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\section{\usemenu{slacpub7063::context::slacpub7063004}{ VERIFICATION OF THE SCALE}}\label{section::slacpub7063004}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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A good way to see the validity of the scale set for the
structurefunction evolution equation is to apply it to moment
evolution and compare the results with experimental data. The moment
analysis of the Fermilab muon deepinelastic scattering data
\cite{11} showed an interesting variation of the effective
$\Lambda_{\text{QCD}}$ with $n$, both at leading and nexttoleading
order. Here we would like to demonstrate that this variation can be
readily explained by the choice of the renormalization scale even at
leading order.
The $n$th ordinary moment of the nonsinglet structure function is defined as:
\begin{equation}
M_n(Q^2) = \int_0^1dxx^{n1}{\cal F}^{\text{NS}}_2(x,Q^2).
\end{equation}
The Nachtman moments will give the same result up to higher twist terms.
Taking the $n$th moment of the structure function evolution equation
(Eq. (\docLink{slacpub7063003.tcx}[scale fixed structure function evolution equation]{17})),
interchanging the order of integration on the righthand side and rearranging,
one can obtain the equation of evolution for the moment:
\begin{equation}
\frac{\partial\ln M_n(Q^2)}{\partial\ln Q^2} = \int_0^1dxx^{n1}
\left(\tilde{P}_{qq}(x, \alpha(Q^*(x)))
\pm \tilde{P}_{\bar q q}(x, \alpha(Q^*(x))\right),
\label{moment evolution}
\end{equation}
which is just the $n$th moment of the evolution kernels.
Let us consider only the leading order prediction of the kernel
function in Eq. (\docLink{slacpub7063003.tcx}[scalefixedPqq]{16a}), then Eq. (\docLink{slacpub7063004.tcx}[moment evolution]{20})
becomes
\begin{equation}
\frac{\partial\ln M_n(Q^2)}{\partial\ln Q^2} = \int_0^1\left(
\frac{\alpha(Q^*(x))}{2\pi}
P_{qq}^0(x) \right)_+
x^{n1}dx,
\label{e13}
\end{equation}
here $Q^*(x)$ is given by Eq. (\docLink{slacpub7063003.tcx}[kernelscale]{14}),
and
\begin{equation}
P_{qq}^0(x) = C_F\frac{1+x^2}{1x}
\end{equation}
the leading order coefficient.
The crucial difference between the conventional calculation, that is, using
the momentum transfer $Q$ as the renormalization scale, and the
calculation here, namely using the BLM scale, is that the
argument of the coupling constant in Eq. (\docLink{slacpub7063004.tcx}[e13]{21}) is now $x$ dependent.
To see the effect of the scale setting on the leadingorder analysis,
expand $\alpha(Q^*)$ in powers of $\beta_0\alpha(Q)$
(similar results can be obtained without the expansion), that is,
\begin{equation}
\alpha(Q^*(x)) = \alpha(Q)\left\{1+
\beta_0\ln \left(\frac{Q}{Q^*(x)}\right) \frac{\alpha(Q)}{2\pi} + \ldots
\right\},
\label{alpha_expansion}
\end{equation}
then Eq. (\docLink{slacpub7063004.tcx}[e13]{21}) becomes
\begin{equation}
\frac{\partial\ln M_n(Q^2)}{\partial\ln Q^2} =
\frac{\alpha(Q)}{2\pi}A_n\left[
1+\beta_0B_n\frac{\alpha(Q)}{2\pi} + \ldots
\right],
\label{e15}
\end{equation}
where
\begin{eqnarray}
A_n &=& \int_0^1dxx^{n1} P_{qq}^0(x)_+, \text{and} \\
B_n &=& \frac{1}{A_n}\int_0^1dxx^{n1} \left[ P_{qq}^0(x)
\ln \left( \frac{Q}{Q^*(x)}\right)
\right]_+.
\end{eqnarray}
Now Eq. (\docLink{slacpub7063004.tcx}[e15]{24}) can be integrated with respect to $\ln Q^2$
by using, to LO (to compare with LO experimental analysis),
\begin{equation}
\alpha(Q) = \frac{4\pi}{\beta_0 \ln (Q^2/ \Lambda^2)},
\label{coupling constant}
\end{equation}
and the solution for $M_n(Q)$ is then
\begin{equation}
M_n(Q)^{1/d_n} =C_n\ln(Q^2/\Lambda^2)
\exp\left[\frac{2B_n}{\ln(Q^2/\Lambda^2)}\right],
\end{equation}
where $d_n = 2A_n/\beta_0$ and $C_n$ are constants.
This is our prediction for the evolution of the moments.
For experimental data taken at large $Q$, the exponential term can
be expanded and we have an equation linear in $\ln(Q^2/\Lambda^2)$:
\begin{equation}
M_n(Q)^{1/d_n} =C_n\left[\ln(Q^2/\Lambda^2) 2B_n\right].
\label{e20}
\end{equation}
Had we set the scale to $Q$, we would have had
\begin{equation}
M_n(Q)^{1/d_n} =C_n\left[\ln(Q^2/{\Lambda_n}^2)\right],
\end{equation}
where $\Lambda_n$ is an ``effective'' $\Lambda$ at LO and
will depend on $n$ as predicted by Eq. (\docLink{slacpub7063004.tcx}[e20]{29}) as
\begin{equation}
\Lambda_n = \Lambda/ \exp (B_n).
\end{equation}
Values of $\exp(B_n)$ for $n$ from 2 to 10 are plotted in
Fig. \docLink{slacpub7063006.tcx}[f3]{2}; these are welldescribed by the form
\begin{equation}
\exp(B_n) = 0.656/n^{1/2}.
\end{equation}
Therefore, we expect the dependence of $\Lambda_n$ on $n$ to be
\begin{equation}
\Lambda_n = \frac{n^{1/2}}{0.656} \Lambda.
\label{e21}
\end{equation}
In fact, the data of Gordon {\it et al} does show the predicted
variation. In Fig. \docLink{slacpub7063006.tcx}[f4]{3} we show the results extracted for $\Lambda_n$
against $n$. A fit to Eq. (\docLink{slacpub7063004.tcx}[e21]{33})
yields $\Lambda_{\overline{\text{MS}}} = 170 \pm 120$ MeV, where the error is
estimated from higherorder terms in Eq. (\docLink{slacpub7063004.tcx}[alpha_expansion]{23}).
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