Conservation Laws

Section 9.4 Conservation Laws

Let's now try to analyze a simple reaction like you might see in a bubble chamber, in which part of the incoming energy is used to create a new particle, say a pion.

\begin{equation} p + p \to p + p + \pi\label{eq_create_pion}\tag{9.4} \end{equation}

To analyze this reaction, we use conservation laws. Three conservation laws of particle physics have analogs in classical physics: conservation of charge, conservation of energy, and conservation of angular momentum. First use the law of conservation of charge, which states that in any reaction the total charge before and after the reaction must be equal. Protons carry a single unit of positive charge, but pions come in three varieties: \(\pi^+\text{,}\) \(\pi^0\text{,}\) and \(\pi^-\text{.}\) We can thus use conservation of charge to determine which variety of pion appears in the reaction (9.4). The answer is the neutral pion, \(\pi^0\text{.}\)

Next, let's see how to apply the conservation of energy. Here is a caution about applying energy conservation to reactions in which two or more particles are present initially. Unless you are told about the incident particles' kinetic energies, energy conservation alone cannot rule out any reaction. But in a decay, when only a single particle is present before the reaction, energy conservation requires that the decay products be less massive than the original particle. In a decay, a single unstable particle disappears to be replaced by two or more less massive particles. (Can you see why at least two particles are needed?)

Here are some typical decays. A positive pion can decay into an antimuon (\(\mu^+\)) and a mu-neutrino (\(\nu_\mu\)) with a mean life of \(2.6 \times 10^{-8}\Xunits{s}\text{:}\)

\begin{equation} \pi^+ \to \mu^+ + \nu_\mu\label{eq_pion_decay}\tag{9.5} \end{equation}

The muon is also unstable and decays on the average after \(2.2 \times 10^{-6}\Xunits{s}\text{:}\)

\begin{equation} \mu^+ \to e^+ + \nu_e + \overline\nu_\mu\label{eq_muon_decay}\tag{9.6} \end{equation}

Finally let's see how to apply angular momentum conservation. In particle physics, angular momentum is quite complicated, since it is a quantized vector quantity that includes contributions from both spin and orbital angular momentum. A complete description of this conservation law would thus require techniques beyond the scope of this course. However, we can derive one simple rule, easy to apply, based on the fact that an even number of half-integral spins always combine to give integral spin.

The Fermion Rule for angular momentum conservation: A reaction can only occur if the number of fermions before the reaction and the number after are either both odd or both even.

These three conservation laws — charge, energy, and angular momentum — are absolute: they apply in all known reactions.

In addition to the absolute conservation laws of charge, energy, and angular momentum, there are two other well-tested conservation laws, those for lepton number and baryon number. Although some modern field theories predict their violation in reactions like proton decay, these violations have never been conclusively observed. So until we discuss grand unified theories in Chapter 11, we will assume that lepton and baryon numbers are conserved.

So what is lepton number? Actually it's quite simple. Each lepton is assigned a lepton number of \(L = +1\text{.}\) The antiparticles of leptons have \(L = -1\text{.}\) All other particles have \(L = 0\text{.}\) Now consider a reaction like pion decay, (9.5). The pion is not a lepton, so it has \(L = 0\text{.}\) The \(\mu^+\) is the antiparticle of the \(\mu^-\text{,}\) a lepton, so \(\mu^+\) has \(L = -1\text{.}\) Finally, a neutrino is a lepton, with \(L = 1\text{.}\) Lepton number is an additive quantity, so simple addition shows that both sides total \(L = 0\text{,}\) and the decay is allowed.

Actually, conservation of lepton number is slightly more complicated than this simple example implies. Consider, for example, the following muon decay

\begin{equation} \mu^+ \to e^+ + \nu_? + \overline\nu_?\text{,}\label{eq_muon_decay_q}\tag{9.7} \end{equation}

and a possible alternative decay

\begin{equation} \mu^+ \to e^+ + \gamma\text{.}\label{eq_not_muon_decay}\tag{9.8} \end{equation}

The second decay seems simpler and also appears to conserve lepton number, but in fact, a decay like (9.8) has never been observed. A situation like this is a puzzle for physicists. Why doesn't (9.8) ever occur? There must be something new going on that prevents it. This “something new” is the existence of a previously undiscovered conservation law!

The new law is this: Each of the three types of leptons is conserved separately. Thus, there are electron neutrinos (\(\nu_e\)), muon neutrinos (\(\nu_\mu\)), and tau neutrinos (\(\nu_\tau\)). The electron and its neutrino each carry \(L_e = +1\text{,}\) \(L_\mu = L_\tau = 0\text{,}\) and similarly for the other leptons. Thus, the muon decay of (9.6) is actually

\begin{equation} \mu^+ \to e^+ + \nu_e + \overline\nu_\mu\tag{9.9} \end{equation}

and the candidate decay (9.8) is ruled out because neither \(L_e\) nor \(L_\mu\) is conserved. So when we say “conserve lepton number,” it really means “conserve all three types of lepton number: \(L_e\text{,}\) \(L_\mu\text{,}\) and \(L_\tau\text{.}\)” REMEMBER: There are THREE separate lepton numbers, and each needs to be conserved separately. E.g., if \(L_e = 1\) and \(L_\mu = L_\tau = 0\) before a reaction, and \(L_e = 0, L_\mu = 1\text{,}\) and \(L_\tau = 0\) after the reaction, then this violates lepton number conservation since \(L_e\) isn't conserved and \(L_\mu\) isn't conserved.

Similar observations lead to the formulation of the law of conservation of baryon number. Consider the two candidate reactions below:

\begin{equation} \pi^0 \to \gamma + \gamma\label{eq_baryon_allowed}\tag{9.10} \end{equation}

and

\begin{equation} n + n \to \gamma + \gamma\label{eq_not_baryon_allowed}\tag{9.11} \end{equation}

Both are allowed by all the conservation laws described so far, but the first occurs all the time, while the second never occurs. Why not? A new conservation law must be operating: conservation of baryon number. Neutrons are baryons, so they each have \(B = 1\text{.}\) Photons and pions are not baryons so \(B = 0\) for them. Any antiparticles of baryons have \(B = -1\text{.}\) Finally, since baryon number is additive, we have that the decay in (9.10) is allowed because \(0=0\text{,}\) while the reaction in (9.11) is forbidden because \(2 \neq 0\text{.}\)

These sorts of ad hoc conservation laws seem to be begging the question. Later on we'll encounter a deeper reason for these laws, especially the law of conservation of baryon number.