Unruh Effect and Thermal Radiation for Accelerated Observers

Accelerate through empty space and it stops being empty. Not because you have disturbed anything — you have not — but because "empty" was never an observer-independent word. A uniformly accelerated detector, moving through the Minkowski vacuum — the state an inertial observer calls nothing at all — clicks. It clicks with a Planck spectrum, at a temperature fixed entirely by its own acceleration:

\boxed{\;k_B T_U \;=\; \frac{\hbar\,a}{2\pi c}\;}

Two observers, one state, and they disagree about how many particles are in it. Nothing settles the disagreement, because particle number is not an invariant. It is defined by a choice of what counts as positive frequency; that requires a clock; and the two of them have different clocks. The derivation below needs no Bogoliubov coefficients and no detector — just one calculation: write the vacuum's two-point function along a hyperbolic worldline and look at it. A piece cancels that had no business cancelling, and what is left is periodic in imaginary time — which is the definition of a thermal state. Then the price. $T_U = \mathbf{4.055\times10^{-21}}$ K per $\mathrm{m/s^2}$ . At $1\,g$ : $4\times10^{-20}$ K — the cosmic microwave background is 69 billion billion times hotter. One kelvin costs $2.47\times10^{20}\ \mathrm{m/s^2}$ . Fifty years on, nobody has seen it.

Key takeaways

Signature $(+,-,-,-)$ : $\mathrm{d}s^2 = c^2\mathrm{d}t^2 - \mathrm{d}x^2$ , timelike positive. The base prints the metric and the interval in mostly-plus; both are re-derived here.
The result. A detector with constant proper acceleration $a$ in the Minkowski vacuum registers a thermal bath at $T_U = \hbar a/2\pi ck_B$ . Not a coordinate artefact: the detector really clicks.
Owned derivation — KMS / imaginary-time periodicity. Evaluate $\langle 0_M|\phi(x)\phi(x')|0_M\rangle$ on the hyperbola. (i) The dependence on where you are along the worldline CANCELS — an accelerating observer sees a stationary state, which is not obvious and is the whole ballgame. (ii) What survives, $\propto 1/\sinh^2(a\Delta\tau/2c)$ , is periodic in imaginary proper time with period $2\pi c/a$ . Match to the KMS period $\hbar\beta$ and read off $T_U$ . No mode expansion. No detector.
STATED, NOT DERIVED — deliberately. The Bogoliubov result ( $\langle \hat n_\omega\rangle = 1/(e^{2\pi c\omega/a}-1)$ ) and the Unruh–DeWitt detector response are quoted, correct, and not derived: a KMS state satisfies detailed balance by definition, so both are already contained in the two-point function.
What it costs. $\hbar/2\pi ck_B = \mathbf{4.055\times10^{-21}}\ \mathrm{K\,s^2/m}$ ; at $1\,g$ , $\mathbf{3.98\times10^{-20}}$ K. One kelvin needs $\mathbf{2.47\times10^{20}\ m/s^2}$ . LEP1's electrons ( $\gamma = 89{,}237$ , $a = 2.31\times10^{23}\ \mathrm{m/s^2}$ ) carry a formal $T_U$ of $\mathbf{937}$ K — a number, not a measurement.
The deep statement, and why the portrait is the vacuum. $|0_M\rangle$ is pure. Trace out what the horizon hides and what remains is mixed and exactly thermal: $\rho_R \propto e^{-2\pi c H_R/\hbar a}$ (Bisognano–Wichmann). The heat is entanglement, not creation. The horizon does not make particles; it makes you ignorant.
Attribution, corrected. The base credits Unruh (1976) alone. It is properly the Fulling–Davies–Unruh effect: Fulling 1973, Davies 1975, Unruh 1976.
⚠ Hawking is gestured at, not annexed. $a \to \kappa = c^4/4GM$ gives $T_H = \hbar\kappa/2\pi ck_B$ exactly. The black hole belongs to the general-relativity track. Special relativity owns the flat-space statement: a horizon is thermal, and you do not need gravity to build one.

The core insight

The vacuum is not a thing that is empty. It is a state, and to count the particles in a state you must first say what a particle is — which means saying which oscillations count as positive-frequency, which means choosing a clock. The two observers' clocks disagree, so their notions of "no particles" disagree, and there is no third clock to arbitrate. What the accelerated observer sees is neither an illusion nor a disturbance: it is the same state, honestly counted, with a different definition of counting. That the answer comes out exactly thermal — Planckian, at a temperature set by nothing but the acceleration — is the surprise, and the reason this result sits underneath black-hole thermodynamics, the information paradox, and every serious attempt at quantum gravity.

Where this lives today

It is the reason "particle" is not fundamental. Quantum field theory in curved spacetime begins by conceding that particle number is observer-dependent, and the Unruh effect is the cleanest proof — because it happens in flat spacetime, where nobody can blame the curvature. The Bunch–Davies vacuum of inflation is thermal for the same reason (de Sitter has a horizon), and the primordial fluctuation spectrum that seeded every galaxy is, at bottom, a horizon-temperature effect.

The experimental programme is alive and has not succeeded. Petawatt lasers can drive electrons to $\sim10^{25}\ \mathrm{m/s^2}$ and $T_U \approx 4\times10^{4}$ K — the leading candidate, but the signature must be dug out from beneath a colossal Larmor background. The effect itself, in the actual vacuum, has never been seen. Fifty years, and counting.

The physical picture

A uniformly accelerated observer has a horizon. `aq-sr-rindler-horizon` establishes it: run the hyperbola $x^2 - c^2t^2 = c^4/a^2$ forever, and a null plane sits behind you from which no signal can ever reach you. You are permanently ignorant of half of Minkowski space — not because it is far away, but because it is causally sealed. Now take the vacuum seriously as a quantum state. $|0_M\rangle$ is not a picture of nothing; it is a highly structured, highly entangled object whose correlators are non-zero across the horizon. The left and right wedges of Minkowski space are entangled in the vacuum, the way two halves of an EPR pair are. Here is the whole mechanism. You cannot access the left wedge, so you must trace it out — and the reduced density matrix of half an entangled pure state is mixed, necessarily, as a theorem of quantum mechanics.

|0_M\rangle\ \text{is PURE.} \qquad \rho_R = \operatorname{Tr}_{\text{L}} |0_M\rangle\langle 0_M|\ \text{is MIXED.}

That is not a heuristic but the Bisognano–Wichmann theorem, and the mixed state it produces is not merely mixed but exactly Gibbsian: $\rho_R \propto e^{-2\pi c H_R/\hbar a}$ — a Boltzmann distribution in the Rindler Hamiltonian, at precisely $\hbar a / 2\pi c k_B$ . So the heat has a source, and it is not the field being stirred up. The heat is your ignorance. The horizon does not create particles; it hides correlations, and hidden correlations are indistinguishable from thermal noise. The energy to run your detector comes from whatever is pushing you. The rocket pays.

Figure

THE KMS MIRACLE — how thermality falls out of a hyperbola with no quantum field theory beyond the vacuum two-point function. LEFT: the accelerated observer's worldline is a hyperbola in Minkowski space, x² − c²t² = c⁴/a², with a horizon (the null line) behind it that seals off half of spacetime forever. Two events on the worldline are labelled by proper times τ₁ and τ₂. Naively the interval between them should depend on BOTH — on their separation Δτ = τ₁ − τ₂ AND on where along the accelerating worldline they sit, Σ = (τ₁ + τ₂)/2, because an accelerating observer has no time-translation symmetry to protect it. CENTRE: it does not. Write the interval out and the Σ dependence CANCELS IDENTICALLY: c²Δt² − Δx² = (4c⁴/a²) sinh²(aΔτ/2c), a function of Δτ alone. The accelerated observer therefore sees a STATIONARY state, and that cancellation is the miracle the whole derivation turns on. RIGHT: what survives, the two-point function G ∝ 1/sinh²(aΔτ/2c), is PERIODIC IN IMAGINARY TIME — because sinh² has period iπ in its argument, G is unchanged under Δτ → Δτ − i·2πc/a. Periodicity in imaginary time with period ħβ is not merely SUGGESTIVE of a thermal state: by the KMS condition it IS the definition of one. Match the periods, ħβ = 2πc/a, and the Unruh temperature is read straight off with no Bogoliubov coefficients and no detector model anywhere in the calculation

The core equations

\boxed{\;T_U \;=\; \frac{\hbar\,a}{2\pi c\,k_B}\;} \qquad\qquad \frac{\hbar}{2\pi c\,k_B} \;=\; \mathbf{4.055\times10^{-21}}\ \mathrm{K\,s^{2}/m}

where: $a$ is the proper acceleration — the invariant, the accelerometer reading (`aq-sr-proper-acceleration-comoving-frame`), not a coordinate acceleration and not a velocity. The observer rides the hyperbola $ct = \tfrac{c^2}{a}\sinh\tfrac{a\tau}{c}$ , $x = \tfrac{c^2}{a}\cosh\tfrac{a\tau}{c}$ (`aq-sr-relativistic-uniform-acceleration`), which comes with a horizon at proper distance $c^2/a$ . Physical reading: that constant is the entire experimental problem.

\boxed{\;\Delta s^{2} \;=\; c^{2}\Delta t^{2} - \Delta x^{2} \;=\; +\,\frac{4c^{4}}{a^{2}}\,\sinh^{2}\!\left(\frac{a\,\Delta\tau}{2c}\right)\;}

where: $\Delta\tau = \tau_1 - \tau_2$ separates two events on the worldline. Physical reading: the load-bearing equation of this article, and the one the base gets wrong. The sign is positive, i.e. timelike — compulsory for two events on one timelike worldline — and $\to c^2\Delta\tau^2$ as $\Delta\tau \to 0$ . The base prints it negative, which in $(+,-,-,-)$ asserts $\Delta\tau^2 < 0$ . Re-derived below, never transcribed.

G(\Delta\tau) \propto \frac{1}{\sinh^{2}\!\big(\tfrac{a}{2c}(\Delta\tau - i\epsilon)\big)} \qquad\quad \boxed{\;G(\Delta\tau - i\,\hbar\beta) = G(\Delta\tau), \quad \hbar\beta = \frac{2\pi c}{a}\;} \qquad\quad \langle \hat n_\omega \rangle = \frac{1}{e^{2\pi c\,\omega/a} - 1}

where: $G$ is the massless Wightman function on the hyperbola, $\beta = 1/k_BT$ , and $\hat n_\omega$ counts Rindler quanta. Physical reading: $G$ depends on $\Delta\tau$ alone — an observer with no time-translation symmetry whatsoever sees a stationary state, and everything hangs on that. KMS periodicity is then the definition of thermality — and mind the units, because the base does not: $\hbar\beta$ is a time, and so is $2\pi c/a$ ; $\beta$ is an inverse energy. The base's card sets $\beta = 2\pi c/a$ , equating an inverse energy to a time. The Planck factor on the right is the Bogoliubov result: stated, not derived, because a state periodic in imaginary time with period $\hbar\beta$ satisfies detailed balance at $1/k_B\beta$ by definition.

Deriving the temperature

Derivation AThe KMS derivation — thermality from a hyperbola, in five lines and no mode expansion

There are three routes to the Unruh temperature — Bogoliubov coefficients, a detector model, and this one. This card takes the third and shortest: write down the vacuum's two-point function, restrict it to the accelerated worldline, and look at what you get. No modes, no detector, no quantisation scheme. The thermality will be forced on us by the geometry of the hyperbola, through a cancellation that has no right to happen — and that cancellation, not the algebra, is what the card is for.

Step 1. Put the observer on the hyperbola — `aq-sr-relativistic-uniform-acceleration`'s result, recalled:

ct(\tau) = \frac{c^{2}}{a}\sinh\frac{a\tau}{c}, \qquad x(\tau) = \frac{c^{2}}{a}\cosh\frac{a\tau}{c}.

Why: $\tau$ is the observer's own clock, the only one they have. "What does this observer see?" can only be asked in the observer's own time.

Step 2. Take two events on the worldline. Write $\Delta\tau = \tau_1 - \tau_2$ and $\Sigma = \tfrac12(\tau_1 + \tau_2)$ , and apply $\sinh A - \sinh B = 2\cosh\frac{A+B}{2}\sinh\frac{A-B}{2}$ , $\cosh A - \cosh B = 2\sinh\frac{A+B}{2}\sinh\frac{A-B}{2}$ :

c\,\Delta t = \frac{2c^{2}}{a}\cosh\!\Big(\frac{a\Sigma}{c}\Big)\sinh\!\Big(\frac{a\Delta\tau}{2c}\Big), \qquad \Delta x = \frac{2c^{2}}{a}\sinh\!\Big(\frac{a\Sigma}{c}\Big)\sinh\!\Big(\frac{a\Delta\tau}{2c}\Big).

Why: look at what just happened. Both factorised into (a function of $\Sigma$ ) × (a function of $\Delta\tau$ ). Nothing guaranteed that.

Step 3. Form the invariant in $(+,-,-,-)$ and watch $\Sigma$ die — because $\cosh^2 - \sinh^2 = 1$ , identically:

\Delta s^{2} = c^{2}\Delta t^{2} - \Delta x^{2} = \frac{4c^{4}}{a^{2}}\sinh^{2}\!\Big(\frac{a\Delta\tau}{2c}\Big)\Big[\cosh^{2}\!\frac{a\Sigma}{c} - \sinh^{2}\!\frac{a\Sigma}{c}\Big] \;=\; \boxed{\;+\,\frac{4c^{4}}{a^{2}}\,\sinh^{2}\!\Big(\frac{a\,\Delta\tau}{2c}\Big)\;}

Why: this is the miracle; everything else is bookkeeping. An accelerating observer has no time-translation symmetry, and there is no reason on earth why the correlation between two events should be blind to where on the trajectory they sit. But $\Sigma$ has cancelled. The interval — and so the correlator — depends on $\Delta\tau$ alone. The accelerated observer sees a STATIONARY state, and stationarity is the precondition for a temperature: you cannot assign one to something still changing. The hyperbola hands it to us free. Note the sign: $\Delta s^2 > 0$ , timelike, compulsory, $\to c^2\Delta\tau^2$ as $\Delta\tau \to 0$ . The base prints it negative, asserting $\Delta\tau^2 < 0$ . Refused.

Step 4. Restrict the vacuum two-point function to the worldline. For a massless scalar the Wightman function goes as the inverse interval (the $i\epsilon$ fixes positive frequency):

G(\Delta\tau) = \langle 0_M|\phi(x_1)\phi(x_2)|0_M\rangle \;\propto\; \frac{1}{\Delta s^{2}}\Big|_{t \to t - i\epsilon} \;\propto\; \frac{1}{\sinh^{2}\!\big(\tfrac{a}{2c}(\Delta\tau - i\epsilon)\big)} .

Why: $1/\Delta s^2$ is the free massless propagator, and it is the only input from quantum field theory in this entire derivation — everything else was trigonometry. We have not chosen a vacuum for the accelerated observer, not expanded in modes, not introduced a detector. We took the inertial observer's global, Poincaré-invariant $|0_M\rangle$ and asked what it looks like along a hyperbola.

Step 5. Read off the temperature. $\sinh^2(z)$ is periodic under $z \to z + i\pi$ ; with $z = a\Delta\tau/2c$ that is $\Delta\tau \to \Delta\tau + i\,2\pi c/a$ , and $G$ is unchanged. The KMS condition says a thermal correlator at inverse temperature $\beta$ is periodic in imaginary time with period $\hbar\beta$ . Ours has period $2\pi c/a$ . Therefore

\hbar\beta = \frac{2\pi c}{a} \qquad\Longrightarrow\qquad \beta = \frac{2\pi c}{\hbar a} \qquad\Longrightarrow\qquad \boxed{\;k_B T_U = \frac{\hbar\,a}{2\pi c}\;}

Why: that is the whole derivation. Not "resembles a thermal state", not "behaves as if". In algebraic quantum field theory KMS periodicity is not evidence of thermality — it is the definition, and the one that survives when the naive $e^{-\beta H}/Z$ does not (here the Rindler Hamiltonian has no normalisable ground state). The Minkowski vacuum on a uniformly accelerated worldline is a thermal state at $T_U$ , exactly. Mind the units, where the base does not: $\hbar\beta$ is a time, $\beta$ an inverse energy. *Verified in verify_aq-sr-unruh-effect.py: the $\Sigma$ -cancellation is checked numerically at fifty digits for arbitrary $\tau_1, \tau_2$ and symbolically with sympy; $\Delta s^2 > 0$ and $\Delta s^2/(c^2\Delta\tau^2) \to 1$ are both asserted (the base's sign is REFUSED); the imaginary-time periodicity is verified in the complex plane; and $\hbar\beta = 2\pi c/a$ is checked against $\hbar/(k_B T_U)$ to fifty digits.*

Go deeper · Derivation B · 2 minWhat one kelvin costs — and why fifty years have produced no detection

Derivation BWhat one kelvin costs — and why fifty years have produced no detection

A formula with $\hbar$ upstairs and $c$ downstairs is going to be small; the only question is how small. This card puts the number on it and is then honest about what the number means: the Unruh effect is, today, entirely unobserved, and every claim to the contrary is either about an analogue system or about a signal so entangled with ordinary radiation-reaction physics that separating them is the whole problem. Five steps, and the last is the one that matters.

Step 1. Assemble the constant. With $\hbar = 1.054572\times10^{-34}$ J s, $c = 2.99792458\times10^{8}$ m/s, $k_B = 1.380649\times10^{-23}$ J/K:

\frac{\hbar}{2\pi c\,k_B} = \frac{1.054572\times10^{-34}}{2\pi (2.99792458\times10^{8})(1.380649\times10^{-23})} = \mathbf{4.0550\times10^{-21}}\ \mathrm{K\,s^{2}/m} .

Why: one number, and the rest is arithmetic. Every $\mathrm{m/s^2}$ buys four parts in $10^{21}$ of a kelvin.

Step 2. Stand on the Earth: $a = g = 9.80665\ \mathrm{m/s^2}$ gives $T_U(1\,g) = \mathbf{3.98\times10^{-20}}\ \mathrm{K}$ . Why: the cosmic microwave background is $2.725$ K — $\mathbf{6.9\times10^{19}}$ times hotter. Sitting in your chair you are in an Unruh bath sixty-nine billion billion times colder than the sky. This is why nobody has noticed.

Step 3. Ask what one kelvin costs. Invert: $a\big|_{T_U = 1\,\mathrm{K}} = 2\pi ck_B/\hbar = \mathbf{2.466\times10^{20}}\ \mathrm{m/s^{2}}$ . Why: that is $2.5\times10^{19}\,g$ , and nothing you can build sustains it. The effect is not weak because the coupling is weak; it is weak because $\hbar/c$ is small, and there is no way to make it bigger.

Step 4. Find the most violently accelerated stable things we have made: electrons in LEP1 at the $Z$ pole, $E = 45.6$ GeV per beam, so $\gamma = 45.6\ \mathrm{GeV}/0.510999\ \mathrm{MeV} = \mathbf{89{,}237}$ ; on a main-dipole bending radius $\rho \approx 3096$ m the proper acceleration is $a = \gamma^2c^2/\rho$ :

a = \frac{(89{,}237)^{2}(2.998\times10^{8})^{2}}{3096} = \mathbf{2.31\times10^{23}}\ \mathrm{m/s^{2}} \qquad\Longrightarrow\qquad T_U = \mathbf{937}\ \mathrm{K} .

Why: a real number attached to a real machine. LEP's electrons formally sat in a vacuum at nearly a thousand kelvin for a decade, in the most scrutinised experiment of its era.

Step 5. Be honest about it. 937 K is a number, not a measurement. The $\gamma^2/\rho$ acceleration is circular, not linear: the worldline is not a hyperbola, the state is not KMS, and the "temperature" is not uniquely fixed by the acceleration. Bell and Leinaas (1983, 1987) argued that the residual polarisation of storage-ring electrons — the Sokolov–Ternov effect, saturating at $8/(5\sqrt3) = \mathbf{92.4\%}$ rather than $100\%$ — carries a genuine Unruh-like contribution. Leinaas' own later verdict on whether that constitutes an observation was "a qualified yes", and the qualifications are severe: the circular-motion temperature depends on the detector, and the correct polarisation curve deviates measurably from any naive Unruh prediction. Nobody has seen the Unruh effect. Why: worth stating flatly, because the literature is full of near-misses. Petawatt lasers at $10^{25}\ \mathrm{m/s^2}$ would give $T_U \approx 4\times10^{4}$ K — but the Larmor radiation is enormous and the signature must be dug out from beneath it. Half a century on, the effect in the vacuum remains unobserved. *Verified in verify_aq-sr-unruh-effect.py: $\hbar/2\pi ck_B$ , $T_U(1g)$ , the one-kelvin acceleration, the CMB ratio, $\gamma$ , the LEP1 proper acceleration and its $937$ K are all recomputed at fifty digits and grepped back out of this file; $8/(5\sqrt3) = 0.92376$ is asserted; artefact guards perturb $\hbar$ , $c$ , $k_B$ and $a$ and confirm the temperature moves linearly in $\hbar$ and in $a$ and inversely in $c$ .*

Go deeper · Derivation C · 2 minWhy the vacuum is the portrait — the heat is entanglement, not creation

Derivation CWhy the vacuum is the portrait — the heat is entanglement, not creation

It is tempting to read the Unruh effect as the acceleration shakes particles out of the vacuum, and every part of that sentence is wrong. Nothing is shaken, nothing is created, and the field is in exactly the state it was in. What changes is what you can see — and this card shows that the temperature is a precise measure of what you cannot. It is the deepest statement in the article, and the reason the portrait is the vacuum state rather than the effect that bears Unruh's name.

Step 1. Split Minkowski space at the horizon. The observer's Rindler wedge is the right wedge $\mathrm R$ ( $x > |ct|$ ); behind the horizon lies the causally inaccessible left wedge $\mathrm L$ , and the Hilbert space factorises, $\mathcal H \simeq \mathcal H_{\mathrm L} \otimes \mathcal H_{\mathrm R}$ . Why: a statement about access, not distance. The left wedge is a finite distance off but causally sealed — a stronger disability. `aq-sr-rindler-horizon` owns the seal.

Step 2. But $|0_M\rangle$ does not factorise: it is not a product $|0\rangle_{\mathrm L}\otimes|0\rangle_{\mathrm R}$ . The vacuum two-point function between a point in $\mathrm L$ and a point in $\mathrm R$ is non-zero, and non-zero correlators across a cut are entanglement. Why: this is the physical content of "the vacuum is not empty". Not that there are particles in it — that it has structure: long-range correlations straddling any surface you draw, including the one you cannot see past.

Step 3. Trace out what you cannot access: $\rho_{\mathrm R} = \operatorname{Tr}_{\mathrm L}|0_M\rangle\langle 0_M|$ . And now the theorem: the reduced state of half an entangled pure state is mixed — exactly, as a consequence of the entanglement being non-zero. Why: $|0_M\rangle$ is pure, with zero entropy. $\rho_{\mathrm R}$ has non-zero von Neumann entropy, and that entropy came from nowhere. It is the entanglement entropy across the horizon, and it is the entire source of the heat.

Step 4. Evaluate the trace. The Bisognano–Wichmann theorem (1975–76), proved from the Wightman axioms alone, says the answer is not merely mixed but exactly Gibbsian:

\boxed{\;\rho_{\mathrm R} \;=\; \frac{1}{Z}\,\exp\!\left(-\frac{2\pi c}{\hbar\,a}\,H_{\mathrm R}\right) \;=\; \frac{1}{Z}\,e^{-H_{\mathrm R}/k_B T_U}\;}

where $H_{\mathrm R}$ generates Rindler time translations (boosts). Why: a Boltzmann distribution, and its temperature is the Unruh temperature. It arrived by a completely different road — no worldline, no correlator, no $\sinh$ — and agrees exactly. Two independent derivations landing on the same $2\pi$ is how you know it is an artefact of neither.

Step 5. The conclusion. The heat is not a property of the field; it is a property of the partition. An inertial observer, seeing both wedges, finds a pure state with zero entropy and no particles — and she is not wrong. The accelerated observer, who cannot, sees a thermal bath — and he is not wrong either. Both are right. There is no fact of the matter about how many particles are present. Why: this is why "particle" cannot be fundamental. A particle is a positive-frequency mode; positive frequency needs a time coordinate; a time coordinate is a choice. The field is real; the particles are an accounting convention. And the horizon is the ledger you are not allowed to read. *Verified in verify_aq-sr-unruh-effect.py: the Bisognano–Wichmann temperature $\hbar a/2\pi ck_B$ read off from $\rho_R \propto e^{-2\pi cH_R/\hbar a}$ is asserted to equal, to fifty digits, the $T_U$ obtained from the KMS route — two independent roads, one number; and a two-mode squeezed state (the finite-dimensional caricature of $|0_M\rangle$ ) is built explicitly, its reduced density matrix computed, and its purity, entropy and thermal occupation checked against $1/(e^{2\pi c\omega/a}-1)$ .*

Limiting cases

Case	Result	Meaning
$\hbar \to 0$	$T_U \to 0$	classically there is no effect at all. The vacuum has no structure to hide
$a \to 0$	$T_U \to 0$	no acceleration, no horizon, nothing traced out, nothing thermal
$a = g$	$T_U = 3.98\times10^{-20}$ K	the CMB is $6.9\times10^{19}\times$ hotter. Why you have never noticed
$a = 2.47\times10^{20}$	$T_U = 1$ K	the price of one kelvin. $2.5\times10^{19}\,g$
$a \to \kappa = c^4/4GM$	$T_U \to T_H$	Hawking, exactly. Gestured at; the GR track owns it

Figure

WHAT ONE KELVIN COSTS — the Unruh temperature is 4.055 × 10⁻²¹ kelvin for every m/s² of proper acceleration, and that single constant is the whole experimental problem. The scale runs across twenty-six orders of magnitude of acceleration. At 1 g — you, sitting in a chair — the Unruh bath is 3.98 × 10⁻²⁰ K, which is 6.9 × 10¹⁹ times colder than the cosmic microwave background at 2.725 K. A thermal photon at that temperature has a wavelength larger than the Milky Way; your detector would have to be a galaxy. To reach ONE KELVIN you need 2.47 × 10²⁰ m/s², which is two and a half times ten to the nineteen gravities, and nothing anyone can build sustains it. The most violently accelerated stable objects humans have made are electrons in the LEP1 storage ring: at 45.6 GeV per beam the Lorentz factor is 89,237, and on a 3096 m bending radius the proper acceleration is 2.31 × 10²³ m/s², for a formal Unruh temperature of 937 K — hotter than a pizza oven. But that is a NUMBER, not a MEASUREMENT: circular motion is not uniform acceleration, the correlator is not KMS, and the temperature is not fixed by the acceleration alone. Petawatt lasers reaching 10²⁵ m/s² would give 4 × 10⁴ K, and are the leading candidate — but the signal must be dug out from beneath a colossal Larmor background. Fifty years after Fulling wrote it down, nobody has observed the Unruh effect

Common trapFour things in the base of this article are wrong, and one of them is load-bearing

1 · The interval has the wrong sign — and this one is load-bearing. The base's KMS card prints

\Delta s^{2} = -c^{2}\Delta t^{2} + \Delta x^{2} = -\frac{4c^{4}}{a^{2}}\sinh^{2}\!\Big(\frac{a\Delta\tau}{2c}\Big).

That is mostly-plus. In $(+,-,-,-)$ it is false, and not harmlessly: it asserts that two events on the same timelike worldline are spacelike-separated, and that $\Delta\tau^2 < 0$ . The correct statement is $\Delta s^2 = +\,(4c^4/a^2)\sinh^2(a\Delta\tau/2c)$ — positive, timelike, reducing to $c^2\Delta\tau^2$ as $\Delta\tau \to 0$ . Re-derived above; never transcribed.

2 · The Rindler metric is mostly-plus. The base prints $\mathrm{d}s^2 = -(1 + a\xi/c^2)^2c^2\mathrm{d}\eta^2 + \mathrm{d}\xi^2$ . In $(+,-,-,-)$ it reads $+(1 + a\xi/c^2)^2c^2\mathrm{d}\eta^2 - \mathrm{d}\xi^2$ . `aq-sr-rindler-metric` owns the chart and states it correctly.

3 · The KMS card is dimensionally false. It writes $\beta = \frac{2\pi c}{\hbar a}\cdot\hbar = \frac{2\pi c}{a} = \frac{1}{k_B T_U}$ . But $2\pi c/a$ is a time and $1/k_BT_U$ is an inverse energy. What equals $2\pi c/a$ is $\hbar\beta$ , not $\beta$ — and the card's own header, $G(\tau) = G(\tau - i\hbar\beta)$ , says so. It contradicts itself in two adjacent lines.

4 · The Bogoliubov occupation has a broken middle step. The base writes $\langle \hat n_\omega\rangle = |\beta_\omega|^2/(1 - |\beta_\omega/\alpha_\omega|^2)$ . With $|\beta/\alpha|^2 = e^{-2\pi c\omega/a} \equiv X$ and $|\alpha|^2 - |\beta|^2 = 1$ , one has $|\beta|^2 = X/(1-X)$ — which already is the Planck factor. Dividing again by $(1-X)$ gives $X/(1-X)^2$ , which is not. Final answer right, middle step wrong. (This article states the result rather than deriving it, so the error is quarantined — but logged.)

Worked examples

1 · The Unruh temperature of a rocket at 1 g

Step 1 — identify. $T_U = a \times 4.0550\times10^{-21}$ K per $\mathrm{m/s^2}$ , with $a = 9.80665\ \mathrm{m/s^2}$ .

Step 2 — apply. $T_U = 9.80665 \times 4.0550\times10^{-21} = \mathbf{3.977\times10^{-20}}\ \mathrm{K}$ .

Step 3 — the wavelength. A thermal photon at that temperature has $\lambda \sim hc/k_BT \sim 4\times10^{5}$ light-years — larger than the Milky Way. Your detector would have to be a galaxy.

Step 4 — compare to the background. $T_{\text{CMB}}/T_U(1g) = 2.725 / 3.977\times10^{-20} = \mathbf{6.85\times10^{19}}$ .

Step 5 — the boundary number. Nothing at $1\,g$ could ever, even in principle, resolve a single quantum of it.

2 · The Hawking temperature of a solar-mass black hole

Step 1 — identify. Substitute the surface gravity for the acceleration: $\kappa = c^4/4GM$ , then $T_H = \hbar\kappa/2\pi ck_B$ — the Unruh formula, unchanged.

Step 2 — the surface gravity. For $M = 1\,M_\odot = 1.989\times10^{30}$ kg,

\kappa = \frac{c^{4}}{4GM} = \frac{8.078\times10^{33}}{4(6.674\times10^{-11})(1.989\times10^{30})} = \mathbf{1.521\times10^{13}}\ \mathrm{m/s^{2}} .

Step 3 — the temperature. $T_H = 1.521\times10^{13} \times 4.055\times10^{-21} = \mathbf{6.17\times10^{-8}}\ \mathrm{K}$ .

Step 4 — check against Hawking's own formula. $T_H = \hbar c^3/8\pi GMk_B$ gives $\mathbf{6.17\times10^{-8}}$ K. Identical — not to within a factor; exactly. The $2\pi$ is the same $2\pi$ .

Step 5 — the boundary number. A solar-mass hole is $6\times10^{-8}$ K — forty million times colder than the CMB; it absorbs far more than it radiates. But that number came out of a flat-space calculation with no gravity in it. That is the whole point of this article.

Origins · 2 minVacuum state and how this was discovered

Vacuum state

1973–1976 · Fulling finds it · Davies confirms it · Unruh explains what it means

The claim. The effect that carries Unruh's name was found by Stephen Fulling in 1973 (Phys. Rev. D 7, 2850), who noticed that quantising a field in Rindler coordinates gives a vacuum inequivalent to the Minkowski one — "the vacuum" was not a single object. Paul Davies (1975, J. Phys. A 8, 609), working the moving-mirror problem, showed the accelerated observer sees a thermal spectrum. William Unruh (1976, Phys. Rev. D 14, 870) supplied the interpretation that made it stick: an accelerating detector — a real device, with real clicks — thermalises at $\hbar a/2\pi ck_B$ , and this is the flat-space skeleton of Hawking's radiation, published two years earlier. It is properly the Fulling–Davies–Unruh effect; the base credits Unruh alone. What was at stake. Hawking's 1974 result was, to many, incomprehensible: a black hole radiating, apparently from nothing. Unruh showed that you do not need a black hole. Take flat, empty, boring Minkowski space, accelerate, and the same thing happens. The gravity was never the point. The horizon was the point — and a horizon is something an accelerometer can build. What survives. Everything downstream: black-hole entropy, the information paradox, holography, the entanglement-entropy programme, the Bunch–Davies vacuum of inflation. The sentence that generalised: a horizon is thermal, and the temperature is the entanglement you cannot see past. The effect itself has never been measured.

History

1973

— Stephen Fulling shows that quantisation in Rindler coordinates yields a vacuum inequivalent to the Minkowski one: "the vacuum" is not one thing (Phys. Rev. D 7, 2850).

1974

— Stephen Hawking announces that black holes radiate thermally at a temperature set by the surface gravity, and nobody quite believes it.

1975

— Paul Davies, working the moving-mirror problem, shows the accelerated observer sees a thermal spectrum (J. Phys. A 8, 609).

1975

— Bisognano and Wichmann prove, from the Wightman axioms alone, that the Minkowski vacuum restricted to a wedge is exactly Gibbsian in the boost generator.

1976

— William Unruh gives the detector interpretation and the temperature ħa/2πck(B), identifying it as the flat-space skeleton of Hawking radiation (Phys. Rev. D 14, 870).

1983

— Bell and Leinaas propose that the incomplete Sokolov–Ternov polarisation of storage-ring electrons carries an Unruh-like contribution (Nucl. Phys. B 212, 131).

2008

— Crispino, Higuchi and Matsas publish the definitive review (Rev. Mod. Phys. 80, 787). The effect remains unobserved.

Going further

Texts. Crispino, Higuchi & Matsas, Rev. Mod. Phys. 80, 787 (2008) — the definitive review · Birrell & Davies, Quantum Fields in Curved Space, §§3.3, 4.5 · Wald, QFT in Curved Spacetime and Black Hole Thermodynamics, ch. 5. Primary: Fulling, Phys. Rev. D 7, 2850 (1973) · Davies, J. Phys. A 8, 609 (1975) · Unruh, Phys. Rev. D 14, 870 (1976) · Bisognano & Wichmann, J. Math. Phys. 16, 985 (1975); 17, 303 (1976) · Bell & Leinaas, Nucl. Phys. B 212, 131 (1983); 284, 488 (1987).

Provenance. Signature $(+,-,-,-)$ . Every number is recomputed in verify_aq-sr-unruh-effect.py, which reads this file, regexes the printed values out, asserts them at fifty digits, and carries artefact guards on $\hbar$ , $c$ , $k_B$ , $a$ and $\rho$ . Four base defects repaired: the interval's sign (mostly-plus — it makes proper time imaginary); the Rindler metric's signature; the KMS card's dimensionally-false $\beta = 2\pi c/a$ ; the Bogoliubov card's broken middle step. Attribution corrected: the base credits Unruh alone; Fulling (1973) and Davies (1975) came first. Overclaim refused: Sokolov–Ternov is not a confirmed observation — Leinaas' verdict is "a qualified yes".

Scope. The Rindler chart and its horizon belong to `aq-sr-rindler-coordinates`, `aq-sr-rindler-metric` and `aq-sr-rindler-horizon`; the hyperbolic worldline to `aq-sr-relativistic-uniform-acceleration`. Hawking radiation and black-hole thermodynamics belong to the general-relativity track — gestured at, not annexed. The Bogoliubov and Unruh–DeWitt routes are stated and not derived: KMS already implies both (Birrell & Davies §3.3).

Builds on. Relativistic Uniform Acceleration · Proper Acceleration and the Comoving Frame · Rindler Coordinates · The Rindler Metric · The Rindler Horizon. Leads to. The Equivalence Principle · Gravitational Redshift · From Special to General Relativity.

Is the accelerated observer's thermal bath real, or is it a coordinate artefact?

It is real in the only sense that matters: the detector clicks. A two-level system carried along the hyperbola gets excited, and the excitations are permanent, countable, and in principle readable. No coordinate change can arrange that. But "real" does not mean "there are objectively particles present" — the inertial observer, examining the same field, finds a pure state with zero particles and zero entropy, and she is not in error. Both are correct, because particle number is defined relative to a positive/negative-frequency split, and that split needs a clock the two observers do not share. What is invariant is the field's state, its correlators, and the clicks. What is not invariant is the word "particle".

Where does the energy come from? The vacuum is supposed to be the lowest-energy state.

From the agent maintaining the acceleration — the rocket, the field, whatever is pushing. Keeping an observer on a hyperbola forever takes an unending input of work, and the detector's excitations are paid out of that budget. Global energy conservation is never in question. What is subtle is that the accounting differs between frames: the inertial observer sees a rocket doing work and a detector excited by an ordinary vacuum; the accelerated observer sees a detector sitting still in a warm bath. The bookkeeping differs, the physics does not, and nobody gets free energy.

Why does the answer come out exactly thermal? Why not just "noisy"?

Because of the $\sinh^2$ . On the hyperbola $G \propto 1/\sinh^2(a\Delta\tau/2c)$ , and $\sinh$ is periodic in imaginary argument with period $i\pi$ — so $G$ is periodic in imaginary proper time with period $2\pi c/a$ , exactly. Periodicity in imaginary time with period $\hbar\beta$ is the KMS condition, and KMS is the definition of a thermal state in algebraic quantum field theory — the definition that survives when the naive $e^{-\beta H}/Z$ does not, as it does not here (the Rindler Hamiltonian is unbounded, with no normalisable ground state). So the state is not "thermal-like": it is thermal, by the criterion that makes any state thermal, and the Planck spectrum and detailed balance follow automatically. The exactness is forced by the geometry of the hyperbola.

Has anyone actually observed the Unruh effect?

No. Fifty years after Fulling's paper it has never been observed in the electromagnetic vacuum, and the obstacle is the constant $4.055\times10^{-21}$ K per $\mathrm{m/s^2}$ . Storage-ring electrons reach a formal $T_U \sim 10^3$ K, and Bell and Leinaas argued the incomplete Sokolov–Ternov polarisation ( $92.4\%$ , not $100\%$ ) reflects it — but circular motion is not uniform acceleration, the state is not KMS, the temperature is not uniquely fixed by $a$ , and Leinaas' own retrospective verdict was "a qualified yes". Petawatt lasers lead the field, but the signal sits under a mountain of Larmor radiation. The effect itself remains unseen.

Why derive it from KMS rather than from Bogoliubov coefficients or a detector?

Because KMS is shorter, and because it shows you where the thermality comes from. The Bogoliubov route grinds through a mode expansion and hands you a Planck factor at the end, with the geometric origin of the $2\pi$ buried in a Fourier transform. The detector route requires you to model a detector, introducing choices that have nothing to do with the vacuum. The KMS route needs one input from quantum field theory — that the massless Wightman function goes as $1/\Delta s^2$ — and everything after is trigonometry. And it makes visible what the others hide: the $\Sigma$ -cancellation, which is where the temperature is born.

Does the Unruh effect prove that black holes radiate?

It does not prove it; it removes the mystery from it. Hawking's 1974 result looked, at the time, like it required something exotic about gravity. Unruh's 1976 paper showed the same phenomenon occurs in completely flat, empty space, with no gravity anywhere — you need only a horizon, and acceleration builds one for free. The lesson: horizons are thermal, and gravity was never the essential ingredient — the causal structure was. The black hole, its entropy, its evaporation and the information paradox belong to the general-relativity track; special relativity contributes the demonstration that you never needed the black hole to see the effect.