Harald Kirsch

genug Unfug.

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}

  1. On the far left we have the term which describes the flow of time in the box with respect to the observer.
  2. 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.