How to Have Unlimited Orthogonal Space and Time Dimensions within 4D Spacetime

ABSTRACT:

By means of a thought experiment and a mathematical proof, it can be shown that unlimited space and time dimensions are possible, and, that only three space dimensions are mutually perpendicular.

Imagine you are throwing a party. You have invited n number of guests. You want to know the following: the starting location and time of each guest and the time each guest arrives at your party. You know the location of the party and you know what time the guests are supposed to arrive. All of this information consists of 2n+1 bits of time and n+1 bits of location. Assuming each location has coordinates x, y and z, the total bits of space information is 3n + 3.

Each bit of information is statistically independent, i.e., orthogonal to all the rest. We can think of any two bits as having a 90 degree separation. In the case of coordinates x, y, and z, such separation can be easily drawn on graph paper. In all other cases, such separation may be purely abstract and unimaginable. In any case, we can argue that 3n+3 space dimensions and 2n+1 time dimensions are necessary. To have all the information you want, you need 5n+4 dimensions. If n = 100 guests, you only need 504 spacetime dimensions!

It's fairly obvious that the above thought experiment only involves three space dimensions that are mutually perpendicular. But is this universally true? Is there, say, a mathematical proof? The following equation suggests there's no upper limit to how many space dimensions you can have:

It seems like the only constraint re: the number of space dimensions is an empirical one. Let's see if we can find a mathematical one. Let's begin with the following premises:

1. w, x, y and z are unit vectors.

2. w is an arbitrary extra space dimension. What is true for w is true for any extra space dimension.

3. A unit vector is consistent with 1D space and points in only one direction.

4. if two unit vectors (x, y) are perpendicular, they define a plane that is consistent with 2D space. Thus plane xy is not perpendicular to plane xy.

Following these premises we have:

At steps 2 through 4 we assume that all four unit vectors are mutually perpendicular. At 5 we assume w is perpendicular to plane xy. At 6 we assume z is also perpendicular to plane xy. At 7 we conclude that either w is parallel to z or if w is perpendicular to z, then plane xy must be perpendicular to plane xy--which violates premise 4. So w is not an extra dimension. According to premise 2, what is true for w is true for any alleged extra space dimension. Thus, we can further conclude there are only three space dimensions that are mutually perpendicular. If such a conclusion is valid, we should be able to falsify the following 7D cross product table:

From this table we can gather that the unit vector e1 is the solution to three cross-products involving 6 dimensions (see equation 8 below). Premise 3 stipulates that a unit vector only points in one direction. It is also obvious that e2e3, e4e5, e6e7 make three planes persuant to premise 4. Vector e1 can't point in just one direction if it is normal to all three planes, unless all three planes are subsets of the same plane. The inevitable conclusion is not all of these dimensions are mutually perpendicular.

Now, let's take a look at the so-called extra dimensions 4 through 7. At each of the equations 9 through 12 below, the unit vectors circled in red contribute to planes with normal vectors pointing in different directions. Thus they can't all have the same normal vector or cross-product solution.

Therefore, it is safe to say that dimensions 4 through 7 do not behave like mutually perpendicular dimensions.

Circling back to our hypercube at equation 1, we can conclude that there is no upper limit to how many dimensions the cube can have, but only three are mutually perpendicular. The rest may or may not be orthogonal in the sense that they are statistically independent.

References:

1. Seven-dimensional Cross Product. Wikipedia

2. Octonian. Wikipedia

Debunking Extra Space Dimensions and Minimum Distance

ABSTRACT:

By means of two thought experiments and some mathematics this paper shows that extra space dimensions are untenable. This paper also shows that the minimum distance is many orders of magnitude shorter than the Planck length.

Imagine a 2D universe on an x-y plane (see diagram below). Imagine a normal vector intersecting this plane at point p. 2D-guy inhabits this universe. He can't see the vector that intersects point p. He can only detect point p, so he has no reason to believe the normal vector exists. Now, to avoid point p, he goes around it (see red arrows).

He knows it's possible to draw an imaginary line through point p that can serve as an axis. He also notices when he goes around point p he's not encircling the x-axis or the y-axis--the two dimensions of his space. Thus, he infers that the imaginary axis he's going around does not belong to his 2D universe. He realizes he has discovered a new dimension!

Now, what happens if we apply 2D-guy's process to 3D space? Will we discover a fourth dimension? Let's try it. First we must scale everything up one dimension: The universe becomes 3D; the normal vector becomes a normal plane; Point p becomes line L. Let's assume there's a fourth dimension w, and let's define the normal plane as wx. Plane wx intersects our universe at line L which runs along the x-axis. We should not be able to detect the w-axis nor the bulk of the wx plane. We illustrate this with broken lines at the diagram below:

To avoid line L, we circle around it (see red circular path). We know we can draw an imaginary plane through line L. We know that x is one dimension of the plane. We know the axis we are circling (to avoid line L) is the plane's other dimension. We note we are not going around the x-axis nor the z-axis. That leaves the w-axis, but notice that the w-axis is indistinguishable from the y-axis. Therefore, our assumption that w is a new dimension and is undectable beyond line L is false. Unlike 2D-guy, we have not discovered a new dimension. However, we learned from 2D-guy that if a new dimension exists, it should be possible to do a rotation around an axis that does not exist in our universe. Until someone demonstrates such a rotation, we can conclude, for now, that the highest dimension of space is 3D.

But what if there are extra dimensions that are very small and curled up? If that's the case we should be able to enter alternate universes and those from alternate universes should be able to enter ours. Let me demonstrate what I mean. Imagine a line and pretend it is 3D space. Extending from it is a small extra curled-up dimension:

Let's introduce an arbitrary red object that is way too big to enter the tiny curled-up dimension:

Because the red object is too big to fit, it is assumed there is no way for the big red object to enter or detect the existence of the curled-up dimension. But didn't Euclid say something about a line existing between any two points? (In this case the line would be 3D.)

There's no reason why the big red object can't follow the path of this new line (3D space)and wind up in an alternate universe adjacent to ours:

As you can see, the big red object still can't enter the small, curled-up dimension, but the curled dimension facilitates access to alternate universes. The fact that big objects don't disappear from our universe and don't seemingly emerge from nowhere is strong evidence that microscopic curled-up dimensions don't exist. But wait! Quantum particles pop into existence and vanish all the time. It is hypothetically believed they enter a curled-up dimension (vanish), then leave that dimension and re-enter our universe. However, there's an alternate hypothesis: particles are really particle-waves. Waves experience constructive and destructive interference. When there's an excitation of a field, a particle pops into existence. That excitation could be or is equivalent to constructive interference. When there's destructive interference, energy vanishes--leaving the impression that the particle has disappeared.

The foregoing arguments seem to kill any notion that there are more than three space dimensions, but what about 4D spacetime? Or, what about the 6D object that can be found in Las Vegas? Let's address the 6D object first.

The 6D object I'm referring to is the die. The die has six orthogonal sides. Each side is statistically independent. We can change the value of a side without impacting the value of the other sides. If we change, say, the one to a seven, the other sides will still be two, three, four, five, and six. The most important point we can take away from the die is it is possible to have more than three orthogonal dimensions within 3D space! The die is a 6D object but it is also a 3D cube.

Spacetime, on the other hand, involves three dimensions of space and one dimension of time. If time is multiplied by a velocity, it has units of distance and is treated as a fourth space dimension. But is it really? Let's see what the math has to say:

Equation 1 represents a photon propagating through dimensions x, y, and z over a period of time t. It covers a distance of ct or r. For the sake of keeping the math simple, at equation 2 we rotate the path r so it is along the x-axis. Equation 3 reveals that space and time are not statistically independent, i.e., orthogonal to each other. The the value of time t depends on how far the photon propagates along x, and the value of x depends on how much time t lapses. This is the consequence of converting t into distance units by multiplying it by velocity c. So ct is not a true space dimension that is orthogonal to x. However, time t without c is a very useful statistically-independent parameter. For example, coordinates x, y, z tell you where to be for your dentist appointment and time t tells you when. A change in location does not have to change the time of the appointment, nor does a change in time have to change the location. So what can be done to make ct orthogonal to x? How about multiplying ct and x by factors of g? (See equation 6.) A change in x still causes a change in t, but g-sub-tt can be adjusted so the term stays constant. By the same token, the other term stays constant if g-sub-xx is adjusted when a change in t changes x.

So can we now credibly argue that (g-sub-tt)ct is a genuine fourth space dimension? Well, no 3D space dimension (x,y,z) has to be a function of (depend on) the others. We can, for example, eliminate y and z and still have x. But we can't eliminate a photon's path (x, y and/or z) and still have ct--the distance along a non-existent path. And, if there's no ct, then there's no (g-sub-tt)ct. Therefore, (g-sub-tt)ct is a pseudo-dimension at best.

So far, it seems we've only debunked a fourth dimension of space. What about dimensions five through infinity? Well, how we label a dimension is arbitrary. Any extra dimension can be labeled the fourth dimension. Thus, all arguments we have made against dimension four apply to any extra space dimension.

Now let's turn our attention to the concept of the shortest distance. The popular choice is the Planck length. In fact some theorists quantize space with Planck-size cubes or Planck-size tetrahedrons or Planck-size strings:

In the above diagram, the cube and tetrahedron have sides that are each one Planck length. However, the red diagonal lines reveal shorter lengths all the way down to a single point. These shorter lengths are absolutely necessary to create the shapes desired. Without a zero-length point, for example, there can be no corners for cubes and tetrahedrons. Additionally, there can be no strings in any string theory, since a string is a 1D object. A 1D object implies a zero cross-section or single point. A minimum-distance-greater-than-zero requirement would be a nightmare for M-theorists, since all D-branes would have to be 10 dimensions (including strings!). To have less than 10 dimensions requires zero distance for one or more dimensions. So it can be argued that the minimum distance is really zero, at least on paper. What about the physical world?

Equation 7 tells us that the shortest wavelength is determined by the highest energy. When the universe was a singularity, how short was the singularity's wavelength? If we only account for the energy in the known universe, that wavelength would be approximately a Planck length of a Plank length of a Planck length! Not exactly zero, but far less than a Planck length. Add energy beyond our known universe, and the distance is even shorter.

From a philosophical standpoint, the very concept of length implies a 1D object in the same manner the concept of area implies a 2D object. To measure length requires that we ignore all but one dimension, i.e., we set all but one dimension to zero. So zero distance is necessary, at least in the mind's eye. Since the mind's eye lives in this universe, we can infer that the minimum distance in this universe is zero.

In conclusion, any extra space dimension would allow rotations around an imaginary axis that is not part of 3D space. It would also allow any object access to an alternate universe. The shortest distance is many orders of magnitude shorter than the Planck length, and the Planck length may only be a lower limit of what we can successfully measure.

References:

1. Greene, Brian. 2003. The Elegant Universe. W. W. Norton

2. Irwin, Klee. 04/23/2017. The Tetrahedron. Quantum Gravity Research.

3. Sutter, Paul. 02/23/2022. Loop Quantum Gravity: Does Space-time Come in Tiny Chunks? Space.com

Introducing Stochastic Trigonometry for Quantum Physics and Statistical Mechanics

In the field of quantum physics, each eigenvalue has an eigenvector, and, when the eigenvector is normalized and squared, we get the probability for the eigenvalue. The normalized eigenvector is sometimes referred to as the probability amplitude.

When all the probability amplitudes are squared and added, the total should be 1. We can represent this with the Pythagorean theorem and the right triangle below:

The above diagram consists of two probability amplitudes: 'a' and 'b.' One is a wave function cos(theta) and the other is a wave function sin(theta).

Now, suppose there are more than two eigenvalues/eigenvectors? The diagram below shows that a and b can be broken up into smaller pieces or smaller and more numerous probability amplitudes. As before, when they are all squared and summed, they give us a total of 1.

It is possible to break up 'a' and 'b' into as many pieces as we like. Below we focus on amplitude 'a':

We can imagine breaking up amplitude 'a' into as many as an infinite number of sub-amplitudes. This can be done in both Euclidean and curved space. Equation 10 below shows how amplitude 'a' and its sub-amplitudes are invariant within flat or curved space.

With a little algebra, we can derive equation 14:

Equation 14 shows amplitude 'a' consists of an infinite number of eigenvalues (eta), each with its own probability (P(eta)). Without the probabilities, the etas would add up to infinity, and that would necessitate some sort of re-normalization technique. If we assume, however, that all quantum numbers have a probability, we will not get infinity; rather, we get the expectation value, i.e., the value actually observed.

What kind of probability values yield a finite result when eta increases linearly to infinity? Probability values that decrease exponentially. Below we derive such a probability function by using the natural-log function and converting eta to 'n':

At 16.2 we have a probability function that will reduce the probability exponentially. It gives us a number between 0 and 1, but we can derive a better function that gives us a number between 0 and 1, and, we can make a substitution. The end game is equation 16.9:

Equation 16.9 claims that if n = Q, the probability of Q (P(Q)) equals the definite integral of the probability function over a range from Q-1 to Q. We can further justify this claim with the diagram below which shows the relation between discrete values (in red) with continuous values (blue line).

Note how the area under the blue line, say, from Q-1 to Q is the same as the area of the red squares from Q-1 to Q. Equation 17 models the fact the the area under the blue line is the same as the area of the red squares over the entire range.

Now, to get a finite expectation value (amplitude 'a') we could combine equations 16.9 and 17, but the math would be more complicated than need be. To simply the math we will encounter later, let's first stretch the above diagram vertically:

Next, we draw a yellow line from zero N+1. This new line is going to make our lives easier and has the same area beneath it as the red line. Wouldn't it be nice if we could nix the red and substitute the yellow? Sure! But first we have to rotate the diagram:

Ah ... now we're in business! Below is the adjusted diagram and equation 18 with a new slope of N/(N+1):

The integral has a new range of zero to N+1, so we give the probability integral the same range:

Let's combine equations 18 and 19 to get 20:

If the limit of N is infinity, equation 20 will always give us the finite probability amplitude 'a.' No re-normalization required.

Using the diagram below and equations 21 and 22, we can derive a formula that finds probability densities:

What we've covered so far allows to find probabilities for integer values. This works fine if the value is, for example, the number of vertices in a Feynman diagram. Albeit, energy can have values of n+.5. Below is the math for that circumstance:

Notice if we divide both sides of equation 26 by Q+.5, and use summation signs, we arrive at equation 23, the formula for finding probability densities.

Now that we have the math the way we want it, let's put it to a test. Let's say we want to add up an infinite number of quantum numbers to get a finite value. Let's assume that the principle of least action applies: the most probable value will be the least action (e.g. least energy, least time, least distance, least resources required, etc.). The least probable value will be the action or event that requires the most resources, time, energy, etc. So we expect the probability to drop exponentially as the value of 'n' increases linearly--this will ensure a finite result.

Let's also assume that experiments confirm that probabilities change according to equation 27:

OK, now we only have to do some complicated math to find the expectation value 'a,' right? Wrong! At 28 and 29 below we convert the right side of 27 to a natural exponent function. If we look at equation 20, it becomes obvious that we can solve this problem by mere inspection. Looking at the exponent, everything to the right of -n is 1/a. Thus equation 31 is our final result.

Here's another test: What is the probability that a particle will travel a distance 'Q' along a pathway 'omega'? Equation 34 below can answer that. At 32 we assume that each pathway has the same probability if the distance traveled is constant, since the action is the same along each pathway (except for the direction, angles, curves, twists, turns, etc.).

Equation 35 gives us a definite answer if we want to know the probability density of a range of distances and pathways the particle can travel:

As you can see, stochastic trigonometry simplifies mathematics that can turn into a complicated, ugly, and infinite mess. It can also improve statistical mechanic's coarse-graining techinique:

Why use squares when you can use triangles?

Update: The following math shows both a convergent series and divergent series can yield a finite number 'q.' First, we start with a divergent series and make it convergent by using the probability function we derived above.

Next, we take a divergent series and assume the coefficients (the c's) don't add up to 1. Each could be any finite size; they could be a random series. The strategy is to factor out 'c' from the coefficients and use one of Ramanujan's techniques:

Another update: The following math generalizes the idea that a finite value can result from any arbitrary convergent or divergent series:

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.