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Re-normalizing Feynman Diagram Amplitudes in a Non-arbitrary Way

Quantum electrodynamics (QED) is perhaps the most precise and successful theory in all of physics. There is, as I've mentioned in pre...

Saturday, October 29, 2016

Deriving Least Action, Least Energy and Matter-Antimatter Broken Symmetry

Today we are going to see why there is a ground-state energy and why there is more matter than anti-matter. Oddly enough, it has to do with Heisenberg's Uncertainty Principle and the Principle of Least Action. At least that's what the mathematics suggest.

We begin our action with the Lagrangian (L): Kinetic energy minus potential energy (m=mass; v=velocity; k=Hooke's coefficient; x=position). We set L to zero:

If we take the integral of L with respect to time (dt, t), we get the action (S).

We set the constant C to zero. Doing so will help us yield the least action physically possible. How do we know this is true? We recognize the action is energy times time (Et). We want the least action, so what is the least (Et) we can think of? How about the right side of Heisenberg's equation?

Thus the action (S) shall be equal to or greater than h-bar (Planck's constant) over two.

From here we can determine the energy of the ground state (f=frequency):

We get an interesting result. Equation (10 shows that L doesn't equal zero after all. There's something rather than nothing. This odd fact of nature could explain why matter and anti-matter didn't totally annihilate each other in the distant past.

If we express matter and anti-matter in terms of energy (+E, -E), we can show that we don't net zero as expected when we subtract anti-matter from an equal amount of matter. Let's commence with Einstein's energy equation (p=momentum; c=light speed):

We have positively charged energy (matter) and negatively charged energy (anti-matter). We get the net energy (J for Jackson) when we add them together. Does J really equal zero?

To find the answer we previously used the action (S). This time we'll use Sj. Sj stands for ... are you ready for this? ... Action Jackson!

The above math is similar to what we did before, and, like before, we end up with something rather than nothing. The amount of matter left over (in terms of energy) is greater than or equal to the ground state energy. Observations tell us it's "greater than" ... enough to fill our universe.

Tuesday, October 18, 2016

Deriving Dirac's Lagrangian

To derive Dirac's Lagrangian, we begin with the Dirac field's adjoint spinor (bar-psi) and spinor (psi). Note that each spinor contains right-handed (psi-R) and left-handed (psi-L) fields. (Right-handed means the spin and momentum of the field particles are in the same direction. Left-handed means spin and momentum are opposite.)

Let's put the fields into an algebraic form:

Next, we multiply them together:

On the right side of the equal sign the first two terms are each zero and add to zero. Here's why:

The mixed terms don't equal zero. Here's why:

The first two terms that equal zero won't equal zero if we multiply them both by Dirac's matrix.

Now we can set those first two terms equal to the mixed terms:

On the left side, we take the derivative of the fields and multiply by -i, Planck's constant, and the speed of light (c). That gives us kinetic energy. On the right side we multiply by mass (m) and c^2. That gives us potential energy.

The Lagrangian is kinetic energy minus potential energy, so we subtract the potential energy from both sides to get the Lagrange (L):

Sunday, October 16, 2016

Why Neutrinos Have Mass

Once upon a time there was a Higgs field. Below it's represented by the greek letter phi and the matrix that contains phi-A and phi-B:

Phi-A is normally zero, but I will show later that it doesn't have to be. Phi-B consists of the Higgs ground state (phi-o) plus a Higgs perturbation (h(x)). Next we have the mass (m) term from Dirac's Lagrangian. This will come in handy later. Within its parentheses are Dirac fields (psi-L for left-handed and psi-R for right-handed).

Let's calculate the masses for the right and left-handed electrons and the left-handed neutrino. We begin with the Lagrange interaction equation (Lint) The Ge variable is the Yukawa coupling which determines the strength between the Higgs field and lepton fields.

Below we solve the first term within the parentheses:

The left-handed Dirac field (psi-L) contains the left-handed neutrino (Ve) along with the left-handed electron (eL). The right-handed Dirac field contains no right-handed neutrino (0) and the right-handed electron (eR).

Time to solve the second term within the parentheses:

Now let's do some algebra gymnastics:

Look at the last two equations above. The first is for electrons and the second is for neutrinos. Now it's time to pull out that mass term from Dirac's Lagrangian. We set it equal to each equation and then calculate the mass for electrons (Me) and the mass for neutrinos (Mv). (In case you were wondering, the h(x) term above was dumped, since it's not a scalar.)

The mass for the electron is the Yukawa coupling times the Higgs ground state. Since phi-A is normally zero, the neutrino's mass is normally zero in mathematical terms. However, such a mathematical result does not agree with experiments. The neutrino, as it turns out, has a little mass.

The challenge before us is can we put together a mathematical model that allows the neutrino to have mass? If the Higgs grounstate, phi-A, could be greater than zero, problem solved? Not necessarily. If we give the the neutrino mass that way, we screw up the math for photons and the weak-force bosons. Here is the model for bosons (Bu = weak hypercharge gauge field; The W's are the weak isospin gauge fields; gb and gw are coupling constants) :

Notice how the right-hand matrices above have zero as the ground state for phi-A. If that value was anything but zero, we could not calculate the proper masses for the weak-force bosons or the photon. Imagine a photon with mass! Not good!

To solve this problem, we need to familiarize ourselves with how the ground states were derived. Here is an example:

The first equation above is the potential. If we take its derivative and set it to zero, we can determine the minima or ground states. For phi-A we definitely get zero. For phi-B we don't get zero; we get the value of phi-o. We add to that the higgs perturbation (h(x)). Here is a crude diagram that illustrates the two ground states:

Imagine a marble at the top of the hill. The marble can roll down either side and end up at zero or v. It seems we are stuck with these two ground states. The good news is we know the neutrino has mass, so the theory (and its crude diagram) is incomplete.

The incompleteness of the theory can be demonstrated with the following derivation of the Higgs field:

The last two equations above show that phi-A and phi-B are absolute values of two complex numbers and four fields (phi-1, phi-2, phi-3, phi-4). Now take a look at this. Here's where it gets interesting:

As you can see we derived the Higgs field, but there's something missing! Can you see it? Yes! It's the third Pauli matrix that's missing. What would happen if it were included? Here we go:

When we add he third Pauli matrix we no longer get zero for the phi-A ground state. We get the variable z2 which can be zero or any quantity other than zero. Ideally we'd like z2 to be zero when we are calculating boson masses, but we'd like it to be more than zero when we calculate the neutrino's mass.

Let's find the ground states for z2 and z1:

The ground state for z1 can be arbitrarily chosen to be zero, so we can get the correct mass for the electron (and the muon and tauon). The variable z2 has two ground states: zero and a mass. This is good news! When z2's ground state is zero, we get the proper masses for bosons. When its absolute value is more than zero, we get a neutrino with some mass! How about another crude diagram to illustrate?

Imagine another marble on top of the big hill. If it rolls to the left it lands on a smaller hill with two more choices: 0+0 if it rolls left, or 0+z2 if it rolls right. Put the marble back on top of the big hill. This time it rolls right. It lands on the tiny hill and has two choices: v-z1 or v-0. Since z1=0, we get v either way. Perfect!

We can trade this old model below (which wrongly predicts zero mass for neutrinos)...

... for a new model that predicts a mass for neutrinos without screwing up the math for bosons.

Thursday, October 6, 2016

The Problem With Modern Theoretical Physics

There's a new theory that explains every mystery in physics. I call it the magic-dust theory. Like other modern theories it is mathematically consistent. But like other modern theories it is currently not testable. To observe the magic dust requires a particle accelerator the size of our galaxy, maybe bigger.

Here's an example of the mathematics of this promising new theory:

Thanks to this new innovative theory, we now know what caused the Big Bang. There is no empirical evidence, but, as I said before, this theory is mathematically consistent. Here is an example:

All kidding aside, the "magic-dust" theory demonstrates what is wrong with modern theoretical physics. If you stop and think about it, there isn't much difference between magic dust, and strings.

Like magic dust, strings have virtually unlimited power: they can vibrate and make the different particles; they can stretch up to infinity in multiple dimensions to give us d-branes. They can do whatever is needed to explain any mystery. As long as the math is consistent, we can call it science. However, Issac Newton would disagree. Below he discusses gravity and his philosophy:

"I have not as yet been able to discover the reason for these properties of gravity from phenomena, and I do not feign hypotheses. For whatever is not deduced from the phenomena must be called a hypothesis; and hypotheses, whether metaphysical or physical, or based on occult qualities, or mechanical, have no place in experimental philosophy. In this philosophy particular propositions are inferred from the phenomena, and afterwards rendered general by induction."

Note how Newton uses the word "hypothesis" rather than "theory." Technically, string theories and other modern theories aren't theories at all. At best, they are hypotheses, since the entities they invoke (e.g. strings, extra dimensions, etc.) have never been observed. Below, William Whewell has this to add:

"What is requisite is, that the hypotheses should be close to the facts, and not connected with them by other arbitrary and untried facts; and that the philosopher should be ready to resign it as soon as the facts refuse to confirm it."

I have to agree. The challenge is to explain the mysteries without magical thinking; i.e., adding extra entities that have never been observed like extra dimensions, strings, branes, magic dust, etc. The whole idea of science, in my opinion, is to reduce mystery, not add to it. If you have to create a new mysterious object to explain an old mystery, then that is the analogue of digging one hole to fill another.

Using new, unobserved, untested entities to explain a mystery has the disadvantage of increasing your burden of proof. This is why the best hypotheses use known facts as much as possible to explain the unknown.

On the flip side it's true that Democritus dreamed up the atom centuries before it was physically observed. And maybe sometime in the distant future, humans will discover strings, extra dimensions, gods, fairies, pixies, and dragons in their basements ... or ... maybe not. The thing I love about science is we are free to change our minds as soon as the evidence becomes available.

Monday, October 3, 2016

Unifying the Dirac, Klein-Gordon and Einstein Equations

Within quantum field theory we find two equations based on Einstein's equation for energy (E = energy; p = momentum; c = light speed; m = mass):

The first is Dirac's equation. The second, the Klein Gordon equation (i = imaginary number; h-bar = Planck's constant; bold y with arrow = Dirac-Pauli matrices; a-sub-u = first derivative; psi = quantum field; t = time; up-side-down triangle = Laplacian).

The goal is to take elements and ideas from the above equations and combine them with Einstein's field equation below (Guv = Einstein's tensor; Gn = Newton's constant; Tuv = the energy-stress tensor; and then there's pi):

Let's first write Guv in quantum terms (psi-sub-r = relativity-field-wave function; Aj = a coefficient element in series j). Note there's a double derivative of psi with respect to an x-vector, and tensor indices (uv) after the closing parenthesis. There is also psi's complex conjugate, which, when multiplied my psi and the rest, gives the expectation value; i.e., a result we expect on the classical scale.

Next, we write Tuv in quantum terms (V = volume). Again we get an expectation value, and the entire term is placed in parenthesis and made a tensor with indices uv.

Now that we have Einstein's equation in quantum terms, let's define each element in greater detail. Below is the definition of psi:

Next we define k. The k represents wave number(s) for one or more particles summed (j=1 to n) by the summation sign. The arrow over k indicates it is a 4-vector. The "a" with a dagger is the creation operator. The zero in the ket represents the field in the ground state.

The variable x is also a 4-vector:

The bold b with an arrow can be any one of three types of spin matrices: Dirac-Pauli matrices (y) for half-spin particles; S-matrices for integer-spin particles; and the number one for zero-spin particles.

Below is a full demo of the Dirac-Pauli matrices:

Here is an example of one Dirac matrix acting on the k 4-vector:

The following are 4X4 matrices I designed for spin-1 particles. I always admired Dirac's equation and matrices, but they are limited to half-spin particles. Now it's possible to include bosons in the field.

We now have a quantized version of Einstein's field equations where we can work with both matter and anti-matter. We can work with zero-spin particles (Higgs field), spin-1 bosons, fermions, atoms, molecules and beyond ... and see how they curve spacetime in terms of the wave number(s) k.

The interesting thing about k^2 is it has the same units as Einstein's field equations: 1/L^2. Thus if we take the double derivative of psi with respect to x, we get k^2. Multiply that by the unit-less coefficient A; multiply psi by its complex conjugate--and we get the equivalent of an Einstein tensor element. Put 16 elements together and we get the complete Einstein tensor.