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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 ...

Tuesday, April 25, 2017

Why Gravity is Weak

Perhaps you are familiar with some of the more exotic and far-out theories that attempt to explain why gravity is so weak compared to, say, electromagnetism. One of my favorites involves gravitons hiding in extra dimensions. The logic is as follows: if gravitons weren't hiding in extra dimensions, gravity would be much stronger. It's analogous to someone screaming at the top of his lungs. The intensity of the sound is very strong until you hide that person away in a broom closet and lock the door.

As much as I like this theory, it has a problem that can't be hidden in a closet or extra dimension: there is no empirical evidence of extra dimensions. That begs the question: is it possible to explain gravity's weakness without using extra dimensions? Yes! When doing physics, I prefer taking an Occam's razor approach and using concepts and ideas that are testable and/or have been tested and confirmed. The end result is not as exciting as extra dimensions--but more likely to be true.

Let's begin with Einstein's field equations:

To simplify matters we equate the left side of equation 1 to K:

In equation 3 above, we have energy density (T44) multiplied by a constant that consists of G and c. Why do we need such a constant? Why can't we get by using equations 4 and 5 below?

The constant in question is a very small number and substantially reduces the value of K, the spacetime curvature. Perhaps equations 6 through 8 can shed some light on the subject:

Imagine energy (E) interacting with a cubic meter of spacetime. Equation 6 shows that spacetime energy density (Ts) is a tiny number. Since curved spacetime causes gravity, and since spactime's energy is so weak, is it any wonder that gravity is weak? Let's see if the math agrees. Equations 7 and 8 are the average radius of a nucleus and the nucleus volume respectively. Why are these numbers important? Imagine energy (E) added to a cubic meter of spacetime. To see how that energy interacts with spacetime, it's easier to take all spacetime's energy and reduce it to a particle. We do the same with the added energy. So we have a particle interacting with another particle within a cubic meter. Spacetime's total energy is equivalent to a proton's. We can imagine it having an approximate radius of e-15 meters and a volume of e-45 cubic meters. Our energy particle will have the same dimensions.

To provide a simple visual demonstration of the interaction between matter and spacetime, let's pretend that the volume of our spacetime proton is .25 cubic meters and that the energy particle we add has a .25 probability of interacting with it.

Because the probability of interaction is .25, the average energy ends up being .25 the total energy. To get the proper value for K, we would need to multiply E/m^3 by a constant of .25. So K is weaker than E/m^3. To illustrate the point further, let's distribute the energy evenly throughout the cubic meter. This time it will be a field of isotropic energy interacting with our spacetime proton:

At any given time, only 25% of the energy field interacts with the spacetime proton. The remaining 75% contributes nothing to the value of K. To drive this point completely home, let's cut up the spacetime proton and evenly distribute it within the cubic meter:

As you can see, it matters not how we distribute the matter energy and spacetime energy. In the case above, only 25% of the total matter energy interacts with the spacetime energy. Our constant of .25 remains constant. Now, let's stop pretending and let's replace the .25 constant with the volume of a nucleus, which is approximately e-45 m^3.

If e-45 is the right constant, then that means only e-45 of the energy contained in T44 contributes to K. Let's check it:

It looks as though it's in the ballpark of the actual constant used in Einstein's field equations. So it's highly plausible that gravity is weak due to energy interacting with very weak spacetime energy.

Thursday, April 20, 2017

Deriving the Schwarzschild Solution to Einstein's Field Equations

Step one: Beginning with Einstein's field equations, derive the Scharzschild radius (equation 13 below):

Next, we call on Pythagoras and a right triangle to derive a basic metric equation (equation 15 below):

Using the same right triangle we derive the Lorentz factor (equation 19 below):

Now check out equation 20:

Because of equation 20, we can make a substitution and derive equations 22 and 23:

Equations 22 and 23 allow us to make more substitutions. The result is something that resembles the Schwarzschild metric (equation 24):

Here's the actual Scharzschild metric:

We can replace equation 24's cdt' with dr (differential radius) to get the following:

It would be great if the middle term (vdt)^2 had a plus sign instead of a minus sign in front of it. With some trigonometric slight of hand we change the minus sign to a plus. The result is equation 29:

So far we've used a triangle with only two space dimensions. We are one dimension short, but we can fix that:

Each dimension in space is a hypotenuse of a right triangle with two other dimensions which can replace the hypotenuse. We make a final substitution and we get the Schwarzschild metric (equation 32). Schwarzschild used spherical coordinates. For clarity and to help you visualize this type of coordinate system, I provide the diagrams below. The first two diagrams show the front and side view of a sphere of spacetime with a mass in the center. The position in spacetime is given by the radius (r), the first angle (top diagram), and a shorter radius (rsin[first angle]) and the second angle (bottom diagram).

The variables used in the Schwarzschild metric, however, are differential--a tiny piece of the radius and each angle. The value of each space variable is indicated in red below:

If we take the limit of these variables we get a point in spacetime indicated by the red dot in the diagram below:

The objective is to figure out the spacetime curvature in that tiny (red) region of space. To solve the field equations, we need to know the metric tensor components; i.e., the g's.

We can find the value of each g component within the Scharzschild metric:

Thus the metric tensor is as follows:

With the information we have, we can derive equation 42 below.

We can solve for R44--the spacetime curvature--by plugging in the mass (m) of a star, planet or black hole; the volume (V) of the space, mass and energy within an imaginary sphere with radius (r); and radius (r).

Tuesday, April 4, 2017

Debunking the Equivalence Principle Thought Experiments

According to the textbooks, the principle of equivalence asserts that gravity and inertia are one and the same--not similar, proportionate or related to each other, but the same. According to the principle, you could kidnap some scientists by smothering each of them with an ether rag, place them in a rocket ship with an acceleration of 9.8 meters per second per second. When they regain consciousness, they won't be able to tell if the gravity they feel is from the earth or the rocket ship--there are no windows for them to look outside.

As the video above says, they can drop a pen and it will fall to the floor like a pen on earth. They are free to do whatever experiment they wish--and they won't be able to distinguish earth's gravity from the rocket's acceleration? Hmmm ... let's take a closer look at how a rocket propels itself:

According to equation 1) above, rocket fuel and oxygen enter the nozzle at a certain mass-flow rate and velocity, then exit the nozzle at a higher velocity and final mass-flow rate. Add to that the pressure difference. The total is the thrust force. Recalling Newton's third law: this is the action. The equal and opposite reaction is the rocket accelerating.

Equation 2) shows that the perceived gravity inside the rocket is acceleration (g). Since action and reaction are equal and opposite, we can set the thrust force equal to the total mass of the rocket (passengers, supplies, fuel included) times the acceleration g:

Doing a bit of algebra gives us equation 4)--the function g. Let's compare equation 4) to equation 5) below:

Equation 5), of course, is Newton's law of gravitation. Let's assume the scientists aboard the rocket are aware of equations 4) and 5). They're also aware of equation 1) that shows fuel mass flowing out the rocket's nozzle. How hard would it be for them to figure out that they are in a rocket ship and not in a sealed box on earth?

Not hard at all! In fact, all they have to do is weigh themselves repeatedly over time. The rocket's total mass (m) is decreasing thanks to the burning fuel exploding out the rocket's back end. This would cause the rocket's acceleration (g) to increase. So when the scientists weigh themselves they notice over time they are gaining weight (scientists' mass * increased g). Either their Weight Watchers diet is failing or they are in an accelerating rocket.

Unlike the rocket, earth's mass is virtually constant. If the scientists weighed themselves there, they would notice little or no change (assuming they didn't quit their diet).

Now, take another look at equations 4) and 5). Notice how mass (m) is in equation 4's denominator, but equation 5) has a mass variable in its numerator. Any change in mass would yield opposite results when comparing the two equations. If mass increases, the acceleration (g) in equation 4) goes down, but equation 5's acceleration (g) increases--and vice versa!

To claim the g in the rocket is indistinguishable from the g on earth requires us to ignore the physical processes involved. Prior to any mass loss, if one drops a pen in a rocket accelerating at 9.8 meters per second per second, the pen's inertia keeps it in place and allows the floor of the rocket to accelerate up to it. A casual observer aboard the rocket will most likely get the impression the pen dropped to the floor. Is this really sufficient evidence to support the claim that inertia and gravity are one and the same? If so, one could also claim that gravity is not a genuine force; it's a byproduct of another force such as the rocket's thrust force.

If gravity and inertia are the same, it should be possible to fool the kidnapped scientists aboard the rocket. Our first attempt failed--the rocket lost some of its mass causing g to increase. Suppose we install some computer software that changes the thrust force in proportion to how much mass is lost or gained. This would allow the rocket's acceleration to remain constant. Perhaps we can rub our hands together in glee: how will the scientists ever figure out they are in a rocket ship and not on earth? Ah, but alas, there's equation 6):

And there's the inequality 7). Equation 6) and inequality 7) show what happens over time (t) if the rocket accelerates at a constant g. The rocket's velocity (v) increases. This causes the the rocket and every person and thing inside the rocket to have more mass than anticipated. When the scientists weigh themselves they once again discover a weight gain. Each starts out with a weight of (m * g) and end up with (m' * g). Compare this result to what happens on earth:

Equations 8) and 9) show that both the initial earth velocity and final earth velocity are pretty much the same. The right side of each equation is the total tangential earth velocity (which includes earth's spin and orbit around the sun) relative to the galaxy and universe's motion. Cutting to the chase, the earth's velocity relative to itself is zero. As a result, equation 12) shows that if the scientists were on earth, their respective weights would not increase over time through no fault of their own. By contrast, the rocket's velocity increases relative to the earth's--and so must the scientists' respective weights.

OK, strike two. We failed to fool the scientists once again. Time for desperate measures. We smother the scientists with ether rags and place them in a rotating, donut-shaped space station. When they regain consciousness, it is hoped they will fail to distinguish between the angular acceleration (g) and the earth's gravity.

The advantage of this new arrangement and equation 13) is the velocity does not change, so relativistic mass increases are avoided. We avoid a mass decrease, due to fuel consumption, by continually topping off the fuel tank (using a maintenance spacecraft). These measures should provide an artificial gravity indistinguishable from earth' gravity, right? But then something very routine happens: the scientists place their garbage and waste in the disposal chute. The waste is jettisoned into space. The scientists are not aware of this. For all they know, the garbage goes to a dumpster here on earth. But once again, when they weigh themselves, their respective weights increase.

Curses! It's that bloody conservation-of-momentum rule! When overall mass decreases due to garbage disposal, the velocity increases, so does the angular acceleration. If the scientists were on earth, jettisoning waste would still decrease the space station's mass, but the overall mass of earth would remain virtually the same, since the garbage ends up in a dumpster on earth. As a consequence, earth's gravity maintains its value. Strike three.

Then again, if the garbage stays on the space station, the perceived gravity there should be the same as earth's, and, we should be able to say with confidence, "Inertia and gravity are one and the same." But before we get too cocky, let's put this claim to another test. Imagine the earth and a red meteor coming together. We can interpret this event in one of two ways: the meteor is falling to earth, or, the meteor is at rest and the earth is moving toward it.

If inertia and gravity are the same, then it should be possible for a blue meteor to be at rest on the other side of the earth. The earth should move toward both meteors:

No force acts on the meteors--they are at rest, floating in space. The force, whatever it may be, is moving the earth in opposite directions to meet the meteors. Then again, maybe it doesn't really work that way. Could it be that the meteors are really falling to earth?

Assuming the meteors are falling to earth, a force must be causing them to accelerate. We call that force gravity. It seems clear at this point that gravity and inertia are not one and the same but are only similar to a casual observer who doesn't ask too many questions.

There is another key difference between inertia and gravity that involves time ... the time it takes a boson, say, a graviton to notify matter of a change in status in other matter. To properly distinguish between gravity and inertia, we need to make Newton's equation and Einstein's field equations time dependent. The same goes for the rocket-thrust equation.

Take a look at equation 15) above. If the sun's mass (m) were to drastically change in the current time (t), it would take about eight minutes (r/c) for us here on earth to feel the gravitational effects. The change will be felt by us at time t+r/c. Equation 16) concurs. Any change in spacetime curvature won't happen instantaneously. It takes time for gravitons to move through space from the sun to the earth.

Equation 17) tells a different story. If the thrust force changes in the rocket ship we discussed earlier, the acceleration (or perceived gravity) changes instantaneously--there is no need for gravitons to notify distant objects of the change in status, and to tell them to accelerate. For it is the rocket floor that is accelerating to objects at rest. Thus, when inertia mimics gravity, it takes less time.

The diagrams below illustrate the time difference between gravity and inertia: