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Zelootma.z64 96

How do I approach this problem using the inbuilt functions in Mathematica?

A:

This is another way with Lattice:
n = 16;
z = zelootma[n];
R = Range[1, z, n];
p = Permutation[R, 29];
t = ReplacePart[z, # -> Round[#, 0.05] & /@ z];
With[{p0 = Partition[p, n]},
p[[1 ;; n]] // Outer[Times, p[[All, 1 ;; n]], p0[[All, 2 ;; n]]]
]
(* {{{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0,
0, 0, 0, 0, 0,

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Why is the mesoscopic limit of classical electrodynamics ill defined?

According to ‘The Standard Model, Vol. 1′, in chap. 12, eq. (12.52), the loop-dependent dynamical factor $R$ in the transition matrix of a mesoscopic loop is
$$ R = |\Delta M| (\frac{a_0}{\xi_0})^2(\frac{k}{2 \pi})^3.$$
But from my understanding, the mesoscopic limit $a_0/\xi_0 \rightarrow \infty$ should be taken first, and $k$ should be small. Then I just obtain the’standard’ result $R \rightarrow 0$.
So what is going on? Why can’t the mesoscopic limit be taken first and then the’small $k$ limit’?

A:

There is a reason that the Lorentz transformation of a mesoscopic loop of side $a$ around a loop of side $a$ is put together before scaling it down in size: the mesoscopic loop is the “smallest loop” that may be relevant (that is, a loop of the size of the smallest scale of interest). Mathematically, putting this together means that the loops should be kept fixed while scaling the transition amplitudes.
This means that the $\Delta M$ on the left side of the expression for $R$ is really a $\Delta M$ at the scale $a$, whereas the $a_0$ and $\

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