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

by Spinnor
Tags: color, force, function, proton, relative, size, strength
P: 367
 Quote by Spinnor 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.
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...

 Quote by The_Duck 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.
Really? we've gone this far? And they keep teaching us the Isospin conservation of the strong forces :P hahahaa
P: 1,204
 Quote by snorkack 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?
Negligible (less than 1%)
P: 353
 Quote by dauto Negligible (less than 1%)
Mass difference is 0,13 % (neutron is bigger), but it is very significant.
So what is the difference in size?
 P: 367 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....
P: 1,360
 Quote by ChrisVer 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... ...
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!
 P: 367 When you introduce the QCD in your Lagrangian $L_{QCD}=\bar{ψ} (i γ^{μ} D_{μ} -m ) ψ$ (Dirac's part) $-G_{μκ}^{a}G^{μκ}_{a}/4$ (the gauge bosons interaction terms) Where you have the: a) Covariant Derivative $D_{μ}= ∂_{μ}+ i g_{3} T^{a} A_{μ}^{a} = ∂_{μ}+ i g_{3} λ^{a} A_{μ}^{a}/2$ where $λ^{a}$ are the generators of $SU(3)$ thus they can be represented as the Gellmann matrices. b) The gauge bosons antisymmetric tensor $G_{μκ}^{a}= ∂_{μ}A_{κ}^{a}-∂_{κ}A_{μ}^{a}-g_{3}f^{abc} A_{μ}^{b}A_{κ}^{c}=A_{μκ}^{a}-g_{3} f^{abc} A_{μ}^{b}A_{κ}^{c}$ $f^{abc}$ the $λ^{a}$ algebra's structure constants... Then you get for your Lagrangian 3 different terms... $L_{QCD}=L_{0}+ L_{g.b.} + L_{int}$ 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 $ψ$ which can be either u,d,c,s,t,b and it can be represented as color triplets... eg $c= [c^{red}, c^{green}, c^{blue}]^{T}$. For the anticolor, you need to work in the adjoint representation of $SU(3)$ Nevermind, to get the color current, you need the interactive Lagrangian: $L_{int}= -g_{3} \bar{ψ} γ^{μ}λ^{a} ψ A_{μ}^{a}/2$ the corresponding conserved current (if you remember from the Dirac's current case) is: $J_{SU(3)}^{μa}= g_{3} \bar{ψ} γ^{μ}(λ^{a}/2) ψ$ 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: $∂_{μ}J_{SU(3)}^{μa}= 0$ and the color charge is: $Q_{c}=\int{d^{3}x J_{SU(3)}^{0a}}$ If they also carry electric charge, you'll get also another current, corresponding to $U(1)_{Q}$ 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: $J_{i}^{μa}= g_{3} \bar{ψ_{i}} γ^{μ}(λ^{a}/2) ψ_{i}$ where $i$ can be each u,d,s,c,b,t quarks, so eg $ψ_{1,2,3,4,5,6}= ψ_{u,d,c,s,t,b}$ In the color representation, then you can also write indices for the $λ$ and $ψ$ such that: $J_{i}^{μa}= g_{3} \bar{ψ_{i}^{k}} γ^{μ}(λ^{a}_{kp}/2) ψ^{p}_{i}$
P: 353
 Quote by ChrisVer 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....
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?
P: 1,360
 Quote by ChrisVer When you introduce the QCD in your Lagrangian $L_{QCD}=\bar{ψ} (i γ^{μ} D_{μ} -m ) ψ$ (Dirac's part) $-G_{μκ}^{a}G^{μκ}_{a}/4$ (the gauge bosons interaction terms) Where you have the: a) Covariant Derivative $D_{μ}= ∂_{μ}+ i g_{3} T^{a} A_{μ}^{a} = ∂_{μ}+ i g_{3} λ^{a} A_{μ}^{a}/2$ where $λ^{a}$ are the generators of $SU(3)$ thus they can be represented as the Gellmann matrices. b) The gauge bosons antisymmetric tensor $G_{μκ}^{a}= ∂_{μ}A_{κ}^{a}-∂_{κ}A_{μ}^{a}-g_{3}f^{abc} A_{μ}^{b}A_{κ}^{c}=A_{μκ}^{a}-g_{3} f^{abc} A_{μ}^{b}A_{κ}^{c}$ $f^{abc}$ the $λ^{a}$ algebra's structure constants... Then you get for your Lagrangian 3 different terms... $L_{QCD}=L_{0}+ L_{g.b.} + L_{int}$ 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 $ψ$ which can be either u,d,c,s,t,b and it can be represented as color triplets... eg $c= [c^{red}, c^{green}, c^{blue}]^{T}$. For the anticolor, you need to work in the adjoint representation of $SU(3)$ Nevermind, to get the color current, you need the interactive Lagrangian: $L_{int}= -g_{3} \bar{ψ} γ^{μ}λ^{a} ψ A_{μ}^{a}/2$ the corresponding conserved current (if you remember from the Dirac's current case) is: $J_{SU(3)}^{μa}= g_{3} \bar{ψ} γ^{μ}(λ^{a}/2) ψ$ 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: $∂_{μ}J_{SU(3)}^{μa}= 0$ and the color charge is: $Q_{c}=\int{d^{3}x J_{SU(3)}^{0a}}$ If they also carry electric charge, you'll get also another current, corresponding to $U(1)_{Q}$ 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: $J_{i}^{μa}= g_{3} \bar{ψ_{i}} γ^{μ}(λ^{a}/2) ψ_{i}$ where $i$ can be each u,d,s,c,b,t quarks, so eg $ψ_{1,2,3,4,5,6}= ψ_{u,d,c,s,t,b}$ In the color representation, then you can also write indices for the $λ$ and $ψ$ such that: $J_{i}^{μa}= g_{3} \bar{ψ_{i}^{k}} γ^{μ}(λ^{a}_{kp}/2) ψ^{p}_{i}$
Who needs books, the physics is right above, very nice, thank you!

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?

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.?

 P: 1,360 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!
P: 367
 Quote by Spinnor Who needs books, the physics is right above, very nice, thank you! 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!
The 8 conserved currents is not the individual colors, but some mixing of colors... The colors come from the components of the fields ψ as I told you, and how those color mix depends on the choice of the λ and so you'll get the associated current... The gluons come because of this λ as you can see from the $L_{inter}$.

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..
P: 1,360
 Quote by ChrisVer The 8 conserved currents is not the individual colors, but some mixing of colors... The colors come from the components of the fields ψ as I told you, and how those color mix depends on the choice of the λ and so you'll get the associated current... The gluons come because of this λ as you can see from the $L_{inter}$. 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..
I'm trying to find the most simple example in principle that has the right physics, can any of my questions be realized in principle in this thought experiment? Color currents in a ring, yes or no? Of the form I suggested in the last post.
 P: 367 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 color-neutral 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) )
P: 1,360
 Quote by ChrisVer 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 color-neutral particles... ...
We are allowing the strong coupling constant to get as weak as we please here.
 P: 367 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
 P: 1,360 I wanted this picture in here somewhere, http://www.upscale.utoronto.ca/Gener...F23/Lect23.htm

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