## Sunday, September 28, 2014

### Nifty papers I wrote that nobody knows about: (Part 2: Quark-Gluon Plasma)

This is Part 2 of the "Nifty papers" series, talking about papers of mine that I think are cool, but that have been essentially ignored by the community. Part 1 is here.

This is the story of my third paper, still as a graduate student (in my third year) at Stony Brook University, on Long Island.

Here's the title:

 Physics Letters B 217 (1989) 5-8
First things first: What on earth is "Charmonium"? To answer this question, I'll give you in what follows a concise introduction to the theory of quarks and gluons, knows as "Quantum Chromodynamics".

Just kidding, of course. The Wiki article I linked above should get you far enough for the purposes of this article right here. But if this is TL;DR for you, here's the synopsis:

There are exactly six quarks in this universe, up (u), down (d), strange (s), bottom (b, also sometimes 'beauty"), and t (top).

These are real, folks. Just because they have weird names doesn't mean you don't carry them in every fibre of your body. In fact you carry only two types of quarks with you, really: the u and d, because they make up the protons and neutrons that make all of you: proton=uud, neutron=udd: three quarks for every nucleon.

The s, c, and b, quarks exist only to annoy you, and provide work for high-energy experimental physicists!

Just kidding again, of course. The fact that they (s,c, and b) exist provides us with a tantalizing clue about the structure of matter. As fascinating as this is, you and I have to move on right now.

For every particle, there's an anti-particle. So there have to be anti-u, and anti-d. They make up almost all of anti-matter. You did know that anti-matter was a real thing, not just an imagination of Sci-Fi movies, right?

 The particles that make up all of our known matter (and energy). The stuff that makes you (and almost all of our known universe) is in the first and 4th column. I'm still on the fence about the Higgs. It doesn't look quite right in this graph, does it?  Kind of like it's a bit of a mistake? Or maybe because it really is a condensate of top quarks? Source: Wikimedia
Right. You did. Good thing that. So we can move on then. So if we have u and d, we also must have anti-u and anti-d. And I'm sure you already did the math on charge to figure out that the charge of u better be +2/3, and the charge of d is necessarily -1/3. Because anti-matter has anti-charge, duh. If you're unsure why, contemplate the CPT theorem.

Yes, quarks have fractional charges. If this blows your mind, you're welcome. And this is how we make one positive charge for the proton (uud), and a neutral particle (the neutron) from udd.

But the tinkerer in you has already found the brain gears in motion: what prevents me from making a (u-anti u), (d-anti d), (u-anti d), (d-anti u ) etc. ?

The answer is: nothing. They exist. (Next time, come up with this discovery when somebody else has NOT YET claimed a Nobel for it, OK?) These things are called mesons. They are important. I wrote my very first paper on the quantization of an effective theory that would predict how nucleons (you remember: protons and neutrons, meaning "we-stuff") interact with pions (a type of meson made only of u,d, anti-u, and anti-d), as discussed here.

What about the other stuff, the "invisible universe" made from all the other quarks, like strange, charm bottom, and top? Well, they also form all kinds of baryons (the world that describes all the heavy stuff, such as protons and neutrons) and mesons (the semi-heavy stuff). But they tend to decay real fast.

But one very important such meson--both in the history of QCD and our ability to understand it-- is the meson called "charmonium".

I did tell yout that it would take me a little time to get you up to date, right? Right. So, Charmonium is a bound state of the charm and anti-charm quark.

(By the way, if there is anybody reading this that still thinks: "Are you effing kidding me with all these quark names and stuff, are they even real?", please reconsider these thoughts, because they are like doubting we landed on the moon. We did, and there really are six quarks, and six anti-quarks. We are discussing their ramifications here. Thank you. Yes, "ramification" is a real word, that's why I linked to a Dictionnary. Yes, those Wiki pages on science are not making things up. Now, let's move on, shall we?)

The reason why we call the ${\rm c}\bar {\rm c}$ meson "charmonium" is because we have a name for the bound state of the electron and positron (also known as the anti-electron): we call it Positronium. Yes, that's a real thing. Not related to the Atomium, by the way. That's a real building, but not a real element.

So why is charmonium important? To understand that, we have to go back to the beginning of the universe.

No, we don't have to do it by time travel. Learning about charmonium might allow you to understand something about what was going on when our universe was really young. Like, less than a second young. Why would we care about these early times? Because they might reveal so us clues about the most fundamental laws of nature. Because the state of matter in the first few milliseconds (even the first few microseconds) might have left clues to decipher for us today.

At that time (before a millisecond), our universe was very different from how we see it today. No stars, no solar systems. We didn't even have elements. We didn't have nuclei! What we had was a jumble of quarks and gluons, which one charitable soul dubbed the "quark gluon plasma" (aka: QGP). The thing about a plasma is that positive and negative charges mill about unhindered, because they have way too much energy to form these laid-back bound states that we might (a few milliseconds later) find everywhere.

So, here on Earth, people have been trying to recreate this monster of a time when the QGP reigned supreme, by shooting big nuclei onto other big nuclei. The idea here is that, for a brief moment in time, a quark gluon plasma would be formed that would allow us to study the properties of this very very early universe first hand. Make a big bang at home, so to speak. Study hot and dense matter.

While contemplating such a possibility at the RHIC collider in Brookhaven, NY (not far from where I was penning the paper I'm about to talk to you about), a pair of physicists (Tetsuo Matsui and Helmut Satz [1]) speculated that charmonium (you know, the $\bar c c$ bound state) might be seriously affected by the hot plasma. In the sense that you could not see the charmonium anymore.

Now, for ordinary matter, the $J/\psi$ (as the lowest energy state of the charmonium system is called for reasons I can't get into) has well known properties. It has a mass of 3.1 GeV (I still know this by heart), and a short but measureable lifetime. Matsui and Satz in 1986 speculated that this $J/\psi$ would look very different if it was born in the midst of a quark gluon plasma, and that this would allow us to figure out whether such a state of matter was formed: all you have to do is measure the $J/\psi$'s properties: if it is much reduced in appearance (or even absent), then we've created a quark gluon plasma in the lab.

It was a straightfoward prediction that many people accepted. The reason why the $J/\psi$ would disappear in a QGP according to Matsui and Satz was the phenomenon of "color screening". Basically, the energy of the collision would create so many additional $\bar c c$ pairs that they would provide a "screen" to the formation of a meson. It is as if a conversation shouted over long distances is disrupted by a bunch of people standing in between, whispering to each other.

For a reason I cannot remember, Ismail Zahed and I came to doubt this scenario. We were wondering whether it was really the "hotness" of the plasma (creating all these screening pairs) that destroyed the $J/\psi$. Could it instead be destroyed even if a hot plasma was not formed?

 Heavy ion collsision in the rest system of the target (above), and in the center of mass rest system (below)

The image we had in our heads was the following. When a relativistically accelerated nucleus hits another nucleus, then in the reference frame of the center of mass of both nuclei each is accelerated (from this center of mass, you see two nuclei coming at you at crazy speed). And when two nuclei are accelerated relativistically, their lateral dimension (in the direction of movement) contracts, while the orthogonal direction remains unchanged. This means that the shapes of the nuclei are affected: instead of spherical nuclei they appear to be squeezed, as the image above suggests.

When looked at from this vantage point, a very different point of view concerning the disappearance of the $J/\psi$ can be had. Each of the nuclei creates around it a colorelectric and color magnetic field, because of all the gluons exchanged between the flat nuclei. Think of it in terms of electrodynamics as opposed to color dynamics: if the two nuclei would be electrical conductors, they would span between them an electric field. Indeed, a set of conducting plates separated by a small distance is a capacitor. So, could it be that in such a collision, instead of all that hot screening, all that happens is the formation of a colorelectric capacitor that simply rips the $J/\psi$ to pieces?

That's the question we decided to check, by doing an old-fashioned calculation. How do you do this? I recall more or less that if I am going to calculate the fate of a bound state within a colorelectric field, I ought to know how to calculate the fate of a bound state in an electric field. Like, for example: who calculated what happens to the hydrogen atom if it is placed between the plates of a capacitor? Today, I would just google the question, but in 1988, you have to really search. But after searching (I spent a lot of time in the library these days) I hit paydirt. A physicist by the name of Cornel Lanczos had done exactly that calculation (the fate of a hydrogen atom in a strong electric field). What he showed is that in strong electric fields, the electron is ripped off of the proton, leading to the ionization of the hydrogen atom.

This was the calculation I was looking for! All I had to do is change the potential (namely the standard Coulomb potential of electrodynamics, and replace it by a the effective potential of quantum chromodynamics.

Now, both you and I know that we if don't have a potential, then we can't calculate the energy of the bound state. And the potential for color-electric flux tubes (as opposed to the exchange of photons, which gives rise to the electromagnetic forces as we all know) ought to be notably different from the Coulonb potential.

No, I'm not known to be sidetracked by engaging in a celebration of the pioneers of quantum mechanics. But the career of Lanczos should give you pause. The guy was obviously brilliant (another one of the Hungarian diaspora) but he is barely remembered now. Spend some time with his biography on Wikipedia: there are others besides Schrödinger, Heisenberg, Planck, and Einstein that advanced our understanding of not just physics, but in the case of Lanczos, computational physics as well.

So I was sidetracked after all. Moving on. So, I take Lanczos's calculation, and just replace the Coulomb potential by the color-electric potential. Got it?

Easier said than done! The Coulomb potential is, as everybody knows, $V(r)=-\frac er$. The color-electric potential is (we decided to ignore color magnetic forces for reasons that I don't fully remember, but that made perfect sense then)
$$V(r)=-\frac43\frac{\alpha_s}r+\sigma r. (1)$$

"All right", you say, "what's all this about?"

Well, I respond, you have to understand that when it comes to color-electric (rather than just electric) effects, the coupling constant is not the electron charge, but 4/3 of the strong coupling constant $\alpha_s$.
"But why 4/3?"

Ok, the 4/3 is tricky. You don't want to see this, trust me. It's not even in the paper. You do want to see it? OK. All others, skip the colored text.

How to obtain the quark-antiquark Coulomb potential
To calculate the interquark potential you have to take the Fourier transform of the Feynman diagram of quark-anti-quark scattering:

The solid lines are quarks or anti-quarks, and the dashed line is the gluon exchanged between them. Because the gluon propagator $D^{-1}_{ab}$ is diagonal, the amplitude of the process is given mostly by the expectation value of $\vec T^2$. $T^{(a)}$ is the generator of the symmetry group of quarks SU(3), given by $T^{(a)}=\frac12\lambda^a$. And $\lambda^a$ is a Gell-Mann matrix. There are eight of them. What the value of $\langle \vec T^2\rangle$ is depends on the representation the pair of quarks is in. A little calculation shows that for a quark-antiquark pair in a singlet state, $\langle \vec T^2\rangle=-4/3$. If the pair is in an octet state, this same expectation value gives you 1/6, meaning that the octet is unbound.

More interesting than the Coulomb term is the second term in the potential  (1), the one with the wavy thing in it. Which is called a "sigma", by the way.

"What of it?"

Well, $\sigma$ is what is known as the "string tension". As I mentioned earlier, quarks and anti-quarks can't just run away from each other (if you gave them enough energy). In strong interactions, the force between a quark and an anti-quark increases in proportion to their distance. In the lingo of strong interaction physics, this is called "asymptotic freedom", because it means that at short distances, quarks get to be free. Not so if they attempt to stray, I'm afraid.

So suppose we insert this modified potential, which looks just like a Coulomb potential but has this funny extra term, into the equations that Lanczos wrote down to show that the hydrogen atoms gets ripped apart by a strong electric field?

Well, what happens is that (after a bit of a song and dance that you'll have to read about by yourself), it turns out that if the color-electric field strength just marginally larger than the string tension $\sigma$, then this is sufficient to rip apart the charmonium bound state. Rip apart, as in disintegrate. The color-electric field generated by these colliding nuclei will kill the charmonium, but it is not because a hot quark gluon plasma creates a screening effect, it is because the cold color-electric field rips the bound state apart!

The observable result of these two very different mechanisms might look the same, though: the number of $J/\psi$ particles you would expect is strongly reduced.

So what have we learned here? One way to look at this is to say that a smoking gun is only a smoking gun if there are no other smoking guns nearby. A depletion of $J/\psi$s does not necessarily signal a quark gluon plasma.

But this caveat went entirely unheard, as you already know because otherwise I would not be writing about this here. Even though we also published this same paper as a conference proceeding, nobody wanted to hear about something doubting the holy QGP.

Is the controversy resolved today? Actually, it is still in full swing, almost 30 years after the Matsui and Satz paper [1], and 25 years after my contribution that was summarily ignored. How can this still be a mystery? After all, we have had more and more powerful accelerators attempt to probe the elusive QGP. At first it was CERN's SPS, followed by RHIC in Brookhaven (not far away from where I wrote the article in question). And after RHIC, there was the LHC, which after basking in the glory of the Higgs discovery needed something else to do, and turned its attention to.... the QGP and $J/\psi$ suppression!

The reason why this is not yet solved is that the signal of $J/\psi$ suppression is tricky. What you want to do is compare the number of $J/\psi$ produced in a collsion of really heavy nuclei (say, lead on lead) with those produced when solitary protons hit other protons, scaled by the number of nucleons in the lead-on-lead collision. Except that in the messy situation of lead-on-lead, $J/\psi$ can be produced at the edge rather than the center, be ripped apart, re-form, etc. Taking all these processes into account is tedious and messy.

So the latest news is: yes, $J/\psi$ is suppressed in these collisions. But whether it is due to "color-screening" as the standard picture of the QGP suggests, or whether it is because a strong color-electric field rips apart these states (which could happen even if there is no plasma present at all as I have shown in the paper you can download from here), this we do not yet know. After all this time.

[1] T. Matsui and H. Satz, “J/ψ Suppression by Quark-Gluon Plasma Formation,” Phys. Lett. 178 (1986) 416.

Now, move over to Part 3, where I awkwardly explain the meaning of the word "Prolegomena".