tag:blogger.com,1999:blog-11295132.post5316387978274908480..comments2017-11-16T08:47:17.857-08:00Comments on A Neighborhood of Infinity: Faster than a speeding photonDan Piponihttps://plus.google.com/107913314994758123748noreply@blogger.comBlogger31125tag:blogger.com,1999:blog-11295132.post-85918521931231640282010-04-28T17:32:09.653-07:002010-04-28T17:32:09.653-07:00My understanding is that you'd need an infinit...My understanding is that you'd need an infinite amount of energy to create the force needed continually accelerate any craft mass to the speed of light in order to match the speed of the intrinsically massless photon in order to stay ahead of it long enough that your finite lifetime in the craft ended.<br /><br />*deep breath*<br /><br />You're not really outrunning the photon, because you're not going faster than it - instead you're keeping it at a steady state arms length.<br /><br />I don't know where you're getting a finite craft with infinite energy (unless your sucking it in the nose of the craft!) or if that craft was an accelerating universe popping into another infinitely larger accelerating universe...<br /><br />Hubble tells us our universe is accelerating, and that we may be that craft though... but doesn't tell us what exactly this accelerating force is. What the force people call dark energy is.<br /><br />Some think it's eternal inflation, others Einstein's cosmological constant, others yet again have labeled it quantum gravity, or now the latest fad everyone loves to call the entropic force.<br /><br />Whatever it is, may the force be with you! ;^)<br /><br />Do you have a favourite?Craighttps://www.blogger.com/profile/12170572784481628483noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-66001910108678709092010-04-28T10:25:56.971-07:002010-04-28T10:25:56.971-07:00doug,
What I just said is a bit misleading. If so...doug,<br /><br />What I just said is a bit misleading. If someone falls into a black hole, people outside never *see* the "victim" cross the event horizon because they will never receive photons emitted as the victim crosses over. It's debatable whether this should be described as "from the reference frame of people outside, the victim never crosses the event horizon".<br /><br />Nonetheless, that's not the situation that applies here. In the black hole case, the victim crosses the event horizon at a finite proper time. In the case here, there is no proper time for people on the spaceship at which they perceive the photon catching up with themselves.sigfpehttps://www.blogger.com/profile/08096190433222340957noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-52688048055490592442010-04-28T09:46:18.149-07:002010-04-28T09:46:18.149-07:00doug,
There is the world-line of the photon and t...doug,<br /><br />There is the world-line of the photon and the world-line of the spaceship. The question is: do these ever intersect? The answer to this question is independent of reference frame. Things like distance, mass and direction are all measured with respect to a reference frame. But whether or not two world-lines meet doesn't change when you change reference frame. If two curves meet in one reference frame, they meet in all reference frames, and vice versa. The world-lines of the photon and the spaceship never meet.<br /><br />Suppose that the photon is the front of an expanding shell of deadly radiation. You could use constant acceleration to outrun this shell and stay alive. But there might be a catch: it might be that the proper time of the hyperbolic curve you follow is finite. Ie. even though you outrun the photon, you only get to experience a finite amount of life. However, it turns out that the proper length of a hyperbola goes to infinity as you travel along the curve. So not only can you outrun the photon, you can do so for what feels like forever in your own frame of reference.<br /><br />Take a look <a href="http://www.mathpages.com/rr/s2-09/2-09.htm" rel="nofollow">here</a> for some technical details. The proper time (ie. the time you experience) remains finite along the full length of the hyperbola.sigfpehttps://www.blogger.com/profile/08096190433222340957noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-30286817237090361662010-04-27T23:02:00.432-07:002010-04-27T23:02:00.432-07:00I believe that you are mistaken. The photon would...I believe that you are mistaken. The photon would only appear to never catch the ship from the reference frame of whatever the ship is accelerating with respect to. From the reference frame of the ship, the photon would still appear to approach at the speed of light and pass the ship. That's just how fast light travels.doughttps://www.blogger.com/profile/16980154668660962867noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-54002531386108527892010-01-06T15:43:05.702-08:002010-01-06T15:43:05.702-08:00It would be interesting while thinking of a proton...It would be interesting while thinking of a proton as a singularity. To produce controllable black holes that can be manipulated to travel in a direction we choose. <br /><br />The next part is building a ship that can safely orbit said miniature black hole and manipulate its direction.<br /><br />Adding this to my to do list.Clarkhttps://www.blogger.com/profile/09990370484774153370noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-15959339682447738592009-05-05T15:48:00.000-07:002009-05-05T15:48:00.000-07:00Sundar, what you are thinking of is constant circu...Sundar, what you are thinking of is constant circular motion, with no angular acceleration. Electrons in a particle accelerator certainly have acceleration in the direction of motion (this just requires a greater force to maintain the circular path).<br /><br />At least, I think that's what you were talking about.<br /><br />If the discussion is constant circular motion, then shouldn't the false centrifugal force provide constant acceleration? However, we do have the problem with our light-beam analysis (light wouldn't travel with the circle). This might not be a problem though, if we let the centripetal force be gravitational, the orbit keep the spaceship from falling (instead of thrusters), and just consider light as radiating from the center. I don't know if I understood this anywhere close to correctly.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-11295132.post-29330780437615350402009-04-29T20:00:00.000-07:002009-04-29T20:00:00.000-07:00"But for a large enough circle, the motion will lo..."But for a large enough circle, the motion will look like approximately linear motion for short intervals at a time"<br /><br />But then, if you are approximating the circular motion as a linear one, don't you lose all the acceleration there? Because by definition, the only acceleration in a circular motion is the change of direction. Once you lose that and assume linear motion, you're just travelling in constant velocity, isn't it so? <br /><br />I understand more rigorous analyses have been performed regarding extending this to circular motion, but it's the 'simplification' here that appears wrong to me...Sundarhttps://www.blogger.com/profile/00415039983973239001noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-16516746028285610442009-04-12T12:58:00.000-07:002009-04-12T12:58:00.000-07:00genneth,I was thinking of giving an account based ...genneth,<BR/><BR/>I was thinking of giving an account based on FFTs but I wanted to try to get some talk of creation and annihilation operators in there and couldn't figure out a "pop science" explanation of them.<BR/><BR/>Good look on the BEC experiments. I seem to remember reading about something similar with researchers looking for Hawking radiation near a BEC "event horizon".sigfpehttps://www.blogger.com/profile/08096190433222340957noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-54968782605414088912009-04-12T09:29:00.000-07:002009-04-12T09:29:00.000-07:00Aha! Finally something on your blog that I underst...Aha! Finally something on your blog that I understand!<BR/><BR/>There is also an entirely classical way to view the Unruh effect: Doppler shift. Imagine you are accelerating into a wave of fixed frequency; due to Doppler shift, you will see it at all sorts of different frequencies. If you do the maths (a Fourier transform) and average over all frequencies, you recover the Planck distribution for a field at the Unruh temperature. If you're familiar with the QFT formalism, then you're simply taking the Fourier transform of the correlation function along the desired trajectory.<BR/><BR/>In the circular case, you do get a similar effect, but the distribution is not Planckian. To a physicist, thermal is really a statement about time-independence of a statistical state. There are quite a few different trajectories which all give a thermal distribution, and all of which can be assigned a temperature in some appropriate sense. In fact, recall that a constant acceleration is really just a constant rotation...<BR/><BR/>A quick plug: my own work has been focussed on whether it's possible to see the Unruh effect in atomic Bose-Einstein condensates. We are hoping to move a detector (atom in various electronic states) in a circle in a BEC. The phonons in the BEC act as a quantum vacuum, and should give rise to an analogous effect. Hopefully in the next couple of years we can get some experimental groups to really take a crack at this.gennethhttps://www.blogger.com/profile/02376760053977600605noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-59147259266448551802009-04-08T09:27:00.000-07:002009-04-08T09:27:00.000-07:00Andrej,I was thinking about a black hole monad, re...Andrej,<BR/><BR/>I was thinking about a black hole monad, related to my <A HREF="http://blog.sigfpe.com/2007/03/shor-quantum-error-correcting-code-and.html" REL="nofollow">heat monad</A>. :-)sigfpehttps://www.blogger.com/profile/08096190433222340957noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-38188301116708304282009-04-07T11:45:00.000-07:002009-04-07T11:45:00.000-07:00Yes, that's right! I completly overlooked the fact...Yes, that's right! I completly overlooked the fact that velocities aproaching c distort the very basic stuff like mass, time and lenght. It makes perfect sense to me now. I sincerely apologize for posting stupid questions like that.Charlienoreply@blogger.comtag:blogger.com,1999:blog-11295132.post-10754882681320730842009-04-07T10:18:00.000-07:002009-04-07T10:18:00.000-07:00Charlie,You ask a good question. Note the represen...Charlie,<BR/><BR/>You ask a good question. Note the representative little loops I drew to represent particle-antiparticle creation-annihilation. These are just a "classical" picture of something quantum mechanical and so the metaphors are straining a bit at the edges. But roughly speaking: you can think of the universe as randomly distributed with lots of these little loops. They are distributed in such a way that the overall distribution looks identical to an observer flying by at constant velocity, no matter what that constant velocity is. It's only when acceleration occurs that this distribution starts looking different. But to go any further requires doing some real <A HREF="http://en.wikipedia.org/wiki/Quantum_field_theory_in_curved_spacetime" REL="nofollow">quantum field theory...</A>sigfpehttps://www.blogger.com/profile/08096190433222340957noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-85383405235073242802009-04-07T09:57:00.000-07:002009-04-07T09:57:00.000-07:00Charlie,Although to an external observer it looks ...Charlie,<BR/><BR/>Although to an external observer it looks like the red path is just approaching a constant velocity it turns out that in the frame of reference of something moving along the red path it feels like constant acceleration. This is not an obvious thing. You can relate it to time dilation. To someone on the accelerating path it feels like they're zipping along much faster than how it looks from the outside. There's some discussion here: http://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity)sigfpehttps://www.blogger.com/profile/08096190433222340957noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-76238435951583161402009-04-07T09:46:00.000-07:002009-04-07T09:46:00.000-07:00I don't think josefs question is silly. It's commo...I don't think josefs question is silly. It's common practice and makes more sense to see position as a function of time (f(x) = y). And how can the acceleration be constant if the velocity never get's greater than c. Youre diagram shows that the accelelration decreases infinetly approaching zero. I don't see how this would be significantly diffrent from a spacehip moving at constant velocity very close to the speed of light. So my question is: What is the relevance of acceleration to observing this unruh effect, other than acceleration is required to reach velocities close to the speed of light?Charlienoreply@blogger.comtag:blogger.com,1999:blog-11295132.post-76061302466074519532009-04-07T04:03:00.000-07:002009-04-07T04:03:00.000-07:00This is a really silly question but is there any p...This is a really silly question but is there any particular reason you put time on the vertical axis instead of the more conventional horizontal axis?Josefhttps://www.blogger.com/profile/13272830598221833253noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-46331260464592764412009-04-06T10:43:00.000-07:002009-04-06T10:43:00.000-07:00When I said "we won't see any Unruh effect" I mean...When I said "we won't see any Unruh effect" I meant to say that it might be just as hard to see in a big ring as in a small one, not that the Unruh effect disappears altogether.<BR/><BR/>For a given speed, say 90% of c, an increased radius will make the acceleration more like linear acceleration, but the magnitude of the acceleration must be smaller.<BR/><BR/>But, to be honest, I don't really know much about how particle accelerators really work so I should have kept my mouth shut :-) I think I understand the principle, but I should stop thinking I can help design a practical experiment :-)Aaron McDaidhttps://www.blogger.com/profile/17204472992759448532noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-50672366242265537982009-04-06T09:09:00.000-07:002009-04-06T09:09:00.000-07:00Aaron,Here's an example of a storage ring with som...Aaron,<BR/><BR/>Here's an example of a storage ring with some figures: http://www.lns.cornell.edu/public/lab-info/cesr.html<BR/><BR/>A big circle and high acceleration!<BR/><BR/>BTW I'm not implying that you require constant acceleration for the Unruh effect to appear. I'm saying that if you want to use the linear acceleration formula with confidence to predict the amount of Unruh effect from circular acceleration you want approximately constant acceleration for reasonably long periods of time. There's another calculation you could carry out specifically for circular motion - but it'll be more difficult. Similarly I sketched out what happens for constant acceleration for all time but you could also do the calculation for something that starts at rest, accelerates for a bit, and then remains at a constant velocity. Again, it's much harder to work out the details, but you still expect to see the effect.sigfpehttps://www.blogger.com/profile/08096190433222340957noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-51000085789411584972009-04-06T07:56:00.000-07:002009-04-06T07:56:00.000-07:00Dan, sigfpe,But with a big enough circle, the acce...Dan, sigfpe,<BR/>But with a big enough circle, the acceleration towards the centre of the accelerator is small and we won't see any Unruh effect. Unless there is to be very high angular acceleration to compensate. Don't we want to see very high acceleration in a constant direction, even if only for a short time?Aaron McDaidhttps://www.blogger.com/profile/17204472992759448532noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-3198955150925663192009-04-05T12:52:00.000-07:002009-04-05T12:52:00.000-07:00Hi, it's [the 2nd] Fergal again,Well, if you t...Hi, it's [the 2nd] Fergal again,<BR/><BR/>Well, if you take 1 um/s^2 per metre, and apply that expansion rate to all the metres between us and the centre of the earth (6000km), you get 6m/s^2, just a little less than g. On the other hand, this effect completely outweighs the acceleration due to the sun (2e30kg, 150e9m -> 6mm/s^2 acc, Vs 1e-6 * 150e9 = 150000m/s^2 total expansion); if true therefore, the earth would fly off with the rushing expansion of space, unless if G was actually a much bigger number than previously thought - i.e. we'd have to rethink gravitation.<BR/><BR/>Much easier is to assume that space just expands in the vicinity of dark energy/matter, which, fortunately, is not here.<BR/><BR/>Fergal.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-11295132.post-65975374888276152042009-04-05T07:34:00.000-07:002009-04-05T07:34:00.000-07:00Dan Doel,The papers on circular accelerators discu...Dan Doel,<BR/><BR/>The papers on circular accelerators discuss the difficulty of applying the theory to circular motion and at least one paper says it doesn't arise in circular motion.<BR/><BR/>But for a large enough circle, the motion will look like approximately linear motion for short intervals at a time. So I'd expect the Unruh effect to look similar.sigfpehttps://www.blogger.com/profile/08096190433222340957noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-14913137916722178502009-04-05T07:30:00.000-07:002009-04-05T07:30:00.000-07:00Anonymous,I believe this paper is on what you're a...Anonymous,<BR/><BR/>I believe this paper is on what you're asking about, particle creation in an expanding universe: http://adsabs.harvard.edu/abs/1968PhRvL..21..562P It's mentioned here: http://www.dartmouth.edu/~physics/news/colloquium.archives/wald_11_3_06.pdfsigfpehttps://www.blogger.com/profile/08096190433222340957noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-54934867570313860202009-04-05T01:27:00.000-07:002009-04-05T01:27:00.000-07:00I don't really see how the explanation in the firs...I don't really see how the explanation in the first part could be applied to electrons in a circular accelerator. Am I correct in thinking that the acceleration is that which is required to keep them traveling in a circular path? If so, then though the magnitude of the acceleration is constant, the direction is continually changing, so the situation seems rather different than the 1-dimensional case, or a space ship firing its thrusters at a constant rate.Dan Doelhttps://www.blogger.com/profile/16761291400347369301noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-32600222425774357862009-04-05T01:22:00.000-07:002009-04-05T01:22:00.000-07:00This is interesting. Could this effect explain th...This is interesting. Could this effect explain the cosmic background radiation? If you need 10^20 m/s^2 to get 1K, then what do you need to get 3K? I read that the expansion of the universe is acclerating (dark energy/matter etc) so if the fabric of space at the boundary of our visible universe was accelerating away so fast, we'd see (a) CMB right? I daresay an acceleration of >10^20mps2 would be quite destructive though.... let's see, 10^20mps2 / 15Gly ~ 1 um/s^2 per metre. Follow? If each metre were to stretch by 1 um/s^2, then the effect 16Gly distant at the edge of the visible universe would amount to 10^20m/s2. Hmm, something tells me 1um/s2 would be enough to pull galaxies apart...<BR/><BR/>[a different] Fergal.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-11295132.post-61997866794982899232009-04-05T00:00:00.000-07:002009-04-05T00:00:00.000-07:00Thank you for a very entertaining reading by my mo...Thank you for a very entertaining reading by my morning coffee! However, where are the monads? ;-) I expect your next post to contain haskell monads for relativity and quantum mechanics. You know, Feynman diagrams liberated from the chains of fortran and the like.Andrej Bauerhttp://andrej.com/noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-8144588601076610002009-04-04T22:42:00.000-07:002009-04-04T22:42:00.000-07:00This is from the wikipedia page: The Unruh effect,...This is from the wikipedia page: <I>The Unruh effect, described in 1976 by Bill Unruh of the University of British Columbia, is the prediction that an accelerating observer will observe black-body radiation where an inertial observer would observe none. In other words, the background appears to be warm from an accelerating reference frame.</I><BR/><BR/>So could we interpret the CMB radiation as some sort of acceleration? That's also blackbody...carlosscheideggerhttp://carlosscheidegger.wordpress.com/noreply@blogger.com