# QCD antiscreening

i'm trying to understand antiscreening in QCD. I'm following Peskin & Schroeder's Introduction to QFT and i'm having trouble to follow their argument. In section 16.17, p542-543, we get an equation that says $$D_i E^{ai} = g \rho^{a},$$ which I'm fine with. then they say that, using $$(D_\mu \phi)^a = \partial _\mu \phi^a + g f^{abc} A_\mu^b \phi^c$$ and $$f^{abc} = \epsilon^{abc}$$ in SU(2), we get $$\partial _i E^{ai} = g \delta^{(3)}(\vec{x})\delta^{a1} + g \epsilon^{abc} A^{bi}E^{ci}.$$ My first question (a) is, I don't understand why there's no minus sign in front of the second right hand term ? Afterwards, they say that the first RHS term gives a $$1/r^2$$ field of type $$a=1$$ radiating from $$\vec{x} = 0$$.

Then, somewhere in space, this field will cross paths with a fluctuation of the vacuum in the form of a bit of vector potential $$A^{ai}$$. "For definiteness, let's assume that this fluctuation has $$a=2$$.

Why can they assume this ? Shouldn't they look at the case where a=3 too ? Then they say the second term on the RHS, with a = 2 for the vector potential and c = 1 from the electric field, will make a field sink where they cross. The new electric field from this sink will be parallel or antiparallel to A depending on which side we are ; there is a source with a=1 closer to the origin and a sink with a=1 farther away.

So my other questions are :

b) how do they choose the values of $$a,b,c$$ ?

c) how do they know there will be a souce on the side of the origin and a sink on the other side ? • It is analyse of equations to understand physical meaning.. – Nikita Jan 1 at 17:09
• thanks i gathered as much. what i don't understand is how they do it ... – Reflets de Lune Jan 1 at 17:16

b) and c) P&S just take these values of the indices as an example. They assume a point charge at the origin of the physical space in the group direction $$a=1$$. They then consider the $$A^{2i}$$ gauge field in the "bubbling'' vacuum. That gauge field then interacts with $$E^{1i}$$ according to the second term in (16.139) and the antisymmetric symbol $$\epsilon^{c12}$$ implies a source field in the $$E^{3i}$$ direction. That electric field will also interact with the $$A^{2i}$$ field, via $$\epsilon^{d23}$$ in the $$E^{1i}$$ direction.
Let us consider the signs - I assume there is a minus sign in (16.139) - and work iteratively. The vacuum disturbance $$A^{2i}$$ implies a contribution to Gauss law in the group direction 3: $$\begin{equation} \partial_i E^{3i} = +\cdots - g \epsilon^{321} A^{2i}E^{1i} = +\cdots - g (-1) A^{2i}E^{1i} = +\cdots + g A^{2i}E^{1i} \end{equation}$$ That that disturbance has an opposite sign of the Gauss Law. This $$E^{3i}$$ then gives a contribution to Gauss Law in direction 1: $$\begin{equation} \partial_i E^{1i} = +\cdots - g \epsilon^{123} A^{2i}(-E^{3i}) = +\cdots + g A^{2i}E^{3i} \end{equation}$$ And so the non-Abelian character of the theory results in an additional electric field contribution on top of the one that comes from the abelian part of the vacuum fluctuations. P&S say that if you work out the details of this - referring to the original literature - then you find that there this vacuum fluctuation results in a strengthening of the field, rather than a weakening. One thus need to work out the dampening effect of the abelian screening and the strengthening effect of the non-Abelian screening.
• Thanks for taking the time to answer. I understand how ti works now. Just a remar k actually it works too if it's really a + sign since we use the equation twice... We just get a sink for $E^3$ in place of a a source. I'll try to check the original article to find out – Reflets de Lune Jan 2 at 16:09