A question I often get, like really often, especially from other physicists, is “How do we know quarks exist?” In particular,
This is a really good question and it has a number of different answers. To a physicist, being able to directly observe an object means being able to isolate it and subsequently measure its properties, for example: electric charge. Due to effects associated with the strong nuclear force, quarks lighter than the top quark will nucleate into other objects (hadrons) in about 3×10-25 seconds. This is pretty fast, much faster than any piece of modern electronics. Consequentially, light quarks cannot be directly observed with present technology. However, this inability to isolate quarks does not imply we cannot directly measure their properties (like electric charge!).
This brings me to today’s post: How physicists measure quarks’ electric charges!
A very typical calculation done by any student in a course on particle physics (undergraduate or graduate) is to calculate the likelihood (called cross section) of an electron and positron annihilating into a virtual photon, which then decays into a muon and anti-muon. (See the diagram to the right.). Since electrons, muons, and their anti-matter partners all have so little mass, it is pretty reasonable to just pretend they are all massless. The calculation becomes considerably easier, trust me on this. When all is said and done, we find that the cross section is equal to a bunch of constants (which I am just going to collectively call σ0), times the square of the electron’s electric charge (Q2e), times the square of the muon’s electric charge (Q2μ):
Likelihood of e+e- → μ+μ- = σ0 × Q2e × Q2μ
However, the electric charges of electrons and muons are both 1 (in elementary units) so the likelihood reduces to just σ0. Convenient, right?
Likelihood of e+e- → qq = 3 × σ0 × Q2q
That’s right: the probability of producing quarks with electrons & positrons is simply three times that for producing muons, scaled by the square of the quarks’ electric charge. This amazing result allows us to then define the quantity “R“, which is just the ratio of the likelihoods:
R = (Likelihood of e+e- → qq) / (Likelihood of e+e- → μ+μ-) = 3 × Q2q
In other words, by measuring the ratio of how likely it is to produce a particular set of quarks to how likely it is to produce muons, we can directly measure quarks’ electric charge! (BOOYA!)
As far as measuring R goes, it is pretty straightforward. However, there has to be some caveat or complication since this is physics we are talking about. Sure enough there are a few and I am just going to ignore them all, all but one.
In order to determine the probability of producing a particular pair of quarks using electron-positron collisions, experimentalists have to make sure the total energy of the collision is large enough. Simply put, no particle can ever be generated if there is not enough energy to make it. It is an example of the Conservation of Energy. The problem is this: if there is enough energy to make a particular set of quarks, then there is sufficient energy to produce any quark pair lighter than the original set. In addition, it is very difficult to isolate different quark-anti-quark pairs (see the top of this post for why that is).
The solution to this issue is to simply measure the likelihood of producing ALL types of quarks for a particular energy. To do so, all we need is to add up all the individual cross sections for each set of quarks. The total cross section simplifies to this:
Likelihood of e+e- → ALL qq = 3 × σ0 × Q2e × Sum Q2q
That is to say, the probability of producing ALL quark-anti-quark pairs in electron-positron collisions is equal to a bunch of constants (σ0) times the square of the electron’s electric charge (Q2e), times the sum of the square of each quark’s electric charge (Q2q). Consequently, R becomes
R = (Likelihood of e+e- → ALL qq) / (Likelihood of e+e- → μ+μ-) = 3 × Sum of all Q2q
R may no longer be a direct measurement of a single quark’s electric charge, but it is still a direct measurement of the electric charge of all the quarks. Without further ado, here are the predictions:
Here are the data. This plot is taken from my favorite particle physics books, Quarks & Leptons:
That Disagreement Near 5-8 GeV is Not Really a Disagreement
Time for a little extra credit. If you look closely at figure 2, you may notice that between 5 GeV and 8 GeV all the data points are uniformly above the R=10/3 line. This feature is actual the result of two things: the first is that quarks really do have masses and cannot be ignored at these energies; the second is that the strong nuclear force surprisingly contributes to this process. I will not say much about the first point other than mention that, in our quick calculation above, we pretended to ignore all masses because electrons and muons were so light. The mass (in natural units) of the charm quark is about 1.3 GeV, and that is hardly small compared to 5 GeV.
Taking a closer look at where the virtual photon produces a quark and anti-quar k pair, we realize that quark and anti-quark are pretty close together. They are actually close enough to emit and absorb gluons, the particle that mediates the strong nuclear force. This has a very important consequence. Previously, the quark and anti-quark pair could only be produced in such a way that the total momentum of the system was conserved. However, if we consider the fact that the quarks can exchange gluons, and hence exchange momenta, then the quark and anti-quark pair can be produced an infinite number of different ways that violate the conservation of total momentum, so long as at least one gluon is exchanged between the two in order to restore total momentum. This amplification in likelihood is highly sensitive to energy but it causes about a 20% increase in R between 5 and 8 GeV. This 20% increase in R is precisely the difference between all the data points and the R = 10/3 line.
- richard (@bravelittlemuon)
P.S. #PhysicsFact should totally be a trend today. Go! Make it trend!