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Size of a proton as a function of relative strength of color force?

  1. Dec 10, 2013 #1
    If we could set the strong force to be ten times weaker (compared with the electromagnetic force) how much would the radius of the proton change if any?

    In the limit that the strength of the strong force goes to zero does the proton radius approach some limit?

    Thanks for any help!
  2. jcsd
  3. Dec 10, 2013 #2
    If the strong force strength was zero there would be no proton (infinite radius).
  4. Dec 10, 2013 #3
    If that is true then if we could slowly reduce the strength of the strong force then matter would undergo a "phase" change where quarks were no longer confined?

    How would a single hydrogen atom in a box change as the strength of the strong force went to zero?

    Two up quarks, a down quark, and an electron could still form a charge neutral bound system in some limit where the strong force goes to zero?

    An excited state of such a bound system could emit a photon or a gluon?

    Thanks for any help!
  5. Dec 10, 2013 #4


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    I am having a weird feeling coming from your questions...
    A zero strong force, as stated above would mean the destruction of nuclei .... A proton would not be a good choice of a system, because what would keep the quarks together? What keeps them together, although they have charges, is the strong interactions... and the same is true for nuclei as well... Without that interaction, they would break up in their constitutes ...

    What is the limit that this can hold? probably depends on what you want to say... For example, you get an idea of the relation between strong forces and electromagnetic, by checking a nuclei... A rough calculation can be that you know that the radius of a nuclei is of magnitude ~1fm, and you just say you'd like at that level the electromagnetic repulse of the protons to be equal to their strong attraction.... then I guess you'll get a "ratio" between strong and electromagnetic interaction strengths.... then by lowering the strong interaction you'd raise the radius, in the limit of zero strong interaction, you'll get that there won't be a bound state....
    I guess there can be more explicit calculations taking more into account...

    How could they make a neutral bound system? that is very unlikely to happen, and if it does I don't think you can have a stable state... In the way that you can't excite it (it would break everything appart). But if you send the strong interaction to zero, then there won't be gluons for sure... Photons will exist because you have charges

    If I am somewhere wrong, let me know...
    Last edited: Dec 10, 2013
  6. Dec 10, 2013 #5
    If we had four charged particles of charge -1, -1/3, 2/3, and 2/3 won't they minimize electrostatic energy and form a bound system (allowing the strong force to get very weak)?
  7. Dec 10, 2013 #6


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    they could.... but as I said even if this happened, it would be terribly unstable.... the fact that in order to have a complete cancel out of the charges you'd have to put them in a very specific way to get the "minimized energy" which of course I am not even sure if it's enough for a bound state...
    To get a complete cancelation of charges you'd have to put in a point A the -1, -1/3 (impossible) and shoot around with enough momentum the 2/3, 2/3... and that's a classical way of seeing it
  8. Dec 10, 2013 #7
    The size of the proton will increase very rapidly as you decrease the strength of the strong force. But as long as the coupling constant is positive, the proton will be bound because of the phenomenon of quark confinement.

    Actually the strength of the strong force depends on the distance scale. The "coupling constant" that defines the strength of the force increases as the distance scale you are examining increases. For instance at a separation of ##10^{-18}## meters the strong force is much weaker than it is at ##10^{-16}## meters. At about ##10^{-15}## meters the coupling constant, by a certain measure, actually blows up to infinity. This length scale where the coupling constant blows up sets the size of hadrons, including the proton: the proton's radius is about ##10^{-15}## meters.

    Now, when we think about "making the strong force ten times weaker," what we presumably mean is that at a fixed length scale--say, ##10^{-18}## meters--we decrease the coupling constant by a factor of 10. This will delay the blow-up of the coupling constant to a larger length scale, and thus will make the proton larger. The dependence of the radius ##r## on the coupling constant ##g## [measured at a fixed length scale] is quite drastic; I think to first order it should look like

    [itex]r \sim e^{c/g}[/itex]

    for some constant ##c##. The exponential means that ##r## increases quite rapidly as ##g## decreases. As ##g \to 0##, ##r \to \infty##.

    No. Quarks will always be confined as long as ##g > 0##. And if you set ##g = 0## then there is no strong force. Confinement has a certain length scale, though--basically, the maximum distance color charges are allowed to separate--which is approximately equal to the proton radius and which will increase as the proton radius increases.

    At first, the radius of the proton would increase without incident. Once the proton radius became comparable to the Bohr radius of the electron wave function things would get interesting. I guess the system would become quite complicated to analyze.

    Yes, in the limit where the strong force is extremely weak [i.e., proton radius much greater than Bohr radius], we could mostly neglect the strong force compared to the electromagnetic force. Then we'd have to analyze the hydrogen atom as an electromagnetic bound state of four particles of charge 2/3, 2/3, -1/3, and -1. This system is definitely bound electromagnetically.

    Yes, but gluons are still confined which essentially sets a maximum distance such a gluon could travel before being absorbed (which could be much greater than the radius of the "hydrogen atom" if the strong force is very weak).
    Last edited: Dec 10, 2013
  9. Dec 11, 2013 #8


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    is there quark confinement without strong interactions?
    Also what kind of state would that be? they are totally asymmetric
  10. Dec 11, 2013 #9

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    It's even stronger, I think: g2 rather than g.
  11. Dec 11, 2013 #10
    Thank you for the details! Interesting!
  12. Dec 11, 2013 #11
    Maybe a follow up question. Let the strong force be small enough so that the following question makes sense (assuming it can make sense),

    Let there be three dipole antennas which occupy nearly the same space with the same orientation. Let these dipole antennas have oscillating color currents of the strong force, red, anti-red, blue, anti-blue, green, and anti-green, with adjustable phase and amplitude (this can't be done but is only a thought experiment to maybe understand the color forces better). When say blue color charge was maximum at one end of the antenna there would be an equal amount of anti-blue color charge at the other end.

    With proper phase and amplitude modulation can we produce the eight types (is it nine if color is not confined) of dipole gluon radiation?

    Thanks for any help!

    Edit, gluons carry color charge so the above won"t work?

    2nd edit, so color and anti color charges must "bleed" off the antennas?
    Last edited: Dec 11, 2013
  13. Dec 11, 2013 #12
    So if say a red charge jumps from the red charged end of the red/anti-red antenna to the green or blue charged end of the green/anti-green or the blue/anti-blue antenna Feynmans diagrams for the color force tell us the red charge can emit a gluon and change color. And like wise for the oppositely charged ends. Charges flow from end to nearby end of the antennas?

    Thanks for any help!
    Last edited: Dec 11, 2013
  14. Dec 12, 2013 #13


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    I'm trying to understand your question... can you please simplify it?
    what do you mean by charged antennas?
  15. Dec 13, 2013 #14
    Compare proton with neutron.
    A neutron is bound by electromagnetic interaction alone. One particle of charge +2/3, and two particles of -1/3 each. Just like a helium atom only with 1/3 the charges.

    Considering the masses of quarks, what would the size of neutron be in absence of strong force? What is the binding energy?

    A proton is not bound by electromagnetic interaction, because it can split up into a particle charge +2/3 and a bound system of +2/3 and -1/3 totalling +1/3. These repel at long distances.

    If strong interaction were of negligible strength compared to electromagnetic then a proton would exist because of colour confinement but would be huge compared to neutron. Whereas a neutron would be only slightly smaller than it would be in complete absence of colour force.

    How much does the size of neutron and proton differ now, at the present strength of colour force?
  16. Dec 13, 2013 #15
    Nope, the neutron - like the proton - is bound by the strong force. The proton and neutron are about the same size.
  17. Dec 13, 2013 #16

    Take Maxwell's classical theory of electromagnetism and change it so we get a similar theory (if possible) using the fundamentals of the color force, three types of charge, red, blue, and green and their opposites anti-red, anti-blue, and anti-green. We would presumably have three conserved currents. Changing color currents give rise to radiation? Such an imaginary classical theory of a very weak strong force would allow us to think of gluon dipole radiation produced by oscillating color currents. My guess was we might have 8 (or 9) types of gluon dipole radiation by proper modulation of the color currents. A big difference with classical electrodynamics is that photons have no charge while gluons have equal amounts of color/anti-color charge. This makes the thought experiment a little harder.
  18. Dec 13, 2013 #17
    Yes, both are bound mainly by colour force.
    However, electromagnetic force mainly works to contract the neutron against the colour force, and it mainly works to expand the proton against the colour force. What is the resulting size difference?
  19. Dec 13, 2013 #18
    There is also a difference between protons and neutrons coming from the fact that up and down quarks have different masses as well as different charges. Both electromagnetic and quark mass difference effects are of order 1% or less. So presumably any size difference between the proton and neutron is of this order. For example, these effects are responsible for the mass splitting between the proton and neutron which is about 0.1% [this is a little smaller than you might expect because there is an approximate cancellation between the quark mass effects and the electromagnetic effects]. Lattice QCD calculations of hadron properties are starting to take into account these "isospin breaking" effects, which had previously been neglected because they were dominated by other errors.
  20. Dec 13, 2013 #19


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    That means that your experiment is theoretical (in principle up to what I know about strong forces)... First of all, as you said the gluons carry color charge. Also you cannot isolate these charges, but they will come in such a configuration that you'll always have neutral things.
    I don't think the strong charge can be in any way compared to that of EM... They have totally different properties in all levels... Even the reason why we chose to consider the existence of that charge (If I recall well, from the reaction ratios of quarks) was because we got triple what we expected...
    Gluons are somewhat difficult to see in analogy to photon because of the above. They are also coming from different symmetries and are subject to different interactions (as a result of the non-abelianity of the symmetry they come from)
    If someone knows anything better than me, I hope they can help you...

    Really? we've gone this far? And they keep teaching us the Isospin conservation of the strong forces :P hahahaa
  21. Dec 13, 2013 #20
    Negligible (less than 1%)
  22. Dec 13, 2013 #21
    Mass difference is 0,13 % (neutron is bigger), but it is very significant.
    So what is the difference in size?
  23. Dec 13, 2013 #22


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    the difference of mass of the neutron to that of the proton is not only associated to the difference of the mass of up and down quark... of course that's I guess the dominating factor... But there are also corrections due to EM interactions on different charge constituents... I.e. they make the neutron less massive I guess....
  24. Dec 14, 2013 #23
    Thank you! How would you even create color currents, time changing color magnetic fields induce time changing color electric fields? Might there be 3 types of magnets?

    Edit, or is it 8 or 9 types of color magnet?

    I guess one way to solve this problem is to compare the Lagrangian for the strong force (very weak strong force) with interaction with the Lagrangian for Electromagnetism with interaction, the latter gives rise to Maxwell's equations via The Euler-Lagrange equations?

    I'm now guessing we might only need one antenna that can simultaneously produce a triplet of currents,

    J_color = j_r + j_b + j_g =

    j_ro*exp(-iωt +iθ_r) + j_bo*exp(-iωt +iθ_b) + j_go*exp(-iωt +iθ_g)

    Is there a continuity equation for J_color?

    Now if the particles that carry color charge also carry electric charge my problem gets even harder? Maybe to simplify things let the particles carry only color charge and no electric charge?

    I think there is bound to be both similarities and differences with electromagnetism.

    Thanks for all the help, I guess I need to study!
    Last edited: Dec 14, 2013
  25. Dec 14, 2013 #24


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    When you introduce the QCD in your Lagrangian
    [itex]L_{QCD}=\bar{ψ} (i γ^{μ} D_{μ} -m ) ψ[/itex] (Dirac's part) [itex]-G_{μκ}^{a}G^{μκ}_{a}/4[/itex] (the gauge bosons interaction terms)

    Where you have the:
    a) Covariant Derivative
    [itex]D_{μ}= ∂_{μ}+ i g_{3} T^{a} A_{μ}^{a} = ∂_{μ}+ i g_{3} λ^{a} A_{μ}^{a}/2[/itex]
    where [itex]λ^{a}[/itex] are the generators of [itex]SU(3)[/itex] thus they can be represented as the Gellmann matrices.

    b) The gauge bosons antisymmetric tensor
    [itex] G_{μκ}^{a}= ∂_{μ}A_{κ}^{a}-∂_{κ}A_{μ}^{a}-g_{3}f^{abc} A_{μ}^{b}A_{κ}^{c}=A_{μκ}^{a}-g_{3} f^{abc} A_{μ}^{b}A_{κ}^{c}[/itex]
    [itex]f^{abc}[/itex] the [itex]λ^{a}[/itex] algebra's structure constants...

    Then you get for your Lagrangian 3 different terms...

    [itex]L_{QCD}=L_{0}+ L_{g.b.} + L_{int} [/itex]

    1st term is the free langrangian term, the 2nd term is the gauge bosons self interaction term, and the last is the interaction of bosons with your quark fields [itex]ψ[/itex] which can be either u,d,c,s,t,b and it can be represented as color triplets...
    eg [itex] c= [c^{red}, c^{green}, c^{blue}]^{T} [/itex]. For the anticolor, you need to work in the adjoint representation of [itex]SU(3)[/itex]
    Nevermind, to get the color current, you need the interactive Lagrangian:

    [itex]L_{int}= -g_{3} \bar{ψ} γ^{μ}λ^{a} ψ A_{μ}^{a}/2 [/itex]

    the corresponding conserved current (if you remember from the Dirac's current case) is:
    [itex] J_{SU(3)}^{μa}= g_{3} \bar{ψ} γ^{μ}(λ^{a}/2) ψ[/itex]

    What can we see from that? That we have 8 conserved currents. Each of them is individually conserved. The continuity relation for the currents, is given by their conservation, thus you have again 8 different continuity relations:
    [itex]∂_{μ}J_{SU(3)}^{μa}= 0 [/itex]

    and the color charge is:
    [itex] Q_{c}=\int{d^{3}x J_{SU(3)}^{0a}}[/itex]

    If they also carry electric charge, you'll get also another current, corresponding to [itex] U(1)_{Q} [/itex] interaction...

    I am not sure for this, if it's wrong someone please correct me:
    If you now leave from the case of a single quark, and you want to put all the quarks in the game, then you should put indices on the quark fields...So the current will:
    [itex] J_{i}^{μa}= g_{3} \bar{ψ_{i}} γ^{μ}(λ^{a}/2) ψ_{i}[/itex]
    where [itex] i [/itex] can be each u,d,s,c,b,t quarks, so eg [itex]ψ_{1,2,3,4,5,6}= ψ_{u,d,c,s,t,b}[/itex]
    In the color representation, then you can also write indices for the [itex]λ[/itex] and [itex]ψ[/itex] such that:
    [itex] J_{i}^{μa}= g_{3} \bar{ψ_{i}^{k}} γ^{μ}(λ^{a}_{kp}/2) ψ^{p}_{i}[/itex]
    Last edited: Dec 14, 2013
  26. Dec 14, 2013 #25
    I see the estimates of 2,8 MeV mass difference of down and up. So the repulsion between the ups in proton makes it 1,5 MeV heavier.

    So, on the limit of no strong force whatsoever, what would be the binding energy of a neutron?
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