---
title: Mass-Energy and Binding
module: Energy
moduleNumber: 3
lessonNumber: 4
order: 304
summary: >
  Relativity puts rest itself on the energy ledger: a mass $m$ carries energy
  $mc^2$ even when it sits still, so weighing a system's separated pieces and
  weighing the assembled whole give different answers, and the gap is binding
  energy. We convert freely between mass units and MeV, compute the energy that
  holds a nucleus together, and read the binding-energy-per-nucleon curve that
  explains why fusing light nuclei and splitting heavy ones both release energy.
  Reaction $Q$ values, thresholds, and recoil then follow from the same
  mass-difference accounting, once the frame and mass convention are fixed.
topics: [Energy]
draft: false
sources:
  - book: Tipler & Mosca
    ref: "Ch. 7 — Conservation of Energy; §7-4 Mass and Energy"
---

## Rest energy and system mass

A body at rest in the selected frame has rest energy:

$$
E_0=mc^2.
$$

The $m$ here is the invariant mass of the whole system, and it counts every internal
energy in the centre-of-momentum frame: kinetic, thermal, chemical, and binding.
Heating a sealed container, compressing a spring inside it, or binding particles
together changes the total energy and therefore the mass by $\Delta m=\Delta E/c^2$.
The shift usually sits far below balance resolution, yet it follows exactly from the
energy-momentum relation.

The consequence runs both ways. A system whose constituents move but sum to zero
total momentum still carries their internal kinetic energy, so its mass exceeds the
sum of the individual rest masses. A bound system releases energy during assembly, so
its mass falls below the separated total. That rest-mass difference is the _mass
defect_. Four-momentum stays conserved throughout.

Energy accounting needs a named system boundary. For an isolated reaction or
assembly, the total four-momentum is the same before and after. In the
centre-of-momentum frame a positive mass difference can emerge as product kinetic
energy, radiation, recoil, thermal motion, or some mix of these. The mass-energy
relation fixes the total available energy; four-momentum conservation and the final
states fix how it splits. Measuring one photon or one product's kinetic energy can
miss recoil or excitation energy elsewhere.

Take the sign convention from initial minus final mass:

$$
Q=(m_{\rm initial}-m_{\rm final})c^2.
$$

Positive $Q$ leaves energy for final kinetic energy, radiation, or extra products;
negative $Q$ demands energy from outside. The definition holds for chemical, nuclear,
and particle processes as long as every mass belongs to one consistently defined side.

## Mass defect and nuclear binding energy

A nucleus containing $Z$ protons and $N$ neutrons has a mass lower than separated
nucleons when it is bound. Its mass defect is

$$
\Delta m=Zm_p+Nm_n-m_{\rm nucleus},
$$

and the binding energy is

$$
B=\Delta m c^2.
$$

By convention $B$ is positive for a bound nucleus: it is the energy needed to pull
the nucleus apart into free protons and neutrons that end at rest in their
centre-of-momentum frame, and the same amount is released when they assemble and the
energy escapes. Binding energy belongs to the whole nucleus. Individual nucleon
separation energies depend on which nucleon leaves and on the final nucleus.

$$
% caption: Nuclear binding-energy accounting. Free protons and neutrons carry a larger combined rest energy than the bound nucleus; the positive binding energy equals the work needed for complete separation and the energy released during assembly.
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Binding energy per nucleon, $B/A$ with $A=Z+N$, compares nuclei of different size.
It is an average separation energy, not the cost of removing one specific proton or
neutron. Larger $B/A$ usually means more energy per nucleon for complete separation,
but reaction energetics still need the masses of the actual initial and final nuclei.
Average binding energy alone does not fix a reaction $Q$ value.

## Units and a helium-4 calculation

Nuclear masses are often given in atomic mass units, with
$1\ \mathrm u=1.66053906660\times10^{-27}\ \mathrm{kg}$. The working conversions are

$$
1\ \mathrm u\,c^2=931.494\ \mathrm{MeV},
\qquad
1\ \mathrm{MeV}=1.602176634\times10^{-13}\ \mathrm J.
$$

Atomic masses include the electrons; nuclear masses exclude them. Either convention
works, but every term in a mass-defect calculation must use the same one. A neutral
atom built from neutral hydrogen atoms and neutrons has its electron masses cancel
across the two sides, whereas mixing a nuclear mass with atomic masses leaves an
electron-mass error.

> **Worked example (helium-4 binding energy).** Take the nuclear masses
> $m_p=1.007276\ \mathrm u$, $m_n=1.008665\ \mathrm u$, and
> $m_{\rm He}=4.001506\ \mathrm u$.
>
> The separated nucleons weigh
> $$
> 2m_p+2m_n=4.031882\ \mathrm u,
> $$
> so the mass defect is $\Delta m=0.030376\ \mathrm u$ and the binding energy is
> $$
> B=(0.030376)(931.494)=28.30\ \mathrm{MeV}.
> $$
> Dividing by four gives $B/A=7.07\ \mathrm{MeV}$ per nucleon, equal to
> $4.53\times10^{-12}\ \mathrm J$ in total. This is the energy for complete
> disassembly of the nucleus.

$$
% caption: Qualitative binding energy per nucleon. The average rises steeply through the light nuclei, reaches a broad maximum among intermediate-mass nuclei, and declines gradually toward the heaviest; specific reaction energies still require the masses of the nuclei involved.
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A mass defect of a few hundredths of a unit already buys tens of MeV, because $c^2$
is large. The release calculation uses only the mass change, not the enormous rest
energy of the bulk material.

The sign of $Q$ decides whether an isolated process releases energy or needs input.
When the reaction carries nonzero total momentum in the laboratory frame, the product
kinetic energies also depend on recoil and cannot come from $Q$ alone; the
centre-of-momentum frame separates the invariant-mass balance from that
frame-dependent split. A nucleus at rest in one frame can recoil in another with no
change to its invariant mass or binding energy.

Mass precision caps energy precision. A six-decimal atomic mass in $\mathrm u$ leaves
a roughly keV-scale uncertainty after the $931.494\ \mathrm{MeV/u}$ conversion, and
subtracting nearly equal masses magnifies the relative error in a small $Q$, so keep
guard digits and cite the mass source. Identify the isotope too: proton number alone
fixes neither neutron number nor nuclear mass, and an abundance-averaged table entry
is useless when a specific isotope is meant.

## Relativistic energy, momentum, and invariant mass

A particle of invariant mass $m$ has energy and momentum forming one four-vector.
The Lorentz factor is $\gamma=(1-v^2/c^2)^{-1/2}$, so in an inertial frame

$$
E=\gamma mc^2,
\qquad
\vec p=\gamma m\vec v.
$$

Eliminating velocity gives the energy--momentum relation

$$
E^2=(pc)^2+(mc^2)^2.
$$

This relation packs rest energy and kinetic energy together without inventing a
velocity-dependent mass. It reduces to $E=mc^2$ at $p=0$ and to $E=pc$ for a massless
particle such as a photon. It is invariant: observers disagree on $E$ and
$\vec p$, but each recovers the same $m$ from $E^2-p^2c^2=m^2c^4$.

$$
% caption: Energy-momentum geometry. The total energy is the hypotenuse built from the momentum term and the rest-energy term; the relation is algebraic rather than a spatial triangle, and every inertial observer recovers the same invariant mass from it.
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$$

Kinetic energy is $K=E-mc^2=(\gamma-1)mc^2$. The nonrelativistic $K\simeq p^2/(2m)$
holds only when $p\ll mc$. At high momentum $pc$ dwarfs $mc^2$, so $E\simeq pc$ and
$K\simeq pc-mc^2$. Forcing $p^2/(2m)$ outside its range can yield an energy below the
rest energy while still implying an impossible speed. Relativistic kinetic energy
depends on momentum and invariant mass, and every massive particle stays below $c$.

System invariant mass is a separate calculation. For a collection with total energy
$E_{\rm tot}$ and total momentum $\vec p_{\rm tot}$,

$$
M^2c^4=E_{\rm tot}^2-(p_{\rm tot}c)^2.
$$

In the centre-of-momentum frame $\vec p_{\rm tot}=0$ and $Mc^2=E_{\rm tot}$, so
$M$ carries the constituent rest energies, their kinetic energies in that frame, and
their interactions. Two particles converging can exceed $m_1+m_2$ from their
centre-of-momentum kinetic energy; a bound nucleus falls below the separated sum
because assembly released energy. Adding individual rest masses is valid only for
well-separated constituents with negligible relative kinetic and interaction energy.

Decay and annihilation bookkeeping runs on total four-momentum, not energy alone. In
a two-body decay of a parent mass $M$ at rest into products $m_1$ and $m_2$, the
products leave with equal and opposite momentum whose magnitude energy and momentum
conservation fix:

$$
pc=\frac{\sqrt{[M^2-(m_1+m_2)^2][M^2-(m_1-m_2)^2]}}{2M}c^2.
$$

The decay is allowed only when $M\ge m_1+m_2$, and the mass difference plus the
product masses then set the kinetic energy and recoil. A parent at rest cannot decay
into one massive product while conserving momentum unless an external body joins in.
For the same reason electron--positron annihilation at rest cannot make a single
photon, since nothing would balance its momentum; the usual final state is two
back-to-back photons of equal energy.

$$
% caption: Electron-positron annihilation in the centre-of-momentum frame. The pair at rest has zero total momentum, so the two photons leave back to back with equal and opposite momenta; each carries half of the 1.022 MeV total rest energy, or 0.511 MeV.
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$$

> **Worked example (electron momentum).** An electron has rest energy
> $m_ec^2=0.511\ \mathrm{MeV}$. Give it kinetic energy $1.00\ \mathrm{MeV}$, so its
> total energy is $E=1.511\ \mathrm{MeV}$.
>
> The invariant relation gives the momentum:
> $$
> pc=\sqrt{E^2-(m_ec^2)^2}
> =\sqrt{(1.511)^2-(0.511)^2}\ \mathrm{MeV}
> =1.42\ \mathrm{MeV}.
> $$
> The nonrelativistic $p=\sqrt{2m_eK}$ would miss here, since the kinetic energy
> exceeds the rest energy.

Annihilation of an electron and positron both at rest releases
$2(0.511)=1.022\ \mathrm{MeV}$, half to each photon in the centre-of-momentum frame.
Give the pair nonzero total momentum in the laboratory and the two photon energies
differ there, though the pair's invariant mass does not.

Experimental accounting needs calibrated momentum or energy measurements and a stated
frame. Magnetic curvature yields momentum only after charge, field, and geometry
calibration; calorimeters yield deposited energy after leakage and response
corrections. Missing momentum can flag an unmeasured particle or radiation, but only
once detector acceptance and the initial four-momentum are fixed. Invariant-mass
reconstruction combines every measured product four-vector, and energy alone suffices
only in the centre-of-momentum frame.

## Reaction thresholds, recoil, and measurement limits

The $Q$ value compares rest masses of complete initial and final systems:

$$
Q=(m_{\rm initial}-m_{\rm final})c^2.
$$

With a stationary target, an endothermic reaction ($Q<0$) needs more laboratory
energy than $|Q|$. The projectile brings laboratory momentum, so the final system
must keep moving to conserve it, and part of the incident energy stays as that
translational kinetic energy instead of building the extra final rest mass.

A projectile $a$ striking a target $A$ at rest has invariant squared energy

$$
s=m_a^2c^4+m_A^2c^4+2m_Ac^2E_a,
$$

where $E_a$ is the projectile's total laboratory energy. At threshold the products
move together in the centre-of-momentum frame, so their invariant mass equals the sum
$M_f$ of their rest masses, and

$$
E_{a,\rm thr}=\frac{M_f^2c^4-m_a^2c^4-m_A^2c^4}{2m_Ac^2},
\qquad T_{a,\rm thr}=E_{a,\rm thr}-m_ac^2.
$$

The nonrelativistic shortcut $T_{\rm thr}\simeq -Q(1+m_a/m_A)$ holds only when the
threshold kinetic energy is small next to the rest energies; the invariant formula
holds in every frame and is the safe choice when that condition is in doubt.

$$
% caption: Threshold energy with a stationary target. The projectile must supply the negative Q value plus the unavoidable centre-of-momentum motion of the combined product, so the laboratory threshold exceeds the magnitude of an endothermic Q value.
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Recoil splits a positive $Q$ among the final products. In a two-body decay of a
parent at rest the product momenta are equal and opposite, and when the kinetic
energies stay small next to the rest energies the nonrelativistic shares are

$$
K_1=Q\frac{m_2}{m_1+m_2},
\qquad
K_2=Q\frac{m_1}{m_1+m_2}.
$$

The lighter product takes the larger kinetic energy, since both share one momentum
magnitude. For larger $Q$, drop the approximation and use the exact energy--momentum
relation per product. A photon emitted by a recoiling atom or nucleus also carries
momentum, so assigning the whole transition energy to the photon skips the emitter's
recoil; the same trap appears when a measured charged product stands in for an unseen
recoil partner.

$$
% caption: Two-body recoil in the parent rest frame. The products leave with equal and opposite momentum; because they share one momentum magnitude, the lighter product carries the larger kinetic energy in the nonrelativistic limit.
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$$

> **Worked example (endothermic threshold).** Fire a projectile of mass
> $1.00\ \mathrm u$ at a stationary target of mass $12.0\ \mathrm u$, and let the
> final rest-mass sum exceed the initial by $2.00\ \mathrm{MeV}/c^2$, so
> $Q=-2.00\ \mathrm{MeV}$.
>
> The low-energy threshold approximation gives
> $$
> T_{\rm thr}\simeq2.00\left(1+\frac{1.00}{12.0}\right)
> =2.17\ \mathrm{MeV}.
> $$
> The extra $0.17\ \mathrm{MeV}$ is the recoil cost of a stationary target.

> **Worked example (two-body recoil).** Let a parent of mass $2.901\ \mathrm u$ decay
> at rest into products of masses $1.000\ \mathrm u$ and $1.900\ \mathrm u$. The mass
> difference is $0.001\ \mathrm u$, so $Q=0.931\ \mathrm{MeV}$.
>
> The recoil approximation splits it as
> $$
> K_1=0.610\ \mathrm{MeV},\qquad K_2=0.321\ \mathrm{MeV},
> $$
> which sum to $Q$ within rounding. Both kinetic energies are small next to the
> corresponding rest energies, so the approximation applies.

Keep the centre-of-momentum and laboratory boundaries apart in data reduction.
Laboratory energy carries the centre-of-mass motion; centre-of-momentum energy strips
that collective motion out and leaves the energy available for final rest masses and
relative motion. A fixed-target run has less invariant energy than colliding beams at
the same total laboratory energy, since the target adds no opposite momentum.
Threshold plots should say whether the horizontal axis is projectile kinetic energy,
total centre-of-momentum energy, or excess above threshold.

At an exact threshold the relative momentum vanishes only in the centre-of-momentum
frame; the final particles still share one laboratory velocity, so zero excess energy
on a centre-of-momentum plot does not put every product at rest in the laboratory.
Just above threshold the allowed relative momentum is small and cross sections can
turn on steeply. A turn-on fitted from a yield curve is not automatically the
kinematic threshold, because target energy loss, beam-energy spread, and detector
acceptance shift or broaden it.

The recoil formulas have their own domain. They assume a two-body final state in the
parent rest frame with nonrelativistic kinetic energies. A third product lets the
first two carry unequal, non-opposite momenta, since it takes part of the vector
balance. A moving parent makes the laboratory product energies depend on emission
angle as well as the centre-of-momentum share. Reconstruct the parent or initial
four-momentum first, transform only if needed, then compare with the right-frame
prediction; a lone laboratory energy cannot fix a two-body recoil split without the
direction or the parent momentum.

Missing-mass reconstruction turns the same invariant relation on unobserved products.
With known initial four-momentum $P_i$ and measured final four-momentum $P_m$, the
missing four-momentum is $P_x=P_i-P_m$ and

$$
M_x^2c^4=(E_i-E_m)^2-|\vec p_i-\vec p_m|^2c^2.
$$

That $M_x$ is the invariant mass of the whole unmeasured system, not necessarily one
particle: it can hold several particles, radiation, or detector losses. Mass
spectrometry carries analogous limits, where charge state, calibration ions, field
scale, time-of-flight path length, dead time, and unresolved isotope peaks all bias
the mass-to-charge ratio. A mass defect below the instrumental resolution is not
recovered by multiplying a noisy mass difference by $c^2$.

Report the resolving power rather than burying it. In a magnetic spectrometer,
same-momentum particles of different charge curve differently, so a wrong charge
assignment gives the wrong mass-to-charge ratio; in time-of-flight work, an error in
flight distance or timing offset shifts the inferred velocity and matters most for
nearby peaks. Reference ions check the scale but do not cancel a drift between
reference and sample runs. A defensible mass difference carries its calibration model,
statistical uncertainty, and systematic resolution limit.

Measurement limits belong in every threshold or recoil claim:

- **Momentum resolution** depends on track length, field calibration, scattering, and alignment.
- **Energy resolution** depends on shower containment, response nonlinearity, and calibration.
- **Angular resolution** controls the vector subtraction used in missing mass.

Validate the full reconstruction against a calibrated reference reaction or mass peak,
and report the masses, momentum convention, target state, frame, missing-system
definition, and uncertainty covariance. Four-momentum bookkeeping constrains the
result without identifying the reaction mechanism or internal structure.

## Binding-energy systematics, fission, fusion, and energy yield

Binding energy is again the energy to separate a nucleus completely into free protons
and neutrons at rest in their common centre-of-momentum frame, read from the mass
defect:

$$
B=\left[Zm_p+(A-Z)m_n-m_{\rm nucleus}\right]c^2.
$$

where $A$ is the mass number and $Z$ the proton number. Larger $B/A$ usually means
less mass per nucleon and tighter binding, but it is not the energy of every reaction
that nucleus can undergo; the reaction energy still comes from the total mass
difference between the complete initial and final systems.

The $B/A$ curve climbs steeply through the light nuclei, peaks near iron and nickel,
then falls off gradually toward the heaviest. Fusing light nuclei or splitting a very
heavy one both move products upward on the curve, cutting total rest mass and opening
a positive $Q$. Nuclei at the peak have little to give in either direction, so the
curve is a bookkeeping guide, not a measure of how readily a reaction runs.

Balance mass number and charge number before any energy calculation: the sums of $A$
and $Z$ must match on both sides. A common fission channel is

$$
{}^{235}_{92}\mathrm U+{}^{1}_{0}\mathrm n
\longrightarrow{}^{141}_{56}\mathrm{Ba}+{}^{92}_{36}\mathrm{Kr}+3{}^{1}_{0}\mathrm n.
$$

with $235+1=141+92+3$ for mass number and $92=56+36$ for charge. Real fission spreads
over many channels, so a quoted fission energy must name the products and say whether
it counts only prompt kinetic energy or every recoverable contribution. The
engineering figure of about $200\ \mathrm{MeV}$ per fission sets the scale but does
not replace a mass-table calculation for a specified channel.

A fusion calculation uses nuclear masses throughout, or neutral-atom masses when the
electron counts cancel. Deuterium--tritium is clean because the neutral atomic masses
balance the two electrons:

$$
{}^{2}_{1}\mathrm H+{}^{3}_{1}\mathrm H
\longrightarrow{}^{4}_{2}\mathrm{He}+{}^{1}_{0}\mathrm n+Q.
$$

> **Worked example (deuterium--tritium fusion).** Take the neutral-atom masses
> $m({}^{2}\mathrm H)=2.01410178\ \mathrm u$,
> $m({}^{3}\mathrm H)=3.01604928\ \mathrm u$,
> $m({}^{4}\mathrm{He})=4.00260325\ \mathrm u$, and
> $m_n=1.00866492\ \mathrm u$, giving a mass difference of $0.01888289\ \mathrm u$.
>
> The released energy is
> $$
> Q=(0.01888289\ \mathrm u)(931.494\ \mathrm{MeV}/c^2\!\!/\mathrm u)c^2
> =17.589\ \mathrm{MeV}.
> $$
> The alpha particle and neutron share it as recoil kinetic energy in the
> centre-of-momentum frame.

This is a mass-difference result; it says nothing about the conditions needed to
sustain fusion at a useful rate.

$$
% caption: Mass and charge bookkeeping for deuterium-tritium fusion. Neutral-atom electron counts cancel, so the tabulated atomic masses give the 17.589 MeV Q value directly; the helium nucleus and the neutron carry the released kinetic energy.
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Energy per unit fuel mass sharpens the scale, but the denominator needs care. At
$200\ \mathrm{MeV}$ per complete fission of a $^{235}\mathrm U$ nucleus, uranium-235
yields about $8.2\times10^{13}\ \mathrm{J\,kg^{-1}}$ consumed; the D--T example yields
about $3.4\times10^{14}\ \mathrm{J\,kg^{-1}}$ of deuterium-plus-tritium consumed.
These are reaction energy densities, not delivered electricity, since blanket mass,
unburned fuel, conversion losses, and auxiliary systems all move a plant-level number.
They still beat chemical fuel densities by orders of magnitude, because the mass
differences are nuclear-scale fractions of the fuel mass rather than chemical-scale
ones.

Write the denominator beside the result. Event energy, reacting-mass energy density,
and apparatus output are distinct reported quantities even when they all start from
one reaction $Q$.

| Scale | Calculation | Boundary |
| --- | --- | --- |
| One reaction | $Q=(m_i-m_f)c^2$ | stated initial and final channel |
| Reacting fuel | $Q$ times events per kilogram | nuclei that actually react |
| Delivered output | retained energy over time | converter, absorber, and losses |

Binding energy per nucleon is not a probability scale. A nucleus can sit on the
favourable side of the curve and still need a projectile energetic enough to clear a
threshold, while an energetically allowed channel can stay rare under given collision
conditions. The curve answers one narrow question, whether the combined rest mass
drops after a specified reshuffling of nucleons, and it does not replace conservation
checks, momentum accounting, or measured yields. "Stable" in a graph caption likewise
means tightly bound relative to neighbours, not a full prediction of a sample's time
history.

Keep the sign convention explicit, but note that the binding-energy comparison and
the direct $Q$ calculation agree only for the same complete set of nuclei. Dropping a
neutron, mixing an atomic mass on one side with a nuclear mass on the other, or
quietly swapping a fragment channel all change the answer, and rounding matters too:
a few thousandths of a unified atomic mass unit is several MeV.

The D--T result checks out in joules. Multiplying by
$1.602176634\times10^{-13}\ \mathrm{J\,MeV^{-1}}$ gives
$2.818\times10^{-12}\ \mathrm J$ per reaction, and dividing by the combined reactant
mass of about $5.030151\ \mathrm u$ returns the stated
$3.4\times10^{14}\ \mathrm{J\,kg^{-1}}$ after rounding. This assumes every reactant
nucleus reacts once; a system figure needs a burn fraction and may have to count
energy carried off by escaping particles. Reaction energy and locally deposited heat
are not interchangeable without a stated boundary around the apparatus.

The quoted $200\ \mathrm{MeV}$ fission energy is an aggregate scale, not the mass
defect of one universal fragment pair. Fragment kinetic energies, emitted neutrons,
and electromagnetic energy separate in an experimental account, and their sum, for a
stated channel and boundary, is fixed by the same initial-final mass difference. So an
energy-per-mass figure must name its mass basis, whether fissile isotope, full fuel
mixture, or entire engineered assembly; changing that denominator matters in practice
without touching the $Q$ of one reaction.

Energy-scale comparisons need both a reaction specification and a mass basis. The
same mass difference feeds several reported quantities, each tied to a different
boundary.

| Reported quantity | Numerator | Denominator or boundary |
| --- | --- | --- |
| Reaction Q value | $(m_i-m_f)c^2$ | one balanced reaction channel |
| Product kinetic energy | Q value minus recoil and internal channels | selected final products |
| Energy per reacting mass | energy per event times event count | participating nuclei only |
| Deposited heat | retained product energy | calorimeter and its stated time window |

A fission chain reaction adds a boundary condition beyond a positive $Q$. Each
generation must leave, on average, one usable neutron for the next to hold a steady
sequence. The effective multiplication factor $k_{\rm eff}$ is below one when
absorption and leakage dominate, one at critical balance, and above one when the
population grows generation over generation. Geometry, fuel distribution, neutron
losses, absorbers, and reflectors all set this balance. It is a property of
propagation through a material assembly; a single fission's positive energy does not
guarantee a self-sustaining sequence.

$$
% caption: Chain-reaction criticality. A material assembly is subcritical when losses leave fewer than one usable neutron per generation, critical when the average is exactly one, and supercritical when the usable-neutron population grows generation over generation.
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Reaction-energy work should state the mass convention, the balanced channel, the
frame, and the energy units, and should keep $Q$ (fixed by initial and final masses)
distinct from a laboratory threshold, a recoil share, energy deposited in one
component, and usable output after losses. Those numbers can coincide in a simple
example yet name different parts of the account.

## Precision, uncertainty, and conservation audits

Mass-energy results often hinge on subtracting nearly equal quantities, so the
numerical precision of the input masses can matter more than the displayed significant
figures. The unified atomic mass unit is defined from a neutral carbon-12 atom, and
tabulated atomic masses refer to neutral atoms unless stated otherwise. Atomic masses
are efficient when the electron counts balance; otherwise electron masses and binding
corrections must enter consistently. Nuclear, ion, and neutral-atom masses are not
interchangeable labels for one table entry.

A mass spectrometer measures an observable tied to mass-to-charge ratio, not a mass in
isolation: a magnetic instrument folds in field scale, trajectory radius, and charge
assignment, while time-of-flight folds in path length and timing offsets. Calibration
references should bracket the mass-to-charge region of interest, since references
spread across it test nonlinear drift, unresolved peaks, and charge-state
contamination that a single scale factor misses. Report the calibration model and the
interval over which it holds, above all for a small mass difference.

A $Q$ value built from independently measured masses has

$$
\sigma_Q=c^2\sqrt{\sigma_{m_i}^2+\sigma_{m_f}^2}.
$$

That form changes when the uncertainties share a calibration. In full,
$\sigma_Q^2=c^4\,\vec a^{\mathsf T}C\vec a$, where $C$ is the covariance matrix
and each entry of $\vec a$ carries the sign with which its mass enters the
difference. Positive covariance partly cancels when two masses are subtracted on the
same scale, though a shared scale error persists when the coefficients do not match.
Separate error bars alone throw that information away.

Small $Q$ values are the ones exposed to cancellation. Let two total rest energies
each sit near $10\,000\ \mathrm{MeV}$ with a difference of $0.020\ \mathrm{MeV}$: a
relative mass-scale uncertainty of a few parts per million, negligible on either mass,
is then comparable with the reported $Q$. Guard digits avoid a rounding slip but do
not fix the calibration. Quote the uncertainty in the same energy convention as $Q$,
and keep the sign of the central result rather than flattening every small difference
to an absolute energy.

Four-momentum closure is a separate audit of a reconstruction. Form the residual
four-vector from every listed initial and final particle:

$$
\Delta P^\mu=P_{\rm initial}^\mu-\sum P_{\rm final}^\mu.
$$

An isolated, fully measured event has residual energy and all three residual momentum
components consistent with zero within their correlated uncertainties. One scalar such
as missing energy is not enough: a calibration shift can hold the energy sum while
leaving a momentum imbalance, and a missing particle can carry little energy but
substantial transverse momentum. Report $\Delta E$, $\Delta p_x$, $\Delta p_y$, and
$\Delta p_z$ together with the frame, the detector-resolution model, and the criterion
for calling the residual consistent with zero.

Detector acceptance sets what "fully measured" means. A particle can slip through a
geometric gap, fall below an energy threshold, overlap another signal, or interact
before reaching the relevant element, and assigning zero to an unrecorded measurement
does not repair those losses. An acceptance correction is a model of which events were
observable, with its own dependence on angle, momentum, particle type, and topology.
Name missing energy as a reconstructed residual, not the energy of one particular
unseen object.

A compact conservation audit can use the deuterium--tritium channel already balanced
by mass number and charge number:

$$
{}^{2}_{1}\mathrm H+{}^{3}_{1}\mathrm H
\longrightarrow{}^{4}_{2}\mathrm{He}+{}^{1}_{0}\mathrm n.
$$

Both sides give $A=5$ and $Z=2$, and the neutral-atom masses give
$Q=17.589\ \mathrm{MeV}$. Take an illustrative centre-of-momentum measurement with
$K_{\mathrm{He}}=3.54\pm0.03\ \mathrm{MeV}$ and $K_n=14.05\pm0.10\ \mathrm{MeV}$: the
sum $17.59\pm0.10\ \mathrm{MeV}$ matches the mass-derived $Q$. If the reconstructed
momentum magnitudes are $162\pm2\ \mathrm{MeV}/c$ for the helium and
$161\pm4\ \mathrm{MeV}/c$ opposite for the neutron, the residual longitudinal momentum
is $1\pm4.5\ \mathrm{MeV}/c$, also consistent with zero. The audit passes within the
stated precision; it does not force the central residual to an exact zero.

That same event would fail the audit if a third energy deposit were dropped from the
final sum or the detector were calibrated for the wrong particle type. A
reconstruction record lists the measured four-vectors, the masses used for each
energy, the sign conventions, the covariance assumptions, and the selection cuts, and
it keeps an event-level closure test apart from a sample-level calibration test. A
sample can show small average residuals while still hiding a biased subset near an
acceptance edge.

Separate a closure residual from a corrected physical quantity. Compare the raw
measurements first against the stated calibration and response model, then list any
correction for inactive material, leakage, or acceptance on its own, with its
uncertainty and correlation to the original measurement. Otherwise a correction can be
tuned until the preferred conservation result appears, leaving no independent test.
Blind control samples and reference reactions test the reconstruction, since their
expected four-momentum balance is known without fitting the same events used to set
the calibration.

The final numerical checks are simple. Use compatible units before adding
four-vectors, commonly MeV for energy and MeV$/c$ for momentum. Square a residual only
after propagating its uncertainty, and keep the correlations when a shared calibration
or common beam-energy measurement feeds several terms. An inferred mass squared
slightly below zero can come from finite resolution, but a large negative value flags
inconsistent inputs or a reconstruction outside its domain. Conservation accounting
bounds what an event can contain without fixing the process that produced the
products.

## System boundaries and energy-channel accounting

An energy statement means nothing until the system boundary is named. A closed system
exchanges no matter across the boundary but can still trade energy by radiation, work,
or heat; an open system also lets matter cross, so the energy that material carries in
and out belongs in the account. A reaction calculation may draw a closed boundary
around every initial and final particle, while an apparatus calculation may enclose a
target, a detector volume, or a calorimeter. These boundaries answer different
questions and need not give the same "released energy."

An isolated complete reaction has its total energy fixed by four-momentum
conservation, with a positive $Q$ emerging as translational kinetic energy, recoil,
radiation, and product internal energy. If the boundary excludes an emitted photon, a
neutral particle, or a recoiling support, that energy has not vanished; it has crossed
the boundary. An unresolved internal-energy contribution is likewise no violation of
conservation, only an unseparated channel. Distinguish energy that is physically
absent from the boundary from energy that stays inside but was never separately
resolved.

The same distinction separates microscopic from macroscopic scales. Nuclear mass
differences run to several MeV per reaction, while chemical changes sit at eV per bond
or MJ per kg of fuel, yet the conversion is always $1\ \mathrm u c^2=931.494\ \mathrm{MeV}$.
The large nuclear energy density comes from a larger fractional change in rest mass,
not a different conservation law. At the apparatus scale a large $Q$ need not become a
large temperature rise in one component: heat capacity, escaping radiation, mechanical
work, incomplete fuel use, and the mass of surrounding material all intervene.

Calorimetry is a boundary-specific closure test. An absorber that retains the relevant
products and reaches a measured temperature change gives a deposited energy
$E_{\rm cal}=C\Delta T$, after correcting for heat loss, baseline drift, and the heat
capacities folded into $C$. A calorimeter transparent to a neutral particle or to
radiation records less than the full reaction energy, yet can still be accurate for
its stated boundary. Comparing it with a mass-derived $Q$ must then add the expected
escape energy, the readout window, and any energy stored in material that has not
equilibrated.

Time matters, because a channel can cross a practical boundary after the nominal
event. A prompt gate records charged-product energy but misses later thermal transfer
to supports, shielding, or gas; a long calorimetric integration recovers some of that
transfer at the cost of environmental heat leakage and baseline uncertainty. Neither
is inherently better. State the time interval beside the boundary, with any delayed
subtraction or extrapolation for energy outside the recorded window. A "total energy"
quoted without those conditions is not reproducible.

Mechanical recoil is another easily hidden channel. In an ideal two-product reaction
the momenta balance in the centre-of-momentum frame, so both products carry kinetic
energy even when one goes undetected. On a target mounted on a support, some recoil
passes into the apparatus as elastic motion or later heating: excluded from the
boundary it is an outgoing channel, included it joins the eventual calorimetric total.
A photon reflected by a wall behaves the same way, staying in the system even after it
leaves the first detector component.

Macroscopic comparisons need a stated mass basis as well as a boundary. Energy per
kilogram of reacting nuclei differs from energy per kilogram of fuel compound, target,
coolant, or whole device, so a reaction with high energy per reacting mass can leave a
much lower assembly-level density once structure and unused fuel count. Chemical and
nuclear systems compare fairly only when numerator and denominator span analogous
boundaries; the underlying $Q$ values stay microscopic, and scaling them to delivered
heat or power adds reaction rate, participation fraction, and capture factors.

An energy-channel table often beats a single residual. List each channel with its sign
convention, value, uncertainty, whether it was measured or inferred, and whether it
crosses the boundary; the signed entries should sum to the stated closure residual.
That exposes an omitted escape term or a double-counted calibration correction before
the result collapses to one number.

| Energy channel | Sign convention | Boundary status |
| --- | --- | --- |
| Charged products | positive deposited energy | measured if stopped in the absorber |
| Photons or neutral particles | positive outward energy when excluded | direct detector or inferred escape term |
| Recoil and support motion | assigned by the selected apparatus boundary | mechanical or later thermal measurement |
| Calibration correction | stated additive or multiplicative sign | retained separately from reaction energy |

Report the boundary and frame, every initial and final mass or measured four-vector,
the mass convention, calibration, and unit conversion, along with the channels inside
the boundary, the expected escape channels, the measurement window, the detector or
calorimeter response, and the statistical and systematic uncertainties including
shared calibration terms. Then give a closure equation such as
$Q=E_{\rm deposited}+E_{\rm escaping}+K_{\rm recoil}+E_{\rm internal}$, each term
marked measured, inferred, or bounded.

That template also keeps rest-energy bookkeeping apart from efficiency. A reactor,
accelerator target, or laboratory source can have a precisely known $Q$ yet deliver
only a fraction as recoverable heat or electrical output, since efficiency layers
engineering boundaries and losses on top of the conservation statement. A
near-complete calorimetric deposit, conversely, does not name every microscopic
channel; it only shows their total stayed inside the boundary. Keeping the two claims
separate makes comparisons across experiments and systems far less ambiguous.