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Difference between the weak & strong nuclear forces?

  1. Oct 22, 2011 #1
    Hi I don't understand my textbook 100% when it comes to this:

    Weak nuclear force: The force that affects leptons and quarks. Uhm, how?

    They tried to illustrate this with an example of a beta-decay, but I didn't get it. Some W- particle was created when an d quark went over to an u quark, and the W particle again turned into an electron and an anti neutrino. Question: Does an u quark have less, uhm, potential energy than a d quark? I think I remember that if protons receive enough energy they'll turn into a positron + a neutron.

    Strong nuclear force: Apparently the force particle, gluon, can "glue" quarks, and thus protons&neutrons, closer together the further away they are, until a certain limit where the gluon has no influence any-more, correct? If the strong nuclear force gets stronger over distance until a certain point, why are the heavy radioactive atoms so unstable? I mean there are neutrons&protons all over the place to counteract the EM force, kind of weird that they would lose the battle of keeping the atom against the EM force.

    Also some more questions: In the model for alpha decay they say that alpha particles are down in an "energy well" waiting to get out and they start talking about stuff like borrowing energy from nowhere and then paying it back later (good thing nature doesn't take rents- haha). Two questions: 1) wouldn't it be more correct to say that the alpha particle has a certain probability of "breaking out", with that probability rising with the particle's kinetic & potential energy?

    2) What if the alpha particle has absolutely zero kinetic or potential energy, and it needs 1 Joule to break out? It borrows 1 joule to break out, and it is out of the alpha particle. But then it has zero kinetic or potential energy, because it has broken out.. What will it pay back the bill with (maybe it has its mass as a security :) )?
  2. jcsd
  3. Oct 22, 2011 #2
    The first thing about the weak force is it's easier at first if you stop thinking of "force" as in Newton's second law. (It does act like that sometime, but that's not its most prominent feature.) Probably better to stay with the term "interaction" when studying particle physics.

    The much more commonly relevant feature of the weak interaction is that it has a component - the "charged current" - which changes the identity of the incoming particles. Hence the down quark changing to the up quark and giving off the (virtual) W- boson.

    The decay of the W- into an electron and an antineutrino is actually equivalent to a W- interacting with an incoming neutrino and changing that to an electron. On a http://en.wikipedia.org/wiki/Feynman_diagram" [Broken], an incoming particle can always be replaced with an outgoing antiparticle (of the corresponding variety) or vice versa. This applies to all Standard Model interactions.

    Note how this interaction preserves electric charge. The initial d quark has -1/3. It gives off a u quark (+2/3) and the W-, which has -1 (hence its superscript). The incoming neutrino (or outgoing anti-) has zero and the electron of course -1. Thus the W- "takes" -1 unit of charge away from the quark and "gives" it to the new electron.

    Moving to the strong interactions, it's important to understand the difference between the attraction between individual quarks within a hadron, which is mediated by the gluons, and the attraction that holds the protons and neutrons together within atomic nuclei. The latter is a more complex phenomenon: it can be described approximately as being mediated by mesons, and only stems somewhat indirectly from the behaviour of the gluons. But what happens when something pushes strongly-bound particles apart depends crucially on which of these two scenarios you are considering!

    When nucleons (or groups thereof) are pushed apart by the electromagnetic forces, as in alpha decay, the strength of the strong force does indeed fall off rapidly beyond distances of the order of the size of the nucleon, and (subject to energy availability - see below) the alpha or neutron can freely escape from its erstwhile nucleus.

    But when quarks within a hadron are pushed apart (eg by a collision in a particle accelerator), something quite different happens. In this case, instead of reducing with distance as per the familiar inverse square laws, the strong force remains roughly constant however far apart the quarks are pushed. This means that to separate them further you have to keep putting in more and more energy. Before long, so much energy has had to be supplied that it is sufficient to promote a new quark/anti-quark pair into reality from the vacuum! By linking up with the quarks being pushed apart, these newly created particles satiate the strong force, and only then can the resulting particles separate freely. This process is termed hadronisation, because it only ever results in multi-quark particles, which are termed hadrons, being able to freely appear. (NB I am omitting a lot of technical detail here.)

    Your answer to Q1 is roughly right. In fact the probability (of an alpha decay) has to be defined over a nominated interval of time, as given infinite time all of the available particles would decay. The probablility is more usually expressed as the lifetime or half-life of the particle. For example, http://en.wikipedia.org/wiki/Uranium-238" [Broken].)

    The above already takes account of the uncertain momentum of the alpha when confined within its larger parent nucleus. By the Heisenberg principle, the limitation to its possible positions (because of being bound within its nucleus) implies a corresponding uncertainty in its momentum.

    Re Q2, the alpha can only break out at all if it has enough energy that it won't be left with a "debt" afterwards. If it has some kinetic energy "in its pocket" beforehand, the "Heisenberg bank" will of course accept that, but it must have at least enough in total to account for its rest energy (m0c2). In practice, alpha decays occur because the total energy bound up in the electromagnetic repulsion between the protons is sufficient to overcome the binding energy of the strong force. More energy is released by not having to hold two extra protons in than is consumed in overcoming the strong force. Thus the final state will always have less rest energy than the original nucleus, the remainder being accounted for by the kinetic energy the particles (mainly, the alpha) gain.

    This is also why alpha particles only travel a few cm in air. The spare electomagnetuc enery available from the nucleus is only limited, so they only end up with a modest amount of kinetic energy, which their collisions with air molecules then steadily dissipates.
    Last edited by a moderator: May 5, 2017
  4. Oct 23, 2011 #3
    thanks for the reply, you really made me understand allot. So the weak force is actually primarily just turning stuff into other stuff?

    Wait.. If the W- particle gives up an anti neutrino, it would need much energy (the rest energy of a neutrino).. While if it merely absorbed a neutrino it would gain energy.. So what is physically happening? Is a W- particle decaying into an anti neutrino + electron or is it turning a neutrino into an electron?

    You mean the potential electric energy + spare kinetic energy simply overcome the the strong force?

    No energy is released when the binding gets broken, the breaking of the binding requires energy,, right?
  5. Oct 24, 2011 #4
    At low energies, mainly. The rest mass of the W bosons is about 80Gev/c2 so even when you get enough energy to create real ones, they still decay pretty rapidly. They always have to decay into a fermion and an anti-fermion, either a quark/antiquark or a lepton/antilepton, and the particle will always be of the opposite "flavour type" of the antiparticle.

    Eg W- can decay into e- + anti-ve or μ- + anti-vμ or τ- + anti-vτ or d + anti-u or d + anti-c or s + anti-u etc... There are more possible quark/antiquark combinations because the quark and antiquark can be of a different "generations". This is not allowed with leptons: the (anti-)neutrino must always be the same generation as the charged lepton.
    It can do either. Remember the neutrino energy is usually small compared to the W rest mass. The interaction vertex always comprises two fermions and the one W boson. But the interaction can comprise either one incoming particle "decaying" into two outgoing ones, or two incoming ones "colliding" to form one outgoing one. As I mentioned, an incoming particle can be replaced with an outgoing antiparticle or vice versa.
    For the strong force, yes, but the electomagnetic energy released is greater and hence the alpha particle can break free.
  6. Oct 24, 2011 #5
    About the beta decay: Where does the energy come from? A down quark turns into an up quark in a neutrino, a W- particle is released and the W- particle decays into an anti neutrino + an electron.

    Same with a proton turning into a neutron and a W+ particle being released.. Where is all the energy coming from? Is there some built up potential energy?
  7. Oct 24, 2011 #6
    Though we don't know their masses very precisely, there is considerable evidence that down quarks are heavier than ups, and the difference is at least 1MeV. The neutron is about 1MeV heavier than the proton, despite the fact that the repulsive (like) electric charges in the neutron are less than the protons. (In neutrons, the strong force has to hold two -1/3 ds together, with help from a +2/3 u; in protons two +2/3 us have to be kept in similar proximity with only a -1/3 d to help out.)

    As the electon weighs in at a mere 0.5MeV, and the neutrino is lighter still, the extra mass of the d as compared to u provides sufficient energy to fuel the decays where beta decays occur.

    But this is not the whole story. If it were, basic isotopes like 2H/3He/4He would not be stable! With some of these, eg 3He or 4He, if a d were to change to u, changing a neutron to a proton, the overall energy of those isotopes would then be increased because the Pauli exclusion principle would force the new proton to have to jump to a higher energy level (spin/orbital state). Nuclei can only have two protons and two neutrons at the lowest energy level (one of each spin up, one of each spin down - this works just like electrons in atomic orbitals).

    In the other cases, such as 2H, if the neutron changed to a proton that would mean you now had two + charges to hold together, and the extra electromagnetic binding energy that would need is enough to offset the mass difference between the d and u.
    Free protons don't spontaneously change to neutrons. If you are talking about "reverse" beta decay, the energy again comes from reduced electromagnetic binding energy due to having one fewer + to hold together. In these isotopes the reduction in the electromagnetic energy evidently more than offsets the quark mass difference.
  8. Oct 25, 2011 #7
    I don't know about that Pauli thing but:

    Well, so what? The Helium nuclei would rip itself apart, is it that much of an issue?

    Though I understand that the d quark has a higher amount of potential energy than an u quark, but it won't just decay randomly into an u quark unless there is something triggering it?

    Anyways, thanks for helping me with this stuff
  9. Oct 26, 2011 #8
    Well, 3He and 4He are stable, so that clearly doesn't happen.
    What happens all depends on the total (ground state) energy level of the nuclear isotopes (or other particles) in question. If one isotope can decay to another with lower ground state energy by converting a neutron into a proton or vice versa, and the difference in energy levels is >= the rest mass of an electon/positron, then that is what happens. The ground state energy levels themselves are of course partly determined by the strong force binding energy and electromagnetic charge confinement energy of the isotopes in question.

    Until you get up to the heavy radioisotopes where the strong force starts struggling to contain the electromagnetic repulsion, once protons and neutrons are combined in a nucleus they do not usually separate spontaneously. (An exception to this is 2He, which does decay into 2 x 1H.)

    So for nuclei where the number N of protos + neutrons is less than 209, the strong force prevents N from being reduced. What you then have to look at is what the lowest energy isotope with N nucleons is. Any heavier one will beta-decay to this providing the differences in energy levels between isotopes with adjacent Z (proton) numbers are each more than an electon mass.

    For N = 2, the lowest energy isotope is 2H. As noted above, 2He spontaneously disintegrates. The other possible combination, which would be 2 neutrons combined with no protons, would beta-decay to 2H.

    For N = 3, there are four potentially possible combinations: 3Li, 3He, 3H and 3 neutrons. The Pauli principle would force the third nucleon of 3Li or 3 x n up into the second 'shell', at some cost in energy terms, so these combinations don't occur in practice. Of the remaining two, tritium is heavier so beta-decays into 3He (with a half-life of about 10 years).

    This is why there are some chemical elements - like Helium - that have more than one stable isotope. It's simply because both/all of those N values have lowest stable isotopes that share the same Z value.

    Free neutrons, where no nuclear binding energy or electromagnetic repulsion between protons are involved, beta-decay to protons with a half-life of about 10 mins. Neutrons are about 1MeV heavier than protons, and the electron mass is about 0.5MeV, so sufficient energy is available for the decay.

    Free d quarks have never been observed, so it's impossible to say for certain what would happen if they existed. But I suspect the quark mass difference itself would suffice to allow them to decay into u's.
    No worries, this is also helping me understand these things better.
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