The New Field Equations

According to the theory of general relativity, mass causes spacetime to curve, and spacetime tells objects how to move. If you look at the equation below, it should be obvious why this is the case.

Well ... OK ... it is not obvious. Perhaps we can derive some field equations that are equivalent but more intuitive? Let's start with diagrams A and B below (r=radius; ct=spacetime portion of the radius; ct'=warped spacetime):

Diagram A shows an arbitrary sphere of space with no mass present. The field lines are straight (or flat) and connect the center with the outer edge. The field lines in diagram B curl like waves and pull all the space inward toward the center, shortening the spacetime wavelengths and creating a smaller sphere with more curvature. (To see more details on how this works, click here.)

The Lorentz equation above is pretty straight forward (G=Newton's constant). It shows how ct' is a function of mass (m). Add mass (m) and ct' shortens. We can use this equation as a model for our new field equations. Let's see what we can come up with:

The last equation above is kind of interesting. All the stuff on the left side must equal one. Let's multiply both sides by 8(pi)r, (the derivative of a sphere area) and do a few more steps (E=energy; T=energy density):

Add some Tensor indices:

We now have something equivalent to Einstein's field equations. Notice how each term contains an 8pi factor. We can do away with it.

What was not obvious before is now more obvious. If the stress-energy tensor (Tuv) changes, the spacetime variable (ct') also changes. Any particle in the vicinity will be affected by the changing ct', and move along a geodesic curve.

Notice there are a couple of 1/r^2's that can be factored. Time to dress this puppy up a little bit more:

We've come full circle. We now have a tensor version of the Lorentz factor we started with. To change things up a bit more, let's bring ct inside the parentheses.

Finally, we can name the ct and ct' tensors S and S', respectively:

Probabilities, Euler's Identity and Super-complex Numbers

Today we are going to examine how super-complex numbers fit in with Euler's identity and probability amplitudes. If you are not familiar with super-complex numbers, read my post entitled: "Introducing Super Complex Numbers."

Euler's identity is as follows:

It has a cosine and one isine or imaginary number. It can be used to model a right triangle:

Or a propagating wave:

Here is the super-complex number version of Euler's identity:

The exponent has i sub-n and theta sub-n. This tells us we will take the cosine of angle theta, and the isine-thetas from 1 to n. In the example below, n = 3.

The super-complex Euler identity is useful if you are working with one cosine and multiple triangles that share the same cosine. Take the exponent i sub-n, theta sub-n; it shows precisely how many isines/triangles are involved. The super-complex Euler identity is a real time saver. Instead of writing a big long string of trig functions we can write one simple exponent.

Since it can represent multiple triangles, it can be used to represent a wing design for a stealth aircraft:

OK, so that wasn't really a stealth aircraft--just more triangles, but you get the idea. Below are multiple waves propagating through space that are modeled by the super-complex Euler ID:

We know that if we multiply Euler's identity with its complex conjugate we get 1. We can also take the inner products of cosine and isine to get probabilities, but notice we are stuck with just two probabilities:

Using the super-complex Euler ID, is it possible to have as many probabilities as we like--and they all add up to one? To answer this question we need to derive the super-complex Euler ID. Start with the normal Euler ID, then split up the isin(theta) into smaller parts. Each part shall have a coefficient of epsilon:

If we examine a unit circle diagram and some trig relations, we discover that each epsilon sub-j isin(theta) equals isin(theta sub-j). We make a substitution:

If we equate i with each i sub-j and i sub-k, we assume the product of any two indexed i's is -1. Unfortunately this will give us non-zero cross product terms.

Let's check to see if the cross product of two arbitrary indexed i's really do give us -1 and not 0. Take the sum of two arbitrary terms and multiply it by its super-complex conjugate. That will give us a real number we label b^2. Like the isine of Euler's ID, we split b into smaller parts. Call those parts c and d.

The last equation above confirms the product of any two indexed i's equals -1. The great thing about Euler's ID is the square of its absolute value yields the same result as a dot product. We want the super-complex Euler's ID to behave the same way--no non-zero cross products! So here's what we do:

At the second equation above, we assume there are no non-zero cross product terms on the right side; however, without the extra cross-product terms, the left side is greater than the right side. We cure this by increasing the value of the isines. We then throw in unit vectors (ej). We now have a multiplication that behaves like an inner or dot product. We make some further refinements below:

Note how each indexed i is converted to an indexed mu, which behaves like a unit vector, giving the result we want:

Each squared sine can represent a probability and the total is 1.

As you can see, the super-complex version of Euler's identity has a great deal more flexibility than the ordinary Euler's identity. It allows us to model complex systems and probability amplitudes with just one simple exponent expression.

Introducing Super Complex Numbers

We know that i and -i are square roots of -1. They are part of the axis of imaginary numbers. Combine them with real numbers and you get complex numbers. Complex numbers are often used to model rotations, spins, oscillations, vibrations. They can even be used to model error margins. 55 +/- 3 can be expressed as 55 +/- 3i.

Complex numbers are useful any time you have a real-number value combined with a number that fluctuates. A good example is a wave. The real number tells you how long the wave is or how far it has moved along the x-axis. The imaginary number tells you the vertical measurement along the i axis. (See diagram below.)

Point A above is the sum of the real part and the imaginary part. Complex numbers work well in the above example, but suppose you want to model, say, an electric wave and a magnetic wave? For that you may want to use super complex numbers. Super complex numbers have an extra imaginary axis so you can model two waves for the price of one:

As you can see, i1 and i2 are both equivalent to i. They are both square roots of -1. Multiplications between them yield -1 or 1 in the same manner as plain old i. Operations of super complex numbers are similar to complex numbers.

Division with super complex numbers is tricky just as it is with complex numbers. You use super complex conjugates ( super complex numbers with the signs reversed) to get a real number solution:

The type of super complex number we've been working with so far is called a type-2 super complex number. It's type-2 because there are two imaginary axes. A complex number is a type-1 super complex number, since it only has one imaginary axis. Real numbers are type-0 for an obvious reason. All this implies we can have type-3 or even type-infinity super complex numbers.

Suppose we have multiple waves propagating through space? We can model the entire system with the following expression:

So far, all our waves have conveniently moved along the x-axis. What about the y and z axes? Or some combination of axes? Suppose we have waves moving along a vector? The imaginary axes would have to become imaginary vectors with the same angular relationship to the real-number vector as they had with the x-axis.

Below is a super complex vector expression:

But why stop at super complex vectors when we can have super complex tensors? The diagram below shows a type-n super complex tensor of rank-2. Below that is a general expression that can fit any super complex number or tensor.

Imagine being able to model a highly complex system filled with fixed values and variations. The weather perhaps? You could also model beams in a building as they vibrate during an earthquake. The beams could make up real-number tensors, and all the different vibrations could make up the imaginary-number tensors. These are but a couple of examples of what you can do with super complex numbers. Their application is only limited by your imagination.

For more information of this topic see "Probabilities, Euler's identity and Super-complex Numbers."

General Relativity's Invariant Tensor Myth

If you took a General Relativity course or read a book on the subject you were probably told the tensors that make up Einstein's field equations are a good thing because they are invariant. What's so special about invariant tensors? Here is a quote from one source:

"As an abstract mathematical entity, tensors have an existence independent of any coordinate system or frame of reference ..."

Wow! Cool! This means I can take a vector (a rank-one tensor), place it in any coordinate system or reference frame (also known as a basis) and it will not be affected by the coordinate system. The only things that will change are the values of the vector's components. The diagrams below show an example of a typical coordinate transformation:

Notice how the vector looks and behaves the same way in the different coordinate systems. However, Einstein's theory of General Relativity would not work if this were really true. If the vector above was a light beam this is what would happen:

The light-beam vector curves if the geometry of the coordinate system is curved. It is not independent of the coordinate system. So technically, it is not a tensor and should not be modeled by tensors. Or, the definition of "tensor" needs to be modified.

The theory of General Relativity claims it is the very geometry of spacetime that causes the light beam to curve. Also, curved spacetime geometry causes other objects that are considered tensors to move along a geodesic or curved trajectory when those objects would move differently in flat spacetime. None of this is consistent with the concept of invariance.

Einstein's Field Equations Simplified

Start with light speed (c) squared equal to velocity (v) squared:

Divide both sides of the equation by c^2:

We know that g is acceleration due to gravity, and it is equal to Gm/r^2. Remove an r from the denominator and we get a velocity squared:

Now we can make a substitution:

If we take the derivative of the sphere area, we get a line with a magnitude of 8(pi)r:

Multiply both sides by 8(pi)r:

We need to convert the mass (m) to energy (E), so we find mass in terms of energy:

Replace m with E/c^2 and divide both sides of the equation by r^3. Doing so gives us a number over an area or r^2. E is replaced by T (energy density). When T is increased, the r^2 on the left side must decrease. If we think of space as an imaginary sphere with radius (r), the smaller r gets, the greater the curvature. An arbitrary distance around the circumference covers a greater angle.

You might have noticed that when r gets smaller, the energy density grows larger, which in turn causes r to shrink even more. Here's where the field equation predicts a star collapsing into a black hole.

If we don't want to dwell on black holes, we need to multiply the left side by a coefficient (g), so when T increases, g increases. The variable g will also be a measure of curvature.

Convert g into the metric tensor and T into the Energy-stress tensor, and we have the equivalent of the field equations in simplified form:

But why settle for simple when you could have complicated? LOL! What about the Einstein tensor which contains the Ricci tensor and all those lovely Christoffel symbols? I show how to derive those here and here.

But what is the logic behind the left side of the field equations? Let's start with the gravitational energy minus the escape velocity energy. When those two values are equal, spacetime is considered flat and gravity is zero.

As you can see, we don't have to think of gravity in terms of spacetime. We could think of gravity as the difference between two energies. But Einstein insisted upon spacetime, we need to express energy in terms of space, or a number over r^2, so we multiply the left side by 8(pi)G/(C^4)(r^3):

Let's replace the energies with R's, then make tensors out of them and add a cosmological constant (since empty space still has energy). In the final step, those are replaced by Gij--the Einstein tensor.

Update: Below is the Lagrangian (L) derived from the field equations--further clarifying the logic behind the left side of the field equations.