, it tends to accelerate toward you. This makes carrying one quite hazardous: you can only pull it (in which case it gets pushed). If you even touch it, it will snap your hand.
Given the above, I think it is safe to say that time turners have positive (and small) mass and don't change their mass when activated.
Replies from: DaFranker↑ comment by DaFranker · 2013-04-17T16:10:55.443Z · LW(p) · GW(p)
To create a user from nothing, the time turner would have to acquire negative mass equal to that of the transported user. To destroy the user at the moment of transport, the time turner would have to have had negative mass prior to the use.
No it doesn't.
The time-turner just has to contain more mass-energy than the user at all times. Pretty simple, really. See it as a magical container and beamer for all that mass that gets displaced (and while creating the user-copy or deleting it, it also removes/generates an appropriate amount of air to equalize the pressure effects).
The process would, by the clock, look like this:
- Time-turner is created, has a mass of 5000 (the mass is not apparent for the same reasons that wingardium leviosa makes objects float / less heavy, whichever reasons those might be).
- Time-turner removes some air (mass 5000 + 2) and simultaneously generates a copy of the user (mass 5002 - 80)
- Time-turner later removes the "old" / time-turned user (mass 4922 + 80) and generates some air to fill in the space (mass back to 5000).
As for how the time-turner itself moves places, well, it presumably either shares the mass among copies somehow (entanglement? wormholes? exotic space geometries?) or just teleports the mass reserve where appropriate, or perhaps the "enchantment" is contained in a separate "server" of sorts (which would be most coherent with the Atlantis hypothesis).
This explanation conserves mass - the mass is just taken from a big rock or something at the creation of the time-turner, and then redistributed/refilled as appropriate to create the apparent effect of creation-from-nothing.
Replies from: shminux, Decius↑ comment by Shmi (shminux) · 2013-04-17T18:54:03.453Z · LW(p) · GW(p)
I agree that if you allowed to screen a fixed amount of energy from GR by magic, you don't have to have negative energy, as you can add an arbitrary and undetectable amount of it to whatever object you want. But then why stop there, GR is already broken.
Replies from: Decius↑ comment by Decius · 2013-04-17T19:08:46.071Z · LW(p) · GW(p)
GR was (apparently) broken by changes in mass; suppose instead that animagi convert between magic, energy, and mass in an analogous manner: When in a smaller form, they are more magical, and vice versa. Brooms convert magic to a combination of energy and/or something which is anti-mass but not anti-inertia, and/or something else which is not directly analogous to anything in GR.
The time turner then remains no harder than apparation combined with future prediction; future prediction is easy enough that it's provided in novelty drinks.
↑ comment by Decius · 2013-04-17T19:01:34.924Z · LW(p) · GW(p)
or perhaps the "enchantment" is contained in a separate "server" of sorts (which would be most coherent with the Atlantis hypothesis)
This also provides a possible mechanism for the 8-hour limit on multiple time turners; the servers can predict up to eight hours in the future, except they cannot accurately predict their own future prediction. Things which do not depend on their future prediction can mostly still be predicted "Do not mess with time" could have originated from there as well.
comment by Luke_A_Somers · 2013-04-16T15:02:14.798Z · LW(p) · GW(p)
Aside from the obvious 'It's magic, what are you talking about' complaint, I'd like to point out that if you have a wormhole and you draw a Gaussian surface around it, you're not done - you also need to draw a surface on the other side of the wormhole. Note that these will have equal and opposite contributions in the time turner scenario.
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-16T15:11:15.648Z · LW(p) · GW(p)
you also need to draw a surface on the other side of the wormhole.
Sure, and this will tell you that the Sun's mass will not suddenly appear out of nowhere some place else, in addition to not disappearing from the Solar system. An easy way to visualize the situation is to think what happens to the electric field lines of an electric charge traveling through a wormhole.
Replies from: Luke_A_Somers↑ comment by Luke_A_Somers · 2013-04-16T22:13:24.006Z · LW(p) · GW(p)
As an electron approaches a wormhole, more and more of its electrical field lines penetrate the wormhole, so its net charge appears to drop on the local side and rise on the far side. The mass and electrical charges do seem to appear and disappear out of nowhere. That's why a wormhole acts as a wormhole instead of a solid wall.
Or is there something going on about wormholes that I'm missing?
Replies from: shminux, Thomas↑ comment by Shmi (shminux) · 2013-04-16T22:27:28.142Z · LW(p) · GW(p)
Or is there something going on about wormholes that I'm missing?
Yes, there is. It's a common error made by those outside of the subject matter, though. The simplest argument is that a wormhole looks like a black hole from outside, and any charge dropped into a black hole is conserved. More visually, all the field lines remain continuous between the electron and whatever points in space the line goes through (otherwise you'll have to have local electric fields sinks and sources, which can only be other charges), so the total number of the field lines per unit area (electric field strength) remain the same. Hence, the measured charge remains the same, even long after the original charge has gone through the wormhole.
John Wheeler originally invented this concept of "charge without charge" and "mass without mass" for other purposes.
So the wormhole end in effect becomes electrically charged. What happens at the other end is equally fascinating.
Replies from: Luke_A_Somers↑ comment by Luke_A_Somers · 2013-04-17T12:13:51.533Z · LW(p) · GW(p)
But that line of reasoning assumes that a wormhole has an event horizon. Aren't wormholes naked singularities?
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-17T14:48:54.067Z · LW(p) · GW(p)
I get a feeling that you did not bother to read past the second line. The argument does not depend on the existence of a horizon. And no, wormholes are not naked singularities, they are not singularities at all, seeing how one can get through unmolested. I wonder if a Discussion post on this topic, with pictures, more details, and maybe some actual calculations would be of general enough interest.
Replies from: Luke_A_Somers↑ comment by Luke_A_Somers · 2013-04-17T16:08:41.869Z · LW(p) · GW(p)
I read the first two sentences, and then you said 'More visually', which suggests that it's a restatement. It's hard to think about dynamics in terms of field lines, and this is about a dynamical event, so I set it aside. Since that seems to be the heart of your argument, though, I'm focusing on it now. And now it's not clear to me that all of the usual identities about electrical field lines apply in curved space.
Where does the notion of field lines come from? In the infinite field lines limit, follow the electrical field outward from each differential angle of each charge. Note that you do this instantaneously in some reference frame. It's not like the field lines trace a traversable path. What you're tracing is offset at different times from the source charge. But if the charge is capable of not being there at some time in the past, then there's no problem with that charge providing no field line - with that field line disappearing without a charge present.
So instead, construct the charge's field lines so that instead of being at a present time, they are along the charge's future light cone. You won't find these field lines disappearing in the middle of nowhere. But you can use this to construct longitudinal electromagnetic waves.
So something is going to break - either we get longitudinal electromagnetic waves, or we get charges without charge. I don't see that it's clear that it has to be the second rather than the first. It depends what electromagnetism IS.
(I seem to have mistaken the definition of 'naked singularity', but either way there's no event horizon.)
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-17T18:17:42.919Z · LW(p) · GW(p)
It's interesting how our intuition differ. After working with curved spacetime, singularities, black holes, worhholes and other exotic constructs for a decade or so, I find it intuitively obvious that global charge conservation is not broken by local effects, so no matter how much weirdness is going on inside some small volume of space, the life must appear normal everywhere else. In particular, the usual Maxwell equations, including the Gauss law must hold in the nearly flat nearly static spacetime around this weirdness, simply by integrating the divergence of the electric field over a closed surface encompassing the region in question. In fact, you can do a little bit of manifold surgery to patch a wormhole solution to the flat spacetime one, and then all your garden variety EM applies in the flat region, with manifest charge conservation and what not.
Replies from: Luke_A_Somers, Luke_A_Somers↑ comment by Luke_A_Somers · 2013-04-17T19:25:17.632Z · LW(p) · GW(p)
Global charge conservation? I don't see charge being globally non-conserved. It just becomes possible to build surfaces that look Gaussian but aren't - they don't completely enclose a volume.
Like, you have a wormhole. Construct spheres A and B around each side of the wormhole. Together, these constitute a Gaussian surface for a charge inside them. Neither one of them does, alone, because there's a way out through the wormhole.
Moving a charge from one side to the other, from a charge conservation point of view, isn't different than if A and B were halves of a single sphere. A or B soaks up more or fewer of the electric field lines.
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-17T23:49:42.422Z · LW(p) · GW(p)
Like, you have a wormhole. Construct spheres A and B around each side of the wormhole. Together, these constitute a Gaussian surface for a charge inside them. Neither one of them does, alone, because there's a way out through the wormhole.
...Still thinking about the best way to show that the other half is irrelevant. The standard proof of the Stokes theorem requires a compact manifold with boundary, but the compactness condition does not hold either for a wormhole, or for a black hole with a singularity. In the latter case because the singularity is not a part of the manifold. The generalization should be trivial, but evades me at the moment.
Replies from: Luke_A_Somers↑ comment by Luke_A_Somers · 2013-04-18T00:14:29.708Z · LW(p) · GW(p)
What does a charge halfway through a wormhole look like?
Are these things one-way?
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-18T02:34:40.624Z · LW(p) · GW(p)
Here is one link informally describing the situation:
If a positive electric charge Q passes through a wormhole mouth, the electric lines of force radiating away from the charge must thread through the aperture of the wormhole. The net result is that the entrance wormhole mouth has lines of force radiating away from it, and the exit wormhole mouth has lines of force radiating toward it. In effect, the entrance mouth has now been given a positive electric charge Q, and the exit mouth acquires a corresponding negative charge -Q. Similarly, if a mass M passes through a wormhole mouth, the entrance mouth has its mass increased by M, and the exit mouth has its mass reduced by an amount -M.
It doesn't matter if the wormhole is one-way or not, but a one-way wormhole appears as a black hole on one end and a white hole on the other. You need some negative energy to support a two-way wormhole, which is, of course, a bit of a problem to obtain, but it does not affect the above argument.
Replies from: Luke_A_Somers↑ comment by Luke_A_Somers · 2013-04-18T13:55:51.700Z · LW(p) · GW(p)
So they're saying that if you start with a charge far from a wormhole on side A and drag it through to side B, then all of its electrical field lines trace their way back through the wormhole to side A again, just to maintain continuity of electrical field lines?
Well, at least I'm sure I understand what they're saying now. It doesn't seem crazy anymore, but I'm not in the least convinced that it's necessarily correct.
↑ comment by Luke_A_Somers · 2013-04-19T00:10:49.563Z · LW(p) · GW(p)
Stepping back, I found what bothers me here:
In particular, the usual Maxwell equations, including the Gauss law must hold in the nearly flat nearly static spacetime around this weirdness
In line with locality, I see the derivative forms of the Maxwell equations as fundamental, and the integral forms as a useful trick. When you go around changing topology, these tricks no longer apply naively.
The analogy between wormhole and black hole is broken. On the way across an event horizon, the charge gets infinitely red-shifted, which immortalizes it (slowing down an electrostatic charge doesn't make it lack charge). Wormholes, lacking an event horizon, lack this feature as well.
Basically, I just don't see any sort of local mechanism for generating this compensating field. Do you have anything that doesn't rely on a nonlocal/integral formation of Maxwell's laws?
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-19T03:23:02.212Z · LW(p) · GW(p)
OK, I'm pretty sure I have the argument.
First, I hope that we agree that what a distant observer thinks the throat charge can be measured by the field strength and correspondingly by the field flux through a Gaussian surface.
Now, if we take this Gaussian surface and shrink it around the charge into almost a point, the flux through it will not change, if it does not cross other charges during shrinking. (That's basically the divergence theorem for zero divergence.)
This shrinking is obviously possible when the charge is in some flat space far from the wormhole throat. It remains possible as the charge starts its trek into the throat. Since the spacetime manifold is continuous everywhere, there are no obstacles to shrinking the Gaussian surface to nothing, no matter how deep inside the wormhole the charge is. Even when it's out the other side. All that happens in that case is that your Gaussian surface has to travel through the throat and out (getting first smaller and then larger again in the process) before finally collapsing onto the charge.
More formally, the conditions for the Stokes theorem hold regardless of the manifold's topology, as long as the manifold is continuous and differentiable. Specifically, the manifold with boundary required by the theorem, which is embedded in a spatial slice of the wormhole spacetime remains topologically a solid sphere with the charge inside it, no matter where you move the charge.
Now, the obvious objection is "if this is so, how come we don't get the field of all other charges on the other side of the throat contributing to the total charge on this side?" And the answer is that they are not inside the original Gaussian surface, i.e. you cannot shrink it to collapse on the point which was not originally in it and did not move inside during the surface collapse process.
This is all rather counter-intuitive, given how imagining a curved 4D spacetime takes some getting used to, but hopefully it makes sense.
Replies from: Luke_A_Somers↑ comment by Luke_A_Somers · 2013-04-19T12:30:37.028Z · LW(p) · GW(p)
First, I hope...
yup
This shrinking is obviously possible when the charge is in some flat space far from the wormhole throat. It remains possible as the charge starts its trek into the throat.
It is approximately correct when the charge is in some flat space far from the wormhole throat and your not-quite-Gaussian surface also includes that throat but doesn't cut it off. The closer the charge gets to the throat, the worse this approximation is. That's because you're going to have some topological defect in your surface as it switches from containing the wormhole to not containing the wormhole. It's not the transition from one side to the other that's the problem. It's the transition from enveloping the wormhole to not enveloping the wormhole that's the problem.
If you never envelop the wormhole, then nothing you're saying ever gets to address the apparent charge on the wormhole.
More formally, the conditions for the Stokes theorem hold regardless of the manifold's topology, as long as the manifold is continuous and differentiable.
You're correct - the rule for Stokes theorem, regardless of the topology, is, you need to divide the space into an inside and an outside, and the inside doesn't get to extend to infinity. By drawing a sphere around one side of a wormhole, you don't accomplish this.
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-19T19:30:29.806Z · LW(p) · GW(p)
It is approximately correct when the charge is in some flat space far from the wormhole throat and your not-quite-Gaussian surface also includes that throat but doesn't cut it off.
No, my point was that it is exactly correct, not approximately. The retraction argument is topological, not geometric, thus the curvature does not matter. As long as the surface can be retracted all the way down to the charge, which it can, at least if you start some place far from the throat.
The closer the charge gets to the throat, the worse this approximation is. That's because you're going to have some topological defect in your surface as it switches from containing the wormhole to not containing the wormhole.
You don't get any topological issues. There is not a single topological defect at any point which prevents the surface to retract onto the charge. This is ensured by the continuity of the manifold.
you need to divide the space into an inside and an outside, and the inside doesn't get to extend to infinity. By drawing a sphere around one side of a wormhole, you don't accomplish this.
As I said, since the retraction is never broken, the charge always remains inside the surface, even when it's on the other side of the wormhole.
Maybe I should make and post a picture for a 2+1 case, now that I have it clear in my mind.
Replies from: Luke_A_Somers↑ comment by Luke_A_Somers · 2013-04-19T21:28:01.812Z · LW(p) · GW(p)
There is not a single topological defect at any point which prevents the surface to retract onto the charge. This is ensured by the continuity of the manifold.
Coffee cups are continuous too, but you can't retract arbitrary figures on them! That aside, you're right that there's no topological defect in the manifold itself. You can deform a 2+1 pair of planes smoothly into a cylinder. The topological problem occurs in respect to the Gaussian surface. What used to be a circle on the plane is turned by the presence of the wormhole into a circle which goes around the cylinder. IT NO LONGER HAS AN INSIDE.
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-19T21:51:23.148Z · LW(p) · GW(p)
IT NO LONGER HAS AN INSIDE.
Yes it does. A continuous deformation does not change topology. I am now pretty sure that you are using a wrong mental picture. The Gaussian surface in question does not wind around the cylinder, because it never has. Think about it this way:
- we start with a very large cylinder and a point on it
- we now draw a small circle around this point.
- the circle is contractible onto the point (by construction)
- we now slide the charge along the cylinder and extend the surface around it to accommodate, this does not affect the retractability in any way
- some part of the cylinder can be pretty narrow, corresponding to the wormhole's body, but it makes no difference, the Gaussian surface will extend through the wormhole and back out, still encompassing the charge in question, never once winding around the cylinder. It is simply connected all the way through.
↑ comment by Luke_A_Somers · 2013-04-19T21:57:43.959Z · LW(p) · GW(p)
I had exactly that mental picture already, thanks. The problem was that I was assuming that you were using a Gaussian surface that actually helped establish what you were trying to establish.
See, what you said works so long as your original Gaussian surface did not contain the wormhole.
If you never have a Gaussian surface containing the wormhole, then how the heck are you using it as an argument concerning the charge of the wormhole?
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-19T22:48:21.767Z · LW(p) · GW(p)
OK, sorry for making an incorrect assumption. It seems that I misunderstood your definition of "contain". According to yours, in 3d the sphere around the wormhole throat does not contain the throat, is this right? Given how it can slide through the throat and into the other side? If the wormhole happens to have the (spherical) event horizon, the event horizon does not "contain" the wormhole throat? What you mean by containing is that both throats must be inside the surface, such that if we excise the inside of the surface and replace it with a solid sphere, the whole of the wormhole disappears?
So, your argument is that, even through the Gaussian surface contains the charge, it does not contain the throat, and so some of the electric field from the charge is free to escape to the other side, as the charge traverses the wormhole? And that, once it is through, almost all of the electric field is on the other side?
Replies from: Luke_A_Somers↑ comment by Luke_A_Somers · 2013-04-20T00:52:00.588Z · LW(p) · GW(p)
Your second paragraph summarizing my position is correct. I don't necessarily understand your first paragraph because there are two senses of contain, but since you got the second paragraph right I trust that you meant the right things in the first.
(to clarify: the 'contain' in my above post was the naive sense of contain, like if the wormhole is hidden and you've got the region of space it's in surrounded -- this sort of surface is still useful for gauging the apparent charge on the wormhole, but it's not a true Gaussian surface)
So... what is wrong with this notion?
Well, before you answer that, I'll say what it occurs to me is wrong with it:
It feels different than other somewhat analogous cases. In particular, magnetic fields sustained by superconducting loops or plasma. Those magnetic fields don't just fade away. Once you thread the loop, it stays there.
On the other hand, the superconducting loop is made of charges that continuously maintain this field!
Once you'e looking at the geometry of space, though, it's not clear what's going to happen. Are the fields continuously radiated by charges and masses, and if you do geometry tricks to remove the charges or masses the fields go away? Or are the fields things that can't slip away like that?
Like... loosey-goosey imagery time! Consider a spherical shell of the electrical field on a charge that's 1 light second away from the charge. Is that a thing, or just a pattern?
If it's a thing, then moving the charge through the wormhole won't make it go through the wormhole too. If it's a pattern, then the charge moving through the wormhole makes the pattern go away.
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-20T03:17:25.618Z · LW(p) · GW(p)
Hmm, static fields are not "radiated" by charges or masses, they basically are charges or masses. That's why you cannot tell a tiny wormhole with a field threaded through it from a dipole, without looking closely down the throat.
I don't think the analogy with a pattern is a good one. Consider one electric force line going between the charge and infinity. As you move the charge, so does the line. But its two ends are firmly fixed at the charge and the infinity correspondingly. As the charge goes through the wormhole, this line is still "attached" to the infinity outside the throat, as it cannot just suddenly discontinuously jump between the two asymptotic regions. As a result, the electric field line gets "caught" in the topology, still going back through the entrance and out even after the charge generating it completely traversed the wormhole.
Here is a quote from an old paper http://arxiv.org/abs/hep-th/9308044:
Replies from: Luke_A_SomersIt is also interesting to consider what happens when a charged particle traverses a wormhole. (Of course, this “pointlike” charge might actually be one mouth of a smaller wormhole.) Suppose that, initially, the mouths of the wormhole are uncharged (no electric flux is trapped in the wormhole). By following the electric field lines, we see that after an object with electric charge Q traverses the wormhole, the mouth where it entered the wormhole carries charge Q, and the mouth where it exited carries charge −Q. Thus, an electric charge that passes through a wormhole transfers charge to the wormhole mouths.
↑ comment by Luke_A_Somers · 2013-04-20T13:45:23.925Z · LW(p) · GW(p)
Static fields sure ACT like they're radiated by charges - same causal structure (see my recent post about causality and static fields), same 1/r^2. And of course we always think of the radiative fields as being radiated by charges. So that covers both cases. Based on what actual lines of reasoning do we not consider static fields to be radiated by charges?
With your example, why do we say the field lines are attached? It certainly acts like it's attached whenever there are no wormholes around and as long as charge is conserved, that's for sure. But that might be a cause or it might be an effect. And when you dicky around with the assumptions connecting them, it may or may not end up being on the fundamental side of things.
Like... forget wormholes for a moment. Let's go to a counterfactual - imagine there was a weak interaction that violated conservation of charge. Assuming it actually happened, against all expectations, what do you think the electrical field would look like? If you trace it causally, what you see is unusual but you have no trouble - it's only in the spacelike cuts that it looks ugly. And physics really really doesn't act like it's implemented on spacelike cuts.
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-20T19:07:07.746Z · LW(p) · GW(p)
Static fields sure ACT like they're radiated by charges
I likely disagree with that, depending on your meaning of "radiated". I'd say they are "attached" to charges, acausally (i.e. not respecting the light cone). That's what the static field approximation is all about.
Then there is the quasi-static case, where you neglect the radiation. The java applet I linked shows what happens there: the disturbance in the static field due to acceleration of charges propagates at the speed of light.
I'll think more about your other arguments.
↑ comment by Thomas · 2013-04-17T05:58:25.019Z · LW(p) · GW(p)
Or is there something going on about wormholes that I'm missing?
Wormholes of cosmic proportions enabling stars to move accross - are just speculative objects, do not forget that!
Replies from: Luke_A_Somers↑ comment by Luke_A_Somers · 2013-04-17T12:16:40.882Z · LW(p) · GW(p)
Yeah. I meant theoretically. I wouldn't even bother to ask if we were talking about a real one, since the answer is obviously going to be 'How would I know?'.
Replies from: Decius↑ comment by Decius · 2013-04-19T03:29:18.735Z · LW(p) · GW(p)
If we were talking about an actualized wormhole, the answer SHOULD be "Let's find out!"
Replies from: Luke_A_Somers↑ comment by Luke_A_Somers · 2013-04-19T12:32:16.307Z · LW(p) · GW(p)
... we'd love to, but it's not as easy as it sounds.
comment by arromdee · 2013-04-16T22:13:42.971Z · LW(p) · GW(p)
To say that something is conserved means that it is the same at one time as it is at another time.
If you cannot time travel, and you have a set of objects at 1 PM, then you can compare them to the same set of objects at another time, such as an hour later.
If you can time travel, you can do the same--but terms such as "at another time" and "an hour later" become tricky.
With time travel, the time as measured in the setting is not the same as the time as measured by the objects themselves. If some of those objects time travel from 2:30 back to 1:30, then at 2 PM, 1 hour passed in the setting, you have some objects for which 1 hour has passed, and you have other objects for which 2 hours have passed. If you add those objects up just because 1 hour has passed in the setting you are adding the wrong things. You should be adding the objects for which one hour has passed (ignoring the objects for which two hours have passed) because conservation means that the amount is preserved one hour later--you can't mix objects from one hour later and from two hours later and expect to get an amount that is conserved. The fact that one hour has passed in the setting is useful in the non time travel case because the time that has passed in the setting is the same as the time that has passed for all of the objects. In the time travel case the fact that one hour has passed in the setting is just a distraction.
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-17T02:26:43.795Z · LW(p) · GW(p)
None of what you wrote makes any sense to me, sorry.
Replies from: MugaSofer↑ comment by MugaSofer · 2013-04-17T18:17:52.456Z · LW(p) · GW(p)
Here's how I read it:
Mass is neither created nor destroyed. It is moved.
Assuming you can tell if energy is conserved by checking if it's still there later, when you have a time machine, is like claiming a bolder has violated mass/energy conservation because somebody broke a piece off.
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-17T18:50:53.171Z · LW(p) · GW(p)
The whole point is that in GR conservation is local, you should be able to trace the movement of mass through space.
Replies from: MugaSofercomment by [deleted] · 2013-04-20T13:20:07.885Z · LW(p) · GW(p)
You should crosspost this to reddit.com/r/hpmor
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-20T18:35:46.819Z · LW(p) · GW(p)
Feel free, I don't really follow any particular subreddit.
comment by DaFranker · 2013-04-16T14:59:49.347Z · LW(p) · GW(p)
No mention of apparating, AKA teleportation. If I had to solve this, I'd theorize that both transmogrification and time turners reduce to the problem of teleportation. Most other magics can probably also be explained if you're able to selectively teleport things like forces, light, chemicals, and various forms of momentum or momentum-producing effects at a nanoscale near-instantaneously (at least as far as the local matter is concerned from the relevant reference frames).
Va bgure jbeqf, Ngynagvf pbagnvaf n Ovt Onqnff Anabgryrcbegre bs Qbbz, be fbzrguvat.
As for how that one would work: Beats me. When I try to work this deeply with relativity, I'm generally in way over my head and end up getting every single detail wrong in some way.
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-16T15:07:36.534Z · LW(p) · GW(p)
Teleportation can work. A targeted short-lived neutrino beam would do the trick, with a bit of work. At least you don't have to break conservation of energy or momentum.
Replies from: wedrifid, Randy_M↑ comment by wedrifid · 2013-04-17T02:59:15.151Z · LW(p) · GW(p)
Teleportation can work. A targeted short-lived neutrino beam would do the trick, with a bit of work.
Once you have teleportation you also have (the appearance of) mass altering transmogrification by simple application of the former to mass unknown to the observers. One you have apparent transmogrification and advanced-prediction-in-advance you have Time Turning. Again, through applying teleportation to carefully synthesized and selected matter that the observers aren't aware of. For example, to a fully synthesized extrapolation of "MoR!Harry!2.hours.from.now". From first personal perspective of Harry this is equivalent to time travel.
This is childs play to Omega, it's nearly exactly what he is contrived for in Newcomb's Problem. It may be a tad more difficult for a mere superintelligence but it remains an engineering problem, not so much a physics one.
At least you don't have to break conservation of energy or momentum.
The above possibile solution isn't "true" time travel. But it is to time travel precisely what targeted short-lived neutrino beams are to perceived "apparition". There is little qualitative difference in either relationship to described magical events or the phisical possibility.
Replies from: Decius↑ comment by Randy_M · 2013-04-16T22:14:13.584Z · LW(p) · GW(p)
My head scratcher with teleporatation is what would happen to momentum? If a Nightcrawler (x-man telporting charater) type jumped off an airplane and teleported mid-drop, would he still continue moving afterwards? I'd assume so, but say he changed position, teleported upside down. Now what?
Basically, is momentum relative to an internal or external frame of reference? I suspect the answer is, that isn't a question because the premises are impossible. Just like you can't divide by zero, you can't multiply by impossible.
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-16T22:43:08.972Z · LW(p) · GW(p)
Energy and momentum conservation is a local law in Relativity. How such conservation laws-compliant Nightcrawler would land depends on the actual mechanism of teleportation. For example, if he is transformed into a neutrino beam for the duration of the transport, the necessary momentum and angular momentum can be imparted by whatever surrounds him at the points of departure and arrival. The airplane or the surrounding air will recoil a bit (or a lot), and so will whatever objects he brakes against upon arrival.
Replies from: Decius↑ comment by Decius · 2013-04-19T03:32:41.762Z · LW(p) · GW(p)
I don't think momentum is conserved in Relativity; how could it be, if mass and therefore inertia are not?
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-19T04:45:21.296Z · LW(p) · GW(p)
Energy-momentum is most emphatically conserved in relativity, just not energy and momentum separately.
Replies from: Decius↑ comment by Decius · 2013-04-20T01:16:59.787Z · LW(p) · GW(p)
What are the units of energy-momentum? Momentum is massdisplacement/time, while energy is length^2mass/time^2, so the conversion ratio would have to have units of length^2/displacement-time. You can use some tricks to make it appear to have units of length/time (speed), but then you need to permit either momentum or energy to be negative.
Is the binding energy variable depending on velocity? Losing mass at rest does conserve momentum, while losing mass in motion does proportionately to the speed of the lost mass. (and notably, NOT depending on whether it was moving towards, away, or lateral to the observer- red/blue shift is irrelevant). Is the amount of energy observed to be released from a given atomic reaction variable depending on the speed of the reactants?
comment by Viliam_Bur · 2013-04-16T09:26:12.442Z · LW(p) · GW(p)
Maybe the person using the Time Turner is just converted to air molecules of the same mass... and in the past, the molecules of air convert to the given person.
Replies from: Vaniver, DaFranker↑ comment by Vaniver · 2013-04-16T14:37:09.543Z · LW(p) · GW(p)
Maybe the person using the Time Turner is just converted to air molecules of the same mass... and in the past, the molecules of air convert to the given person.
So, people are about 700 times as dense as air- a 45 kilo girl will go from occupying about .042 m^3 to about 30 m^3. If done slowly, this isn't a problem- but if you do the swap instantaneously, you need to have all of those molecules in that small volume. The pressure required (i.e. the pressure that it will expand outward with) without changing the temperature is 700 atm; this is comparable to the maximum chamber pressure of a firing pistol. The overpressure from the Oklahoma City bombing explosion was only about 40% that large.
In the reverse direction, 30 m^3 (the size of a room that's 12 feet by 11 feet by 8 feet, or a sphere with a 3.1m radius) of air will be required. If done too quickly, this could cause similar problems.
↑ comment by DaFranker · 2013-04-16T14:54:56.877Z · LW(p) · GW(p)
To summarize what Vaniver said: This is exactly equivalent to the person exploding into bits, except with even more pressure and the bits are solid compressed air flying around like shrapnel and killing everyone nearby with sheer pressure.
comment by DanielLC · 2013-04-16T16:45:04.864Z · LW(p) · GW(p)
It could be that she turns into neutrinos, which would go unnoticed, or tachyons, allowing her to physically travel back in time, or both if a certain explanation of the neutrino anomaly is to be believed.
Replies from: army1987, MugaSofer↑ comment by A1987dM (army1987) · 2013-04-16T20:12:15.871Z · LW(p) · GW(p)
The OPERA anomaly was due to... well...
↑ comment by MugaSofer · 2013-04-17T18:32:12.881Z · LW(p) · GW(p)
Who downvotes just because someone hasn't heard the latest news about the OPERA anomaly? Upvoted to neutral.
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-17T18:43:59.855Z · LW(p) · GW(p)
Don't sweat the vote. In the last few weeks I get flash-downvote spells (5-10 points drop in a matter of minutes on unrelated old comments) nearly every day, presumably from some anonymous coward(s) who are quietly pissed at me, and Merlin knows I said enough on this forum to piss off some otherwise rational people. Others complained of the same treatment before, hence no downvote button on the user page, but there is little one can do against a determined attacker.
Anyway. I find it best to stay away from karma discussions other than asking for feedback on unusually high upvotes or downvotes.
comment by Zaine · 2013-04-16T07:51:47.400Z · LW(p) · GW(p)
The time turners don't work as they are described; can you make them work? Assume you're operating in HP:MoR-verse, if it helps.
(Thank you for the treat.)
Replies from: shminux, Kawoomba↑ comment by Shmi (shminux) · 2013-04-16T16:48:55.799Z · LW(p) · GW(p)
I tried in this reply. Not sure if you find it to your satisfaction.
comment by [deleted] · 2013-04-23T12:15:29.503Z · LW(p) · GW(p)
Now another, deceptively similar question: if the Sun disappears this instant, how long before the Earth will stop orbiting the point where it used to be? The common answer: gravity travels with the speed of light, so also 8.5 min. This answer is obvious, simple and wrong. Yes, dead wrong. Why? because static gravity is not like light, it's more like electric field, only worse.
I think you're calling this more wrong than you should be. You follow it up by arguing that if the sun instantly disappears, there'll never be a gravity change, so the 8.5min is wrong. But really, what you're doing is showing that the sun can't instantly disappear in the sense that people mean the question. The natural follow-up is to modify the question to something like the "what if the sun split into two halves that rapidly moved apart" version, and in that case, unless I'm mistaken, it would be 8.5 minutes before the earth's orbit changed.
comment by orthonormal · 2013-04-18T13:35:14.969Z · LW(p) · GW(p)
Nice application of the point of Universal Fire to HPMoR. (Eliezer gets a pass on this, IMO, because it's someone else's fictional universe he's working in.)
comment by Armok_GoB · 2013-04-16T20:25:26.284Z · LW(p) · GW(p)
What if we turn the suns mass into two lasers that shot out equally from the poles?
Replies from: Thomas, shminux↑ comment by Thomas · 2013-04-16T20:55:16.662Z · LW(p) · GW(p)
Maybe we could do that. But the laser pulses short and narrow and therefore dense enough are two black holes made of pure light. Two black holes, leaving the scene with the exact speed of light.
Not much different from the already mentioned solution, albeit even cooler.
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-16T21:30:49.278Z · LW(p) · GW(p)
Two black holes, leaving the scene with the exact speed of light.
Black holes have rest mass, made of light or not, so this is impossible. Can you spot the error?
Hint: bar cubgba unf ab erfg znff, juvyr n ohapu bs cubgbaf va n obk (cubgba tnf) qbrf.
Replies from: wedrifid, Thomas↑ comment by wedrifid · 2013-04-16T22:41:20.816Z · LW(p) · GW(p)
Black holes have rest mass, made of light or not, so this is impossible. Can you spot the error?
It is true that black holes have rest mass, regardless of the form of mass-energy used to create them. As such it is impossible for them to travel at the speed of light. However the error-hint you give is misleading.
one photon has no rest mass, while a bunch of photons in a box (photon gas) does.
This is true but innaplicable. Making a black hole out of photons requires that the photons are not all moving in the same direction. For an explanation see here. Alternately, simply consider the case where all the energy is concentrated into a single (ridiculously) high energy photon. As you observe, "one photon has no rest mass". No black hole is formed by a single (unconfined) photon. The explanation is somewhat similar to why you don't form a black hole if you travel really fast.
↑ comment by Thomas · 2013-04-16T21:52:54.069Z · LW(p) · GW(p)
There is no error here.
http://en.wikipedia.org/wiki/Kugelblitz_%28astrophysics%29
You CAN have a black hole made of only photons. Of neutrinos even.
Replies from: wedrifid, shminux↑ comment by wedrifid · 2013-04-16T22:54:31.114Z · LW(p) · GW(p)
There is no error here.
I tentatively suggest that there may be an error in a detail here, albeit not the one shminux suggests.
http://en.wikipedia.org/wiki/Kugelblitz_%28astrophysics%29
You CAN have a black hole made of only photons. Of neutrinos even.
You can indeed (and such black holes are even more cool). However I believe the formation of a black hole from photons requires that the photons aren't all propogating in the same direction. This is a good thing because if you did have a black hole traveling at the exact speed of light then, well, reality is all broken down. Black holes have rest mass; speed of light travel is off limits. In general it doesn't really matter (so to speak) whether the mass-energy that they are created from is photons or a mix of paperclips and left over Babyeater offspring.
↑ comment by Shmi (shminux) · 2013-04-16T22:06:06.867Z · LW(p) · GW(p)
Of course you can. But they cannot travel at light speed.
Replies from: Thomas↑ comment by Thomas · 2013-04-17T17:13:44.515Z · LW(p) · GW(p)
The so called Kugelblitz can of course travel with the speed of light.
www.en.wikipedia.org/wiki/Kugelblitz_(astrophysics)
Imagine a laser beam pulse, 1 nanosecond long. A foot long and an inch tick cylinder of light is traveling with the speed of light, into the darkness of space. When it passes by an atom, there is a brief gravity effects between the two. The atom and the beam.
Now imagine two parallel beams, 1 kilometer long. The gravity effect is even larger. You can pile as many laser beams together, as you wish. You can make them (light) years long. Eventually, the passing atom can't escape the beam's gravity field, for the escape velocity is greater than c. The beam is a black hole.
Traveling with the speed of light!
Replies from: wedrifid, Luke_A_Somers, Decius↑ comment by Luke_A_Somers · 2013-04-18T23:30:55.049Z · LW(p) · GW(p)
In that case it didn't form a black hole until it interacted with the atom, and at that moment it slowed down from light speed. How do I know?
Until the atom came along, you're free to use any old reference frame without complicating matters terribly. So you can arbitrarily red-shift that laser pulse to the point that the energy density is trivial.
If you do that, you shift energy into the atom. So, it's the collision that allows it to condense into a black hole.
Replies from: Thomas↑ comment by Thomas · 2013-04-19T06:35:53.107Z · LW(p) · GW(p)
It is rather interesting question, what happens when the pure light black hole interacts with an atom. I have no idea, except that is interesting to consider this kind of questions. What happens under some peculiar circumstances. Do the current theories work there or break down?
I don't know the answer, who does?
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-19T07:40:24.242Z · LW(p) · GW(p)
As I mentioned before, there is no such thing as a "pure light black hole". All uncharged black holes are pure vacuum, regardless of how they formed. There are also no beam-shaped black holes (all black holes are spherical in shape, this is a well known result in General Relativity), though a regular black hole can theoretically form from light alone (kugelblitz) under certain rare circumstances.
Your model has a number of technical errors which prevent it from working. For example, you cannot form a black hole by shooting a light pulse from a laser, except maybe by focusing it, in which case the center-of-mass velocity will be sublight and you end up either creating multiple black holes or feeding a single one, depending on the details. Moreover, while light does curve spacetime, it's not as simple as the gravitational attraction between massive bodies. There are many known solutions which include perfect null fluid (that's what a continuous beam of light is), none are trivial.
If you still think that you can think up something that 100 years of GR research by the geniuses like Einstein, Hawking and others did not notice, consider learning the subject seriously first.
Replies from: Thomas↑ comment by Shmi (shminux) · 2013-04-16T20:47:37.214Z · LW(p) · GW(p)
That certainly works, in theory, though to emit the sun's mass as light in a short enough time, you will have to outshine a quasar by orders of magnitude (try a Fermi estimate on this). It will probably heat up the dust around the Sun enough to bake and evaporate all of the Solar system in its heat and scattered laser light. Also note that this would not contradict my assertion that stuff has to travel outward, not just disappear.
comment by [deleted] · 2013-04-16T15:37:07.681Z · LW(p) · GW(p)
So what happens after the Vogons have finished sucking the mass through their wormhole and close it off? Does the earth end up orbiting a patch of empty space that still exerts gravity, or was that section actually an argument that they can't open/close a wormhole and suck the sun through it?
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-16T15:48:56.427Z · LW(p) · GW(p)
It will not be just empty space, it will be a black hole as massive as the Sun was. Rather counter-intuitive, I know. But note that black hole is "a patch of empty space that [...] exerts gravity".
comment by Vladimir_Nesov · 2013-04-16T11:06:53.055Z · LW(p) · GW(p)
But if you give up on General Relativity, quite a few things will unravel, like all four Newton's laws.
"Weasley says that rockers use a special kind of science called opposite reaction, so the plan is to develop a jinx which will prevent that science from working around Azkaban."
comment by tgb · 2013-04-18T12:19:42.452Z · LW(p) · GW(p)
I'm disappointed in this thread - not because there are mistakes here but because posters are not completing the due diligence of admitting their own ignorance on the subject. General relativity is known to not be simple and some claims made here wouldn't even pass muster in the much simpler Newtonian world. We should all be practicing the skill of saying 'this subject is deemed complex and so I am quite likely to be making a mistake given that I haven't studied it in depth.' Usually at this point we have some grad students studying the subject stepping in to clear up misconceptions, but I don't see any here so I'm speaking up even though I have no training in GR. (And to take my own advice: I could be wrong about this all! please let me know if I am. But calling out an argument takes less than making one so my job here is easier.)
Okay, so for some specifics: How about the over-application of Gauss's Law as a tool to say "it doesn't matter what happens, the net effect is the same on Earth'. Gauss's Law only says that about spherically symmetric mass distributions and quite a few of the situations here are not that. (Otherwise the three body problem would be trivial to solve!) Or how about some of the comments here that involve accelerating massive bodies to near light-speed very rapidly. Sure this is fine in Newtonian mechanics, but in GR the acceleration of a large body creates gravitational waves, etc. It's not so simple as it's being made to seem!
I expect more diligence and caution from a community that values being correct over sounding correct! And here I will applaud those who has shown uncertainty in this, such as RichardKennedy.
(Note: I wasn't one of the ones who originally downvoted shminuxl.)
Replies from: Luke_A_Somers, shminux↑ comment by Luke_A_Somers · 2013-04-18T23:23:14.675Z · LW(p) · GW(p)
Upvoted for first paragraph. Nearly took it back for confidently asserting false statements about Gauss's Law, in direct contradiction of point in first paragraph.
Replies from: tgb↑ comment by tgb · 2013-04-19T12:46:54.636Z · LW(p) · GW(p)
Please note the unconfidence already stated in the first paragraph and the point that I should expect lower standards for myself to be able to say "hey I don't think this is right" versus "here's a strong argument I created whole-cloth." Secondly, I am still reasonably confident about the statements about Gauss's law that I tried to make, though there is a good chance I have miscommunicated them. See my reply to shminux for an attempt at clarification.
Replies from: Luke_A_Somers↑ comment by Luke_A_Somers · 2013-04-19T13:29:11.520Z · LW(p) · GW(p)
First, I did say 'nearly'. The only weakening was by that first paragraph. If you're not so familiar with Gauss's Law that you can restate the differential and integral forms without half thinking about it, and you hear something like the above, maybe it would have been a good idea to visit the following:
http://en.wikipedia.org/wiki/Gauss_Law
Which would have avoided all of this. For further information, see:
↑ comment by Shmi (shminux) · 2013-04-18T21:15:59.499Z · LW(p) · GW(p)
As an expert in the area:
Gauss's Law only says that about spherically symmetric mass distributions and quite a few of the situations here are not that.
False, it applies to any closed surface. Spherical symmetry is just the simplest problem where the Gauss law can be usefully applied.
(Otherwise the three body problem would be trivial to solve!)
The three-body problem has no relation to the Gauss law, except insofar as it is equivalent to the inverse square law for point masses in flat space.
Or how about some of the comments here that involve accelerating massive bodies to near light-speed very rapidly. Sure this is fine in Newtonian mechanics, but in GR the acceleration of a large body creates gravitational waves
This is true in most cases, but not in spherical symmetry (there are no spherically symmetric gravitational waves). That's why I could write "an amount of matter equivalent to her mass has to travel inwards out of nowhere and coalesce into a person" and not worry about energy losses to gravitational radiation. Though there would be, of course, a lot of losses to EM radiation if the matter in question consisted of charged particles.
Hope this makes sense.
Replies from: tgb↑ comment by tgb · 2013-04-19T12:46:48.583Z · LW(p) · GW(p)
False, it applies to any closed surface. Spherical symmetry is just the simplest problem where the Gauss law can be usefully applied.
Oh I absolutely don't deny that. What I do assert is that you can't ignore what happens on the inside of a closed surface if you don't know that the mass is spherically symmetric and you want to calculate the force at a specific force. If this is not what you are saying, then this is a miscommunication not a disagreement. What I took you to mean when you made comments like "This still conserves mass, since black holes have mass. It also obeys the Gauss Law" is to mean not just that Gauss's law holds, but that it's integral form can still be used to calculate the force on the Earth in a trivial matter depending only upon the mass inside the closed surface and hence then the force on the Earth would stay the same as if the sun had not split in two.
I hope that we both agree that Gauss's law could not be used in such a manner (as it only gives the integral of flux over the surface and so without symmetry this force is not constant over the surface). It believe that you would also agree that splitting the sun in two and sending them in opposite directions as described would result (in the case of Newtonian gravity) in a decrease in the gravitational force the Earth experienced. Given this, I am not sure how to charitably interpret the comment I quoted. Could you elaborate on what you meant if you didn't mean what I thought you meant? I will at this point certainly admit and apologize for the fact that I interpreted your comments uncharitably since it sounded like you were making a common beginner's error.
My three-body comment meant this: if we could use Gauss's law to say that the force from inside of an enclosed area is directly proportional to the mass inside, then we would be able to draw a region in space around two of the three bodies and calculate the force on the third. This force would then depend only upon the total mass which is constant and would point to the center of the enclosed area. So then we'd be able to solve for that one body's motion independent of the other two. This is patently absurd and not possible. I chose to give it as an example of how such a naive application of Gauss' law would give obviously incorrect results but the illusion of transparency grabbed me and I failed to make myself clear.
This is true in most cases, but not in spherical symmetry (there are no spherically symmetric gravitational waves).
That comment was not directed at your original post but at some of the comments on splitting the sun in two and shooting them apart. Sorry for the confusion!
Replies from: shminux↑ comment by Shmi (shminux) · 2013-04-19T19:21:09.423Z · LW(p) · GW(p)
What I took you to mean when you made comments like "This still conserves mass, since black holes have mass. It also obeys the Gauss Law" is to mean not just that Gauss's law holds, but that it's integral form can still be used to calculate the force on the Earth in a trivial matter depending only upon the mass inside the closed surface and hence then the force on the Earth would stay the same as if the sun had not split in two.
This would be a novice mistake I have been correcting countless times as a tutor and TA. The Gauss law holds even when there is no symmetry, but it is much less useful to calculating electric or gravitational field at a given point. It is, however, can be profitably used to argue other points, like the one I made.
I will at this point certainly admit and apologize for the fact that I interpreted your comments uncharitably since it sounded like you were making a common beginner's error.
No need for an apology, I'm glad we cleared that up.
if we could use Gauss's law to say that the force from inside of an enclosed area is directly proportional to the mass inside, then we would be able to draw a region in space around two of the three bodies and calculate the force on the third.
I understand what you mean now, given what you said previously and I certainly agree that this would make no sense.
Inferential distance is a b****.
comment by Thomas · 2013-04-16T09:34:23.644Z · LW(p) · GW(p)
mass cannot just disappear, it has to spread out.
This is not entirely true. At least in principle, the Sun can be divided into two black holes. One consists the Sun's north half, the other one of the Sun's south half, each going in the opposite direction with the nearly light speed, perpendicular to the ecliptic plane.
Earth would feel a rapidly fading gravitational pull and the darkness.
The question is, can we do it even better? Can the Sun disappear in the opposite direction, away from us with the (nearly) speed of light, without spraying us with some deadly rays?
Yes, it can. Can anybody figure it out how?
Replies from: shminux, Mestroyer, wedrifid↑ comment by Shmi (shminux) · 2013-04-16T14:36:25.505Z · LW(p) · GW(p)
each going in the opposite direction with the nearly light speed, perpendicular to the ecliptic plane.
This still conserves mass, since black holes have mass. It also obeys the Gauss Law
Replies from: Luke_A_Somers↑ comment by Luke_A_Somers · 2013-04-18T23:09:40.717Z · LW(p) · GW(p)
Who's downvoting this? It's obviously correct, while the post it's responding to is completely wrong on its first, most important statement.
↑ comment by Mestroyer · 2013-04-16T15:47:43.488Z · LW(p) · GW(p)
Would this work?
↑ comment by Thomas · 2013-04-16T16:02:53.276Z · LW(p) · GW(p)
It would work, to a degree. But for the majority of the Sun's mass to escape from us with the nearly light speed, it wouldn't. The small pieces should be very close to the speed of light and thus very heavy for an observer here on Earth. That would mean an even bigger gravitational pull toward the ex-Sun, for a while.
↑ comment by wedrifid · 2013-04-16T10:19:33.782Z · LW(p) · GW(p)
The question is, can we do it even better? Can the Sun disappear in the opposite direction, away from us with the (nearly) speed of light, without spraying us with some deadly rays?
Sure, it just has to make sure it's rays (or relativistically accelerated mass packets) miss us. Shooting two (or more) simultaneous streams of mass on appropriate vectors that add up to directly away from us should work fine... or it would in a true vacuum. Depending on the details of the implementation I'd be a little worried about indirect rays hitting earth due to collisions with dust.
Yes, it can. Can anybody figure it out how?
What is your solution? The obvious one (above) requires the Sun sacrificing rather a lot of mass in the process of running away. Does your solution avoid this?
Replies from: Thomas, ikrase↑ comment by Thomas · 2013-04-16T10:47:04.464Z · LW(p) · GW(p)
I have a solution, which by its effect, is nearly indistinguishable from the "instantly disappearing Sun".
You have to convert all the Sun's mass into a Dyson sphere at the Mercury's orbit. For the Sun's now missing light, it can be a hot spot somewhere on the Dyson. Arbitrary perfect imitation of the Sun shinning on the sky for us.
Now, the sphere has started to inflate at the rate of a few km per second. Our orbit would be just the same until the moment the Dyson sphere reaches us. At the last few million kilometers the sphere's inflation accelerates and we are enveloped inside otherwise empty sphere in a matter of seconds. We are in zero gravity of a hollow sphere, shortly after everything still seemed normal. Of course, there is a hole in the surface of the DS just big enough that no collision takes place.
Replies from: wedrifid, shminux↑ comment by wedrifid · 2013-04-16T11:35:18.350Z · LW(p) · GW(p)
We are in zero gravity of a hollow sphere
Oh, that's what the gravity from a hollow sphere all adds/multiplies out to? Uniform zero (net) gravity inside, normal outside the sphere? Neat.
Replies from: iDante, Vaniver↑ comment by iDante · 2013-04-17T00:29:39.069Z · LW(p) · GW(p)
An aside for those curious about the Gauss Law argument. The law in its integral form states that the flux of the gravitational field inward through any closed surface encompassing the Sun is proportional to the Sun's mass.
As long as the mass distribution is spherically symmetric the gravity outside of the sun is the same as if the mass was all located at the center. It's the same for electrostatic force since that goes like 1/r^2 too :D.
Oh, that's what the gravity from a hollow sphere all adds/multiplies out to? Uniform zero (net) gravity inside, normal outside the sphere? Neat.
adds
↑ comment by Vaniver · 2013-04-16T14:16:15.353Z · LW(p) · GW(p)
Yep! Newton had a proof in Principia, but here's a more recent, geometric one.
Replies from: Decius↑ comment by Decius · 2013-04-17T00:09:56.831Z · LW(p) · GW(p)
At the center there is no net, but inside the sphere the net effect is the same as if all the mass were concentrated at the center. (Suppose that there was a normal net effect everywhere at and just above the surface and zero net effect just below the surface: that would require that the gravitational field not be continuous, which is not the case)
Replies from: Vaniver, Luke_A_Somers↑ comment by Vaniver · 2013-04-17T12:42:59.067Z · LW(p) · GW(p)
At the center there is no net, but inside the sphere the net effect is the same as if all the mass were concentrated at the center.
Remember that this is a spherical shell, not a sphere. When calculating the force of gravity inside the Earth, for example, you ignore all mass at higher radius that the location you're interested in, but not the mass at lower radius.
Why does this work? Imagine opposite cones that originate at the center of the spherical shell. The intersection of that cone with the shell will have surface area that increases with r^2, but the inverse square law decreases with r^2, and so the gravitational effect only depends on the angle of the cone, regardless of distance- but the two cones are pointed in opposite directions, and so cancel out. (This is obvious at the center, but works just as well elsewhere inside the spherical shell.)
that would require that the gravitational field not be continuous, which is not the case
For an infinitely thin shell, the drop in gravitation is infinitely steep. But shells with real thickness will have a gradual dropoff that corresponds to that thickness.
Replies from: Decius↑ comment by Decius · 2013-04-17T18:54:05.229Z · LW(p) · GW(p)
... So pressure in a gaseous celestial body doesn't increase linearly; delta-p falls off to zero in the center?
Replies from: Vaniver↑ comment by Vaniver · 2013-04-17T19:14:33.036Z · LW(p) · GW(p)
It's not clear to me where that model came from. We shouldn't expect the density of celestial bodies to be uniform unless they're made of something incompressible, and it's important to separate out the net force of gravity and the weight of mass above you. In steady state, all of the mass in the celestial body is being pulled towards the center by gravity, pushed upwards by the mass below it, and has to push upwards on the mass above it (with the net force being 0). I haven't done the math to see what the pressure function would look like for a gaseous celestial body, and it seems like the full calculation will have lots of complications, but we can note that the mass below you has to push up harder than the mass above you is pushing down, suggesting the pressure is highest at the center.
Replies from: Decius↑ comment by Decius · 2013-04-17T20:03:56.501Z · LW(p) · GW(p)
delta-P is the first derivative of pressure; it would have to be zero at the center for there to be a pressure maximum at zero.
I would expect a gaseous body to have a roughly spherically symmetric mass distribution, which is all we need. Treat it as an infinite number of infinitely thin spheres each of uniform density, and we can do calculus on it.
We can also do this though experiment with a perfect liquid of uniform density; at least it will have a surface that we can stop at. Pressure is still highest at the center and reality is continuous, meaning dP/dR is zero at the center and approaches zero as R approaches zero.
Surface gravity of a sphere of constant density and radius R is proportional to R? Mass is proportional to volume (R^3) and surface gravity is proportional to mass/R^2, or R^3/R^2, or R.
Okay, I've got a new respect for the problems involved with using barometric pressure to measure altitude, and the advantages of using barometric pressure directly for navigational purposes at high altitudes.
↑ comment by Luke_A_Somers · 2013-04-18T23:13:01.402Z · LW(p) · GW(p)
Can you please use questions instead of confident contradiction when you're up against Newton?
Replies from: Decius↑ comment by Decius · 2013-04-19T03:20:25.100Z · LW(p) · GW(p)
At the time of writing, I thought I was paraphrasing Newton. Downbranch I realized that the behavior of being attracted to the center of a shell while inside the shell would, in a body composed entirely of fluid, yield a maximum first derivative of pressure at the center, and no maximum of pressure within the body.
↑ comment by Shmi (shminux) · 2013-04-16T14:37:47.132Z · LW(p) · GW(p)
This will indeed work, but it has no relation to your original puzzle:
Replies from: ThomasCan the Sun disappear in the opposite direction, away from us with the (nearly) speed of light, without spraying us with some deadly rays?
↑ comment by ikrase · 2013-04-16T10:34:46.927Z · LW(p) · GW(p)
Why bother with black holes?
Replies from: Thomas, wedrifid↑ comment by wedrifid · 2013-04-16T11:02:35.194Z · LW(p) · GW(p)
Why bother with black holes?
Better question: why not bother with black holes? I mean, since we're talking about the implications doing crazy impossible stuff to reality anyway...
Mind you, I can't find any reference or allusion to black holes in the grandparent so I'm not entirely sure why you are asking me.
comment by MrMind · 2013-04-17T07:28:54.048Z · LW(p) · GW(p)
I suspect that someone is systematically downvoting every comment made to this post, but what the heck: it's interesting, I'll add my two cents and gladly take the downvote. 'gimmie your best shot pal, I can take it.
About a possible way for a time-turner to work...
IIRC, in order to make an Alcubierre wave more efficient, a space-time metric was proposed that in a sense isolated the ship from the outside universe, exposing only a very tiny surface. This allowed the wave to carry away the object without requiring the energy of all stars in our galaxy.
I can assume that Time-Turners, since they can work only hour-by-hour, deploy a very tiny wormohole every hour (!!) in a connected chain, and when the user activate it, it encloses him/her in this bubble and transport him/her through one of the point of the chain.
Of course I cannot calculate it, but I suspect that the problem lies in the bubble formation: I don't know if the energy spent to form it can only be directed inwards or through the holes...
But hey, it's magic!
↑ comment by wedrifid · 2013-04-17T08:59:48.990Z · LW(p) · GW(p)
I suspect that someone is systematically downvoting every comment made to this post, but what the heck: it's interesting, I'll add my two cents and gladly take the downvote. 'gimmie your best shot pal, I can take it.
It is clearly not the case that someone is sytematically downvoting every comment made to this post. It is a prediction that is easily tested by hovering the mouse pointer over the karma of the comments. This will show you that many comments, including some of the older ones, are 100% positive and many more are 0% positive with 0 karma. Both of those states preclude them having been downvoted, systematically or otherwise.
I am about to downvote this comment without reading further. I encourage others to do likewise (unless you edit out the bluster). As a rule of thumb nearly any comment that conveys "I'm going to get downvoted but " is worth downting.
Replies from: MrMind, MrMind↑ comment by MrMind · 2013-04-17T12:36:59.481Z · LW(p) · GW(p)
It is clearly not the case that someone is sytematically downvoting every comment made to this post. It is a prediction that is easily tested by hovering the mouse pointer over the karma of the comments. This will show you that many comments, including some of the older ones, are 100% positive and many more are 0% positive with 0 karma. Both of those states preclude them having been downvoted, systematically or otherwise.
Ah, I didn't knew about that function: in that case, your analysis is absolutely correct! I retract my statement.
I am about to downvote this comment without reading further.
That, on the other side, seems wrong. You should downvote and read forward.
I encourage others to do likewise (unless you edit out the bluster)
That seems very wrong: in an environment that predicates rationality, stating that you're wrong should be encouraged, not punished. It's not about the downvoting, as I've written, it's about downvoting unless I cancel the original statement. The shift seems to have moved, at least in your comment, from being less wrong to being right.
I refuse such a drift. I won't edit out the statement (but I'll gladly admit I was wrong).
Replies from: wedrifid, MugaSofer↑ comment by wedrifid · 2013-04-17T13:34:49.621Z · LW(p) · GW(p)
That seems very wrong: in an environment that predicates rationality, stating that you're wrong should be encouraged, not punished. It's not about the downvoting, as I've written, it's about downvoting unless I cancel the original statement.
Pardon me, I wasn't clear. Retraction and (noted, transparent) editing can sometimes be even better than removal so of course can be included in the exception to the suggestion.
The shift seems to have moved, at least in your comment, from being less wrong to being right.
Note that wrongness isn't the issue. It is the combination "needless provocation && wrong". For example if there wasn't available proof that the claim was wrong then I would perhaps still have downvoted just because the karma-drama is annoying but I would be unlikely to have also made a comment expressing criticism of the behavior.