Cosmology That ClicksEpisode 9 · what i really is, Part 2 — ambiguity becomes temperature

Episode 8: localize i at each point and it becomes a force → This time: rotate the same i onto the imaginary axis of time?

Rotate i onto the Imaginary Axis
and It Becomes Temperature Turn Schrödinger's i ninety degrees onto the imaginary axis of time. Oscillation turns into decay,
and the period of imaginary time becomes temperature itself — and the expansion of c·t=constant cools the universe.

Tools you'll need: the i from Episode 8, the exponential function, the BBN of Episode 7 t → −iτ ／ τ = ℏ/k_BT

In Episode 8, localizing the phase ambiguity of \(i\) at each point gave birth to the electromagnetic force. This time we rotate the same \(i\) in an entirely different direction — ninety degrees onto the imaginary axis of time. Apply this operation (the Wick rotation) and the quantum oscillation turns into thermal decay, and the "length of the period" of imaginary time becomes temperature itself. \(i\) was the hinge: localize it and it gives a force (Episode 8); rotate it onto the imaginary axis and it gives temperature (this time). And overlay this imaginary-time temperature with the expansion of \(c\cdot t=\text{constant}\), and — the temperature drops as \(1/t\), and Episode 7's BBN (the universe cools, neutrons freeze out, helium forms) can be rewritten in the language of \(i\). \(i\), temperature, and the cooling of the universe join into one thread.

01Rotate i onto the imaginary axis of time — oscillation turns into decay

Quantum time evolution has the \(i\)-containing form \(e^{-iEt/\hbar}\) we saw in Episodes 3 and 8. A state of energy \(E\) spins round and round (oscillates) in the complex plane at angular speed \(E/\hbar\). Now apply the operation of rotating time from the real axis ninety degrees onto the imaginary axis. Set \(t\to-i\tau\) — this is the Wick rotation.

Try it — substitute t → −iτ

Quantum oscillation

$$e^{-iEt/\hbar}\qquad(\text{spinning round the complex plane = oscillation})$$

Rotate time onto the imaginary axis: t → −iτ

$$e^{-iE(-i\tau)/\hbar} = e^{-E\tau/\hbar}\qquad(i\ \text{gone, just decay})$$

Since \(i\times(-i)=1\), the \(i\) cleanly disappears and the exponent becomes real. The round-and-round oscillation has turned into a smooth decay. This is no coincidence — \(e^{-E\tau/\hbar}\) has the very form of the Boltzmann factor \(e^{-E/k_BT}\) that represents thermal equilibrium at temperature \(T\) in statistical mechanics.

Compare them: imaginary-time evolution \(e^{-E\tau/\hbar}\) and thermal equilibrium \(e^{-E/k_BT}\). Set the exponents equal and —

The heart of this episode — the period of imaginary time becomes temperature
$$\frac{\tau}{\hbar}=\frac{1}{k_BT}\qquad\Rightarrow\qquad \tau=\frac{\hbar}{k_BT}$$

"Roll imaginary time \(\tau\) up into a period \(\hbar/k_BT\) and make it periodic," and the system becomes completely equivalent to one immersed in a heat bath at temperature \(T\). Short period = hot, long period = cold. Through the imaginary axis of time, \(i\) was a device that gives birth to temperature.

A contrast with Episode 8 — the two faces of i Episode 8 localized the ambiguity of \(i\) to space, to each point → the electromagnetic force (the partner \(A\)) was born. This time we rotate the same \(i\) onto the imaginary axis of time → temperature (the period of imaginary time) is born. One device, \(i\), is the hinge that pours out physics in both directions: the real axis (phase per location = force) and the imaginary axis (rotating time to imaginary = heat). The "something folded into the time axis" you sensed in \(i\) was — both force and heat. This is the single answer running through Episodes 8 and 9.

02Overlay c·t=constant — expansion stretches the period, the universe cools

Now we overlay this series' \(c\cdot t=\text{constant}\). As Bonus 4 showed, \(c\cdot t=\text{constant}\) is the coordinate that pushes expansion onto the time side (conformal time), with the expansion going as \(a\propto t\) (straight-line expansion). And as a known fact of standard cosmology, in an expanding universe the temperature drops inversely with the scale factor (\(T\propto1/a\)). Since \(a\propto t\) for \(c\cdot t=\text{constant}\) —

Try it — how temperature and the imaginary-time period move with expansion

The time dependence of temperature (substitute a ∝ t)

$$T\propto\frac{1}{a}\propto\frac{1}{t}$$

The imaginary-time period, being the inverse, stretches

$$\tau=\frac{\hbar}{k_BT}\propto t$$

As the age of the universe advances, the temperature \(T\) drops as \(1/t\), and the imaginary-time period \(\tau\) stretches in proportion to \(t\). Exactly the same \(1/t\) family as Episode 1's speed of light \(c\propto1/t\) and Bonus 1's Hubble \(H=1/t\) — expansion, the slowing of light, the drop in temperature, and the stretching of the imaginary-time period are all different faces of one and the same phenomenon.

In the figure below, move the age of the universe \(t\). At the left (young universe), the temperature is high and the imaginary-time period is short — a "hot" state where the complex-plane spiral winds densely. At the right (aged universe), the temperature drops and the period stretches — a "cold" state where the spiral winds loosely and wide. You can watch expansion stretch out the imaginary-time spiral of \(i\).

Figure: age of the universe t (horizontal slider) → temperature T∝1/t drops → the imaginary-time period τ=ℏ/k_BT stretches. Left = hot young universe (short period), right = cooled universe (long period). Green line = BBN threshold k_BT ~ Q.
temperature T (∝1/t) imaginary-time period τ (∝t) BBN threshold k_BT ~ Q

03Episode 7's BBN is the physical manifestation of this picture

Here we connect to Episode 7, where "the universe cools, neutrons freeze out, and helium forms." Rewritten in this episode's language, this is precisely the process of the imaginary-time period stretching.

Characterize the moment BBN happened in the language of imaginary-time temperature — it's when the temperature dropped and \(k_BT\) cooled to around the leftover energy of neutron decay, \(Q\approx0.8\) MeV (the star of Episode 7). Cool further and thermal fluctuations can no longer maintain the neutron–proton conversion, and the ratio freezes out. After that, only the neutron's free decay (Episode 7's Sargent rule \(\Gamma\propto Q^5\)) matters, and the helium abundance is fixed.

BBN = the moment the imaginary-time period crosses the threshold

The moment the imaginary-time period \(\tau=\hbar/k_BT\) stretches past the neutron's characteristic scale \(\hbar/Q\) — that is, the moment \(k_BT\) drops below \(Q\) — thermal equilibrium can no longer be held and the neutron ratio freezes. This is the imaginary-time-temperature characterization of the "BBN freeze-out" that set Episode 7's helium abundance.

The reveal — your original intuition finds its rightful place At the start of the proton-decay discussion, you offered the picture of "a system error when the universe's resolution drops below a threshold." That was rejected for proton decay (face-value substitution into \(E=mc^2\)). But in this episode's language of imaginary-time temperature — BBN freeze-out is realized correctly, precisely as "a systemic transition at the moment the imaginary-time period crosses the threshold \(\hbar/Q\)." The intuition of a systemic transition across a threshold found its rightful place inside the trace-leaving-no gauge (this episode's road).
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04The verdict — what it predicts and what it does not

As in Episode 7, we judge honestly by Bonus 4's equivalence condition.

What it does not predict (the part that leaves no trace in observation). Both the Wick rotation (\(t\to-i\tau\)) and the conformal time of \(c\cdot t=\text{constant}\) are pure relabelings of coordinates and variables. If \(\alpha\) is invariant, overlaying them yields zero new observational predictions. Both "\(T\propto1/t\)" and "the imaginary-time period stretches" merely rewrite standard cosmology on a different time axis, not differing from the standard one iota — as Episode 7's verdict says, it's safe as long as you don't run \(c\cdot t=\text{constant}\) at face value, and this episode doesn't, so it survives.

What it does predict (or rather, clarify). This episode's value is not new predictions but a unifying viewpoint. Expansion, the slowing of light, cooling, the quantum phase, heat — phenomena that looked scattered become one picture as "the folding of a single complex time axis, its real part (oscillation, expansion) and imaginary part (temperature)." The same spirit as Episode 4's "relabel to log time and the computation gets lighter" — fold the time axis cleverly without discarding information and the view improves dramatically. This is the deepest version of that folding.

The honest line — the condition under which this beauty stands

This unifying picture all stands on the condition that "\(\alpha\) is invariant." When the imaginary-time temperature drops, what moves is the temperature (a dimensionful quantity), not the atomic ratio \(\alpha\). The moment you reread it as "the drop in imaginary-time temperature runs \(\alpha\)," you fall onto Episode 8's "dying branch" (\(\alpha\) runs) and are rejected by Episode 7's neutron and BBN.

This episode survives only insofar as it folds only the dimensionful quantity of temperature and touches the dimensionless \(\alpha\) not at all. The Wick rotation itself is established standard physics (the imaginary-time / Matsubara formalism of statistical mechanics), and the quantum–heat correspondence is textbook material. The connection with \(c\cdot t=\text{constant}\), read as an \(\alpha\)-invariant gauge, leaves no trace in observation and is fully consistent with Bonus 4's equivalence condition.

Practice problems (solvable with just this episode's formulas)
  1. When you substitute \(t\to-i\tau\) into the oscillation \(e^{-iEt/\hbar}\), why does \(i\) vanish and give decay \(e^{-E\tau/\hbar}\)?
    Show answer
    The exponent is \(-iE(-i\tau)/\hbar\); since \(i\times(-i)=1\), this becomes \(-E\tau/\hbar\). The \(i\) vanishes and the exponent becomes a real decay — the same form as the Boltzmann factor \(e^{-E/k_BT}\).
  2. With the imaginary-time period \(\tau=\hbar/k_BT\), how does the period change when the temperature doubles? Confirm "hot = short period."
    Show answer
    Since \(\tau\propto1/T\), doubling \(T\) halves \(\tau\). The higher the temperature, the shorter the imaginary-time period — hot = short period, cold = long period.
  3. For \(c\cdot t=\text{constant}\) (\(a\propto t\)) with \(T\propto1/t\), if the age of the universe increases tenfold, to what fraction does the temperature drop? Is this the same behavior as Episode 1's \(c\propto1/t\)?
    Show answer
    Since \(T\propto1/t\), a tenfold \(t\) gives 1/10 the temperature. The same \(1/t\) family as Episode 1's \(c\propto1/t\) and Bonus 1's \(H=1/t\) — expansion, the slowing of light, and cooling are different faces of the same time dependence.

Episode 9 wrap-upi had both force and heat folded into it

Rotate Schrödinger's \(i\) onto the imaginary axis of time (\(t\to-i\tau\)) and the quantum oscillation \(e^{-iEt/\hbar}\) turns into thermal decay \(e^{-E\tau/\hbar}\), with the imaginary-time period becoming temperature itself, \(\tau=\hbar/k_BT\) (Step 01). Overlay the expansion of \(c\cdot t=\text{constant}\) (\(a\propto t\)) and the temperature drops as \(T\propto1/t\) while the imaginary-time period stretches — the slowing of light, Hubble, and cooling all the same \(1/t\) family (Step 02).

And Episode 7's BBN is the physical manifestation of this "moment the imaginary-time period crosses the threshold \(\hbar/Q\)" (Step 03). Your original intuition of "a system error across a threshold" found its rightful place inside the trace-leaving-no gauge. The verdict is the same as Episode 7 — no new observational prediction (safe, being an \(\alpha\)-invariant gauge), but the most powerful thinking tool for folding expansion, cooling, quantum, and heat onto one axis (Step 04). Together with Episode 8: \(i\) is the hinge — localize it and it gives a force, rotate it onto the imaginary axis and it gives temperature. The "something folded into time" you sensed in \(i\) was both force and heat.

This is Episode 9 of "Cosmology That Clicks," a reading piece for curious high-schoolers and undergraduates. That the Wick rotation (\(t\to-i\tau\)) makes quantum-mechanical time evolution \(e^{-iHt/\hbar}\) correspond to the statistical-mechanical partition function \(e^{-H\tau/\hbar}\), with the imaginary-time period \(\tau=\hbar/k_BT\) becoming inverse temperature (the Matsubara / imaginary-time formalism, the KMS condition), is established standard physics. That the radiation temperature in an expanding universe drops inversely with the scale factor (\(T\propto1/a\)) is also a known fact, and for linear expansion \(a\propto t\) (\(c\cdot t=\text{constant}\), \(R_h=ct\)) it gives \(T\propto1/t\). That BBN freeze-out occurs near \(k_BT\sim Q\) was treated in Episode 7. The figure is a schematic of the concept, visualizing the relationship between temperature, imaginary-time period, and BBN threshold in normalized units (the actual radiation-dominated temperature–time relation is more complex, e.g. \(T\propto t^{-1/2}\)). The Wick rotation and conformal time are relabelings of coordinates and variables and yield no new observational predictions under invariance of \(\alpha\) (Bonus 4). Interpretations that run \(\alpha\) are rejected by Episode 7 and atomic clocks. — To print, use your browser's Print → Save as PDF (in the printed version, the slider and answers are static/hidden).

Print / save as PDF: ⌘+P (Ctrl+P on Windows). On screen, moving the age-of-universe slider drops the temperature, stretches the imaginary-time period, and shows the moment it crosses the BBN threshold. "Show answer" opens the solutions.