
#19
Dec1313, 01:08 PM

P: 376

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 nonabelianity of the symmetry they come from) If someone knows anything better than me, I hope they can help you... 



#20
Dec1313, 01:50 PM

P: 1,272





#21
Dec1313, 02:56 PM

P: 360

So what is the difference in size? 



#22
Dec1313, 03:54 PM

P: 376

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....




#23
Dec1413, 08:03 AM

P: 1,362

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 EulerLagrange 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! 



#24
Dec1413, 09:19 AM

P: 376

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] 



#25
Dec1413, 12:08 PM

P: 360

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



#26
Dec1513, 11:52 AM

P: 1,362

8 currents then. I thought we gain 1 current because color is no longer confined? Edit, color is no longer confined up to some radius r? Follow up, Can Feynman diagrams for the process colored quark emits a gluon (8 types) and changes color (or does not change color?) be grouped to represent the 8 fundamental currents of the color force? Edit, is there a graphical representation of the 8 fundmental currents involving colored quarks? Can I produce the 8 types of fundamental currents with the J I wrote down above, 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) ? What I wrote down has the same group properties as the 3D harmonic osscilator, SU(3)? We can vary three phases and three amplitudes of the J above as we can do with a 3D H.O.? Thanks your help! 



#27
Dec1513, 01:04 PM

P: 1,362

Let the current I gave represent osscilating color current in a ring, does that produce gluon radiation?
Now let the currents in the ring above not change in time, let the total current be the sum of three indpendant color currents, j_r, j_b, and j_g, is there a static color magnetic field produced? How many types of color magnetic dipoles are there (assuming such a question makes sense)? Thanks for any help! 



#28
Dec1513, 01:33 PM

P: 376

I am trying to understand why are you trying oscilation solution for your color current :P The current is the one I wrote above, you can try to choose a particular λ, and maybe the up quark with the three colors, do the matrix multiplication and find it out. Of course you will get: J~ Jr +Jb+ Jg, but each of them is not individually conserved, rather their whole configuration that is a colorless mixing.. 



#29
Dec1513, 01:56 PM

P: 1,362





#30
Dec1513, 02:05 PM

P: 376

I don't know about that...QCD works differently to EM... so you can't speak of "radiation" of the bosons... This radiation works for the EM because it has infinite range...
QCD doesn't have infinite range, and that's why we never count quarks or gluons... We just count jets of colorneutral particles... As for the harmonic oscilator's SU(3) you stated, I would tell that you should avoid mixing the SU(3) symmetry of the isotropic harmonic oscillator, which is a global symmetry, to that of color which is a local one... To get the idea of what a global SU(3) would mean, I'd say you should think that the reason you need the covariant derivative instead of the partial derivative alone is because your Lagrangian would not be locally SU(3) invariant (although it would always be globally). A global SU(3) would not need the existence of the gauge bosons... By inserting the local SU(3) you get gluon fields (as you also did with U(1) ) 



#31
Dec1513, 02:09 PM

P: 1,362





#32
Dec1513, 02:27 PM

P: 376

You'll always have confinement, even if you have very weak strong coupling constant g due to beta function's negativeness
http://en.wikipedia.org/wiki/Asympto...ptotic_freedom 



#33
Dec1513, 02:57 PM

P: 1,362




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