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3 hours ago by whatshisface

>The dynamics of quarks and gluons can be described perturbatively in hard processes thanks to the smallness of the strong coupling constant at short distances, but the spectrum of stable hadrons is affected by non-perturbative effects and cannot be computed from the fundamental theory. Though lattice QCD attempts this by discretising spaceā€“time in a cubic lattice, the results are time consuming and limited in precision by computational power. Predictions rely on approximate analytical methods such as effective field theories.

I'm glad this was mentioned, non-perturbative effects are not well understood and this is a big part of why it's worthwhile to study bound states of the strong force.

an hour ago by munchler

I assume that if we ever unify QCD with General Relativity, the resulting theory would be able to predict these hadrons from first principles?

23 minutes ago by frutiger

No. The reason perturbation theory doesnā€™t work as well for QCD as it does for QED is because of two reasons:

1. The coupling constant of QCD is much higher than QED so contributions to the overall result from Feynman diagrams that have more vertices (the multiplicative factor of each element in the sum is proportional to the power of the number of vertices) donā€™t vanish as quickly as they do for QED

2. The gauge bosons in QCD (i.e. gluons) themselves have colour charge whereas those in QED (i.e. photons) do not have electrical charge.

18 minutes ago by whatshisface

You can't give a definite no to that because, since gravitons have stress-energy and are non-perturbative, a field theory advance that worked for them could also help with the strong force.

an hour ago by undefined
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an hour ago by ur-whale

> the resulting theory would be able to predict these hadrons from first principles?

Not sure how bringing GR into the fray would help solve what essentially seems to be a computational complexity problem. Might actually make things worse.

an hour ago by whatshisface

It's not a computational complexity problem, it's an undefinedness problem. Proving that the lattice simulations converge has been estimated as well beyond this century's mathematics by the pair of people (Glimm and Jaffe) that have done the most to study it. In any case it is beyond today's.

3 hours ago by addaon

Are there a lot of missing overbars in this article, or some other typographic marker for antiquarks? I assume the hexaquark descriptions early on are supposed to be (using Q for q-overbar) "QQQqqq or qqqqqq", where it reads to me as "qqqqqq or qqqqqq".

an hour ago by dukwon

They are there in the print version (page 26) https://cerncourier.com/wp-content/uploads/2024/12/CERNCouri...

3 hours ago by cwillu

ā€œOther manifestly exotic hadrons followed, with two exotic hadrons Tcccc(6600) and Tcccc(6900) observed by LHCb, CMS and ATLAS in the J/ĻˆJ/Ļˆ spectrum. They can be interpreted as a tetraquark made of two charm and two anti-charm quarks ā€“ a fully charmed tetraquark.ā€

Not sure if it was deliberate or not, but yeah.

3 hours ago by timthorn

As I wrote somewhere else, I rather like the cuddly hadrons from The Particle Zoo: https://www.particlezoo.net/collections/particle-packs

2 hours ago by linsomniac

My dyslexic brain: "Exotic whatnow?"

an hour ago by vvpan

Same...

2 hours ago by grahamj

what, you have something against beastiality hardons?

(HN has no sense of humour lol)

2 hours ago by snapetom

"Comments should get more thoughtful and substantive"

Please go to reddit if you want to post easy one liners for quick karma.

43 minutes ago by saagarjha

If youā€™re going to quote the guidelines at least do it with the appropriate context.

2 hours ago by undefined
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an hour ago by anyfoo

They do, but you usually need to put some effort in.

2 hours ago by addaon

"also implies the existence of a Tbb state, with a bbud quark content, that should be stable except with regard to weak decays"

Can someone explain this to me?

Tcc(3875)+ can decay to a D0 and a D+, yes? And this is a strong decay?

I guess the reason Tbb doesn't have an equivalent strong decay to B mesons because of the sign difference -- that is, B0 and B+ would have anti-bs, not bs; and anti-B0 and anti-B+ would have negative charge?

And so the only major decay pathway is for the b itself to decay to a K+ (plus lepton noise), giving a temporary bu\s\u\d pentaquark, that then has uninhibited decays?

I guess what I'm asking is... is this the right way to think about this?

2 hours ago by adrian_b

In strong decays, the products will contain the same quarks and antiquarks that have existed in the original particle, possibly with the addition of one or more quark-antiquark pairs that have been generated during the decay.

In weak decays, one or more of the original quarks or antiquarks will be converted in a quark or antiquark with a different flavor, which is a process that has a low probability of happening, so the weak decays happen less frequently, therefore the hadrons that can decay only through weak decays have a lifetime that is many orders of magnitude greater than the hadrons that can decay through strong decays (or electromagnetic decays, i.e. annihilation of quarks with the corresponding antiquarks).

D+ is c quark + d antiquark, D0 is c quark + u antiquark

Tcc(3875)+ is 2 c quarks + d antiquark + u antiquark

Therefore the 4 quarks/antiquarks in Tcc(3875)+ are the same as the 4 quarks/antiquarks in D0 + D+.

So this is a strong decay, because no quark or antiquark is converted into another kind of quark or antiquark.

For the Tbb- tetraquark, its composition would allow a similar strong decay into two b-quark + u or d antiquark hadrons, except that its binding energy is so great that the mass of the Tbb- tetraquark is smaller than the sum of the masses of the hadrons that would be produced during a strong decay (it is also smaller than the sum of masses of the hadrons that could be produced by an electromagnetic decay, see https://www.sciencedirect.com/science/article/pii/S037026931... ).

This forbids the strong decay and the electromagnetic decay, so the only admissible decay must be weak, where one of the b quarks will be converted into another kind of quark.

an hour ago by dukwon

The strong decay would just be forbidden from conservation of energy. If the mass of the Tbb state is less than the sum of the B+ and the B0 masses, then that decay isn't allowed.

2 hours ago by FabHK

Do we have anomalies accumulating here that indicate the early phase of a scientific revolution in Thomas Kuhn's terminology, that is, a replacement of the standard model/QCD? Or is it still "so far, so good"?

an hour ago by undefined
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an hour ago by undefined
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2 hours ago by drpossum

Do you feel like those two options would cover all possible scenarios for "the state of the field"?

an hour ago by anyfoo

Well, either the standard model is right, or it isn't, isn't it? They asked for indication of an "early phase", not that we're ready to throw the standard model out (which, boringly, held up extremely well so far).

an hour ago by whatshisface

The standard model Lagrangian is a sum of many terms, and changing one of them, adding a new one or even a radical revolution in our understanding of the results of integrals taken over it would not count as a Kuhnian revolution. Physics has not had one of those since Newton.

an hour ago by Mistletoe

ā€œAll science is either physics or stamp collecting.ā€ -Ernest Rutherford

Is this stamp collecting? Do these exotic hadrons mean anything?

43 minutes ago by iterance

That quote isn't real; it was a metaphor Rutherford purportedly once used, posthumously recalled by John Bernal. It was incorrectly converted into a direct quotation by later writers. But even then, you're misunderstanding the quote, by which is meant that physics has supremacy and all other sciences are collecting specific instances of physics; the LHC is decidedly doing physics.

However, even if you take the quote to mean what you imagined, it is unnecessarily cynical. LHC has advanced our understanding of physics.

40 minutes ago by whatshisface

Learning about the properties of exotic hadrons clarifies our understanding of nuclear forces.

44 minutes ago by ur-whale

>Do these exotic hadrons mean anything?

Given their horribly short lifespans, probably not much other than the fact that they manage to exist for however short a time might vindicate QFT a tad more (I'm assuming that QFT somewhat predicts their likelihood to show up).

Or maybe they'll bring a deeper understanding of the strong force.

But generally speaking, I feel you: lots of work and energy spent to create these exotic things, but that may or may not have an actual use or even meaning.

A lot of science is like this these days, it looks like we're hitting exponentially diminishing returns (as in: useful applications) in some areas of science.

38 minutes ago by undefined
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