2
$\begingroup$

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 ? figure 16.11 of P&S.

$\endgroup$
  • $\begingroup$ It is analyse of equations to understand physical meaning.. $\endgroup$ – Nikita Jan 1 at 17:09
  • $\begingroup$ thanks i gathered as much. what i don't understand is how they do it ... $\endgroup$ – Reflets de Lune Jan 1 at 17:16
1
$\begingroup$

a) In my notes on P&S I have a minus sign as well. Maybe this is just a typo?

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.

I must admit that I don't find P&S very clear about the details although the general gist is. Maybe you should revert to the original article?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Reflets de Lune Jan 2 at 16:09
  • $\begingroup$ yes - it works with both signs indeed. $\endgroup$ – Oбжорoв Jan 2 at 16:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.