2015-03-17

# A mechanistic explanation of Time

$
\def\Vec#1{\mathbf{#1}}
\def\vt#1{\Vec{v}_{#1}(t)}
\def\v#1{\Vec{v}_{#1}}
\def\av{\bar{\Vec{v}}}
$

If you search the Internet for an explanation of what
*time* is, there are a lot of
sites with a
lot of words basically saying that they cannot say anything about
time. Here I have a different approach in that I ask you to compare two
simple formulas and let you draw your own conclusions.

## The speed of light

Given an observer whose local time is $t$ and an observed
object passing by with a velocity of $\Vec{v}(t)\in\mathbb{R}^3$, Special
Relativity tells us that for the proper time $\tau$ of the object the following holds:
\begin{equation}\label{eq:srt}
c^2 = |\Vec{v}(t)|^2 + \left(\frac{c\,d\tau(t)}{dt} \right)^2
\end{equation}
where $c$ is the speed of light and $|\cdot|$ denotes the absolute
value of a vector.

The term $d\tau/dt$ could be called the speed of aging of the
object with respect to the observer. If the object is at rest with
respect to the observer, i.e. $|\Vec{v}(t)|=0$, then $d\tau/dt =
1$, meaning the observer and the object age with the same
speed.

The other extreme case is where the relative velocity
approaches the speed of light $c$ and $d\tau/dt\to 0$ meaning that
with respect to the observer, the object is not aging anymore.

## Average velocity of an ensemble of points

The second formula describes the average velocity of an ensemble
of $n$ points. For each point particle $0\leq i<n$, its
velocity shall be $\vt{i}$. Further, all absolute values of the
velocities shall be identical, namely $|\vt{i}|=c$. The average velocity of
the ensemble is then
\begin{equation}
\av(t) = \frac{1}{n}\sum_{i=0}^{n-1} \vt{i} .
\end{equation}
For brevity we leave out the dependence on $t$ for now as we
compute the difference $c^2-\av^2$.
\begin{align}
c^2-\av^2 &= c^2 - \frac{1}{n^2} \left(\sum_{i=0}^{n-1} \v{i}\right)^2 \\
&= c^2 - \frac{1}{n^2} \sum_{i,j=0}^{n-1} \v{i}\v{j} \\
\end{align}
The last sum is symmetric in $i$ and $j$ and therefore contains
each pair $\v{i}\v{j}$ twice, except where $i=j$. The latter,
quadratic terms are extracted from the sum to arrive at:
\begin{align}
c^2-\av^2&= \underbrace{c^2 - \frac{1}{n^2}\sum_{i=0}^{n-1} \v{i}^2}_{(A)}
- \frac{1}{n^2}\sum_{i<j} 2\v{i}\v{j}
\end{align}
Since $\v{i}^2=|\v{i}|^2=c^2$, the terms denoted as $(A)$ in the
formula can be written as
\begin{align}
(A)&= c^2 -1/n^2\cdot n c^2 \\
&= c^2(1-1/n) \\
&= \frac{n-1}{n}c^2 \\
&= \frac{1}{2} n (n-1) \frac{2c^2}{n^2}
\end{align}
The term $1/2 n(n-1)= 1/2 (n^2-n)$ is the number of elements in the
lower right half of a square matrix without the diagonal elements
and is therefore the same as $\sum_{i<j} 1$, which we
multiply by $\frac{2c^2}{n^2}$ and plug it into $(A)$ to get
\begin{align}
c^2 -\av^2 &= \sum_{i<j}\frac{2c^2}{n^2}
- \frac{1}{n^2}\sum_{i<j} 2\v{i}\v{j}\\
&= \frac{1}{n^2} \sum_{i<j} c^2-2\v{i}\v{j}+c^2 \\
\scriptsize \text{(since $c^2=\v{i}^2$)}\qquad
&= \frac{1}{n^2} \sum_{i<j} (\v{i}-\v{j})^2 .
\label{eq:final}
\end{align}
What does this last equation mean? Remember, we are talking
about an ensemble of $n$ point particles all moving at the speed of
light. Lets envisage the particles trapped in a box with $n$
sufficiently large and their movements sufficiently random, then
the velocity of the box, $\Vec{v}_B$ is a close approximation to
$\av$, the average over the particle velocities.

## Putting it together

Suppose the box is the observed object we talked about in the
first section. Then, by comparing equation \eqref{eq:srt} with the last
equation \eqref{eq:final}, we get
\begin{equation}
\left(\frac{c\,d\tau(t)}{dt} \right)^2
= c^2- |\Vec{v}_B|^2
\approx c^2 - \av^2
= \frac{1}{n^2} \sum_{i<j} (\v{i}-\v{j})^2
\end{equation}

- On the far left we have the term which describes the flow of
time in the box with respect to the observer.
- On the far right we have a purely mechanical term, the
average delta velocities of the point particles in the box.

## Does this make sense?

Here comes the part that I have to leave to the reader, except
that I throw in my opinion anyway: it does make a lot of sense to
me. Consider the box moving faster and faster, until it reaches the
speed of light. What happens to the point particles in the box?
Since they move also at the speed of light along with the whole
box, the delta velocities $\v{i}-\v{j}$ have to converge to
zero. Consequently the relative positions of the points inside the
box become constant: the content of the box is kind of frozen,
nothing changes anymore. This is the situation where also
$c\,d\tau/dt=0$, meaning that proper time comes to a halt.

My conclusion is that (proper) time is nothing but the
integrated average of change happening in a closed system. For a
(closed) system of point particles, change is simply the delta
velocity between points.

## Open questions

Of course there are many. My priority one question is whether a
similar derivation is possible for a system "filled" with a
changing field, such that $d\tau/dt$ turns out to be equal to some
measure of change of the field.