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 Entropy Happens

In the beginning, there was order, a very hot singularity, but as time progressed the universe expanded and cooled--and became more disorderly. Scientists predict a "big freeze." It's all due to entropy. As you read this blog, entropy continues. Why? That's what we will explore below. First, let's define the variables we will use:

We begin with the partition function, which has the Boltzmann factor, an exponent with a thermodynamic beta power over the base e:

If we want to determine the probabilities of the energies in a system, we make sure the probabilities add up to 1, so we normalize the partition function by dividing it by itself (Z):

However, if we want to model the universe's evolution, we need to make a slight change to the partition function. Instead of using the thermodynamic beta, we use its reciprocal. We also change the i index to a time (t) index:

Also, we want the universe's total energy to be conserved. We know dark energy is increasing and radiation energy is decreasing, so we put together an energy-conservation equation:

At equation 5, notice how an increase in the universe's volume (V) reduces the radiation energy (Er) but increases the dark energy (pV). Multiply the two energies, add a little dark and baryonic matter, and take the square root and we get a constant energy (E).

We define temperature as follows:

As volume (V) increases, the universe's temperature decreases. Equation 7 below gives us the probability of the temperature at a given time t:

A high temperature has a low probability. A low temperature has a high probability. So there is a high probability the universe's temperature will continue to decrease, and a low probability the temperature will increase. Thus, an expanding universe has a higher probability.

Now, let's take a look at entropy. We define it as follows:

We see that entropy increases as temperature decreases--so it has the same probability as temperature:

So why does entropy happen? Greater entropy has a higher probability than lower entropy. We can also say that reverse entropy is possible but less probable. A good example is the one Tyson discussed in the above video. There are pockets of order caused by star energy, so life is possible.

An Amplitude Squared Equals a Probability--a Mathematical Proof

Start with a wave:

Add more waves. Waves are in and out of phase with each other (constructive and destructive interference):

The range is from zero degrees out of phase to 180 degrees out of phase. On average, any pair of waves is 90 degrees out of phase, making a sine wave and a cosine wave:

Each has an amplitude (Ao). We can add the waves together, making a complex number (equation 1). Since our intention is to isolate and square the amplitude, we need a complex conjugate (equation 2). Using Euler's identity we write the following equations:

When we square the the left sides and square the right sides of equations 1 and 2, here's what we get:

It's entirely possible we don't get a probability. Instead, we could get a value greater than one. Plus we have distance units--so we need to normalize the amplitude:

Now, let's express our oscillating wave(s) in terms of Hooke's law and derive an energy (E):

Below we make a couple of substitutions to derive equation 10:

At 11 we define the probability of E. It could be from zero to one:

Multiply both sides of equation 10 by the probability P(E) then add a pinch of algebra to get equation 16:

Equation 16 shows the probability is equal to the square of the reduced normalized amplitude. Of course we are not limited to just energy eigenvalues. We can show that any type of eigenvalue has a probability equal to the square of a normalized amplitude:

In fact, if we express energy E in terms of momentum (p), mass (m), position (x), time (t) and wave number (k), we discover that the momentum, mass, position, time and wave number each have the same probability as the energy: the square of the amplitude A'. This is due to the energy being dependent on a specific value of each of these other eigenvalues. Each term below contains one eigenvalue and constants (which don't change). So the energy eigenvalue correlates with each of these other eigenvalues, and, likewise, the probabilities correlate.

In conclusion it is safe to say a wave amplitude squared does not guarantee a probability; however, it looks as though the square of a normalized amplitude does.

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.

How to Solve Complex Equations When Perturbation Theory Fails

Perturbation theory is useful to any mathematician or physicist who needs to solve a seemingly unsolvable equation such as the one below:

Equation 1 may be hard to solve, but its structure looks simple. Perturbation theory works fine in its case. Later, we will try to solve a much more complex equation, where perturbation theory ceases to be practical, and we will use an alternative method. But for now, let's explore the benefits and limitations of perturbation theory. To solve equation 1 we first reformat it as follows:

Then we expand x and x^2:

Next, we make substitutions:

Let's assume the following terms equal zero:

Using a little algebra and more substitutions we solve for x0, x1, and x2. When we add these x's up we get the value of the original x:

Not bad! But check out this monster:

OK, let's set some values for y and z and solve for x. We derive equation 19:

Should we solve it with perturbation theory? Well, let's see. We first have to expand x^5:

I don't know about you, dear reader, but I don't even want to think about expanding x^5! In our previous example we only had to expand x^2 to only the second power of epsilon. Can you imagine the can of worms we will open if we expand x^5 to two or more powers of epsilon? There's got to be an easier way ... and there is.

To solve equation 19 we use a systematic guess-and-check method. The idea is to choose a value for x that will give us an answer that is close to zero if not exactly zero. We first test -1, 0, and 1. We see that 1 is closer to zero. We try 10. If 10 is better, we try 100, and higher powers of 10 as we get closer to zero. After all, the value for x could be any real number from negative infinity to positive infinity.

Now, after checking 10, we see it's no good. So we now know x is between 1 and 10. We try 2. Whoops, 1 was better. So we try 1.1 and see an improvement. Then we try 1.2--even better. We increment by tenths to 1.3. At 1.4, we overshoot. So the answer is around 1.3, so we try 1.29 and 1.31, i.e., we increment in hundredths. Our final answer is 1.29. If we want more accuracy, we increment in thousandths, and ever smaller powers of 10. Below is a table of our results:

The chief advantage of this method is it is possible to write a computer program that will yield a quick answer.

Update: Click here for a proof of perturbation theory

Resolving the Cosmological Constant Problem

Imagine you're a quantum physicist and you just got through calculating the energy density of the vacuum. To your horror you find that your calculation is around a hundred orders of magnitude greater than vacuum energy density measurements! This, in a nutshell, is the cosmological constant problem. According to your numbers, the cosmological constant should be huge. According to WMAP measuremments, the cosmological constant is too small to tweet about. In this post we shed some light on this problem. First, here are the variables we will be using:

Below we derive the cosmological constant value based on the WMAP vacuum energy density measurement of approximately 10^-10 Joules per cubic meter (see equations 3 and 4):

The next order of business is to consider the limitations caused by the Compton wavelength formula and the Heisenberg uncertainty principle--so we derive equation 8:

There are two ways we can interpret equation 8: 1. If delta-x grows smaller, energy (delta-E) grows bigger. This makes perfect sense when you consider that smaller spaces contain shorter wavelengths, and shorter wavelengths correspond to higher energies. 2. If delta-x grows smaller, delta-E becomes more uncertain. In any case, we use equation 8 as our means to make sense of the cosmological constant problem.

Equation 9 below is equation 8 modified. On the left side we have the vacuum energy measurement (E) and one meter (x). Of course the product of these values is far greater than Planck's constant, so the right side has the scale factor n:

OK, here's what we need to do: we need to reduce the vacuum's volume while maintaining a constant energy density. This can be done, but only up to a point. The question becomes, up to what point? Or, if you prefer, down to what point? How small can we make the energy and space before things get weird? Here's the answer:

Equation 10 above is the formula for finding the cutoff, or, the Heisenberg uncertainty limit. The product of the energy and space can't go below h-bar/2 without violating the uncertainty principle.

At 11 and 12 below, we plug-n-chug:

At the cutoff, the distance is around 10^-4 meters. The volume is 10^-12 cubic meters. The energy is approximately 10^-22 Joules. If the space is reduced further, the energy value will explode! With an upper limit of infinity! Obviously, if the energy skyrockets, or becomes more uncertain, the energy density won't be the same--and any calculation done at that level will yield ridiculous results.

The diagram below shows vacuum energy density matches observations at or above the cutoff. Any meaningful measurement is taken above that point. But if you prefer infinities and uncertainties, feel free to go smaller than the Heisenberg limit.

Thus, the cosmological constant problem appears to be a natural consequence of doing calculations and measurements below the cutoff limit.