Harald Kirsch

genug Unfug.

$\def\v#1{\mathfrak{#1}} \def\vx{\v{x}} \def\vy{\v{y}} \def\vz{\v{z}} \def\mA#1#2{a_{#1 #2}} \def\ma#1#2{a^{#1}_{#2}} \def\t#1{\tilde #1} \def\tx{\t{x}} \def\ty{\t{y}} \def\d#1{\partial #1} \def\dd#1{\partial_{#1}} \def\pderiv#1{\frac{\partial}{\partial #1}} $

2017-11-11

An Explanation of Time

Contents

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

Watching a Movie

Movies are comprised of a sequence of pictures. Showing pictures in quick succession creates the impression of a continuous sequence of events taking place. Slow motion reduces the frequency of pictures. Down to a certain limit motion still looks smooth, but then we start noticing that there are individual pictures shown. Yet for the following argument this is not important. The point is that we can slow down the frame rate ever more, even down to zero. At this point a last picture is shown, frozen on the screen. The next picture of course still exists, but will not be shown.

What is the time coordinate in the film. It is discrete and given by the sequence numbers of the pictures. The sequence of pictures defines the sequence of events taking place. When it comes to the last frame shown, where we decided to stop the film, local time in the film does not proceed anymore, simply because time is equivalent to frames shown and the next frame is never shown.

For observers in the movie theater it is irrelevant that there are potentially more pictures to show. There was a last frame and that was that. More will not happen.

Time in Physics

The movie with its slow motion and arbitrary stop can help to understand time in physics. The bold transfer from the discrete time co-ordinate of the film to continuous time in physics works as follows:

  1. Time is a sequence of events, a sequence of things happening.
  2. If events follow each other slower, time runs slower.
  3. If no more events follow, if things stop happening, time does not proceed.

The following paragraphs try to explain why this can be a viable way of looking at time in physics.

Time in physics, as far as we know, is continuous, not discrete. But there are many examples in mathematics and physics where the limit of discrete points results in a continuum. In the same way we can imagine that points in time is are individual events, yet they comprise a continuum.

In this view, time is implicitly given by a continuous sequence of events or states of the world. You could even say that time has no existence in itself. It is "only" a measure of how fast the states of the world follow each other.

Special and General Relativity tell us that time slows down in moving objects or under high gravity. How does this translate into the "continuous sequence of states of the world" view? Easy: the events follow slower onto each other than before. Similar to a film in slow motion. The time between the frames is longer than before.

Oh wait, what? Time between frames gets longer for slow motion, but how does this translate into the time in physics? How can there be more time between states of a continuum of states if the continuum of state is time?

The answer is the same as for the movie: we are talking about two time axes. In the movie, one axis is the frame number. The other is the observer's time axis as given by the observer's stop watch that shows a longer time between frames in slow motion. The same is done in physics, where a difference is made between the local time on a moving object or an object under gravity and the observer's time.

So we have two time axes, the one of the observer and the one of the object being observed. For the film it is easy to see how things happen slower. What about special or general relativity? What constitutes the longer time between world states, what makes world states follow each other slower — at least from the vantage point of the observer?

What are individual world states anyway? This is a difficult question, so we start with simplified world. Take a game of chess. A state is defined by the positions of all the pieces on the board. Look at the board now and look at the board some time later during which the players are thinking, not moving any piece. The state is the same — because nothing has changed. The important word is "change". While it is difficult to fully describe states of the world, it is easy to grasp that they are different due to change.

How do states of the world change? Again we fall back to a simple model. Consider a group of $n$ photons-in-a-mirrored-box and assume that they are point particles. Each photon has the (three dimensional) velocity $\Vec{v}_i$ with the absolute value being the the speed of light $c=|\Vec{v}_i|$. The average value of these velocities is \begin{align} \av &= \frac{1}{n}\sum_{i=0}^{n-1} \Vec{v}_i . \end{align} If the box moves at the speed of light, all photons move in the same direction and their $\Vec{v}_i$ are all identical. In this case $\av$ is also the same and in particular $|\av|=c$. On the other hand, if the box does not move we have $|\av|=0$.

When all the velocities are identical, does the world state of this small world change? Well, the photons are moving at the speed of light through space, so this constitutes a change, doesn't it? The anser is "no", because this takes the world around the photons into account. If you blind out the world around and only look at the group of photons now and 5 minutes later, nothing has changed. Their positions relative to each other are the same. Like on the unchanged chess board we can say that the world state of the group of photons did not change at all.

Put another way: the state changes only if the photon velocities are not all identical. Intuitively, if one photon of a billion has a slightly different direction than all the others, the rate of change is smaller than when the box has stopped and the photons are bouncing about in arbitrary directions within the box. The more the velocities differ, the bigger is the rate of change. One way to measure this totalled difference of pairwise photon velocities are the averaged squared differences \begin{equation} \frac{1}{n^2} \sum_{i<j} (\v{i}-\v{j})^2 \;, \end{equation} which is the same as the variance of the set of velocities.

For the following, note that a sum $\sum_{i,j=0}^{n-1}$, if symmetric in $i$ and $j$, can be viewed as summing the elements $(i,j)$ of a symmetric matrix. As such it can be taken apart into twice the sum over the lower left triangle of the matrix plus the diagnoal like $\sum_{i<j} \dots + \sum_{i}\cdots$. And remember that $|\v{i}|^2=\v{i}\v{i}=c^2$.

Lets add the average velocity squared of the photons in the box and proposed measure for the rate of change, the variance of the velocities: \begin{align} \av^2 + \frac{1}{n^2} \sum_{i<j} (\v{i}-\v{j})^2 &= \frac{1}{n^2}\sum_{i,j=0}^{n-1} \v{i}\v{j} &+ \frac{1}{n^2} \sum_{i<j} c^2-2\v{i}\v{j}+c^2\\ &= \frac{1}{n^2} \left( \sum_{i<j} 2\v{i}\v{j} + \sum_{i=0}^{n-1} \v{i}^2 \right) &+ \frac{1}{n^2} \sum_{i<j} c^2-2\v{i}\v{j}+c^2\\ &= \frac{1}{n^2} \left( \sum_{i=0}^{n-1} c^2 + \sum_{i<j} 2c^2\right)\\ &= c^2 \label{eq:start} \end{align} There is a tradeoff between moving by average velocity, $\av^2$, and changing state as measured by velocity variance, $1/n^2 \sum_{i<j} (\v{i}-\v{j})^2$, because together they always add up to $c^2$. The two extreme cases are:

  1. If all $\v{i}$ are identical, the average speed is $c$, and the rate of change is zero.
  2. If the average is zero, the rate of change is at its maximum, $c^2$.

Lets compare this to the formula from Special Relativity according to which 4 dimensional velocity through space-time is equal to the speed of light $c$: \begin{equation} \left(\frac{c\,d\tau(t)}{dt} \right)^2 + \Vec{v}^2 = c^2. \label{eq:v4} \end{equation} The $\tau$ denotes local time on the moving object, while $t$ is the time for the observer and $\Vec{v}$ is the velocity of the object with respect to the observer. For the ensemble of photons in a box we have $\Vec{v}=\av$, so we can equate (\ref{eq:start}) and (\ref{eq:v4}). Dropping the $\av^2=\Vec{v}^2$ we get \begin{equation} \left(\frac{c\,d\tau(t)}{dt} \right)^2 = \frac{1}{n^2} \sum_{i<j} (\v{i}-\v{j})^2 \end{equation}

The left side describes how time proceeds locally on a moving object, in this case the ensemble of photons. On the right we have the variance of these photons' velocities as a measure of their state change.

For fun and formula simplification, we note that the square root of the variance is the standard deviation $\sigma$ of an ensemble, so we can also write \begin{equation} c\frac{d\tau}{dt} = \sigma(\v{0},\dots,\v{n-1})\;. \end{equation}

It is tempting to speculate that the ensemble of photons is a good role model for the real world, by assuming that

  1. time is (nothing but) a measure of the (rate of) change in physical systems,
  2. the rate of change is governed by all the physical processes that proceed at the speed of light,
  3. all change in a physical system can ultimately be attributed to processes with a direction and proceeding at the the speed of light, i.e. can meaningfully have an $\v{i}$ with $|\v{i}|=c$ ascribed to them.

To summarize: \begin{align} \text{rate of state change} &= \text{photon velocity's standard deviation}\\ &= \text{progression of local time} \end{align}

Open Ends

For a photons-in-a-box world we have seen that the progression of local time is given by how different the photon velocities (i.e. their directions) are. Smaller differences in the velocities intuitively mean a smaller rate of change. And this is formally equal to a slower progression of local time.

Does this have any meaning for the real world?

  • photon is role model for change
  • photon as a point particle is only half the truth, can we do the above with phtons as a wave?
  • max rate of change given by $c$
  • SR: moving reduces possibility for change
  • GR: gravitation reduces possibility for change
  • check entropy$\lt$variance bound: which, with the above remotely looks like the progresseion of time $d\tau/dt$ limits entropy (hmm, of what)