Natural units

Ask a particle physicist how heavy a proton is and the answer comes back "about a GeV" — not $1.67\times10^{-27}$ kilograms. Ask how long a muon lives and you may hear "a couple of microseconds" or, just as readily, "a decay width of $3\times10^{-10}$ eV". The same physicist will write the electron's size, the range of the weak force, and a scattering cross section all as powers of one energy unit. This is not sloppiness or shorthand: it is a deliberate, exact change of units called natural units, in which the two constants that define relativistic quantum mechanics — the speed of light $c$ and the reduced Planck constant $\hbar$ — are set equal to one. The move is worth making because $c$ and $\hbar$ are not really telling you about physics; they are telling you about the metre, the second and the kilogram, human-chosen rulers with no special standing in the subatomic world. Special relativity says mass and energy are the same thing ( $E=mc^2$ ); the factor $c^2$ is only there because we weigh in kilograms and count energy in joules. Quantum mechanics says energy and inverse time are the same thing ( $E=\hbar\omega$ ); the $\hbar$ is only there because we time things in seconds. Set $c=\hbar=1$ and those artificial bridges disappear. Mass, energy and momentum collapse into a single unit — the giga-electronvolt, GeV — while length and time become $\text{GeV}^{-1}$ . Nothing is lost: the missing constants are still bookkept in the dimensions of every quantity, and any answer can be pushed back to metres or seconds by reinserting them through $\hbar c=0.197\ \text{GeV}\,\cdot\,\text{fm}$ . What is gained is that the algebra of a Feynman diagram, a decay rate, or a cross section stops carrying a fog of $\hbar$ 's and $c$ 's, and the physics stands out.

Key takeaways

Natural units set $\hbar=c=1$ — an exact choice of units, not an approximation — so every quantity is a power of one energy unit, the GeV.
Mass, energy and momentum all become GeV; length and time become $\text{GeV}^{-1}$ ; area (cross section) becomes $\text{GeV}^{-2}$ .
The energy–momentum relation collapses from $E^2=p^2c^2+m^2c^4$ to $E^2=p^2+m^2$ , and $E=mc^2$ to $E=m$ .
No information is lost: reinsert factors of $\hbar$ and $c$ by dimensional analysis, anchored by $\hbar c=0.197\ \text{GeV}\,\cdot\,\text{fm}$ and $\hbar=6.582\times10^{-25}\ \text{GeV}\,\cdot\,\text{s}$ .
Heaviside–Lorentz units add $\varepsilon_0=\mu_0=1$ , so the fine-structure constant is the pure number $\alpha=e^2/4\pi\approx1/137$ and the charge is $e=\sqrt{4\pi\alpha}\approx0.303$ .
Key idea: setting $\hbar=c=1$ fixes the conversions between mass, energy, momentum, length and time, leaving a single unit — formulas simplify dramatically, and any result is translated back to S.I. by reinserting the hidden powers of $\hbar$ and $c$ .

The core insight

By setting $\hbar=c=1$ you fix the conversions between mass, energy, momentum, length and time, leaving a single unit (the GeV). Formulas lose their clutter of $\hbar$ and $c$ , and any answer can be translated back to metres, seconds or barns by reinserting the missing constants, which dimensional analysis identifies uniquely.

Where this lives today

Natural units are not a pedagogical curiosity — they are the working notation of every modern paper in particle physics, quantum field theory and cosmology. A cross section computed from a Feynman diagram emerges in $\text{GeV}^{-2}$ and is reported to experimenters in femtobarns; the LHC's integrated luminosity is quoted in inverse femtobarns precisely so that (events) = (cross section) × (luminosity) needs no stray constants. Lattice QCD simulates quarks and gluons on a grid whose spacing $a$ is an inverse energy, so results emerge as dimensionless ratios like $a m_\pi$ and are converted to MeV only at the end through $\hbar c$ . In cosmology the convention is pushed further by also setting Boltzmann's constant $k=1$ , so temperatures are energies: the cosmic microwave background sits at $T\approx2.35\times10^{-4}\ \text{eV}$ . And the road not taken by particle physics — setting $G=1$ as well — defines the Planck mass $M_{\text{Pl}}=1.22\times10^{19}\ \text{GeV}$ , whose enormous distance above the electroweak scale of $\sim100\ \text{GeV}$ is the hierarchy problem, one of the sharpest open questions in the field. That this gap is visible at all as a plain ratio of two numbers, rather than buried under powers of $\hbar$ and $c$ , is itself a dividend of natural units.

The physical picture

Two constants stand between the human unit system and the particle world. The speed of light $c$ is the exchange rate between the metre and the second; the reduced Planck constant $\hbar$ is the exchange rate between energy and inverse time. Neither number is "fundamental physics" in the way a mass or a coupling is — each is an accident of measuring lengths in metres and times in seconds. Setting them to one is the statement that we will stop paying those exchange rates explicitly and instead measure everything with a single currency. The first figure shows what happens to the units when we do: the three S.I. base quantities collapse onto a single ladder whose rungs are powers of the GeV.

Figure

Setting ħ = c = 1 collapses the S.I. triple kg, m, s onto a single ladder of GeV powers: mass, energy and momentum sit at GeV⁺¹ tied by E² = p² + m², velocity and action are dimensionless at GeV⁰, length and time fall to GeV⁻¹, and a cross-sectional area to GeV⁻²

The ladder is not arbitrary: which rung a quantity lands on is fixed by its ordinary dimensions. A quantity that is $M^{a}L^{b}T^{c}$ in the kilogram–metre–second system becomes $\text{GeV}^{\,a-b-c}$ once $\hbar$ and $c$ are absorbed, because $c$ trades a length for a time and $\hbar$ trades an energy for an inverse time. Velocity ( $L\,T^{-1}$ ) lands on $\text{GeV}^{0}$ — it is dimensionless, the familiar $\beta=v/c\le1$ . Energy, mass and momentum all land on $\text{GeV}^{+1}$ ; length and time drop to $\text{GeV}^{-1}$ ; an area falls to $\text{GeV}^{-2}$ . The single unit that fills this ladder is chosen to be the GeV because it is the natural scale of the field: it is roughly the rest energy of a proton, and it sits at the centre of the mass spectrum of the known particles, shown in the second figure.

Figure

The GeV is the natural rung of particle physics: rest masses run from the electron at 0.511 MeV = 5.11×10⁻⁴ GeV up through the muon 0.106 GeV and proton 0.938 GeV to the electroweak scale, the W boson 80.4 GeV and the top quark 173 GeV — and via ħc = 0.197 GeV·fm one inverse-GeV of length equals 0.197 fm

The core equations

Every simplification of natural units begins with special relativity, where the factor $c^2$ that separates mass from energy, and $c$ that separates momentum from energy, are the first casualties. The energy–momentum relation is the cleanest illustration: three physically identical quantities forced to carry different units in S.I. become one unit here.

E^2 = p^2c^2 + m^2c^4 \;\xrightarrow{\;c=1\;}\; E^2 = p^2 + m^2 \qquad (1)

where: $E$ — total energy; $p$ — momentum magnitude; $m$ — rest mass; $c$ — speed of light. Reading: with $c=1$ the rest energy $E=mc^2$ becomes simply $E=m$ , and a massless particle obeys $E=p$ . Mass, energy and momentum are now three names for quantities measured in the same unit, the GeV. Which unit a quantity lands on is not a matter of taste — it is fixed by its ordinary dimensions through a single rule. Absorbing $\hbar$ (dimension $M L^2 T^{-1}$ ) and $c$ (dimension $L T^{-1}$ ) turns any mechanical quantity into a pure power of energy.

[\,Q\,] = M^{a}\,L^{b}\,T^{c} \;\xrightarrow{\;\hbar=c=1\;}\; \text{GeV}^{\,a-b-c} \qquad (2)

where: $a,b,c$ — the powers of mass, length and time in the S.I. dimension of $Q$ . Reading: mass ( $a=1$ ) is $\text{GeV}^{+1}$ ; length ( $b=1$ ) is $\text{GeV}^{-1}$ ; time ( $c=1$ ) is $\text{GeV}^{-1}$ ; area ( $b=2$ ) is $\text{GeV}^{-2}$ ; velocity ( $b=1,c=-1$ ) is $\text{GeV}^{0}$ , dimensionless. The exponent $a-b-c$ is the "mass dimension" of $Q$ . Because the powers of $\hbar$ and $c$ were only hidden, not deleted, every quantity can be returned to S.I. units by multiplying by the right combination of the two anchor constants below. These three numbers do all the practical work of natural units.

\hbar c = 0.1973\ \text{GeV}\,\cdot\,\text{fm}, \qquad \hbar = 6.582\times10^{-25}\ \text{GeV}\,\cdot\,\text{s}, \qquad (\hbar c)^2 = 0.3894\ \text{GeV}^2\text{mb} \qquad (3)

where: $1\ \text{fm}=10^{-15}\ \text{m}$ ; $1\ \text{mb}=10^{-3}\ \text{barn}=10^{-31}\ \text{m}^2$ . Reading: a length in $\text{GeV}^{-1}$ becomes metres by multiplying by $\hbar c$ ; a time in $\text{GeV}^{-1}$ becomes seconds by multiplying by $\hbar$ ; a cross section in $\text{GeV}^{-2}$ becomes millibarns by multiplying by $(\hbar c)^2$ . Electromagnetism carries one more convention. Particle physics rationalizes Coulomb's law (Heaviside–Lorentz units, $\varepsilon_0=\mu_0=1$ ), so that the fine-structure constant — already the same pure number in every unit system — takes its cleanest algebraic form.

\alpha = \frac{e^2}{4\pi\varepsilon_0\hbar c} \;\xrightarrow{\;\hbar=c=\varepsilon_0=1\;}\; \frac{e^2}{4\pi} \approx \frac{1}{137.036} \qquad (4)

where: $e$ — magnitude of the electron charge; $\varepsilon_0$ — vacuum permittivity; $\alpha$ — fine-structure constant, dimensionless. Reading: because $\alpha$ is a pure number, the charge in these units is fixed, $e=\sqrt{4\pi\alpha}\approx0.303$ — no coulombs, no $\varepsilon_0$ , just a dimensionless coupling.

Derivation AWhy one unit suffices for mass, length and time

The goal is to show, from the dimensions of $\hbar$ and $c$ alone, that setting both to one forces every mechanical quantity onto a single ladder of powers of energy, and to derive the exact rule that a quantity of S.I. dimension $M^{a}L^{b}T^{c}$ carries mass dimension $a-b-c$ . We do this by using the two equations $c=1$ and $\hbar=1$ as two constraints that eliminate the length and time units in favour of the mass (energy) unit, then reading off the surviving power.

Step 1. Start from three independent base dimensions: mass $M$ , length $L$ , time $T$ . Any mechanical quantity is $M^{a}L^{b}T^{c}$ for some powers. With three base units we would normally need three independent measurements to fix a quantity's units. Why: natural units are ultimately a statement about how many independent units the subatomic world really needs — and the answer, we will find, is one.

Step 2. Impose $c=1$ . Since $c$ has dimension $L\,T^{-1}$ , the statement " $c$ is a pure number" means $L$ and $T$ now have the same dimension: one second is a length. Length and time merge, and the energy–momentum relation $E^2=p^2c^2+m^2c^4$ becomes $E^2=p^2+m^2$ . Why: $c=1$ is exactly the relativistic statement that space and time are measured in the same units — the metre and the second differ only by the exchange rate $c$ .

Step 3. Impose $\hbar=1$ . Since $\hbar$ has dimension $M L^2 T^{-1}=$ (energy)(time), setting it to one means energy and inverse time are the same dimension. Combined with Step 2, inverse length is also an energy. So $L\sim T\sim M^{-1}$ : everything is now a power of a single unit, the energy, which we choose to be the GeV. Why: $\hbar=1$ is the quantum statement $E=\hbar\omega$ read as a units identity — a frequency is an energy, hence a time is an inverse energy.

Step 4. Read off the rule. Replace every $L$ by $M^{-1}$ and every $T$ by $M^{-1}$ in $M^{a}L^{b}T^{c}$ : the result is $M^{\,a-b-c}$ , i.e. $\text{GeV}^{\,a-b-c}$ . Test it: energy $M L^2 T^{-2}$ gives $1-2+2=1$ ✓; area $L^2$ gives $-2$ ✓; velocity $L T^{-1}$ gives $-1+1=0$ ✓, dimensionless as required. Why: this single exponent is all the "unit information" left after the two constraints, and it is what tells you exactly which powers of $\hbar$ and $c$ to reinsert later.

Go deeper · Derivation B · 1 minReinserting ħ and c to recover S.I. units

Derivation BReinserting ħ and c to recover S.I. units

The aim is to make precise the claim that no information is lost when $\hbar=c=1$ : given any result in natural units we can recover its metres, seconds or barns uniquely, because the required powers of $\hbar$ and $c$ are fixed by matching dimensions. We build the recipe once, then apply it to a cross section — the quantity where the bookkeeping bites hardest — to turn a number in $\text{GeV}^{-2}$ into an area in $\text{m}^2$ .

Step 1. Write the natural-units result as a pure number times $\text{GeV}^{n}$ , where $n$ is the mass dimension found from the previous derivation. For a cross section $\sigma$ , $n=-2$ , so $\sigma = (\text{number})\times\text{GeV}^{-2}$ . Why: separating the numerical value from its energy power is the whole game — the power tells you the unique factor of $\hbar$ and $c$ to attach.

Step 2. Find the product $\hbar^{p}c^{q}$ that converts $\text{GeV}^{n}$ into the desired S.I. dimension. For an area we need $\text{GeV}^{-2}\to L^{2}$ . Since $\hbar c$ has dimension (energy)(length), $(\hbar c)^2$ has (energy) $^2$ (length) $^2$ ; multiplying $\text{GeV}^{-2}$ by $(\hbar c)^2$ cancels the $\text{GeV}^{-2}$ and leaves length $^2$ . Why: $\hbar c$ is the single constant carrying one power of energy and one of length, so it is the natural "converter" between inverse energy and length.

Step 3. Use the numerical anchor $(\hbar c)^2 = 0.3894\ \text{GeV}^2\text{mb}$ . Then $\sigma[\text{mb}] = \sigma[\text{GeV}^{-2}]\times0.3894$ . For example $1\ \text{GeV}^{-2}=0.3894\ \text{mb}=3.894\times10^{-32}\ \text{m}^2$ . Why: this is the everyday conversion a collider physicist uses to turn a computed amplitude into a measured event rate in barns.

Step 4. Confirm uniqueness. The powers $p,q$ in $\hbar^{p}c^{q}$ are fixed by requiring the final S.I. dimension to come out right — two dimension equations (for the length and time content) for the two unknowns $p,q$ — so there is exactly one valid reinsertion. Why: this is precisely why "setting $\hbar=c=1$ loses no information" is a theorem, not a hope — the inverse map exists and is unique.

Go deeper · Derivation C · 1 minThe fine-structure constant in Heaviside–Lorentz units

Derivation CThe fine-structure constant in Heaviside–Lorentz units

The purpose here is to show why the same coupling strength that reads $\alpha=e^2/4\pi\varepsilon_0\hbar c$ in S.I. becomes the tidy $\alpha=e^2/4\pi$ in particle-physics units, and why the numerical value $1/137$ is untouched by any of this — it is the charge $e$ , not $\alpha$ , that changes its number. We track the two separate conventions folded into "natural units for electromagnetism": setting $\hbar=c=1$ , and rationalizing the fields so $\varepsilon_0=1$ .

Step 1. In S.I., the Coulomb energy of two electrons a distance $r$ apart is $e^2/4\pi\varepsilon_0 r$ , and the dimensionless ratio measuring quantum-electrodynamic coupling strength is $\alpha=e^2/4\pi\varepsilon_0\hbar c\approx1/137.036$ . This number is dimensionless, so it is the same in every unit system. Why: any honest coupling strength must be a pure number; $\alpha$ is what survives after all the units are stripped away, which is why it is the same for a chemist and a string theorist.

Step 2. Set $\hbar=c=1$ . The $\hbar c$ in the denominator becomes $1$ , leaving $\alpha=e^2/4\pi\varepsilon_0$ . Why: this removes the quantum and relativistic exchange rates but keeps the electromagnetic one, $\varepsilon_0$ , which is still lurking.

Step 3. Adopt Heaviside–Lorentz (rationalized) units, in which the $4\pi$ is moved out of Maxwell's equations and into Coulomb's law, and $\varepsilon_0$ is set to $1$ (so $\mu_0=1$ automatically, since $\varepsilon_0\mu_0=1/c^2=1$ ). Now $\alpha=e^2/4\pi$ . Why: rationalization is a convenience that keeps the field equations free of $4\pi$ at the cost of a $4\pi$ in the potential — the same trade Heaviside made for classical electromagnetism.

Step 4. Solve for the charge: $e=\sqrt{4\pi\alpha}=\sqrt{4\pi/137.036}\approx0.303$ . This is the dimensionless number that appears at every QED vertex. Why: in these units the electron charge is not $1.6\times10^{-19}$ coulombs but the pure number $0.303$ — a Feynman diagram with $V$ vertices carries a factor $e^{V}$ , i.e. a power of $\alpha^{V/2}$ , which is why cross sections come out as powers of $\alpha$ .

Limiting cases

The "cases" of natural units are the rungs of the dimension ladder — each physical quantity, and the factor you reinsert to bring it back to S.I.

Case	Result	Meaning
Velocity	$\beta=v/c$ , dimensionless, $\le1$	speeds are pure fractions of $c$ ; nothing exceeds $1$
Action, angular momentum, spin	dimensionless	measured in units of $\hbar$ ; electron spin is $\tfrac12$ , photon $1$
Mass, energy, momentum	$\text{GeV}^{+1}$	$E^2=p^2+m^2$ ; reinsert $c^2$ for kg via $E=mc^2$
Length, time	$\text{GeV}^{-1}$	reinsert $\hbar c=0.197\ \text{GeV}\,\cdot\,\text{fm}$ (length) or $\hbar=6.58\times10^{-25}\ \text{GeV}\,\cdot\,\text{s}$ (time)
Cross section (area)	$\text{GeV}^{-2}$	reinsert $(\hbar c)^2=0.389\ \text{mb}\,\cdot\,\text{GeV}^2$
Decay rate, width	$\text{GeV}^{+1}$	a width $\Gamma$ is an inverse lifetime, $\tau=1/\Gamma$ ; reinsert $\hbar$
Newton's constant $G$	$G=M_{\text{Pl}}^{-2}$ , $\text{GeV}^{-2}$	gravity defines the Planck mass $M_{\text{Pl}}=1.22\times10^{19}$ GeV

Common trap“Setting ħ = c = 1 throws information away, or is only an approximation.”

Neither is true, and the confusion is worth dismantling. Setting $\hbar=c=1$ is an exact change of units, on the same footing as deciding to measure road distances in light-years instead of kilometres. It is fully reversible: the constants have not vanished, they have moved into the dimensions of each quantity, where the mass-dimension exponent $a-b-c$ records exactly which powers of $\hbar$ and $c$ to put back. A cross section written as $2.2\times10^{-6}\ \text{GeV}^{-2}$ contains precisely as much physics as the same number written as $8.7\times10^{-34}\ \text{cm}^2$ ; you convert between them by multiplying by $(\hbar c)^2$ , and the map is unique. Nor is it an approximation like " $v\ll c$ " or " $\hbar\to0$ ": those are physical limits that change the answer, whereas $\hbar=c=1$ changes only the bookkeeping. The tell that someone has misunderstood is the phrase "but what if $\hbar$ isn't really $1$ ?" — $\hbar$ is never numerically $1$ in metres and seconds; the statement is that we have chosen units in which its numerical value is $1$ , which we are always free to do.

Worked examples

1 · One mass, read in five units at once (confidence builder)

The electron rest energy is $m_e = 0.511\ \text{MeV} = 5.11\times10^{-4}\ \text{GeV}$ . Express its mass in kilograms, and its inverse mass $1/m_e$ as a length and as a time.

Step 1. Mass in kilograms. In natural units $m_e$ is an energy; to get a mass in S.I. divide the energy by $c^2$ . Numerically $m_e c^2 = 0.511\ \text{MeV}=0.511\times1.602\times10^{-13}\ \text{J}=8.19\times10^{-14}\ \text{J}$ , so $m_e=8.19\times10^{-14}/(2.998\times10^{8})^2=\mathbf{9.11\times10^{-31}\ kg}$ — the tabulated electron mass, recovered by reinserting one factor of $c^2$ .

Step 2. Inverse mass as a length. The quantity $1/m_e$ has mass dimension $-1$ , so it is a length; reinsert $\hbar c$ . Then $1/m_e = \hbar c/(m_e c^2) = (197.3\ \text{MeV}\,\cdot\,\text{fm})/(0.511\ \text{MeV}) = \mathbf{386\ fm}=3.86\times10^{-13}\ \text{m}$ . This is the electron's reduced Compton wavelength, where relativistic quantum effects take over.

Step 3. Inverse mass as a time. The same $1/m_e$ read as a time needs $\hbar$ instead: $1/m_e = \hbar/(m_e c^2) = (6.582\times10^{-22}\ \text{MeV}\,\cdot\,\text{s})/(0.511\ \text{MeV}) = \mathbf{1.29\times10^{-21}\ s}$ . The single natural-units object $1/m_e$ is at once $386$ fm and $1.29\times10^{-21}$ s — the light-crossing time of the Compton wavelength — with $c$ tying the two together.

2 · A cross section from GeV⁻² to nanobarns (fully worked)

The lowest-order cross section for $e^+e^-\to\mu^+\mu^-$ is $\sigma=4\pi\alpha^2/3s$ , where $s$ is the squared centre-of-mass energy. Evaluate it at $\sqrt{s}=10\ \text{GeV}$ and convert to nanobarns.

Step 1. Work entirely in natural units first. With $\alpha=1/137.036$ and $s=(10\ \text{GeV})^2=100\ \text{GeV}^2$ , $\sigma = \dfrac{4\pi\alpha^2}{3s} = \dfrac{4\pi}{3\times137.036^2\times100\ \text{GeV}^2} = \mathbf{2.23\times10^{-6}\ GeV^{-2}}.$ Note the units: $\alpha$ is dimensionless and $s$ is $\text{GeV}^2$ , so $\sigma$ comes out in $\text{GeV}^{-2}$ — an area, exactly as the dimension rule demands.

Step 2. Reinsert $(\hbar c)^2$ to get millibarns: $\sigma = 2.23\times10^{-6}\times0.3894\ \text{mb}=8.69\times10^{-7}\ \text{mb}$ . Convert to nanobarns ( $1\ \text{mb}=10^{6}\ \text{nb}$ ): $\sigma=\mathbf{0.869\ nb}$ .

Step 3. Cross-check against the standard collider formula $\sigma = 86.8\ \text{nb}/(s\,[\text{GeV}^2])$ : at $s=100\ \text{GeV}^2$ this gives $0.868\ \text{nb}$ , agreeing to the rounding. σ(e⁺e⁻→μ⁺μ⁻) at 10 GeV ≈ 0.87 nb, a number a physicist reads off in seconds because the $\text{GeV}^{-2}$ result converts through one constant.

3 · A lifetime as a decay width

The muon lives $\tau_\mu = 2.197\times10^{-6}\ \text{s}$ . What is its decay width $\Gamma$ in natural units?

Step 1. A width is an inverse lifetime, $\Gamma=\hbar/\tau$ , and in natural units $\Gamma=1/\tau$ once $\tau$ is expressed as a $\text{GeV}^{-1}$ . Use $\hbar=6.582\times10^{-25}\ \text{GeV}\,\cdot\,\text{s}$ : $\Gamma = (6.582\times10^{-25}\ \text{GeV}\,\cdot\,\text{s})/(2.197\times10^{-6}\ \text{s}) = \mathbf{3.00\times10^{-19}\ GeV}=3.00\times10^{-10}\ \text{eV}$ .

Step 2. The tiny width reflects the muon's long life; the same arithmetic run backwards turns the $W$ boson's width $\Gamma_W=2.09\ \text{GeV}$ into a lifetime $\tau_W=\hbar/\Gamma_W=(6.582\times10^{-25})/2.09=\mathbf{3.15\times10^{-25}\ s}$ . Width and lifetime are one quantity ( $\text{GeV}^{+1}$ ) read two ways — the reason particle tables quote a "width" for short-lived states and a "lifetime" for long-lived ones interchangeably.

Physics of the portrait

Natural units encode a genuine idea about physics: that the constants of nature, not human artifacts, should set the scale of measurement. The person who first made that idea explicit was Max Planck, in the same burst of work that opened the quantum era.

Origins · 1 minMax Planck and how this was discovered

Max Planck

1858–1947 · the origin of natural units

In 1899, a year before his radiation law, Max Planck pointed out something quietly radical. Having identified a new constant of nature — the quantum of action later called $\hbar$ — he noticed that together with the speed of light $c$ , Newton's constant $G$ , and Boltzmann's constant $k$ , it allowed a system of units built entirely from nature, "independent of special bodies or substances", that "would necessarily retain their meaning for all times and for all cultures, even extraterrestrial and non-human ones". These are the Planck units, and they were the first system to take the fundamental constants themselves as the rulers. Planck's full set fixes $\hbar=c=G=k=1$ , which pins down an absolute length ( $1.6\times10^{-35}$ m), time, mass ( $M_{\text{Pl}}=1.22\times10^{19}$ GeV) and temperature — the scales where quantum gravity must matter. Particle physics adopts a deliberate subset of Planck's vision: it keeps $\hbar=c=1$ (and $k=1$ for temperature) but leaves $G$ alone, because gravity is negligible at accelerator energies, and it fixes the remaining freedom not with $G$ but with a laboratory unit, the electronvolt. The result is the working language of the field — one in which Planck's own constant has, fittingly, been set to one. Planck thus supplied both the first quantum and the first natural units, two ideas that particle physics fused into a single convention.

History

1874

— George Johnstone Stoney proposes the first natural-unit system, built from e, c and G

1893

— Oliver Heaviside rationalizes Maxwell's equations, moving the 4π into Coulomb's law

1899

— Max Planck defines units from the constants h, c, G and k, independent of human artifacts

1905

— Einstein's E = mc² makes mass and energy one quantity, the physical content of c = 1

1927

— Dirac's quantum theory of radiation makes ħ = c = 1 the natural language of field theory

Going further

Texts. Thomson, Modern Particle Physics, §2.1 (units in particle physics) and §2.1.1 (natural and Heaviside–Lorentz units), Table 2.1 · Griffiths, Introduction to Elementary Particles, 2nd ed., §1.3 and Appendix A (units) · Peskin & Schroeder, An Introduction to Quantum Field Theory, front-matter "Notation and Conventions" · Halzen & Martin, Quarks and Leptons, §1 and Appendix A · Schwartz, Quantum Field Theory and the Standard Model, §1. The conversion anchors follow from the CODATA constants: $\hbar c = 197.327\ \text{MeV}\,\cdot\,\text{fm}$ , $\hbar = 6.582\times10^{-25}\ \text{GeV}\,\cdot\,\text{s}$ , and $(\hbar c)^2 = 0.3894\ \text{mb}\,\cdot\,\text{GeV}^2$ (equivalently $1\ \text{GeV}^{-2}=0.3894\ \text{mb}$ ). Key numerical results in this article: the dimension rule $\text{GeV}^{a-b-c}$ applied to energy, area and velocity; $m_e=0.511\ \text{MeV}\Rightarrow9.11\times10^{-31}$ kg, reduced Compton wavelength $386$ fm, and light-crossing time $1.29\times10^{-21}$ s; $\sigma=4\pi\alpha^2/3s=2.23\times10^{-6}\ \text{GeV}^{-2}=0.869$ nb at $\sqrt{s}=10$ GeV, consistent with the collider formula $86.8\ \text{nb}/s$ ; $\Gamma_\mu=\hbar/\tau_\mu=3.00\times10^{-10}$ eV; and $e=\sqrt{4\pi\alpha}=0.303$ . Particle masses ( $m_p=0.938$ , $m_W=80.4$ , $m_t=173$ GeV) and $M_{\text{Pl}}=1.22\times10^{19}$ GeV are standard tabulated values.

Prerequisites. Special Relativity Kinematics · Non-relativistic Quantum Mechanics Review. Leads to. Four-vector Notation · Interaction Cross Sections · Introduction to QED.

Ask the tutor: Ask me to convert any particle-physics quantity — a mass, a lifetime, a cross section — back and forth between natural units and S.I., or to show how the $\hbar$ 's and $c$ 's reappear in a formula you know.

Isn't setting ħ = c = 1 just a lazy shorthand that a careful physicist should avoid?

It is the opposite of lazy — it is a precise choice of units that most theorists consider the careful way to work. Carrying explicit $\hbar$ 's and $c$ 's through a multi-line quantum-field-theory calculation is a reliable source of algebra errors, and those constants convey nothing physical: they are exchange rates between the metre, the second and the joule. By choosing units in which they equal one, you let the physics — masses, momenta, couplings — stand unobscured, and you restore metres or seconds only at the very end, uniquely, through dimensional analysis. Every standard graduate text works this way for exactly this reason. The one discipline it demands is that you keep track of a quantity's mass dimension, which is easy once you internalize the rule $a-b-c$ .

Why the GeV specifically, rather than joules, ergs or the electronvolt?

Because the GeV sits at the natural scale of the subject. The proton mass is $0.938$ GeV, the electroweak bosons are tens of GeV, and hadron and quark masses populate the range in between, so masses and energies come out as convenient numbers of order one to a few hundred, rather than the tiny fractions you would get in joules ( $1\ \text{GeV}=1.6\times10^{-10}$ J) or the unwieldy large numbers you would get in electronvolts for heavy states. The choice is pure convenience: the physics is identical whether you call the base unit the eV, MeV, GeV or TeV — nuclear physics often prefers MeV, collider phenomenology increasingly uses TeV. What matters is that there is a single energy unit, not which prefix you attach.

How do I convert a cross section in GeV⁻² to a real area in cm² or barns?

Multiply by $(\hbar c)^2$ . The clean anchor is $(\hbar c)^2 = 0.3894\ \text{GeV}^2\text{mb}$ , which says $1\ \text{GeV}^{-2}=0.3894\ \text{mb}=3.894\times10^{-28}\ \text{cm}^2=3.894\times10^{-32}\ \text{m}^2$ . So a cross section of, say, $2.2\times10^{-6}\ \text{GeV}^{-2}$ becomes $2.2\times10^{-6}\times0.3894\ \text{mb}\approx8.6\times10^{-7}\ \text{mb}=0.86\ \text{nb}$ . The barn hierarchy is worth memorizing: $1\ \text{barn}=10^{-24}\ \text{cm}^2$ , and the common subunits are the millibarn ( $10^{-3}$ ), microbarn, nanobarn, picobarn and femtobarn — the last being the currency of LHC results.

Is setting Boltzmann's constant k = 1 part of natural units too?

It is a natural extension, used routinely in cosmology and thermal field theory, though not always bundled into the phrase "natural units" as used at colliders. Setting $k=1$ identifies temperature with energy, so a plasma at temperature $T$ has typical particle energies $\sim T$ measured directly in eV or GeV. This is why cosmologists say the early universe was "at 100 GeV" at the electroweak epoch or that the CMB is " $2.35\times10^{-4}$ eV" today — those are temperatures expressed as energies. To recover kelvin you reinsert $k$ via $1\ \text{eV}=1.16\times10^{4}\ \text{K}$ . The pattern is the same as for $\hbar$ and $c$ : a constant that is only a unit-conversion factor is set to one, and its job is taken over by the dimensions.

What is the difference between these units and Planck units?

Both are natural-unit systems, but they fix a different number of constants. Particle physics sets $\hbar=c=1$ (two constants), which leaves one unit still free — filled by a chosen energy scale, the eV or GeV. Planck units go further and also set Newton's constant $G=1$ (and often $k=1$ ), which uses up the last freedom and leaves no dimensionful units at all: length, time, mass and temperature all take absolute values, the famous Planck length $1.6\times10^{-35}$ m and Planck mass $1.22\times10^{19}$ GeV. Particle physicists deliberately stop short of $G=1$ because gravity is irrelevant at accelerator energies, so pinning the scale to $G$ would give absurdly large or small numbers. Planck units are the tool when gravity matters — quantum gravity, black-hole thermodynamics, the very early universe.

How do I know which powers of ħ and c to put back when converting an answer?

You never guess — the powers are forced by dimensional analysis. Write your natural-units result as a number times $\text{GeV}^{n}$ , where $n$ is the quantity's mass dimension (energy $+1$ , length or time $-1$ , area $-2$ , and so on). Then find the unique product $\hbar^{p}c^{q}$ whose dimensions convert $\text{GeV}^{n}$ into the S.I. dimension you want. Because you are matching two independent things — the length content and the time content — you get two equations for the two unknowns $p$ and $q$ , so the solution is unique. In practice you rarely solve equations: you just reach for the right anchor — $\hbar c$ for a length, $\hbar$ for a time, $(\hbar c)^2$ for a cross section, $c^2$ to turn an energy into a mass — because those cover almost every case.