TD Lee: Is the vacuum a physical medium? (contd.)

0:00 Now, this is a curve, almost like a plateau, with peaks, so these are this, it starts off at low end, there are 1 jab with a ratio of 2, except for the low peak, omega peak, 5 peak, remains 2 until about 4 jab, then jumps, then jumps to 3 and 1 third, then remains almost flat until about 10 jab, then jumps to 3 and 1 third. So let's understand that. All of these have to be greater than or less than 2 times the mass of charm. In which case, you may recall that we will have up fork, down fork, and strange fork. Up fork charge is 2 thirds, down is 1 third, minus 1 third, and strange is 1 third, also minus 1 third. So 2 thirds squared plus 1 third squared plus 1 third squared, that's the sum of the Q squared. Times three, that's a three different kind of color, that gives you three times six nines, which is two. And this is a very well-founded. Then, when it passes around the spectacular discovery of J.F. Tsai's residence, then it jumps to, it increases the amount by delta r. The increase amount is the three times the chunk force charge, two-thirds squared, so three times four-ninths, that gives you four-thirds, jumps two into three and one-third. Then up to the tenth, Jeff, and that is the bottom fork and ending fork production, the epsilon, and it jumps to three and two. So you see, and this then gives you the following information. It is true that these are fractional chunks. Second, is it true that there is a bigger freedom if you multiply the charge square by three? Is it true that their masses are small because whenever they jump, that gives you the two times the mass threshold? And, is it true, except for the resonance, of course, at the resonance, you have 90 degree phase shift, you of course cannot neglect the interaction. But outside the resonance, they essentially act like a free particle. This formula is correct, only if you neglect strong interaction. So all these things are true. Now, how can it be, yet, a particle whose masses are really not high, you will go across the spectrum, and you know there is no strong interaction, otherwise the formula will not be true.

2:30 But they are not produced. Being produced in a direct way, you don't see them. And that, of course, is a mystery. Why don't you see them? And this is the so-called, quote, confinement. We say we don't see them because they are confined. Now, what does that mean? And this is the topic that we shall... To resolve this, we need first to introduce so-called quantum chromodynamics. But don't be frightened. So we introduce quantum chromodynamics. Quantum chromodynamics is like quantum electrodynamics. The basic field itself, the bulb field, has a color. Color will be three kinds, one, two, three. Has flavor. Flavor will be up, down, and straight. The photons and cell photons, we now have the glue on. The gluons are the photons of quantum chromodynamics. The gluons have colors. There are three colors for the quarks. The gluons have... Why eight? Because eight is three times three minus one, as you can count. Why? Because the quarks interact... The source of the gluons is the quark Q bar Q, so Q of C prime. Now, there are three kinds of antiquarks, three kinds of quarks, so all together there are nine different varieties, but the gluons have color. Of these nine, one is colorless, that we have already said. That's the red bar red, plus blue bar blue, white bar white. So you must subtract the colorless one out, because that one does not produce the color gluon. So that is why there are eight gluons. Then technically, we call this is a SU3. Now I'm giving gauge group, and the generators of SU3 group, there are eight of them.

5:00 And these are governed by the familiar Gaumont. Now, a dynamics of this is called QCD. Now QCD has, however, a very serious... So we shall first remedy this by saying that the universe is really finite, if we put the whole universe in a finite volume, then of course infrared is cut off because it's a natural cut off of the radius of the universe, and that's what we are going to consider. So let me consider the quantum chromodynamics in a finite universe such as the universe that we are in. So then we now have a very good theory, very normalizable, and the charge. There is a difference between the quark and the gluon, called G, for this volume of sub-capital L. This is a technique called renormalized coupling constant for wavelengths which is as long as this. Then we can prove a theory which is valid to all orders. That is, if I consider two such volumes, one big, one small, but otherwise identical, then the big one, the renormalized constant, is always bigger than the smaller one. Now, if I make the universe bigger, and I always constantly respect the nuance of this wavelength, then it becomes stronger and stronger, and this is valid to all. Now, I shall give you a proof only for those who have heard of asymptotic freedom, but I will give you a little bit of explanation. It was discovered by Pollitzer and Gross and Wojcik. That is, if I consider the QCD for very small volume, that is, at very high energy, Then the coupling goes to zero. This is called asymptotic freedom. For your purpose, that's the magic, it says already. And the reason this is very good, then it explains why the clocks are easy. Because at very high energy, which means very small distance, when you have zero couplings, then they are like three particles.

7:30 Because that gives you, you see, the ratio that we just count our value, we must regard it as three particles. And you get a strong interaction. So this is called asymptotic freedom. Now, since in quantum thermodynamics there is no scale except over or cut off, so if you make the coupling become small, then the same argument gives you at large distance the coupling becomes large. So, that then proves the theory to all. So, therefore, we know that if I make the coupling small, at high energy the coupling will be small, and that if you apply high-energy epoxide free, it will become big. We do live in an even or finite, but really, all or two. Therefore, g of infinity, most likely, will be infinite, or even at best condition will be greater. So we are living in an ultra-strong coupling universe. Now our modern technology, for example, is ultra-strong coupling, we don't know what to do. Therefore, it is difficult for us even to conclude that QCD is not true. We don't even know how to deal with it. It is this difficult yet. That we will try to resolve and will become the main discussion of the remaining part of this lecture. A bunch of definitions. The definitions are the following. That will be the, let me introduce the definition of color. Dye of the back. Define the top part. In order to do this, let me refresh you, your memory about the constant in the ordinary QED. If I consider two charts B, then in a medium that was the way original over kappa, for the vacuum we call kappa equals one. That's the definition.

10:00 E squared is a unit of 1 over 127. We may call kappa over the vacuum two, in which case the E squared will be two times. Now, so therefore, the dielectric constant by itself has no meaning. It's a product of E squared over kappa. Then the relative dielectric constant. Anticipating that the g squared is infinity, so the vacuum is not a good standard to use, so the ratio to zero is zero. So let me introduce a convention. Consider the vacuum for a small volume. A volume, let's say, of microscopic distance. Let me call that radius a proton, and call that volume the dielectric constant 1, and the g for that volume is g without substitute. Then I define g for any other volume. So we simply introduce r is equal to r equal to proton, the g of r is g, defined to be g equal to any other square is defined to be this g squared. So this is a redefinition from the coupling constant to the dielectric constant. So far we have not done anything. But now, using this standard, I can rephrase the theorem which we have just solved. It's plausible, but can be true. If I make the volume of the universe, the G becomes bigger, dielectric constant, to be one, or less, I will call it, the vacuum is a perfect dielectric, I will call the vacuum to be a nearly perfect dielectric, that's what we have to say. Let me assume that the vacuum is a nearly perfect dielectric medium, dielectric constant, much less than one.

12:30 All the vacuum is a perfect color diode electric medium, in which case the kappa will be zero, which is now proven here. Whenever there are physical quarks, anti-quarks, gluons present, there will be spatial, inhomogeneous solutions, cracks, holes, solid on the back will be developed. Because the rationale will be quite elementary. In classical electricity, let me assume we have a medium. In classical electricity, of course, we always take the cup of the vacuum. Let me assume there is an electric medium whose dielectric constant is a direct electron. In nature, such a medium does not exist, not for static electricity. Let me give a proof why it does not happen. Then we shall argue that if it does happen, what kind of, or if it did happen, what kind of physical phenomena would impact the electrodynamics. All mediums of the world have a dielectric constant bigger than one. This is because if you recall elementary physics course, I was told that at least people have taken freshman physics in this audience. So the displacement vector equals the electric vector plus 4 pi of polarity. So if I put a positive charge here, then I will produce the polarization around the medium, but since positive and negative attract each other, positive and positive recurse, all the diapositives produced will be lined up in this way along the electric field.

15:00 That will mean P will be parallel to E. That means the cutoff, the dielectric constant, will be bigger than 1. This, of course, is the proof why all material with dielectric constant bigger than one, that is, it is always screened. But in quantum chromodynamics, it so happened, or rather, let's say, that somehow we have it whose dielectric constant, that will be the time totally admissible in the 19th century before we have atomic field. Or if we do want to use electricity, then we can assume we have a frequency. Then, in fact, experiments have been done of such a thing with the at a certain frequency the dielectric constant is less than one, the medium is. So for the moment, let's imagine there is such a medium, which is anti-screening, is less than one. Let me put now a charge in there, see what happens. So if the person pose to the gentleman, we imagine we are in classical electricity. Now, if I have a chart with a little extension, then the positive electricity will tell you that I will produce a field outside, and the field stress will be given by epsilon of kappa r squared, with r within that, if this chart is infinitesimal. It is true when Kappa is bigger than 1. It is not true when Kappa is less than 1. When Kappa is less than 1, no matter how small this charge is, provided it is finite, a hole will be developed around it.

17:30 If you put it in a material, this hypothetical material, a crack will open, and inside this crack, this is what we want to prove. In order to prove this, let me balance the energy content with this homogeneous solution. And finally, we have the lowest energy state with any charge E in classical electricity in a medium which is an energy screening, so this is really a barrier in the nature, and we do not have one to add to that. But if that is the case, it must produce a hole in the medium so that the whole will not Outside, we have the medium whose dielectric constant is less than 1. Inside, it is a hole with kappa equals 1. Why? Let's compare the energy of this configuration with this configuration, which has low. Let's ask, which has low energy? Now that can be calculated very easily. Because you have a charge here, the medium is anti-screening, so around the wall of this hole, you have the same sign of the charge as this one. So if I now shrink this form to reach that configuration, I have to do work. So it's obvious this has lower energy than that. Is that very obvious to you? If not, maybe I will pause for a minute, because that is the catch. Kappa is bigger than this. They are screaming. That means this charge will be negative. That means that you can't have this form because you release it. It will collapse. The positive and negative will affect it. And that reaches the usual classical textbook. But if it's energy screening, the minute you put into a medium, the hole will be developed, just because I have low energy. If this is very clear to you, then we will capture what will be the energy difference.

20:00 The energy difference will be the electric energy, energy you have to make a crack. The energy to make a crack, although this depends on the medium. But if the practice is very big, presumably the bigger the hole is, the harder you have to dig, so the energy will be proportional to the volume of the hole. That seems to be very reasonable. So let's call that an U-R. Is energy required to produce a hole of radius R? So U-R is energy required to produce a hole of radius R, and that must be an increasing function like shown here. Some function u, plotting against r, and asymptotically will be provided r cubed, the volume, but can have r squared, can have linear r, all. Then we want to calculate electric energy. Now d, the displacement field, is continuous. The Gauss theorem, d in, at this point, is e over r squared. It's the same as d outside. At the outside, the couple is less than one, so e in the outside region is this field divided by a couple. Electric energy is d times e, so it's e squared over kappa r. Now really, we should compare this with vacuum, so it will be e squared over kappa r minus, yet ignore that, if kappa is very small. And that will give you a curve like this. With r small, the electric energy becomes big. So it then gives you a radius which is never zero. The minimum energy. This just reaffirms what we have got intuitively. If kappa is less than 1, you always will have a finite radius. Conversely, if kappa will be bigger than 1, then we really have minus e squared r for the vector. I have not put this on because this transparency is a little bit small. Then the curve will become this way. So then, of course... Is that very clear to everyone?

22:30 If by dm dr equals zero, because u increases with r, it gives you the height. If I only take the volume part, by differentiation of this, the r will be in kappa to the minus one-fourth, the total energy will be one over kappa to the three-fourth. When kappa goes to zero, this bottom will go to infinity. In other words, this is because electric energy, kappa goes to zero and electric energy will go to infinity. In other words, what we will find is this. First, if I have a dielectric medium whose dielectric constant is less than one, a dielectric, no matter how small a charge there, I will produce holes. Furthermore, if the total charge is not zero, if the medium is a dielectric, then the total energy of this charged particle will be infinite. If I put a dipole in there, notice the energy goes, becomes very large, it's because of the Gaussian. It leaks out to the medium, that's why you have a large electric field. If I have a dipole, this is an N-confined transparency, now if I have a dipole, then if the Gaussian of the cup is very small, in the limit of zero, all the electric field inside will be parallel, because there's no reason for the minimum energy to leak out. If it leaks out, it costs you lots of energy, so it goes. Therefore, the mass of a dipole becomes finite. So the first thing is, no matter whether you have a single charged particle or a dipole, once you have a dielectric medium, any screening medium, the minute you put a charged particle, you will make holes. If it's a total charge non-zero, then the mass of that particle will be empty. If it's a dipole, that means the bias will be finite. It also means that you can never take them apart. Now QCD we know is an anti-screening. Its dialectic concept is less than one. We are going to not know it's truly zero, but it's at least nearly a good color dialect to me.

25:00 Therefore, if we can have a cork, or anti-cork, or cork, anti-cork, and color singlet, or non-singlet, for whatever it is, you have both. It's a perfect dialect to read. Otherwise, we can also make the argument the other way around. The very fact we have not seen free in the SLAC experiment, we can give a limit of the mass of a free core. It must be bigger than 5G by free means with the whole of a single colored object. From that, we can conclude that the color die electric constant of the vacuum must be less than 0.013 times e to the power 4 pi, which is about half or a quarter, depending on how you determine it. Now we say, in what way have we solved this difficulty of ultra-square coupling? We have solved this ultra-square coupling because You see, let's look at in physics what kind of problem we have strong interaction and does not give us technical difficulties. A very good example is the process between helium and helium. If I plot the helium atom versus helium atom, plot against radius, it's a potential, like this. At the central region, there's a strong repulsive core. Then it's a very shallow, attractive repulsive core. In fact, once it's repulsed, the storm goes back, because if the repulsed core is very strong, then the height is very high, then you really do not care how high it is, because the helium atom is never there. The wave only starts to appear, because it's repulsed, it pushes the thing away from the storm. Is that very clear? Now, so is the case of the fork, you see. Once the medium, let's assume the vacuum is a perfect die-hard electric medium, color-wise, you have a single quark, you see, quark, copper is one, but it's a single quark or a quark-and-quark.

27:30 So in other words, the place where your strong copper and your slow copper is this, that's here, but the quark is there. So once you recognize this, then there is no difficulty in expansion. The only thing is that you should not make expansion which has a Fourier transform to cover all the forbidden. Just like here, you should not expand. You cover the repulsive core. That would be a terrible mistake to make. To do this, you simply use the hard core. That's easy. To do that is also not typical. It's simply the soliton, where you have to make the relativistic formulation, that you have the soliton of that solution. But the main thing is that there is no difficulty to begin with, because this is the action of the vacuum. Even though in nature we don't have electricity, we do not have a real dielectric medium at zero frequency, but at Columbia they have performed experiments with laser beams of dielectric constant for a given frequency which is less than one. Then by simulating for the beam interaction, you can see this traffic thing that I talked about. But we can draw another analogy, that is superconductivity. Now, in nature, quantum electrodynamics, there exist superconductors, which has the magnetic susceptibility equal to zero. It's a perfect die of magnet. So we change magnetic field in superconductivity to the color of electric field in quantum chromodynamics. We change the magnetic susceptibility of a superconductor to the kappa of the vacuum. Superconductor, this is superconductor, outside the vacuum,