Intro to Particle Physics

Below the nucleus

In the last lesson you treated a nucleus as a bag of protons and neutrons, and you got a long way with nothing more than mass differences and $E = mc^2$. But a free neutron, left alone, turns into a proton in about fifteen minutes. Something inside it changed. A bag of two fixed ingredients cannot do that.

Push a proton hard enough into another proton and you do not get proton fragments. You get a spray of particles that were not there beforehand: pions, kaons, lambdas, muons, things that live for a hundredth of a billionth of a second and then vanish. Hundreds of them have been catalogued. The question this lesson answers is how anyone makes sense of that list, and how you decide (on paper, before any experiment) whether a proposed reaction can happen at all.

The answer has two halves. One half is bookkeeping: a small set of numbers that every particle carries and that must balance on both sides of a reaction. The other half is energy and momentum, done relativistically, which decides whether you can afford the products even when the bookkeeping works out. Almost every problem in this unit is one of those two, or both.

Twelve fermions and four forces

Start with the matter particles. There are twelve of them, in two families of six.

Quarks come in six flavours, arranged in three pairs of increasing mass: up and down, charm and strange, top and bottom. Each pair is a generation. The first member of each pair (up, charm, top) carries electric charge $+\tfrac{2}{3}e$, where $e$ is the elementary charge. The second (down, strange, bottom) carries $-\tfrac{1}{3}e$. Those thirds are the whole reason the quark model was believed before anyone saw a quark: they explain the charges of everything else exactly.

Leptons also come in three generations: the electron, the muon and the tau, each paired with its own neutrino. The charged leptons have charge $-1$ in units of $e$; the neutrinos have zero. A muon is an electron that happens to be 207 times heavier, identical in every other respect.

Six quarks plus six leptons is twelve fundamental fermions. Every one has an antiparticle with the same mass and opposite charge, written with a bar ($\bar{u}$) or a flipped sign ($e^+$).

Four forces act between them, each carried by an exchanged boson. The strong force is carried by gluons, acts only on quarks, and reaches about $10^{-15}$ m: one nucleus wide. Electromagnetism is carried by the photon, acts on electric charge, and has infinite range. The weak force is carried by the massive W and Z bosons, acts on every fermion including neutrinos, and reaches only about $10^{-18}$ m. Gravity is about $10^{-38}$ times the strong force between two protons and you can ignore it entirely here.

The pattern to hold onto: a massless carrier gives infinite range, a heavy carrier gives short range and a slow reaction. Beta decay is weak, and that is exactly why a free neutron takes minutes rather than nanoseconds.

Building hadrons out of quarks

Quarks are never found alone. They only appear in bound groups, and there are two kinds you need.

A baryon is three quarks. A meson is one quark and one antiquark. That is the entire rule, and it lets you get the charge of any hadron by adding fractions.

A proton is $uud$: $\tfrac{2}{3} + \tfrac{2}{3} - \tfrac{1}{3} = +1$. A neutron is $udd$: $\tfrac{2}{3} - \tfrac{1}{3} - \tfrac{1}{3} = 0$. A $\pi^+$ is $u\bar{d}$, and since the antiparticle of the down quark carries $+\tfrac{1}{3}$, the total is $\tfrac{2}{3} + \tfrac{1}{3} = +1$. The $\Omega^-$ is $sss$, three strange quarks at $-\tfrac{1}{3}$ each, giving $-1$.

The numbers that must balance

Charge, energy and momentum you already conserve out of habit. Three more quantum numbers matter here, and they are what most of the "is this reaction allowed" problems turn on.

Baryon number $B$. Every quark counts $+\tfrac{1}{3}$ and every antiquark $-\tfrac{1}{3}$. So any baryon has $B = +1$, any antibaryon $B = -1$, and every meson $B = 0$. No process has ever been seen to change $B$. That single fact is why the proton appears to be stable: it is the lightest particle with $B = 1$, so there is nothing lighter for it to turn into without violating the rule. The proposed decay $p \to e^+ + \gamma$ balances charge perfectly and is still never seen, because it would change $B$ from $1$ to $0$, a change of magnitude $1$.

Lepton number $L$. Each lepton counts $+1$, each antilepton $-1$, and it is conserved separately for each generation, so there are really three numbers $L_e$, $L_\mu$, $L_\tau$. This is what forces the antineutrino into beta decay. Write $n \to p + e^-$ and the left side has $L_e = 0$ while the right has $L_e = +1$. Add an electron antineutrino, $L_e = -1$, and it balances: $n \to p + e^- + \bar{\nu}_e$. The same rule kills $\mu^- \to e^- + \gamma$, which is fine on charge and energy but takes $L_\mu$ from $1$ to $0$ and $L_e$ from $0$ to $1$.

Strangeness $S$. Each strange quark counts $-1$ and each anti-strange $+1$ (the sign is a historical accident, but it is the convention). So $\Lambda^0 = uds$ has $S = -1$, $K^- = s\bar{u}$ has $S = -1$, and $K^+ = u\bar{s}$ has $S = +1$. Here is the twist: strangeness is conserved by the strong and electromagnetic forces but not by the weak force, which can change it by one unit. That asymmetry explains the whole behaviour of strange particles. They are made in pairs by the strong force, fast, with $\Delta S = 0$ overall; then each one has to decay weakly, slowly, because only the weak force can get rid of the strangeness. In $\Lambda^0 \to p + \pi^-$, strangeness goes from $-1$ to $0$, so $\Delta S = +1$ and the decay must be weak, which is why it takes $2.6 \times 10^{-10}$ s instead of $10^{-23}$ s.

Worked example 1: counting broken laws

Is $p + \pi^- \to K^+ + \pi^-$ allowed? Take the three quantities one at a time.

Charge: the left side is $(+1) + (-1) = 0$. The right side is $(+1) + (-1) = 0$. Charge is conserved.

Baryon number: the proton is a baryon, $B = +1$; the pion is a meson, $B = 0$. So the left side has $B = 1$. The right side is a kaon and a pion, both mesons, so $B = 0$. Not conserved.

Strangeness: the left side has no strange quarks, $S = 0$. The $K^+$ is $u\bar{s}$, so $S = +1$, and the pion contributes nothing. The right side has $S = +1$. Not conserved.

Two of the three fail. The baryon number failure alone is fatal: no amount of energy fixes it.

Relativistic bookkeeping: the invariant

Now the energy half. Every problem below uses one object, so it is worth building carefully.

For a single particle, energy and momentum are tied together by $E^2 = (pc)^2 + (mc^2)^2$ which you can rearrange to define the mass from the other two. For a system of several particles, add the energies and add the momenta as vectors, then form the same combination:

This quantity $s$ has a property that makes it the most useful number in the subject: it is the same in every reference frame. Different observers disagree about the individual energies and momenta, but they all compute the same $s$. And because momentum is zero in the centre-of-mass frame, $\sqrt{s}$ is exactly the total energy available there: the energy you can actually spend on making new particles. Everything else is locked up in the motion of the system as a whole.

Two consequences. First, $s$ is conserved through a reaction, because energy and momentum are. So you can compute it from the incoming particles and then demand something of the outgoing ones. Second, at threshold (the minimum energy for which a reaction is possible) the products come out at rest relative to one another, so $\sqrt{s}$ is just the sum of their rest energies.

Worked example 2: the momentum in a lambda decay

A $\Lambda^0$ at rest, rest energy $M c^2 = 1115.7$ MeV, decays to a proton ($m_1c^2 = 938.3$ MeV) and a $\pi^-$ ($m_2c^2 = 139.6$ MeV). How fast do they come out?

Momentum started at zero, so the two products fly back to back with equal momentum magnitudes $p$. Energy gives

One unknown, one equation. Isolating a square root and squaring twice gives the standard result

where every mass is in the same units. Put the numbers in, using MeV throughout. The sum is $938.3 + 139.6 = 1077.9$ and the difference is $938.3 - 139.6 = 798.7$. Then

$M^2 = 1115.7^2 = 1244786$

$(m_1+m_2)^2 = 1077.9^2 = 1161869$

$(m_1-m_2)^2 = 798.7^2 = 637922$

So the first bracket is $1244786 - 1161869 = 82917$ and the second is $1244786 - 637922 = 606864$. Their product is $5.032 \times 10^{10}$, whose square root is $2.243 \times 10^{5}$. Divide by $2M = 2231.4$:

$pc = \frac{2.243\times 10^{5}}{2231.4} = 100.5 \text{ MeV}$

So each product carries about $101$ MeV/c. Sanity check: the mass released is $1115.7 - 1077.9 = 37.8$ MeV, and a momentum of $100$ MeV/c is comfortably larger than that, which is right: the pion is light, so most of the released energy goes into its kinetic energy rather than the heavy proton's.

The same formula handles a photon by setting one mass to zero. For $\Sigma^0 \to \Lambda^0 + \gamma$, with $M = 1192.6$ and $m_1 = 1115.7$, both brackets collapse to $M^2 - m_1^2$, so

$E_\gamma = \frac{M^2 - m_1^2}{2M}c^2 = \frac{1422295 - 1244786}{2385.2} = \frac{177509}{2385.2} = 74.4 \text{ MeV}$

And if both products are massless and identical, as in $\pi^0 \to \gamma\gamma$, symmetry alone gives each photon half the rest energy: $135/2 = 67.5$ MeV. Electron–positron annihilation at rest is the same picture: each photon takes $0.511$ MeV.

Thresholds, and why colliders exist

Run the same invariant backwards and you get every threshold problem in the unit.

Take a beam particle of total energy $E$ hitting a target of rest energy $m_Tc^2$ sitting still. The target contributes energy $m_Tc^2$ and no momentum; the beam contributes $E$ and momentum $pc = \sqrt{E^2 - (m_bc^2)^2}$. Then

The beam energy enters only linearly, and it is multiplied by the target's rest energy, which is small. At high energy $\sqrt{s} \approx \sqrt{2m_Tc^2E}$: quadruple the beam energy and you only double what you can spend.

Now collide two beams of energy $E$ head-on instead. The momenta cancel, so $s = (2E)^2$ and $\sqrt{s} = 2E$. Every joule you paid for is available. That is the whole argument for building colliders, and the gap is enormous: a $7.00$ TeV beam on stationary hydrogen gives

$\sqrt{s} \approx \sqrt{2(0.9383)(7000)} = \sqrt{13136} = 114.6 \approx 115 \text{ GeV}$

while two $6.50$ TeV beams meeting head-on give $\sqrt{s} = 13.0$ TeV. More than a hundred times more, for less beam energy.

Worked example 3: making antiprotons

Antiprotons are produced by firing protons at stationary hydrogen: $p + p \to p + p + p + \bar{p}$. Baryon number forces this: you cannot make an antiproton ($B = -1$) alone, so a proton must come with it, and the three final protons plus one antiproton give $B = 3 - 1 = 2$, matching the two you started with. What beam energy do you need?

At threshold all four products sit at rest together, so $\sqrt{s} = 4m_pc^2$ and $s = 16(m_pc^2)^2$. From the fixed-target formula with $m_b = m_T = m_p$:

$2(m_pc^2)^2 + 2m_pc^2E = 16(m_pc^2)^2$

Divide through by $2m_pc^2$:

$m_pc^2 + E = 8m_pc^2 \quad\Rightarrow\quad E = 7m_pc^2$

That $E$ is total energy, and the question asks for kinetic energy, so subtract the rest energy: $K = E - m_pc^2 = 6m_pc^2 = 6(0.9383) = 5.63$ GeV. You spend six proton masses to create one antiproton's worth of new mass, and the factor of six is pure wasted centre-of-mass motion.

The same two lines do photoproduction. For $\gamma + p \to p + \pi^0$ the photon has no rest mass, so $s = (m_pc^2)^2 + 2m_pc^2E_\gamma$, and at threshold $s = (m_pc^2 + m_\pi c^2)^2$. Therefore

$E_\gamma = \frac{(m_p + m_\pi)^2 - m_p^2}{2m_p}c^2 = \frac{(1073.3)^2 - (938.3)^2}{2(938.3)} = \frac{1151973 - 880407}{1876.6} = 145.0 \text{ MeV}$

If instead the two particles hit head-on and both are ultra-relativistic, the momenta subtract rather than add and $s \approx 4E_1E_2$. That is the GZK calculation: a cosmic-ray proton meeting a microwave background photon of energy $E_\gamma$ makes a $\Delta^+$ when $4E_pE_\gamma \approx (m_\Delta c^2)^2 - (m_pc^2)^2$, which is the formula quoted in that problem.

Lifetimes, widths and flight distances

Two more numbers show up constantly.

A particle that lives for a time $\tau$ cannot have a perfectly sharp mass. The uncertainty principle in its energy–time form, $\Delta E\,\Delta t \sim \hbar$, says a state that exists only briefly is spread out in energy. Experimenters see that spread as the width $\Gamma$ of a resonance peak, and the relation is

$\Gamma \tau = \hbar$

So a $\rho$ meson whose peak is $150$ MeV wide lives for $\tau = \hbar/\Gamma = (6.582\times 10^{-22})/150 = 4.39\times 10^{-24}$ s. That is a strong-force lifetime, as the enormous width already told you. Run it the other way and a narrow peak means a long-lived particle.

Second, a moving unstable particle travels further than $c\tau$ before decaying, because of time dilation. Its mean flight distance in the lab is

$L = \beta\gamma c\tau = \frac{pc}{mc^2}\,c\tau$

The combination $\beta\gamma = p/(mc)$ is the useful one, because momentum is what a detector measures.

Worked example 4: how far does a kaon get?

A $K^0_S$ has rest energy $497.6$ MeV and proper lifetime $0.895\times 10^{-10}$ s, and is made with momentum $5.00$ GeV/c. First the boost factor:

$\beta\gamma = \frac{pc}{mc^2} = \frac{5000}{497.6} = 10.05$

Then $c\tau = (3.00\times 10^8)(0.895\times 10^{-10}) = 2.685\times 10^{-2}$ m, about $2.7$ cm. Multiply:

$L = 10.05 \times 2.685\times 10^{-2} = 0.270 \text{ m}$

Twenty-seven centimetres: far enough to leave a visible gap between the production point and the decay point in a detector, which is exactly how these particles are identified.

Recognising which tool a problem wants

Four signatures, and nearly every problem in this unit shows one of them.

A reaction written out with an arrow, and the question "is this allowed" means the conservation checklist. Charge, baryon number, lepton number by generation, strangeness, and then the mass budget as a separate last step. Get the quark content first if you are unsure of $B$ or $S$: three quarks means $B = 1$, each $s$ quark means $S = -1$.

A single particle at rest turning into two means the two-body decay formula. You have no choices to make: the momenta are equal and opposite, and the answer is fixed by the three masses. If one product is a photon, set its mass to zero and the formula shortens to $(M^2 - m^2)/2M$.

The words "threshold", "minimum energy", "stationary target" or "centre of mass" mean compute $s$ before and set $\sqrt{s}$ equal to the sum of final rest energies. Watch which energy the question wants at the end (total or kinetic) because that off-by-one-rest-mass is the most common lost mark in the whole unit.

A lifetime, a width, or a distance travelled means $\Gamma\tau = \hbar$ or $L = (pc/mc^2)c\tau$. If you are given a momentum and asked for a distance, you are being asked about time dilation, whether or not the problem says so.

The one habit that pays everywhere: keep every energy in MeV and every momentum in MeV/c, so that $c$ never appears in your arithmetic and factors of $10^8$ never get squared by accident. Convert at the very end if the answer is wanted in joules or metres.

Next comes data analysis and uncertainty, which is how you decide whether a bump in a histogram is a new particle or a statistical accident: the question that stands behind every number quoted above.