Hunting δ, and the Absolute Mass Scale · Elementary Particle Physics

Two questions remain, and neither can be answered by watching neutrinos oscillate. One needs a beam run twice with opposite charge; the other needs a beta spectrum’s last electronvolt, or the whole Universe.

🎯 Why this matters

Matter and δ produce the same signature, so no single baseline can separate them however long it runs. The answer needs two experiments at different distances — which is why the field is building DUNE and Hyper-Kamiokande rather than one larger detector.

Two questions are left that no oscillation experiment so far has answered.

Is there CP violation in the lepton sector? §10.1 quoted $\delta = 238^{+41}_{-33}$ degrees, which is a hint and not a measurement. §10.5 is the two experiments trying to turn it into one.

How heavy are they? Everything in this chapter has measured differences of squared masses. §10.6 is the two completely different techniques — a beta-decay end point and the large-scale structure of the Universe — that go after the scale itself.

Only appearance can see δ

💡 What this really says — a CPT argument, in three lines

A disappearance probability is $1$ minus a survival probability. A survival process — $\nu_\mu \to \nu_\mu$ — is its own time reverse: swap initial and final state and you have the same process back. So CPT invariance forces the survival probability to be even in $\delta$ , and every disappearance measurement, at any precision, is blind to CP violation.

That is why §10.2b’s beautiful $\nu_\mu$ disappearance data contribute nothing here, and why the entire CP programme lives in the $\nu_\mu \to \nu_e$ appearance channel — which exists only because $\theta_{13}$ turned out not to be zero. Daya Bay’s 2012 result is the reason §10.5 has any subject matter at all.

\Delta = \Delta m^2,\quad \Delta m^2_{21} = \htmlClass{t-a}{\alpha}\Delta \simeq 0.03\Delta,\quad \Delta m^2_{32} = (1-\htmlClass{t-a}{\alpha})\Delta \qquad\qquad P(\nu_\mu\to\nu_e) = \htmlClass{t-0}{P_0} + \htmlClass{t-s}{P_{\sin\delta}} + \htmlClass{t-c}{P_{\cos\delta}}

(10.60, 10.61)

Bettini p. 463. A series expansion in the hierarchy parameter, due to Freund (2001) — the same α of (10.9) that let §10.2 use two flavours.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

💡 What this really says — run at the maximum and run both beams — the design follows from three terms

Three terms, and they do three different jobs. $P_0$ is the size of the signal. $P_{\cos\delta}$ moves both beams the same way, so it is a nuisance. Only $P_{\sin\delta}$ flips sign, and a sign flip between two beams is exactly what a CP-violating observable looks like.

The design consequence is immediate and it is why both experiments look the way they do: run at the oscillation maximum, and run both beams. At the maximum the phase is $\pi/2$ , so the $\cos(\text{phase})$ in $P_{\cos\delta}$ is zero and the nuisance term switches itself off. Then the difference between the two beams is the CP-odd term alone.

\begin{aligned} \htmlClass{t-0}{P_0} &= \sin^2\theta_{23}\sin^2 2\theta_{13}\,\sin^2\!\left(1.27\Delta\tfrac{L}{E}\right)\\[3pt] \htmlClass{t-s}{P_{\sin\delta}} &= \mp\,\alpha\sin\delta\,\cos\theta_{13}\sin2\theta_{12}\sin2\theta_{23}\sin2\theta_{13}\,\sin^2\!\left(1.27\Delta\tfrac{L}{E}\right)\\[3pt] \htmlClass{t-c}{P_{\cos\delta}} &= \alpha\cos\delta\,\cos\theta_{13}\sin2\theta_{12}\sin2\theta_{23}\sin2\theta_{13}\,\htmlClass{t-k}{\cos\!\left(1.27\Delta\tfrac{L}{E}\right)}\sin^2\!\left(1.27\Delta\tfrac{L}{E}\right) \end{aligned}

(10.62, 10.63, 10.64)

Bettini p. 464. The upper sign in P_sinδ is for neutrinos — WITH THE SIGN ASSIGNMENT FLIPPED from the book's, so that it agrees with p. 465 and with the data; see the erratum below.

Every symbol, one at a time

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🔢 Worked example — which sign does what, at T2K’s baseline

P(ν_e) and P(ν̄_e) at δ = ±π/2

import numpy as np
# T2K at the first oscillation maximum: L = 295 km, E = 0.6 GeV
Dm2, L, E = 2455e-6, 295.0, 0.6
D = 1.27*Dm2*L/E
s23s, s13s, s12s, alpha = 0.551, 0.0214, 0.304, 0.0299
s2t12 = 2*np.sqrt(s12s*(1-s12s)); s2t23 = 2*np.sqrt(s23s*(1-s23s))
s2t13 = 2*np.sqrt(s13s*(1-s13s)); c13 = np.sqrt(1-s13s)

P0   = s23s*s2t13**2*np.sin(D)**2                                    # (10.62)
Psin = alpha*c13*s2t12*s2t23*s2t13*np.sin(D)**2                      # (10.63), magnitude
Pcos = alpha*c13*s2t12*s2t23*s2t13*np.cos(D)*np.sin(D)**2            # (10.64), at delta = 0
print(f"phase 1.27 dm2 L/E = {D:.3f} rad = {D/(np.pi/2):.3f} x (pi/2)")
print(f"cos(phase) = {np.cos(D):+.4f}  ->  the cos-delta term (10.64) is {Pcos:.5f}, "
      f"{100*abs(Pcos)/P0:.1f} % of P0\n")
print(f"P0 = {P0:.5f}    |P_sin| = {Psin:.5f}  ({100*Psin/P0:.0f} % of P0)\n")

for name, sgn in (("as printed: + for nu", +1), ("flipped:    - for nu", -1)):
    for d_deg, lab in ((-90, "delta = -pi/2"), (+90, "delta = +pi/2")):
        sd = np.sin(np.deg2rad(d_deg))
        Pnu, Pnb = P0 + sgn*sd*Psin, P0 - sgn*sd*Psin
        who = "nu_e enhanced" if Pnu > Pnb else "nu_e SUPPRESSED"
        print(f"{name}, {lab}:  P(nu)={Pnu:.5f}  P(nubar)={Pnb:.5f}  -> {who}")
    print()

prints

phase 1.27 dm2 L/E = 1.533 rad = 0.976 x (pi/2)
cos(phase) = +0.0378  ->  the cos-delta term (10.64) is 0.00030, 0.6 % of P0

P0 = 0.04609    |P_sin| = 0.00782  (17 % of P0)

as printed: + for nu, delta = -pi/2:  P(nu)=0.03827  P(nubar)=0.05391  -> nu_e SUPPRESSED
as printed: + for nu, delta = +pi/2:  P(nu)=0.05391  P(nubar)=0.03827  -> nu_e enhanced

flipped:    - for nu, delta = -pi/2:  P(nu)=0.05391  P(nubar)=0.03827  -> nu_e enhanced
flipped:    - for nu, delta = +pi/2:  P(nu)=0.03827  P(nubar)=0.05391  -> nu_e SUPPRESSED

Two things worth carrying away, before the erratum.

The nuisance term really does switch itself off. T2K’s phase is 0.976 of $\pi/2$ , so $\cos(\text{phase}) = 0.038$ and $P_{\cos\delta}$ contributes 0.6 % of $P_0$ . The experiment does not have to model it away; the geometry does.

The asymmetry is big. $P_{\sin\delta}$ is 17 % of $P_0$ , so between the two beams the rates differ by up to 34 %. Against a signal of a few hundred events that is a measurable difference — which is why a hint exists at all after a decade of running.

Erratum — (10.63)‘s sign assignment contradicts p. 465 and the data

The book writes $P_{\sin\delta} = \pm\alpha\sin\delta\,(\ldots)$ and says ”+ is for neutrinos and − for antineutrinos”. Two pages later it says that values of the phase near $\delta = -\pi/2$ enhance $\nu_\mu\to\nu_e$ and suppress $\bar\nu_\mu\to\bar\nu_e$ .

Those cannot both be true. With the + assignment, $\sin(-\pi/2) = -1$ makes $P_{\sin\delta}$ negative for neutrinos, so $\delta = -\pi/2$ suppresses $\nu_e$ appearance — as the snippet above prints, 0.0383 against 0.0539 for antineutrinos.

The argument is clean at T2K’s baseline precisely because the other first-order term is switched off there: at 0.976 of the oscillation maximum $P_{\cos\delta}$ is 0.6 % of $P_0$ , so $P_{\sin\delta}$ alone decides the sign of the asymmetry.

p. 465 is the one that matches nature. T2K reports an excess of $\nu_e$ and a deficit of $\bar\nu_e$ candidates, with a best fit at $\delta \approx -1.89$ rad — near $-\pi/2$ . So the sign in (10.63) should be − for neutrinos and + for antineutrinos, which is how the equation is written above.

(Two smaller slips in the same passage. p. 463 attributes α to “Eq. (10.7)”; α is defined in (10.9), and (10.7) defines Δm². And the same sentence promises “second-order effects in the hierarchy parameter” where the next line, and (10.61), are explicitly first order.)

P(ν_μ → ν_e)
P(ν̄_μ → ν̄_e)

Eqs. (10.61)–(10.64) at T2K's L and E, with the corrected sign. The two curves cross at δ = 0° and 180° — the CP-conserving points, where the beams must agree — and separate most at 90° and 270°. The global fit sits at 238°, in the half where ν_e appearance is enhanced. This plot is the entire experiment: measure two rates, read off where you are on the horizontal axis.

Matter gets in the way, a little

x = \pm\frac{2\sqrt2\,G_F \htmlClass{t-n}{N_e} E}{\Delta m^2}

(10.65)

Bettini p. 464 — the matter parameter for a beam crossing the Earth's crust. The book calls Nₑ the 'neutrino number density'; it is the ELECTRON density, as (10.34) and (10.38) both say.

Every symbol, one at a time

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💡 What this really says — the Earth is a small correction that carries the mass ordering in its sign

This is $A/\Delta m^2$ from §10.3 — the same matter parameter, written as a dimensionless ratio. For T2K it is 0.05 and for NOvA 0.17, so the Earth is a small correction rather than the whole story it was in the Sun.

Small, but not ignorable, and it carries a sign that depends on the mass ordering. That is genuinely useful: matter effects and $\delta$ both make the $\nu$ and $\bar\nu$ rates differ, so the two are entangled — which is why NOvA at 810 km, with three times T2K’s matter effect, adds information T2K cannot supply, and why the global fit prefers normal ordering partly because of these experiments.

\sin^2\!\left(1.27\Delta\tfrac{L}{E}\right) \Rightarrow \frac{\sin^2\!\left((1-x)1.27\Delta\tfrac{L}{E}\right)}{(1-x)^2} \qquad\qquad \sin^2\!\left(1.27\Delta\tfrac{L}{E}\right) \Rightarrow \frac{\sin\!\left(x\,1.27\Delta\tfrac{L}{E}\right)\sin\!\left((1-x)1.27\Delta\tfrac{L}{E}\right)}{x(1-x)}

(10.66, 10.67)

Bettini p. 464. Two substitutions: the first in P₀, the second in both first-order terms.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

💡 What this really says — matter fakes the very asymmetry it is being used to measure

Matter stretches the oscillation. Replacing the phase $\varphi$ by $(1-x)\varphi$ is a change of effective wavelength, and dividing by $(1-x)^2$ keeps the amplitude right — exactly the structure of §10.3’s effective mixing angle, expanded for small $x$ .

Since $x$ flips sign between $\nu$ and $\bar\nu$ , matter by itself makes the two rates differ even at $\delta = 0$ . Separating that fake asymmetry from the real one is the central difficulty of the whole programme.

The off-axis trick

Both experiments point their beams slightly away from the far detector. It sounds perverse and it is the reason they work.

🔢 Worked example — how pointing away narrows the beam

A pion decaying in flight gives $E_\nu = 0.427\,E_\pi/(1+\gamma^2\theta^2)$ . On axis that is proportional to $E_\pi$ , so a broad pion spectrum gives a broad neutrino spectrum. Off axis, the $\gamma^2$ in the denominator cancels the growth.

On-axis versus off-axis neutrino energy

import numpy as np
m_pi, m_mu = 0.13957, 0.10566            # GeV
E_star = (m_pi**2 - m_mu**2)/(2*m_pi)    # neutrino energy in the pion rest frame
k = 2*E_star/m_pi
print(f"E*(nu) in the pion frame = {E_star*1e3:.2f} MeV,  so E_nu(0 deg) = {k:.3f} E_pi\n")

def E_nu(E_pi, deg):                     # E_nu = k E_pi / (1 + gamma^2 theta^2)
    g = E_pi/m_pi
    return k*E_pi/(1 + (g*np.deg2rad(deg))**2)

E = np.linspace(0.5, 20, 20000)
print(" off-axis   peak E_nu   at E_pi    spread of E_nu over E_pi = 2-10 GeV")
for deg in (0.0, 1.0, 2.5, 14e-3*180/np.pi):
    y = E_nu(E, deg)
    band = E_nu(np.linspace(2, 10, 400), deg)
    lo, hi = band.min(), band.max()
    print(f"  {deg:5.2f} deg  {y.max():7.3f} GeV  {E[y.argmax()]:6.2f} GeV"
          f"      {lo:.2f} - {hi:.2f} GeV  (factor {hi/lo:.1f})")
print("\nT2K sits 2.5 deg off axis at 0.6 GeV; NOvA 14 mrad off axis at 2 GeV")

prints

E*(nu) in the pion frame = 29.79 MeV,  so E_nu(0 deg) = 0.427 E_pi

off-axis   peak E_nu   at E_pi    spread of E_nu over E_pi = 2-10 GeV
 0.00 deg    8.538 GeV   20.00 GeV      0.85 - 4.27 GeV  (factor 5.0)
 1.00 deg    1.707 GeV    8.00 GeV      0.80 - 1.71 GeV  (factor 2.1)
 2.50 deg    0.683 GeV    3.20 GeV      0.40 - 0.68 GeV  (factor 1.7)
 0.80 deg    2.128 GeV    9.97 GeV      0.82 - 2.13 GeV  (factor 2.6)

T2K sits 2.5 deg off axis at 0.6 GeV; NOvA 14 mrad off axis at 2 GeV

Read the last column. On axis, pions from 2 to 10 GeV make neutrinos spanning a factor 5 in energy. At 2.5° off axis the same pions make neutrinos spanning a factor 1.7, all piled up near 0.68 GeV — which is T2K’s quoted 600 MeV. NOvA’s 14 mrad gives a peak at 2.13 GeV, its quoted 2 GeV.

You give up flux and you buy a narrow band, and a narrow band is what lets you sit exactly on the oscillation maximum, where §10.2b showed K2K and MINOS could not. It is also why the beam energy is no longer a free parameter: the angle sets it.

⚙️ Engineer’s bridge — the off-axis beam is a stationary-phase filter

The mechanism is one you have used. $E_\nu(E_\pi)$ rises, turns over and falls, so near its maximum $dE_\nu/dE_\pi = 0$ — a wide range of inputs maps onto a narrow range of outputs. That is a stationary point of the transfer function, and it concentrates the output the same way a stationary point of phase concentrates an integral, or a well-chosen operating point linearises away a nuisance variable.

Concretely: the source spectrum is broad and you cannot narrow it; instead you choose a mapping that is flat over the part you care about. Zero derivative means zero first-order sensitivity to the input spread.

Where it breaks: the price is real and it is not a free lunch. Off-axis flux falls steeply with angle, so T2K needs a megawatt-class proton beam to make up for the neutrinos it throws away. Stationary-phase filtering costs signal, here as everywhere.

Bettini §10.5. The two experiments hunting δ, and the two being built to finish the job.

Experiment↕	Beam↕	L↕	E↕	L/E↕	x↕	Far detector↕
T2K	J-PARC 30 GeV, 2.5° off axis	295 km	0.6 GeV
NOvA	FNAL NuMI 700 kW, 14 mrad off axis	810 km	2 GeV	405
DUNE	FNAL, under construction		few GeV	—	large	40 kt liquid argon TPC
Hyper-Kamiokande	J-PARC, under construction	295 km	0.6 GeV	492	0.05

x is the matter parameter of (10.65). The atmospheric first maximum is at L/E = 504 km/GeV.

Erratum — the quadrant labels on p. 465 disagree with §10.1’s own

p. 465 says that values of the phase “in the second quadrant (maximally for $\delta = -\pi/2$ ) enhance $\nu_\mu\to\nu_e$ ”, and those “in the first quadrant (maximally for $\delta = +\pi/2$ )” suppress it.

The physics is right; the labels are not. $\delta = -\pi/2 \equiv 270^\circ$ sits on the third/fourth boundary and is in no sense the second quadrant; $\delta = +\pi/2 = 90^\circ$ is on the first/second boundary. With the book’s own convention $\delta \in [0^\circ, 360^\circ)$ from (10.5), the $\nu_e$ -enhancing half is 180°–360° with its maximum at 270°, and the suppressing half is 0°–180° with its maximum at 90°.

§10.1 uses that convention correctly, calling the best fit $\delta = 238^\circ$ “close to the maximum in the third quadrant”. The two sections label the same region differently.

🔬 Experiment card — T2K, and the first hint of leptonic CP violation

Apparatus

A 30 GeV proton beam at J-PARC makes pions; the beamline is aimed 2.5° away from Super-Kamiokande, 295 km across Japan, which narrows the neutrino spectrum to a band around 600 MeV. A near-detector complex at 280 m measures the flux, angular distribution and energy spectrum before any oscillation. The horns are reversed periodically so the same apparatus runs a $\nu_\mu$ beam and a $\bar\nu_\mu$ beam.

What is measured

Electron-like rings in Super-Kamiokande, from $\nu_e n \to e^-p$ and $\bar\nu_e p \to e^+n$ — the appearance channel. The observable is the difference between the neutrino and antineutrino rates, because that is where $\delta$ lives and nothing else does.

Two features of the design do the work. The process has no threshold, so the beam can be soft, which suppresses the $\pi^0 \to 2\gamma$ background from non-oscillated $\nu_\mu$ . And sitting at 98 % of the oscillation maximum makes the CP-even term of (10.64) vanish.

The result

An excess of $\nu_e$ candidates and a deficit of $\bar\nu_e$ . Fitting, for normal ordering, $\delta = -1.89^{+0.70}_{-0.58}$ rad; for inverted, $-1.38^{+0.48}_{-0.54}$ . In both cases the CP-conserving points $\delta = 0$ and $\delta = \pi$ are excluded at 95 % confidence.

What it proved

Nothing yet, in the strict sense — 95 % is roughly $2\sigma$ against a field standard of $5\sigma$ , and the measurement is limited by statistics rather than by systematics. What it established is that the measurement is possible: the asymmetry is large enough to see with a few hundred events, which is why Hyper-Kamiokande and DUNE are being built rather than argued about.

The T2K result, for normal ordering, is $\delta = -1.89^{+0.70}_{-0.58}$ rad, and for inverted $-1.38^{+0.48}_{-0.54}$ . In both cases the CP-conserving points $\delta = 0$ and $\delta = \pi$ are excluded at 95 % confidence. That is a strong hint from a field that calls $5\sigma$ a discovery, and it is limited by statistics — which is precisely what Hyper-Kamiokande and DUNE are for.

The absolute scale: three ways in, none of them easy

Oscillation is blind to the mass scale, because the scale lives in the part of the Hamiltonian proportional to the identity (§10.2a). Three observables reach it, and they are three different functions of the same three masses.

Bettini p. 466. Three probes of the absolute mass, and they do not measure the same thing.

Probe↕	Observable↕	Kind of sum↕	Present limit↕
β-decay end point	$m_{\nu_e} = \left(\sum \|U_{ei}\|^2 m_i^2\right)^{1/2}$		< 800 meV
cosmology	$\sum_{i=1}^{3} m_i$	plain sum
neutrinoless double β	$M_{ee} = \left\|\sum U_{ei}^2 m_i\right\|$

💡 What this really says — three weightings of the same three masses, and only one of them cannot cancel

Three probes, three different weightings of $(m_1, m_2, m_3)$ , and that is a feature. If two of them ever disagree, the three-neutrino picture is wrong somewhere — a sterile state, a non-standard interaction, or a cosmology that is not ΛCDM.

Note especially what the middle column says about cancellation. Beta decay sums squares with positive weights, so nothing can hide. Double beta decay sums signed complex numbers, so a null result never excludes the Majorana hypothesis. Same three masses, opposite epistemic status.

Cosmology: the neutrinos that would not clump

📐 Physics you need first — the cosmological budget, in one box

This is the only place in the book that leans on cosmology, and it assumes vocabulary the site has not taught. Four ideas are enough.

1. The critical density. General relativity ties the expansion rate to the total density. There is one density, $\rho_c = 3H_0^2/8\pi G$ , for which space is exactly flat. Every component is then quoted as a fraction of it: $\Omega_X \equiv \rho_X/\rho_c$ . Because $\rho_c$ itself carries $H_0^2$ , every $\Omega$ carries a hidden $1/h^2$ — which is the whole content of the erratum below.

2. $h$ . The Hubble constant is written $H_0 = 100\,h$ km s⁻¹ Mpc⁻¹ with $h \approx 0.7$ , precisely so that results can be quoted as $\Omega h^2$ and stay correct while $h$ is being argued about.

3. The budget. Measurements say the total is critical to within 1.5 % — space is flat — and splits as 5 % ordinary matter, 27 % cold dark matter, 68 % dark energy. Nobody knows what the last two are. Neutrinos are a fourth, much smaller, entry, and only an upper limit on it exists.

4. Where the information comes from. When the Universe was 380 000 years old it cooled enough for atoms to form and became transparent; that snapshot is the cosmic microwave background, and the sizes of its temperature fluctuations encode the contents. Later structure — the distribution of millions of galaxies, and a 150 Mpc shell called the baryon acoustic oscillation — extends the reach to lower redshift. Neutrino mass shows up in how those structures failed to grow.

\Omega_\nu\,\htmlClass{t-h}{h^2} = \frac{\sum_{i=1}^{3} m_i}{\htmlClass{t-k}{93.14\ \text{eV}}} \qquad\Longrightarrow\qquad \Omega_\nu \approx \frac{\sum_{i=1}^{3} m_i}{45.6\ \text{eV}}

(10.68)

Bettini p. 467, WITH h² MOVED to the left-hand side — the book prints it in the numerator on the right; see the erratum below. h ≈ 0.7 is the Hubble constant in units of 100 km s⁻¹ Mpc⁻¹.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

💡 What this really says — the tightest limit of the three comes from never detecting one

Relic neutrinos are the second most abundant particle in the Universe after photons — 339.5 of them per cubic centimetre, everywhere, left over from one second after the Big Bang. They outnumber protons by a billion to one.

So even a milli-electronvolt of mass each is a real contribution to the mass–energy budget, and not seeing that contribution is a limit. That is why cosmology gives the tightest bound of the three probes despite never detecting a single neutrino.

🔢 Worked example — deriving 93.14 eV, and finding the h² in the wrong place

Check (10.68) against the book's own relic density

# Where the 93.14 eV of (10.68) comes from, and where the h^2 belongs.
n_nu   = 339.5          # cm^-3, the book's own total relic density
rho_c  = 1.05371e4      # eV cm^-3, the critical density DIVIDED by h^2
h      = 0.7

per_species = n_nu/3    # Sum m_i pairs each mass with ITS OWN species density
coeff = rho_c/per_species
print(f"relic density per species = {per_species:.2f} cm^-3")
print(f"rho_c / h^2 = {rho_c:.4g} eV cm^-3")
print(f"=> Omega_nu h^2 = Sum m / {coeff:.2f} eV      (the book prints 93.14)\n")

Sm = 0.130              # eV, the cosmological limit of (10.69)
correct = Sm/(coeff*h**2)
printed  = h**2*Sm/93.14
print(f"for Sum m = {Sm*1e3:.0f} meV:")
print(f"  correct   Omega_nu = Sum m /(93.14 h^2) = {correct:.5f}")
print(f"  as printed Omega_nu = h^2 Sum m / 93.14 = {printed:.5f}"
      f"   -> too small by a factor {correct/printed:.2f} = 1/h^4")
print(f"\nand the book's own substitution h^2 -> 0.5 in the NUMERATOR should be")
print(f"1/h^2 -> {1/h**2:.2f} in the denominator, i.e. Sum m / {coeff*h**2:.1f} eV")

prints

relic density per species = 113.17 cm^-3
rho_c / h^2 = 1.054e+04 eV cm^-3
=> Omega_nu h^2 = Sum m / 93.11 eV      (the book prints 93.14)

for Sum m = 130 meV:
correct   Omega_nu = Sum m /(93.14 h^2) = 0.00285
as printed Omega_nu = h^2 Sum m / 93.14 = 0.00068   -> too small by a factor 4.17 = 1/h^4

and the book's own substitution h^2 -> 0.5 in the NUMERATOR should be
1/h^2 -> 2.04 in the denominator, i.e. Sum m / 45.6 eV

The book’s own $n_\nu = 339.5\ \mathrm{cm^{-3}}$ , divided by three because $\sum m_i$ pairs each mass with its own species, and divided into the critical density, gives 93.11 eV — its own printed constant, to four digits. The constant is right; the $h^2$ is on the wrong side.

Erratum — (10.68) has h² in the numerator, and the Hubble constant in kpc⁻¹

The $h^2$ . (10.68) is printed as $\Omega_\nu = h^2\sum m_i/(93.14\ \mathrm{eV})$ , then evaluated by substituting $h^2 \approx 0.5$ in that numerator. The standard relation — and the one the book’s own $n_\nu$ reproduces — is $\Omega_\nu h^2 = \sum m_i/(93.14\ \mathrm{eV})$ , so $h^2$ belongs in the denominator. As printed the result is too small by $1/h^4 \approx 4.2$ : for $\sum m_i = 130$ meV it gives $\Omega_\nu = 0.00068$ where the answer is 0.00285.

The tell is dimensional in spirit rather than in units: $\Omega$ is a ratio to the critical density, and the critical density itself carries $H_0^2$ . Anything normalised to it must carry $1/h^2$ .

The units of $h$ . The same sentence defines $h \approx 0.7$ as the Hubble constant in units of 100 km (s kpc)⁻¹. It is Mpc⁻¹. With kpc, $H_0$ would be a thousand times larger and the age of the Universe about 14 million years. The surrounding text uses Mpc correctly four times — 150 Mpc for the BAO scale, hundreds of Mpc for free streaming, 100 Mpc for a supercluster — and even uses kpc correctly for a galaxy, so this is a slip and not a misunderstanding.

💡 What this really says — why cosmology can weigh a neutrino at all

Free streaming is the whole mechanism. When the Universe cooled enough for neutrinos to become non-relativistic, they were still moving far faster than the escape velocity of a galaxy-sized overdensity, so they leaked out of small structures instead of falling in. Heavier neutrinos slow down earlier and leak less.

The consequence is a suppression of the matter power spectrum below a characteristic scale — hundreds of Mpc for masses around 100 meV. Cosmology does not weigh a neutrino by catching one; it weighs it by noticing which structures failed to form. That is why the limit is tight and why it is model-dependent: it is a statement about $\Lambda$ CDM as much as about neutrinos.

Tritium: the last electronvolt of a beta spectrum

\htmlClass{t-d}{{}^3\mathrm{H} \to {}^3\mathrm{He} + e^- + \bar\nu_e} \qquad\qquad \frac{dN_e}{dE_e} \sim F(Z,E_e)\,p_e^2\,\htmlClass{t-p}{E_\nu p_\nu} = F(Z,E_e)\,p_e^2\,\htmlClass{t-q}{(Q-E_e)^2}

(10.70, 10.71)

Bettini p. 468. Tritium is chosen for one reason: Q = 18.6 keV, the smallest useful end point in nature.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

💡 What this really says — everything is in the last few eV out of 18 600

Everything the experiment can see lives in the last few eV out of 18 600. The neutrino mass never appears as a peak or a resonance — only as a slight change in how the electron spectrum reaches its end.

The Kurie plot is what makes it tractable: divide out the phase space and the nuclear physics, and a massless neutrino gives a straight line. A straight line is a thing you can see a deviation from; a steeply falling spectrum is not.

Bettini Fig. 10.17, with the masses drawn far larger than reality so the structure is visible. The dotted line is what a massless neutrino gives; the solid curve carries one step per mass eigenstate, at Q − m_i, with heights |U_ei|². The real separations are milli-electronvolts on an 18 600 eV axis, so what is actually measured is the single weighted average of (10.73).

\frac{dN_e}{dE_e} \approx p_e^2(Q-E_e)\sum_{i=1}^{3}\htmlClass{t-u}{|U_{ei}|^2}\sqrt{(Q-E_e)^2-m_i^2} \qquad \htmlClass{t-m}{m_{\nu_e}} = \left(\sum_{i=1}^{3}|U_{ei}|^2 m_i^2\right)^{1/2} \approx \left(0.68m_1^2+0.30m_2^2+0.02m_3^2\right)^{1/2}

(10.72, 10.73)

Bettini p. 469. Three steps in principle; one weighted average in practice.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

💡 What this really says — there is no such thing as the mass of the electron neutrino

The electron neutrino is not a mass eigenstate, so “the mass of the electron neutrino” is not a well-defined quantity — and the experiment nevertheless measures something clean. The three steps are far too close to resolve, so what survives is the effective electron neutrino mass $\sqrt{\sum|U_{ei}|^2 m_i^2}$ , weighted by exactly the probabilities that a $\nu_e$ is each mass state.

That is the honest way to read every “neutrino mass limit” you will see quoted: it is a limit on a specific weighted average, not on any one particle.

🔢 Worked example — the floors, and whether KATRIN can reach them

Fig. 10.18, computed: m_νe against Σm for both orderings

import numpy as np
dm2, Dm2 = 73.4, 2455.0                 # meV^2
U2 = np.array([0.68, 0.30, 0.02])       # |U_ei|^2 from (10.73)

def masses(lightest, order):
    if order == "NO":
        m1 = lightest; m2 = np.sqrt(m1**2 + dm2); m3 = np.sqrt((m1**2+m2**2)/2 + Dm2)
    else:
        m3 = lightest; half = m3**2 + Dm2
        m1 = np.sqrt(half - dm2/2); m2 = np.sqrt(half + dm2/2)
    return np.array([m1, m2, m3])

print(" ordering  lightest   m1    m2    m3    Sum m   m_nu_e   (all meV)")
for order in ("NO", "IO"):
    for lightest in (0.0, 30.0, 100.0):
        m = masses(lightest, order)
        mve = np.sqrt(U2 @ m**2)
        print(f"   {order}     {lightest:6.1f}  {m[0]:5.1f} {m[1]:5.1f} {m[2]:5.1f}"
              f"  {m.sum():6.1f}   {mve:6.1f}")

print("\nfloors, with the lightest state massless:")
for order in ("NO", "IO"):
    m = masses(0.0, order)
    print(f"  {order}: Sum m > {m.sum():.1f} meV   and   m_nu_e > {np.sqrt(U2 @ m**2):.1f} meV")
print("\nKATRIN: design sensitivity 200 meV, present limit 800 meV")
print("cosmology (10.69): Sum m < 130 meV")

prints

 ordering  lightest   m1    m2    m3    Sum m   m_nu_e   (all meV)
 NO        0.0    0.0   8.6  49.9    58.5      8.5
 NO       30.0   30.0  31.2  58.2   119.4     31.2
 NO      100.0  100.0 100.4 111.8   312.1    100.4
 IO        0.0   49.2  49.9   0.0    99.1     48.9
 IO       30.0   57.6  58.2  30.0   145.8     57.4
 IO      100.0  111.4 111.8 100.0   323.2    111.3

floors, with the lightest state massless:
NO: Sum m > 58.5 meV   and   m_nu_e > 8.5 meV
IO: Sum m > 99.1 meV   and   m_nu_e > 48.9 meV

KATRIN: design sensitivity 200 meV, present limit 800 meV
cosmology (10.69): Sum m < 130 meV

Two conclusions the book leaves for the reader to draw from Fig. 10.18.

KATRIN cannot reach either floor. Its design sensitivity is 200 meV; the inverted-ordering floor for $m_{\nu_e}$ is 49 meV and the normal-ordering floor is 8.5 meV. A factor of four short of the more optimistic case. So a null KATRIN result at design sensitivity excludes nothing that oscillation has not already excluded — and a positive one would be a crisis, because it would sit above the cosmological limit and demand that something in the three-neutrino picture is wrong.

That is not an argument against the experiment. KATRIN is the only one of the three probes that assumes nothing — no cosmological model, no Majorana hypothesis, no nuclear matrix elements. Its value is that it is model-independent, and in the quasi-degenerate region, where all three masses are nearly equal, it is directly competitive.

And notice the last column collapses onto the first. Once the lightest mass exceeds about 100 meV, $m_{\nu_e} \approx m_1 \approx \sum m_i/3$ and the three probes measure the same thing. The interesting region — where they differ, and where the ordering is visible — is exactly the region below 100 meV that cosmology has already reached and beta decay has not.

inverted ordering, Δm² < 0
normal ordering, Δm² > 0

Bettini Figs. 10.16 and 10.18 combined and computed from (10.8) and (10.73). The two curves separate only below Σm ≈ 300 meV; above it the spectrum is quasi-degenerate and the ordering is invisible. The region between the two dashed lines — everything cosmology allows and KATRIN cannot see — is where the answer almost certainly lies.

🔬 Experiment card — KATRIN, weighing a neutrino with no model at all

Apparatus

A windowless gaseous tritium source, and the world’s largest electrostatic spectrometer — 23 m long, 10 m in diameter — at Karlsruhe. Electrons from $^3\mathrm{H} \to {}^3\mathrm{He} + e^- + \bar\nu_e$ are guided magnetically into it without changing their energy, and the spectrometer rejects all but the last few electronvolts of an 18.6 keV spectrum.

What is measured

The shape of the electron spectrum in its final few eV, where a neutrino mass moves the end point to $Q - m$ and turns the approach vertical. The observable is the incoherent weighted average $m_{\nu_e} = (\sum|U_{ei}|^2m_i^2)^{1/2}$ — the three steps of (10.72) are far too close together to resolve.

The result

Running since 2018. The present limit is $m_{\nu_e} < 800$ meV; the design sensitivity is 200 meV.

What it proved

That the mass scale can be bounded kinematically, with no cosmological model, no Majorana hypothesis and no nuclear matrix elements. That is the whole value of the measurement, because the two competing probes each rest on an assumption. It will not reach the floors — 49 meV for inverted ordering, 8.5 meV for normal — so a null result at design sensitivity settles nothing that oscillation has not, and a positive one would contradict cosmology and be a crisis.

KATRIN is the experiment that gets there: a 23 m long, 10 m diameter electrostatic spectrometer at Karlsruhe, guiding electrons from a tritium source without changing their energy and measuring them to below an electronvolt. It started in 2018; its present limit is $m_{\nu_e} < 800$ meV.

🔑 If you remember only three things

Only appearance can see δ, and appearance needed θ₁₃ ≠ 0. A single reactor result in 2012 is what gave the entire CP programme something to measure.
Pointing the beam away from the detector is what makes it usable. The energy mapping turns over off axis, so a broad spread of pion energies lands in one narrow neutrino band.
A weaker limit that assumes nothing can beat a tighter one that assumes a cosmology. KATRIN will not reach either floor, and it is still the only probe with nothing to withdraw.

Where this goes next

Two open questions, two very different states of play. $\delta$ has a $3\sigma$ hint and a clear route to a measurement — more statistics, at DUNE and Hyper-Kamiokande. The absolute mass has three probes converging on a region between about 10 and 130 meV, and the tightest of them depends on cosmology being right.

§10.7 asks the question underneath both: is the neutrino its own antiparticle? If it is, there are two more physical phases nobody has ever measured, lepton number is not conserved, and a third mass observable — the coherent, cancellable $M_{ee}$ — becomes accessible through a decay that has never been seen.

✅ Check yourself — δ and the absolute mass scale

0/6 answered · 0 correct

1.Why can no disappearance experiment, however precise, ever measure CP violation in neutrinos?
2.T2K deliberately sits at 98 % of the first oscillation maximum. What does that buy, beyond the obvious large signal?
3.An off-axis beam at 2.5° gives neutrinos spanning a factor 1.7 in energy from pions spanning a factor 5. Why?
4.The three probes of the absolute mass scale weight the masses differently. Which statements follow? (Select all that apply.)
5.KATRIN's design sensitivity is 200 meV. The inverted-ordering floor for m_νe is 49 meV and the normal-ordering floor 8.5 meV. What does that mean for the experiment?
6.Eq. (10.68) as printed gives Ω_ν = h²Σm/(93.14 eV). What is wrong with it, and how does the book itself show that?