Many elementary particles decay, for instance a charm quark (according to wikipedia) will decay into a strange quark (and I assume some other elementary particles but I don't know what they are). Does this imply that on some fundamental level both charm quarks and strange quarks share composition?

My (probably naive) reasoning is that in chemistry if I have substance A isolated in a vacuum, and it decays into B and C, then A almost 100% composed of some ratio of B and C, as an example: $$\rm CaCO_3\rightarrow CaO + CO_2$$ which can be used to demonstrate that calcium carbonate consists of the same substances that compose calcium oxide and carbon dioxide. Does this logic not apply on a quantum level?

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    $\begingroup$ Good question tox123. You might want to ask a similar question about proton-antiproton annihilation. See Wikipedia: "The newly created mesons are unstable, and will decay in a series of reactions that ultimately produce nothing but gamma rays, electrons, positrons, and neutrinos”. You can annihilate the electrons with the positrons. Then all the "elementary" quarks and gluons you started with have gone, and all you're left with is photons and neutrinos. $\endgroup$ Commented Apr 16, 2018 at 16:05

3 Answers 3


In particle physics we have a list of "quantum numbers" that describe a particle. Different types of interactions may conserve, or may not conserve, different quantum numbers.

You give the example of decays which change quark flavor. Baryons and mesons (which we model as being made of quarks, even though individual quarks are confined) are assigned "flavor quantum numbers": the $D$ mesons have charm quantum number $C=\pm1$, the $K$ mesons have strangeness $S=\pm1$ but charm $C=0$, and so on. The strong and electromagnetic interactions do not change the flavor quantum numbers in a system, but the weak interaction does. So the stoichiometry skills that you learned in chemistry work for strong-interaction scattering, such as the strangeness-conserving production of hypernuclei, but not for weak decays which change those quantum numbers.

(In fact, you might say that it only makes sense for us to talk about flavor quantum numbers because the interaction that changes them is weak.)

A few of these quantum numbers are conserved by all known interactions. Those include

  • electric charge: the number of positive charges minus the number of negative charges
  • baryon number: the number of protons, neutrons, and hyperons, minus the number of antiprotons, antineutrons, and antihyperons. In the quark model, each quark has baryon number 1/3 (and antiquark $-$1/3), so you can use baryon stoichiometry to analyze reactions where mesons are produced.
  • lepton number: the number of electrons, muons, taus, and neutrinos, minus the number of their antiparticles.

When you do chemical stoichiometry, like in your calcium carbonate decomposition reaction, you're conserving electric charge, the number of electrons, and the number of protons and neutrons. Your baryon number conservation is constrained because there's no interaction at the energies chemists care about which allows protons to change into neutrons or vice-versa, so you have to conserve proton and neutron numbers separately. Furthermore there's no interaction, at the energies that chemists care about, which allows a nucleon to hop from one nucleus to another, so you have to separately conserve the number of calciums, the number of carbons, etc.

It's tempting and useful to take these conservation laws and use them to conclude that a calcium nucleus is "made of" twenty protons and twenty-ish neutrons. But that approach breaks down when you start to consider the flavor-changing weak interactions. The muon decays by the weak interaction into a neutrino, an antineutrino, and an electron; but there's evidence against any model where the muon "contains" those decay products in the way that we can say a nucleus "contains" nucleons.

  • $\begingroup$ This is what I wanted to say, but said better. Luckily I still have a description of non-commuting quantum numbers (in neutrino oscillations), unless you go and trump that as well. $\endgroup$ Commented Apr 16, 2018 at 8:00
  • $\begingroup$ I think I always want quark flavor oscillations (where the Cabbibo mixing angle tells you both about strangeness-changing beta decays and about strangeness-changing kaon oscillations) and neutrino flavor oscillations (where the charged weak current strictly conserves the generational quantum numbers, and the flavor oscillations happen in the neutral sector) to have more parallels than they actually do. So, I wrote my way around it for this answer. $\endgroup$
    – rob
    Commented Apr 16, 2018 at 11:31

No, that logic does not apply. Radioactive decays $-$ particularly those mediated by the weak force $-$ do not need to conserve the numbers of the different types of particles.

What this means is that there are several important quantities, like

  • the number of electrons (minus the number of positrons),
  • the number of muons,
  • the number of strange quarks,
  • the number of charm quarks,

and so on (with the number of antimuons subtracted from the number of muons, etc.) which are generally conserved by time evolution, particularly under that imposed by electromagnetism and the strong nuclear force, but which are not conserved under the weak nuclear force. Thus, when you have a weak-force decay, you really do swap out a strange quark for a charm quark; to the extent that either of those two terms makes sense (i.e. beyond just saying "a quark"), they cannot be said to share a composition.

That said, I should point out that there are indeed some situations some particles really can't (or shouldn't) be said to have a type at all (with neutrino oscillations being a prominent example) but they do not apply to the $c\to s$ decay you mentioned, where the start and end states do have well-defined (different) types.


The two answers are quite correct, describing particle decays. I want to address

can be used to demonstrate that calcium carbide consists of the same substances that compose calcium oxide and carbon dioxide.

In chemistry one can use the classical concept that masses, as measured by scales , are invariant for each substance in the periodic table of elements , and that mass is an additive quantity.

The periodic table of elements though, when looked at by adding the constituent protons and neutrons of each element leads to special relativity,SR, the last entry in the table.

In SR masses are not invariant under Lorenz transformations. The invariant quantity accompanying each particle is called invariant mass , and is the "length" of a four vector given by the energy $E$ and momentum $p$ of the particle. As with usual vectors, lengths are not additive in vector addition, so masses are not additive if in the realm of special relativity, where the study of nuclei belongs.

So one can add the masses of atoms in the periodic table, and consider them classical in behavior, because any special relativity effects due to the electromagnetic bindings of van der Waals and similar forces, are very small to be measurable in chemical reactions. (less than $eV$ chemical versus $MeV$ nuclear in energy ranges)

In elementary particle decays and interactions the energy and momentum fourvectors play a crucial role together with conservation of a number of quantum numbers, as described in the other answers.

The sum of the invariant masses of the particles coming from the decay cannot exceed the mass of the decaying particles, because of energy conservation in the center of mass, so the addition of invariant masses is an upper limit to what a particle can decay into.


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