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.
Reality says, "It's finite," but your math says,"It's infinite." Why? And, is there a simple way to solve the problem? Today we present an incredibly simple formula for removing unwanted infinities. First, let's define the variables we will use:
Now that we got that out of the way, let's examine why infinities occur. One way to get infinity is to add up an infinite number of numbers as the following integral-summation shows:
You may have noticed something missing in equation 1 above. The thing that's missing makes the infinity finite:
The missing ingredient is none other than dx -- a really tiny number. When the infinity is multiplied by dx it magically becomes a finite number.
So one way to rid ourselves of an unwanted infinity is to multiply it by dx or something equivalent. Equation 5 below is a general formula that always provides a finite value by canceling infinity should one arise.
The first factor is equivalent to dx. The second factor could be equivalent to an infinite sum. Let's apply equation 5 to a classical example. Suppose we have a lattice with N cells. Each lattice cell contains some fraction (epsilon) of the total energy (E):
We could just add up all the parts to get the total energy (E):
We could calculate the average energy per cell (bar-epsilon*E) and multiply that value by the number of cells (N) to get the total energy:
But let's see what happens when we apply our new formula:
At equations 13 and 14, the dx equivalent is simply 1 which is often true in a classical case. It's a bit redundant but yields the right answer as do the more familiar methods above. In this example, it's like counting bricks to get the total. Very simple but effective. It's effective because the average energy per cell is simply the total energy divided by N--the number of cells. Thus, the bar-epsilon in our formula is the reciprocal of N--and the two variables cancel, leaving the total energy E. But look what happens when bar-epsilon doesn't equal the reciprocal of N:
In the above example we have infinite or infinitely uncertain energy in each cell, but the bar-epsilons and N's cancel, the uncertainty and/or infinity is nullified and we end up with the finite energy E.
The following example shows how infinity can rear its ugly head in a seemingly classical lattice.
Here's a data table and chart for different values of n:
Notice how the density increases as n decreases. As n increases the density levels off. The same is true for a 3D lattice:
When n is increased to infinity, the density is virtually constant and behaves like bricks. Increasing and decreasing the number of bricks does not change their mass density.
When n is reduced to zero, the density blows up to infinity! No more brick-density behavior:
We can translate the lattice equation into something resembling perturbation theory:
At equation 29 above we take epsilon to infinity (which is equivalent to taking n to zero) and the answer is 1 + infinity. We can use a famous ad hoc method to rid ourselves of the infinity: simply subtract it or ignore it. By pretending it isn't there, we get a legitimate finite energy value. On the other hand, it would make more mathematical sense to reduce epsilon to zero:
Let's apply our new formula to this problem. We begin with the original lattice equation and reduce n to zero to get the infinite result. At equation 33 we multiply the infinite density by the lattice volume to get the total energy--which is also infinite.
At 33b we restate the formula for convenience. At 34 we plug in the values we get from equation 33. At 35 we see the total energy is no-longer infinite, but the correct finite value.
Now, let's consider an even more classical setup. Here's the lattice diagram:
Surely the density should always behave like bricks, since we have one dot per cell. Sixteen dots per sixteen cells has the same density as one dot per one cell:
But even bricks have a limit. If you divide a brick up into smaller and smaller pieces, eventually you end up with something that no-longer resembles a brick. We can simulate this by reducing n to 1, and k to 0 (see 39 and 40). Once again we get the dreaded infinity--but we can fix that using the new formula (see 41 through 44).
How about a real-world example? At 45 below we set the total energy of two electrons equal to their Coulomb energy at a fixed radius (ro). You will note the total energy (2Ee) is quantized at a fixed integer value (no).
If the radius were to shrink to zero, one would think the energy would explode to infinity. To solve this problem we do some algebra to derive equation 52 below:
At 51 and 52 we see the radius cancelling itself. The total energy of the electrons is determined by the quantum number n. Apparently it is possible to have zero to infinite Coulomb energy (or force) without impacting the overall energy? The crude diagrams below sheds some light on this:
The diagrams show two Coulomb energies: E1 and E2 with radii of r1 and r2, respectively. There is a unit area A. Over a large surface area there are more units of A and vice versa. There are two circles; each representing two different surface areas. Using this information we derive equation 58:
Equation 58 tells us that a large Coulomb energy over a small surface area is equal to a small Coulomb energy over a large surface area, and the total energy is the Coulomb energy over unit area A times the surface area. The total energy is, of course, the total energy of the two electrons--and that energy is finite and conserved! This result is consistent with our formula:
If we plug in the following values and do the math, we get the finite energy of the two electrons:
Thus, we now have an incredibly simple and redundant formula that shows the connection between the parts and the whole at the classical and quantum levels.
There once was a physics student named Richard Feynman who asked his professor (re: the double-slit experiment), "What happens if you increase the number of slits to infinity?" This question led to summing the infinite number of paths a particle can take from point A to point B. If each path has a finite energy and we add up all those energies, we should get infinite energy! But that can't be right, since we started with one particle with a finite energy. Energy should be conserved, so what's wrong here? That's what we shall address in this post. First, let's define the variables we will use:
To tame this infinity problem we borrow a great idea from calculus: the integral. The integral is used to find the area beneath a curve (see diagram below). Equations 1 and 2 define the integral in terms of a summation.
At equation 2, notice how we can take N (the number of y's) to infinity and end up with a finite area beneath the curve. A finite number is definitely what we are after. Another way to get the finite area under the curve is to determine the average y (see equation 3), then map our curve to a simple rectangle (see diagram below) that has the same area as the curve shape. Equation 4 shows that the integral is equal to 'a' (the x value) times the average y value.
Now, let's apply what we know to the slit experiment. At the bottom of the next diagram, a gun fires an electron with energy Ee. Let's assume that all lights and detectors are off and the electron goes through all the slits simultaneously. The energy at each slit is some multiple or fraction of Ee. These energies can vary due to constructive and destructive interference. As with our y's above, we determine the average epsilon or coefficient for each Ee.
We don't know the average energy per slit. We do know it must be greater than zero according to the Heisenberg uncertainty principle and the Compton wavelength formula (see equations 5 and 5b). To have zero energy requires infinite time or an infinite wavelength. Obviously the space we use for the experiment is less than infinity, and, the longest time on record is the age of the universe--also less than infinity, so yes, the energy per slit must be greater than zero.
If N equals infinity, we might naively multiply N by the average energy at each slit to get infinite energy:
But here's a better approach. First let's reset our dimension variables:
Let's express x and y in terms of the electron's initial energy:
Next, let's express x and y in terms of the average energy per slit:
To get dx we divide by the number of slits (N).
At 12 we sum all the y's. At 13 we convert that sum to its energy equivalent:
We solve the integral at 14. At 16 the infinities and epsilons cancel. That brings us to 17, the area under the curve, but expressed in terms of energy.
Multiply both sides by E^2/xy, find the square root which gives a finite energy for the electron.
Thus energy is conserved. Our final energy is the same as our initial energy instead of being absurdly infinite.
