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Tension matrix expansion

The tension matrix (using the `dilation' weighting function ) in the scaling eigenfunction basis has elements

$\begin{displaymath} \tilde{F}_{\mu\nu}(k) \ = \ \oint \! \frac{ds}{r_n} \, \psi_\mu(k_\mu + \delta_\mu) \, \psi_\nu(k_\nu + \delta_\nu), \end{displaymath}$

(I.19)

where the wavenumber shifts for the two states involved are $\delta_\mu \equiv k-k_\mu$ and $\delta_\nu \equiv k - k_\nu$ . As usual I define the wavenumber difference $\kappa \equiv k_\mu - k_\nu = \delta_\nu - \delta_\mu$ . Substitution of (I.18) gives a power series in the $\delta$ 's, which we will study and make estimates for unknown boundary integrals to grasp the general structure. We will assume for simplicity that the billiard size ${\mathsf{L}}\approx 1$ , and remain in

. We will ignore differences in $k_\mu$ and $k_\nu$ from

when appropriate, and look only for the dominant off-diagonal terms.

The $O(\delta^2)$ term in $\tilde{F}$ comes from the lowest order in (I.18),

$\begin{displaymath} \tilde{F}^{(2)}_{\mu\nu}(k) \ = \ \frac{\delta_\mu \delta_\nu}{k_\mu k_\nu} \oint \! ds \,r_n \, \psi_\mu^n \psi_\nu^n \end{displaymath}$

(I.20)

which we can write as $2\delta_\mu \delta_\nu M_{\mu\nu}$ (see Section 6.1.1), where

is quasi-diagonal. The diagonal is $M_{\mu\mu} = 1$ , and we have from the considerations of Section 6.1.2 small off-diagonal elements of size $M_{\mu\nu} \sim k^{-1/2} \kappa^2 O(1)$ , for

The $O(\delta^3)$ term in $\tilde{F}$ is, after some rearrangement,

$\begin{displaymath} \tilde{F}^{(3)}_{\mu\nu}(k) \ = \ -\frac{2 \delta_\mu^3}{k}... ..._t \, (\psi_\nu^n \psi_\mu^{nt} - \psi_\mu^n \psi_\nu^{nt}) , \end{displaymath}$

(I.21)

where an integral of $r_n r_t \partial_t(\psi_\mu^n \psi_\nu^n)$ was performed by parts. This required use of $\partial_t(r_n r_t) = r_n + \alpha(r_n^2 - r_t^2)$ , a different expression than found by VS. It can be proved easily using $\partial_t r_n = \alpha r_t$ and $\partial_t r_t = 1 - \alpha r_n$ , which follow from (I.8) and (I.9). Remarkably, the $\alpha$ dependence then cancels out, giving the same $\tilde{F}^{(3)}$ diagonal term as that of VS. We believe that the integral in (I.21) does not have any quasi-orthogonal property, so can be estimated using random waves. The estimate gives $k^{5/2} O(1)$ for this integral, and shows that this term dominates over any off-diagonal contribution from the first term (involving

). The factor of $\kappa$ in this term means that there is a weak form of quasi-diagonality at this order. Importantly, for $\kappa \ll 1$ the off-diagonal $O(\delta^3)$ error renders the $O(\delta^2)$ error insignificant. Hence we expect the quasi-diagonality property of

to play no role in errors in the scaling method.

The $O(\delta^4)$ and higher terms in become very messy. I believe that the dominant $O(\delta^4)$ terms, both on and off the diagonal, are

$\begin{displaymath} \tilde{F}^{(4)}_{\mu\nu}(k) \ = \ -\frac{\delta_\mu\delta_\... ...} \oint \! ds \, r_n^2 r_t^2 \, \psi_\mu^{nt} \psi_\nu^{nt} . \end{displaymath}$

(I.22)

This can be seen by comparing powers of

and using random-wave estimates. A random-wave estimate of the integrals then gives $\tilde{F}^{(4)} \approx \delta_\mu^4 O(1)$ on the diagonal and $\delta_\mu^4 k^{-1/2} O(1)$ off-diagonal. Note that the off-diagonal has no quasi-diagonality property at this or higher orders. For higher orders $O(\delta^m)$ for

even, we expect $\tilde{F}^{(m)} \approx \delta_\mu^m O(1)$ on the diagonal and $O(\delta^m) k^{-1/2} O(1)$ off-diagonal. For

odd, the leading diagonal terms are down by a factor of

which renders them insignificant.

To summarize, the diagonal of the tension matrix has the form given in Eq.(6.27), and for $\delta_\mu, \delta_\nu \ll 1$ the dominant off-diagonal terms are

$\begin{displaymath} \tilde{F}_{\mu\nu}(k) \ \approx \ \frac{\delta_\mu\delta_\... ... O(1) + \delta_\mu\delta_\nu O(1) \rule{0in}{0.2in} \right] . \end{displaymath}$

(I.23)

Here contributions from orders 3 and 4 were included because it may be that the 4th order (the lowest order with no quasi-diagonality, i.e. no powers of $\kappa$ ) contributes most to errors in the scaling method. It is important to note that the 2nd order term (due to off-diagonal strength of

) is negligible. It is clear that more research is needed on the properties of the higher-order terms, especially if an explanation of the tension error $\delta^6$ power-law growth (Section 6.3.1) is sought.

Next: Useful geometric boundary integrals Up: Appendix I: Scaling expansion Previous: Applying boundary conditions and

Alex Barnett 2001-10-03