Threshold Energy for Particle Production

In 1949 the Radiation Laboratory at Berkeley had to commit to a number. They were going to build a proton synchrotron to hunt for the antiproton, and the machine — ten thousand tons of steel, a magnet the size of a house — would be built once, to one energy. Too low and it would find nothing, and everyone would know why. So: how much energy does it take to make an antiproton? The naive answer is easy, and it is wrong. You need to create a proton–antiproton pair, so you need their rest energy, $2m_pc^2 = 1.88\ \mathrm{GeV}$ . Build a machine that can push a proton to $1.88\ \mathrm{GeV}$ of kinetic energy, aim it at a copper target, and collect your Nobel Prize. That machine would have found nothing. The true threshold is $5.63\ \mathrm{GeV}$ — exactly three times larger. The factor of three is not a fudge or a correction; it is momentum conservation, and it is unavoidable. A moving proton hits a stationary one, and the products cannot be created at rest, because the pair has momentum and momentum must go somewhere. The kinetic energy locked into that unavoidable forward motion is energy you paid for and cannot spend. At this particular threshold, exactly half of the beam's energy is unavailable. Berkeley got the number right. The Bevatron was designed for $6.2\ \mathrm{GeV}$ — comfortably above $5.63$ — and in 1955 Chamberlain, Segrè, Wiegand and Ypsilantis found the antiproton with it. The machine's name is the number: Bevatron, billions of electron volts. The tool that gets this right, every time, in one line and with no algebra, is the invariant $s$ . And once you have it, it tells you something else — something that reshaped every accelerator built after 1971. In a fixed-target machine, the useful energy grows only as the square root of what you pay for. Push a beam ten times harder and you get three times the reach. That is a losing race, and the only way to stop losing it is to stop using a fixed target.

Key takeaways

Signature $(+,-,-,-)$ : $p\cdot p = +m^2c^2$ . $s \equiv \big(\sum E\big)^2 - \big|\sum\vec p\,\big|^2c^2$ is in energy² units, so $\sqrt{s}$ is an energy — the total energy in the CM frame.
The threshold condition, and the only one you need:

\boxed{\;\sqrt{s} \;\geq\; \sum_{f} m_f c^2\;}

The reaction runs iff the invariant CM energy covers the rest energies of everything you want to make. Never compare a lab energy sum to $\sum m_f c^2$ — that is the classic error, and it is wrong by a factor of 3 for the antiproton.

Why the naive lab sum fails: the products must carry the initial momentum, so they cannot all be at rest. That forced kinetic energy is unavailable. Only the fraction $1/\gamma_{\mathrm{cm}}$ of the total energy is usable.
Fixed target: $s = m_1^2c^4 + m_2^2c^4 + 2m_2c^2E_1$ , so at high energy $\sqrt{s} \approx \sqrt{2m_2c^2E_1} \;\propto\; \sqrt{E_1}$ .
Collider (equal beams, head-on): $\sqrt{s} = 2E_{\mathrm{beam}} \;\propto\; E_{\mathrm{beam}}$ .
The punchline: $\sqrt{E}$ versus $E$ . At the LHC's $7\ \mathrm{TeV}$ per beam, colliding gives $\sqrt{s} = 14\ \mathrm{TeV}$ ; the same beam on a fixed target gives $114.6\ \mathrm{GeV}$ — 122 times less. To reach $14\ \mathrm{TeV}$ with a fixed target you would need a beam of $\mathbf{1.04\times10^{8}\ GeV}$ , about $10^5\ \mathrm{km}$ of LHC.
Antiproton: $p+p \to p+p+p+\bar p$ . Baryon number forces four baryons out, so $\sqrt{s} \geq 4m_pc^2 = 3.753\ \mathrm{GeV}$ , giving $T_{\mathrm{thr}} = 6m_pc^2 = 5.630\ \mathrm{GeV}$ . The Bevatron was built to $6.2$ .
Key idea: a threshold is a statement about the INVARIANT $s$ , never about a lab energy. Compute $s$ , take its square root, compare it to the rest energies you want to create. Everything else — the factor of three, the $\sqrt{E}$ scaling, the existence of colliders — falls out of that one line.

The core insight

At threshold, the products are created at rest in the centre-of-momentum frame — they have exactly enough energy to exist and not a joule more, so they cannot be moving relative to one another. The total CM energy is then just $\sum_f m_f c^2$ . But $\sqrt{s}$ is an invariant: it has the same value in the lab, where nothing is at rest and the algebra looks hard. So evaluate $\sqrt{s}$ in the frame where it is easy (the lab, from the beam energy), impose the condition in the frame where it is meaningful (the CM, where it reads $\sqrt{s} \geq \sum_f m_f c^2$ ), and equate. That single move — one quantity, two frames — is the whole subject. Everything below is bookkeeping.

Where this lives today

Every collider that exists is an argument about the square root, and the formula still sets budgets: a future $100\ \mathrm{TeV}$ machine is specified by the masses it must be able to create, run backwards through $\sqrt{s} \geq \sum_f m_fc^2$ , with the circumference and the magnet field following from that. The same reasoning, in reverse, is why the highest-energy collisions ever observed are not made by us at all — and why they are less impressive than they sound. A cosmic ray striking the upper atmosphere is a fixed-target event, so it pays the same square-root tax we do. At each step of the last fifty years the collider did not merely save money; it made possible a measurement that was otherwise off the table entirely, and the reason is the single factor of $\tfrac12$ in the exponent of equation (3).

The physical picture

Fire a proton at a proton sitting still. You want to make an antiproton. Antiprotons cannot be made alone — baryon number is conserved, so creating a $\bar p$ (baryon number $-1$ ) requires creating an extra $p$ (baryon number $+1$ ) alongside it. The cheapest reaction is therefore

p + p \;\longrightarrow\; p + p + p + \bar p

with four baryons in the final state, not two. So you must conjure $2m_pc^2 = 1.877\ \mathrm{GeV}$ of new rest energy. Now the naive calculation. "Give the beam proton $1.877\ \mathrm{GeV}$ of kinetic energy and it can pay for the new pair." This is wrong, and it is wrong for a reason that has nothing to do with relativity. The incoming proton carries momentum. Momentum is conserved. So the four outgoing particles must, between them, carry that same momentum — they cannot all be sitting still afterwards. And particles that are moving have kinetic energy. That kinetic energy came out of the beam, and it is not available to build mass. This is the whole difficulty, and it is already present in Newtonian mechanics: you cannot bring a moving object to rest by hitting a stationary one, because the pair's momentum is not zero. What relativity adds is that the wasted fraction is large, and computable, and that it grows with energy. How large is the waste? Derivation 2 makes it exact: at the antiproton threshold half the energy is stuck in centre-of-mass motion, and the factor of three between the naive answer and the truth is precisely that inefficiency. And it gets worse as you push harder — which is the second, larger point of this article. The available energy in a fixed-target collision grows only as the square root of what you pay, so doubling your reach means quadrupling your beam. There is one way out: collide two beams head-on, so that the lab frame is the CM frame, nothing is wasted, and $\sqrt{s}$ grows linearly. That is why every energy-frontier machine since 1971 has been a collider.

The core equations

The invariant, built from the total four-momentum of the system, in energy² units:

s \;\equiv\; \Big(\sum_i E_i\Big)^{\!2} - \Big|\sum_i \vec p_i\Big|^{2}c^2 \qquad\text{so}\qquad \sqrt{s} \;=\; E_{\mathrm{cm}} \qquad (1)

where: $s$ has units of energy², and $\sqrt{s}$ is an energy — it is the total energy of the system as measured in the CM frame, where the momenta cancel. Physical reading: $s$ is the same number in every frame, which is precisely what makes it useful. It is the LHC's " $14\ \mathrm{TeV}$ " and the Bevatron's " $3.75\ \mathrm{GeV}$ ". The threshold condition. It is one line, and it is the only thing in this article you must remember:

\boxed{\;\sqrt{s} \;\geq\; \sum_{f} m_f c^2\;} \qquad (2)

where: the sum runs over the rest masses of every particle in the final state. Physical reading: at threshold the products are born at rest in the CM frame, so the CM energy is exactly their total rest energy; above threshold there is some left over as kinetic energy. Note what is not here: no lab energies, no beam momentum, no angles. Conservation laws (charge, baryon number, lepton number) tell you what the final state must contain; equation (2) then tells you what it costs. For a beam of energy $E_1$ striking a stationary target of mass $m_2$ , expanding $s$ gives the fixed-target workhorse:

\boxed{\;s = m_1^2c^4 + m_2^2c^4 + 2m_2c^2E_1\;} \qquad\Longrightarrow\qquad E_1^{\mathrm{thr}} = \frac{\big(\sum_f m_f\big)^2c^4 - m_1^2c^4 - m_2^2c^4}{2m_2c^2} \qquad (3)

where: $E_1$ is the beam's total energy (rest plus kinetic). Physical reading: $s$ is linear in $E_1$ , so $\sqrt{s}$ goes like $\sqrt{E_1}$ . That square root is the tyranny of the fixed target, and everything about the shape of modern physics follows from it. Note also the $2m_2$ : a heavier target buys more $\sqrt{s}$ for the same beam, which is why fixed-target experiments use heavy nuclei when they can. For two equal beams of energy $E_{\mathrm{beam}}$ meeting head-on, the momenta cancel and $s$ is trivial:

\boxed{\;\sqrt{s} = 2E_{\mathrm{beam}}\;} \qquad (4)

where: nothing is wasted, because the lab frame is the CM frame. Physical reading: linear, not square-root. Compare (3) and (4) and you have the entire argument for colliders — and the reason the LHC exists.

Deriving the threshold

Derivation AThe threshold condition: why it is a statement about the invariant s, and never about a lab energy

We want the minimum beam energy for a reaction $1 + 2 \to \sum_f f$ . The temptation is to write "beam energy $\geq$ total rest energy created" and be done. We will derive the correct condition instead, and then see exactly how much the naive version misses by and why.

Step 1. Ask what "threshold" means physically. At the minimum energy, the products are made with nothing to spare: they have enough energy to exist and none left over for relative motion. So in the CM frame, all products are at rest — they move as one lump. Why: any relative motion of the products costs kinetic energy, and at threshold, by definition, there is none to spend. This is the only place a frame is chosen, and the CM frame is chosen because it is the only one in which "at rest" can be said of everything at once.

Step 2. Evaluate $s$ in the CM frame at threshold. With every product at rest there, $\sum \vec p_f = \vec 0$ and $\sum E_f = \sum m_f c^2$ , so directly from the definition,

s_{\mathrm{thr}} = \Big(\sum_f m_f c^2\Big)^2 - 0 = \Big(\sum_f m_f c^2\Big)^{2} \qquad\Longrightarrow\qquad \boxed{\sqrt{s} \geq \sum_f m_f c^2}

with the inequality because any extra energy shows up as products in motion, which only increases $s$ . Why: this is equation (2), and it is exact and completely general — any number of products, any masses, any reaction.

Step 3. Now use the invariance. $s$ is a Lorentz scalar, so the number computed in the CM frame is the same number computed in the lab. In the lab, with a beam of total energy $E_1$ hitting a target at rest,

s = (E_1 + m_2c^2)^2 - |\vec p_1|^2c^2 = E_1^2 + 2m_2c^2E_1 + m_2^2c^4 - (E_1^2 - m_1^2c^4) = m_1^2c^4 + m_2^2c^4 + 2m_2c^2E_1

using the mass shell $|\vec p_1|^2c^2 = E_1^2 - m_1^2c^4$ . Why: this is the whole trick. The condition is meaningful in the CM frame; the data are available in the lab frame; $s$ is the same in both. We never had to transform a single four-vector.

Step 4. Equate and solve. Setting the lab expression $\geq$ the CM threshold value:

\boxed{\,E_1^{\mathrm{thr}} = \frac{\big(\sum_f m_f\big)^2c^4 - m_1^2c^4 - m_2^2c^4}{2m_2c^2}\,}

Why: this is the number you build a machine to. Everything on the right is a mass.

Step 5. Now diagnose the naive answer. It says $T_1 \geq \big(\sum_f m_f - m_1 - m_2\big)c^2$ — pay for the new mass, no more. This is a lab-frame energy sum, and it is wrong, because it tacitly assumes the products can be created at rest in the lab. They cannot: the initial momentum $\vec p_1 \neq \vec 0$ must be carried away by something. Why: the naive condition is not merely imprecise, it violates conservation of momentum. It is not an approximation to the right answer; it is an answer to a different, impossible problem.

Step 6. Quantify what is lost. The products, at threshold, move together with the CM velocity, and their total energy in the lab is $\gamma_{\mathrm{cm}}\sqrt{s}$ while only $\sqrt{s}$ of it is available to make mass. So the available fraction is $1/\gamma_{\mathrm{cm}} = \sqrt{s}/E_{\mathrm{tot}}$ , and the rest is locked in the unavoidable forward motion of the debris. Why: this is the factor of three, in general form. It is not a relativistic correction — a Newtonian collision wastes energy the same way — but relativity makes it enormous, because $\gamma_{\mathrm{cm}}$ grows without bound. *Verified in verify_aq-sr-threshold-energy.py: $s = m_1^2c^4 + m_2^2c^4 + 2m_2c^2E_1$ is derived symbolically from the definition of $s$ plus the mass shell; $E_1^{\mathrm{thr}}$ is inverted from it symbolically; both are then checked to be dimensionally an energy² and an energy respectively; and the resulting antiproton numbers printed in this article are reproduced from PDG masses.*

Go deeper · Derivation B · 2 minThe antiproton: the factor of three, and why the Bevatron was built to 6.2 GeV

Derivation BThe antiproton: the factor of three, and why the Bevatron was built to 6.2 GeV

Now the canonical case, worked twice — once naively, once correctly — so that the gap between them is unmistakable and its origin is exactly locatable. The reaction is $p + p \to p + p + p + \bar p$ , forced to four baryons in the final state by conservation of baryon number.

Step 1. Get the final state right, because this is where the physics is. Baryon number is conserved; $\bar p$ carries $B = -1$ ; the initial state has $B = +2$ . So the final state must have $B = +2$ , and with a $\bar p$ present that means three protons plus the antiproton. Why: you cannot make an antiproton alone. The lightest allowed final state contains four baryons, and this is a conservation law doing physics before any kinematics is written down.

Step 2. Apply equation (2). All four products have mass $m_p$ , so

\sqrt{s} \;\geq\; 4m_pc^2 = 4 \times 938.272\ \mathrm{MeV} = \mathbf{3.753\ GeV}

Why: one line. This is the entire threshold, and it is frame-independent — it is as true at the LHC as at the Bevatron.

Step 3. Convert to a beam energy with equation (3), $m_1 = m_2 = m_p$ :

(4m_pc^2)^2 = 2m_p^2c^4 + 2m_pc^2E_1 \qquad\Longrightarrow\qquad E_1^{\mathrm{thr}} = \frac{16m_p^2c^4 - 2m_p^2c^4}{2m_pc^2} = 7m_pc^2

so $E_1^{\mathrm{thr}} = 7 \times 938.272 = \mathbf{6.568\ GeV}$ of total energy, and the kinetic energy the machine must actually supply is

\boxed{\,T_1^{\mathrm{thr}} = E_1 - m_pc^2 = 6m_pc^2 = \mathbf{5.630\ GeV}\,}

Why: the sevens and sixes are exact — the masses cancel completely. $T_{\mathrm{thr}} = 6m_pc^2$ is one of the prettiest results in the subject: to make one antiproton you must supply six proton rest energies.

Step 4. Compare with the naive answer. Naively you need only the new pair's rest energy, $2m_pc^2 = \mathbf{1.877\ GeV}$ . The truth is $6m_pc^2$ . The naive lab-energy sum is wrong by a factor of exactly 3. Why: and it errs on the fatal side — it tells you to build a machine three times too small. A laboratory that trusted it would have built a $2\ \mathrm{GeV}$ synchrotron and found nothing, with no way of knowing whether the antiproton did not exist or the machine was simply too weak.

Step 5. Find where the missing energy went. At threshold, $E_{\mathrm{tot}} = E_1 + m_pc^2 = 8m_pc^2$ while $\sqrt{s} = 4m_pc^2$ , so

\gamma_{\mathrm{cm}} = \frac{E_{\mathrm{tot}}}{\sqrt{s}} = \frac{8m_pc^2}{4m_pc^2} = \mathbf{2}\quad\text{exactly} \qquad\Longrightarrow\qquad \text{available fraction} = \frac{1}{\gamma_{\mathrm{cm}}} = \mathbf{50\%}

The four products emerge as a single lump moving at $\beta_{\mathrm{cm}} = \sqrt{3}/2 = 0.866c$ , carrying $3.753\ \mathrm{GeV}$ of kinetic energy that the beam paid for and no experiment can use. Why: exactly half wasted, exactly at threshold. $5.630 = 1.877 + 3.753$ : of the beam's kinetic energy, $1.877\ \mathrm{GeV}$ becomes the new pair and $3.753\ \mathrm{GeV}$ becomes forward motion. That ledger is the factor of three, itemised.

Step 6. Build the machine. $5.630\ \mathrm{GeV}$ is a floor, and a machine sitting exactly on a threshold produces nothing at a useful rate (the cross-section vanishes there, and nuclear targets smear the effective energy). Berkeley specified $6.2\ \mathrm{GeV}$ , clearing the threshold by $570\ \mathrm{MeV}$ . Why: they got it right, and the antiproton appeared in 1955. The name Bevatron is the number: billions of electron volts. *Verified in verify_aq-sr-threshold-energy.py: $\sqrt{s} = 4m_pc^2$ , $E_1 = 7m_pc^2$ , $T_1 = 6m_pc^2$ and $\gamma_{\mathrm{cm}} = 2$ are proved symbolically — the proton mass cancels identically, so the integers 4, 7, 6 and 2 are exact — and the energy ledger $6m_pc^2 = 2m_pc^2 + 4m_pc^2$ is checked to close.*

Go deeper · Derivation C · 2 minFixed target versus collider: root-E against E, and why every modern machine collides

Derivation CFixed target versus collider: root-E against E, and why every modern machine collides

The threshold formula contains a second, deeper result, and it is the one that determined the architecture of every accelerator built in the last fifty years. It is visible in a single feature of equation (3): $s$ is linear in the beam energy, so $\sqrt{s}$ is not.

Step 1. Take the ultra-relativistic limit of the fixed-target formula. For $E_1 \gg m_1c^2,\, m_2c^2$ , the mass terms in $s = m_1^2c^4 + m_2^2c^4 + 2m_2c^2E_1$ are negligible and

\boxed{\;\sqrt{s} \;\approx\; \sqrt{2m_2c^2 E_1} \;\propto\; \sqrt{E_1}\;}

Why: the useful energy grows as the square root of the energy you pay for. To double your reach you must quadruple your beam; to gain a factor of ten you must build a machine one hundred times more powerful. This is a losing race and it gets worse the harder you run.

Step 2. Now collide two equal beams head-on. The total momentum is zero by construction, so the lab frame is the CM frame, and $s = (2E_{\mathrm{beam}})^2 - 0$ :

\boxed{\;\sqrt{s} = 2E_{\mathrm{beam}} \;\propto\; E_{\mathrm{beam}}\;}

Why: linear. Not one joule is spent on centre-of-mass motion, because there is none. Every scrap of the energy you supply is available to create mass. This is the entire case for colliders, and it is a kinematic case — no dynamics, no cross-sections, no technology.

Step 3. Put numbers on the gap, at the LHC's beam energy of $7\ \mathrm{TeV}$ on protons. Colliding: $\sqrt{s} = 14\ \mathrm{TeV}$ . The same beam on a stationary hydrogen target: $\sqrt{s} = \sqrt{2 \times 0.938 \times 7000}\ \mathrm{GeV} = \mathbf{114.6\ GeV}$ — less than the mass of a Higgs boson. A factor of 122. Why: the world's most powerful accelerator, pointed at a target instead of at itself, could not make the particle it was built to find.

Step 4. Ask the question the other way round: what fixed-target beam would be needed to reach $\sqrt{s} = 14\ \mathrm{TeV}$ ? Invert equation (3):

E_1 = \frac{s - 2m_p^2c^4}{2m_pc^2} = \frac{(14{,}000\ \mathrm{GeV})^2}{2 \times 0.938\ \mathrm{GeV}} = \mathbf{1.04\times10^{8}\ GeV} = 104\ \mathrm{PeV}

Why: fifteen thousand times the LHC's beam energy. The LHC's ring is $27\ \mathrm{km}$ ; at the same bending technology this machine's would be about $10^5\ \mathrm{km}$ — roughly a quarter of the way to the Moon. The collider is not a convenience. It is the only option that exists.

Step 5. Read the history off the same formula. The ISR reached $\sqrt{s} = 62\ \mathrm{GeV}$ in 1971 — a fixed-target machine would have needed a $2.05\ \mathrm{TeV}$ beam, which did not exist. The $\mathrm{Sp\bar pS}$ reached $540\ \mathrm{GeV}$ and found the $W$ and $Z$ in 1983 — the fixed-target equivalent is $155\ \mathrm{TeV}$ . Why: at every step, the collider bought a reach that no plausible fixed-target machine could reach. The square root is why the energy frontier moved into storage rings and never came back.

Step 6. State the residual honestly. Fixed targets did not die: they give enormous rates (a solid target is $\sim 10^{24}$ times denser than a counter-rotating bunch), and for rare-decay and neutrino experiments — where you need statistics, not $\sqrt{s}$ — they remain the right tool. Why: the square root is fatal only at the energy frontier. Elsewhere, the trade is luminosity against reach, and it often goes the other way. *Verified in verify_aq-sr-threshold-energy.py: the $\sqrt{s} \propto \sqrt{E}$ scaling is confirmed numerically (quadrupling the beam energy multiplies $\sqrt{s}$ by $1.993$ , approaching 2); the whole fixed-target/collider table printed below is recomputed from PDG masses; and the $1.04\times10^8$ GeV figure is inverted from the exact formula, not the approximation.*

Figure

THE PUNCHLINE OF THE CHAPTER — collider reach grows LINEARLY with beam energy, fixed-target reach only as the SQUARE ROOT. Both axes are logarithmic, so the collider line has slope 1 and the fixed-target line slope 1/2, and they diverge without limit. Every real machine is marked. At the LHC's 7 TeV per beam, colliding gives 14 TeV of CM energy while the SAME beam on a stationary target gives only 114.6 GeV — a factor of 122. To reach 14 TeV with a fixed target you would need a 1.04e8 GeV beam. This gap is why every energy-frontier machine since 1971 has been a collider

Limiting cases

Case	Result	Meaning
products at rest in CM	$\sqrt{s} = \sum_f m_fc^2$	the threshold — the definition, not an approximation
ultra-relativistic fixed target	$\sqrt{s} \approx \sqrt{2m_2c^2E_1} \propto \sqrt{E_1}$	quadruple the beam to double the reach
collider, equal beams	$\sqrt{s} = 2E_{\mathrm{beam}} \propto E_{\mathrm{beam}}$	nothing wasted: the lab frame is the CM frame
massless projectile ( $m_1 = 0$ )	$s = m_2^2c^4 + 2m_2c^2E_\gamma$	photoproduction: $\gamma p \to p\pi^0$ at $E_\gamma = 144.7$ MeV
heavy target ( $m_2 \to \infty$ )	$\sqrt{s} \to \infty$ for fixed $E_1$	more mass to recoil against — heavier targets buy reach
non-relativistic limit	$T_{\mathrm{thr}} \to \big(1 + \tfrac{m_1}{m_2}\big)\,Q$	the familiar Newtonian threshold; for $m_1 = m_2$ , twice $Q$
$\gamma_{\mathrm{cm}} \to 1$	available fraction $\to 1$	the fixed-target limit that never happens

Common trapNever compare a lab-frame energy to the rest energies you want to create

This is the single most common error in the subject, and it is the one that would have sunk the Bevatron. The threshold condition is $\sqrt{s} \geq \sum_f m_fc^2$ — an invariant on the left. It is not $T_{\mathrm{beam}} \geq \sum_f m_fc^2 - \sum_i m_ic^2$ , and the two are not close: for the antiproton the naive form gives $1.877\ \mathrm{GeV}$ against a true $5.630\ \mathrm{GeV}$ , wrong by a factor of 3, and wrong on the side that leaves you with no antiprotons. The reason is momentum conservation, not relativity: a lab-frame energy sum silently assumes the products can be created at rest in the lab, which is impossible whenever the initial momentum is non-zero. The fix is never a correction factor — it is to use the invariant. Write $s$ from the lab data, take $\sqrt{s}$ , compare it to $\sum_f m_fc^2$ . It is a shorter calculation than the wrong one.

Worked examples

1 · The LHC beam, pointed at a wall

The LHC accelerates protons to $7\ \mathrm{TeV}$ . Compare the CM energy obtained by colliding two such beams with that obtained by firing one at a stationary proton.

Step 1 — collider. The momenta cancel, so equation (4): $\sqrt{s} = 2E_{\mathrm{beam}} = \mathbf{14\ TeV}$ . Every joule is available.

Step 2 — fixed target. Equation (3) with $E_1 = 7000\ \mathrm{GeV} \gg m_pc^2$ : $\sqrt{s} \approx \sqrt{2 \times 0.938272 \times 7000}\ \mathrm{GeV} = \sqrt{13{,}136} = \mathbf{114.6\ GeV}$ .

Step 3 — the gap. $14{,}000/114.6 = \mathbf{122}$ . The same beam, the same protons, the same magnets — and $122$ times less reach. At $114.6\ \mathrm{GeV}$ the LHC could not make a Higgs boson, whose mass is $125\ \mathrm{GeV}$ .

Step 4 — the boundary number. To reach $14\ \mathrm{TeV}$ on a fixed target: $E_1 = s/(2m_pc^2) = (14{,}000)^2/(2 \times 0.938272) = \mathbf{1.04\times10^{8}\ GeV}$ — about 15,000 times the LHC's beam energy, needing a ring some $10^5\ \mathrm{km}$ around. This is why the LHC is a collider. Not for economy, not for elegance: because the alternative does not exist.

2 · Pion photoproduction, with a massless projectile

Find the minimum photon energy to produce a $\pi^0$ off a proton at rest: $\gamma + p \to p + \pi^0$ . Take $m_{\pi^0}c^2 = 134.977\ \mathrm{MeV}$ .

Step 1 — the threshold. $\sqrt{s} \geq (m_p + m_{\pi^0})c^2 = 1073.25\ \mathrm{MeV}$ .

Step 2 — the lab expression. Equation (3) with $m_1 = 0$ : $s = m_p^2c^4 + 2m_pc^2E_\gamma$ . Setting these equal and solving,

E_\gamma^{\mathrm{thr}} = \frac{(m_p + m_{\pi^0})^2c^4 - m_p^2c^4}{2m_pc^2} = m_{\pi^0}c^2\left(1 + \frac{m_{\pi^0}}{2m_p}\right) = \mathbf{144.7\ MeV}

Step 3 — the boundary number. The naive answer is $m_{\pi^0}c^2 = 134.98\ \mathrm{MeV}$ ; the truth is $144.7\ \mathrm{MeV}$ , 7.2 % higher, and the excess $\big(m_{\pi^0}/2m_p\big)$ is exactly the proton's recoil. Here the naive answer is nearly right — because the target is heavy compared with what you are making. That is the whole story of the factor of three: it is small when $\sum_f m_f \ll m_2$ , and ruinous when it is not.

Figure

WHY THE NAIVE LAB-ENERGY SUM FAILS — at threshold the four products are at rest in the CM frame (left), which is the definition of threshold. But the CM frame is itself moving at 0.866c in the lab, so in the LAB (right) those products stream forward as a single lump carrying 3.753 GeV of kinetic energy that the beam paid for and no experiment can spend. Of 5.630 GeV of beam kinetic energy, only 1.877 GeV becomes the new proton-antiproton pair. gamma_cm = 2 exactly, so exactly HALF the energy is unavailable — and that is the factor of 3 between the naive answer and the truth

Physics of the portrait

Threshold energy is the rare piece of theory that gets poured in concrete. Most calculations can be revised; this one is spent, once, on a magnet.

Origins · 2 minThreshold energy and how this was discovered

Threshold energy

1949–1955 · the number the Bevatron was built to · a factor of three · and the antiproton at last

By the late 1940s the antiproton was theoretically inevitable — Dirac's equation had demanded an antielectron in 1928 and Anderson had found it in 1932 — and yet nobody had made one. The reason was energy: the machines of the day could not reach the threshold, and the threshold was not where intuition put it. The naive estimate said a proton beam of about $2\ \mathrm{GeV}$ should suffice, since that is the rest energy of the pair you are trying to create. The invariant calculation said $5.63\ \mathrm{GeV}$ of kinetic energy, because baryon number forces four baryons into the final state and momentum conservation locks half the beam's energy into the forward motion of the debris. A factor of three, and a laboratory that got it wrong would have built a machine that found nothing and learned nothing. Ernest Lawrence's Radiation Laboratory got it right, and built the Bevatron to $6.2\ \mathrm{GeV}$ — deliberately above the true threshold, with margin for a nuclear target and a workable production rate. It came into operation in 1954. Within a year Owen Chamberlain, Emilio Segrè, Clyde Wiegand and Thomas Ypsilantis had separated antiprotons from a flood of $\sim 40{,}000$ background pions per antiproton, by simultaneously measuring momentum and velocity so that the mass followed. They published in Physical Review in 1955; Chamberlain and Segrè took the 1959 Nobel Prize. The whole enterprise rested on a threshold formula, and the formula is one line: $\sqrt{s} \geq \sum_f m_fc^2$ .

History

1928

— Dirac's relativistic electron equation implies a negative-energy partner for every particle, and by extension an antiproton.

1932

— Anderson finds the positron in cosmic rays, confirming that antiparticles are real and turning the antiproton into a target.

1949

— Berkeley's Radiation Laboratory sets the Bevatron's design energy at 6.2 GeV, chosen to clear the 5.63 GeV antiproton threshold with margin.

1954

— The Bevatron comes into operation, the first accelerator built to a relativistic threshold calculation.

1955

— Chamberlain, Segre, Wiegand and Ypsilantis identify the antiproton, publishing in Physical Review; Chamberlain and Segre share the 1959 Nobel Prize.

1971

— CERN's Intersecting Storage Rings collide protons on protons for the first time, reaching a CM energy that no fixed-target machine of the era could have approached.

Going further

Texts. Griffiths, Introduction to Elementary Particles, 2nd ed., §3.5 · Rindler, Introduction to Special Relativity, 2nd ed., §32 · Perkins, Introduction to High Energy Physics, 4th ed., §2.5 · Chamberlain, Segrè, Wiegand & Ypsilantis, "Observation of Antiprotons", Physical Review 100, 947 (1955) · Particle Data Group, Review of Particle Physics, "Kinematics".

Provenance. Signature $(+,-,-,-)$ ; $p\cdot p = +m^2c^2$ ; $s$ in energy² units, so $\sqrt{s}$ is an energy. Every threshold in this article is computed from the invariant $s$ and never from a lab-frame energy sum. verify_aq-sr-threshold-energy.py parses this article and re-derives every number it prints: $\sqrt{s} = 4m_pc^2 = 3.753$ GeV, $E_1 = 7m_pc^2 = 6.568$ GeV, $T_1 = 6m_pc^2 = 5.630$ GeV, $\gamma_{\mathrm{cm}} = 2$ , the 50 % available fraction, the ledger $5.630 = 1.877 + 3.753$ , the factor of 3, the Bevatron's $6.2$ GeV and its $570$ MeV margin, the fixed-target table ( $4.53$ , $13.76$ , $43.34$ , $114.6$ GeV), the $\times122$ gap, $1.04\times10^8$ GeV, and the photoproduction threshold $144.7$ MeV.

Scope. This article gives the kinematic threshold: the energy below which a reaction is impossible. It says nothing about the cross-section — how likely the reaction is — which is dynamics, and which in practice vanishes at threshold and rises above it, so that real machines are always built with margin. Luminosity, the second half of any accelerator's specification, is not treated. Thresholds in a medium (Cherenkov, Greisen–Zatsepin–Kuzmin) use the same invariant but a different final state.

Builds on. Centre-of-Momentum Frame · Invariant Mass · Conservation of Four-Momentum · Relativistic Inelastic Collisions · Mandelstam Variables. Leads to. Particle Decay Kinematics · Mandelstam Invariants · Colliding-Beam Experiments · Relativistic Energy.

Why can't I just say "the beam needs enough energy to make the new particles"?

Because momentum has to go somewhere. The incoming beam carries momentum; the total momentum after the collision must equal it; so the products cannot all be at rest in the lab. Whatever momentum they carry brings kinetic energy with it, and that energy is not available to build mass. Your naive condition tacitly assumes the products can be at rest in the lab, which conservation of momentum forbids — so it is not an approximation to the right answer, it is the answer to an impossible problem. For the antiproton it is wrong by a factor of exactly 3. The correct statement puts an invariant on the left: $\sqrt{s} \geq \sum_f m_fc^2$ . That version is right in every frame, and it is a shorter calculation than the wrong one.

Where exactly does the "wasted" energy go?

Into the forward motion of the debris. At threshold the products are at rest in the CM frame — but that frame is itself hurtling through the lab, so in the lab the products stream forward together as a single lump whose kinetic energy no experiment can spend on making particles. For the antiproton $\gamma_{\mathrm{cm}} = 2$ , so the available fraction is $1/\gamma_{\mathrm{cm}} = 50\%$ : of $5.630\ \mathrm{GeV}$ of beam kinetic energy, $1.877\ \mathrm{GeV}$ becomes the new pair and $3.753\ \mathrm{GeV}$ becomes motion. Nothing is destroyed — it is simply not available, which for an experimenter is the same thing.

Is this a relativistic effect? Would a Newtonian collision waste energy too?

Yes it would, because the mechanism is momentum conservation, not relativity: even in Newtonian mechanics only the fraction $m_2/(m_1+m_2)$ of the beam's kinetic energy is available in the CM frame, which is why the Newtonian threshold is $T_{\mathrm{thr}} = (1 + m_1/m_2)\,Q$ and not simply $Q$ . What relativity changes is the scale. The available fraction is $1/\gamma_{\mathrm{cm}}$ , and $\gamma_{\mathrm{cm}}$ grows without bound as the beam energy rises — so the waste, mild at low energy, becomes overwhelming at high energy. The square-root law is that growth, written as a formula.

Why do we bother with fixed-target experiments at all?

Because $\sqrt{s}$ is not the only thing that matters — rate matters too. A solid or liquid target is some $10^{24}$ times denser than a counter-rotating bunch of protons, so a fixed-target beam produces vastly more interactions per second than a collider does. If your physics needs statistics rather than reach — rare kaon decays, neutrino beams, nuclear structure — a fixed target is the right instrument and a collider would be a poor one. The square root is fatal only at the energy frontier; everywhere else, luminosity wins.

Does a heavier target really help?

Yes: $s = m_1^2c^4 + m_2^2c^4 + 2m_2c^2E_1$ grows with the target mass $m_2$ , so at fixed beam energy a heavy nucleus gives more $\sqrt{s}$ than hydrogen does — physically, the beam is recoiling against more inertia, so less energy is squandered on centre-of-mass motion. But the gain is only $\sqrt{m_2}$ , and it costs you a target that is not a single free nucleon, which introduces nuclear binding, Fermi motion and a smeared effective energy. That is one reason Berkeley wanted margin above $5.63\ \mathrm{GeV}$ rather than a machine sitting exactly on the threshold.

If cosmic rays reach 10²⁰ eV, why are they not far beyond the LHC?

Because a cosmic ray hitting the atmosphere is a fixed-target event, and the square root is merciless. A $10^{20}\ \mathrm{eV}$ proton — $10^{11}\ \mathrm{GeV}$ , about $10^{7}$ times the LHC's beam energy — striking a stationary nucleon gives $\sqrt{s} \approx \sqrt{2 \times 0.938 \times 10^{11}}\ \mathrm{GeV} \approx 430\ \mathrm{TeV}$ : real, remarkable, and only about $30$ times the LHC's $14\ \mathrm{TeV}$ . Ten million times the energy buys thirty times the reach. Nature runs fixed-target experiments and pays the same tax we do — which is exactly why we build colliders instead of waiting for the sky.