Count again where c actually appeared in this series ── wherever it did, there was a horizon
After the main series was finished, a reader asked: "the speed of light shows up in both Temperature and Tunneling ── wouldn't looking at things in a \(c\cdot t=\)constant frame make the explanations easier?" It is a good question, so I started by counting again. The result is surprising ── c is genuinely doing work in only one episode of this series: Episode 7. It does not appear at all in Episodes 1 through 6. Not even the imaginary-time period \(\beta\hbar\) contains it. So is the \(c\cdot t\) view useless? Not at all. There is exactly one place where it works, and there it works dramatically: wherever there is a horizon. And a horizon, by definition, is something built out of a light cone ── i.e. out of \(c\cdot t\) itself. Three temperatures collapse into one line.
| Episode | The starring expression | Any c? |
|---|---|---|
| 1, \(k_B\) is a conversion factor | \(E/k_BT\) | none |
| 2, how energy is shared out | \(\beta=d\ln\Omega/dE\) | none |
| 3, heat is motion you stopped tracking | \(\hbar\omega/k_BT\) | none |
| 4, the fluctuation theorem | \(\sigma/k_B\) | none |
| 5, negative temperature | \(\Delta E/k_BT\) | none |
| 6, a period in imaginary time | \(\beta\hbar=\hbar/k_BT\) | none |
| 7, the Unruh effect | \(T=\hbar a/2\pi ck_B\) | yes |
The absence in Episode 6 is what matters. The imaginary-time period \(\beta\hbar\) is a "time," not a "length." So multiplying by c to turn it into a length normally gains nothing ── you have only changed units.
The same held in the Tunneling series. The Gamow factor, the BCS gap, the crossover temperature \(T_0=\hbar\omega_b/2\pi k_B\) ── none of them contain c. Naturally enough: they are non-relativistic.
"c appears, so let's look at it in \(c\cdot t\)" does not go through as stated, because across both series c appears in only a handful of places.
The question worth asking is ── do those few places have something in common? They do.
Line up the places where c is essential and they all share one feature.
The Unruh effect has a Rindler horizon, Hawking radiation has an event horizon, and an expanding universe has a cosmological horizon.
And what is a horizon? The boundary beyond which light no longer reaches you. Its very definition is a light cone ── that is, \(c\cdot t\).
So only when there is a horizon do "distance" and "time" become one and the same quantity. It is the one place where multiplying by c stops being a unit conversion.
So write the light-travel time to the horizon as \(t_{\rm hor}=L_{\rm hor}/c\). Then:
The form on the right shows the link to Episode 6 ── "the imaginary-time period is \(2\pi\) times the light-travel time to the horizon."
Episode 6 said "temperature is the reciprocal of the circumference of a circle in imaginary time." This bonus is its sequel ── what sets the circumference of that circle is the light-travel time to the horizon.
| System | distance to the horizon \(L_{\rm hor}\) | \(t_{\rm hor}\) | resulting temperature |
|---|---|---|---|
| an accelerating observer | \(c^2/a\) | \(c/a\) | \(T=\dfrac{\hbar a}{2\pi ck_B}\) ✓ |
| a black hole | \(2r_s=4GM/c^2\) | \(4GM/c^3\) | \(T=\dfrac{\hbar c^3}{8\pi GMk_B}\) ✓ |
| an expanding universe (de Sitter) | \(c/H\) | \(1/H\) | \(T=\dfrac{\hbar H}{2\pi k_B}\) ✓ |
Let us verify the black-hole row. Since \(r_s=2GM/c^2\), \(t_{\rm hor}=2r_s/c=4GM/c^3\). Substituting,
$$k_BT=\frac{\hbar}{2\pi\cdot 4GM/c^3}=\frac{\hbar c^3}{8\pi GM}\quad\Longrightarrow\quad T=\frac{\hbar c^3}{8\pi GMk_B}\ \checkmark$$You no longer need to remember three formulas. There is one thing to remember: the light-travel time to the horizon.
More precisely still, all three can be written using the surface gravity \(\kappa\) as \(T=\hbar\kappa/2\pi ck_B\), with \(t_{\rm hor}=c/\kappa\) ── acceleration, surface gravity and the Hubble rate all sit in the same seat.
The figure below is that formula. The horizontal axis is \(t_{\rm hor}\), the vertical is temperature. On log–log axes it is a single straight line of slope −1.
What is fun is that moving the slider produces five readings at once ── the same \(t_{\rm hor}\) can be read as "this acceleration," "a black hole of this mass," or "this Hubble rate." Three physically unrelated situations are specified by one number.
This is the most satisfying use of the formula.
That is, \(t_{\rm hor}\approx1\) year. Put it straight into the formula:
$$T=\frac{\hbar}{2\pi k_B\times 1\ \text{year}}\approx 4\times10^{-20}\ \mathrm{K}$$This matches the "Unruh temperature of Earth's gravity, \(4\times10^{-20}\) K" from Episode 7. It comes straight out of the number "one year."
About one light-year behind the chair you are sitting in, there is a horizon that belongs to you. (Strictly, that is for an acceleration continued forever ── see the honest line.)
It works backwards too. How far away must a horizon be for you to feel room temperature, 300 K?
The corresponding acceleration is \(a=c^2/L=7.4\times10^{22}\ \mathrm{m/s^2}\) ── exactly the number in Episode 7's table.
And one more thing ── Episode 6 said "the imaginary-time period at room temperature is 25 femtoseconds." Multiply the \(t_{\rm hor}=4.1\) fs found here by \(2\pi\) and you get 25 fs. The same single number, seen from a different side.
Here is the real subject. Why does \(c\cdot t\) only help when there is a horizon?
Recall the \(2\pi\) from Episode 7. It came from the Euclideanised spacetime becoming polar coordinates. But making polar coordinates has a prerequisite.
Perform the Wick rotation on a relativistic spacetime and the metric changes like this:
$$-c^2dt^2+dx^2\ \xrightarrow{\ t=-i\tau\ }\ +c^2d\tau^2+dx^2$$The right-hand side is an ordinary Euclidean plane. But it reads that way only because \(c\tau\) and \(x\) are in the same units, on the same plane. Being able to treat \(c\cdot t\) as a length is itself the precondition for turning imaginary time into an "angle."
Because there is a plane you can build polar coordinates; because there is an angle it must close after \(2\pi\); and the no-conical-singularity condition then fixes the period ── that whole derivation stands on \(c\cdot t\).
Non-relativistic systems cannot use this. For a particle in a potential there is no metric mixing imaginary time \(\tau\) with space \(x\). With no plane, there is no angle either.
In Episode 3 of the sister series "Tunneling That Clicks" we derived the crossover between heat and tunneling, \(T_0=\hbar\omega_b/2\pi k_B\). It looks exactly like the Unruh temperature. There we called it "the same \(2\pi\)"; with the \(c\cdot t\) view, we can now be more precise.
| Unruh, \(T=\hbar(a/c)/2\pi k_B\) | Tunneling Ep. 3, \(T_0=\hbar\omega_b/2\pi k_B\) | |
|---|---|---|
| where the \(2\pi\) comes from | geometry (one turn of an angle) | dynamics (one round trip of harmonic motion) |
| is \(c\cdot t\) needed? | yes (no plane, no angle) | no |
| is there a horizon? | yes | no |
| the \(\omega\) in \(2\pi/\omega\) is | \(a/c\) (\(=1/t_{\rm hor}\)) | \(\omega_b\) (the frequency at the barrier top) |
| verdict | grade 2 ── a different implementation of the same framework (a period of \(2\pi/\omega\) in imaginary time). Not grade 3 (neither follows from the other) | |
Established: the Unruh temperature \(T=\hbar a/2\pi ck_B\), the Hawking temperature \(T=\hbar c^3/8\pi GMk_B\), and the Gibbons–Hawking temperature \(T=\hbar H/2\pi k_B\) (1977, for the static patch of de Sitter spacetime); that all three can be written uniformly as \(T=\hbar\kappa/2\pi ck_B\) using the surface gravity \(\kappa\); that the Rindler horizon lies at a distance \(c^2/a\) from an accelerating observer; the Schwarzschild radius \(r_s=2GM/c^2\); and the numbers quoted (\(4\times10^{-20}\) K for Earth's gravity, \(6.2\times10^{-8}\) K for a solar mass, 1.7 K for a lunar mass, \(2.7\times10^{-30}\) K for our universe now). All standard results.
Points to note: (1) This bonus episode is a rewriting of three known formulas, not new physics. Writing them via the surface gravity is textbook material, and this merely re-reads that as a "light-travel time." What it buys is memory and intuition; the content has not grown. (2) The de Sitter horizon temperature (\(2.7\times10^{-30}\) K) is an entirely different thing from the CMB temperature (2.7 K). They differ by thirty orders of magnitude. The former is heat emitted by a horizon; the latter is radiation diluted by expansion. Do not mix them. (3) Relatedly, "\(c\cdot t=\)constant so the temperature goes as \(1/t\)" applies to the horizon temperature. In the radiation era the CMB temperature goes as \(T\propto t^{-1/2}\) (since \(a(t)\propto t^{1/2}\)), not \(1/t\). (4) There are conventions for "distance to the horizon." The Rindler value \(c^2/a\) is exact, but setting \(t_{\rm hor}=2r_s/c\) for a black hole is a choice made to align the formulas (it is fixed uniquely by the surface gravity, but note that it is the light-crossing time of the diameter, not of the radius). (5) The Gibbons–Hawking temperature is a result about the static patch of de Sitter spacetime; the "horizon temperature" of a general expanding universe is considerably subtler (during epochs when the expansion is not accelerating, a stationary horizon temperature is hard to define at all). (6) Neither the Unruh effect nor Hawking radiation has been directly observed (the same caveat as Episode 7). (7) "A horizon one light-year behind you" is an idealisation assuming that acceleration is maintained from the infinite past to the infinite future. Merely standing on the Earth does not actually form that horizon.
Counting again, c is genuinely doing work in Episode 7 alone. Zero in Episodes 1 through 6 ── not even the imaginary-time period \(\beta\hbar\) contains it, because that is a "time," not a "length."
But wherever c does appear, there is always a horizon. A horizon is the boundary beyond which light no longer reaches you ── its very definition is \(c\cdot t\). It is the one place where distance and time become the same quantity. Writing \(t_{\rm hor}=L_{\rm hor}/c\) then gives ──
\(k_BT=\dfrac{\hbar}{2\pi\,t_{\rm hor}}\) i.e. \(\beta\hbar=2\pi\,t_{\rm hor}\)
Unruh, Hawking and de Sitter all come out of this one line. It is the sequel to Episode 6's "temperature is the reciprocal of a circle's circumference in imaginary time" ── what set that circumference was the light-travel time to the horizon. Your horizon standing on the Earth is about one light-year away, i.e. one year, and \(4\times10^{-20}\) K follows immediately.
And why does it work? Because being able to treat \(c\cdot t\) as a length is the precondition for turning imaginary time into an "angle." A relativistic metric puts \(c\tau\) and \(x\) on the same plane, so polar coordinates exist and \(2\pi\) appears. Non-relativistic systems have no such plane, which is why Tunneling Episode 3's \(2\pi\) is a different implementation coming from dynamics ── grade 2.
Print / make a PDF: ⌘+P (Ctrl+P on Windows). On screen, one slider gives you the time to the horizon, its distance, the temperature, the equivalent acceleration, the equivalent black-hole mass and the equivalent Hubble rate all at once. The buttons jump to real examples. "See the answer" opens each solution.