Bonus ④ nearly concluded that gravity cannot see the phase of a mass. Quantise, and there turns out to be exactly one way through.
Bonus ④ chased "does the electron's mass have an imaginary part" and found that the phase has two knobs (Weyl and chiral), that a lone phase is bookkeeping while phase differences are physics. This time we follow where that phase goes in gravity, all the way. The answer comes in three stages — classical gravity does not see the phase; quantum gravity sees it through an anomaly; and it is visible only when spacetime is rotating. Along the way it also turns out that there is no such thing as a single "strength of gravity".
Apply episode 10's decision procedure to the most famous constant of all. Newton's \(G\) has units \(\mathrm{m^3\,kg^{-1}\,s^{-2}}\) — emphatically dimensionful. Left column: bookkeeping.
So what is the physics? Build a dimensionless quantity out of \(G\), \(\hbar\), \(c\) and a mass, and there is only one answer.
Exactly the same shape as the fine structure constant \(\alpha=e^2/(4\pi\varepsilon_0\hbar c)\) — except that a mass appears in it.
That is the decisive difference. Charge is quantised, so \(\alpha\) is the same number for every particle; mass is not quantised. So \(\alpha_G\) is a different number for every particle.
| \(m/M_{\rm Pl}\) | \(\alpha_G=(m/M_{\rm Pl})^2\) | |
|---|---|---|
| neutrino (\(m_3\)) | \(4.10\times10^{-30}\) | \(1.68\times10^{-59}\) |
| electron | \(4.19\times10^{-23}\) | \(1.75\times10^{-45}\) |
| muon | \(8.65\times10^{-21}\) | \(7.49\times10^{-41}\) |
| proton | \(7.69\times10^{-20}\) | \(5.91\times10^{-39}\) |
| tau | \(1.46\times10^{-19}\) | \(2.12\times10^{-38}\) |
| top quark | \(1.41\times10^{-17}\) | \(2.00\times10^{-34}\) |
| Planck mass | \(1\) | \(\mathbf{1}\) |
From the neutrino to the top that is a spread of \(1.19\times10^{25}\). Which means:
"Gravity is weak" is a sentence with no meaning until you name the particle.
There is no single strength of gravity. \(G\) is bookkeeping; the physics is a per-particle \(\alpha_G\). Against electromagnetism, \(\alpha/\alpha_G(e)=4.17\times10^{42}\) — but that too needs the rider "for the electron".
This number has a far more intuitive meaning. Every particle comes with two lengths: its size as a black hole (the Schwarzschild radius) and its size as a quantum (the Compton wavelength).
The two lengths
$$r_s=\frac{2Gm}{c^2},\qquad \lambda_C=\frac{\hbar}{mc}$$Divide, and every unit cancels
$$\frac{r_s}{\lambda_C}=\frac{2Gm}{c^2}\cdot\frac{mc}{\hbar}=\frac{2Gm^2}{\hbar c}=2\alpha_G$$The mass appears squared because one length grows with \(m\) and the other shrinks as \(1/m\).
\(\alpha_G\) simply is "how close this particle is to being a black hole".
For the electron, \(r_s/\lambda_C=3.50\times10^{-45}\): compared with its quantum spread, its horizon is \(10^{-45}\) of nothing. Conversely, look for the mass at which \(\alpha_G=1\) and you find the Planck mass — the point where a particle's Compton wavelength becomes its own horizon. The definition of "the quantum gravity scale" falls out of this single line.
Where does \(\alpha_G(e)=1.75\times10^{-45}\) come from? In the style of episode 4, split the mass into "dimensionless coupling × scalar vacuum value" by substituting \(m_e=y_e v/\sqrt2\). That is all.
Left is the flavour problem (why is \(y_e\sim10^{-6}\)?), right the hierarchy problem (why is \(v\ll M_{\rm Pl}\)?). Two of the great unsolved problems in physics, sitting side by side in a product.
In orders of magnitude, 11.4 and 33.4: three quarters of the reason gravity is weak for an electron is the hierarchy problem, one quarter the flavour problem. And recall bonus ④ — the left factor had a face: \(\delta\) stands \(2.27^\circ\) short of the zero at \(135^\circ\). The sensitivity there was \(d\ln m_e/d\delta=0.90\) per degree, so
$$\frac{d\ln\alpha_G(e)}{d\delta}=1.80\ \text{per degree}\qquad\Longrightarrow\qquad \text{move } \delta \text{ by } 1^\circ \text{ and gravity on an electron changes by a factor of 6}$$Now the main business. Feed bonus ④'s mass \(m=m_1+im_2\gamma_5\) to gravity. Gravity couples to exactly one thing: the stress-energy tensor \(T_{\mu\nu}\).
But as section 04 of bonus ④ showed, the chiral rotation \(\psi\to e^{i\alpha\gamma_5/2}\psi\) maps the theory exactly onto the theory with mass \(|m|=\sqrt{m_1^2+m_2^2}\). Every observable is the same — including \(T_{\mu\nu}\). Therefore:
Gravity sees only \(|m|\). The phase is completely invisible.
This is a completely independent second confirmation of bonus ④'s "② a lone chiral phase is bookkeeping". That argument was in the language of field theory; this one is in the language of geometry, and they land in the same place.
Section 02's \(\alpha_G\) says the same thing: \(\alpha_G=(|m|/M_{\rm Pl})^2\) has no slot for a phase. To gravity, the imaginary part of a mass may as well not exist.
── And there it does not end. The step "the chiral rotation maps it away" is exactly where quantum mechanics bites.
Recall episode 8. A knob that is free classically leaves a trace when you quantise — an anomaly. The chiral rotation has one too, and in fact two kinds.
The first term is the electromagnetic field, the second the gravitational one. \(R\tilde R=\tfrac12\epsilon^{\mu\nu\alpha\beta}R_{\mu\nu\rho\sigma}R_{\alpha\beta}{}^{\rho\sigma}\) is called the Pontryagin density.
What it means: when you rotate away the mass phase \(\theta\), that \(\theta\) is not annihilated — it moves into the coefficient of \(R\tilde R\) in the action (a gravitational theta term). A re-entry in the ledger, not a deletion.
That makes three of these across the series.
| Knob free classically | Trace left by quantisation | Where it shows up | Episode |
|---|---|---|---|
| Weyl \(\Omega\) | conformal anomaly \(T^\mu_\mu=\tfrac{\beta}{2g}F^2\) | \(\beta\) function, 99% of the proton mass | ep. 8 |
| chiral \(\alpha\) (EM) | \(F\tilde F\) = \(\theta_{\rm QCD}\) | neutron EDM, \(\bar\theta<10^{-10}\) | bonus ④ |
| chiral \(\alpha\) (gravity) | \(R\tilde R\) = \(\theta_{\rm grav}\) | this episode | bonus ⑤ |
So where does \(R\tilde R\) actually appear? This is the prettiest part.
In a vacuum solution the curvature information condenses into a single complex number, the Weyl scalar \(\Psi_2\) (for the class known as type D). And the two curvature invariants are precisely its real and imaginary parts.
Kretschmann invariant (the familiar "size of the curvature")
$$R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}\ \propto\ \mathrm{Re}\!\left(\Psi_2^2\right)$$Pontryagin density (the one in the anomaly)
$$R\tilde R\ \propto\ \mathrm{Im}\!\left(\Psi_2^2\right)$$Real and imaginary parts of the same complex number. One is how much it bends, the other how much it twists.
And for the Kerr solution (a rotating black hole) that complex number has a closed form.
$$\Psi_2=-\frac{M}{(r-i\,a\cos\theta)^3}$$where \(a=J/Mc\) is the rotation parameter. The imaginary part comes only from \(a\cos\theta\). Therefore:
The gravitational field has an imaginary part only when spacetime is rotating.
At \(a=0\) (Schwarzschild) \(\Psi_2=-M/r^3\) is real and \(R\tilde R=0\). FRW is conformally flat, so the whole Weyl tensor vanishes and again \(R\tilde R=0\). Neither the expanding universe nor a quiet black hole reflects the phase.
| Spacetime | \(\mathrm{Re}(\Psi_2^2)\) | \(\mathrm{Im}(\Psi_2^2)\) | Is the phase visible? |
|---|---|---|---|
| FRW (expanding universe) | 0 | 0 | no (conformally flat) |
| Schwarzschild | \(\ne0\) | \(\mathbf{0}\) | no |
| Kerr (rotating black hole) | \(\ne0\) | \(\boldsymbol{\ne0}\) | yes |
Put numbers in. Dividing out the \(1/r^6\) radial falloff and evaluating at the reference point \(r=2M\), \(\cos\theta=0.5\):
| \(a/M\) | real part (normalised) | imaginary part | \(|\mathrm{Im}/\mathrm{Re}|\) |
|---|---|---|---|
| 0 (static) | 1.0000 | 0.0000 | 0 |
| 0.3 | 0.8858 | 0.4270 | 0.48 |
| 0.6 | 0.5863 | 0.7289 | 1.24 |
| 0.9 | 0.2074 | 0.8370 | 4.04 |
| 0.998 (near-extremal) | 0.0864 | 0.8298 | 9.60 |
At \(a/M\approx0.53\) the imaginary part overtakes the real one. Near a rapidly spinning black hole the "imaginary" curvature invariant is ten times the "real" one. As a mirror for the phase of a mass, there is no brighter place in the universe.
For \(a\gtrsim0.9\) a sign reversal appears in the right-hand picture, because \(\Psi_2^2\) goes as \((r-ia\cos\theta)^{-6}\) and the argument gets multiplied by \(6\): move towards the pole and the argument passes \(180^\circ\), flipping the sign of the imaginary part. At \(a/M=0.998\) it winds to \(6\varphi=251^\circ\) just outside the horizon.
Time to hit the brakes honestly. The picture above is pretty, but as it stands it is not an observable. Two reasons.
Reason ①: it averages to zero over a sphere. \(\mathrm{Im}(\Psi_2^2)\) is an odd function of \(\cos\theta\) — opposite signs in the northern and southern hemispheres — so integrating over a sphere kills it. Numerically,
$$\left\langle \mathrm{Im}(\Psi_2^2)\right\rangle_{r=3M}\ \sim\ 10^{-17}\quad(\text{zero to numerical precision}),\qquad \max\left|\mathrm{Im}\right|=0.837$$Reason ②: a constant \(\theta_{\rm grav}\) is a total derivative. The integral of the Pontryagin density is
$$\int R\tilde R\,\sqrt{-g}\,d^4x=32\pi^2\times(\text{Pontryagin number})$$which on a closed manifold is an integer. So if \(\theta_{\rm grav}\) is a constant, all it can affect is the topology of spacetime; in an ordinary asymptotically flat spacetime nothing happens at all. Exactly the situation of the QED theta term, which is unobservable in the absence of magnetic monopoles.
So how could it ever matter? There is only one answer — make \(\theta_{\rm grav}\) a field rather than a constant.
This has a name — dynamical Chern–Simons gravity (Jackiw & Pi, 2003). And what happens there is dramatic.
The non-rotating black hole is unharmed; only the rotating one breaks. Section 06's statement about \(\mathrm{Im}(\Psi_2)\) translates directly into the existence or non-existence of a solution.
Together with bonus ④, here is every place the phase of a mass can surface in an observation.
| Where | Mechanism | Status |
|---|---|---|
| electron EDM | phase difference between mass and Yukawa → asymmetric charge distribution | \(d_e/(e\lambda_C)<1.1\times10^{-19}\) |
| strong CP | \(\arg\det M_q\) → \(F\tilde F\) → neutron EDM | \(\bar\theta<10^{-10}\) |
| rotating spacetime | gravitational anomaly → \(R\tilde R\propto\mathrm{Im}(\Psi_2^2)\) | only if \(\theta\) is a field |
All three share a feature: not one dimensionful quantity appears. The EDM is a ratio of lengths, \(\bar\theta\) is an angle, the Pontryagin number is an integer. Exactly as episode 10's procedure demands, the physics of phases was dimensionless from beginning to end.
Dynamical Chern–Simons gravity in section 07 is a proposal for modified gravity, not established physics. "Kerr ceases to be a solution" is a consequence within that framework. The observational bounds are still loose, and there is no evidence that this effect exists in nature.
The anomaly coefficient \(1/384\pi^2\) is convention-dependent (Dirac versus Weyl fermion, and whether a factor of \(1/2\) is folded into the definition of \(R\tilde R\)); other sources write \(768\pi^2\) and so on. Likewise the proportionality constant and sign in \(R\tilde R\propto\mathrm{Im}(\Psi_2^2)\) depend on conventions; this article uses only the qualitative fact that it comes from the imaginary part.
The closed form for \(\Psi_2\) holds only for vacuum type-D solutions. With matter present there are five Weyl scalars and this simple restatement fails. The meridional slice in the figure is a schematic: Boyer–Lindquist coordinates plotted directly on a plane, not proper distance.
"The dilaton is \(\theta_{\rm grav}\)" at the end of section 07 is this article's storytelling. The closeness of \(\rho_\Lambda^{1/4}\) and \(m_\nu\) in section 08 is an unexplained numerical coincidence. And the remarks about \(\delta\) in section 03 presuppose bonus ④'s Koide relation — an empirical formula with no derivation that holds only for pole masses.
First, \(G\) was bookkeeping, because it carries units. The physics is \(\alpha_G=(m/M_{\rm Pl})^2\), which differs per particle and spans \(10^{25}\) from neutrino to top. "Gravity is weak" is meaningless until you name the particle. And \(\alpha_G=\frac12(r_s/\lambda_C)\) — \(\alpha_G\) simply is how close that particle is to being a black hole, with \(\alpha_G=1\) marking the Planck mass.
The electron's \(\alpha_G=1.75\times10^{-45}\) splits exactly into \((y_e^2/2)\times(v/M_{\rm Pl})^2\) — the flavour problem (11.4 orders) times the hierarchy problem (33.4 orders). Joined to bonus ④, the left factor means "\(\delta\) is \(2.27^\circ\) short of a zero". Move \(\delta\) by \(1^\circ\) and gravity on an electron changes sixfold.
The phase itself is entirely invisible to classical gravity, because \(T_{\mu\nu}\) is chiral-rotation invariant — an independent second confirmation of bonus ④'s conclusion. But quantise, and that chiral rotation carries an anomaly: the bill goes to the coefficient of \(R\tilde R\). That is the third instance, after episode 8's conformal anomaly and strong CP's electromagnetic chiral anomaly.
And we located where \(R\tilde R\) lives. In vacuum type D, \(R\tilde R\propto\mathrm{Im}(\Psi_2^2)\), and the imaginary part of Kerr's \(\Psi_2=-M/(r-ia\cos\theta)^3\) comes only from \(a\cos\theta\). Zero for the expanding universe (conformally flat) and zero for Schwarzschild; nonzero only for Kerr. At \(a/M\approx0.53\) the imaginary part overtakes the real one, and at \(a/M=0.998\) it is 9.6 times larger. ── Ask whether a mass has an imaginary part, and the only mirror is the imaginary part of the gravitational field; and the gravitational field has an imaginary part only when spacetime rotates. Though it averages to zero over a sphere, and a constant \(\theta\) is a total derivative. To make it bite, \(\theta\) must be a field — and then Schwarzschild survives while Kerr alone stops being a solution.
Print / save as PDF: ⌘+P (Ctrl+P on Windows). On screen, the slider changes the rotation parameter and you can watch the right-hand picture — the imaginary part — grow out of nothing. "Show the answer" opens each solution.