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First-principles predictions of properties of baryons are only in their infancy. Even the best lattice QCD calculations of mesonic properties at the physical p
by evanb 9y ago
First-principles predictions of properties of baryons are only in their infancy. Even the best lattice QCD calculations of mesonic properties at the physical pion mass are good to 1 part in 10^(4ish), nothing approaching the accuracy needed to detect new physics in the baryon sector.
Calculations with baryons suffer from a severe signal-to-noise problem, and the latest calculation of a substantially-easier-to-compute quantity, the nucleon isovector axial coupling, is only at ~1% precision: https://arxiv.org/abs/1710.06523 https://arxiv.org/abs/1710.06523 (full disclosure: one author of said calculation).
I don't know of a calculation of the proton magnetic moment with physical parameters.
- nine_k 9y agoWhy is the precision so low? I assume it's not due to lack of computing resources. Is it due to the limited accuracy of inputs? inherent instability of the model, like with weather prediction?
- danbruc 9y agoNot an expert but I think it is actually due to a lack of computing resources, lattice QCD calculations seem to be extremely demanding based on what I read.
- phkahler 9y agoI seem to recall reading that some calculations complexity were dramatically reduced with a new method having something to do with volumes of n-dimensional polyhedra. Can't find it with google because I don't recall enough.
- danbruc 9y agoMaybe you are thinking of the amplituhedron [1] which can make calculations of interaction amplitudes much easier than summing up a large number Feynman diagram. [1] https://en.wikipedia.org/wiki/Amplituhedron https://en.wikipedia.org/wiki/Amplituhedron
- phkahler 9y agoThat's it I think. No infinite nesting of diagrams for virtual particles.
- deleted 9y ago[deleted]
- semi-extrinsic 9y agoIt's a classic example of combinatorial explosion making something untractable. Going to one more order in perturbation theory is such an incredible increase in computational expense. Solving chess (in the strict sense) is a similar example. E.g. all 6-piece endgames were known in 2005; by 2012 we had the complete 7-piece endgame tablebase, but the 8-piece endgame tablebase is such a hard problem that we're nowhere close even with Moore's law; the sheer size of that tablebase is estimated to be ~10 petabytes.
- evanb 9y agoLattice QCD does not rely on perturbation theory, and so it doesn't suffer from a combinatorially explosion in the way you think. However, we do attempt to perform many-millions-dimensional (scaling with the spacetime volume grid size) integrals stochastichally.
- semi-extrinsic 9y agoHuh, can you explain more? I always assumed this stuff was done by expanding the action or the S-matrix or something using some relevant diagrams to some loop order? (I must confess it's too long since I did QFT.)
- ahh 9y ago(Disclaimer: not a QCD expert, but I did read a paper or two back in grad school from the perspective of someone who liked numerical linear algebra. Don't rely on this advice.) I believe the core difference here between QCD and QED (the classical perturbative QFT) is that QED has a very small coupling constant (it's the fine structure constant, so ~1/137) whereas QCD has a large one (comparable to or greater than one.) It turns out that a perturbative expansion is effectively in powers of the coupling constant, so QED expansions converge rapidly...and QCD expansions converge not at all.
- ISL 9y agoThis explanation ought to be correct at low energy, just where most LQCD work occurs (attempting to predict even lower-energy observables, like the proton mass, etc.). At high (collider) energies, it is okay to work perturbatively.
- Filligree 9y agoParticle physics simulations are one of the areas where quantum computers would be exceedingly useful, dropping many calculations from exponential-time to, in the best case, linear. That's if you're trying to find exact solutions. Which you can't, but even the approximations we've got are frankly awful. So, basically a lack of computing resources. QCs aren't generally that much faster than classical ones, but in this particular case they are--understandably so, given the universe runs on quantum mechanics. Turing machines are not good enough for this, and no number of Intel CPUs will let us overcome that barrier. Physicists try really hard anyway, but the approximations are... Well. As I said, awful.
- eru 9y ago> Turing machines are not good enough for this, [...] Usually we only bring up Turing machines for decidability, ie when the only thing that matters is whether it takes finite or infinite amounts of time. (Yes, I know, you can also calculate exact runtimes with Turing machines, but nobody really cares, I think?) I guess you mean to say classical computer or so? > [...] and no number of Intel CPUs will let us overcome that barrier. Or at least that's everybody's best guess. On similar footing to the empirical evidence for P != NP. It's still theoretically possible that BQP = BPP, if memory serves right. (https://en.wikipedia.org/wiki/BQP https://en.wikipedia.org/wiki/BQP vs https://en.wikipedia.org/wiki/BPP_(complexity) https://en.wikipedia.org/wiki/BPP_(complexity))
- cgmg 9y agoThe best 'physical' model of computation is probably the reversible 3D mesh (R3M) model: http://iopscience.iop.org/article/10.1088/0957-4484/9/3/005/meta http://iopscience.iop.org/article/10.1088/0957-4484/9/3/005/....
- eru 9y agoOf course, it depends a bit on what you are trying to model. Most of the time the different ways to model computing only have a polynomial slowdown to emulate each other. In that sense, quantum computers are different from Turing machines: the best way we know to emulate quantum computers on Turing machines has exponential slow downs.
- evanb 9y agoIt is due to computing resources. LQCD calculations consume a large fraction of all the world's supercomputing, and there's never enough to go around. The procedure is essentially an evaluation of a millions-of-dimensions-large integral via Markov-chain Monte Carlo. Each step in the Markov chain requires inverting extremely large ill-conditioned matrices, which we can make better-conditioned by moving away from physical parameters (and then trying to extrapolate). People are glad to have only hundreds or thousands of independent samples in an ensemble. You need ensembles for multiple lattice spacings and spatial volumes in order to ensure you can control your discretization and finite-volume artifacts when you extrapolate to the continuum and to infinite volume. What is generated in the Markov chain is essentially the configurations of the gluons and quark loops. With these configurations in hand we can measure physical observables on each. Observables with baryons have an exponentially bad signal-to-noise problem in the temporal direction (time on the grid, not Monte Carlo time) and you either need an exponentially large computer, an exponentially long time, or an exponentially wealthy funding agency. There are attempts to circumvent this issue, we'll see how successful they prove...
- effie 9y agoIs it really necessary to invert those matrices? Often that is possible, but not the best way to find the solution of the problem: https://www.johndcook.com/blog/2010/01/19/dont-invert-that-matrix/ https://www.johndcook.com/blog/2010/01/19/dont-invert-that-m...
- msds 9y agoPresumably, an ensemble of nuclear theorists, particularly those dedicating their lives to making big LQCD calculations, would avoid inverting large ill-conditioned matrices, if that was a clever thing to do. I know a few QFT types - they tend to be decently intelligent and mathematically capable. Ok, they're more like wickedly smart...
- evanb 9y agoIt depends on the computational step. Often we "just" need to compute the determinant, in which case we use a trick. In some other cases, we do only need linear solves. Exact inversion of the whole thing is often prohibitive.