Why a Discrete Minimum Distance Fails

In the previous post entitled "Why Strings Don't Exist", we showed it is possible to have a length shorter than the famed Planck length. The question becomes, what is the minimum discrete distance, assuming there is such a thing? Normally, we use infinitesimal points to build geometrical objects like lines, planes, rectangles, etc. What happens if we use a line with the smallest magnitude possible that is greater than zero? Before we delve into these questions, let's define the variables:

OK, let's assume the fundamental building block is a line somewhere between zero and the Planck length. We'll call it d:

Let's try building a square with d:

So far, so good. All the distances appear to be no less than d. But what is the distance along the diagonal (or hypotenuse)?

The diagonal distance is a little bit more than d. The additional distance is marked in red. Notice this distance is not an integer multiple of d. To make this distance, we need a distance d plus a distance less than d. There can be no distance less than d, so we can't draw the above square. Here's an idea: draw a rectangle with 3X4 d-units. Thanks to Pythagoras, the diagonal will be 5 d-units:

Because all distances must be integer units of d, our geometry does not include squares, rectangles or triangles that have diagonals and sides that fail to have magnitudes that are integer multiples of d. But at least we found one rectangle that works--or maybe not:

If we draw lines from each d to every other d, we once again have distances that are not integer multiples of d. In the example above, we have a distance (c) of 3.6d. To have that distance, we need 3d plus a 0.6d. In our geometry, there's no such thing as 0.6d. Thus we can't draw the 3X4 rectangle. In fact, every rectangle and triangle we draw will have some distance that includes a fraction of d.

Perhaps we'll have better luck with circles? Check this out:

If we take distance d and shape it into a circle circumference (C), the diameter (D) will be less than d! D = d/pi (where C=d). If we try to draw any circle with d-units, we hit a brick wall. You see, you calculate the circumference with pi * D. If D is an integer (n) multiple of d, pi * n won't equal a circumference with an integer multiple of d. Pi is an irrational number.

OK, so perfect circles are out. How about imperfect circles? Perhaps we can replace pi with something more rational. Even if we do, circles have the same problem we encountered earlier:

There's always a line between two d's, that is not an integer multiple of d. This is also true with any shape imaginable:

So far, things look pretty hopeless for our geometry based on d. But unfortunately, there's more pain. Let's go back to the beginning and reexamine d:

Distance d is a line along the x-axis. But what is it along the y and z axis, i.e., what is its cross section?

Line d has a cross section of zero magnitude! That zero magnitude is just a single point in space with zero distance! Zero distance is not allowed, so a distance-d line is not allowed. Perhaps we can convert the line into a cylinder, so the cross-section will have a d-magnitude. Let's look at our new cross section:

Oops! The cross section is shaped like a circle. We can easily draw a red line from one d to the other that isn't an integer multiple of d. Thus it is now abundantly clear that a geometry based on distance d (instead of a point) is a dismal failure.

Update: Here is a formal proof that shows that, at any two non-parallel adjacent unit lengths (in this case the Planck length), it is possible to draw lines (lines a, b) shorter than the unit length. The following diagram is an example of an arbitrary shape:

Note that the shape is made up of connected unit lengths. Here are two examples of adjacent-unit lengths blown up to a convenient size for inspection:

Now let's take one arbitrary pair of connected unit lengths and label the lines and angles:

Now we are ready to do the proof. Here it is:

Update: The following proof demonstrates why a curved Planck-length string creates distances shorter than the so-called shortest distance. Below are some examples of curved strings:

The examples above clearly make the point, but is the point generally true? Here's the proof:

Taming Infinities--Introducing n-space

Each line has an infinite number of points. We tame this infinity by creating an arbitrary finite unit. For example, take the set of real numbers. Between 0 and 1 there are an infinite number of numbers:

Normally, we count using integers: 1, 2, 3, etc. But we don't have to do it that way. We could count like this: 1-infinity, 2-infinity, 3-infinity--all the way up to infinity-infinity. If 3 is greater than 2, than 3-infinity is greater than 2-infinity. So what we have are different magnitudes of infinity that make up our finite numbers. With this in mind, it seems reasonable to assume we could add up an infinite set of numbers and get a finite number. For example, we could take the entire line of positive real numbers ...

...and shape it into a circle:

Now infinity is equal to zero and 2pi radians, i.e., finite numbers.

Let's imagine we are extremely naive, we don't know the first integer greater than zero. So we decide to add up all the real numbers from zero to the next integer point. That gives us an infinity:

The vertical lines represent the infinite quantity of real numbers between zero and the question mark. They make a nice 2D drawing of a triangle. The average real number is at the half-way point. If we take this number (.5) and multiply it by 2, we get the right answer: not infinity, but 1. This is the basic logic behind n-space. We take an infinite number of points in space of any number of dimensions and map them to a 2D space. The average value (expectation value) becomes our vertical axis. We multiply this value by the horizontal axis to get the total area which is a finite value.

The above diagram shows how each point in the original lattice space is mapped to n-space. Each point in the original space becomes a vertical line in n-space. So an infinite number of points, lines, planes or cubes (lattice cells) become an infinite number of vertical lines. The average vertical line (bar-np) is multiplied by the horizontal line (nx) to get the area--which is the correct finite answer.

Why is the n-space area the correct answer--and not infinity? Consider the following diagram:

Max Planck found that if he added up a set of finite discrete energies, he got the correct finite value. The above diagram shows we can also add up an infinite set of continuous energies and get the same finite value! Whether the energies are discrete or continuous, the area under the curve is the same. Thus, finding the area under the n-space curve is a way to find the correct answer. (Take note that, throughout this post, we take the energy term normally reserved for a single particle and use it to represent any energy. Sometimes the frequency and Planck's constant are set to one.)

The following relationships show us how to get the values in n-space we need to calculate the correct, finite answer:

Now, we want n-space to help us solve infinity problems in the quantum as well as the classical realm. This is why n-space was derived from Heisenberg's uncertainty principle. Here are the variables involved:

Here is the derivation:

Equation 12 shows the n-space area is always greater than or equal to 1/2--or the ground-state:

According to equation 13, the total energy in a system, like Planck's constant, has two components, dimensions, or factors (nx, np). The horizontal dimension (nx) is derived from position, and the vertical dimension is derived from momentum. The total energy is equal to or greater than the ground-state energy. Using equation 12 we can derive equation 14:

At 14, k is a constant, so if equation 14 represents the total energy in the system, that energy is conserved. It does not matter how big or small the average energy is at any given point in the original space or lattice. Nor does it matter if there are an infinite number of such points. That energy or quantum number (np) is offset by quantum number (nx), giving the conserved quantity.

Now that we've laid the groundwork for n-space, let's attempt to solve a classic problem: calculating the total energy in a sphere, where each point in that sphere has a given amount of energy, momentum, and/or mass. And, of course, there are an infinite number of points in the sphere.

Immediately we run into a problem: if we know the exact energy at each point, we know the exact momentum if we divide the energy by c (light speed), and, it is obvious we know the exact position as well--a clear violation of the uncertainty principle. If we zero in on a point in space, according to Heisenberg, we should be totally uncertain about the energy and momentum. According to the de Broglie wavelength formula, if we reduce a wavelength to zero, i.e., a single point, we should have infinite energy! And, we can only know that if we have no clue where that point is located!

Below is the relevant math:

Realistically, each point of energy is not a point at all, but a wavelength with a one-dimensional magnitude. If the average wavelength is greater than zero, then we have a finite energy at each wavelength.

We can think of each wavelength as a line. Assuming we know the energies and momentums, we don't know the positions, but we can make this fact unimportant if we calculate the average energy/momentum. Then we know that at any randomly chosen position the average energy is always the same. We can then map each energy/momentum to n-space.

Using equations 21 and 22 we find the average vertical factor (np).

We use the following equations to find the horizontal factor (nx):

At equation 23 we see a problem. To find nx we must first know nt--the total that we are trying to calculate! So we move on to equation 24. We know the total volume but we don't know this thing called the unit volume. We get a unit volume by taking the volume of another system, where we know all the variable values, and multiplying that volume by a factor of np/nt (see equation 25). Once we have our unit volume, we can plug that into equation 24 to get the nx value for the instant problem.

We do the final steps below:

The strategy we used works as long as the following is true:

Suppose we have a scenario where we have a volume of energy, say, a star. The energy is conserved as follows:

The star collapses into a black hole. All wavelengths allegedly shrink to a zero limit. That forces the average momentum factor np to blow up to infinity:

At equation 31, the star's radius also shrinks to a zero limit. We should end up with a singularity that has a position unknown to us, assuming we know the total energy, mass, and momentum. We can imagine the singularity being anywhere within the Schwarzschild radius. Nevertheless, we can crudely map the star to n-space as follows:

As explained earlier, the positions of the wavelengths and the position of the singularity become irrelevant when we determine the average np for each wavelength. Now, let's assume we don't know the star's total energy. We want to find it, so we need to find the value of nx. The star volume is its radius cubed times 4/3 pi. The Schwarzschild radius cubed times 4/3 pi shall serve as the unit volume. We divide the volume by the unit volume--the 4/3 pi's cancel:

When we do the math we see that the star's total energy is finite and is equal to the black hole's (assuming energy is constant and none was transferred).

So in the case of the black hole, np had an infinite limit, but nx had a zero limit--so the total energy ended up being finite and conserved.

Why the Ground State Ain't Zero

If you are familiar with Max Planck's work or Albert Einstein's photo-electric effect, then you know energy comes in discrete packets called quanta. These discrete energy levels are represented by nice, neat integers, where n = 1, 2, 3 ... But then there is this equation:

The above equation's second term is the ground state--but why isn't the frequency (f) and Planck's constant (h) multiplied by a nice, neat integer? Why 1/2? That's odd. And why ain't the ground state zero? Surely if you remove everything there should be nothing, nada, zip! Definitely not 1/2.

To understand what's going on, let's try a thought experiment. Imagine you want to push a couch a distance of x. You put your hands on the couch and apply zero force or energy. You gradually increase the energy applied until you reach a critical value where the couch begins to move. Let's label that critical value 1E.

Now suppose you want to push two couches along distance x. You gradually increase the applied energy to, say, 1.7E:

Unfortunately, 1.7E is not enough energy to do the work, so gradually increase the applied energy to 2E:

Assuming the couches are identical, the critical values of energy needed to push one or two couches are 1E and 2E. The critical coefficients are integers. If we want to move n couches we need nE. Below is a diagram of our thought experiment:

Note that the energy you applied is continuous, but the critical values (in red) are discrete. Also note the lowest energy is zero, so again, where does the (1/2)fh come from? Consider a single die. It has six discrete states: 1, 2, 3, 4, 5, 6. If we add up these states, we get 21:

This time, instead of a continuous line, the following die diagram has discrete steps:

Now, let's imagine a die with an infinite number of sides (or states) ranging from zero to six. Its states are no-longer discrete, but continuous. However, like the couch experiment, there are critical values marked in red (see diagram below).

Once again note the ground state is zero, but also note when we add up all those energies (see equation 3) we don't get 21 like before. We get 18! The following diagram illustrates the difference between the six-sided die and the infinite-sided die. The six-sided die clearly has more area under the curve. That extra area is indicated in gray:

So how do we fix this discrepancy? Let's include a y-intercept as they say in calculus parlance:

At equation 5 we use an intercept of 1/2. Doing so gives us a sum of 21--equal to the six-sided die! The diagram below illustrates this point. Note the gray areas cancelling each other:

Also note the ground state is not zero--it's 1/2.