The heat hidden in a Doppler shift

The wall of silence behind a rocket that never stops found that an observer with constant acceleration has a horizon, a surface behind it from which no signal can ever arrive. The ship that never arrives at c followed that observer’s journey, and the clock that does not feel the turn established that its clock keeps proper time regardless of the acceleration. The temperature of an acceleration then reported the strangest property of all: empty space, to such an observer, is a thermal bath at a temperature proportional to the acceleration, $kT = \hbar a/2\pi c$, and a detector carried along will register particles.

That result is usually derived by rewriting a quantum field in the coordinates natural to the accelerating observer and computing how its ground state looks in them — a calculation of several pages that makes the thermal spectrum appear almost by accident. It leaves the question of where the heat comes from. There is a shorter route, which takes almost nothing but the Doppler shift and a Fourier transform, and it shows that the thermal form is not quantum at all. It is a property of what an exponentially receding observer sees of any single wave.

One wave, seen from a rocket #

An observer with constant proper acceleration $a$ follows a hyperbola in spacetime. In the coordinates of an inertial frame its position and time are $x = (c^2/a)\cosh(a\tau/c)$ and $t = (c/a)\sinh(a\tau/c)$, where $\tau$ is the observer’s own time. Its speed approaches that of light and never reaches it, and behind it is the horizon, the light line it approaches asymptotically.

Send a plane wave of one frequency after it, travelling in the same direction. Its crests are lines at 45 degrees in the diagram, evenly spaced. The observer meets them, but at intervals of its own time that grow and grow: each crest has to chase a receding observer that is moving ever closer to the speed of light, and takes longer to catch it. After some proper time the observer meets no further crest at all — every later crest lies on or beyond its horizon, and the observer will never see them.

In terms of the wave’s phase the calculation is one line. The phase of a wave travelling in the $+x$ direction is $\omega(t - x/c)$. Along the hyperbola, $t - x/c = -(c/a)e^{-a\tau/c}$, so the phase the observer sees is $-(\omega c/a)\,e^{-a\tau/c}$ and the frequency it measures, the rate of change of that phase, is $\omega e^{-a\tau/c}$. The wave is redshifted, and the redshift grows exponentially in the observer’s time. That is the shift that survives at right angles and every other Doppler shift in this subject, applied to an observer whose speed keeps growing, and the exponential comes from the fact that uniform acceleration adds rapidity at a steady rate — speeds that refuse to add found rapidity to be the quantity that does add, and the Doppler factor is its exponential.

The same curve for every frequency #

The next observation is the one on which everything turns.

On a logarithmic axis, the frequency seen falls along a straight line of slope $-a/c$, and waves of different original frequencies give the same straight line shifted sideways. A wave ten times higher in frequency passes through every seen frequency exactly $\ln 10 = 2.3$ units of $c/a$ later. So the observer, watching the fading signal, can tell nothing about the original frequency from its shape — only from when the fading happens. Anything that depends only on the shape of the signal, such as the distribution of its power over frequency, must be the same for every wave sent.

That is already surprising. A spectrum that does not depend on the frequency of the source is a property of the observer, not of the source. The only scale in the observer’s situation is $a/c$, a rate, so the spectrum, when it is computed, can depend on frequency only through $\nu c/a$.

A Planck spectrum from a Fourier transform #

The spectrum of the fading signal is its Fourier transform in the observer’s time, and it can be done exactly. The transform of $\exp[-i(\omega c/a)e^{-a\tau/c}]$ at frequency $\nu$ reduces, by a change of variable, to Euler’s gamma function, and its squared magnitude is

$$|F(\nu)|^2 = \left(\frac{c}{a}\right)^2 \frac{2\pi}{(\nu c/a)\left(e^{2\pi\nu c/a} - 1\right)}$$

for positive $\nu$ — with no $\omega$ anywhere in it, as the argument from the shifted curves required.

Multiply by the frequency and divide by $2\pi$ to turn power per unit frequency into the number of quanta per mode, and what is left is $1/(e^{2\pi\nu c/a} - 1)$. That is the Bose–Einstein distribution, the occupation of a mode of frequency $\nu$ in a bath at temperature $T$, with $\hbar\nu/kT = 2\pi\nu c/a$: that is, $kT = \hbar a/2\pi c$. The figure computes the transform from the gamma function’s own integral and lays the Planck occupation over it; they are the same function. The curve that would not come down found that Planck’s formula was forced on physics in 1900 by the failure of classical equipartition in a cavity. Here the same formula falls out of a Doppler shift, with no cavity, no equipartition and no quantisation used in obtaining it.

What the classical calculation does not supply is the source. A single classical wave has to be sent after the rocket by somebody. Quantum theory says that the vacuum is full of such waves — the zero-point fluctuations of every mode of the field, one for each frequency and direction — and that the accelerating observer sees every one of them through the same exponential redshift. The spectrum each produces is thermal and identical, and together they make the thermal bath of the Unruh effect.

Why it has to be an exponential #

It is worth asking what other motions would do, because the answer shows that the Planck shape is not a coincidence of this one calculation. An observer moving at a constant speed sees a wave with a constant Doppler factor — the $k$ of everything from an exchange of pulses — and a wave of constant frequency has a spectrum that is a single sharp line. Nothing thermal about it. An observer whose speed changes for a while and then settles sees a frequency that drifts and then stops, and its spectrum is a smear with no particular shape. Only a Doppler factor that changes exponentially for ever, which is to say a rapidity that grows linearly for ever, has no preferred moment: shifting the observer’s clock by any amount maps the signal onto the same signal with a different starting frequency. That invariance under time shifts, combined with a redshift that runs through every frequency, is what forces the spectrum into a form that depends only on $\nu c/a$, and the particular form is set by the gamma function.

A familiar relative of this is the connection between a decay and a line width. The width that is a lifetime found that an oscillation dying away as $e^{-t/\tau}$ has a spectrum of a particular shape, the Lorentzian, with a width fixed by the lifetime and nothing else. Here it is the frequency rather than the amplitude that falls exponentially, and the transform of an exponentially falling frequency is not a Lorentzian but a Planck factor. In both cases a single rate in the time domain fixes a single scale in the frequency domain, and the exponential is what makes the shape universal.

More power going down than up #

A thermal bath has one further property that distinguishes it from a signal that merely happens to have a Planck-shaped spectrum: detailed balance.

A detector with two levels, carried by the observer, is excited by the part of the signal at positive frequency, which can supply energy, and de-excited by the part at negative frequency. In a bath at temperature $T$ the de-exciting rate exceeds the exciting rate by exactly $e^{\hbar\nu/kT}$ — the Boltzmann factor — so that the detector settles into its two levels in thermal proportion. The fading wave’s spectrum has precisely that asymmetry: the power at $-\nu$ exceeds the power at $+\nu$ by $e^{2\pi\nu c/a}$, as the figure confirms at every frequency. The detector, if carried by the observer, does not merely see a spectrum shaped like Planck’s; it comes to equilibrium with it at the temperature $\hbar a/2\pi ck$. That is the operational content of saying the accelerated observer is in a heat bath, and the Doppler shift supplies all of it.

The same fading at a black hole #

The argument never used anything about acceleration except the exponential redshift near a horizon, and a black hole has one too.

Light emitted by something falling into a black hole reaches a distant observer ever more redshifted, and near the horizon the redshift grows exponentially in the distant observer’s time, with an e-folding time of $4GM/c^3$ for a non-rotating hole — twenty microseconds for a hole of one solar mass. Run the same argument backwards: a wave that leaves the region near the horizon late is a wave that was squeezed by that exponential, and the distant observer receives each mode of the field with the Planck factor at the temperature $\hbar/2\pi k$ times the fading rate. That is Hawking’s temperature, $\hbar c^3/8\pi GMk$, $6\times10^{-8}$ kelvin for a solar-mass hole. Hawking’s 1974 derivation is a calculation of exactly this exponential distortion of modes, and the hole that outlives everything and then does not followed its consequence: a black hole radiates, and loses mass.

The difference between the two horizons is where the energy comes from. The accelerating observer’s bath is paid for by whatever keeps it accelerating: a detector that absorbs a quantum from the bath is, seen from an inertial frame, a detector that emitted one while being pushed. The black hole’s radiation is paid for by the hole’s mass. The river that sound cannot swim up found the same exponential peeling of waves at a sonic horizon in a flowing fluid, with a temperature set by how fast the flow changes there, and the laboratory experiments on such analogue horizons are, in effect, tests of this argument: an exponential redshift at a rate $\kappa$ makes a spectrum at $\hbar\kappa/2\pi k$, whatever the waves are.

An expanding universe does it too #

There is a third horizon of the same kind, and it surrounds every observer in an expanding universe whose expansion accelerates. In a universe dominated by a cosmological constant, distances grow as $e^{Ht}$, light from distant sources is redshifted exponentially as it travels, and each observer has a horizon at a distance $c/H$ beyond which no signal sent now will ever arrive. Gary Gibbons and Stephen Hawking showed in 1977 that such an observer should see a thermal bath at $kT = \hbar H/2\pi$ — the same formula with the expansion rate in place of the acceleration.

For the expansion rate the universe appears to be approaching, the temperature is about $2\times10^{-30}$ kelvin, far below anything measurable and far below the microwave background’s 2.7 kelvin, which it will not dominate for an unimaginably long time. But it means that an accelerating universe, even when every particle in it has been diluted away, is not at absolute zero. The vacuum energy that drives the acceleration — whose estimated size the estimate that misses by a hundred and twenty found to be one of the worst predictions in physics — also sets a floor on the temperature of everything its horizon encloses.

Why the answer is so cold #

The temperature is $4\times10^{-21}$ kelvin for each metre per second squared of acceleration. For the thermal bath to be detectable, the acceleration has to be enormous — around $10^{20}$ metres per second squared for a kelvin — and no macroscopic object has been accelerated so hard for long enough. The Fourier picture says why: the fading has to be fast compared with the frequencies a detector responds to. A detector sensitive at a gigahertz sees a thermal spectrum only if the observer’s Doppler factor changes by a factor $e$ in a fraction of a nanosecond, which requires reaching a large fraction of the speed of light within a nanosecond.

Electrons in the strongest laser fields and in storage rings do reach accelerations of $10^{20}$ to $10^{25}$ metres per second squared, briefly or in circles. John Bell and Jon Leinaas argued in 1983 that the incomplete polarisation of electrons in storage rings — they settle at 92 per cent aligned rather than 100 — can be read as the effect of a thermal bath at the electrons’ acceleration, modified because circular motion is not uniform acceleration. Whether such readings are tests of the Unruh effect or merely a different description of well-understood radiation physics has been argued ever since, because the same numbers come out of ordinary quantum electrodynamics in the laboratory frame.

One more consequence ties this to an older result. Observers riding at different distances from the same horizon, in a rigid accelerating frame, have different proper accelerations — the one a distance $\rho$ from the horizon has $a = c^2/\rho$ — and so see different temperatures, $\hbar c/2\pi k\rho$, hotter nearer the horizon. The column that is hotter at the bottom found that a column of gas in equilibrium in a gravitational field is hotter lower down, by exactly the factor by which clocks there run slow, and that the product of temperature and clock rate is what is uniform. The Unruh temperatures of the rigid frame obey the same rule: temperature times the local clock rate is the same everywhere. The accelerated vacuum is a column in thermal equilibrium, heated from below by the horizon.

What the calculation does not show #

The derivation here uses one plane wave in one spatial dimension, travelling in the direction of acceleration. A wave arriving from another direction is Doppler shifted differently at early times, but its late-time fading near the horizon has the same exponential form, and the full quantum calculation, summing over all directions and polarisations, gives the same temperature. The classical argument also treats the transform over all of the observer’s time, from the infinite past to the infinite future, as a real accelerating rocket cannot — a finite period of acceleration gives a spectrum that is thermal only for frequencies well above the inverse of that period.

Nor does the calculation say what the observer’s thermometer is made of. The temperature of an acceleration stressed that the detector’s response is local and unambiguous while the description of the state is observer-dependent; the Fourier argument is a way of seeing why the description comes out thermal, not a replacement for the calculation of a detector’s response, which needs the quantum field.

Still open: whether the effect will ever be measured directly #

No experiment has detected the Unruh effect in a way that could not be explained otherwise. Proposals include electrons accelerated by the most intense lasers, where the acceleration during each optical cycle is enormous but brief; atoms falling past mirrors, or placed in cavities that enhance their coupling to the modes the acceleration shifts; and analogue systems in which the role of the accelerating frame is played by a moving medium. Each faces the difficulty that the signal is small and that conventional radiation from the accelerated particle is large and has a related spectrum. Whether a clean, unambiguous detection is possible — and whether, when made, it will count as a test of quantum field theory in accelerated frames or as a confirmation of electrodynamics already known — is actively debated.

The habit worth carrying away is to ask of a thermal spectrum what exponential produced it. An observer whose Doppler factor grows exponentially sees any single wave fade as e^(−aτ/c), and the spectrum of that fading is exactly the Planck distribution at kT = ħa/2πc, independent of the wave — so the heat of the accelerated vacuum, and the heat of a black hole, are both the Fourier transform of a redshift that never stops. Quantum theory decides that the vacuum has waves to be seen; the horizon decides that they look warm.