Do baryons only decay via the weak force?

[Silversonic]

Provided they aren't in an excited state (which would allow a decay via EM or strong). I had a question asking why the neutral lambda particle, consisting of uds quarks, has a lifetime characteristic of the weak. The answer being that it must decay via the weak to change the s-quark flavour into a down or an up.

But this has me thinking, don't all baryons decay only by the weak force? I don't have it confirmed in my notes anywhere, and a google search revealed to me that most baryons do (but didn't mention why there was an exception). If a decay process must happen through the constraint that the constiuent produced particles have less rest mass-energy than the decaying particle, then surely weak is the only way? Because a strong decay (emission of a gluon by a quark) or a electromagnetic decay (emission of a photon) of a hadron (consititing of only three quarks, no antiquarks) can only hope to pair-produce more quarks but not change any of the quarks of the original hadron. Hence increasing mass-energy, rather than lowering, contradicting the original assumption.

So a baryon can only reduce the rest mass-energy by the weak force, the strong/EM just increase it for a baryon. Surely? It seems like something that should've been mentioned if it was true. Maybe I'm over-simplifying the decay process (and there isn't something I've considered).

Case isn't the same for a meson of course. The neutral pion can self annihilate and decay into two photons via the EM force.

 

[snorkack]

I think because not being able to decay via strong force is the definition of a baryon - that is the distinguishing line between hyperon and resonance. Or meson and resonance.

Charged pions have no lower mass photon or meson states to decay into - all possible paths are weak. Neutral pion can decay via EM force, but again no strong force paths - no lower mesons. The first meson resonances are rho and omega However, since eta meson can decay to pions, not sure why it is not resonance.

In baryon spectrum, the first resonance is delta, which decays by strong force. But there is also the neutral sigma, which is uds - yet somehow different from lambda, also uds. Neutral sigma cannot decay by strong force to lambda - not enough energy to create a pion - which is why it is not a resonance but a baryon.

 

[The_Duck]

snorkack, your distinction between "resonances" and "baryons" seems unfamiliar to me: in my experience, an excited state of a baryon is still called a baryon, with a different name. For example Wikpedia calls the delta the "delta baryon": http://en.wikipedia.org/wiki/Delta_baryon , and searching Google for "strong decays of baryons" turns up this paper, titled "Strong decays of charmed baryons": http://prd.aps.org/abstract/PRD/v75/i9/e094017

 

[Bill_K]

snorkack, your distinction between "resonances" and "baryons" seems unfamiliar to me: in my experience, an excited state of a baryon is still called a baryon, with a different name. For example Wikpedia calls the delta the "delta baryon":

It's also been referred to as the (3,3) resonance. (Meaning I = 3/2, S = 3/2.)

 

[Vanadium 50]

You seem to be adopting a non-standard definition of baryon, and that definition is equivalent to saying it decays by the weak force. So what you suppose is true, tautologically so.

 

[kokolovehuh]

I am not so sure all baryon decay weakly.
First, consider the classical example of positive pions scattering with protons which can produce neutral lambdas. Then, for instance, it has been experimentally proven that such lambda can decay strongly b/c we can trace its track within reasonable distance. In this case, strangeness is also conserved in production of the pion, meaning this is either EM or strong.
Second, the idea of lifetime is based on its means. Following the exponential pattern of lifetimes, it is almost certain that some of the lifetimes can be short if the production of the baryon is large.

 

[Bill_K]

First, consider the classical example of positive pions scattering with protons which can produce neutral lambdas. Then, for instance, it has been experimentally proven that such lambda can decay strongly b/c we can trace its track within reasonable distance.

The Λ⁰ has a lifetime of about 10^-10 sec, typical of a weak decay. Being neutral, there is no visible track. The vertices for production and decay will be a few cm apart. If it was a strong decay, the lifetime would be more like 10^-23 sec and the track length 10^-13 cm, too short to detect.

In this case, strangeness is also conserved in production of the pion, meaning this is either EM or strong.

Strangeness is conserved in the production process, which is strong, and since the Λ⁰ has strangeness S = -1, Λ's can only be produced in association with other strange particles, e.g. a K⁺ meson with S = +1. The Λ⁰ decay does not conserve strangeness - it's a nonleptonic weak decay. In terms of quarks a strange quark turns into a down quark.

 

[mfb]

Provided they aren't in an excited state (which would allow a decay via EM or strong).

There is the problem. You exclude all baryons which can decay via the strong and electromagnetic interaction, and ask why the others cannot decay via those two interactions. Well, if they could, they would be an excited state.

This is not exact - you could imagine processes like $ucc \to udc + c\bar{d}$. But the masses of the corresponding particles are dominated by the quark masses, plus some binding energy for the new meson. Therefore, I don't think those processes are possible (and baryons with two heavy quarks are hard to study anyway).

 

[snorkack]

This is not exact - you could imagine processes like $ucc \to udc + c\bar{d}$. But the masses of the corresponding particles are dominated by the quark masses, plus some binding energy for the new meson. Therefore, I don't think those processes are possible (and baryons with two heavy quarks are hard to study anyway).

These processes are eminently possible.

Δ resonances include $uuu$ (Δ++) and $ddd$ (Δ-). These are NOT "excited states" of anything - there are no lower states for $uuu$ or $ddd$. Yet they are resonances. Why? Because their mass is about 1232 MeV, which is about 293 MeV above the lowest available states of $uud$ (namely p) and 292 MeV above the lowest available state of $udd$ (namely n). Whereas creating a $d\bar{d}$ pair to emit a $u\bar{d}$ (π+) takes mere 139 MeV. So Δ are resonances.

By contrast, the lowest state of $sss$ is at 1672 MeV - it is Ω. The lowest state of $uss$ is at 1315 MeV - it is Ξ°. This means 357 MeV difference - but the problem is, creating a $u\bar{u}$ pair to emit a $s\bar{u}$ (K-) takes 494 MeV. Therefore, omega hyperon, unlike either delta, is stable against all strong processes and is not a resonance.

Does anyone know the exact masses of $ccc$ and $bbb$ lowest states, compared to states like $ucc$ and $ubb$, and whether or not they are stable against strong emissions of D and B mesons respectively?

 

[mfb]

Oh, a nice example, thanks.

Does anyone know the exact masses of $ccc$ and $bbb$ lowest states, compared to states like $ucc$ and $ubb$, and whether or not they are stable against strong emissions of D and B mesons respectively?

3 heavy quarks? I hope you don't look for experimental results. There is http://pdglive.lbl.gov/Rsummary.brl?nodein=S065 for a doubly charmed baryon (and some contradicting evidence against that), and nothing else.