4 ms·
These are great questions! > Is it the case that a given particle is trying to settle into a "lowest energy state" possible? Not exactly. Energy is conserved
by lisper 4y ago
These are great questions!
> Is it the case that a given particle is trying to settle into a "lowest energy state" possible?
Not exactly. Energy is conserved during these decays. In fact, energy is conserved during all physical processes, so the "lowest energy state possible" is a little bit of a white lie. What makes it a white lie is that it is a very good approximation to the truth for thermodynamic systems, i.e. systems consisting of large numbers of particles. But for quantum systems, it is no longer a good approximation. In quantum systems, what happens is that you have a wave function that describes all of the possible states a system can be in. The more mass the system contains, the more possible states there are in its wave function, and so the more likely it is to end up in some state other than the one it started out in.
It is even possible for the process of decay to reverse itself, and for the constituent particles to come back together and reconstruct the original, but for that to happen all the constituents have to be brought back together, so as a practical matter this never happens spontaneously in nature. In fact, that is the whole reason for building the LHC -- to make particles (protons) come together and make high-mass systems which then decay in interesting ways.
> are these particles, due to the number of options available to them, decaying into the lightest stable variant allowed by the laws of physics?
Not the lightest stable variant, just to one of the possibilities described by that particle's wave function. These will always be subject to the constraints of conservation laws, so the decay products will always be lighter than the original. But which particular set of possible decay products is actually produced in any given decay event is fundamentally random.
> if that is the case, then could we perhaps find ways to engineer structures within which these particles last for a whole lot longer than they should (on a human timescale)?
No. The wave functions for particles are fixed by nature. They are what give particles their identities. They cannot be engineered. The only thing that we can engineer is the arrangement of particles. Particles are like Lego bricks. You can stick them together in lots of different ways, but you can't change the shape of a given brick. Sometimes quantum Lego bricks fall apart spontaneously, but there is no way to control that.
- dotopotoro 4y agoIf neutron is likely to decay into. Why protons dont decay while being part of an atom? (Wouldn’t this be example of a structure that prevents decaying?)
- at_a_remove 4y agoNo. Protons weigh less than neutrons. What will your proton decay into? Be careful to remember your conservation of baryon number when listing your options!
- dotopotoro 4y agoEdit: If neutron is likely to decay into. Why *neutrons* dont decay while being part of an atom? (Wouldn’t this be example of a structure that prevents decaying?)
- lisper 4y agohttps://profmattstrassler.com/articles-and-posts/particle-physics-basics/mass-energy-matter-etc/the-energy-that-holds-things-together/neutron-stability-in-atomic-nuclei/ https://profmattstrassler.com/articles-and-posts/particle-ph...
- davrosthedalek 4y agoNeutrons don't decay while being part of a stable atom, because the atom has actually less energy than the sum of the constituents -- the difference is the binding energy. Look at deuterium, for example. It has a mass of 2.0141 u. A proton alone is 1.0073 u, and a neutron is 1.0087 u. Deuterium is lighter than the mass of proton + neutron. It's also slightly lighter than two protons, so the neutron cannot decay without external energy input.
- DiggyJohnson 4y agoThis comment caps off an exceedingly educational thread of questions and answers.
- wyager 4y agoThis is a direct counter-example to the claim that we cannot engineer a situation where a particle lasts longer than it would in its free state.
- 4y ago
- Gerard0 4y agoI wished I had a friend like you so I could hear these things!
- ngcc_hk 4y agoI thought in q field theory there is on a fundamentally level no particle. They are artificial excite state in a field. Hence the all possible state possibly as it is not the particle as this is already in a state, but a wave. The interaction of fields … wonder if we reframe the q as CSB we have or find new operators to …
- lisper 4y ago> I thought in q field theory there is on a fundamentally level no particle. They are artificial excite state in a field. That is correct. "Particle" is another one of those "white lies." https://arxiv.org/abs/1204.4616 https://arxiv.org/abs/1204.4616
- wesammikhail 4y agoThis was of great help and a fun read. Thank you for the detailed reply.
- lisper 4y agoMy pleasure.