# Renormalization

From: Niclas Thisell <niclas.domain.name.hidden>
Date: Wed, 29 Dec 1999 13:15:34 +0100

Here's a half-cooked thought for you to chew on.

One of the initial gripes I had with the turing approach to the
plenitude (vs e.g. the grand ensemble) was that I figured it would favor
a finite resolution lattice and finite resolution calculations. And I
find it very hard to believe that we will ever actually detect an
'absolute' lattice. And it's even harder to believe that we will ever
notice that nature uses a finite number of decimals. This issue has been
discussed before and some of you found that a turing machine won't
properly handle true 'reals'.

The answer is, of course, that the number of computations in the
Schmidhuber plenitude using an insanely high number of decimals is a lot
higher than the ones that use a specific measurable but life-permitting
precision. The measure of a computation using 10^10^10 decimals is
roughly the same as one using 10^10^10^10 decimals. And the computations
themselves will most likely remain virtually identical throughout the
history of the universe and the observer-moments will be identical. The
same goes for grid spacing (and grid extent, for that matter). Therefore
observer-moments in a universe using precision indistinguishable from
'reals' and a lattice indistinguishable from a continuum seem to be
favoured.

Now, in Quantum Field Theory we have this annoying renormalization
problem. One could perhaps state that the equations become
trivial/divergent in the limit of a 4-dimensional continuous spacetime.
There are various tricks to get around this, like doing the complete
calculation in _almost_ 4 dimensions and then look at the result as you
push the dimensionality towards 4. Another trick is to do the
calculations using a finite lattice. This has lead people to believe
that the universe is 'really' using a finite resolution lattice. And it
must have an absolute resolution, mustn't it? (Not really). The
Planck-distance and Planck-time have been suggested.

I guess the Schmidhuber plenitude would predict that any experiment
trying to detect this lattice would fail. Even assuming we could argue
that there must be a finite resolution lattice, all one can conclude
after an experiment is that the resolution apparently must be higher
than was previously thought. And there need not be an absolute coupling
constant etc. - it simply depends on the lattice resolution of the
computation.

This claim is, however, unfortunately fairly vacuous and merely a matter
of retrofitting known theories. Also, the reasoning assumed that there
are no deeper reasons for the necessity of an absolute grid, which there
may be I guess. But still :-).

(Obviously I realise that many of you probably already have come to the
same conclusion. It is a lot more obvious in regard to grid extent. But
I don't recall having seen it on this list so I thought I'd mention it.)

/Niclas
Received on Wed Dec 29 1999 - 04:09:56 PST

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