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Archive for January, 2014

## A Remark based on False Claim 1

So I was thinking that, the statistical description of a dynamical system (as one described by a generator Pasquali patch) really does give us a lot of power in computing the probable "position" of a particle (photon, electron) moving in space at different (integer) time intervals. Â If the quantum mechanical supposition of time having a minimum discreteness (Planck-time) is correct, we can find the "finest" Pasquali patch generator that will give a complete description of the dynamical system. Â Any Pasquali patch generator descriptive of the system which is not this "first" will generate an accurate, yet less refined ("coarser") version of the system (this is what we mean by Claim 3 of the previous post, in that such Pasquali patch will be "contained" in the finest description, yet is not the finest), and in fact either system of course converges to the same steady state (this is what is meant by Claim 4). Â If we are able to find a continuous description (like the Shrodinger equation, via a "Pasqualian") of such system then we are in luck (this description would be the finest, though non-discrete, description), and I speculate though I cannot be sure yet that either discrete descriptions will be contained in such.

Whatever the description of the dynamical system via a generator Pasquali patch (or a Pasqualian), each Pasquali patch represents the transition (position) probabilities of a particle (photon, electron) moving within that system. Â If we suppose that the particle moves with same velocity (take photons in vacuum as an example), then each Pasquali patch power is descriptive of the transition-position probability at equally spaced spacial or distance intervals. Â Though I've remarked about this before, if it were the case that, for a particle with a particular (steady) velocity, the Pasquali patch power is not exactly equally spaced in distance intervals, it must mean that the arrow of time is bent (time is moving faster for smaller-spaced intervals, slower for longer-spaced). Â We have not yet described accelerating particles but at present that is not of our interest.

However, we can tell if time is passing so long as each Pasquali patch description is different at each (equal or unequal) interval. Â If a single Pasquali patch were to describe the system at EACH distance interval, there is no way to know if time is moving at all. Â Take for example the False Claim 1 of the previous post. Â We had the collection $\mathbb{Q} = \{Q^1, Q^2, \ldots, Q^k, \ldots\}_{k \in \mathbb{Z}^+}$ with $Q = q(x,y)$ being an explicit function of $y$ and converging to $Q^\infty = 1$. Â We could track the time-distance interval via the Pasquali patch power, so that $1$ was the first distance interval from start (we take it as given that time is not being bent, so that a fixed distance implies the passage of 1 unit of time), $2$ was the second distance interval (2 units of time), and so on, and we could tell if time were bent if each power were descriptive of different distance intervals. Â Furthermore, since each $Q \in \mathbb{Q}$ is different, this implies each position transition probability is different and the system is in movement.

This is definitely not the case with the collection $\mathbb{P} = \{P^1 = 1, P^2 = 1, \ldots, P^j, \ldots \}_{j \in \mathbb{Z}^+}$ which also converges to $P^\infty = 1$. Â Since at each distance interval the movement probability is the same (uniform), one cannot be convinced that each power represents a distance interval equal to equally spaced time intervals or different-spaced time intervals. Â Where we could with $\mathbb{Q}$ ascertain that time was moving, we cannot with $\mathbb{P}$. Â The statistical description cannot tell if the system is frozen.

When a system has reached the steady state (which, is the highest entropy state!), there is no way to tell if time flows, as the statistical description is and forever will be unchanging. Â Recall that the steady state for a Pasquali patch is always a function of $x$ alone, say $p(x)$. Â Furthermore recall that any power of $p(x)$ is always $p(x)$ itself (see Compendium). Â We reach an impasse: is time flowing normally, faster than what is conventional, slower? At such a point it is impossible to say, at least from the statistical point of view. Â We would have to track particles individually in order to ascertain if they have deviated their path at all (it could be the case that they shifted to all positions with equal probability or in the shape of $p(x)$, e.g., but we cannot be sure of either situation).

## More On Collections of Powers (via Starring) of Functions

These proofs really belong to the Compendium, but I have no access to the tex file at the moment, so I am using this post to jot down ideas.

If we accept powering of functions via the defintions of the star product, then the following will make sense. Â On the other hand if we view this from the standpoint of simple matrices (and replace the star product by matrix multiplication), the following claims may also make sense.

Definition 1.Â  For a countably infinite collection $\mathbb{F} = \{F^1, F^2, \ldots, F^j, \ldots \}_{j \in \mathbb{Z}^+}$, the "generator" isÂ $F^1$ since all other elements in the collection are powers of such. Â (As an aside, we may contemplate finite collections in the event of powering periodicity or nilpotency, but not presently).

Claim 1. Take the countably infinite collectionsÂ $\mathbb{F}$ andÂ $\mathbb{G}$ with generators $F^1$ and $G^1$ respectively. Â If $F^1 = G^1$, then $\mathbb{F} = \mathbb{G}$.

Proof. Â By hypothesis we have thatÂ $F^1 = G^1$, and let us assume thatÂ $F^k = G^k$. Then:

Â Thus we have shown that Â $\mathbb{F} \subset \mathbb{G}$ and $\mathbb{G} \subset \mathbb{F}$ and so naturally we concludeÂ $\mathbb{F} = \mathbb{G}$.

Claim 2.Â Take the countably infinite collectionsÂ $\mathbb{F}$ andÂ $\mathbb{G}$ with generators $F^1$ and $G^1$ respectively. Â If $F^1 = G^n$ for some $n \in \mathbb{Z}^+$, then $\mathbb{F} \subset \mathbb{G}$.

Proof.Â By hypothesisÂ $F^1 = G^n$ for some positive integer $n$. Â Also,

using Claim 1.10 of the Compendium. Â Next suppose thatÂ $F^k = G^{kn}$. Â Then:Â

Â Thus $\mathbb{F} \subset \mathbb{G}$ since all elements of the collection $\mathbb{F}$ are contained in $\mathbb{G}$.

Definition 2.Â Â For a countably infinite collectionÂ $\mathbb{F} = \{F^1, F^2, \ldots, F^j, \ldots \}_{j \in \mathbb{Z}^+}$, we call the subset $\mathbb{F}_{n} \subset \mathbb{F}$ the "subsequent power subset at $n$", and it is composed of all those powers greater than or equal to $n \in \mathbb{Z}^+$.

Claim 3. Take the countably infinite collectionsÂ $\mathbb{F}$ andÂ $\mathbb{G}$ with generators $F^1$ and $G^1$ respectively. Â If $F^m = G^n$ for some $m, n \in \mathbb{Z}^+$ and $m (since we haven't defined "roots" it must also be true that $\frac{n}{m} \in \mathbb{Z}^+$), then $\mathbb{F}_m \subset \mathbb{G}_n \subset \mathbb{G}$. (At this point we don't really care about containment of powers less than $n$, since we haven't established a mechanism to obtain them by "unpowering" or "taking roots").

Proof. Â By hypothesisÂ $F^m = G^n$. Â In particular, we have the statement $F^{m+1} = G^{n + p}$ for some integer $p \in \mathbb{Z}^+$. Â (Advancing one element in the collection $\mathbb{F}$ starting from $m$, advances us $p$ units in the collection $\mathbb{G}$ starting from $n$.) Â Taking this statement to the logical consequence, $F^m \star F^1 = G^n \star G^p \Rightarrow F^1 = G^p$. Â (Notice that since we have not defined negative powers of functions, $p$ is necessarily a positive integer).

Next let us assume thatÂ $F^{m+k} = G^{n + kp}$. Â Then:Â

It can be seen now thatÂ Â $\mathbb{F}_m \subset \mathbb{G}_n \subset \mathbb{G}$.

Corollary 1. Â Take the countably infinite collectionsÂ $\mathbb{F}$ andÂ $\mathbb{G}$.Â Â $\mathbb{F}_m = \mathbb{G}_n \subset \mathbb{G}$ if and only if $m = n$ and $p = 1$.

Proof.

$\Rightarrow$ SinceÂ $\mathbb{F}_m = \mathbb{G}_n$ there is no element of either collection that does not match one in the other collection. Â Thus the step between elements must be 1, and $p = 1$. Â Furthermore the elements in a collection can be put in bijective correspondence (by definition of equality of sets) with the elements in the other collection, and the indices must exactly coincide by the ordering being the usual order on the positive integers (we order by power). Â Thus $m=n$.

$\Leftarrow$ Since $m = n$ we have that $F^m = G^n \Rightarrow F^m = G^m$ for all $m \in \mathbb{Z}^+$, and also $F^{m+1} = G^{n + p} \Rightarrow F^{m+1} = G^{m + 1}$Â since $p = 1$. Â So now assume that this works for the $k$th element $k>m+1$, so that $F^{m+k} = G^{m+k} \Rightarrow F^k = G^k$. Â Then we have that:

and we are done.

The following claims and proofs are applicable to Pasquali patches as they are to Â Markov matrices, where we shift from a function point-of-view to a matricial point-of-view using in one the star product and in the other normal matrix multiplication.

Claim 4. Take the countably infinite collections of Pasquali patch self-powers $\mathbb{P} = \{ P^1, P^2, \ldots, P^j, \ldots \}_{j \in \mathbb{Z}^+}$ andÂ $\mathbb{Q} = \{ Q^1, Q^2, \ldots, Q^k, \ldots \}_{k \in \mathbb{Z}^+}$ with stationary limiting $P^\infty$ and $Q^\infty$ respectively. If $\mathbb{P}$ and $\mathbb{Q}$ coincide at ($m, n$) so that $P^m = Q^n$ ($\frac{n}{m} \in \mathbb{Z}^+$) then $P^\infty = Q^\infty$.

Proof.

The infinite collections coincide at $(m,n)$ for $m, n, \frac{n}{m} \in\mathbb{Z}^+$, thus we have that $P^m = Q^n$. Â By Claim 3 we have thatÂ $\mathbb{P}_m \subset \mathbb{Q}_n$. Â Now the limit $\lim_{n \to \infty} Q_n = Q^\infty$ by hypothesis of the existence of the stationary patch, and sinceÂ $\mathbb{P}_m$ is a subset of $\mathbb{Q}_n$, it follows that$\lim_{m \to \infty} P_m = Q^\infty$. Â We also know thatÂ $\lim_{m \to \infty} P_m = P^\infty$ however (again by hypothesis of the existence of the stationary patch), so it must therefore be true thatÂ Â $P^\infty = Q^\infty$.

The reverse claim is not (always) true:

False Claim 1. Â IfÂ $P^\infty = Q^\infty$, thenÂ $\mathbb{P}$ and $\mathbb{Q}$ coincide at ($m, n$) so that $P^m = Q^n$ ($\frac{n}{m} \in \mathbb{Z}^+$).

Disproof. Â Take the collection of Pasquali patches $\mathbb{P} = \{P^1 = 1, P^2 = 1, \ldots, P^j = 1, \ldots\}_{j \in \mathbb{Z}^+}$ with $P^\infty = 1$, and any collection of Pasquali patches $\mathbb{Q} = \{Q^1, Q^2, \ldots, Q^k, \ldots\}_{k \in \mathbb{Z}^+}$ so that the generator $Q^1 = q(x,y)$ is an explicit function of $y$ and $Q^\infty = 1$, as in an example in the Compendium. Â By a Claim in the Compendium, all elements in the collectionÂ $\mathbb{Q}$ are explicit functions of $y$ (even as $Q^\infty$ is not). Â Clearly, no $P \in \mathbb{P}$ is equal to an element $Q \in \mathbb{Q}$ (all $P$ are constant where all $Q$ vary with $y$), yet they have the same stationary patch.