The Last Theory
The Last Theory
The Last Theory
26 May 2022

# Are Wolfram’s graphsthree‑dimensional?

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Are Wolfram’s graphs three-dimensional?

In my article How to measure the dimensionality of the universe, I introduced a mathematically-minded crab, which was able to determine the dimensionality of its universe by measuring how much space it covered moving different distances in every possible direction.    Now I’m going to use the same crabby method to determine the dimensionality of graphs generated by Wolfram Physics.

I’m finally going to answer the question: how many dimensions are there in this universe? And the answer’s going to be unexpected.

Here’s a hint: it’s not two and it’s not three.

## The measure of the graph Now, I’m pretty sure I know the dimensionality of this graph.

It’s two-dimensional.

From our God’s-eye perspective, that’s pretty obvious.

But let’s do this from a crab’s-eye perspective, from inside the graph.

Let’s start from the node in the middle of the graph: Remember that our crab determined the dimensionality of its universe by measuring how much space it covered moving different distances in every possible direction.

To use the crab’s method, first, we need to decide the distance we’re going to move.

The only obvious measure of distance in the crab’s universe was one crab length. The only obvious measure of distance in our graph is one edge.

Remember that the edges of the graph don’t have any particular length. I mean, I’ve drawn all the edges of this graph the same length, about half an inch long, depending on what kind of screen you’re looking at right now. But I could have drawn all the edges different lengths, or I could have drawn them half a micron long or half a mile long, and it’d still be the same graph. So the only way to measure distance in the graph is to count numbers of edges.

Following the crab’s example, we’re going to move one edge in every possible direction: You can see how a graph is different from the crab’s continuous universe. The crab was able to move in an infinite number of directions, tracing out a circle. In this graph, however, we can only move in four directions from that middle node, along the four edges shown in red.

Next, we need to decide how much space we’ve just covered.

The only obvious measure of the amount of space the crab covered in its universe was the amount of space the crab took up itself. The only obvious measure of the amount of space we’ve covered in our graph is the number of nodes we’ve reached.

Again, you can see how a graph is different from the crab’s continuous universe. The amount of space the crab covered was measured in numbers of crabs, including fractions of a crab: the crab might cover 3.14 crab’s worth of space, or 12.57 crab’s worth of space. In the graph, however, we always reach a whole number of nodes.

In this case, we’ve covered five nodes’ worth of space, including the one we started from and the four nodes we’ve just reached: So by moving 1 edge in every possible direction, we’ve covered 5 nodes’ worth of space:

r = 1 → V = 5

## Edge trip

Now let’s go two edges in every possible direction. By doing so, we reach 13 nodes: the 5 we reached within 1 edge, now shown in darker red, plus another 8 we reach within 2 edges, shown in brighter red: r = 2 → V = 13

We can keep doing this, moving 3, 4 and 5 edges in every possible direction: r = 3 → V = 25 r = 4 → V = 41 r = 5 → V = 61

## I’ve got the power

Remember how the two-dimensional crab found that the amount of space it covered increased with the distance it moved to the power of 2, and the three-dimensional crab found that it increased with the distance it moved to the power of 3?

Well, we’re going to do the same thing the crabs did, to find out whether this graph is two-dimensional or three-dimensional.

I’m going to spare you the mathematics, but every time we move one more edge, we can calculate whether V is increasing with r to the power of 2, or 3 or some other number.

I’m going to call that number D, the dimensionality of the graph.

If you’re interested in the formula I use to calculate D, check out the footnote below.

## Cut to the chase

So now we can measure the dimensionality of our simple graph.

I’m going to start again from the node in the middle of the graph, and every time we move one more edge in every direction, I’m going to calculate D, the dimensionality of the graph.

It’s... D = 0

Well, that’s disappointing. All that work, and our calculations tell us that this is a zero-dimensional graph? That’s obviously not right.

Let’s move another edge in every direction: D = 1.38

Well, that’s a bit better, but it’s still wrong. After this second step, our calculations are telling us that this is a 1.38-dimensional graph, but we can clearly see that it’s a two-dimensional graph.

Let’s go another edge: D = 1.61

Closer, but still not right. What’s happening here? Why can’t our calculations get it right? This is not a 0- or a 1.38- or a 1.61-dimensional graph, it’s a two-dimensional graph.

Well, here’s the problem.

The crabs’ universes, remember, were continuous. They could move in an infinite number of directions.

This graph, however, is not continuous. We can only move along edges.

On a large scale, the graphs of Wolfram Physics are going to look continuous: But when we’re starting out with our measurement of the dimensionality of the graph, we’re operating on a small scale, at which the graph is not continuous.

The solution is to keep going. As we move ever more edges in every direction, the graph is an ever closer approximation to a continuous space, so our calculation of D is an ever more accurate measurement of the dimensionality of the graph on a large scale. D = 1.72 D = 1.78

Now we run into another problem: the edge of the graph.

If the two-dimensional crab ran into the edge of it’s universe, it would no longer be tracing out a circle – it’d be a squared off circle – and the amount of space it covered would no longer increase with the distance it moved to the power of 2. It’s the same for our graph: D = 1.56 D = 1.17

Our calculation of the dimensionality of the graph is now falling away from two.

So, unfortunately, for this small graph, we don’t have a definitive number of dimensions.

Our numbers start out wrong – 0, 1.38, 1.61 – because the graph is not continuous. They get closer to the right answer – 1.72, 1.78 – as we move an increasing number of edges in every direction. The numbers go wrong again – 1.56, 1.17 – as we run up against the edge of the graph.

To get the most accurate possible number of dimensions, we’re going to have to ignore the numbers at the start and the end of our measurement, and go with the value we converge on in the middle of the measurement.

And we’re not going to be able to find the dimensionality of a graph that’s too small.

## Go big or go home

So let’s try measuring the dimensionality of a larger graph: This starts exactly as the smaller graph did, with numbers that are too small, because the graph is not continuous; but the numbers do trend towards the right answer: r = 5 → V = 61 → D = 1.78

With this larger graph, however, we don’t run into the edges so soon, and the number of dimensions, D, edges ever closer to 2: r = 10 → V = 221 → D = 1.9 r = 15 → V = 481 → D = 1.93

D = 1.93 isn’t the answer we were expecting, D = 2, but it’s close.

When we do run into the edges of the graph, the numbers go wrong again: r = 20 → V = 741 → D = 1.19

Still, as long as we go big and stay away from the edges, we can get a reasonable approximation of the dimensionality of the graph.

## Now in 3D

Now let’s level up a dimension.

I might be jumping the gun here, but this graph looks kinda three-dimensional to me: We’re going to use same crabby method as ever: starting at the node in the middle of the graph, we’re going to measure how much space we cover moving different distances in every possible direction, and determine whether that amount of space increases with the distance to the power of 2, or to the power of 3, or to some other power.

As ever, we start out with the wrong answer, because the graph is not continuous: r = 1 → V = 7 → D = 0

D = 0 is wrong. This is not a zero-dimensional graph.

But we soon start trending towards the right answer: r = 3 → V = 63 → D = 2.28

Until, that is, we run into the edges of the graph, when it all goes wrong again: r = 5 → V = 195 → D = 2.07

We didn’t get to the answer we were expecting, D = 3, because we didn’t go big enough.

So let’s go larger: With a bigger three-dimensional grid, we come much closer to D = 3: r = 5 → V = 231 → D = 2.61 r = 10 → V = 1561 → D = 2.83

## Over to Wolfram

Now, at last, we can apply our method for measuring the dimensionality of a graph to one that’s been generated by one of the rules of Wolfram Physics: Any guesses what the answer will be?

Remember, we’re going to have to ignore the numbers at the start and the end of our measurement, and go with the value we converge on in the middle of the measurement.

It really doesn’t matter which node we start with, since there’s no middle to this graph, so I’m going to start with the one at the bottom: After a few steps, the number of dimensions goes over 2: r = 5 → V = 44 → D = 2.03

A couple of steps further, and it goes over 3: r = 7 → V = 116 → D = 3.37

Then we hit the limits of the graph, and the number goes back down: r = 9 → V = 196 → D = 1.31

You can see that it doesn’t matter whether the graph has edges, like the grids we were looking at earlier, or whether it wraps back on itself, like this one. Either way, when we run out of edges to move along, the calculation goes wrong.

Let’s go back to that number we reached:

D = 3.37

It’s a small graph, so this is just an approximation, but it looks like this graph is at least 3.37-dimensional.

## What does this mean?

How can dimensionality be fractional? I mean, I can imagine what a two-dimensional universe might look like, and I know for sure what a three-dimensional universe looks like, but what would a 3.37-dimensional universe look like?

And is it possible that we might live in a universe that’s more than three-dimensional?

In my next article, I’m going to ask a deeper question to try to make sense of this 3.37-dimensional graph: what are dimensions?

Here’s the formula I apply, every time we move one more edge in every direction, to determine the dimensionality of the graph:

D = (log(Vr+1) – log(Vr)) / (log(r+1) – log(r))

It’s a formula Stephen Wolfram proposes on page 165 of his book A project to find the Fundamental Theory of Physics.

All it does is determine whether V increases with r to the power of 2, 3 or some other number.

For example:

• when we step from r = 1 → V = 5 to r = 2 → V = 13, we can plug the numbers r = 1, Vr = 5, r + 1 = 2 and Vr+1 = 13 into the formula to give D = (log 13 – log 5) / (log 2 – log 1) = 1.38
• when we step from r = 2 → V = 13 to r = 3 → V = 25, we can plug the numbers r = 2, Vr = 13, r + 1 = 3 and Vr+1 = 25 into the formula to give D = (log 25 – log 13) / (log 3 – log 2) = 1.61
• and so on

Why does this formula work?

Well, remember that the two-dimensional crab found that moving a distance r in every possible direction covered an amount of space V = π r2.

Plugging the 2D crab’s formula into Wolfram’s formula gives:

D

= (log(Vr+1) – log(Vr)) / (log(r+1) – log(r))

= (log(π (r+1)2) – log(π r2)) / (log(r+1) – log(r))

= (log(π) + 2 log(r+1) – log(π) – 2 log(r)) / (log(r+1) – log(r))

= 2 (log(r+1) – log(r)) / (log(r+1) – log(r))

= 2

And the three-dimensional crab found that moving a distance r in every possible direction covered an amount of space V = 4/3 π r3.

Plugging the 3D crab’s formula into Wolfram’s formula gives:

D

= (log(Vr+1) – log(Vr)) / (log(r+1) – log(r))

= (log(4/3 π (r+1)3) – log(4/3 π r3)) / (log(r+1) – log(r))

= (log(4/3 π) + 3 log(r+1) – log(4/3 π) – 3 log(r)) / (log(r+1) – log(r))

= 3 (log(r+1) – log(r)) / (log(r+1) – log(r))

= 3

So you can see that the formula does exactly what we want it to: it determines whether V increases with r to the power of 2, 3 or some other number.

In other words, it tells us the dimensionality of the universe.

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