Is there the smallest particle that can be guaranteed to be unable to be broken down into smaller particles?
In particle physics, you have to be specific about what you mean by breaking down a particle: A neutron in an atom can decay into a proton, an electron, and an electron-antineutrino. But this does not mean that a neutron is made of a proton, an electron and an electron-antineutrino. What the neutron is made of are three quarks (one up and two down).
If by unable to break down into smaller particles you mean that a particle has no internal structure, then our current understanding is that electrons, muons, taus, their corresponding neutrinos, all the quarks, as well as the photon, the gluon, the Z- and the W-boson, together with all the anti-particles, are all elementary particles. ,
Can this be guaranteed? No. Nothing really can be guaranteed in physics. Ever finer measurements mean that we are ever more sure about the elementary nature of, e.g., the electron. We can conclude that at the length-scales currently accessible for probing, the electron has no internal structure. This does not rule out that a structure might be found at tinier scales. This is what happened with protons and neutrons, where scattering experiments gave results that were inconsistent with assuming a point-particle with no internal structure.
Note that, although the aforementioned particles are counted as elementary particles, they can decay: See Muon Decay
A particle can only decay into particles with smaller (rest) masses than itself. So the lowest-mass particle can't decay.
Since there are zero-mass particles, that argument by itself doesn't suffice to show that there are any massive particles that can't decay, but we can strengthen the argument a bit to show that there are. Because charge is conserved, the decay products of a charged particle must include at least one charged particle. So the lowest-mass charged particle can't decay. That's why electrons are stable.
The same argument applies to any conserved quantity, not just electric charge. In (some? all?) supersymmetric theories, there is a conserved quantity called R-parity. The lightest particle with nontrivial R-parity must be stable. That means that such theories always have an additional stable particle, besides the ones we know about. That particle is often called the lightest supersymmetric particle (LSP). In these theories, particles like this should have been produced in the early Universe. Since the LSP can't decay, it should still be around today. The LSP is one of the most popular candidates to be the dark matter in the Universe.
I would like to add to the previous answers that compositeness of the elementary particles of the standard model is an ongoing research question, not favored by theoriticians.
CMS has a preprint out where they are searching for compositeness in dijet angular distributions.
The measured dijet angular distributions can be used to set limits on quark compositeness represented by a four-fermion contact interaction term in addition to the QCD Lagrangian.
They have set limits.
That depends on what you call a particle.
And what you mean by breaking down, exactly, In the context of this definition of particle, of course.
And, most importantly, what you mean by guarantee.
Or it can not be answered in a meaningful way.
(I see no obvious definition of "guarantee" that causes the trivial case to work out)
There will never be the last tiny particle. The particle is only as tiny as we can break it otherwise if we have instruments to break it further, there will never be a tiniest particle it is only as tiny as our limitation of instrument of breaking it more.
protected by Qmechanic♦ Feb 3 '13 at 20:42
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