Suppose you want to solve Schrodinger's equation for the hydrogen atom. It's the simplest atom in the universe: only one proton ...
Monday, August 29, 2016
Saturday, August 27, 2016
The universe is getting bigger and colder. We are told that one day it could suffer a heat death. By contrast, the early universe was smaller and hotter. All this implies an inverse correlation between the universe's average temperature and its size.
We might be tempted to model it with the Ideal Gas equation. We could think of the universe as a big balloon filled with gas. As the volume of the balloon increases, the pressure decreases along with the temperature. Below is the equation (P=pressure; T=temperature; V=volume; R=universal gas constant; n=kilomoles of gas):
Right away the equation bursts our balloon. When volume (V) increases, pressure (P) decreases but the temperature can remain constant. So why isn't our universe the same temperature as it was in its early years? Its volume is increasing; its pressure is decreasing; yet, the temperature isn't remaining constant.
So what's the problem with the Ideal Gas Law? We know that pressure is equivalent to energy density, so let's plug energy density (E/V) into the equation in lieu of pressure (P):
The second equation above reveals the problem: The volume (V) in the denominator cancels the volume in the numerator. Change in volume in this case has no effect on the temperature.
We know that entropy increases as the universe expands. Perhaps lower temperature is a function of increased entropy? Let's take a look at the entropy equations (E=heat or energy; s=change in entropy; T=temperature; k=Boltzmann's constant; omega=number of different arrangements of particles):
If we equate the two equations above, we can derive a temperature equation:
Temperature appears to be a function of energy (E) over the number of ways particles can arrange themselves (omega). Once again, temperature is not a function of volume, but it seems like it should be. We know from experience that a candle can heat up and maintain the temperature of pickle jar, but a candle in an aircraft hanger can be a much colder space.
Let's explore the omega variable and see if we can make it a function of volume. Imagine a single particle with one state in a single space (see first diagram below). In that case, omega equals one. Now double the space. The particle will have two spaces it can occupy (see second diagram below). Omega now equals two.
Entropy has increased due to increased space. We can define omega in the following way:
Below are some examples that include the above diagrams, a coin toss, and dice:
Consider the coin-toss example above. Normally we think of a coin as having two states: heads or tails, so omega equals two, right? Well that's only true if the coin lands on the same spot on your floor because there is no other spot it can land. But suppose you have a big floor and you divide it into a grid of 20 spaces where the coin can land. Omega would then be 2^2 * 20 = 40 states.
Now suppose you have the same floor space, but you now have two coins. Omega becomes 1520 states! We are now ready to put together a temperature equation:
Notice the temperature is now a function of volume because entropy is now a function of volume. If the volume increases, the temperature decreases and so does the pressure (Boyle's Law) if the other omega variables and E are held constant. Remember pressure is E/V. The difference here is that E/V is no longer canceled by V*E/V.
Taking the temperature of the Universe or any other system can be done by this system of equations. The first equation is for systems that have more space than particles. The second is for singularities, high pressure systems, where the energy density is high. The third equation allows you to make a quick-and-dirty calculation. It is similar to the Ideal Gas equation albeit it includes entropy.
Friday, August 26, 2016
To derive the Planck length we start with the Schwarzschild radius (x=length; m=mass; G=Newton's constant; E=energy; c=light speed). (To find out how the Schwarzschild radius is derived, click here.)
We also need Heisenberg's Uncertainty Principle (p=momentum; x=position; h-bar=Planck's constant). (To find out how to derive the Uncertainty Principle, click here.) For our convenience we convert momentum (p) to energy (E) by multiplying both sides by light speed (c):
Divide both sides by E:
Multiply the above result by the Schwarzschild radius, then do some algebra to get the Planck length (x with a sub p):
To get the Planck time (t with a sub p), take the Schwarzschild radius and divide it by c:
Shall we call the above result the Schwarzschild time? We also need the other Heisenberg Uncertainty Principle, the one with energy (E) and time (t). Now do some more algebra to get the Planck time:
To get the Planck mass, set the Schwarzschild time equal to the Heisenberg time, then solve for E-squared:
Let's assume E is kinetic energy. Square it, make a substitution, do some more algebra, and get the Planck mass (m with a sub p):
Thursday, August 25, 2016
This post will further address the problem discussed in a previous post. To read that post, click here.
How much energy is there in the universe? According to one source, there is zero energy. (To read the related article, click here.) Zero energy would make sense if the escape-velocity energy of matter exactly matched the gravitational potential energy. But the universe is expanding. This implies that the positive energy outweighs the negative gravitational energy. Most importantly, the zero-energy theory doesn't match energy measurements. According to the Wilkinson Microwave Anisotropy Probe, the universe's energy density is E-10 joules per cubic meter. Multiply that figure by the volume of the universe and you don't get zero.
In any case, the universe's non-zero energy is allegedly conserved. We have Einstein's famous energy equation below, and the gravitational-energy equation (E=energy; P=momentum; c=light speed; G=Newton's constant; r=radius m=mass):
We might be tempted to add them together (or subtract one from the other).
The first equation below models a singularity with an infinitesimal radius. The total energy (Et) is infinite (or minus infinity). The second equation below models an expanded universe. It's total energy is finite. Clearly energy is not conserved.
So how can energy be conserved? Let's try this:
The first equation above is not a conserved quantity. Energy increases or decreases when the radius changes. The second equation above is a conserved quantity--changes in the radius and energy always make the equation's right side equal to one.
Let's multiply Einstein's equation with the second equation above.
The universe's total energy is now conserved. Changes in the universe's radius cancel the changes in the gravitational energy. Using the above technique, we can factor in as many inverse-square energies as we like. Just for fun, let's factor in electromagnetism (k is a constant; q=charge):
As you can see, when all is said and done, the universe's total energy is equal to the right side of Einstein's equation--and it doesn't have to be zero, so it agrees with the Wilkinson probe's findings.
Sunday, August 21, 2016
More than a hundred years ago Albert Einstein set forth his relativity theory. He had this weird notion that time and space change in magnitude and shape when mass/energy is present or when your spaceship is traveling at high velocity. Many equations have been put together to model this phenomenon, including the ones listed below (t' and t=change in time; M=mass; G=Newton's constant; v=velocity; ds=spacetime; c=light speed; x=distance; r=radius):
The first equation above is the good old spacetime metric. The middle equation is the Lorentz factor. The third shows how mass affects time. These equations tell us that more mass or more velocity cause the change in time to shrink. Like time, distance (x) also shrinks, so we can replace t and t' with x and x' in the last two equations.
Thanks to E=mc^2 we know that mass, momentum and energy are different aspects of the same thing. Thus, the above equations are telling us that whenever we introduce energy into a system, time and space are affected. What the equations don't tell us is how or why energy causes time and space to shrink or curve. Let's see if we can figure it out.
To simplify the process, let's convert space and time to spacetime (ct and ct'). Doing so will make both space and time one variable:
If we call upon Pythagoras we can model spacetime with some simple trigonometry and even derive the Lorentz factor:
Notice in the diagrams, above and below, how ct' shrinks when vt grows. When vt is zero, ct = ct'.
Now let's change the variables to something even Schrodinger could appreciate:
We can now model spacetime with a wave function:
So what happens when a wave's energy increases? Its wavelength shortens. In the case below, the wavelength is ct, but becomes ct' when energy is increased. Notice how the circles or endpoints (particles) of each wave move closer together. Gravity?
In the diagram below we start with an arbitrary sphere of spacetime. The lines inside the first sphere (A) are straight or flat long waves of spacetime. The only energy present is zero-point energy:(1/2)hf. The second sphere (B) has a small circle at the center. This shall represent, say, a black hole. There is more energy present (nfh)--the spacetime wavelengths grow shorter, making sphere B smaller. (Variables: s=angular distance; r=radius or wavelength; phi=angle.)
Sphere B's spacetime is more curved than sphere A's as evidenced by the formula phi = s/r. Distance s covers a greater angle around B than A.
So far, for the sake of simplicity, we've depicted spacetime as a single one-dimensional wave as the first two diagrams below illustrate. The third diagram below is probably more realistic. It shows a complex hodge-podge of waves whose sum and average yield expectation values of vt and ct'.
The relativity theory is consistent with Heisenberg's uncertainty principle: Et >= h-bar/2. If the change in energy (E) is large, then the change in time (t) is small if their product is to equal Planck's constant (h-bar) divided by two. For spacetime we can modify this to Ect >= ch-bar/2.
If spacetime's energy is one half frequency (f) times Planck's constant (h), it follows there would be less spacetime if energy is increased to nfh (n=any positive integer) or some amount greater than 1/2fh. At high energies, there is simply less ground-state energy or spacetime.
Saturday, August 20, 2016
Maybe you've read somewhere that particles can escape a black hole via quantum tunneling, Hawking radiation or pair production; however, according to classical physics, there is no way light can escape--or is there?
Light escapes a gravitational potential if it has sufficient escape velocity. In terms of energy, a photon's kinetic energy must exceed the gravitational potential energy. A black hole's event horizon, whose distance from the center is the Schwarzschild radius (r), is the place where kinetic energy equals potential energy.
Theoretically, if a photon passes through the event horizon toward the black hole's center, it can't escape. You will note that the second equation above assumes a photon has mass (bar-m). This should be converted to a mass equivalent: fh/c^2 (f=frequency; h=Planck's constant; c=light speed; G=Newton's constant; m=black hole mass). We can now derive the Schwarzschild radius:
Houston, we have a problem! The second equation above assumes that a photon's kinetic energy is (1/2)fh. Electromagnetism has kinetic energy (photons) and potential energy (which involves charge). If photons are kinetic energy, then it only makes sense to use a photon's total energy (fh) when deriving the radius (r).
If (1/2)fh equals the gravitational potential energy, then fh must be greater than the gravitational potential energy--and the photon should be able to escape.
The math above indeed shows that fh is greater than the gravitational potential. There is another consideration: does the photon have more than zero energy? The last equation above shows that fh would equal (1/2)fh if fh is zero. When a photon enters a gravitational field it gains energy or frequency (f). When it tries to leave, it loses energy or frequency (f). The equations below demonstrate this (d=distance the photon travels; g=gravitational acceleration):
The total frequency (ft) of the photon is the sum of the initial frequency (fo) plus or minus the change in frequency (triangle f). How much energy the photon gains or loses depends on its angle of entry or exit relative to the unit normal line (n). If the photon is parallel to the unit normal (n), cos(0) equals 1, the photon will lose a maximum amount of energy when exiting a black hole. If the photon leaves at an angle, it will lose less energy.
So, if a photon enters the event horizon it will gain energy. When it tries to leave, it will lose energy. If it has more than zero energy, will it escape? Let's compare the escape velocity (c) to the gravitational equivalent. As you recall, we derived this inequality:
Let's multiply both sides by C^2/fh, then simplify:
Since c is greater than .707c, the photon will escape. In fact it will escape from any point that is outside the radius of Gm/c^2--one half the Schwarzschild radius.