Ask an experimental physicist

35 Post author: RolfAndreassen 08 June 2012 11:43PM

In response to falenas108's "Ask an X" thread. I have a PhD in experimental particle physics; I'm currently working as a postdoc at the University of Cincinnati. Ask me anything, as the saying goes.

This is an experiment. There's nothing I like better than talking about what I do; but I usually find that even quite well-informed people don't know enough to ask questions sufficiently specific that I can answer any better than the next guy. What goes through most people's heads when they hear "particle physics" is, judging by experience, string theory. Well, I dunno nuffin' about string theory - at least not any more than the average layman who has read Brian Greene's book. (Admittedly, neither do string theorists.) I'm equally ignorant about quantum gravity, dark energy, quantum computing, and the Higgs boson - in other words, the big theory stuff that shows up in popular-science articles. For that sort of thing you want a theorist, and not just any theorist at that, but one who works specifically on that problem. On the other hand I'm reasonably well informed about production, decay, and mixing of the charm quark and charmed mesons, but who has heard of that? (Well, now you have.) I know a little about CP violation, a bit about detectors, something about reconstructing and simulating events, a fair amount about how we extract signal from background, and quite a lot about fitting distributions in multiple dimensions. 

Comments (294)

Comment author: mfb 12 July 2012 09:53:08PM 0 points [-]

Just wondering: Apart from the selection that D* should come from the primary vertex, did you do anything special to treat D* from B decays? I found page 20, but that is a bit unspecific in that respect. Some D° happen to fly nearly in the same direction as the B-meson, and I would assume that the D°/slowpi combination cannot resolve this well enough.

(I worked on charm mixing, too, and had the same issue. A reconstruction of some of these events helped to directly measure their influence.)

Comment author: Cyan 04 July 2012 01:18:13AM *  0 points [-]

Is there any redeeming value in this article by E.T. Jaynes suggesting that free electrons localize into wave packets of charge density?

The idea, near as I can tell, is that the spreading solution of the wave equation is non-physical because "zitterbewegung", high-frequency oscillations, generate a net-attractive force that holds the wave packet together. (This is Jaynes holding out the hope of resurrecting Schrödinger's charge density interpretation of the wave equation.)

Comment author: RolfAndreassen 09 July 2012 04:17:10AM 2 points [-]

I don't have time to read it right now, but I suggest that unless it accounts for how a charge density can be complex, it doesn't really help. The problem is not to come up with some physical interpretation of the wave mechanics; if that were all, the problem would have been solved in the twenties. The difficulty is to explain the complex metric.

Comment author: MrMind 22 June 2012 01:51:05PM 1 point [-]

I always wondered why there is so little study/progress on plasma Wakefield acceleration, given that there's such a need of more and more powerful accelerator to study presently unaccessible energy regions. Is that because there's a fundamental limit which cannot be used to create giant plasma based accelerator or it's just a poorly explored avenue?

Comment author: Dreaded_Anomaly 12 July 2012 10:13:03PM 0 points [-]

It's a growing field. One of my roommates is working on plasma waveguides, a related technology.

Comment author: RolfAndreassen 25 June 2012 10:15:16PM 4 points [-]

Sorry, I missed your post. As shminux says, new concepts take time to mature; the first musket was a much poorer weapon than the last crossbow. Then you have to consider that this sort of engineering problem tends intrinsically to move a bit slower than areas that can be advanced by data analysis. Tweaking your software is faster than taking a screwdriver to your prototype, and can be done even by freshly-minted grad students with no particular risk of turning a million dollars of equipment into very expensive and slightly radioactive junk. It is of course possible for an inexperienced grad student to wipe out his local copy of the data which he has filtered using his custom software, and have to redo the filtering (example is completely hypothetical and certainly nothing to do with me), thus costing himself a week of work and the experiment a week of computer-farm time. But that is tolerable. For engineering work you want experienced folk.

Comment author: TimS 26 June 2012 12:11:39AM *  0 points [-]

no particular risk of turning a million dollars of equipment into very expensive and slightly radioactive junk.

<smirk> Nice turn of phrase there.

Comment author: shminux 22 June 2012 05:16:02PM 0 points [-]

I'm not an experimental physicist, but from what I know, the whole concept is relatively new and it takes time to get it to the point where it can compete with the technologies that had been perfected over many decades. With the groups at SLAC, CERN and Max Planck Institute (among others) working on it, we should expect to see some progress within a decade or so.

Comment author: Peter_de_Blanc 17 June 2012 08:18:59AM 3 points [-]

What happens when an antineutron interacts with a proton?

Comment author: RolfAndreassen 17 June 2012 04:16:39PM 2 points [-]

Very complicated things.

Both the antineutron and the proton are soups of gluons and virtual quarks of all kinds surrounding the three valence quarks Dreaded_Anomaly mentions; all of which interact by the strong force. The result is exceedingly intractable. Almost anything that doesn't actually violate a conservation law can come out of this collision. The most common case, nonetheless, is pions - lots of pions.

This is also the most common outcome from neutron-proton and neutron-antiproton collisions; the underlying quark interactions aren't all that different.

Comment author: Dreaded_Anomaly 17 June 2012 11:17:25AM *  4 points [-]

There are various possibilities depending on the energy of the particles.

An antineutron has valence quarks , , . A proton has valence quarks u, u, d. There are two quark-antiquark pairs here: u + and d + . In the simplest case, these annihilate electromagnetically: each pair produces two photons. The leftover u + becomes a positively-charged pion.

The pi+ will most often decay to an antimuon + muon neutrino, and the antimuon will most often decay to a positron + electron neutrino + muon antineutrino. (It should be noted that muons have a relatively long lifetime, so the antimuon is likely to travel a long distance before decaying, depending on its energy. The pi+ decays much more quickly.)

There are many other paths the interaction can take, though. The quark-antiquark pairs can interact through the strong force, producing more hadrons. They can also interact through the weak force, producing other hadrons or leptons. And, of course, there are different alternative decay paths for the annihilation products that will occur in some fraction of events. As the energy of the initial particles increases, more final states become available. Energy can be converted to mass, so more energy means heavier products are allowed.

Edit: thanks to wedrifid for the reminder of LaTeX image embedding.

Comment author: wedrifid 17 June 2012 01:57:32PM *  3 points [-]

An antineutron has valence quarks u¯, d¯, d¯. (The bar should really be directly above the letter to indicate antiparticles, but Markdown does not have an overline syntax as far as I know.)

Piece of cake:

![](http://www.codecogs.com/png.latex?\\bar\{u\},%20\\bar\{d\},%20\\bar\{d\})
Comment author: kpreid 17 June 2012 11:10:11PM *  3 points [-]

Another approach is to use actual combining overlines U+0305: u̅, d̅, d̅. This requires no markup or external server support; however, these Unicode characters are not universally supported and some readers may see a letter followed by an overline or a no-symbol-available mark.

If you wish to type this and other Unicode symbols on a Mac, you may be interested in my mathematical keyboard layout.

Comment author: wedrifid 17 June 2012 09:19:48AM *  0 points [-]

What happens when an antineutron interacts with a proton?

Good question.

I'm going to tender the guess that you get a kaboom (energy release equivalent to the mass of two protons) and a left over positron and neutrino spat out kind of fast.

Comment author: DanielLC 17 June 2012 04:18:57AM *  0 points [-]

I'm confused about part of quantum encryption.

Alice sends a photon to Bob. If Eve tries to measure the polarization, and measures it on the wrong axis, there's a chance Bob won't get the result Alice sent. From what I understand, if Eve copies the photon, using a laser or some other method of getting entangled photons, and she measures the copied photon, the same result will happen to Bob. What happens if Eve copies the photon, and waits until Bob reads it before she does?

Also, you referred to virtual particles as a convenient fiction when responding to someone else. I assumed that they were akin to a particle being in a place with more potential energy than there is energy in a system during quantum tunneling. The particle is real. It's just that due to the fact that the kinetic energy is negative, it behaves in a way that makes the waveform small at any real distance. Was I completely off base?

Also, should I have just edited my old post instead of adding a new one?

Comment author: RolfAndreassen 17 June 2012 04:38:24PM 0 points [-]

What happens if Eve copies the photon, and waits until Bob reads it before she does?

Not my field, but it seems to me that it should be the same thing that happens if Bob tries to read the photon after Eve has already done so. You can only read the quantum information off once. Now, an interesting question is, what happens if Eve goes off into space at near lightspeed, and reads the photon at a time such that the information "Bob has read the photon" hasn't had time to get to her spaceship? If I understand correctly, it doesn't matter! This scenario is just a variant of the Bell's-inequality experiment.

Also, you referred to virtual particles as a convenient fiction when responding to someone else. I assumed that they were akin to a particle being in a place with more potential energy than there is energy in a system during quantum tunneling. The particle is real. It's just that due to the fact that the kinetic energy is negative, it behaves in a way that makes the waveform small at any real distance. Was I completely off base?

So firstly, in quantum tunneling the particle never occupies the forbidden area. It goes from one allowed area to another without occupying the space between; hence the phrase "quantum leap". Of course this is not so difficult to imagine when you think of a probability cloud rather than a particle; if you think of a system with parts ABC, where B is forbidden but A and C are allowed, then there is at any time a zero probability of finding the particle in B, but a nonzero probability to find it in A and C. This is true even if at some earlier time you find it in A, because, so to speak, the wave function can go where the particle can't. So, yes, if you ever found the particle in B its kinetic energy would be negative, but in fact that doesn't happen. So now we come to matters of taste: The wave function does exist within B; is this a mathematical fiction, because no experiment will find the particle there, or is it real since it explains how you can find the particle at C?

Then, back to virtual particles. The mass of a virtual particle can be negative; it is really unclear to me what it would even mean to observe such a thing. Therefore I think of them as a convenient fiction. But they are certainly a very helpful fiction, so, you know, take your choice.

Also, should I have just edited my old post instead of adding a new one?

I don't think so, the number of comments here is so large that it would be very easy to miss an edit.

Comment author: DanielLC 17 June 2012 05:23:21PM 0 points [-]

You can only read the quantum information off once.

Bob knows the right way to polarize it, though. If Eve tries to read it but polarizes it wrong, it would mess with the polarization of Bob's particle, so there's a chance he'd notice. If Bob polarizes it the way Alice did, and then Eve polarizes it wrong when she reads it, will Bob notice? If Bob notices, he just predicted the future. If he does not, then he can tell whether or not when Eve reads it constitutes "future", violating relativity of simultaneity.

So firstly, in quantum tunneling the particle never occupies the forbidden area.

If you solve Schroedinger's time-independent equation for a finite well, there is non-zero amplitude outside the well. If you calculate kinetic energy on that part of the waveform, it will come out negative. You obviously wouldn't be able to observe it outside the well, in the sense of getting it to decohere to a state where it's mostly outside the well, without giving it enough energy to be in that state. That's just a statement about how the system evolves when you put a sensor in it. If you trust the Born probabilites and calculate the probability of being in a configuration space with a particle mid-quantum tunnel, it will come out finite.

... it is really unclear to me what it would even mean to observe such a thing.

I don't really care about observation. It's just a special case of how the system evolves when there's a sensor in it. I want to know how virtual particles act on their own. Do they evolve in a way fundamentally different from particles with positive kinetic energy, or are they just what you get when you set up a waveform to have negative energy, and watch it evolve?

Comment author: RolfAndreassen 19 June 2012 04:03:09AM 0 points [-]

Bob knows the right way to polarize it, though. If Eve tries to read it but polarizes it wrong, it would mess with the polarization of Bob's particle, so there's a chance he'd notice. If Bob polarizes it the way Alice did, and then Eve polarizes it wrong when she reads it, will Bob notice? If Bob notices, he just predicted the future. If he does not, then he can tell whether or not when Eve reads it constitutes "future", violating relativity of simultaneity.

Good point. My initial answer wasn't fully thought through; I again have to note that this isn't really my area of expertise. There is apparently something called the no-cloning theorem, which states that there is no way to copy arbitrary quantum states with perfect fidelity and without changing the state you want to copy. So the answer appears to be that Eve can't make a copy for later reading without alerting Bob that his message is compromised. However, it seems to be possible to copy imperfectly without changing the original; so Eve can get a corrupted copy.

There is presumably some tradeoff between the corruption of your copy, and the disturbance in the original message. You want to keep the latter below the expected noise level, so for a given noise level there is some upper limit on the fidelity of your copying. To understand whether this is actually a viable way of acquiring keys, you'd have to run the actual numbers. For example, if you can get 1024-bit keys with one expected error, you're golden: Just try the key with each bit flipped and each combination of two bits flipped, and see if you get a legible message. This is about a million tries, trivial. (Even so, Alice can make things arbitrarily difficult by increasing the size of the key.) If we expected corruption in half the bits, that's something else again.

I don't know what the limits on copying fidelity actually are, so I can't tell you which scenario is more realistic.

As I say, this is a bit out of my expertise; please consider that we are discussing this as equals rather than me having the higher status. :)

If you solve Schroedinger's time-independent equation for a finite well, there is non-zero amplitude outside the well. If you calculate kinetic energy on that part of the waveform, it will come out negative. You obviously wouldn't be able to observe it outside the well, in the sense of getting it to decohere to a state where it's mostly outside the well, without giving it enough energy to be in that state. That's just a statement about how the system evolves when you put a sensor in it. If you trust the Born probabilites and calculate the probability of being in a configuration space with a particle mid-quantum tunnel, it will come out finite.

You are correct. It seems to me, however, that you would not actually observe a negative energy; you would instead be seeing the Heisenberg relation between energy and time, \Delta E \Delta t >= hbar/2; in other words, the particle energy has a fundamental uncertainty in it and this allows it to occupy the classically forbidden region for short periods of time.

I don't really care about observation. It's just a special case of how the system evolves when there's a sensor in it. I want to know how virtual particles act on their own. Do they evolve in a way fundamentally different from particles with positive kinetic energy, or are they just what you get when you set up a waveform to have negative energy, and watch it evolve?

Your original question was whether virtual particles are real; perhaps I should ask you, first, to define the term. :) However, they are at least as real as the different paths taken by the electron in the two-slit experiment; if you set things up so that particular virtual-particle energies are impossible, the observed probabilities change, jsut like blocking one of the slits.

Well, as they can have negative mass you have to assume that their gravitational interactions are, to coin a phrase, counterintuitive. (That is, even for quantum physicists! :) ) But, of course, we don't have any sort of theory for that. As far as interactions that we actually know something about go, they are the same, modulo the different mass in the propagator. (That is, the squiggly line in the Feynman diagram, which has its own term in the actual path integral; you have to integrate over the masses.)

Comment author: [deleted] 16 June 2012 04:28:37AM 1 point [-]

What is your opinion of the Deutsch-Wallace claimed solution to the probability problems in MWI?

Also are you satisfied with decoherence as means to get preferred basis?

Lastly: do you see any problems with extending MWI to QFT (relativity issues) ?

Comment author: RolfAndreassen 17 June 2012 04:52:50PM 0 points [-]

What is your opinion of the Deutsch-Wallace claimed solution to the probability problems in MWI?

Now we're getting into the philosophy of QM, which is not my strength. However, I have to say that their solution doesn't appeal to that part of me that judges theories elegant or not. Decision theory is a very high-level phenomenon; to try to reason from that back to the near-fundamental level of quantum mechanics - well, it just doesn't feel right. I think the connection ought to be the other way. Of course this is a very subjective sort of argument; take it for what it's worth.

Also are you satisfied with decoherence as means to get preferred basis?

I'm not really familiar enough with this argument to comment; sorry!

Lastly: do you see any problems with extending MWI to QFT (relativity issues) ?

Nu, QM and QFT alike are not yet reconciled with general relativity; but as for special relativity, QFT is generally constructed to incorporate it from the ground up, unlike QM which starts with the nonrelativistic Schrodinger equation and only introduces Dirac at a later stage. So if there's a relativity problem it applies equally to QM. Apart from that, it's all operators in the end; QFT just generalises to the case where the number of particles is not conserved.

Comment author: epigeios 12 June 2012 03:28:02AM *  0 points [-]

I've got a lot of questions I just thought of today. I am personally hoping to think of a possible alternative model of quantum physics that doesn't need anything more than the generation 1 fermions and photons, and doesn't need the strong interaction.

  • What is the reason for the existence of the theory of the charm quark (or any generation 2-3 quark)? What are some results of experiments that necessitate the existence of a charm quark?
  • Which of the known hadrons can be directly observed in any way, as opposed to theorized as a mathematical in-between or as a trigger for some directly observable decay?
  • Am I right in thinking that the tau lepton is only theorized in order to explain an in-between decay state? If you don't know, do you know of anything related to any other fermions (or hadrons) that only exist as a theoretical in-between?
  • How were the masses of the tau lepton and the top quark determined? If the methods are different for the charm quark, how was the mass of the charm quark determined?
  • Does the weak interaction cause any sort of movement, or hold anything together, or does it only act as a trigger for decay? Why is it considered a field energy?
  • When detecting gamma radiation, how much background is there to extract from? Does the process of extracting from the background require performing hundreds of iterations of the experiment?
  • Since you know quite a lot about it, and since the majority of my knowledge comes from Wikipedia, what does "fitting distributions in multiple dimensions" mean? What is the possibility of error of this process?
  • Oh, and lastly, do you know of any chart or list anywhere that details the known possible decay paths of bosons and fermions?

That's all for now. I SO hope you can answer any of these questions; because Wikipedia can't :'( (as someone who enjoys theory, I find it annoying when Wikipedia can neither confirm nor deny my conjectures, despite the fact that the information is certainly out there somewhere, and someone knows it.)

Comment author: RolfAndreassen 12 June 2012 06:06:35AM *  4 points [-]

I had to split my answer in two, and clumsily posted them in the wrong order - some of this refers to an 'above' which is actually below. I suggest reading in chronological rather than page order. :)

Am I right in thinking that the tau lepton is only theorized in order to explain an in-between decay state?

Well no, you get a specific resonance in hadron energy spectra, as described above.

If you don't know, do you know of anything related to any other fermions (or hadrons) that only exist as a theoretical in-between?

There's the notorious sigma and kappa resonances, which are basically there only to explain a structure in the pion-pion and pion-kaon scattering spectrum. Belief in these as particles proper, rather than some feature of the dynamics, is not widespread outside the groups that first saw them. (I have a photoshopped WWII poster somewhere, captioned "Is YOUR resonance needed? Unnecessary particles clutter up the Standard Model!) I see the PDG doesn't even list them in its "needs confirmation" section. I'm aware of them basically because I used them in my thesis just as a way to vary the model and see how the result varied - I had all the machinery for setting up particles, so a more-or-less fictional particle with some motivation from what others have seen was a convenient way of varying the structure.

How were the masses of the tau lepton and the top quark determined? If the methods are different for the charm quark, how was the mass of the charm quark determined?

So quark masses are a vexed subject. The problem is that you cannot catch a quark on its own, it's always swimming in a virtual soup of gluons and quarks. So all quark masses are determined, basically, by taking some model of the strong interaction and trying to back-calculate the observed hadron and meson masses. And since the strong interaction is insanely computationally intractable, you can't get a very good answer.

For the tau lepton it's rather simpler: Wait for one to decay to charged hadrons, calculate the four-momentum of the mother particle, and get the peak of the mass distribution as described above.

Does the weak interaction cause any sort of movement, or hold anything together, or does it only act as a trigger for decay?

I don't believe anyone has observed a bound state mediated purely by the weak force. In fact one of the particles in such a state would have to be a neutrino, since otherwise there would be other forces involved; and observing a neutrino is hard enough without adding the requirement that it be a bound state. However, I suppose that in inverse-beta-decay, or neutrino capture, the weak force causes some movement at the final movement, to the extent that it's meaningful to speak of movement at these scales.

Why is it considered a field energy?

Because it can be quantised into carrier bosons, presumably.

When detecting gamma radiation, how much background is there to extract from?

This is really hard to give a general answer for. In the BaBar detector, photons are reconstructed by the EMC, the electromagnetic calorimeter. My rule of thumb for this instrument is that photons with energy less than 30 MeV are worthless; such energies can easily be faked by the electronic noise and ambient radiation. Above 100 MeV you have to be fairly unlucky for an EMC hit to be background. I don't know if this is helpful; perhaps you can give me a better idea of the context of your question?

Does the process of extracting from the background require performing hundreds of iterations of the experiment?

Again, this is really dependent on context. Can you be more specific about what sort of experiment you're asking about?

Since you know quite a lot about it, and since the majority of my knowledge comes from Wikipedia, what does "fitting distributions in multiple dimensions" mean? What is the possibility of error of this process?

Have a look at my answer to magfrump. As for errors, our search algorithm does rely on the log-probability function being reasonably smooth, and can give misleading answers if that's not true. It can get caught in local minima; we try to avoid this by starting from several different points and checking that we converge to the same place. In some cases the assumption of symmetric errors can mislead you, so we often look at asymmetric errors as well. Most insidiously, of course, you can get the physics just wrong, but right enough to mimic the data within the limits of the fit's accuracy.

Oh, and lastly, do you know of any chart or list anywhere that details the known possible decay paths of bosons and fermions?

You could try the PDG's summary tables.

Comment author: RolfAndreassen 12 June 2012 06:06:23AM 5 points [-]

Ok, that's a lot of questions. I'll do my best, but I have to tell you that your quest is, in my opinion, a bit quixotic.

What is the reason for the existence of the theory of the charm quark (or any generation 2-3 quark)? What are some results of experiments that necessitate the existence of a charm quark?

Basically the strange quark is motivated by the existence of kaons, charm quarks by the D family of mesons (well, historically the J/psi, but I'm more familiar with the D mesons), and beauty quarks by the B family. As for truth quarks, mainly considerations of symmetry. Let's take kaons, the argument being the same for the other families. If the kaon were to decay by the strong force, it would be extremely short-lived, because it could go pretty immediately to two pions; there would certainly be no question of seeing it in a tracking detector, the typical timescale of strong decays being 10^-23 seconds. Even at lightspeed you don't get far in that time! We therefore conclude that there is some conservation principle preventing the strong decay, and that the force by which the kaon decays does not respect this conservation principle. Hence we postulate a strange quark, whose flavour (strangeness) is conserved by the strong force (so, no strange-to-up (or down) transition at strong-force speeds) but not by the weak force.

I should note that quark theory has successfully predicted the existence of particles before they were observed; you might Google "Eightfold Path" if you're not familiar with this history, or have a look at the PDG's review. (Actually, on closer inspection I see that the review is intending for working physicists familiar with the history - it's not an introduction to the Eightfold Path, per se. Probably Google would serve you better.)

Which of the known hadrons can be directly observed in any way, as opposed to theorized as a mathematical in-between or as a trigger for some directly observable decay?

For this I have to digress into cross-sections. Suppose you are colliding an electron and a positron beam, and you set up a detector at some particular angle to the beam - for example, you can imagine the detector looking straight down at the collision point:

_detector

e+ -----> collision <------- e-

Now, the cross-section (which obviously is a function of the angle) can be thought of as the probability that you'll see something in the detector. If electron and positron just glance off each other without annihilating (at relativistic speeds this can easily happen - they have to get pretty close to interact, and our control of the beams is only so good), we call that Bhabha scattering, and it has a particular cross-section structure. For obvious reasons, the cross-section is highest at small angles; that is, it is really quite unlikely for the electron and positron to dance past each other in the exact way that throws them out at a ninety-degree angle to their previous paths; but it's pretty easy for them to give each other a one-degree kick. If you calculate the cross-section at some particular angle as a function of the total beam energy, you'll see that the higher the energy, the lower the cross-section, and indeed experiment confirms this.

What if the electron and positron do annihilate, creating a virtual photon that then decays to some other pair of particles - for example, a charm-anticharm pair? Well, again, the cross-section is highest near the beam (basically to conserve the angular momentum - you have to do spin math) and decreases with energy.

So we have this cross-section that decreases monotonically with energy. However, as you run your beam energy up, at very specific energies you will see a sharp increase and drop-off, in a classic Breit-Wigner shape. In other words, at some particular energy it suddenly becomes much more likely that your decay products get kicked away from the beam. Why is that? We refer to these bumps in the spectrum as resonances, and explain them by appealing to bound states - particles, in other words. What happens is that with an intermediate bound state, there are additional Feynman paths that open up between the initial state "electron and positron" and the final state "hit in detector at angle X". Additional paths through parameter space gives you additional probability unless you're very unlucky with the phases, hence the bump in the cross-section - the final state becomes more likely. (Additionally, for reasons of spin math that I won't go into here, the decay products from a bound state of two quarks are produced much more isotropically than back-to-back quark-antiquark pairs.)

Here's a different way of looking at it. Suppose you have a detector that encloses the collision space, so you can reconstruct most of the decay products; and you decide to take all pion pairs and calculate "If these two particles came from a common decay, what was the mass of the particle that decayed?" Then this spectrum will basically be flat, but you will get an occasional peak at specific masses. Again, we explain this by appeal to a bound state.

It occurs to me that this may not actually differ from what you call "mathematical in-betweens"; I have answered as though this phrase refers to virtual particles, which are indeed a bit of a convenient fiction. Anyway, this is why we believe in the various hadrons and mesons.

(I'm getting "comment too long" errors; splitting my answer here.)

Comment author: DanielVarga 11 June 2012 09:56:15PM 1 point [-]

Can photon-photon scattering be harnessed to build a computer that consists of nothing but photons as constituent parts? I am only interested in theoretical possibility, not feasibility. If the question is too terse in this form, I am happy to elaborate. In fact, I have a short writeup that tries to make the question a bit more precise, and gives some motivation behind it.

Comment author: RolfAndreassen 12 June 2012 04:54:16AM *  1 point [-]

Well, it depends on what you mean by "nothing but". You can obviously (in principle) make a logic gate of photon beams, but I don't see how you can make a stable apparatus of nothing but photons. You have to generate the light somehow.

NB: Sometimes the qualifier "in principle" is stronger than other times. This one is, I feel, quite strong.

Comment author: DanielVarga 12 June 2012 12:21:29PM 0 points [-]

What I mean by "in principle" is not that different from what Fredkin and Toffoli mean by it when talking about their billiard ball computer. The intuition is that when you figured out that some physical system can be harnessed for computation in principle, then you can start working on noise tolerance and energy consumption, and usually it turns out that those are not the show-stopper parts. And when I eventually try to link "in principle" to "in practice", I am still not talking about the scale of human engineering. You say you need to generate light for the system, and a strong gravitational field to trap the photons? I say, fine, I'll rearrange these galaxies into laser guns and gravitational photon traps for you.

Comment author: RolfAndreassen 12 June 2012 04:50:50PM 1 point [-]

Fair enough. I'm just saying, the galaxies aren't made purely of light, so you still don't have a computer of "nothing but" photons. But sure, the logic elements could be purely photonic.

Comment author: shminux 11 June 2012 10:12:58PM 0 points [-]

It's an intriguing idea, a pure photon-based gate based on elastic scattering of photons, however I don't see how such a system would function, even in principle. Feel free to elaborate. Also, presumably constructing an equivalent electron- or neutron-based gate would be easier.

Comment author: DanielVarga 12 June 2012 12:01:50AM 0 points [-]

It's an intriguing idea, a pure photon-based gate based on elastic scattering of photons, however I don't see how such a system would function, even in principle.

I have no idea either. All that I have is a flawed analogy: We could in principle build a computer consisting of nothing but billiard balls as constituent parts. This would work even if meeting billiard balls, instead of bouncing off each other, just changed their trajectories slightly, with a very small probability. I'd like to know whether this crude view of photon-photon scattering is A. a simplification that helps focus on the interesting part of the question, or B. a terrible misunderstanding.

Now I'll tell the original motivation behind the question. As an old LW regular, you have probably seen some phrase like "turn our future light cone into computronium" tossed out during some FAI discussion. What I am interested in is how to actually do that optimally, if you are limited by nothing but the laws of physics. In particular, I am interested in whether the optimal solution involves light-speed (or asymptotically light-speed) expansion, or (for entropy or other considerations) does not actually end up eating the whole light cone.

Obviously this is not my home turf, so maybe it is not even true that the scattering question is relevant at all when we try to answer the computronium question. I would appreciate any insights about either of them or their relationship.

Comment author: pengvado 12 June 2012 02:49:40PM 2 points [-]

I am interested in whether the optimal solution involves light-speed (or asymptotically light-speed) expansion, or (for entropy or other considerations) does not actually end up eating the whole light cone.

The form of the expansion has very little to do with the form of the computronium.

Launch von Neumann probes at c-ε. They can be tiny, so the energy cost to accelerate them is negligible compared to the energy you can harvest from a new star system. When one arrives, it builds a few more probes and launches them at further stars, then turns all the local matter into computers. The computers themselves don't need to move quickly, since the probes do all the long-distance colonization.

Comment author: DanielVarga 12 June 2012 03:13:31PM 0 points [-]

You are right. Originally I became interested in purely photon-based computation because I had an even more speculative idea that seemed to require it. If you have a system that terraforms everything in its path and expands with exactly the speed of light, then you are basically unavailable to outside observation. You can probably see where this line of thought leads. I am aware of the obvious counterargument, but as I explained there, it is a bit weaker than it first appears.

Comment author: Dreaded_Anomaly 12 June 2012 09:34:33AM 2 points [-]

The difference I see between photons and your example with billiard balls is that billiard balls have a rest frame. In other words, you can set them up so that they have no preexisting motion relative to you, and any change in their positions is due to your inputs. You can't do this with photons in a vacuum; they are massless, and must always move at c.

Photon-photon scattering is also a rare process in quantum electrodynamics. If you look at the Feynman diagram:

Photon-photon scattering

It has four vertices. Each vertex gives the cross-section of the process another factor of the fine structure constant α, which is a small number, about 1/137. A process like electron-electron or electron-positron scattering, on the other hand, has diagrams with only two vertices, so only two factors of α. (Of course, cross-sections also depend on mass, momentum, and so forth, but this gives a very simple heuristic for comparing processes.) The additional factor of α² ~ 0.00005 makes the cross section tiny compared to common QED processes.

If you want to use photons for computing, photonic crystals are your best bet, although the technology is still in early stages of development.

Comment author: DanielVarga 12 June 2012 12:21:05PM 1 point [-]

I don't know much about photon-photon scattering, but I do know that the cross section is very small. I see this as something that does not make a difference from a strictly theoretical point of view, but that might be because I don't understand the issues. Photonic crystals are not really relevant for my thought experiments, because you definitely can't build computers out of them that expand with the asymptotic speed of light. Maybe if you can turn regular material into photonic crystal by bombarding it with photons.

Comment author: Dreaded_Anomaly 12 June 2012 01:18:52PM 1 point [-]

If two billiard balls come to occupy an overlapping volume in space at the same time, they will collide with probability (1 - ε) for ε about as small as we can imagine. However, photons will only scatter off each other rarely. Photons are bosons, so the vast majority of the time, they will just pass right through each other. That doesn't give you a dependable logic gate.

Comment author: DanielVarga 12 June 2012 01:48:08PM 1 point [-]

Maybe you are right, but it is not immediately obvious to me that small cross-section is a deadly problem. You shouldn't look at one isolated photon-photon encounter as a logic gate. Even an ordinary electronic transistor would not work without error correction. Using error correction, you can build complex systems that seem like magic when you attempt to understand them at the level of individual electrons.

Comment author: shminux 12 June 2012 04:05:01AM 2 points [-]

We could in principle build a computer consisting of nothing but billiard balls as constituent parts.

I am quite sure that would be impossible without the balls being constrained by some other forces, such as gravity or outside walls.

Comment author: DanielVarga 12 June 2012 12:21:12PM 0 points [-]

You can build outside walls out of billiard balls. Eventually, such a system will disintegrate, but this is no different from any other type of computer. The important thing is that for any given computation length you can build such a system. The size of the system will grow with required computation length, but only polynomially.

Comment author: shminux 12 June 2012 03:55:35PM *  1 point [-]

I would be interested in seeing a metastable gate constructed solely out of billiard balls. Care to come up with a design?

Comment author: DanielVarga 12 June 2012 05:58:22PM 0 points [-]

Ah, now I see your point. I had this misconception that if you send a billiard ball into a huge brick-wall of billiard balls, it will bounce back. Okay, I don't have a design.

Comment author: shminux 12 June 2012 09:51:53PM 0 points [-]

if you send a billiard ball into a huge brick-wall of billiard balls, it will bounce back.

It sure will, after imparting some momentum to the wall. My point is that I do not know how to construct a gate out of components interacting only through repulsive forces. I am not saying that it is impossible, I just do not see how it can be done.

Comment author: DanielVarga 12 June 2012 10:59:54PM 0 points [-]

It sure will

How much momentum will it lose before it bounces back? If a large enough wall can make this arbitrarily small, then I think the Fredkin and Toffoli billiard gates can be built out of a thick wall of billiard balls. Lucky thing, in this model there is no friction, so gates can be arbitrarily large. Sure, the system might start to misbehave after the walls move by epsilon, but this doesn't seem like a serious problem. In the worst case, we can use throw-away gates that are abandoned after one use. That model is still as strong as Boolean circuits.

Comment author: Stuart_Armstrong 10 June 2012 09:20:17AM 3 points [-]

May be slightly out of your area, but: do you believe the entropy-as-ignorance model is the correct way of understanding entropy?

Comment author: RolfAndreassen 10 June 2012 04:14:04PM 4 points [-]

Well no, it seems to me that there is a real physical process apart from our understanding of it. It's true that if you had enough information about a random piece of near-vacuum you could extract energy from it, but where does that information come from? You sort of have to inject it into the problem by a wave of the hand. So, to put it differently, if entropy is ignorance, then the laws of thermodynamics should be reformulated as "Ignorance in a closed system always increases". It doesn't really help, if you see what I mean.

Comment author: DanielLC 17 June 2012 04:09:39AM 0 points [-]

What I've heard seemed to indicate that, if you assigned a certain entropy density function to classical configuration space, and integrated it over a certain area to get entropy at the initial time, then let the area evolve, and integrated over that area to get the entropy at the final time, the entropy would stay constant.

This would mean that conservation of entropy is the actual physical process. Increase in entropy is just us increasing the size at the final time because we're not paying close enough attention to exactly where it should be.

Also, the more you know about the system, the smaller the area you could give in configuration space to specify it, and thus the lower the entropy.

Is this accurate at all?

Comment author: Manfred 10 June 2012 07:39:19PM 0 points [-]

It's not really any more "unhelpful" than the statement that the number of bits of information needed to pick out a specific state of a system always increases. And that one's just straight Shannon entropy.

Comment author: RolfAndreassen 10 June 2012 07:41:53PM 4 points [-]

Sure; the point is that we have lots of equivalent formulations of entropy and I don't see the need to pick out one of them as the correct way of understanding it. One or another may be more intuitively appealing to particular students, or better suited to particular problems, but they're all maps and not territories.

Comment author: Manfred 10 June 2012 07:52:24PM *  1 point [-]

Given a quantum state, you can always tell me the entropy of that specific quantum state. It's 0. If that's the territory, then where is entropy in the territory?

Comment author: wnoise 17 June 2012 07:03:05AM 0 points [-]

Given a quantum state, you can always tell me the entropy of that specific quantum state. It's 0.

Only for pure states. Any system you have will be mixed.

Comment author: Manfred 17 June 2012 09:19:47AM 0 points [-]

I believe you mean "you will have incomplete information about any system you could really have."

Comment author: wnoise 17 June 2012 09:24:29AM 0 points [-]

Operationally, it's a distinction without a difference.

Comment author: Manfred 17 June 2012 09:34:50AM 0 points [-]

Since the way this whole nest of comments got started was whether it makes sense to identify entropy with incomplete information, I'd say my reply to you was made with loaded language :P

Comment author: RolfAndreassen 10 June 2012 09:20:02PM 1 point [-]

How much work can I extract from a system in that state? It's often useful to keep the theoretical eyes on the thermodynamical ball.

Comment author: Manfred 10 June 2012 11:11:37PM *  1 point [-]

Helmholtz free energy (A, or F, or sometimes H) = E - TS in the thermodynamic limit, right? So A = E in the case of a known quantum state.

Comment author: wnoise 17 June 2012 07:24:03AM 0 points [-]
  1. Wouldn't Gibbs free energy be more appropriate? pV should be available for work too.

  2. I find myself slightly confused by that definition. Energy in straight quantum mechanics (or classical Newtonian mechanics) is a torsor. There is no preferred origin, and adding any constant to all the states changes the evolution not at all. It therefore must not change the extractable work. So the free energies are clearly incorrectly defined, and must instead be defined relative to the ground state. In which case, yes, you could extract all the energy above that, if you knew the precise state, and could manipulate the system finely enough.

Comment author: Manfred 17 June 2012 09:30:00AM *  0 points [-]

1) Meh.

2) Right. I clarified this two posts down: "the free energy change between two states is the work you can extract by moving between those two states." So just like for energy, the zero point of free energy can be shifted around with no (classical) consequences, and what really matters (like what comes out of engines and stuff) is the relative free energy.

Comment author: RolfAndreassen 11 June 2012 01:06:54AM 1 point [-]

So statistical mechanics was my weakest subject, and we're well beyond my expertise. But if you're really saying that we cannot extract any work from a system if we know its quantum state, that is highly counterintuitive to me, and suggests a missed assumption somewhere.

Comment author: Manfred 11 June 2012 02:06:06AM 2 points [-]

Helmholtz free energy (A) is basically the work you can extract (or more precisely, the free energy change between two states is the work you can extract by moving between those two states). So if A = E, where E is the energy that satisfies the Schroedinger equation, that means you can extract all the energy.

Sort of like Maxwell's demon.

Comment author: RolfAndreassen 11 June 2012 02:10:32AM 1 point [-]

Excuse me, the thought somehow rotated 180 degrees between brain and fingers. My point from a couple of exchanges up remains: How did you come to know this quantum state? If you magically inject information into the problem you can do anything you like.

Comment author: [deleted] 10 June 2012 08:21:32PM 2 points [-]

There's something subtle about what's map and what's territory in density matrices. I'd like to think to the territory as a pure quantum state and to maps as mixed states, but... If John thinks the electron in the centre of this room is either spin-up or spin-down but he has no idea which (i.e. he assign probability 50% to each), and John thinks the electron in the centre of this room is either spin-east or spin-west but he has no idea which, then for any possible experiment whatsoever, the two of them would assign the same probability distribution to the outcome. There's something that puzzles me about this, but I'm not sure what that is.

Comment author: [deleted] 10 June 2012 04:59:12AM 4 points [-]

What will happen if we don't find super-symmetry at the LHC? What will happen if we DO find it?

Comment author: RolfAndreassen 10 June 2012 04:16:36PM 7 points [-]

Well, if we do find it there are presumably Nobel prizes to be handed out to whoever developed the correct variant. If we don't, I most earnestly hope we find something else, so someone else gets to go to Stockholm. In either case I expect the grant money will keep flowing; there are always precision measurements to be made. Or were you asking about practical applications? I can't say I see any, but then they always do seem to come as a surprise.

Comment author: [deleted] 13 June 2012 10:34:32AM 3 points [-]

In either case I expect the grant money will keep flowing; there are always precision measurements to be made.

I somehow fear that if LHC finds the Higgs boson but no beyond-the-Standard-Model physics it'll become absurdly hard to get decent funding for anything in particle physics.

Comment author: RolfAndreassen 13 June 2012 06:00:41PM 3 points [-]

For large-scale projects like the LHC that may be true, but that's not the only way to do particle physics. You can accomplish a lot with low energies, high luminosities, and a few hundred million dollars - pocket change, really, on the scale of modern governments.

That said, it is quite possible that redirecting funding for particle physics into other kinds of science is the best investment at this point even taking pure knowledge as valuable for its own sake. There's such a thing as an opportunity cost and a discount rate; the physics will still be out there in 50 years when a super-LHC can be built for a much smaller fraction of the world's economic resources. If you have no good reason to believe that there's an extinction-risk-reducing or Good-Singularity-Causing breakthrough somewhere in particle physics, you shouldn't allow sentiment for the poor researchers who will, sob, have to take filthy jobs in some inferior field like, I don't know, astronomy, or perhaps even have to go into industry (shudder), to override your sense of where the low-hanging fruits are.

Comment author: [deleted] 13 June 2012 06:30:11PM 0 points [-]

you shouldn't allow sentiment for the poor researchers

The problem is that I've been planning to be such a researcher myself! (I'm in the final year of my MSc and probably I'm going to apply for a PhD afterwards. I'm specializing in cosmic rays rather than accelerators, though.)

Comment author: RolfAndreassen 13 June 2012 09:33:45PM 5 points [-]

Well, I am such a researcher, and so what I say to you applies just as much to myself: Sucks to be you. The privilege of working on what interests us in a low-pressure academic environment is not a god-given right; it depends on convincing those who pay for it - ultimately, the whole of the public - that we are a good investment. In the end we cannot make any honest argument for that except "Do you want to know how the universe ticks, or not?" Well, maybe they don't. Or maybe their understanding-the-universe dollars could, right now, be spent in better places. If so, sucks to be us. We'll have to go earn six-figure wages selling algebra to financiers. Woe, woe, woe is us.

Comment author: [deleted] 09 June 2012 10:44:11PM 5 points [-]

I have three pretty significant questions: Are you a strong rationalist (good with the formalisms of Occams Razor)? Are you at all familiar with String Theory (in the sense of Doing the basic equations)? If yes to both, what is your bayes goggles view on String Theory?

What on earth is the String Theory controversy about, and is it resolvable at a glance like QM's MWI?

Comment author: Mitchell_Porter 10 June 2012 11:03:06AM *  15 points [-]

There isn't a unified "string theory controversy".

The battle-tested part of fundamental physics consists of one big intricate quantum field theory (the standard model, with all the quarks, leptons etc) and one non-quantum theory of gravity (general relativity). To go deeper, one wishes to explain the properties of the standard model (why those particles and those forces, why various "accidental symmetries" etc), and also to find a quantum theory of gravity. String theory is supposed to do both of these, but it also gets attacked on both fronts.

Rather than producing a unique prediction for the geometry of the extra dimensions, leading to unique and thus sharply falsifiable predictions for the particles and forces, present-day string theory can be defined on an enormous, possibly infinite number of backgrounds. And even with this enormous range of vacua to choose from, it's still considered an achievement just to find something with a qualitative resemblance to the standard model. Computing e.g. the exact mass of the "electron" in one of these stringy standard models is still out of reach.

Here is a random example of a relatively recent work of string phenomenology, to give you an idea of what is considered progress. The abstract starts by saying that certain vacua are known which give rise to "the exact MSSM spectrum". The MSSM is the standard model plus minimal supersymmetry. Then they point out that these vacua will also have to have an extra electromagnetism-like force ("gauged U(1)_B-L"). We don't see such a force, so therefore the "B-L" photons must be heavy, and the gist of the paper is to point out that this can be achieved if one of the neutrino superpartners acts like a Higgs field (by "acquiring a vacuum expectation value"). In fact this paper doesn't contain string calculations per se; it's an argument at the level of quantum field theory, that the field-theory limit of these string models is potentially consistent with experiment.

That might not sound exciting, but in fact it's characteristic, not just of string phenomenology, but of theoretical particle physics in general. Progress is incremental. Grand unified theories don't explain the masses of the particles, but they can explain the charges. String theory hasn't yet explained the masses, but it has the potential to do so, in that they will be set by the stabilized size and shape of the extra dimensions. The topology of the extra dimensions is (currently) a model-building choice, but once that choice is made, the masses should follow, they're not free parameters as in field theory.

As for what might determine the topology of the extra dimensions, anthropic selection is a popular answer these days - and that has become another source of dissatisfaction for string theory's critics, because it looks like another step back from predictivity. Except in very special cases like the cosmological constant, where a large value makes any kind of physical structure impossible, there's enormous scope for handwaving explanations here... Actually, there are arguments that the different vacua of the "landscape" should be connected by quantum tunneling, so the vacuum we are in may be a long-lived metastable vacuum arrived at after many transitions in the primordial universe. But even if that's true, it doesn't tell you whether the number of metastable minima in the landscape is one or a googol. This is an aspect of string theory which is even harder than calculating the particle masses in a particular vacuum, judging by the amount of attention it gets. The empirical side of string theory is still dominated by incrementally refining the level of qualitative approximation to the standard model (including the standard cosmological model, "lambda CDM") that is possible.

As for quantum gravity, the situation is somewhat different. String theory offers a particular solution to the problems of quantum gravity, like accounting for black hole entropy, preserving unitarity during Hawking evaporation, and making graviton behavior calculable. I'd say it is technically far ahead of any rival quantum gravity theory, but none of that stuff is observable. So approaches to quantum gravity which are much less impressive, but also much simpler, continue to have supporters.

Comment author: [deleted] 10 June 2012 12:25:22PM 3 points [-]

Great reply, thank you for clearing up my confusion.

Comment author: shminux 10 June 2012 04:32:57AM 10 points [-]

What on earth is the String Theory controversy about, and is it resolvable at a glance like QM's MWI?

I wonder how you resolve the MWI "at a glance". There are strong opinions on both sides, and no convincing (to the other side) argument to resolve the disagreement. (This statement is an indisputable experimental fact.) If you mean that you are convinced by the arguments from your own camp, then I doubt that it counts as a resolution.

Also, the Occam's razor is nearly always used by physicists informally, not calculationally (partly because Kolmogorov complexity is not computable).

As for the string theory, I don't know how to use Bayes to evaluate it. On one hand, this model gives some hope of eventually finding something workable, since it provided a number of tantalizing hints, such as the holographic principle and various dualities. On the other hand, every testable prediction it has ever made has been successfully falsified. Unfortunately, there are few other competing theories. My guess is that if something better comes along, it will yield the string theory in some approximation.

Comment author: RolfAndreassen 10 June 2012 03:55:11AM 14 points [-]

I don't do formal Bayes or Kolmogorov on a daily basis; in particle physics Bayes usually appears in deriving confidence limits. Still, I'm reasonably familiar with the formalism. As for string theory, my jest in the OP is quite accurate: I dunno nuffin'. I do have some friends who do string-theoretical calculations, but I've never been able to shake out an answer to the question of what, exactly, they're calculating. My basic view of string theory has remained unchanged for several years: Come back when you have experimental predictions in an energy or luminosity range we'll actually reach in the next decade or two. Kthxbye.

The controversy is, I suppose, that there's a bunch of very excited theorists who have found all these problems they can sic their grad students on, problems which are hard enough to be interesting but still solvable in a few years of work; but they haven't found any way of making, y'know, actual predictions of what will happen in current or planned experiments if their theory is correct. So the question is, is this a waste of perfectly good brains that ought to be doing something useful? The answer seems to me to be a value judgement, so I don't think you can resolve it at a glance.

Comment author: [deleted] 10 June 2012 12:23:51PM 0 points [-]

This is roughly what I can discern from outside academia in general (I'm 19 years old and at time of posting about to graduate the local equivalent of high-school).

Comment author: pleeppleep 09 June 2012 10:35:27PM 4 points [-]

When and why did you first start studying physics? Did you just encounter it in school, or did you first try to study it independently? Also, what made you decide to focus on your current area of expertise?

Comment author: RolfAndreassen 10 June 2012 04:09:52AM 7 points [-]

I took a physics course in my International Baccalaureate program in high school - if you're not familiar with IB, it's sort of the European version of AP - and it really resonated with me. There's just a lot of cool stuff in physics; we did things like building electric motors using these ancient military-surplus magnets that had once been installed in radars for coastal fortresses. Then when I went on to college, I took some math courses and some physics courses, and found I liked the physics better. In the summer of 2003 (I think) I went to CERN as a summer student, and had an absolute blast even though the actual work I was doing wasn't so very advanced. (I wrote a C interface to an ancient Fortran simulation program that had been kicking around since it was literally on punchcards. Of course the scientist who assigned me the task could have done it himself in a week, while it took me all summer, but that saved him a week and taught me some real coding, so it was a good deal for both of us.) So I sort of followed the path of least resistance from that point. I ended up doing my Master's degree on BaBar data. Then for my PhD I wanted to do it outside Norway, so it was basically a question of connections: My advisor knew someone who was looking for a grad student, wrote me a recommendation, and I moved to the US and started my PhD. Then, when it was time to choose a thesis topic, I actually, at first, chose something completely different, involving neutrinos and reconstructing a particular decay chain from missing energy and some constraints. It turned out we couldn't get a meaningful measurement with the data we had, there were too many random events that would fake the signal. So I switched to charm mixing, which (with perhaps the teensiest touch of hindsight bias) I now actually find more interesting anyway.

As you can see, 'decide' may be a somewhat strong word in this context; I've basically worked on what my advisors have suggested, and found it interesting enough not to quit. I suspect I could have worked on practically any problem with much the same results.

Comment author: shminux 10 June 2012 04:15:39AM 3 points [-]

As you can see, 'decide' may be a somewhat strong word in this context; I've basically worked on what my advisors have suggested, and found it interesting enough not to quit. I suspect I could have worked on practically any problem with much the same results.

Yep, sunk cost is not always a fallacy.

Comment author: Vaniver 10 June 2012 07:14:05PM 11 points [-]

There's a better way to put that: switching costs are real. Sunk costs, properly identified, are fallacious.

Comment author: magfrump 09 June 2012 08:12:38PM 2 points [-]

How often do you invoke spectral gap theorems to choose dimensionality for your data, if ever?

If you do this ever, would it be useful to have spectral gap theorems for eigenvalue differences beyond the first?

(I study arithmetic statistics and a close colleague of mine does spectral theory so the reason I ask is that this seems like an interesting result that people might actually use; I don't know if it is at all achievable or to what extent theorems really inform data collection though.)

Comment author: RolfAndreassen 09 June 2012 10:44:44PM 2 points [-]

How often do you invoke spectral gap theorems to choose dimensionality for your data, if ever?

I have never done so; in fact I'm not sure what it means. Could you expand a bit?

Comment author: magfrump 11 June 2012 12:50:52AM 2 points [-]

Given a graph, one can write down the adjacency matrix for the graph; its first eigenvalue must be positive; scale the matrix so that the first eigenvalue is one. Now there is a theorem, known as the spectral gap theorem (there are parallel theorems that I'm not totally familiar with) which says that the difference between the first and second eigenvalue must be at least some number (on the order of 5% if I recall; I don't have a good reference handy).

I went to a colloquium where someone was collecting data which could be made to essentially look like a graph; they would they test for the dimensionality of the data by looking at the eigenvalues of this matrix and seeing when the eigenvalues dropped off such that the variance was very low. however, depending on the distribution of eigenvalues the cutoff point may be arbitrary. At the time, she said that a spectral gap for later eigenvalues would be useful, for making cutoff points less arbitrary (i.e. having a way to know if the next eigenvalue is definitively NOT a repeated eigenvalue because it's too far).

This isn't exactly my specialty so I'm sorry if my explanation is a little rough.

Comment author: RolfAndreassen 11 June 2012 01:04:32AM 1 point [-]

Ok, I've never used such an approach; I don't think I've ever worked with any data that could reasonably be made to look like a graph. (Unless perhaps it was raw detector hits before being reconstructed into tracks; and I've only brushed the edge of that sort of thing.) As for dimensionality, I would usually just count the variables. We are clearly talking about something very different from what I usually do.

Comment author: magfrump 12 June 2012 02:57:08AM 2 points [-]

The graph theory example was the only thing I thought of at the time but it's not really necessary; on recounting the tale to someone else in further detail I remembered that basically the person was just taking, say, votes as "yes"es and "no"s and tallying each vote as a separate dimension, then looking for what the proper dimension of the data was--so the number of variables isn't really bounded (perhaps it's 100) but the actual variance is explained by far fewer dimensions (in her example, 3).

So given a different perspective on what it is that fitting distributions means; does your work involve Lie groups, Weyl integration, and/or representation theory, and if so to what extent?

Comment author: RolfAndreassen 12 June 2012 04:48:20AM 2 points [-]

I don't understand how you get more than two dimensions out of data points that are either 0 or 1 (unless perhaps the votes were accompanied by data on age, sex, politics?) and anyway what I usually think of as 'dimension' is just the number of entries in each data point, which is fixed. It seems to me that this is perhaps a term of art which your friend is using in a specific way without explaining that it's jargon.

However, on further thought I think I can bridge the gap. If I understand your explanation correctly, your friend is looking for the minimum set of variables which explains the distribution. I think this has to mean that there is more data than yes-or-no; suppose there is also age and gender, and everyone above 30 votes yes and everyone below thirty votes no. Then you could have had dimensionality two, some combination of age and gender is required to predict the vote; but in fact age predicts it perfectly and you can just throw out gender, so the actual dimensionality is one.

So what we are looking for is the number of parameters in the model that explains the data, as opposed to the number of observables in the data. In physics, however, we generally have a fairly specific model in mind before gathering the data. Let me first give a trivial example: Suppose you have some data that you believe is generated by a Gaussian distribution with mean 0, but you don't know the sigma. Then you do the following: Assume some particular sigma, and for each event, calculate the probability of seeing that event. Multiply the probabilities. (In fact, for practical purposes we take the log-probability and add, avoiding some numerical issues on computers, but obviously this is isomorphic.) Now scan sigma and see which value maximises the probability of your observations; that's your estimate for sigma, with errors given by the values at which the log-probability drops by 0.5. (It's a bit involved to derive, but basically this corresponds to the frequentist 66%-confidence limits assuming the log-probability function is symmetric around the maximum.)

Now, the LessWrong-trained eye can, presumably, immediately see the underlying Bayes-structure here. We are finding the set of parameters that maximises the posterior probability of our data. In my toy example you can just scan the parameter space, point by point. For realistic models with, say, forty parameters - as was the case in my thesis - you have to be a bit more clever and use some sort of search algorithm that doesn't rely on brute force. (With forty parameters, even if you take only 10 points in each, you instantly have 10^40 points to evaluate - that is, at each point you calculate the probability for, say, half a million events with what may be quite a computationally expensive function. Not practical.)

The above is what I think of when I say "fitting a distribution". Now let me try to bring it back into contact with the finding-the-dimensions problem. The difference is that your friend is dealing with a set of variables such that some of them may directly account for others, as in my age/vote toy example. But in the models we fit to physics distributions, not all the parameters are necessarily directly observed in the event. An obvious example is the time resolution of the detector; this is not a property of the event (at least not solely of the event - some events are better measured than others) and anyway you can't really say that the resolution 'explains' the value of the time (and note that decay times are continuous, not multiple-choice as in most survey data.) Rather, the observed distribution of the time is generated by the true distribution convolved with the resolution - you have to do a convolution integral. If you measure a high (and therefore unlikely, since we're dealing with exponential decay) time, it may be that you really have an unusual event, or it may be that you have a common event with a bad resolution that happened to fluctuate up. The point, however, is that there's no single discrete-valued resolution variable that accounts for a discrete-valued time variable; it's all continuous distributions, derived quantities, and convolution integrals.

So, we do not treat our data sets in the way you describe, looking for the true dimensionality. Instead we assume some physics model with a fixed number of parameters and seek the probability-maximising value of those parameters. Obviously this approach has its disadvantages compared to the more data-driven method you describe, but basically this is forced upon us by the shape of the problem. It is common to try several different models, and report the variance as a systematic error.

So, to get back to Lie groups, Weyl integration, and representation theory: None of the above. :)

Comment author: magfrump 12 June 2012 03:36:27PM 0 points [-]

I definitely agree that the type of analysis I originally had in mind is totally different than what you are describing.

Thinking about distributions without thinking about Lie groups makes my brain hurt, unless the distributions you're discussing have no symmetries or continuous properties at all--my guess is that they're there but for your purposes they're swept under the rug?

But yeah in essence the "fitting a distribution" I was thinking is far less constrained I think--you have no idea a priori what the distribution is, so you first attempt to isolate how many dimensions you need to explain it. In the case of votes, we might look at F_2^N, think about it as being embedded into the 0s and 1s of [0,1]^N, and try to find what sort of an embedded manifold would have a distribution that looks like that.

Whereas in your case you basically know what your manifold is and what your distribution is like, but you're looking for the specifics of the map--i.e. the size (and presumably "direction"?) of sigma.

I don't think "disadvantages" is the right word--these processes are essentially solving for totally unrelated unknowns.

Comment author: RolfAndreassen 12 June 2012 05:05:59PM *  0 points [-]

Thinking about distributions without thinking about Lie groups makes my brain hurt, unless the distributions you're discussing have no symmetries or continuous properties at all--my guess is that they're there but for your purposes they're swept under the rug?

That is entirely possible; all I can tell you is that I've never used any such tool for looking at physics data. And I might add that thinking about how to apply Lie groups to these measurements makes my brain hurt. :)

Comment author: magfrump 13 June 2012 06:10:07AM 1 point [-]

tl;dr: I like talking about math.

Fair enough :)

I just mean... any distribution is really a topological object. If there are symmetries to your space, it's a group. So all distributions live on a Lie group naturally. I assume you do harmonic analysis at least--that process doesn't make any sense unless it lives on a Lie group! I think of distributions as essentially being functionals on a Lie group, and finding a fitting distribution is essentially integrating against its top-level differentials (if not technically at least morally.)

But if all your Lie groups are just vector spaces and the occasional torus (which they might very well be) then there might be no reason for you to even use the word Lie group because you don't need the theory at all.

Comment author: RolfAndreassen 13 June 2012 06:38:58PM 1 point [-]

any distribution is really a topological object.

I find this interesting, but I like to apply things to a specific example so I'm sure I understand it. Suppose I give you the following distribution of measurements of two variables (units are GeV, not that I suppose this matters):

1.80707 0.148763 1.87494 0.151895 1.86805 0.140318 1.85676 0.143774 1.85299 0.150823 1.87689 0.151625 1.87127 0.14012 1.89415 0.145116 1.87558 0.141176 1.86508 0.14773 1.89724 0.149112

What sort of topological object is this, or how do you go about treating it as one? Presumably you can think of these points in mD-deltaM space as being two-dimensional vectors. N-vectors are a group under addition, and if I understand the definition correctly they are also a Lie group. But I confess I don't understand how this is important; I'm never going to add together two events, the operation doesn't make any sense. If a group lives in a forest and never actually uses its operator, does it still associate, close, identify, and invert? (I further observe that although 2-vectors are a group, the second variable in this case can't go below 0.13957 for kinematic reasons; the subset of actual observations is not going to be closed or invertible.)

I'm not sure what harmonic analysis is; I might know it by another name, or do it all the time and not realise that's what it's called. Could you give an example?

Comment author: jsteinhardt 13 June 2012 06:46:57AM 1 point [-]

You can do harmonic analysis on any locally compact abelian group, see e.g. Pontryagin duality.

Comment author: alex_zag_al 09 June 2012 04:41:42PM 3 points [-]

Of the knowledge of physics that you use, what of it would you know how to reconstruct or reprove or whatever? And what do you not know how to establish?

Comment author: RolfAndreassen 09 June 2012 09:32:56PM 8 points [-]

It depends on why I want to re-prove it. If I'm transported in a time machine back to, say, 1905, and want to demonstrate the existence of the atomic nucleus, then sure, I know how to run Rutherford's experiment, and I think I could derive enough basic scattering theory to demonstrate that the result isn't compatible with the mass being spread out through the whole atom. Even if I forgot that the nucleus exists, but remembered that the question of the mass distribution internal to an atom is an interesting one, the same applies. But to re-derive that the question is interesting, that would be tough. I think similar comments apply to most of the Standard Model: I am more or less aware of the basic experiments that demonstrated the existence of the quarks and whatnot, although in some cases the engineering would be a much bigger challenge than Rutherford's tabletop setup. Getting the math would be much harder; I don't think I have enough mathematical intuition to rederive quantum field theory. In fact I haven't thought about renormalisation since I forgot all about it after the exam, so absent gods forbid I should have to shake the infinities out. I think my role would be to describe and run the experiments, and let the theorists come up with the math.

Comment author: Luke_A_Somers 09 June 2012 11:33:15AM 2 points [-]

Experimental condensed matter postdoc here. Specializing in graphene and carbon nanotubes, and to a lesser extent mechanical/electronic properties of DNA.

Comment author: epigeios 12 June 2012 04:18:37AM 0 points [-]

This might be out in left field, but:

Can water be pumped through carbon nanotubes? If so, has anyone tried? If they have, has anyone tried running an electric current through a water-filled nanotube? How about a magnetic current? How about light? How about sound?

Can carbon nanotubes be used as an antenna? If they can be filled with water, could they then be used more effectively as an antenna?

Comment author: Luke_A_Somers 27 July 2012 01:48:14PM *  3 points [-]

Sorry for the delayed response - I don't see a mechanism for reply notifications.

You can definitely cram water into carbon nanotubes, but they're hydrophobic, so it's not easy.

You can run an electric current through carbon nanotubes whether they've got water in them or not.

Spin transport is possible in perfect carbon nanotubes (magnetic current).

Carbon nanotubes are strong antennas, so they strongly interact with light. However, they are way way way too small to be waveguides for optical wavelengths, and EM radiation with an appropriate wavelength is way way way too penetrating. Water within them would just cause more scattering, not help carry current. Water carries ionic currents, which are orders of magnitude slower than electron or hole currents in nanotubes.

You can definitely carry sound with carbon nanotubes - google 'nanotube radio'.

Comment author: shokwave 27 July 2012 02:01:07PM 2 points [-]

I don't see a mechanism for reply notifications.

On the right, beneath your name and karma bubbles, there is a grey envelope. It will turn orange-red if you have replies. Click it to be taken to your inbox.

Comment author: Tripitaka 09 June 2012 02:28:40PM 2 points [-]

Carbon nanotubes in space elevators: Nicolas Pugno showed that the strenght of macroscale CNs is reduced to a theoretical limit of 30 gigapascal, with a needed strenght of 62 GPa for some desings... Whats the state of the art in tensile strenght of macro-scale CNs? Any other thoughts related to materials for space elevators?

Comment author: Luke_A_Somers 27 November 2012 04:03:24PM 1 point [-]

I just read an article raising a point which is so obvious in retrospect that I'm shaking my head that it never occurred to me.

Boron Nitride nanotubes have a very similar strength to carbon nanotubes, but much much stronger interlayer coupling. They are a much better candidate for this task.

Comment author: Luke_A_Somers 11 June 2012 05:10:26PM 1 point [-]

I'm not really up to speed on that, being more on the electronics end. Still, I've maintained interest. Personally, every year or so I check in with the NASA contest to see how they're doing.

http://www.nasa.gov/offices/oct/early_stage_innovation/centennial_challenges/tether/index.html

Last I heard, pure carbon nanotube yarn was a little stronger by weight than copper wire. Adding a little binder helps a lot.

Pugno's assumption of 100 nm long tubes is very odd - you can grow much longer tubes, even in fair quantity. Greater length helps a lot. The main mechanism of weakness is slippage, and having longer tubes provides more grip between neighboring tubes.

This is more in the realm of a nitpick, though. If I were to ballpark how much of a tensile strength discount we'd have to swallow on the way up from nanoscale, I would have guessed about 50%, which is not far off from his meticulously calculated 70%.

I'd love for space elevators to work; it's not looking promising. Not on Earth, at least. Mars provides an easier problem: lower mass and a reducing atmosphere ease the requirements on the cable. My main hope is, if we use a different design like a mobile rotating skyhook instead of a straight-up elevator, we could greatly reduce the required length, and also to some extent the strength. That compromise may be achievable.

Comment author: witzvo 09 June 2012 10:58:28AM 2 points [-]

When I read about quantum mechanics they always talk about "observation" as if it meant something concrete. Can you give me an experimental condition in which a waveform does collapse and another where it does not collapse, and explain the difference in the conditions? E.g. in the two slit experiment, when exactly does the alleged "observation" happen?

Comment author: RolfAndreassen 09 June 2012 09:35:46PM 8 points [-]

'Observation' is a shorthand (for historical reasons) for 'interaction with a different system', for example a detector or a human; but a rock will do as well. I would actually suggest you read the Quantum Mechanics Sequence on this point, Eliezer's explanation is quite good.

Comment author: witzvo 10 June 2012 05:33:11AM *  0 points [-]

Thanks.

Edit:

'Observation' is a shorthand (for historical reasons) for 'interaction with a different system', for example a detector or a human; but a rock will do as well.

I'm still confused. This seems to imply that there is no physical meaning to the term "observation," only a meaning relative to whatever model we're entertaining in a given instance. Specifically (as far as I know) there's only one system of relevance, the Universe (or the Universe of Universes, if multiple worlds stuff means anything and we insist on ruining another perfectly clear English word), so it can't interact with a different system except from the point of view of a particular mathematical model of a subset of that system. Edit: or is the word system a technical term too. Sigh.

Comment author: RolfAndreassen 10 June 2012 03:48:26PM *  6 points [-]

Indeed, your point is well taken; it is precisely this sort of argument that makes the MWI (sorry if you dislike the phrase!) attractive. If we prepare an electron in a superposition of, say, spin-up and spin-down, then it makes good sense to say that the electron eventually interacts with the detector, or detector-plus-human, system. But hang on, how do we know that the detector doesn't then go into a superposition of detecting-up and detecting-down, and the human into a superposition of seeing-the-detector-saying-up and seeing-the-detector-saying-down? Well, we don't experience a superposition, but then we wouldn't; we can only experience one brain state at a time!

Push this argument out to the whole universe and, as you rightly say, there's no further system it can interact with; there's no Final Observer to cause the collapse. (Although I've seen Christians use this as an argument for their god.) So the conclusion seems to be that there is no collapse, there's just the point where the human's wave function splits into two parts and we are consciously aware either of the up or down state. Now, there's one weakness to this: It is really not clear why, if this is the explanation, we should get the Born probabilities.

So, to return to the collapse postulate, one popular theory is that 'observation' means "the system in superposition becomes very massive": In other words, the electron interacts with the detector, and the detector-plus-electron system is in a superposition; but of course the detector is fantastically heavy on the scale of electrons, so this causes the collapse. (Or to put it differently, collapse is a process whose probability per unit time goes asymptotically to one as the mass increases.) In other words, 'observation' is taken as some process which occurs in the unification of QM with GR. This is a bit unsatisfactory in that it doesn't account for the lack of unitarity and what-have-you, but at least it gives a physical interpretation to 'observation'.

Comment author: witzvo 10 June 2012 05:01:55PM 1 point [-]

Indeed, your point is well taken; it is precisely this sort of argument that makes the MWI (sorry if you dislike the phrase!) attractive.

Yay! The rest of your argument seems sensible, but I'm too giddy to really understand it right now. I'll just ask this: can you point me to a technical paper (Arxiv is fine) where they explain, in detail, exactly how they get a certain electron "in a superposition of, say, spin-up and spin-down"?

Comment author: RolfAndreassen 10 June 2012 06:53:28PM 3 points [-]

Well, I don't know that I need to point you to arxiv, because I can describe the process in two sentences. Take a beam of electrons and pass it through a magnetic field which splits it into two beams, one going left and one going right. The ones which went left are spin-left, or to put it differently, they are spin-up with respect to the left-right axis; conversely the ones that went right have the opposite spin polarisation on that axis. Now rotate your axis ninety degrees; the electrons in both beams are in a perfect up-down superposition with respect to the new axis. If you rotate the axis less than ninety degrees you will get a different superposition.

Comment author: witzvo 10 June 2012 08:31:55PM *  1 point [-]

describe the process in two sentences.

Well, that's helpful, but of course, I don't know how you know that the electrons have such and such spin or what superposition has to do with anything. Neither could I reproduce the experiment (someone competent could, I'm sure). Maybe there was a first experiment where they did this and spin was discovered?

EDIT: anyway, I'm tapping out of here and will check out the sequences. Thanks All

Comment author: Dreaded_Anomaly 10 June 2012 10:27:03PM 4 points [-]

I don't know how you know that the electrons have such and such spin

Electrons have both electric charge and spin (which is a form of angular momentum), and in combination, these two properties create an intrinsic magnetic moment. A magnetic field exerts torque on anything with a magnetic moment, which causes the electron to precess if it is subjected to such a field. Because spin is quantized and has only two possible values for electrons (+1/2 or -1/2), they will only precess in two discrete ways. This can be used to separate the electrons by their spin values. The first experiment to do this was the Stern-Gerlach experiment, a classic in the early development of QM, and often considered to be the discovery of spin.

Comment author: witzvo 10 June 2012 10:58:58PM 1 point [-]

Thanks.

Comment author: Alicorn 10 June 2012 07:20:42PM 0 points [-]

That was four sentences! D:

Comment author: RolfAndreassen 10 June 2012 07:39:16PM 2 points [-]

Four is equal-ish to two for large values of two, at least in the limit where four is small. Besides, the last sentence is a comment, not a description of the process, so it doesn't count. :)

Comment author: Ezekiel 10 June 2012 01:24:01AM 1 point [-]

Eliezer's explanation hinges on the MWI being correct, which I understand is currently the minority opinion. Are we to understand that you're with the minority on this one?

Comment author: RolfAndreassen 10 June 2012 02:43:38AM 7 points [-]

Well, yes. But if you don't like MWI, you can postulate that the collapse occurs when the mass of the superposed system grows large enough; in other words, that the explanation is somewhere in the as-yet-unknown unification of QM and GR. Of course, every time someone succeeds in maintaining a superposition of a larger system, you should reduce your probability for this explanation. I think we are now up to objects that are actually visible with the naked eye.

Comment author: witzvo 10 June 2012 05:43:06AM *  1 point [-]

When I hear the phrase "many worlds interpretation," I cringe. This is not because I know something about the science (I know nothing about the science), it's because of confusing things I've heard in science popularizations. This reaction has kept me from reading Eliezer's sequence thus far, but I pledge to give it a fair shot soon.

Above you gave me a substitute phrase to use when I hear "observation." Is there a similar substitute phrase to use for MWI? Should I, for example, think "probability distribution over a Hilbert space" when I hear "many worlds", or is it something else?

Edit: Generally, can anyone suggest a lexicon that translates QM terminology into probability terminology?

Comment author: Grognor 10 June 2012 06:39:06PM 5 points [-]

You could go with what Everett wanted to call it in the first place, the relative state interpretation.

To answer your "Edit" question, no, the relative state interpretation does not include probabilities as fundamental.

Comment author: witzvo 10 June 2012 08:16:10PM *  2 points [-]

You could go with what Everett wanted to call it in the first place, the relative state interpretation.

Thanks! Getting back to original sources has always been good for me. Is that "Relative state" formulation of quantum mechanics?

Comment author: RolfAndreassen 10 June 2012 03:59:16PM 4 points [-]

I think it is necessary to exercise some care in demanding probabilities from QM. Note that the fundamental thing is the wave function, and the development of the wave function is perfectly deterministic. Probabilities, although they are the thing that everyone takes away from QM, only appear after decoherence, or after collapse if you prefer that terminology; and we Do Not Know how the particular Born probabilities arise. This is one of the genuine mysteries of modern physics.

Comment author: Douglas_Knight 10 June 2012 03:03:01PM 8 points [-]

I'm not sure I'm addressing your question, but I advocate in place of "many worlds interpretation" the phrase "no collapse interpretation."

Comment author: shminux 10 June 2012 06:33:48PM 0 points [-]

Personally, I advocate "no interpretation", in a sense "no ontology should be assigned to a mere interpretation".

Comment author: Viliam_Bur 11 June 2012 02:40:20PM *  1 point [-]

I am curious how exactly would this aproach work outside of quantum physics, specifically in areas more simple or more close to our intuition.

I think we should be use the same basic cognitive algorithms for thinking about all knowledge, not make quantum physics a "separate magisterium". So if the "no interpretation" approach is correct, seems to me that it should be correct everywhere. I would like to see it applied to a simple physics or even mathematics (perhaps even such as 2+2=4, but I don't want to construct a strawman example here).

Comment author: shminux 11 June 2012 02:58:19PM *  2 points [-]

I was describing instrumentalism in my comment, and so far it has been working well for me in other areas as well. In mathematics, I would avoid arguing whether a theorem that is unprovable in a certain framework is true or false. In condensed matter physics, I would avoid arguing whether pseudo-particles, such as holes and phonons, are "real". In general, when people talk about a "description of reality" they implicitly assume the map-territory model, without admitting that it is only a (convenient and useful) model. It is possible to talk about observable phenomena without using this model. Specifically, one can describe research in natural science as building a hierarchy of models, each more powerful than the one before, without mentioning the world "reality" even once. In this approach all models of the same power (known in QM as interpretations) are equivalent.

Comment author: witzvo 10 June 2012 08:10:59PM *  1 point [-]

Personally, I advocate "no interpretation", in a sense "no ontology should be assigned to a mere interpretation".

Can you elaborate on this? (I'm not voting it down, yet anyway; but it has -3 right now)

I'm guessing that your point is that seeing and thinking about experimental results for Themselves is more important than telling stories about them, yes?

Comment author: witzvo 10 June 2012 04:11:59PM *  1 point [-]

I'm not sure I'm addressing your question, but I advocate in place of "many worlds interpretation" the phrase "no collapse interpretation."

That's very helpful. It will help me read the sequence without being prejudiced by other things I've heard. If all we're talking about here is the wavefunction evolving according to Schr\:odinger's equation, I've got no problems, and I would call the "many worlds" terminology extremely distracting. (e.g. to me it implies a probability distribution over some kind of "multiverse", whatever that is).

Comment author: witzvo 10 June 2012 06:49:25AM 3 points [-]

I was reflecting on this, and considering how statistics might look to a pure mathematician:

"Probability distribution, I know. Real number, I know. But what is this 'rolling a die'/'sampling' that you are speaking about?"

Honest answer: Everybody knows what it means (come on man, it's a die!), but nobody knows what it means mathematically. It has to do with how we interpret/model the data that we see that comes to us from experiments, and the most philosophically defensible way to give these models meaning involves subjective probability.

"Ah so you belong to that minority sect of Bayesians?"

Well, if you don't like Bayesianism you can give meaning to sampling a random variable X=X(\omega) by treating the "sampled value" x as a peculiar notation for X(\omega), and if you consider many such random variables, the things we do with x often correspond to theorems for which you could prove that a result happens with high probability using the random variables.

"Hmm. So what's an experiment?"

Sigh.

Comment author: witzvo 10 June 2012 04:50:10PM *  3 points [-]

I was reflecting on this, and considering how statistics might look to a pure mathematician: "Probability distribution, I know. Real number, I know. But what is this 'rolling a die'/'sampling' that you are speaking about?"

Reflecting some more here (I hope this schizophrenic little monologue doesn't bother anyone), I notice that none of this would trouble a pure computer scientist / reductionist:

"Probability? Yeah, well, I've got pseudo-random number generators. Are they 'random'? No, of course not, there's a seed that maintains the state, they're just really hard to predict if you don't know the seed, but if there aren't too many bits in the seed, you can crack them. That's happened to casino slot machines before; now they have more bits."

"Philosophy of statistics? Well, I've got two software packages here: one of them fits a penalized regression and tunes the penalty parameter by cross validation. The other one runs an MCMC. They both give pretty similarly useful answers most of the time [on some particular problem]. You can't set the penalty on the first one to 0, though, unless n >> log(p), and I've got a pretty large number of parameters. The regression code is faster [on some problem], but the MCMC let's me answer more subtle questions about the posterior.

Have you seen the Church language or Infer.Net? They're pretty expressive, although the MCMC algorithms need some tuning."

Ah, but what does it mean when you run those algorithms?

"Mean? Eh? They just work. There's some probability bounds in the machine learning community, but usually they're not tight enough to use."

[He had me until that last bit, but I can't fault his reasoning. Probably Savage or de Finnetti could make him squirm, but who needs philosophy when you're getting things done.]

Comment author: TheOtherDave 10 June 2012 05:33:21PM 5 points [-]

who needs philosophy when you're getting things done

Well, among others, someone who wonders whether the things I'm doing are the right things to do.

Comment author: witzvo 10 June 2012 06:13:46PM *  1 point [-]

Fair point. Thanks, that hyperbole was ill advised.

Comment author: Luke_A_Somers 09 June 2012 11:39:28AM 1 point [-]

The different cases of an observation are different components of the wavefunction (component in the vector sense, in a approximately-infinite dimensional space called Hilbert Space). Observation is the point where the different cases can never come back together and interfere. This normally happens because two components differ in ways that are so widespread that only a thermodynamically small (effectively 0) component of each of them will resolve and contribute to interference against the other.

This process is called Decoherence.

Comment author: witzvo 09 June 2012 08:01:11PM *  0 points [-]

This normally happens because two components differ in ways that are so widespread that only a thermodynamically small (effectively 0) component of each of them will resolve and contribute to interference against the other.

What? I'm looking for a specific experimental condition where collapse happens and where it doesn't. E.g. suppose an electron (or rather the waveform that represents it) is impinging on a sheet of some fluorescent material. I'm guessing it hasn't collapsed yet, right? Then the waveform interacts with the sheet and causes a specific particle of the sheet it to eject a photon. Is that collapse? Or does collapse not happen until some "observer" comes along? Or is collapse actually more subtle and can be partial?

Comment author: Luke_A_Somers 11 June 2012 05:18:10PM *  1 point [-]

Then the waveform interacts with the sheet and causes a specific particle of the sheet it to eject a photon. Is that collapse?

The waveform interacts with the sheet such that a small part of many many different parts of the sheet interact, and only exactly one in each case. Since it's fluorescent, and not simply reflective, the time scale of the rerelease is finely dependent on local details, and going to wash out any reasonable interference pattern anyway.

This means that it is thermodynamically unlikely for these different components to 'come back together' so they could interfere. That's also when it loses its long-range correlations, which is the mathematical criterion for decoherence.

Due to the baggage, I personally avoid the term 'collapse', but if you're going to use it, then it's attached to the process of decoherence. Decoherence can be gradual, while 'collapse' sounds abrupt.

A partially decoherent system would be one where you have a coherent signal passing repeatedly around a mirror track. Each lap, a little bit of the signal gets mixed due to imperfections in the mirrors. The beam becomes decreasingly coherent.

So, where in there is a collapse? Eh. It would be misleading to phrase the answer that way.

Comment author: witzvo 09 June 2012 08:33:43PM *  0 points [-]

What? I'm looking for a specific experimental condition where collapse happens and where it doesn't.

Wikipedia seems to indicate that the answer is that we don't know when or if collapse happens. This is interesting, because when I was taught quantum mechanics, the notion seemed to be "of course it happens.... when we observe it... now back to Hilbert spaces" which rather soured me on the enterprise. I don't mind Hilbert spaces by the way, I just want to know how they relate to experiment. So is wikipedia right?

Comment author: evand 10 June 2012 12:18:35PM 2 points [-]

"It doesn't" is a decidedly possible interpretation of the data. It's called the Many Worlds Interpretation, and is the interpretation advocated by the Less Wrong sequence on QM. Have you read that sequence?

Comment author: witzvo 10 June 2012 04:09:30PM *  0 points [-]

No. I've been thrown off by the terminology "many worlds" and nonsense I've heard elsewhere (see below). Hope to give the sequence a fair shot soon.

Comment author: jacob_cannell 09 June 2012 08:31:12AM 7 points [-]

Rolf, I'm curious about the actual computational models you use.

How much is or can be simulated? Do the simulations cover only the exact spatial-temporal slice of the impact, or the entire accelerator, or what? Does the simulation environment include some notion of the detector?

And on that note, the Copenhagen interpretation has always bothered me in that it doesn't seem computable. How can the collapse actually be handled in a general simulation?

Comment author: Dreaded_Anomaly 10 June 2012 01:59:59AM 10 points [-]

I am a graduate student in experimental particle physics, working on the CMS experiment at the LHC. Right now, my research work mainly involves simulations of the calorimeters (detectors which measure the energy deposited by particles as they traverse the material and create "showers" of secondary particles). The main simulation tool I use is software called GEANT, which stands for GEometry ANd Tracking. (Particle physicists have a special talent for tortured acronyms.) This is a Monte Carlo simulation, i.e. one that uses random numbers. The current version of the software is Geant4, which is how I will refer to it.

The simulation environment does have an explicit description of the detector. Geant4 has a geometry system which allows the user to define objects with specific material properties, size, and position in the overall simulated "world". A lot of work is done to ensure the accuracy of the detector setup (with respect to the actual, physical detector) in the main CMS simulation software. Right now, I am working on a simplified model with a less complicated geometry, necessary for testing upgrades to the calorimeters. The simplified geometry makes it easier to swap in new materials and designs.

Geant4 also has various physics lists which describe the various scattering and interaction processes that particles will undergo when they traverse a material. Different models are used for different energy ranges. The choice of physics list can make a significant difference in the results of the simulation. Like the geometry setup, the physics lists can be modified and tuned for better agreement with experimental data or to introduce new models. The user can specify how long the program should keep track of particles, as well as a minimum energy cutoff for secondary particles (generated in showers).

An often frustrating part of Geant4 simulations is that the computing time scales roughly linearly with the number of particles and the energy of the particles. One can mitigate this problem to some extent by running in parallel, e.g. submitting 10 jobs with 1000 events each, instead of one job with 10000 events. (Rolf talks about parallelization here.) However, as we keep getting more events with higher energies at the LHC, computing time becomes more of an issue.

Because of this, there is an ongoing effort in "fast simulation." To do a faster simulation than Geant4, we can come up with parameterizations that reproduce some essential characteristics of particle showers. Specifically, we parameterize the distribution of energy deposited in the material in both the longitudinal and transverse directions. (For example, the longitudinal distribution is often parameterized as a gamma distribution.) The development of these parameterizations can be complicated, but once we have an algorithm, the simulation just requires evaluating the functions at each step. Fast simulation essentially occurs above the particle level, which is what makes it faster. A caveat: this is much easier for electromagnetic showers (which involve only electrons and photons, and only a few main processes for high energies) than for hadronic showers (which involve numerous hadrons and processes, because the strong force plays a crucial role, and therefore the energy distributions fluctuate quite a bit).

What I have given here is an overview of the simulation study of detectors; in all of this, we send single particles through the detector material. We do the same thing in real life, with a "test beam", so that we can compare to data. The actual collisions at the LHC, however, produce events far more complex than a single particle test beam. We simulate those events, too (Rolf discusses some of that below), and there are even more complications involved. I am not as knowledgeable there (yet), and this post is long enough as it is, so I will hold off on elaborating. I hope this has given you some insight into modern particle simulations!

Comment author: RolfAndreassen 09 June 2012 10:00:54PM *  6 points [-]

So the reason we simulate things is, basically, to tell us things about the detector, for example its efficiency. If you observe 10 events of type X after 100k collisions, and you want to know the actual rate, you have to know your reconstruction efficiency with respect to that kind of event - if it's fifty percent (and that would be high in many cases) then you actually had 20 physical events (plus or minus 6, obviously) and that's the number you use in calculating whatever parameter you're trying to measure. So you write Monte Carlo simulations, saying "Ok, the D* goes to D0 and pi+ with 67.4% probability, then the D0 goes to Kspipi with 5% probability and such-and-such an angular distribution, then the Ks goes to pions pretty exclusively with this lifetime, then the pions are long-lived enough that they hit the detector, and it has such-and-such a response in this area." In effect we don't really deal with quantum mechanics at all, we don't do anything with the collapse. (Talking here about experiments - there are theorists who do, for example, grid calculations of strong-force interactions and try to predict the value of the proton mass from first principles.) Quantum mechanics only comes in to inform our choice of angular distributions. (Edit: Let me rephrase that. We don't really simulate the collapse; we say instead, "Ok, there's an X% chance of this, so roll a pseudorandom number between zero and one; if less than X, that's the outcome we're going with. We don't deal with the transition, as it were, from wave functions to particles.) The actual work is in 'swimming' the long-lived decay products through our simulation of the detector. The idea is to produce information in the same format as your real data, for example "voltage spike in channel 627 at timestamp 18", and then run the same reconstruction software on it as on real data. The difference is that you know exactly what was produced, so you can go back and look at the generated distributions and see if, for example, your efficiency drops in particular regions of phase space. Usually it does, for example if one particle is slow, or especially of course if it flies down the beampipe and doesn't hit the active parts of the detector.

Calibrating these simulations is a fairly major task that consumes a lot of physicist time and attention. We look at known events; at BaBar, for example, we would occasionally shut off the accelerator and let the detector run, and use the resulting cosmic-ray data for calibration. It helps that there are really only five particles that are long-lived enough to reach the detector, namely pion, kaon, neutron, electron, and proton; so we can study how these particles interact with matter and use that information in the simulations.

Another reason for simulating is to do blind studies. For example, suppose you want to measure the rate at which particle X decays to A+B+C. You need some selection criteria to throw away the background. The hihger your signal-to-noise ratio, the more accurately you can measure the rate, within some limits - there's a tradeoff in that the more events you have, the better the measurement. So you want to find the sweet spot between 0 data of 100% purity and 100% of the data at 2% purity. (Purity, incidentally, is usually defined as signal/(signal+background).) But you usually don't want to study the effects of your selections directly on data, because there's a risk of biasing yourself - for example, in the direction of agreement with a previous measurement of the same quantity. (Millikan's oil drops are the classic example, although simulations weren't involved.) So you tune your cuts on Monte Carlo events, and then when you're happy with them you go see if there's any actual signal in the data. This sort of thing is one reason physicists are reasonably good about publishing negative results, as in "Search for X"; it could be very embarrassing to work three years on a channel and then be unable to publish because there's no signal in the data. In such a case the conclusion is "If there had been data of such-and-such a level, we would have seen it (with 95% probability); we didn't; so we conclude that the process, if it occurs, has a rate lower than X".

Comment author: kilobug 09 June 2012 08:17:48AM 1 point [-]

Not sure you're the right person to ask that to, but there have been two questions which bothered me for a while and I never found any satisfying answer (but I've to admit I didn't take too much time digging on them either) :

  1. In high school I was taught about "potential energy" for gravity. When objects gain speed (so, kinetic energy) because they are attracted by another mass, they lose an equivalent amount of potential energy, to keep the conservation of energy. But what happens when the mass of an object changes due to nuclear reaction ? The mass of sun is decreasing every second, due to nuclear fusion inside the sun (I'm not speaking of particles escaping the sun gravity, but of the conversion of mass to energy during nuclear fusion). So the potential energy of the Earth and all other planets regarding to gravity is decreasing. How is this compatible with conversation of energy ? It can't be the energy released by the nuclear reaction, the fusion of hydrogen doesn't release more energy just because Earth and Jupiter are around.

  2. Similarly for conservation issue, I always have been bothered with permanent magnet. They can move things, so they can generate kinetic energy (in metal, other magnets, ...). But where does this energy comes from ? It's stored when the magnet is created and depleted slowly as the magnet does it's work ? Or something else ?

Sorry if those are silly questions for a PhD physicist as you are, but I'm a computer scientist, not a physicist and they do bother me !

Comment author: RolfAndreassen 09 June 2012 10:41:06PM 1 point [-]

They are not silly questions, I asked them myself (at least the one about the Sun) when I was a student. However, it seems army1987 got there before I did. So, yep, when converting from mass-energy to kinetic energy, the total bending of spacetime doesn't change. Then the photon heads out of the solar system, ever-so-slightly changing the orbits of the planets.

As for magnets, the energy is stored either in their internal structure, ie the domains in a classic iron magnet; or in the magnetic field density. I think these are equivalent formulations. An interesting experiment would be to make a magnet move a lot of stuff and see if it got weaker over time, as this theory predicts.

Comment author: [deleted] 10 June 2012 07:26:08PM 2 points [-]

An interesting experiment would be to make a magnet move a lot of stuff and see if it got weaker over time, as this theory predicts.

If you're not thinking of moving a lot of stuff at once, every time you pull a piece of the stuff back off the magnet where it was before you're returning energy back to the system, so the energy needn't eventually be exhausted. (Though I guess it still eventually be if the system is at a non-zero temperature, because in each cycle some of the energy could be wasted as heat.)

Comment author: [deleted] 09 June 2012 10:43:39AM *  3 points [-]

The mass of sun is decreasing every second, due to nuclear fusion inside the sun (I'm not speaking of particles escaping the sun gravity, but of the conversion of mass to energy during nuclear fusion).

IMO “conversion of mass to energy” is a very misleading way to put it. Mass can have two meanings in relativity: the relativistic mass of an object is just its energy over the speed of light squared (and it depends on the frame of reference you measure it in), whereas its invariant mass is the square root of the energy squared minus the momentum squared (modulo factors of c), and it's the same in all frames of references, and coincides with the relativistic mass in the centre-of-mass frame (the one in which the momentum is zero). The former usage has fallen out of favour in the last few decades (since it is just the energy measured with different units -- and most theorists use units where c = 1 anyway), so in recent ‘serious’ text mass means “invariant mass”, and so it will in the rest of this post.

Note that the mass of a system isn't the sum of the masses of its parts, unless its parts are stationary with respect to each other and don't interact. It also includes contributions from the kinetic and potential energies of its parts.

The reason why the Sun loses mass is that particles escape it; if they didn't, the loss in potential energy would be compensated by the increase in total energy. The mass of an isolated system cannot change (since neither its energy nor its momentum can). If you enclosed the Sun in a perfect spherical mirror (well, one which would reflect neutrinos as well), from outside the mirror, in a first approximation, you couldn't tell what's going on inside. The total energy of everything would stay the same.

Now, if the Sun gets lighter, the planets do drift away so they have more (i.e. less negative) potential energy, but this is compensated by the kinetic energy of particles escaping the Sun... or something. I'm not an expert in general relativity, and I hear that it's non-trivial to define the total energy of a system when gravity is non-negligible, but the local conservation of energy and momentum does still apply. (Is there any theoretical physicist specializing in gravitation around?)

As for 2., that's the energy of the electromagnetic field. (The electromagnetic field can also store angular momentum, which can leading to even more confusing situations if you don't realize that, e.g. the puzzle in The Feynman Lectures on Physics 2, 17-4.)

Comment author: Dreaded_Anomaly 09 June 2012 11:44:41PM 3 points [-]

I'm not an expert in general relativity, and I hear that it's non-trivial to define the total energy of a system when gravity is non-negligible, but the local conservation of energy and momentum does still apply. (Is there any theoretical physicist specializing in gravitation around?)

Sean Carroll has a good blog post about energy conservation in general relativity.

Comment author: gjm 09 June 2012 10:26:31AM *  2 points [-]

I'm not Rolf (nor am I strictly speaking a physicist), but:

  1. There isn't really a distinction between mass and energy. They are interconvertible (e.g., in nuclear fusion), and the gravitational effect of a given quantity of energy is the same as that of the equivalent mass.

  2. There is potential energy in the magnetic field. That energy changes as magnets, lumps of iron, etc., move around. If you have a magnet and a lump of iron, and you move the iron away from the magnet, you're increasing the energy stored in the magnetic field (which is why you need to exert some force to pull them apart). If the magnet later pulls the lump of iron back towards it, the kinetic energy for that matches the reduction in potential energy stored in the magnetic field. And yes, making a magnet takes energy.

[EDITED to add: And, by the way, no they aren't silly questions.]

Comment author: kilobug 09 June 2012 10:59:41AM 1 point [-]

Hum, that's a reply to both you and army1987; I know mass and energy aren't really different and you can convert one to the other; but AFAIK (and maybe it's where I'm mistaken), if massless energy (like photons) are affected by gravity, they don't themselves create gravity. When the full reaction goes on in the Sun, fusing two hydrogen into an helium, releasing gamma ray and neutrinos in the process, the gamma ray doesn't generate gravity, and the resulting (helium + neutrino) doesn't have as much gravitational mass as the initial hydrogen did.

The same happen when an electron and a positron collide, they electron/positron did generate a gravitation force on nearby matter, leading to potential energy, and when they collide and generate gamma ray photons instead, there is no longer gravitation force generated.

Or do the gamma rays produce gravitation too ? I've pretty sure they don't... but I am mistaken on that ?

Comment author: shminux 09 June 2012 07:33:00PM *  3 points [-]

Or do the gamma rays produce gravitation too ? I've pretty sure they don't... but I am mistaken on that ?

There is a lot of potential (no pun intended) for confusion here, because the subject matter is so far from our intuitive experience. There is also the caveat "as far as we know", because there have not been measurements of gravity on the scale below tenths of a millimeter or so.

First, in GR gravity is defined as spacetime (not just space) curvature, and energy-momentum (they are linked together in relativity) is also spacetime curvature. This is the content of the Einstein equation (energy-momentum tensor = Ricci curvature tensor, in the units where 8piG/c^2=1).

In this sense, all matter creates spacetime curvature, and hence gravity. However, this gravity does not have to behave in the way we are used to. For example, it would be misleading to say that, for example, a laser beam attracts objects around it, even though it has energy. Let me outline a couple of reasons, why. In the following, I intentionally stay away from talking about single photons, because those are quantum objects, and QM and GR don't play along well.

  • Before a gravitational disturbance is felt, it has to propagate toward the detector that "feels" it. For example, suppose you measure the (classical) gravitational field from an isolated super-powerful laser before it fires. Next, you let it fire a short burst of light. What does the detector feel and when? If it is extremely sensitive, it might detect some gravitational radiation, mostly due to the laser recoiling. Eventually, the gravitational field it measures will settle down to the new value, corresponding to the new, lower, mass of the laser (it is now lighter because some of its energy has been emitted as light). The detector will not feel much, if any, "pull" toward the beam of light traveling away from it. The exact (numerical) calculation is extremely complicated and requires extreme amounts of computing power, and has not been done, as far as I know.

  • What would a detector measure when the beam of light described above travels past it? This is best visualized by considering a "regular" massive object traveling past, then taking a limit in which its speed goes to the speed of light, but its total energy remains constant (and equal to the amount of energy of the said laser beam). This means that its rest mass is reduced as its speed increases. I have not done the calculation, but my intuition tells me that the effects are reduced as speed increases, because both the rest mass and the amount time the object remains near the detector go down dramatically. (Note that the "relativistic mass" stays the same, however.)

There is much more to say about this, but I've gone on for too long as it is.

EDIT: It looks like there is an exact solution for a beam of light, called Bonnor beam. This is somewhat different from what I described (a short pulse), but the interesting feature is that two such beams do not attract. This is not very surprising, given that the regular cosmic strings do not attract, either.

Comment author: [deleted] 09 June 2012 08:11:52PM *  1 point [-]

8piG

How comes no-one has come up with a symbol (say G-bar) for that, as they did with ħ for h/2pi when they realized ħ was a more ‘natural’ constant than h? (or has anybody come up with a single symbol for 8piG?)

Comment author: Alejandro1 10 June 2012 12:44:44AM *  1 point [-]

The notation kappa = 8 pi G is sometimes used, e.g. in this Wiki article. However, it is much less universal than ħ.

Comment author: shminux 09 June 2012 10:05:23PM *  1 point [-]

There aren't many people who do this stuff for a living (as is reflected in exactly zero Nobel prizes for theoretical work in relativity so far), and different groups/schools use different units (most popular is G=1, c=1), so there is not nearly as much pressure to streamline the equations.

Comment author: Alejandro1 09 June 2012 02:48:26PM 5 points [-]

Or do the gamma rays produce gravitation too ? I've pretty sure they don't... but I am mistaken on that ?

They do. In Einstein's General Relativity, the source of the gravitational field is not just "mass" as in Newton's theory, but a mathematical object called the "energy-momentum tensor", which as it name would indicate encompasses all forms of mass, energy and momentum present in all particles (e.g. electrons) and fields (e.g. electromagnetic), with the sole exception of gravity itself.

Comment author: bogdanb 03 July 2012 08:18:55PM *  1 point [-]

with the sole exception of gravity itself.

I’ve seen this said a couple of times already in the last few days, and I’ve seen this used as a justification for why a black hole can attract you even though light cannot escape them. But black holes are supposed to also have charge besides mass and spin. So how could you tell that without electromagnetic interactions happening through the event horizon?

Comment author: Alejandro1 03 July 2012 09:10:48PM 1 point [-]

That is a good question. There is more than one way to formulate the answer in nonmathematical terms, but I'm not sure which would be the most illuminating.

One is that the electromagnetic force (as opposed to electromagnetic radiation) is transmitted by virtual photons, not real photons. No real, detectable photons escape a charged black hole, but the exchange of virtual photons between a charge inside and one outside results in an electric force. Virtual particles are not restricted by the rules of real particles and can go "faster than light". (Same for virtual gravitons, which transmit the gravitational force.) The whole talk of virtual particles is rather heuristic and can be misleading, but if you are familiar with Feynman diagrams you might buy this explanation.

A different explanation that does not involve quantum theory: Charge and mass (in the senses relevant here) are similar in that they are defined through measurements done in the asymptotic boundary of a region. You draw a large sphere at large distance from your black hole or other object, define a particular integral of (respectively) the gravitational or the electromagnetic field there, and its result is defined as the total mass/charge enclosed. So saying a black hole has charge is just equivalent to saying that it is a particular solution of the coupled Einstein-Maxwell equations in which the electromagnetic field at large distances takes such-and-such form.

Notice that whichever explanation you pick, the same explanation works for charge and mass, so the peculiarity of gravity not being part of the energy-momentum tensor that I mentioned above is not really relevant for why the black hole attracts you. Where have you read this?

Comment author: bogdanb 05 August 2012 11:42:46AM *  0 points [-]

Hi Alejandro, I just remembered I hadn’t thanked you for the answer. So, thanks! :-)

I don’t remember where I’ve seen the explanation (that gravity works through event horizons because gravitons themselves are not affected), it seemed wrong so I didn’t actually give a lot of attention to it. I’m pretty sure it wasn’t a book or anything official, probably just answers on “physics forums” or the like.

For some reason, I’m not quite satisfied with the two views you propose. (I mean in the “I really get it now” way, intellectually I’m quite satisfied that the equations do give those results.)

For the former, I never really grokked virtual particles, so it’s kind of a non-explanatory explanation. (I.e., I understand that virtual particles can break many rules, but I don’t understand them enough to figure out more-or-less intuitively their behavior, e.g. I can’t predict whether a rule would be broken or not in a particular situation. It would basically be a curiosity stopper, except that I’m still curious.)

For the latter, it’s simply that retreating to the definition that quickly seems unsatisfying. (Definitions are of course useful, but less so for “why?” questions.)

The only explanation I could think of that does make (some) intuitive sense and is somewhat satisfactory to me is that we can never actually observe particles crossing the event horizon, they just get “smeared”* around its circumference while approaching it asymptotically. So we’re not interacting with mass inside the horizon, but simply with all the particles that fell (and are still falling) towards it.

( * : Since we can observe with basically unlimited precision that their height above the EH and vertical speed is very close to zero, I can sort of get that uncertainty in where they are around the hole becomes arbitrarily high, i.e. pretty much every particle becomes a shell, kind of like a huge but very tight electronic orbital. IMO this also “explains” the no-hair theorem more satisfyingly than the EH blocking interactions. Although it does get very weird if I think about why should they seem to rise as the black hole grows, which I just dismiss with “the EH doesn’t rise, the space above it shrinks because there are more particles pulling on it”, which is probably not much more wrong than any other “layman” explanation.)

Of course, all this opens a different** can of worms, because it’s very unintuitive that particles should be eternally suspended above an immaterial border that is pretty much defined as no-matter-how-hard-you-try-you'll-still-fall-through-it. But you can’t win them all, and anyway it’s already weird that falling particles see something completely different, and for some reason relativity always seemed to me more intuitive than quantum physics, no matter how hairy it gets.

(**: Though a more accurate metaphor would probably be that it opens the same can of worms, just on a different side of the can...)

Comment author: Alejandro1 08 August 2012 03:11:24AM *  0 points [-]

OK, here is another attempt at explanation; it is a variation of the second one I proposed above, but in a way that does not rely on arguing by definition.

Imagine the (charged, if you want) star before collapsing into a black hole. If you have taken some basic physics courses, you must know that the total mass and charge can be determined by measurements at infinity: the integral of the normal component of the electric field over a sphere enclosing the star gives you the charge, up to a proportionality constant (Gauss's Law), and the same thing happens for the gravitational field and mass in Newton's theory, with a mathematically more complicated but conceptually equivalent statement holding in Einstein's.

Now, as the star begins to collapse, the mass and charge results that you get applying Gauss's Law at infinity cannot change (because they are conserved quantities). So the gravitational and electromagnetic fields that you measure at infinity do not change either. All this keeps applying when the black hole forms, so you keep feeling the same gravitational and electric forces as you did before.

Comment author: bogdanb 10 August 2012 07:23:55PM *  0 points [-]

Thanks for your perseverance :-)

Yeah, you’re right, putting it this way at least seems more satisfactory, it certainly doesn’t trigger the by-definition alarm bells. (The bit about mass and charge being conserved quantities almost says the same thing, but I think the fact that conservation laws stem from observation rather than just labeling things makes the difference.)

However, by switching the point of view to sphere integrals at infinity it sort of side-steps addressing the original question, i.e. exactly what happens at the event horizon such that masses (or charges) inside it can still maintain the field outside it in such a state that the integral at infinity doesn’t change. Basically, after switching the point of view the question should be how come those integrals are conserved, after the source of the field is hidden behind an event horizon?

(After all, it takes arbitrarily longer to pass a photon between you and something approaching an EH the closer it gets, which is sort of similar to it being thrown away to infinity the way distant objects “fall away” from the observable universe in a Big Rip, it doesn’t seem like there is a mechanism for mass and charge to be conserved in those cases.)

Comment author: shminux 10 August 2012 10:36:43PM 0 points [-]

how come those integrals are conserved, after the source of the field is hidden behind an event horizon?

First, note that there are no sources of gravity or of electromagnetism inside a black hole. Contrary to popular belief, black holes, like wormholes, have no center. In fact, there is no way to tell them apart from outside.

Second, electric field lines are lines in space, not spacetime, so they are not sensitive to horizons or other causal structures.

it takes arbitrarily longer to pass a photon between you and something approaching an EH

This is wrong as stated, it only works in the opposite direction. It takes progressively longer to receive a photon emitted at regular intervals from someone approaching a black hole. Again, this has nothing to do with an already present static electric field.

Comment author: Alejandro1 06 August 2012 03:26:28AM 0 points [-]

I'm sorry that my explanations didn't work for you; I'll try to think of something better :).

Meanwhile, I don't think it is good to think in terms of matter "suspended" above the event horizon without crossing it. It is mathematically true that the null geodesics (lightray trajectories) coming from an infalling trajectory, leaving from it over the finite proper time period that it takes for it to get to the event horizon, will reach you (as a far-away observer) over an infinite range of your proper time. But I don't think much of physical significance follows from this. There is a good discussion of the issue in Misner, Thorne and Wheeler's textbook: IIRC, a calculation is outlined showing that, if we treat the light coming from the falling chunk of matter classically, its intensity is exponentially suppressed for the far-away observer over a relatively short period of time, and if we treat it in a quantum way, there are only a finite expected amount of photons received, again over a relatively short time. So the "hovering matter" picture is a kind of mathematical illusion: if you are far away looking at falling matter, you actually do see it disappear when it reaches the event horizon.

Comment author: [deleted] 03 July 2012 08:52:37PM 0 points [-]

Interesting question, I never though about if there is any way to test a black holes charge. My guess is that we only can assume if it is there based on the theory right now

Comment author: wedrifid 03 July 2012 09:20:25PM 0 points [-]

Interesting question, I never though about if there is any way to test a black holes charge.

Calculate the black hole's mass. Put a charged particle somewhere in the vicinity of the black hole. Measure acceleration. Do math.

Comment author: [deleted] 03 July 2012 10:15:37PM 0 points [-]

That much is obvious given an assumption that charged fields work proberly through a blackhole, which was not obvious particularily given aljandro's statement. After confirming that the charge of a blackhole can interact with being impeded by the singularity, there are a lot of obvious ways to check the charge

Comment author: JulianMorrison 03 July 2012 09:30:06PM -1 points [-]

Will that work? Or to put it particle-ish-ly, how is the information about a charge inside an event horizon able to escape?

Comment author: [deleted] 03 July 2012 09:05:44PM 0 points [-]

found a relevant answer at http://www.astro.umd.edu/~miller/teaching/questions/blackholes.html "black holes can have a charge if they eat up too many protons and not enough electrons (or vice versa). But in practice this is very unusual, since these charges tend to be so evenly balanced in the universe. And then even if the black hole somehow picked up a charge, it would soon be neutralized by producing a strong electric field in the surrounding space and sucking up any nearby charges to compensate. These charged black holes are called "Reissner-Nordstrom black holes" or "Kerr-Newman black holes" if they also happen to be spinning." -Jeremy Schnittman

Comment author: Risto_Saarelma 09 June 2012 07:37:23AM 2 points [-]

More of a theoretical question, but something I've been looking into on and off for a while now.

Have you ever run into geometric algebra or people who think geometric algebra would be the greatest thing ever for making the spatial calculation aspects of physics easier to deal with? I just got interested in it again through David Hestenes' article (pdf), which also features various rants about physics education. Far as I can figure out so far, it's distantly analogous to how you can use complex numbers to do coordinate-free rotations and translations on a plane, only generalizable to any number of dimensions you want.

Comment author: RolfAndreassen 09 June 2012 09:24:17PM 2 points [-]

Have you ever run into geometric algebra or people who think geometric algebra would be the greatest thing ever for making the spatial calculation aspects of physics easier to deal with?

I can't say I have, no. Sorry! I'm afraid I couldn't make much of the Wiki article; it lost me at "Clifford algebra". Both definitions could do with a specific example, like perhaps "Three-vectors under cross products are an example of such an algebra", supposing of course that that's true.

Comment author: Risto_Saarelma 10 June 2012 12:01:17AM 2 points [-]

Linking to Wikipedia on an advanced math concept was probably a bit futile, those generally don't explain much to anyone not already familiar with the thing. The Hestenes article, and this tutorial article are the ones I've been reading and can sort of follow, but once they get into talking about how GA is the greatest thing ever for Pauli spin matrices, I have no idea what to make of it.

Comment author: RolfAndreassen 10 June 2012 02:37:15AM 2 points [-]

The tutorial article is much easier to follow, yes. Now, it's been years since I did anything with Pauli spinors, and one reason for that is that they rather turned me off theory; I could never understand what they were supposed to represent physically. This idea of seeing them as a matrix expression isomorphic to a geometric relation is appealing. Still, I couldn't get to the point of visualising what the various operations were doing; I understand that you're keeping track of objects having both scalar and vector components, but I couldn't quite see what was going on as I can with cross products. That said, it took me a while to learn that trick for cross products, so quite possibly it's just a question of practice.

Comment author: DanielLC 09 June 2012 06:47:45AM 2 points [-]

Why can't you build an electromagnetic version of a Tipler cylinder? Are electromagnetism and gravity fundamentally different?

How does quantum configuration space work when dealing with systems that don't conserve particles (such as particle-antiparticle annihilation)? It's not like you could just apply Schrödinger's equation to the sum of configuration spaces of different dimensions, and expect amplitude to flow between those configuration spaces.

A while ago I had a timelss physics question that I don't feel I got a satisfactory answer to. Short version: does time asymmetry mean that you can't make the timeless wave-function only have a real part?

Comment author: RolfAndreassen 09 June 2012 09:18:30PM 4 points [-]

Why can't you build an electromagnetic version of a Tipler cylinder? Are electromagnetism and gravity fundamentally different?

Well yes, to the best of our knowledge they are: Electromagnetic charge doesn't bend space-time in the same way that gravitational charge (ie mass) does. However, finding a description that unifies electromagnetism (and the weak and strong forces) with gravity is one of the major goals of modern physics; it could be the case that, when we have that theory, we'll be able to describe an electromagnetic version of a Tipler cylinder, or more generally to say how spacetime bends in the presence of electric charge, if it does.

How does quantum configuration space work when dealing with systems that don't conserve particles (such as particle-antiparticle annihilation)? It's not like you could just apply Schrödinger's equation to the sum of configuration spaces of different dimensions, and expect amplitude to flow between those configuration spaces.

You have reached the point where quantum mechanics becomes quantum field theory. I don't know if you are familiar with the Hamiltonian formulation of classical mechanics? It's basically a way of encapsulating constraints on a system by making the variables reflect the actual degrees of freedom. So to drop the constraint of conservation of particle number you just write a Hamiltonian that has number of particles as a degree of freedom; in fact, the number of particles at every point in position-momentum space is a degree of freedom. Then you set up the allowed interactions and integrate over the possible paths. Feynman diagrams are graphical shorthands for such integrals.

A while ago I had a timelss physics question that I don't feel I got a satisfactory answer to. Short version: does time asymmetry mean that you can't make the timeless wave-function only have a real part?

I'm afraid I can't help you there; I don't even understand why reversing the time cancels the imaginary parts. Is there a particular reason the T operator should multiply by a constant phase? That said, to the best of the current knowledge the wave function is indeed symmetric under CPT, so if your approach works at all, it should work if you apply CPT instead of T reversal.

Comment author: bogdanb 01 September 2012 10:47:46AM *  0 points [-]

Why can't you build an electromagnetic version of a Tipler cylinder? Are electromagnetism and gravity fundamentally different?

Well yes, to the best of our knowledge they are: Electromagnetic charge doesn't bend space-time in the same way that gravitational charge (ie mass) does. However, finding a description that unifies electromagnetism (and the weak and strong forces) with gravity is one of the major goals of modern physics; it could be the case that, when we have that theory, we'll be able to describe an electromagnetic version of a Tipler cylinder, or more generally to say how spacetime bends in the presence of electric charge, if it does.

There’s something very confusing to me about this (the emphasized sentence). When you say “in the same way”, do you mean “mass bends spacetime, and electromagnetic charge doesn’t”, or is it “EM change also bends spacetime, just differently”?

Both interpretations seem to be sort-of valid for English (I’m not a native speaker). AFAIK it’s valid English to say “a catapult doesn’t accelerate projectiles the way a cannon does”, i.e., it still accelerates projectiles but does it differently, but it’s also valid English to say “neutron stars do not have fusion in their cores the way normal stars do”, i.e., they don’t have fusion in their cores at all. (Saying “X in the same way as Y” rather than the shorter “X the way Y” seems to lean towards the former meaning, but it still seems ambiguous to me.)

So, basically, which one do you mean? From the last part of that paragraph (“if it does”), it seems that we don’t really know. But if we don’t, than why are Reissner-Nordström or Kerr-Newman black holes treated separately from Schwarzschild and Kerr black holes? Wikipedia claims that putting too much charge in one would cause a naked singularity, doesn’t the charge have to bend spacetime to make the horizon go away?

I encountered similar ambiguity problems with basically all explanations I could find, and also for other physics questions. One such question that you might have an answer to is: Do superconductors actually have really, trully, honest-to-Omega zero resistance, or is it just low enough that we can ignore it over really long time frames? (I know superconductors per se are a bit outside of your research, but I assume you know a lot more than I do due to the ones used in accelerators, and perhaps a similar question applies to color-superconducting phases of matter you might have had to learn about for your actual day job.)

Comment author: shminux 07 September 2012 05:23:31PM *  1 point [-]

Re Tipler cylinder (incidentally, discovered by van Stockum). It's one of those eternal solutions you cannot construct in a "normal" spacetime, because any such construction attempt would hit the Cauchy horizon, where the "first" closed timelike curve (CTC) is supposed to appear. I put "first" in quotation marks because the order of events loses meaning in spacetimes with CTCs. Thus, if you attempt to build a large enough cylinder and spin it up, something else will happen before the frame-dragging effect gets large enough to close the time loop. This has been discussed in the published literature, just look up references to the Tipler's papers. Amos Ori spent a fair amount of time trying to construct (theoretically) something like a time-machine out of black holes, with marginal success.

Comment author: RolfAndreassen 02 September 2012 01:56:20AM 3 points [-]

Superconductor resistance is zero to the limit of accuracy of any measurement anyone has made. In a similar vein, the radius of an electron is 'zero': That is to say, if it has a nonzero radius, nobody has been able to measure it. In the case of electrons I happen to know the upper bound, namely 10^-18 meters; if the radius was larger than that, we would have seen it. For superconductors I don't know the experimental upper limit on the resistance, but at any rate it's tiny. Additionally, I think there are some theoretical reasons, ie from the QM description of what's going on, to believe it is genuinely zero; but I won't swear to that without looking it up first.

About electromagnetic Tipler cylinders, I should have said "the way that". As far as I know, electromagnetism does not bend space.

Comment author: bogdanb 02 September 2012 07:27:18AM 0 points [-]

Thank you for the limits explanation, that cleared things up.

About electromagnetic Tipler cylinders, I should have said "the way that". As far as I know, electromagnetism does not bend space.

OK, but if so then do you know the explanation for why:

1) charged black holes are studied separately, and those solutions seem to look different than non-charged black holes?

2) what does it mean that a photon has zero rest mass but non-zero mass “while moving”? I’ve seen calculations that show light beams attracting each other in some cases (IIRC parallel light beams remain parallel, but “anti-parallel” beams always converge), and I also saw calculations of black holes formed by infalling shells of radiation rather than matter.

3) doesn’t energy-matter equivalence imply that fields that store energy should bend space like matter does?

What am I missing here?

Comment author: RolfAndreassen 02 September 2012 05:54:54PM 0 points [-]

2) what does it mean that a photon has zero rest mass but non-zero mass “while moving”? I’ve seen calculations that show light beams attracting each other in some cases (IIRC parallel light beams remain parallel, but “anti-parallel” beams always converge), and I also saw calculations of black holes formed by infalling shells of radiation rather than matter.

A moving photon does not have nonzero mass, it has nonzero momentum. In the Newtonian approximation we calculate momentum as p=mv, but this does not work for photons, where we instead use the full relativistic equation E^2 = m^2c^4 + p^2c^2 (observe that when p is small compared to m, this simplifies to a rather more well-known equation), which, taking m=0, gives p = E/c.

As for light beam attracting each other, that's an electromagnetic effect described by high-order Feynmann diagrams, like the one shown here. (At least, that's true if I'm thinking of the same calculations you are.)

1) charged black holes are studied separately, and those solutions seem to look different than non-charged black holes?

3) doesn’t energy-matter equivalence imply that fields that store energy should bend space like matter does?

Both good points. I'm afraid we're a bit beyond my expertise; I'm now unsure even about the electromagnetic Tipler cylinder.

Comment author: MixedNuts 01 September 2012 08:52:07PM 3 points [-]

Do superconductors actually have really, trully, honest-to-Omega zero resistance, or is it just low enough that we can ignore it over really long time frames?

It's for-real zero. (Source: conference La supraconductivité dans tous ses états, Palaiseau, 2011) Take a superconductive loop with a current in it and measure its resistance with a precise ohmeter. You'll find zero, which tells you that the resistance must be less than the absolute error on the ohmeter. This tells you that an electron encounters a resistive obstacle at most every few ten kilometers or so. But the loop is much smaller than that, so there can't be any obstacles in it.

Comment author: bogdanb 02 September 2012 07:12:55AM 1 point [-]

It's for-real zero. (Source: conference La supraconductivité dans tous ses états, Palaiseau, 2011)

Man, that is so weird. I live in Palaiseau—assuming you’re talking about the one near Paris—and I lived there in 2011, and I had no idea about that conference. I don’t even know where in Palaiseau it could have taken place...

Comment author: MixedNuts 02 September 2012 10:46:31AM 1 point [-]

That one talk was at Supoptique. There were things at Polytechnique too, and I think some down in Orsay.

Comment author: James_Miller 09 June 2012 06:25:46AM *  6 points [-]

Might life in our universe continue forever? Does proton decay and the laws of thermodynamics, if nothing else, doom us?

Comment author: RolfAndreassen 09 June 2012 07:34:46PM 12 points [-]

Proton decay has not been observed, but even if it happens, it needn't be an obstacle to life, as such. For humans in anything remotely like our present form you need protons, but not for life in general. Entropy, however, is a problem. All life depends on having an energy gradient of some form or other; in our case, basically the difference between the temperature of the Sun and that of interstellar space. Now, second thermo can be stated as "All energy gradients decrease over a sufficiently long time"; so eventually, for any given form of life, the gradient it works off is no longer sharp enough to support it. However, what you can do is to constantly redesign life so that it will be able to live off the gradients that will exist in the next epoch. You would be trying to run the amount and speed of life down on an asymptotic curve that was nevertheless just slightly faster than the curve towards total entropy. At every epoch you would be shedding life and complexity; your civilisation (or ecology) would be growing constantly smaller, which is of course a rather alien thing for twenty-first century Westerners to consider. However, the idea is that by growing constantly smaller you never hit the wall where the gradient just cannot support your current complexity anymore, and instantly collapse to zero. An asymptote that never hits zero is, presumably, better than a curve of any shape that hits the wall and crashes - at least this is true if your goal is longevity; of course, pure survival is not the only goal of humans, so there's a value judgement to be made there. You might decide that it's better not to throw anyone out of the lifeboat and all starve together, rather than keep going at the price of endless sacrifice and endless shrinking. And, of course, if we can extrapolate to such incredibly distant beings at all, there's going to be quarrels over exactly who gets thrown out, and the resulting conflict might well make the asymptote shrink drastically, or collapse, as resources are used to fight instead of survive. To survive literally forever you need to be lucky every time; entropy only needs to be lucky once.

That said, even with total entropy you get the occasional quantum fluctuation that creates a small, local gradient again - in fact, arbitrarily large gradients if you wait arbitrarily long times; if somehow you were able to survive the period between such events, you could indeed live for ever. In fact, if you are able to wait long enough you will see a quantum fluctuation the size of the Big Bang. The problem is, of course, that a human, and probably life more generally as well, is extremely low-entropy compared to the sort of universe you get at 10^1000 years. In fact, interstellar space from our era would look rather low-entropy compared to that stuff. So the difficulty is to protect yourself against the, as it were, sucking vacuum that tries to rip the low entropy out of your body, without using up your reserves of energy on self-repair.

Overall, I'd say it doesn't look utterly hopeless, although it is subject to a Fermi paradox: If survival over arbitrary timescales is possible, why don't we see any survivors from previous BB-level events? If my account is correct, it seems unlikely that ours is the first such fluctuation.

Comment author: DanielLC 10 June 2012 10:20:53PM 1 point [-]

You would be trying to run the amount and speed of life down on an asymptotic curve that was nevertheless just slightly faster than the curve towards total entropy.

Is the total subjective time finite or infinite?

That said, even with total entropy you get the occasional quantum fluctuation that creates a small, local gradient again - in fact, arbitrarily large gradients if you wait arbitrarily long times;

Does the expansion of space pose a problem? If you had a universe of a constant size, you'd expect fluctuations in entropy to create arbitrarily large gradients in energy if you wait long enough, but if it keeps spreading out, the probability of a gradient of a given size ever happening would be less than one, wouldn't it?

Also, wouldn't we all be Boltzmann brains if it worked like that?

Comment author: RolfAndreassen 11 June 2012 07:43:11PM 0 points [-]

Is the total subjective time finite or infinite?

The intention was to make it infinite, otherwise there's no use to the process. You'll notice that the laws of thermodynamics don't say anything about the shape of the downward trend, so it is at least conceivable that it allows a non-convergent series.

If you had a universe of a constant size, you'd expect fluctuations in entropy to create arbitrarily large gradients in energy if you wait long enough, but if it keeps spreading out, the probability of a gradient of a given size ever happening would be less than one, wouldn't it?

This doesn't look obvious to me. You get more vacuum to play with; the probability per unit volume should remain constant.

Also, wouldn't we all be Boltzmann brains if it worked like that?

Could be. Do you know we aren't? :)

Comment author: DanielLC 11 June 2012 07:54:52PM 0 points [-]

This doesn't look obvious to me. You get more vacuum to play with; the probability per unit volume should remain constant.

I was assuming that there has to be stuff in space for stuff to happen. I guess I was wrong.

Do you know we aren't? :)

There's a chance that our experiences are just random, which we can't do much to reduce. All we can do is look at the probability of physics working a certain way given that we are not random. That cosmology would be ridiculously unlikely given that we are not random, because that would require that we not be Boltzmann brains, which is extraordinarily unlikely.

Comment author: James_Miller 09 June 2012 06:20:14AM 11 points [-]

How good of an understanding of physics is it possible to acquire if you read popular books such as Greene's but never look at the serious math of physics. Is there lots of stuff in the math that can't be conveyed with mere words, simple equations and graphs?

Comment author: RolfAndreassen 09 June 2012 08:08:42PM *  14 points [-]

I guess it depends on what you mean by 'understanding'. I personally feel that you haven't really grasped the math if you've never used it to solve an actual problem - textbook will do, but ideally something not designed for solvability. There's a certain hard-to-convey Fingerspitzggefühl, intuition, feel-for-the-problem-domain - whatever you want to call it - that comes only with long practice. It's similar to debugging computer programs, which is a somewhat separate skill from writing them; I talk about it in some detail in this podcast and these slides.

That said, I would say you can get quite a good overview without any math; you can understand physics in the same sense I understand evolutionary biology - I know the basic principles but not the details that make up the daily work of scientists in the field.

Comment author: satt 09 June 2012 08:47:36PM 1 point [-]

Podcast & slide links point to the same lecture9.pdf file, BTW.