Collider vs Fixed-Target Energetics

Fire a $7\,$ TeV proton at a proton sitting still and you have spent $7\,$ TeV. You get to keep about $115\,$ GeV of it. The other $98\%$ is not lost, not converted, not wasted on heat — it is simply unavailable, locked up in the forward motion that momentum conservation forces the debris to keep. You cannot make new mass out of it. You paid for a hammer and you swung it at a target that ran away. Turn the target around and fire it back, and the same two protons give you $14\,$ TeV, all of it. That single fact — that available energy grows as $\sqrt{E_{\text{beam}}}$ against a fixed target and linearly in a collider — is why every machine at the energy frontier for the last fifty years has been a collider, and why no fixed-target machine will ever get there again. It is not an engineering preference. It is a theorem about four-momentum, and the price of ignoring it can be quoted exactly: to match the LHC by hitting a stationary proton you would need a beam of $\mathbf{9.9\times10^{4}\ TeV}$ , steered by the LHC's own $8.33\,$ T dipoles around a ring $\mathbf{6.2}$ times the circumference of the Earth.

Key takeaways

Signature $(+,-,-,-)$ : $\mathrm{d}s^2 = c^2\mathrm{d}t^2 - \mathrm{d}x^2$ , $p\!\cdot\!p = +m^2c^2$ , timelike positive. Convention locked so that $\sqrt{s}$ is an energy: $s \equiv (E_1+E_2)^2 - |\mathbf{p}_1+\mathbf{p}_2|^2c^2$ .
The only energy available to make new particles is the invariant $\sqrt{s}$ — the total energy in the frame where the total momentum vanishes. Everything else is bulk motion of the debris and is untouchable.
Fixed target: $\sqrt{s} = \sqrt{(m_bc^2)^2 + (m_tc^2)^2 + 2m_tc^2E_{\text{beam}}} \to \sqrt{2m_tc^2 E_{\text{beam}}}$ . Available energy grows as the square root of what you pay for. Each doubling of $\sqrt{s}$ costs a quadrupling of beam energy.
Symmetric collider: $\sqrt{s} = 2E_{\text{beam}}$ . Linear. Every joule you buy is a joule you can spend.
The waste is exactly $E_{\text{lab}} - \sqrt{s} = (\gamma_{\text{CM}} - 1)\sqrt{s}$ , and it goes to zero — identically, not approximately — when the total momentum is zero. A collider is not a better hammer; it is the unique configuration with nothing to waste.
Owned result: the ring you would have to build. Matching the LHC's $\sqrt{s} = 13.6$ TeV against a stationary proton needs $E_{\text{FT}} = s/2m_pc^2 = \mathbf{9.86\times10^{4}\ TeV}$ . At the LHC's own $8.33\,$ T dipole field the magnetic rigidity fixes the bending radius at $\boldsymbol{\rho = 39\,470\ \mathrm{km}}$ — $\mathbf{94\%}$ of the way out to geostationary orbit, a ring $\mathbf{6.2\times}$ the Earth's circumference. That number is the whole argument.
Fixed target is not obsolete — it wins on luminosity. A litre of liquid hydrogen has more protons in it than any beam ever built. The frontier is colliders; the statistics are fixed targets.
Key idea: momentum conservation, not energy conservation, is what taxes you. Zero the total momentum and the tax is zero.

The core insight

Energy is conserved, but energy is not invariant. Momentum is conserved, and that is the trap: whatever momentum walks into the collision must walk out, so the debris is condemned to keep moving, and the kinetic energy of that motion can never become mass. The invariant $\sqrt{s}$ is precisely the part of the energy that is not hostage to momentum conservation. Building a collider is nothing more subtle than arranging for there to be no hostage.

Where this lives today

The energy frontier is closed to fixed targets, permanently. Not a statement about current technology: equation (6) says the required beam energy grows as $s$ , while every practical bending magnet is bounded near $10$ – $20$ T. Together those fix a hard ceiling on fixed-target $\sqrt{s}$ for any ring that fits on the planet, and the LHC passed it decades ago. FCC-hh, muon colliders, linear $e^+e^-$ machines — they differ in many ways and agree on exactly one: they collide.

$\sqrt{s}$ is the only fair currency for comparing machines across eras. Fermilab's Tevatron fixed-target programme ran $800$ GeV protons for $\sqrt{s} = 38.8$ GeV — less reach than the $\sqrt{s} = 62.8$ GeV CERN's ISR had reached a decade earlier with $31.4$ GeV beams, using twenty-five times less beam energy.

And nature runs the experiment anyway — which proves the other half of the argument. Cosmic rays are fixed-target collisions, and the very highest of them beat the LHC outright: the Fly's Eye "Oh-My-God" particle of 15 October 1991, at $3.2\times10^{20}$ eV, striking a stationary atmospheric nucleon delivers $\sqrt{s} = \sqrt{2m_pc^2E} = \mathbf{775\ TeV}$ — some $\mathbf{57\times}$ the LHC's $13.6$ TeV, achieved by a single proton and a nitrogen molecule. Above $10^{20}$ eV such particles arrive at roughly one per square kilometre per century. So the square-root law is not what stops us using nature's beam. Luminosity is. The two halves of this article are the two things a collider buys, and the sky demonstrates both by supplying one and withholding the other.

The physical picture

Two cars meet head-on at $50$ km/h each. Now park one and hit it at $100$ km/h. The second crash is not twice as violent — it is far less violent, because the wreckage flies off down the road carrying most of the kinetic energy with it, and only the remainder goes into crumpling metal. The sharpening is relativistic. Non-relativistically, a beam of kinetic energy $T$ hitting an equal stationary mass wastes exactly half of it: the reduced-mass factor $m_t/(m_b+m_t) = \tfrac12$ . A fixed, tolerable tax. But at high energy the beam's momentum is $p \approx E/c$ , so the centre of momentum must itself move at nearly $c$ — and the faster it must move, the more of your energy is consumed carrying it. The tax is not a fixed fraction. It grows without bound. One line contains the whole subject. For a beam of energy $E$ on a target of mass $m_t$ at rest,

s \;=\; (E + m_tc^2)^2 - p^2c^2 \;=\; (m_bc^2)^2 + (m_tc^2)^2 + 2m_tc^2 E \;\xrightarrow[\;E\;\text{large}\;]{}\; 2m_tc^2E .

Everything you add to $E$ arrives at the collision divided by a square root. Turn the target around and the momenta cancel, the CM stands still, $\gamma_{\text{CM}} = 1$ , and $\sqrt{s} = 2E$ — every joule delivered.

Figure

THE WHOLE ARGUMENT IN ONE PLOT. Available energy √s against proton beam energy, log-log. The blue line is a symmetric proton collider: √s = 2E, a straight line of slope 1 — every joule you buy, you get to spend. The red curve is a proton beam on a stationary proton: √s → √(2m_p c² E), slope 1/2 — the returns are square-root and they never recover. The gap is not a constant; it OPENS without bound. At the LHC's operating point (blue dot, 6.8 TeV per beam, √s = 13.6 TeV) the red curve has to be run all the way out to 9.86×10⁴ TeV (red dot) to reach the same √s — a factor of 14,500 in beam energy. The dashed grey line marks the antiproton threshold √s = 4 m_p c² = 3.75 GeV, which the Bevatron cleared in 1955 with a 6.2 GeV beam. Fixed target is the flatter curve, and no amount of money changes its slope

The plot is the whole article. Two lines on log-log paper with different slopes diverge for ever; no budget and no cleverness closes a gap in slope. The ratio of what the two machines must be fed, for the same reach, is

\frac{E^{\text{FT}}_{\text{beam}}}{E^{\text{coll}}_{\text{beam}}} \;=\; \frac{\sqrt{s}}{m_tc^2} ,

not a constant but the collision energy itself, measured in units of the target's rest energy. At the LHC that is $13\,600/0.938 = 14\,500$ .

The core equations

The master invariant. $s$ is an energy-squared here, so $\sqrt{s}$ is an energy and compares directly with a mass in GeV.

\boxed{\;s \;\equiv\; \left(\textstyle\sum_i E_i\right)^{2} - \left|\textstyle\sum_i \mathbf{p}_i\right|^{2}c^{2}\;} \qquad (1)

where: $E_i$ , $\mathbf{p}_i$ — the lab energy and three-momentum of each incoming particle. Physical reading: $s$ is the squared length of the total four-momentum, so it is the same number in every frame. Evaluate it in the CM frame, where $\sum\mathbf{p}_i = 0$ , and it collapses to $s = E_{\text{CM}}^2$ . Hence $\sqrt{s}$ is the CM total energy — literally, not by analogy. The two-body value, for any geometry:

s \;=\; (m_1c^2)^2 + (m_2c^2)^2 + 2\big(E_1E_2 - \mathbf{p}_1\!\cdot\!\mathbf{p}_2\,c^2\big) \qquad (2)

where: the cross term carries all the geometry. Physical reading: head-on, $\mathbf{p}_1\!\cdot\!\mathbf{p}_2 = -|\mathbf{p}_1||\mathbf{p}_2|$ and the term adds; stationary target, $\mathbf{p}_2 = 0$ and it contributes only $2m_2c^2E_1$ . The entire subject is the difference between those two lines. Fixed target, exact and then asymptotic:

\boxed{\;\sqrt{s}_{\text{FT}} = \sqrt{(m_bc^2)^2 + (m_tc^2)^2 + 2\,m_tc^2\,E_{\text{beam}}} \;\xrightarrow[\;E_{\text{beam}} \gg mc^2\;]{}\; \sqrt{2\,m_tc^2\,E_{\text{beam}}}\;} \qquad (3)

where: $m_t$ is the target mass — the only mass surviving the limit. Physical reading: the target's rest energy is the lever arm the beam pushes against, and the $\sqrt{\;}$ is not an approximation you can improve on. It is the exact high-energy behaviour. Symmetric collider — exact, no limit needed:

\boxed{\;\sqrt{s}_{\text{coll}} = 2E_{\text{beam}}\;} \qquad (4)

where: equal masses, equal and opposite momenta. Physical reading: the lab is the CM frame. There is no boost to pay for. The wasted energy, in closed form:

\boxed{\;E_{\text{wasted}} \;=\; \Big(\textstyle\sum_i E_i\Big) - \sqrt{s} \;=\; (\gamma_{\text{CM}} - 1)\sqrt{s}\;},\qquad \gamma_{\text{CM}} = \frac{\sum_i E_i}{\sqrt{s}} \qquad (5)

where: $\gamma_{\text{CM}}$ is the Lorentz factor of the centre-of-momentum frame in the lab. Physical reading: the waste is the CM frame's own kinetic energy. Zero if and only if $\gamma_{\text{CM}} = 1$ — if and only if the total momentum vanishes. A collider does not waste less; it wastes nothing. And the equivalent-energy law:

\boxed{\;E^{\text{FT}}_{\text{beam}} = \frac{s}{2\,m_tc^2}\;} \qquad\Longrightarrow\qquad \frac{E^{\text{FT}}_{\text{beam}}}{E^{\text{coll}}_{\text{beam}}} = \frac{\sqrt{s}}{m_tc^2} \qquad (6)

where: equate (3) and (4). Physical reading: quadratic in $\sqrt{s}$ . Doubling a fixed-target machine's reach costs four times the beam; doubling a collider's costs two. Over six decades of climbing, that difference compounds into the factor of $14\,500$ on the plot.

Why the fixed target throws it away

Derivation AThe square-root law and the exact size of the loss

The claim to be established is not that a fixed target is worse — everyone believes that — but that it is worse in a specific functional form that no engineering can escape. We compute $\sqrt{s}$ for a beam on a stationary target exactly, extract its high-energy behaviour, and then show that the missing energy is precisely the kinetic energy of the centre-of-momentum frame, forced to move by momentum conservation and by nothing else. The output is a closed form for the waste, and a closed form for the price.

Step 1. Write the two four-momenta. In the lab, target at rest: $p_1 = (E_{\text{beam}}/c,\ \mathbf{p}_b)$ and $p_2 = (m_tc,\ \mathbf{0})$ . On shell, $p_1\!\cdot\!p_1 = (m_bc)^2$ and $p_2\!\cdot\!p_2 = (m_tc)^2$ in our signature $(+,-,-,-)$ , where a timelike square is positive. Why: the on-shell condition is the only physical input. Everything below is bookkeeping on it.

Step 2. Square the sum. $s/c^2 = (p_1+p_2)^2 = p_1^2 + p_2^2 + 2p_1\!\cdot\!p_2$ , and the cross term is

p_1\!\cdot\!p_2 = \frac{E_{\text{beam}}}{c}\,m_tc - \mathbf{p}_b\!\cdot\!\mathbf{0} = m_t E_{\text{beam}} .

Restoring factors of $c$ so that $s$ has units of energy squared,

s = (m_bc^2)^2 + (m_tc^2)^2 + 2\,m_tc^2\,E_{\text{beam}} .

Why: the target's vanishing momentum kills the spatial part of the dot product. That is the whole mechanism — the beam's enormous momentum has nothing to cancel against, so it survives into the final state.

Step 3. Take the beam energy large. Both rest-energy terms are constants; only the last term grows. So

\sqrt{s} \to \sqrt{2\,m_tc^2\,E_{\text{beam}}} \qquad\Longrightarrow\qquad \sqrt{s} \propto \sqrt{E_{\text{beam}}} .

Why: the square root is the fingerprint of a CM frame racing toward $c$ . As $E_{\text{beam}}$ grows, $\gamma_{\text{CM}} \to \sqrt{E_{\text{beam}}/2m_tc^2}$ grows too, and the energy you add is increasingly spent on the boost rather than on the collision.

Step 4. Identify the loss exactly. The lab energy is $E_{\text{beam}} + m_tc^2$ ; the usable energy is $\sqrt{s}$ . Their difference, using $\gamma_{\text{CM}} = (E_{\text{beam}} + m_tc^2)/\sqrt{s}$ , is

E_{\text{wasted}} = (E_{\text{beam}} + m_tc^2) - \sqrt{s} = (\gamma_{\text{CM}} - 1)\sqrt{s} .

Why: this is not an estimate. $(\gamma_{\text{CM}}-1)\sqrt{s}$ is literally the kinetic energy of a body of rest energy $\sqrt{s}$ moving with Lorentz factor $\gamma_{\text{CM}}$ — the debris, considered as a single object. Momentum conservation compels that motion, so the energy is not merely hard to reach: it is spoken for.

Step 5. Set the two machines equal and read off the price. Equating $s_{\text{FT}} = 2m_tc^2E^{\text{FT}}_{\text{beam}}$ with $s_{\text{coll}} = (2E^{\text{coll}}_{\text{beam}})^2$ ,

\boxed{\;E^{\text{FT}}_{\text{beam}} = \frac{s}{2m_tc^2} = \frac{2\,(E^{\text{coll}}_{\text{beam}})^2}{m_tc^2}\;}

Why: quadratic. The required fixed-target energy grows as the square of the collider energy it is trying to imitate. That is the sentence that closed the fixed-target frontier — not a cost overrun, an exponent. *Verified in verify_aq-sr-collider-fixed-target-energetics.py: sympy confirms $s$ is invariant under a boost applied to both four-momenta; the high-energy limit and the wasted-energy identity $(E_1+E_2) - \sqrt{s} = (\gamma_{\text{CM}}-1)\sqrt{s}$ are checked symbolically; and $\gamma_{\text{CM}} \to \sqrt{E/2m_tc^2}$ is confirmed. At $E = 1$ TeV on a proton, $\sqrt{s} = 43.34$ GeV and $95.7\%$ of the lab energy is untouchable.*

Go deeper · Derivation B · 2 minThe ring you would have to build to match the LHC with a stationary target

Derivation BThe ring you would have to build to match the LHC with a stationary target

The equivalent-energy law (6) stays abstract until it is spent. This card converts it into a machine — an actual synchrotron, with an actual radius, obeying the actual magnetic rigidity of the actual dipoles CERN already owns — and the answer is this article's owned structural result. The goal is a single length that makes the case for colliders unanswerable; the method is to take the LHC's own hardware and ask it to do the job the wrong way round.

Step 1. Fix the target. The LHC collides protons at $E_{\text{beam}} = 6.8$ TeV per beam, so $\sqrt{s} = 13.6$ TeV and $s = (13\,600\ \mathrm{GeV})^2 = 1.850\times10^{8}\ \mathrm{GeV}^2$ . Why: Run 3's real operating point, not the $14$ TeV design value the magnets were never trained to.

Step 2. Apply (6) with a proton target, $m_pc^2 = 0.9383$ GeV:

E^{\text{FT}}_{\text{beam}} = \frac{s}{2m_pc^2} = \frac{1.850\times10^{8}}{2 \times 0.9383} = 9.86\times10^{7}\ \mathrm{GeV} = \mathbf{9.86\times10^{4}\ TeV} .

Why: a beam of a hundred thousand TeV — $9.86\times10^{16}$ eV. Nature does supply such particles: the "Oh-My-God" cosmic ray recorded by Fly's Eye on 15 October 1991 carried $3.2\times10^{20}$ eV, some $3\,200$ times more. Nature simply does not supply them often — see the closing section.

Step 3. Ask what magnet holds it. A charged particle of momentum $p$ in a field $B$ bends with radius $\rho$ given by the magnetic rigidity $B\rho = p/q$ , which in accelerator units is

p\,[\mathrm{GeV}/c] = 0.299792 \cdot B\,[\mathrm{T}] \cdot \rho\,[\mathrm{m}] .

Check it on the machine itself: the LHC's dipoles reach $B = 8.33$ T over a bending radius $\rho = 2804$ m, giving $p = 0.299792 \times 8.33 \times 2804 = \mathbf{7.00\ TeV}$ — the design energy, to three figures. Why: the formula has just been validated against a real accelerator, so what it says next can be trusted.

Step 4. Invert it for the fixed-target beam, at the same $8.33$ T:

\rho = \frac{9.86\times10^{7}}{0.299792 \times 8.33} = 3.95\times10^{7}\ \mathrm{m} = \boxed{\;\mathbf{39\,470\ km}\;}

Why: that is $6.2$ Earth radii. The ring's circumference is $2\pi\rho = 2.48\times10^{5}$ km, or $\mathbf{6.2}$ times the circumference of the Earth, and its radius is $\mathbf{94\%}$ of the way out to geostationary orbit ( $42\,164$ km). The machine would not fit on the planet; it would have to encircle it, at the altitude where we park communications satellites.

Step 5. Note what would happen if you built it. Nothing good: at that rigidity the beam radiates, the stored energy is astronomical, and — the point — you would have spent all of it to reproduce a collision the LHC gets for $6.8$ TeV per beam in a $27$ km tunnel. Why: the factor is $14\,500$ in energy and roughly $14\,000$ in radius, and it comes from a square root. No material, no field, no funding line, and no century of engineering can change the exponent in equation (3). This is what it means for a result to be kinematic rather than technical. *Verified in verify_aq-sr-collider-fixed-target-energetics.py: the rigidity coefficient $0.299792$ is recomputed from $c\times10^{-9}$ ; the LHC self-check $0.299792 \times 8.33 \times 2803.95 = 7.002$ TeV is asserted against the real design energy; $E^{\text{FT}} = 9.856\times10^{4}$ TeV, $\rho = 39\,469$ km, $2\pi\rho/C_\oplus = 6.19$ and $\rho/R_{\text{geo}} = 0.936$ are each recomputed at 50 dps and grepped out of this article.*

Figure

The waste, priced. A proton beam of TOTAL energy E hits a stationary proton. The green bar is the part you can spend — the invariant √s — and the grey bar is the part momentum conservation locks into bulk motion of the debris, E_wasted = (γ_CM − 1)√s. At E = 1 GeV the beam is barely moving, the collision is nearly non-relativistic, and 98.4% of the energy is still usable. By 1 TeV, 95.7% of it is untouchable and the usable √s has crawled only to 43.34 GeV. The waste fraction does not level off — it climbs toward 100%, because γ_CM ∼ √(E/2m_p c²) grows without bound. A symmetric collider is the single configuration with γ_CM = 1 exactly, hence zero grey bar at any energy whatsoever

Limiting cases

Case	Result	Meaning
$E_{\text{beam}} \gg mc^2$ , fixed target	$\sqrt{s} \to \sqrt{2m_tc^2E_{\text{beam}}}$	square-root returns. The frontier closes
beam at rest ( $E_{\text{beam}} = m_bc^2$ )	$\sqrt{s} = (m_b + m_t)c^2$	nothing happens; $\sqrt{s}$ is just the total rest energy. A useful sanity check on (3)
non-relativistic beam	$\sqrt{s} \approx (m_b{+}m_t)c^2 + \tfrac{m_t}{m_b+m_t}T_{\text{beam}}$	the classical reduced-mass tax — a fixed fraction. Relativity is what makes it unbounded
symmetric collider	$\sqrt{s} = 2E_{\text{beam}}$ , $\gamma_{\text{CM}} = 1$	linear, and zero waste. Exact at all energies
asymmetric collider	$\sqrt{s} \approx 2\sqrt{E_1E_2}$	the geometric mean. PEP-II: $2\sqrt{9.0 \times 3.1} = 10.56$ GeV, on the $\Upsilon(4S)$
$c \to \infty$	$\sqrt{s} - (m_b{+}m_t)c^2 \to \dfrac{m_t}{m_b+m_t}T_{\text{beam}}$	Galilean: a fixed reduced-mass fraction, no square root, no unbounded boost tax. The catastrophe is entirely relativistic
$m_t \to \infty$	$\sqrt{s} - m_tc^2 \to E_{\text{beam}}$	an infinitely heavy target absorbs momentum for free — it is a wall, and a wall wastes nothing

Common trapThe base cannot decide whether s is a momentum or an energy — and it mislabels the antiproton threshold

Two defects in the AllQFT base, and the second one has been repeated in textbooks for seventy years. (i) A unit split. The base's first equation defines $s \equiv \left(\tfrac{E_1+E_2}{c}\right)^2 - |\mathbf{p}_1+\mathbf{p}_2|^2$ , which has units of momentum squared, and then its very next equation writes $\sqrt{s} = \sqrt{(m_1c^2)^2 + \dots}$ , which is an energy. Both cannot be true. This article locks the energy convention, $s = (E_1+E_2)^2 - |\mathbf{p}_1+\mathbf{p}_2|^2c^2$ , throughout. (ii) The antiproton threshold is quoted as a "beam kinetic energy of about $6.5$ GeV". It is not. Solving $\sqrt{s} = 4m_pc^2$ for $p + p \to p+p+p+\bar p$ gives a beam total energy of $7m_pc^2 = 6.57$ GeV and therefore a beam kinetic energy of $6m_pc^2 = \mathbf{5.63\ GeV}$ . The base printed the total energy and called it kinetic. The distinction is exactly one proton mass and it is the difference between a correct story about the Bevatron and a wrong one.

Common trapDo not read "colliders always win" off this page

The theorem is about energy, and energy is not the only currency. Event rate is $\dot N = \mathcal{L}\sigma$ , and the luminosity of a fixed target is enormous: a beam crossing a metre of liquid hydrogen meets $\sim4\times10^{24}$ protons per cm $^2$ , while two crossing bunches present each other with $\sim10^{11}$ particles in a spot of order $10^{-5}$ cm $^2$ and miss almost every time. For any process whose $\sqrt{s}$ requirement is already met, the fixed target wins on statistics by orders of magnitude — which is why neutrino beams, kaon factories, hyperon physics and nuclear structure are all fixed-target, and why the bottom quark was found in 1977 by firing $400$ GeV protons at a stationary target. Colliders own the frontier. Fixed targets own everything behind it.

Worked examples

1 · The Bevatron, and whether it was really built for the antiproton

In 1955 Berkeley's Bevatron delivered protons of about $6.2$ GeV kinetic energy onto a copper target, and the antiproton was found. Was the machine's energy a coincidence, or was it designed for exactly this?

Step 1 — identify. Antiprotons cannot be made alone: baryon number forces $p + p \to p + p + p + \bar{p}$ , so four nucleon masses must appear in the final state. The threshold condition is that $\sqrt{s}$ reaches the total rest energy of the lightest allowed final state, all four particles at rest in the CM frame:

\sqrt{s}_{\min} = 4\,m_pc^2 = 4 \times 0.9383 = \mathbf{3.753\ GeV} .

Step 2 — apply (3). With beam and target both protons, $s = 2(m_pc^2)^2 + 2m_pc^2E_{\text{beam}}$ . Setting this equal to $(4m_pc^2)^2 = 16(m_pc^2)^2$ :

2m_pc^2 E_{\text{beam}} = 14\,(m_pc^2)^2 \qquad\Longrightarrow\qquad E_{\text{beam}} = 7\,m_pc^2 = \mathbf{6.568\ GeV\ (total)} .

Step 3 — convert to what an accelerator actually quotes. Machines are specified by kinetic energy. Subtract the rest energy:

T_{\text{beam}} = E_{\text{beam}} - m_pc^2 = 6\,m_pc^2 = \mathbf{5.630\ GeV\ (kinetic)} .

This is the number, and it is not $6.5$ . The famous "six proton masses" is the kinetic threshold; the total is seven.

Step 4 — the boundary number. The Bevatron's $6.2$ GeV clears $5.63$ GeV with about $10\%$ to spare — enough for a usable cross-section, not so much that the machine was over-built. The answer is yes, deliberately: the energy was chosen from this kinematic calculation before a magnet was poured. (The effective threshold in a copper target is lower still, because the struck nucleon is not at rest — Fermi motion gives it a head start of a few hundred MeV/c. The design did not rely on that, and did not need to.)

Step 5 — the moral. A four-line invariant-mass calculation fixed the size, cost and field of one of the largest machines built to that date, and it was right. Chamberlain, Segrè, Wiegand and Ypsilantis published Observation of Antiprotons in Phys. Rev. 100, 947 (1955); Segrè and Chamberlain took the 1959 Nobel, and Wiegand and Ypsilantis did not.

2 · Why PEP-II ran its two beams at different energies on purpose

The PEP-II B-factory at SLAC collided $9.0$ GeV electrons with $3.1$ GeV positrons. A symmetric machine at $\sqrt{s}/2 = 5.28$ GeV per beam would have reached the same collision energy with less total power. Why throw away the symmetry?

Step 1 — identify. Use (2) for two head-on ultrarelativistic beams, where the masses are negligible and $\mathbf{p}_1\!\cdot\!\mathbf{p}_2 = -|\mathbf{p}_1||\mathbf{p}_2| \approx -E_1E_2/c^2$ :

s \approx 4E_1E_2 \qquad\Longrightarrow\qquad \sqrt{s} \approx 2\sqrt{E_1E_2} \quad\text{— the \textbf{geometric} mean.}

Step 2 — evaluate. $\sqrt{s} = 2\sqrt{9.0 \times 3.1} = \mathbf{10.56\ GeV}$ , which sits on the $\Upsilon(4S)$ resonance at $10.58$ GeV — the lightest state that decays to a $B\bar{B}$ pair and nothing else. The asymmetry costs nothing in reach: the geometric mean of $9.0$ and $3.1$ is what a symmetric pair of $5.28$ GeV beams would have given.

Step 3 — find what the asymmetry buys. From (5), the CM frame is not at rest in the lab. Its boost is

\beta\gamma_{\text{CM}} = \frac{|E_1 - E_2|}{\sqrt{s}} = \frac{9.0 - 3.1}{10.56} = \mathbf{0.559} .

The $B$ mesons are therefore produced moving, with $\beta\gamma \approx 0.56$ in addition to their small momentum in the CM frame.

Step 4 — the boundary number. A $B$ meson lives $\tau = 1.52$ ps. On a symmetric machine the $\Upsilon(4S)$ is produced at rest, the two $B$ s crawl away with almost no momentum, and their decay vertices are microns apart — unresolvable. With the $0.56$ boost the pair separates by $\beta\gamma\, c\tau \approx \mathbf{250\ \mu m}$ , comfortably resolved by a silicon vertex detector. That separation is the entire experiment: measuring the time-ordering of the two $B$ decays is how BaBar and Belle established CP violation in the $B$ system.

Step 5 — the moral. Equation (5) says bulk CM motion is wasted energy. It does not say it is useless. PEP-II bought $\gamma_{\text{CM}} = 1.15$ deliberately, paid for it in beam energy, and got a measurable decay length back. The lesson is not "always zero the momentum" — it is know exactly what zeroing it is worth, and what it costs not to.

Physics of the portrait

There is no single person behind this article, and that is the point: the square-root law was known to anyone who could square a four-vector, and it was ignored for twenty years because nobody could build the alternative.

Origins · 2 minParticle accelerator and how this was discovered

Particle accelerator

1928–1983 · Cockcroft and Walton split the atom · Lawrence's cyclotron · phase stability makes the synchrotron · storage rings make the collider

The kinematics were never in doubt. That $\sqrt{s}$ grows as $\sqrt{E}$ against a stationary target is a two-line consequence of four-momentum conservation, and was understood as soon as four-momentum was. The difficulty was never knowing what to do; it was doing it. A collider needs two beams held in the same place at the same time for hours, dense enough to actually hit each other — and until the 1960s nobody knew how to store a beam at all. Three inventions closed the gap. Ernest Lawrence's cyclotron (1931) showed a beam could be accelerated by a modest voltage applied many times. Vladimir Veksler (1944) and Edwin McMillan (1945) independently found phase stability, which lets a bunch survive an accelerating cavity for millions of turns instead of a few — the discovery that made the synchrotron possible. And Courant, Livingston and Snyder's strong focusing (Phys. Rev. 88, 1190 (1952)) — anticipated, unpublished, by Nicholas Christofilos in 1950 — shrank the beam pipe from a metre to centimetres and made large rings affordable. Then the argument won. Gerard O'Neill proposed storage rings in 1956; the first electron–electron collisions came around 1965; in 1971 CERN's ISR collided protons on protons. The decisive demonstration was the SPS $p\bar{p}$ collider, which in 1983 found the $W$ and $Z$ at $\sqrt{s} = 540$ GeV — a reach a fixed-target machine would have needed a $155\,000$ GeV beam to match. Rubbia and van der Meer took the 1984 Nobel. Since then the energy frontier has been colliders and nothing else, for a reason written in equation (6) and not open to argument.

History

1928

— Rolf Widerøe's thesis describes the linear resonance accelerator; Ernest Lawrence reads it and conceives the cyclotron.

1931

— Lawrence and Livingston operate the first cyclotron, reaching 80 keV; the multi-pass principle is established.

1944

— Vladimir Veksler publishes the principle of phase stability (Compt. Rend. Acad. Sci. U.R.S.S. 43, 444).

1945

— Edwin McMillan independently proposes the synchrotron on the same principle (Phys. Rev. 68, 143).

1952

— Courant, Livingston and Snyder publish strong focusing (Phys. Rev. 88, 1190), anticipated in Christofilos's unpublished 1950 patent application.

1955

— The Bevatron, at 6.2 GeV, exceeds the 5.63 GeV antiproton threshold; Chamberlain, Segrè, Wiegand and Ypsilantis observe antiprotons (Phys. Rev. 100, 947).

1956

— Gerard O'Neill proposes colliding-beam storage rings, making the linear-scaling regime physically reachable.

1971

— CERN's Intersecting Storage Rings collide protons on protons, the first hadron collider.

1977

— Lederman's E288 finds the Upsilon at 9.5 GeV with 400 GeV protons on a fixed target (Phys. Rev. Lett. 39, 252) — the last great fixed-target discovery at the frontier.

1983

— UA1 and UA2 find the W and Z at the SPS proton-antiproton collider, sqrt(s) = 540 GeV (Phys. Lett. B 122, 103 and B 122, 476).

2022

— The LHC begins Run 3 at sqrt(s) = 13.6 TeV, an energy no fixed-target machine could reach with a ring smaller than the Earth's orbit about it.

Going further

Texts. Griffiths, Introduction to Elementary Particles, 2nd ed., ch. 3 · Halzen & Martin, Quarks and Leptons, ch. 4 · Wiedemann, Particle Accelerator Physics, 4th ed. · Particle Data Group, Review of Particle Physics, "Kinematics" review · Chamberlain, Segrè, Wiegand & Ypsilantis, Phys. Rev. 100, 947 (1955) · Arnison et al. (UA1), Phys. Lett. B 122, 103 (1983).

Provenance. Signature $(+,-,-,-)$ , $s$ in energy-squared units. Every number is recomputed at 50 dps in verify_aq-sr-collider-fixed-target-energetics.py, which reads this file and asserts each printed value against what it computed. Base defects repaired: (i) the base defines $s$ with momentum-squared units in eq. 0 and energy-squared units in eq. 1 — the convention is locked here to energy-squared; (ii) the base states the antiproton threshold as "a beam kinetic energy of about $6.5$ GeV" — $6.57$ GeV is the beam total energy ( $7m_pc^2$ ); the kinetic threshold is $6m_pc^2 = 5.63$ GeV. The base's equivalent-beam-energy ratio $E_{\text{FT}}/E_{\text{coll}} \approx 2E_{\text{coll}}/m_tc^2$ was checked and is correct.

Scope. This article owns the energetics: $\sqrt{s}$ , the square-root law, the waste, and the equivalent-energy price. The invariant $s$ as a member of the Mandelstam triple, and its partners $t$ and $u$ , belong to `aq-sr-mandelstam-variables`. The construction of the centre-of-momentum frame and the invariant mass of a general system belong to `aq-sr-center-of-momentum-invariant-mass`. The full apparatus of two-body final states, decay kinematics and thresholds belongs to `aq-sr-relativistic-collision-kinematics`. The beams themselves — how you steer, focus and store them — are `aq-sr-relativistic-beam-dynamics`, and what they radiate while you do it is `aq-sr-synchrotron-radiation`.

Builds on. Four-Momentum · Relativistic Collision Kinematics · The Invariant Interval. Leads to. Relativistic Beam Dynamics · Synchrotron Radiation · Mandelstam Variables.

If the collider gives 2E and the fixed target gives about √(2 m E), where did the beam energy actually GO in the fixed-target case?

Nowhere — it is all still there, and energy is exactly conserved. It is simply in the wrong form. Momentum conservation forces the debris to carry away the beam's original momentum $\mathbf{p}_b$ , and a system with momentum $\mathbf{p}$ and invariant mass $\sqrt{s}/c^2$ necessarily has energy $\sqrt{s + |\mathbf{p}|^2c^2} > \sqrt{s}$ . That excess, $(\gamma_{\text{CM}} - 1)\sqrt{s}$ , is kinetic energy of the whole final state moving down the beam pipe. It shows up in your detector as forward-going particles with large momenta. It just cannot be converted into rest mass, because doing so would violate momentum conservation. The tax collector is momentum, not energy.

Does a heavy nucleus target help? Equation 3 says √s grows as √m_t — so why not just use lead?

Because at high energy you do not hit the nucleus. You hit one nucleon inside it. The $\sqrt{m_t}$ scaling in equation (3) assumes the target recoils as a single coherent body of mass $m_t$ , and a lead nucleus does that only for very soft, long-wavelength probes. A TeV projectile has a de Broglie wavelength far smaller than a nucleon, so the hard collision that could make new particles is a collision with a single nucleon of mass $m_pc^2 = 0.938$ GeV — Fermi-smeared by a few hundred MeV/c, and no more. The $\sqrt{207}$ never arrives. What a heavy target genuinely buys is areal density, which is the luminosity argument, not the energy argument. Lead helps you with rate. It does not help you with reach. That is the distinction this whole article turns on, and nuclear targets are the sharpest place to get it backwards.

Is there any regime where the fixed target is genuinely better?

Yes, and it is most of experimental particle physics. The reach argument is about $\sqrt{s}$ ; almost every measurement is about rate, $\dot N = \mathcal{L}\sigma$ . Once you are above threshold for the process you care about, a dense stationary target delivers orders of magnitude more events than two bunches that mostly miss each other. That is why neutrino beams, kaon and hyperon experiments, hadron spectroscopy and nuclear physics are all fixed-target — and why the bottom quark was found in 1977 by shooting $400$ GeV protons into a stationary target.

Why is the antiproton threshold four proton masses and not two?

Baryon number. The reaction $p + p \to p + \bar{p}$ would take baryon number from $+2$ to $0$ , which is forbidden. To make an antiproton you must make a proton alongside it, so the lightest allowed final state is $p + p + p + \bar{p}$ — four particles, total rest energy $4m_pc^2 = 3.753$ GeV. That is the $\sqrt{s}$ you must reach; against a stationary proton it costs $6m_pc^2 = 5.63$ GeV of kinetic beam energy. The conservation law sets the factor of four; the energetics only tell you what it costs.

Could a fixed-target machine ever reach LHC energies with better magnets?

No, and the number is worth carrying. To reach $\sqrt{s} = 13.6$ TeV on a stationary proton you need a beam of $9.86\times10^{4}$ TeV; at the LHC's own $8.33$ T dipoles that fixes $\rho = 39\,470$ km — a ring $6.2$ times the Earth's circumference, its radius $94\%$ of the way to geostationary orbit. Now suppose you found a magnet ten times stronger than anything that exists, at $83$ T. The radius shrinks to $3\,950$ km: still a ring spanning a continent, to do what a $27$ km tunnel already does. The scaling is $\rho \propto s$ , so every factor of $2$ in reach costs a factor of $4$ in radius. Magnets improve linearly; the requirement grows quadratically. You lose.

Is the invariant s the same thing as the invariant mass of the final state?

Yes — and saying so is the cleanest way to see why $\sqrt{s}$ is the budget. Four-momentum conservation makes the final state's total four-momentum equal to the initial state's, so their invariant squares agree: $s_{\text{initial}} = s_{\text{final}} = M_{\text{final}}^2c^4$ . The final state's invariant mass therefore cannot exceed $\sqrt{s}/c^2$ , and equals it only when every product is at rest in the CM frame. $\sqrt{s}$ is not "roughly" the energy available for mass — it is exactly the maximum invariant mass the final state may have. The generalization to the full Mandelstam triple is `aq-sr-mandelstam-variables`.