Is This Why the Speed of Light is Universally Invariant?

Question

Please could you tell me if the following is an original thought or whether this is already understood. I ask because I am undertaking a piece of writing on the nature of spacetime. What I discuss below is actually my own epiphany but I am curious to know if I am stating something that is already commonly understood:

Time flows at the same rate for subjects in the same frame of reference. Thus, if two observers are completely still, relative to each other, no matter how far apart they might be (they could be hundreds of light years apart) then time will flow at exactly the same rate for both.

If they are still, then they will both also be travelling through time at the speed of light. Now,

$$distance = ct$$

So in 1 second, our subjects will have travelled $3$x$10^8 metres$ through time. This implies $3$x$10^8$ metres is equivalent to 1 second.

How fast we flow into the future when we are completely still may govern the physics of our motion through space.

If we think about it, this has to be true! If we accept that we exist within the fabric of spacetime where time itself is woven into space as a fourth dimension (HG Wells gives a great explanation of why this must be so in his book, ‘The Time Machine’), then simple maths will show that time has to slow down the moment we start to move from rest. This is simply from the consequence that time is woven into space and we flow through time (into the future) at a specific rate -a universal rate when we are at the same stationery frame of reference.

If we plot the world line of a stationery person, we get a vertical line upwards (when time runs along the vertical axis and space along the horizontal axis). When that person wants to move (spatially), then his world line slants forward. If we consider that his motion along his world line must remain constant at all times, such that he covers the same distance directly along his world line, in a given period, as he did when he was still, then as he gains spatial motion, the component of his motion along the vertical (time) axis is shortened for the same distance along his gradient. Effectively, his movement through space is traded from his movement through time. Time therefore has to slow down. At the speed of light, all of his motion through spacetime is along the spatial direction and he will have no component along the vertical time axis. This means, at the speed of light spatially, time stops.

Therefore, we have a picture where we are either stood still and moving through time at the speed of light or we are moving through space at the speed of light and time stands still for us or our motion through the fabric of spacetime lies somewhere in between where time slows down in order to allow us to move. I hope I have made sense.

I have an obvious model in my head that provides this result, naturally: Imagine spacetime is quantised (perhaps is at the Planck level). Think of space as a cube of pixels (similar to a video monitor). Imagine lots of cubes. Each cube represents our space at any given moment in time. Let’s now arrange these cubes of space along a fourth dimension, the dimension of time such that we create a pixelated tesseract to represent our universe. Let’s consider each ‘cube’ of space as a frame (like a frame in a video (or hologram)), then contiguous frames across the fourth dimension could represent different moments in time. Motion across spacetime would then be the displacement of our subject between each contiguous frame. When we are still, we experience flow into the future only as our awareness flicks from frame to frame at the speed of light (temporally). Now if our motion between pixels in any direction, is fixed (ie: constant), then any subsequent spatial movement (displacement) will mean that we are moving slower across frames in order to accommodate our spatial motion such that absolute magnitude of our motion remains fixed.

We can only move one quanta/pixel at a time in any direction, whether that be a purely temporal direction, purely spatial direction or shared between temporal and spatial components of our spacetime. Thus, if we remain completely still, all our motion is through time. If we now wish to be displaced spatially instead (ie, move through space), then the logic of geometry dictates that the new spatial advance must arise from a loss in temporal component of our advance.1

Or in plain English, if you want to move through space, time has to slow down to allow you to do that because we are flowing through spacetime at a fixed rate overall!

Let me put it another way: Our motion through the fabric of spacetime is fixed! This speed is conserved, if you like. This is because the experience of our existence is a result of an awareness of our co-ordinates in spacetime at any given moment (or ‘frame’ in my model). As we move, our co-spacetime ordinates are displaced one quanta (‘pixel’) at a time between each temporal frame. So, for any motion through spacetime: You cannot go slower. You cannot go faster. Our motion through the fabric of spacetime may be dictated simply by how fast we move through time when we are still. If we accept this as a postulate, then it follows that any gain in spatial displacement between frames can only arise out of a reduction of temporal displacement between frames as a consequence of this conservation of flow through spacetime.

If our subjects then begins to move through space, accelerating towards the speed of light, all his temporal flow gets converted to spatial flow.

For this reason, for a photon which has no rest mass, all of its motion is spatial (as there is no mass to inhibit its motion through space). The photon therefore HAS to move at the conserved rate of flow through spacetime. However, as all of its motion is spatial, and that rate of flow is conserved, it HAS to travel spatially at the speed of light. As a result, it also has no temporal flow and time is still for the photon.

Is this the reason why we can never exceed the speed of light? Because this speed is decided by the rate at which we flow through time when we are "still".

Is what I have presented above, of common awareness amongst physicists? Am I stating the obvious? Or is this a unique perspective which providing a bit of clarity?  

Answer 1

It has nothing to do with string theory. You should never try to explain a prediction of a scientific theory in terms of a more recent one without first understanding the argument in the original, in this case special relativity. But until then, your reasoning was actually pretty much correct.

Before Einstein, we would have said the squared length of the path between two points in space was the sum of the squares of the differences between the points' Cartesian coordinates $x,\,y,\,z$. If we represent a change in a variable $v$ as $\Delta v$, the squared length is$$(\Delta x)^2+\Delta y^2+\Delta z^2,$$where I've dropped the implicit brackets after the first term. But if you rotate your choice of axes, the length is invariant.

In special relativity, we include time as well, so the invariant quantity is$$\Delta x^2+\Delta y^2+\Delta z^2-c^2\Delta t^2,$$where the speed $c$ makes the units right (you can't subtract a squared time from a squared length). Well, that works provided the speed doesn't change. In general, we need to talk with calculus about$$ds^2=dx^2+dy^2+dz^2-c^2dt^2$$being invariant. In a moment, I'll rearrange that equation with some colours added.

So, let's move at speed $\beta c$, with $\beta$ just a number. By Pythagoras, $dx^2+dy^2+dz^2=\beta^2c^2dt^2$, so $ds^2=-(1-\beta^2)c^2dt^2$. (Don't worry about $ds$ being imaginary-valued; we could have multiplied the definition by $-1$ without changing the gist of what follows, but physicists have been known to use both conventions.) From calculations like this, you can deduce time dilation, length contraction etc. But the formula for $ds^2$ conveys how a speed of $c$ gets Pythagoras-shared between $\color{red}{\text{time}}$ and $\color{blue}{\text{space}}$ viz.$$c^2=\color{red}{-\left(\frac{ds}{dt}\right)^2}+\color{blue}{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2}.$$