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Wilson loops (srednicki eqn. 82.37)

  1. Aug 20, 2009 #1
    In eqn. (82.37) Srednicki derives for the string tension between two quarks as [tex]\tau=\frac{c(g)}{a^2} [/tex] where [tex]c(g)=ln(g^2)[/tex], 'g' is the coupling, and 'a' is the lattice spacing. He later goes on to say that the string tension should be independent of the lattice spacing 'a', and using this condition, calculated the string tension for small 'g' and small lattice spacing 'a' (82.41): [tex]\tau=Ce^{\frac{-1/(b_1g^2)}{a^2}} [/tex]. But didn't he just contradict himself? If the string tension must be the same, then which is it, [tex]\tau=\frac{ln(g^2)}{a^2} [/tex], or [tex] \tau=Ce^{\frac{-1/(b_1g^2)}{a^2}} [/tex]?

    Also, it seems to me that the result eqn. (82.35), [tex](\frac{1}{g^2})^{\frac{A}{a^2}} [/tex], the area law (A=area of the Wilson loop), doesn't depend on its derivation whether g is large or small. You can always expand an exponential [tex]e^x [/tex] about x=0 whether x is big or small! And in order for the Wilson loop to not be zero via eqns (82.31), (82.32), and (82.33), then (82.35) must hold!
     
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  3. Aug 20, 2009 #2

    Avodyne

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    The first formula applies when [itex]g\gg 1[/itex] and [itex]a\gg 1/\Lambda[/itex], where [itex]\Lambda\sim 1\,\rm GeV[/itex] is the characteristic energy scale of QCD, and the second when [itex]g\ll 1[/itex] and [itex]a\ll 1/\Lambda[/itex].
    You can expand the exponential for any [itex]g[/itex], but after you do the integrals over [itex]U[/itex], the resulting series may have a finite radius of convergence. This is what is supposed to happen for U(1) lattice gauge theory.
     
  4. Aug 20, 2009 #3
    But how can such a sharp, discontinuous contrast exist? Even if both of those formulas (on their respective regimes) give the same value of the string constant, surely such discontinuity indicates some nth order phase transition?

    I wish Srednicki would have mentioned that. I didn't realize after you integrate a convergent power series that the resulting series could be divergent. So whether the resulting power series has a finite or infinite radius of convergence depends on the size of 'g'?
     
  5. Aug 20, 2009 #4

    Avodyne

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    It's not discontinuous, it changes smoothly from one limiting case when [itex]g\gg 1[/itex] to the other when [itex]g\ll 1[/itex].
    Sure. As a trivial example, consider
    [tex]I=\int_{0}^{\infty}dx\;e^{-x} \;e^{gx}.[/tex]
    Expand in powers of [itex]g[/itex] and do the integral term by term; the result is
    [tex]I=1+g+g^2+g^3+\ldots\;,[/tex]
    which only converges if [itex]|g|<1[/itex]. Of course, since the integrand can be written as [itex]e^{-(1-g)x}[/itex], it's obvious that the integral only makes sense if [itex]g<1[/itex].
     
  6. Aug 22, 2009 #5
    So I guess what Srednicki is saying is that the formula for the string constant for strong coupling is [tex]\tau=\frac{ln(g^2)}{a^2} [/tex]. It would be great if [tex]g(a)=\sqrt{e^{Ca^2}}[/tex], because plugging that back in, the string constant will always remain the constant 'C' for all 'a'. However the relationship between 'g' and 'a' is governed by the beta function and not the string equation [tex]\tau=\frac{ln(g^2)}{a^2} [/tex].

    So out of nowhere, to resolve this problem, Srednicki says instead of [tex]ln(g^2(a))[/tex], let's call it c(g(a)), and determine a function c(g(a)) that keeps the string constant the same.

    So Srednicki is assuming that the string constant can always be written as [tex]\frac{c(g(a))}{a^2} [/tex], despite the fact that the 1/a^2 dependence was derived only in a special case.

    This is not really satisfying.

    edit: I can actually believe the [tex] \frac{c(g(a))}{a^2} [/tex] area law now, even though technically it was only derived in the strong coupling limit case. Although the power series expansion gets messed up (after integration) in the weak coupling limit, I guess the same physical arguments hold, that you have to have every plaquette inside the loop.
     
    Last edited: Aug 22, 2009
  7. Aug 22, 2009 #6

    Avodyne

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    The form [itex]\tau=c(g(a))/a^2[/itex] follows just from dimensional analysis; [itex]a[/itex] is the only dimensionful parameter in the theory, and [itex]\tau[/itex] has dimensions of inverse length squared (with [itex]\hbar=c=1[/itex]).

    Then, we know from the strong coupling analysis that [itex]c(g)=\ln g^2[/itex] when [itex]g\gg 1[/itex], and from the weak coupling analysis (that is, the perturbative computation of the beta function) that [itex]c(g)=C\exp(-1/b_1 g^2)[/itex] when [itex]g\ll 1[/itex], where [itex]C[/itex] is just a numerical constant.
     
  8. Aug 23, 2009 #7
    Why is [itex]c(g)=C\exp(-1/b_1 g^2)[/itex] so difficult that we have to use computers instead? As you say we have the beta function for perturbative computation to one loop (i.e., order [tex]g^3[/tex]). Can we not integrate this beta function to get g(a), plug this g(a) into [itex]c(g)=C\exp(-1/b_1 g^2)[/itex], and then take the limit as 'a' goes to zero? Srednicki comments that the exponential [tex]\exp(-1/b_1 g^2) [/tex] is not analytic at g=0 so that the exponential can't be expanded in Taylor series, but why expand in Taylor series when you can just plug it directly into the exponential? Also if you were to do a lattice calculation, can you plug in a Wilson loop of any shape to figure out [tex]\tau [/tex]?
     
  9. Aug 23, 2009 #8

    Avodyne

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    We need computers to check that c(g) has the expected form at weak coupling, and to compute the constant C. What we really want to compute is dimensionless numbers that characterize the physics, such as the string tension in units of the mass of the lightest glueball. The mass of the lightest glueball would be given in the lattice theory by a formula like m = d(g)/a. At small g, we expect d(g) = D exp(-1/2 b1 g^2), so that tau/m^2 = C/D^2 when g goes to zero.
     
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