# An Introduction to Theoretical Physics (Part 4 of 5)

Posted on 2011/10/20 by

## Introduction to Part 4

Now we have a solution for the tesseract, we could ask about still higher dimensions.  Rather than a case by case study, we might now aim at a general equation that tells you the resistance of a cube of identical resistors of any dimension.  We have a little experience with this when defining ‘opposite vertices’ for any-dimensional cubes – or n-cubes.

Let’s try to solve this:

What is the total resistance between two opposite vertices of a n-cube of identical resistors?

To progress to a solution, we’ll need to consider the distribution of resistors in an n-cube and see if we can apply the same method that has so far been successful to with the cube and tesseract.  This involved collapsing the vertices to equipotentials and then adding parallel components of the circuit.  It will therefore be useful to know how many vertices and edges make up any n-cube.

## Properties of n-Cubes

### Opposite Vertices

From our previous work:

Opposite vertices on an n-cube are any two vertices where the minimum path connecting them requires moving through n edges.

### Number of Vertices

Let’s consider how we built the tesseract from the lower n-cubes in Part 3.  Starting with a point, each consecutive n-cube required two copies of the previous.  In for example:

$V_3=2V_2$

Where “V” represents the number of vertices.    The indices of “3″ or “2″ tells us that we are talking about the vertices of the 3-cube, or the 2-cube.  Notice that as our goals become more abstract, and require a greater number of variables, the notation must become more compact.

The number of vertices can also be found by considering that we start with one vertex and double each time.  An so they are powers of two:

$V_n=2^n$

### Number of Edges

If we knew how many edges an n-cube has, we’d know how many resistors we’d need to represent an n-cube resistor network.

Again, thinking about how to build the n-cubes from lower dimensions we can quite easily find an equation.  We joined one vertex of the first cube to one (and only one) vertex in the second.  So, for a tesseract, we needed the same number of edges as two cubes and an additional number of edges to join them up.  How many?  As many vertices in the cube.

Mathematically, we could write this as:

$E_4=2E_3 + V_3$

Where ‘E’ represents the number of edges and, as before, ‘V’ the number of vertices.

More generally, we could write:

$E_{n+1}=2E_n + V_n$

But we know that:

$V_n=2^n$

And so we have:

$E_{n+1}=2E_n + 2^n$

This tells us how many vertices an (n+1) cube has, if we know how many make up an n-cubes.

But we can think of the same idea in another way.  For the cube, each vertex has 3 edges – one for each dimension.  The tesseract has 4 edges per vertex – it’s a 4 dimensional object.  So, the n-cube has n edges joining at each vertex.  However, each edge is joined to two vertices, so we must ensure not to count them twice.

$E_n=\frac{nV_n}{2}$

But, as before:

$V_n=2^n$

Which means that:

$E_n=\frac{n2^n}{2}$

Finally:

$E_n=n2^{n-1}$

The first equation we derived is categorized as recursive.  This means that if you know one value in the sequences, you can figure out the next value using the equation.  So if you knew n=4, it would take 5 steps of calculation to figure out n=9.

The second equation is categorised explicit because it is possible to pick any n and figure out the number of edges on the n-cube, without needing to know any other values in the sequence.

Obviously, the latter if these equation is the easier, although the first is perhaps a little more intuitive, considering how we first thought to build the tesseract.

But, intuition can often be misleading, especially in physics or mathematics.  So, we must first check that these two equation agree, as they are describing the same thing.  Let’s take the first few terms:

 Dimension (n) Edges (Recursive) Edges (Explicit) 0 0 0 1 1 1 2 4 4 3 12 12 4 32 32 5 80 80 6 192 192

The equations agree for the first 6 dimensions, but do they agree for all of them?

### Proof by Induction for Number of Edges

The explicit equation can be shown to be equivalent to the recursive by a mathematical procedure called induction.  It requires two components:

1. The explicit equation agrees with the recursive equation for the first term.
2. The (n+1)th explicit equation can be derived from the nth explicit equation using the recursive equation.

The first is obvious, and need not be expressed mathematically.  Both equations yeild zero edges for a zero-dimensional cube (a point).

The second statement requires a little more work and ingenuity.  Here explicit equations:

$E_n=n2^{n-1}$

Let’s substitute it into the recursive equation:

$E_{n+1}=2[n2^{n-1}]+2^n$

And rearrange the brackets:

$E_{n+1}=n[2\times2^{n-1}]+2^n$

Tidying up:

$E_{n+1}=n2^{n}+2^n$

Now here is the difficult part.  We need to rewrite the equation so that it looks like the explicit equation for an (n+1)-cube.  There are no rules to achieve this, just mathematical insight.  Here’s the first step:

$E_{n+1}=(n+1)2^n$

Lastly. is a very sneaky trick, which is difficult to see because it would normally be considered an insane use of algebra.  However, for this purpose, it finishes shaping our equation into the form we want:

$E_{n+1}=(n+1)2^{(n+1)-1}$

The above is now an explicit equation where “n” has been substituted from “n+1″.  This mathematically proves that the recursive equation is equivalent to the explicit equation, and therefore, the proof by induction is complete.

## Solutions So Far

 n Total Resistance 0 0 1 R 2 R 3 5R/6 4 2R/3

This doesn’t seem to provide us with enough examples to guess the equation for the total resistance of an n-cube.  And after noticing that the next n-cube (the 5-cube) has 80 resistors, it seems unlikely that the solution will be found with our present understanding alone.

## Further Solutions

Thinking about the first 5 solutions to our new problem, we can try to think of what quantities were necessary for the end result.

We kept track of how many vertices collapsed to each equipotential, and how many resistors there were between them.  However, it was only the latter of these quantities that were needed for the end calculation.  Diagrammatically, we can represent the collapsed circuit diagram for a general n-cube as follows:

Using the method for calculating identical resistors in parallel (from Part 2), the solution to the above diagram can be written algebraically as follows:

$R_{Tn}=\sum\limits_{i=0}^{n-1}\frac{R}{N_i}$

The sigma is a notation to show summation.  The “N”s represent the number of resistors leaving the “i”th set of vertices in an equipotential.

As the R is constant (the resistors are identical), it can be taken out of the summation.

$R_{Tn}=R\sum\limits_{i=0}^{n-1}\frac{1}{N_i}$

Obviously, this is not a solution, but is a scheme for finding one.

### Restrictions

We also have already learned two additional restrictions upon the solutions scheme.  Firstly, the Ns must be whole numbers.  Secondly:

$R_{Tn}=\sum\limits_{i=0}^{n-1}N_i=E_n$

All this says is that the total number of resistors must (by definition) equal the number of edges on the n-cube.  And so, using the above work on n-cube edges, it must be the case that:

$R_{Tn}=\sum\limits_{i=0}^{n-1}N_i=n2^{n-1}$

This does not seem useful yet, but we can use it to test any potential solutions we find.

## New Notation

With the knowledge that an ever increasing, undrawable, number of resistors are involved with higher dimensional n-cubes, we might consider a new notation.

We presently have numbers associated with 5 solutions.  They are of two kinds: the number of vertices that have been collapsed to each equipotential, and the number of resistors between them.  It is only the latter of these numbers that are used in the final calculation, but we do not yet know the pattern of these numbers so it might be wise to keep track of the equipotentials too.

However, what is clear is that the resistors for the higher dimensions cannot be well represented with the conventional resistor symbols.  Instead, we’ll omit the symbols and just write the number of resistors in each parallel branch.  We obtain the following diagrams for the first five n-cubes.

The 0-cube:

The 1-cube:

The 2-cube:

The 3-cube:

The 4-cube:

As with the previous diagrams, the blue numbers represent the number of vertices that have collapsed into equipotentials and the red numbers are the number of resistors in the parallel components between them.

## Combinations and Pascal’s Triangle

This is what the first 5 diagrams look like in a slightly different presentation.  You might already notice the pattern for the vertices.  They represent the beginning of Pascal’s Triangle.  Notice that each number can be found by adding the two number above it (joined with lines).

The immediate question for a theoretical physicist is to ask, “why?”  Why are the number of verticies in each equipotential of an n-cube equal to the nth row of Pascal’s Triangle?

The nth row of Pascals Triangle is also used for the calculation of binomial coefficients.  For example:

When we find all the possible ways to choose “a” twice and “b” twice from a fourth powered binomial expansion, we find there are six combinations.  The 4th row of Pascal’s triangle allows us to find all the coefficients for this expansion.  For example, this is the full expansion of the above expression:

$(a+b)^4=a^4+4a^3 b+6a^2 b^2+4ab^3+b^4$

Compare the 4th row (the bottom row, as the top is the 0th row!) of Pascal’s Triangle and you will see the correspondence.

Another way to think of the binomial expansion is to impose a meaning upon the algebra that applies to our problem.  We can think of it as follows:

Thinking about the binomial expansion in this way allows us to understand how Pascal’s Triangle is connected to our problem.  When traversing the n-cube, equipotentials are vertices that are the same number of dimensions away from our start vertex.  More formally:

Equipotentials of an n-cube are vertices that have equal minimum paths connecting them with the start vertex.

So the “binomial” has been replaced with the logical statement such “dimension traversed” or “dimension not traversed”.  Vertices with the same number of dimensions traversed are equipotentials.

To calculate how many vertices are in each equipotential, we need two pieces of information.  How many dimension there are in the n-cube (n) and how many dimensions the equipotential has traversed.  Note it does not matter which dimensions were traversed, just how many.  This is a consequence of the physics of equipotentials.

How many vertices can we arrive at after traversing a minimum of “i” out of “n” dimensions?  We can write this question in the following notation:

${n \choose i}$

To figure out how many possibilities there are, or combinations, we can first recognise that we have n dimension to choose from and only i to traverse.    So, we have n possibilities the for the first choice, n-1 for the second, and so on until we have chosen i dimensions.  The number of possibilities is then:

$\frac{n!}{(n-i)!}$

n! means “n factorial“.

However, we have not considered repetitions.  For example there are 2 ways to get to the same vertex on the cube after two dimensions.  Just take the other route around the same square face and you can arrive at the same vertex.  So the order of the dimensions chosen doesn’t matter.  To account for this, we must notice that there are i! repetitions possible for each combination.  And so we arrive at:

${n \choose i}=\frac{n!}{i!(n-i)!}$

Which if we let i=0 to n, gives us the coefficients of the nth binomial expansion, the ith row Pascal’s Triangle, and the equipotentials for our n-cube resistor problem!

### Vertex and Row Totals

It is also easy to see that just as the vertices of an n-cube are powers of 2, so are the totals of each row of Pascal’s Triangle.  We start at the top with 1, which is added twice, following the lines down the triangle, to form the second row.  Then, those two values are added twice each to the make the next row, and so on.  We start with a power of 2 (2 to the power of zero) and then double the result each time.  Hence, each row is a power of two, demonstrating that the rows sum to the total of vertices for each n-cube.

Another very neat proof of this can be provided, if we assume the Binomial Theorem is true.

$2^n=(1+1)^n=\sum\limits_{i=0}^{n}{n \choose i}i^{n-i} 1^i = \sum\limits_{i=0}^{n}{n \choose i}$

## Hasse Diagrams

As often with discoveries such as this, it turns out this type of thinking has already been done elsewhere.  Hasse diagrams are diagrams representing the same sort of logical statements that were used with “dimension traversed”, and so on.

We don’t require this for our investigation and the mathematics requires a background in set theory.  So, I won’t teach this from the beginning.  However, I will comment that the equipotential diagrams we have drawn so far are equivalent to graded partially ordered power sets of n elements.  Look it up!

Briefly explaining that previous sentence, we can partially order the power set of n elements by inclusion (if one set is a subset of another, lacking one element only).  Then, we can grade each subset by how many elements it contains.  The grade corresponds to how many vertices collapse to each equipotential (ordered by the grade) and the partial order relations corresponds to the edge connections between vertices of the n-cube.

If you wish to learn more, Hasse Diagrams also represent Boolean Algebras which is a more formal description of our interpretation of the binomial expansions.

Here are the Hasse Digarams for the first 5 n-cubes (omitting the diagram for the 0-cube).  You will hopefully see the similarity:

The 1-cube

The 2-cube:

The 3-cube

The 4-cube:
The (yet to solve) 5-cube:

## Summary

I have delibrately left the solution until our final part.  Perhaps try to see if you can solve it yourself before tomorrow.  Mostly everything you need to do it is contained in the previous 4 parts.