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A Simple Schrodinger-Equation Solution for the Periodic-Table Elements

Suppose you want to solve Schrodinger's equation for the hydrogen atom. It's the simplest atom in the universe: only one proton ...

Monday, September 26, 2016

Proving Anti-time Mathematically?

The CPT theorem consists of three symmetries: charge conjugation (C), parity reversal (P), and time reversal (T). Taken together they provide the symmetries needed to make the laws of physics invariant. However, time reversal causes paradoxes that mess up the laws of physics. To find out more details, click here.

When we think of reverse time, we think of going back in time or time travel and the paradoxes that come with it. That's why I prefer the term anti-time. Anti-time is symmetrical with time, but it's not about going back in time to the past. It makes the past, so it doesn't create the paradoxes as reverse time does. (To learn more about anti-time, click here.)

Let's see if we can at least mathematically prove that anti-time exists. We begin our proof with a very famous equation (E=energy; p=momentum; c=light speed; m=mass):

Notice the plus and minus signs in front of the radical. Einstein's equation predicts both matter and anti-matter, since the square root of a positive number can be negative or positive. Anti-particles allegedly have negative mass and energy--so the negative root corresponds with anti-matter.

With a little algebra we can derive the following:

We can take the square roots of the last equation and fit them in a Lorentz right triangle (t and t' = time; v = velocity). Such a triangle is used to derive the Lorentz Factor. (Click here for more details.)

By matching the Einstein equation terms with the Lorentz terms we can derive something equivalent to the Lorentz Factor.

It looks like we ended up with a Lorentz-type equation with an extra v^2 added to the c^2. We can fix that when we realize c is the top speed limit and that added velocity (v) has no effect. Just to be sure, we do the math below (where 0 <= epsilon <= 1):

In the steps above we take the square root of v^2 + c^2 to be equal to c + (epsilon)c. We then plug it into the velocity-addition formula. We get c. Good! Let's square it and put it back into our Lorentz equation:

Since we were able to derive the Lorentz time equation from an equation that predicts matter and anti-matter, does it then follow that the time equation predicts time and anti-time? Note the plus and minus sign in front of the radical. A negative square root is just as mathematically valid as a positive one.

One could assume that matter corresponds with time and anti-matter corresponds with anti-time, but let's look at this issue from another angle. We go back a few steps and derive the following:

The last equation shows we can only get anti-time if we either have a positive numerator divided by a negative denominator or vice versa. If we divide negative energy by negative energy or positive energy by positive energy, we get positive time!

Imagine two universes: one made entirely of matter and the other entirely of anti-matter. They both would have positive time and no anti-time to put the present moment into the past. Now imagine a universe with both matter and anti-matter. There we can have both time and anti-time. To get anti-time, we need to be able to divide matter by anti-matter (or vice versa) in the last equation above. All this implies that positive time is caused by matter or anti-matter, and anti-time is caused by a combination of matter and anti-matter.

Friday, September 16, 2016

How Entangled Particles Communicate

When we last checked in on Alice and Bob they were observing the spins of two entangled particles. According to their findings, particles A and B can send information between them instantaneously--no matter how far apart they are.

Alice noticed when A's spin changed, Bob reported that B's spin also changed. From Alice's point of view, particle A notified B. Bob has a different take: It was B's spin that changed first, then B notified A. The fact that Alice and Bob can't agree whether A notified B or vice versa is a major clue that tips the hand on how entangled particles communicate.

When two observers can't agree on the order of events, we can look to relativity to sort it out. We know that space and time are relative. In the case of particles A and B we are interested in the space or distance between them (x' and x). Perhaps the Lorentz factor below can model the relationship between the observed distance between the particles (x) and the distance the particles experience (x').

If A and B moved at light speed (c), then v^2 in the above equation would equal c^2. Distance x' would be zero. It would be as though A and B never separated. If they never separated, they would not need to send information back and forth. All their information would be contained in particle A-B, so to speak.

According to Alice and Bob, though, A and B are at rest, so their distance would be x, the distance Alice and Bob observe. Then again, a mass (m) can be at rest and, along with its radius (r), can produce a velocity-squared equivalent (G=Newton's constant):

If mass (m) is big enough, distance x' would drop to zero. Unfortunately, particles A and B have tiny masses. The question then arises: Do entangled particles have a velocity-squared equivalent of some kind that is equal to c^2? And what of particles that aren't entangled (separable)? If a velocity-squared equivalent exists, what would it be for them?

We know that all particles have mass or a mass equivalent (m), and said mass is coupled with c^2. Particles A and B have the following parameters (E=energy):

Separable particles also have an mc^2. If we use c^2 as our v^2 equivalent, there seems to be no distinction between entangled and separable particles. Entangled particles are like particles that have never separated and they are statistically dependent on each other. Separable particles, by contrast, behave like they are far apart and are statistically independent; i.e., are orthogonal.

Dot products (inner products) could be used to model orthogonal separable particles and statistically dependent entangled particles. Below i and i are not orthogonal and yield a scalar value of one; whereas, i and j are orthogonal and yield zero.:

Using the entangled particles' parameters and the dot product (inner product) concept we formulate the following equation:

Let's put the above equation to work and see what happens. We shall first calculate the distance x' between two separable particles:

The distance x' between two separable particles is simply x--what Alice and Bob perceive and what the two particles experience--since none of the aforementioned are part of an entangled system. Note that the separable particles' v^2 equivalent is zero. How can that be when they each have an mc^2? Answer: The v^2 does not represent the individual particles; it is the v^2 of the entangled system. Since separable particles are not entangled, the entangled system is nonexistent and the v^2 equivalent is zero.

Now we determine the distance x' between entangled particles:

The distance x' between two entangled particles is zero. From A and B's point of view, they never separated. They don't send information back and forth like Alice and Bob imagine. Poor Alice and Bob are still not part of the entanglement and only see distance x and believe the particles are far apart. As a result, they scratch their heads and wonder how A and B can send information faster than the speed of light.

Wednesday, September 14, 2016

Would Chance Exist If We Knew All There is to Know?

This post is a continuation of the previous post "What We Didn't Know About Time Travel Could Surprise Us."

It is commonly believed that if we went back in time, history would repeat itself exactly as before if we started with the same initial conditions. It is also commonly believed that chance is a function of our ignorance. If we knew everything that goes into a coin toss, we could predict the outcome 100% of the time. Today we put these beliefs to a rigorous test.

For illustrative purposes, and to keep our first test simple, let's cut our 3D universe down to 2D. Imagine a pencil balanced on its tip. It can fall either right or left. A hand grabs the eraser end and tilts the pencil to the right at some unknown angle that's less than 90 degrees. Which way will it fall? Notice we are not completely informed. We don't know the exact angle of the tilt. Any estimate we make can be way off; yet, we can predict the outcome with 100% accuracy.

The pencil will always fall to the right. This demonstration shows that our ability to make 100% accurate predictions is not dependent on perfect knowledge. There are many things we don't know, like the tilt's angle, but our ignorance does not deter us. More importantly, it does not create a chance environment. Thus the so-called correlation between ignorance and chance breaks down here.

Since it is possible to be ignorant and make 100% accurate predictions, is it also possible to know all there is to know and be uncertain about an outcome?

Imagine a simple universe that contains a particle called B. As time (t) passes, B can move either right or left. Which way will it move? We don't know--so B's movements seem random to us. If we knew what causes B to move right or left, we could predict B's moves and the randomness would disappear, right?

We get all the scientists together and give them a big fat grant. Their mission (if they choose to except it) is to find the initial condition(s) that cause B to move as it does. They discover particle A. When particle A moves right, B moves right. When particle A moves left, B moves left. Perfect! We simply watch A to see how it moves, then we can predict how B moves.

The randomness is gone ... or is it? What causes A to move right or left? Well, nothing. "A" is the "initial condition." Nothing causes it. Nothing precedes it--or it would not be initial. So when A moves, it is completely random.

The above example demonstrates that you can have complete knowledge; i.e., know the initial condition(s) and still have randomness. But suppose the initial condition A moved left the first time history unfolded. We go back in time and A moves left again. Will history unfold the same way?

Let's see. If A has gone left, then B will move left. So far, so good. But which way will B move next? We can't look at A; it has done its job of initiating the universe's evolution. B moves on its own--either right or left. We don't know which direction now that A is out of the picture. So we can't predict with certainty that history will repeat itself because it doesn't have to--even when we know the initial condition with absolute certainty!

Since we no longer can predict B's movements, does that mean we are ignorant? Not if our thought-experiment universe consists only of A, B, time, space, and the rule that particles can go right or left. If that's all there is to know then we know it all--except what B will do next.

We don't know what B will do next because chance is not a function of ignorance in this case--it is a function of more than one option. If, for example, B could only move left, we could be as ignorant as a retarded cockroach and still predict B's movements with 100% precision. On the other hand, we could be geniuses with access to supercomputers and have no clue what B will do next if B can move in a gazillion number of ways (or exist in a gazillion number of varied states).

Finally, history need not repeat itself even if the initial conditions are the same as before and we have perfect knowledge of them.

Tuesday, September 13, 2016

What We Didn't Know About Time Travel Could Surprise Us

According to Hollywood and even some scientists, going back in time means seeing history repeat itself. We worry that if we interfere with the past, the future might unfold differently, unexpectedly and even disastrously.

Imagine Hitler winning World War II or your parents never getting together, so you are never born--all because you spit on the sidewalk in the year 1933, which set off a chain of events that changed the future.

What I described above is rather nonsensical if we take entropy into account (s=change in entropy; k=Boltzmann's constant; ln is the natural log; omega=number of ways particles can arrange themselves):

If our most common belief about time travel is correct, then entropy would have to shut down, so history could repeat itself exactly as before. But then we might ask, why should history repeat itself exactly as before (assuming we don't interfere)?

To satisfy our curiosity, what would happen if entropy remains intact during our time-travel excursion? Imagine a universe with one particle in it. It can go either right or left as it passes through time. The probability (p) is 0.5 for each move just like a coin toss:

The particle makes three moves, following a path that is one out of eight possible paths. The path it takes is outlined in red:

The red path represents its history. The entropy starts with a single possibility and evolves into eight possible paths.

Now let's find a wormhole and travel back in time to the particle's starting point and watch its history unfold again:

Oh wow! History failed to repeat itself. We didn't interfere. We were good. We watched from afar ensconced in our time-travel VW bug. Relax, it's not our fault. Entropy was just doing its job. There was only a one-in-eight chance history would repeat itself and a seven-out-of-eight chance it would not. I mentioned earlier that we didn't interfere, but we did go back in time and that in itself is a kind of interference that could possibly alter history.

OK, so maybe going through a wormhole is not the best way to go back in time. Sometimes going back in time is described as a movie going in reverse. If we flip a cosmic projector switch and make the particle go back in time, it should follow the red path back to its starting point:

Oh no! What happened? The particle didn't behave like a backward-running film. It took a different path back. And why not? It can go left or right at any node along the path. To get the equivalent of a backward-running flick, the particle's probability distribution would have to exit the theater.

Backward time travel, as we imagine it, requires an entropy of zero: one path the particle can take and not the usual eight. If we treat the particle's present as a single starting point, the change in entropy would evolve as follows:

Just for fun, let's assume the particle's probabilities of going left or right stay intact when time is reversed. Its past would most likely be different than expected. The past it remembers is only one of eight possible paths to its present time.

Now apply what we've learned to your past. Imagine the implications! You go back in time, but you see a different past than you remember. You think to yourself, "This is not my life!" Your past self looks different, like a brother or sister--because you may not be the same person or sex! A different sperm cell beat all the others to your mother's egg. Your parents still gave you the same name (assuming your sex is the same) but you're watching the life of a different person who took your place because you just had to try out that bloody time machine!

Since your past is messed up anyway, you might as well interfere with it, i.e., take advantage. You read in a history book that the racehorse, Fools-errand, won the Kentucky Derby, so you decide to place a bet, but don't place that bet--it's a fool's errand.

For more on this topic, click here or here or here.

Saturday, September 10, 2016

Why Gravity Attracts: Quantizing Spacetime

This is part three of the gravity series. To read part one, click here. To read part two, click here.

When we increase the energy of a particle wave, its wavelength shortens; its frequency increases to more cycles per second or one cycle per fraction of a second:

It's apparent that more energy corresponds to shorter space and shorter time, i.e., shorter spacetime.

Below is a diagram with an equation that gives us the velocity of a falling body at any point (x) along the radius (r). (G=Newton's constant; m=mass; v=velocity.)

The thin horizontal lines can be thought of as different potential energy levels. The gaps between the horizontal lines increase as the body falls. The gaps correspond to increasing velocity and kinetic energy. The kinetic energy levels increase and potential energy levels decrease as the body falls. The body in turn accelerates.

OK, so the body accelerates when it acquires more kinetic energy. Why should it fall down? Why not up? Or sideways?

Imagine a 2D universe with nothing but space and time. We place in the middle of that universe a micro-black hole. We label it "B." There are concentric circles surrounding it. Each circle is a different energy level whose kinetic energy decreases as the distance from the black hole's center increases.

We introduce particle A. There is a field line of spacetime between A and B. A comes into contact with B's outer most energy circle. The spacetime field line shortens bringing A and B closer together. Potential energy is lost. Particle A gains kinetic energy.

Now that A and B are closer together, A is exposed to a higher energy circle or level. The spacetime field line shortens again. A's kinetic energy increases again and more potential energy is lost:

Below are diagrams modeling the spacetime wave. Variables ct' equal the wavelength. Variables vt equal the amplitude. As velocity (v) and kinetic energy increase, the wavelength shortens and the amplitude increases--and vice versa for decreased potential energy.

If it's true there is a connection between shorter wavelengths and spacetime contraction, then it should be possible to reconcile quantum physics with relativity; i.e., it should be possible to reconcile the two equations below (t' and t = time; h-bar=Planck's constant; E=energy; v=velocity; c=light speed):

If the above two equations are connected, then we can derive one from the other.

In the steps below, we introduce the variable "u^2." It is the difference between light speed squared (c^2) and velocity squared (v^2). You can also think of it as potential energy where the mass (m) is set to one kilogram. There is also the Greek symbol lambda for the wavelength and a coefficient k.

Excellent! It appears quantum physics and relativity are connected. Shorter wavelengths correspond with spacetime contraction. If we know the energy (E) we can estimate the spacetime wave's change in time (t') and wavelength (lambda). But what is the energy (E) exactly? And what time units should we use if we want to know the exact quantum wavelength?

Let's steal Max Planck's idea and use time units equal to Planck's constant (not the Planck time, however). If nothing else, it will make the math so much easier. We can use h-bar divided by J (Joules) to make a time unit. Time to calculate energy (E). We can start with the first equation in the above frame:

So the energy (E) of the spacetime wave is the total energy (c^2) divided by the potential energy (u^2). Divide E into h-bar and you get the quantum time (t'). Multiply t' by c to get the quantum wavelength of the spacetime wave.

To set up a realistic spacetime wave function, we need to find time (t):

Now we have all the numbers we need for the wave function. There is, however, one more issue: photons. They have zero time (t') but don't really have infinite energy:

This seems to drill a big hole in our nice, neat little math formula. Ironically, a photon's wavelength can be determined using a t' that is greater than zero. (Where f=frequency.)

We know that photon's have zero mass, but have a mass equivalent. Photon's apparently also have zero time and a time equivalent we can use to calculate the wavelength. (The little e subscript below stands for equivalent.)

We can now calculate spacetime wavelengths and energies that involve massless, time-free bosons as well as other particles. Time to set up the wave function. First we plug in our numbers into the variables below that make up the right triangle:

If velocity (v) is unknown, we can find it using v = ({(ct)^2 - (ct')^2}^.5)/t. We then determine the angle phi:

If we set phi equal to my relativity wave function's angle, we can bridge between this spacetime wave function and Einstein's field equations. To find out more details, click here.

And the final spacetime wave function (and its complex conjugate) is (drum roll):

Monday, September 5, 2016

Is the Universe's Net Energy Really Zero?

The above video makes the case that the universe's net energy is zero. The argument is as follows: The universe isn't raised in a gravitational field; therefore, it has no potential energy. The universe isn't moving within a larger reference frame; therefore, it has no kinetic energy. Thus, it has a total net energy of zero joules.

Here is a claim that is equally fatuous: a box full of energy has no energy because it rests on the ground at the lowest point, and it isn't moving.

The video also claims the amount of negative energy perfectly cancels the universe's positive energy. However, according to the Wilkinson Microwave Anisotropy Probe, the universe's energy density is E-10 joules per cubic meter. So why does this zero-energy claim persist? Here's another reason I stumbled upon while working with gravity mathematics (g=acceleration due to gravity; c=light speed; t=time):

The above equation shows the relation between gravity and time, but there's no mention of energy, so here's the relativistic version (E=universe's energy; G=Newton's constant; r=universe's radius):

Now let's see what we can derive:

The final equation above gives us the universe's total energy. Now it's plain why so many insist it is zero. Notice how the terms inside the parentheses are multiplied by the radius (r). If the stuff inside the parentheses isn't zero, then when r changes, so does the universe's energy. Such change is a violation of energy conservation. To make the result conform to energy conservation, we set g^2t^2 equal to c^2.

But if there's zero energy, then energy multiplied by time is also zero. This violates Heisenberg's Uncertainty Principle:

If we take the derivative with respect to time, we can determine the minimum amount of energy of the universe:

The net energy is the ground state or zero-point energy. We tack it on to the energy equation above. Notice that it is unaffected by changes to r, so energy is conserved. But can we just tack it on? After all, we didn't derive it from the gravity equation we started with ... but we could. It's just a matter of making a minor adjustment in our starting equation.

If we do the algebra over again using the second equation above, we will get the right answer: something greater than or equal to the ground state energy.

Here are a couple of more reasons to shy away from zero-energy claims. Consider entropy (s) and temperature (T).

If there's zero energy, then there's zero entropy! And zero temperature. Scotty, beam me up!