
Nuclear, particle & quantum-information sequence · Interactive lesson
Inertial confinement and the state of fusion energy
The magnetic bottle is gone; what holds the fuel together is its own mass, for as long as a sound wave takes to cross it. Learn why that makes ρR the number that matters, why a thousandfold compression follows, what Rayleigh–Taylor does to an imploding shell, and how far NIF's gain of 1.5 sits from a power station.
UPV-16.09 · Lesson 9 of 9 in this unit. Position is not a completion score.Nuclear, particle & quantum-information sequence. These units continue into nuclear and particle physics and quantum information. Follow your university’s prerequisites, mathematical depth and assessment; this independent pathway is not a universal programme specification.
Build the model
Confinement by inertia: ρR sets the burn, and the wall plug delivers the verdict.
The key idea
Inertial confinement fusion compresses a millimetre capsule of deuterium-tritium to hundreds of g cm⁻³ so that its own inertia holds it together for the ~100 ps needed to burn, with the areal density ρR setting the burn fraction and alpha self-heating igniting a central hot spot.
Example. NIF, 5 December 2022: 2.05 MJ of laser light, 3.15 MJ of fusion yield, target gain 1.54. Full burn of the 0.2 mg of D-T would have been 67 MJ, so 4.7 % burned, which f = ρR/(ρR + 7) reads as ρR ≈ 0.35 g cm⁻² — while the laser drew about 300 MJ from the wall.
The complete physical model and assumptions
Inertial confinement replaces the magnetic field with nothing at all: a compressed sphere of D-T is held together only by its own inertia while a rarefaction wave runs in from its surface at the sound speed, for a mass-averaged time R/(4cₛ) of tens of picoseconds. Lawson's triple product nTτ is unchanged, so a confinement time 10¹¹ times shorter than a tokamak's is paid for with a density 10¹¹ times higher. Two consequences follow from the same algebra. The burn fraction is f = ρR/(ρR + HB) with HB ≈ 7 g cm⁻², because the disassembly time grows as R and the reaction rate per ion as ρ, so only their product enters; and at fixed ρR the fuel mass is (4π/3)(ρR)³/ρ², so a thousandfold compression turns a 1.8 kg charge into a 1.8 mg pellet. Compression is an ablation-driven rocket implosion, unstable to Rayleigh–Taylor growth while the light ablation plasma pushes the dense shell and again while the light hot spot decelerates it, so the shell must survive some nine e-folds from nanometre seeds. The cost is the ladder: NIF's 3.15 MJ from 2.05 MJ of laser light is target gain 1.5 against roughly 300 MJ drawn from the wall, one shot a day, no tritium bred, and a plant condition ηL G ≳ 10 that sits three orders of magnitude away.
1Inertia is the bottle: the fuel burns while it disassembles
Take away the magnetic field and nothing holds the plasma but its own mass. A hot sphere of radius R lets go from the outside in, as a rarefaction wave running inward at the sound speed cₛ = √(2kT/mᵢ), and the useful burn lasts about τ ≈ R/(4cₛ) — with cₛ ≈ 1.5 × 10⁸ cm s⁻¹ at 30 keV and R ≈ 120 μm that is 20 ps. For equimolar D-T with nD = nT = n/2 the ion density obeys dn/dt = −(n²/2)⟨σv⟩, which integrates to 1/n − 1/n₀ = ⟨σv⟩t/2. The fraction burned by time τ is f = x/(1 + x) with x = n₀⟨σv⟩τ/2. Substitute n₀ = ρ/mᵢ and τ = R/(4cₛ) and x becomes ρR/HB with HB = 8mᵢ cₛ/⟨σv⟩: at 30 keV, 8 × 4.15 × 10⁻²⁴ g × 1.5 × 10⁸ cm s⁻¹ / 6.6 × 10⁻¹⁶ cm³ s⁻¹ ≈ 7.5 g cm⁻², and the tabulated HB(T) sits near 7 g cm⁻² across the 30–50 keV of a burning hot spot, the value this topic uses. The whole burn has collapsed into one areal density, because the confinement time grows as R/cₛ and the reaction rate per ion as ρ⟨σv⟩, so only their product ρR survives. The Maxwell-averaged ⟨σv⟩ is the one from the Gamow-peak topic; it enters only through HB.
2Why a thousandfold compression: mass falls as 1/ρ² at fixed ρR
The burn wants ρR ≈ 3 g cm⁻² for f ≈ 30 %. A uniform sphere carries M = (4π/3)ρR³ = (4π/3)(ρR)³/ρ², so at fixed ρR the fuel mass — and with it the yield and the pressure the chamber must swallow — falls as 1/ρ². At cryogenic solid density, 0.25 g cm⁻³, ρR = 3 means R = 12 cm and M = 1.8 kg, whose 30 % burn is 1.8 × 10¹⁴ J, forty kilotonnes: a weapon, not a power plant. Compress a thousandfold to 250 g cm⁻³ and the same ρR needs R = 120 μm and M = 1.8 mg, releasing 180 MJ, a pulse a chamber can take ten times a second. Compression is done by ablation: the laser, or the X-rays it makes inside a gold hohlraum, boils off the outer ablator, and the rocket reaction drives the shell inward at 350–400 km s⁻¹; its kinetic energy stagnates into hundreds of gigabar at the centre. Keeping the cold fuel on a low adiabat, near the Fermi-degenerate minimum energy for that density, is what the shaped multi-shock laser pulse is for.
3Hot-spot ignition is Lawson at 10¹¹ times the density
Heating all 1.8 mg to burn temperature would cost far more drive energy than the yield repays, so the design ignites a central hot spot and lets the alphas carry the burn outward into the cold, dense shell. The hot spot needs (ρR)ₕₛ ≳ 0.3 g cm⁻², because that is the range of a 3.5 MeV alpha in 10 keV D-T (it grows roughly as T⁵⁄⁴) — thinner and the alphas leave before heating — and Tₕₛ ≳ 5 keV, so that alpha heating outruns bremsstrahlung and electron conduction. This is the tokamak's triple product with τ = R/cₛ. Put ρ = 100 g cm⁻³, so n = 2.4 × 10³¹ m⁻³, with R = 30 μm and T = 5 keV: cₛ = 6.2 × 10⁵ m s⁻¹, τ = 48 ps, and nTτ = 5.8 × 10²¹ keV s m⁻³, above the 3 × 10²¹ of the previous topic. The August 2021 NIF shot (1.35 MJ) was the first to pass the ignition form of the criterion, alpha heating outrunning all of the hot spot's losses, after shots of late 2020 and early 2021 had reached a burning plasma, alpha heating exceeding the compression work; the 5 December 2022 shot, 3.15 MJ from 2.05 MJ, was the first with target gain above one, and later shots have raised both yield and gain.
4Rayleigh–Taylor: the shell must survive its own acceleration
A light fluid pushing a heavy one is unstable, and an implosion does it twice. During acceleration the low-density ablation plasma pushes the dense shell; at stagnation the light hot spot decelerates the dense shell, the inner surface goes unstable, and cold fuel and carbon ablator are pushed into the hot spot, where Z² bremsstrahlung radiates its heat away. A ripple of wavenumber k on a classical interface grows as exp(γt) with γ = √(Akg), A the Atwood number, so short wavelengths grow fastest. Ablation rescues the outer surface: material is stripped from the front at velocity vₐ, and Takabe's fit γ = 0.9√(kg) − 3kvₐ has a cutoff at kc = 0.09g/vₐ² and its worst mode at kc/4. With g = 10¹⁴ m s⁻² (400 km s⁻¹ in 4 ns), vₐ = 3 μm ns⁻¹ and R = 1 mm, the cutoff is ℓc ≈ 1000, the worst mode ℓ ≈ 250 grows at 2.3 × 10⁹ s⁻¹, and 4 ns of acceleration gives nine e-folds, ×8000, against an in-flight shell 30–40 μm thick. That is why capsule roughness is specified in nanometres, why the fill tube and support tent are studied, and why low-mode hohlraum asymmetry is the other budget line.
5Reading NIF's numbers: the ladder from wall plug to yield
Write the ladder down for 5 December 2022. About 300 MJ of electrical energy charged the capacitor banks; 2.05 MJ of ultraviolet light reached the hohlraum, a wall-plug efficiency ηL ≈ 0.7 %; roughly a tenth of that was absorbed by the capsule, and of order 20 kJ ended as kinetic energy of the fuel. The 3.15 MJ of yield gives target gain G = 3.15/2.05 = 1.54, a capsule gain near 15, and an engineering gain of 3.15/300 ≈ 0.01. Full burn of the 0.2 mg of D-T would have been 67 MJ, so the burn fraction was 4.7 %, which f = ρR/(ρR + 7) reads as ρR ≈ 0.35 g cm⁻² for the fuel that burned. For a plant, electrical output Pₑ = ηₜₕ Mb G EL fᵣₑₚ must repay the laser's draw PL = EL fᵣₑₚ/ηL with a recirculating fraction ε = 1/(ηL ηₜₕ Mb G); ε ≤ 0.25 with ηₜₕ = 0.4 and Mb = 1.1 needs ηL G ≳ 10. NIF stands at 0.01. A 10 %-efficient diode-pumped laser would need G ≈ 100, sixty times NIF's, fired at 10 Hz rather than once a day, at targets costing cents rather than the hand-built cryogenic assemblies NIF fires today.
6What no device has done: tritium, materials, net electricity
Three rungs remain untouched by any device, magnetic or inertial. Tritium: D-T consumes one triton per 17.6 MeV, so a 1 GW-fusion plant runs 3.5 × 10²⁰ reactions per second and burns 56 kg of tritium a year, against a world civil inventory of a few tens of kilograms. The plant must breed its own from n + ⁶Li → T + ⁴He + 4.8 MeV and the endothermic n + ⁷Li → T + ⁴He + n, with a breeding ratio comfortably above one after neutron leakage, and no blanket has yet closed that cycle. Materials: the 14.1 MeV neutron carries 80 % of the energy through the first wall, and no structural steel has been qualified against decades of that spectrum. Net electricity: JET's 59 MJ in 2021 was 0.33 of its heating power; ITER is built for Q = 10 in the plasma, with D-T operation not scheduled before the late 2030s; NIF's target gains above one are a physics result, not an energy source. Check every headline against the ladder: plasma gain, target gain, engineering gain, and finally electricity on the grid.
Key equations
Read the units, assumptions and limits before substituting.
Burn fraction from areal density
fb = ρR / (ρR + HB), HB = 8 mᵢ cₛ / ⟨σv⟩ ≈ 7 g cm⁻²
Doubling ρR from 0.35 to 0.7 g cm⁻² lifts the burn from 4.8 % to 9.1 %; a plant wants 3 g cm⁻² for 30 %. Temperature enters only through HB.
ρR in g cm⁻²; mᵢ = 2.5 u the mean ion mass, cₛ = √(2kT/mᵢ) the sound speed in cm s⁻¹, ⟨σv⟩ the D-T reactivity in cm³ s⁻¹Fuel mass at fixed areal density
M = (4π/3) ρ R³ = (4π/3) (ρR)³ / ρ²
The whole case for compression: at ρR = 3 g cm⁻², 1.8 kg at solid density becomes 1.8 mg at 250 g cm⁻³, a factor 10⁶ from a factor 10³ in ρ.
ρ in g cm⁻³, ρR in g cm⁻², M in g; a uniform sphereD-T yield and target gain
Y = fb M × 3.37 × 10¹¹ J g⁻¹, G = Y / EL
0.2 mg burning at 4.8 % gives 3.2 MJ, and against 2.05 MJ of laser light that is G ≈ 1.6, NIF's reported 1.54 to within rounding.
M the fuel mass in g, so Y is in J; 17.6 MeV per reaction over 5.03 u of fuel consumed; EL the laser energy delivered to the target, in JHot-spot ignition
(ρR)ₕₛ ≳ 0.3 g cm⁻², Tₕₛ ≳ 5 keV, with τ = R/cₛ in nTτ
Lawson's product survives unchanged: ICF meets it at n ≈ 10³¹ m⁻³ for 50 ps instead of 10²⁰ m⁻³ for seconds, and the alphas must stop inside the hot spot.
n the ion density in m⁻³, T in keV, R the hot-spot radius; 0.3 g cm⁻² is the range of a 3.5 MeV alpha in 10 keV D-T; τ is the disassembly time in s, tens of picosecondsAblative Rayleigh–Taylor growth (Takabe)
γ = 0.9 √(k g) − 3 k vₐ, k = ℓ/R, kc = 0.09 g / vₐ²
Ablation kills modes above kc; the worst sits at kc/4 with γₘₐₓ = 0.0675 g/vₐ, and e-folds = γt over the acceleration. Raising vₐ buys stability.
γ the growth rate in s⁻¹, k the wavenumber in m⁻¹ on a shell of radius R in m; g the shell-frame acceleration in m s⁻², vₐ the ablation velocity (1 μm ns⁻¹ = 10³ m s⁻¹), ℓ the spherical mode numberPlant condition on the gain ladder
ε = PL / Pₑ = 1 / (ηL ηₜₕ Mb G) ⇒ ηL G ≳ 10
NIF: ηL ≈ 0.007, G ≈ 1.5, so ηL G ≈ 0.01 — three orders of magnitude short before repetition rate, target cost or tritium enter.
PL the laser's electrical draw and Pₑ the plant's electrical output, in W; G the target gain; ηL laser wall-plug efficiency, ηₜₕ ≈ 0.4 thermal-to-electric, Mb ≈ 1.1 blanket multiplication, ε the recirculating fractionInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Start at NIF's numbers, ρR 0.35, 0.2 mg, ηL 0.7 %: G ≈ 1.6 and ηL⋅G ≈ 0.01. Push ρR to 3 and the mass to 2 mg and G passes 90, yet the yield bar reaches the plant tick only once ηL nears 10 % — every rung has to move, and laser efficiency is the rung NIF was never built to climb.
Live results
- BURN FRACTION f
- 4.8 %
- FUSION YIELD Y
- 3.21 MJ
- TARGET GAIN G = Y/2.05 MJ
- 1.57
- ηL⋅G (PLANT NEEDS ≥ 10)
- 0.011
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Burning-fuel areal density ρR: 0.35 g/cm²; DT fuel mass reaching that ρR: 0.20 mg; Laser wall-plug efficiency ηL: 0.7 %
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The error is to read “3.15 MJ out for 2.05 MJ in” as net energy. The 2.05 MJ is the ultraviolet light that reached the hohlraum; producing it drew about 300 MJ from the grid, because NIF's flashlamp-pumped glass amplifiers convert under one percent of what they are fed. Against the wall plug the ratio is 0.01, and the 3.15 MJ is heat, worth about 1.3 MJ of electricity after a turbine. The result is a genuine milestone — the first target gain above one, with alpha heating dominating the hot spot's energy balance — but it clears the plasma and target rungs of the ladder, not the engineering one. A cousin of the same error reads ignition as complete burn: at ρR ≈ 0.35 g cm⁻² only 4.7 % of the 0.2 mg of D-T burned, because f = ρR/(ρR + 7) is set by areal density, not by whether the hot spot lit.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A compressed D-T assembly burns with areal density ρR = 1.0 g cm⁻² and holds 1.5 mg of fuel. Using HB = 7 g cm⁻², find the burn fraction and the fusion yield, and express the yield in kilograms of TNT (4.184 MJ kg⁻¹).
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- Burn fraction from the areal density alone: f = ρR/(ρR + HB) = 1.0/(1.0 + 7) = 0.125, so one eighth of the fuel reacts before the sphere disassembles.
- Energy content of D-T: each 17.6 MeV reaction consumes one D and one T, 5.03 u = 8.35 × 10⁻²⁴ g, so 17.6 × 1.602 × 10⁻¹³ J / 8.35 × 10⁻²⁴ g = 3.375 × 10¹¹ J g⁻¹ = 337.5 MJ mg⁻¹.
- Yield: Y = f M × 337.5 MJ mg⁻¹ = 0.125 × 1.5 × 337.5 = 63.3 MJ.
- In explosive terms 63.3/4.184 = 15.1 kg of TNT — a pulse a chamber can absorb, released in about a hundred picoseconds.
Answer. f = 12.5 %, Y ≈ 63 MJ (about 15 kg of TNT equivalent).
MediumExample 2
A reactor design wants ρR = 3.0 g cm⁻² so that 30 % of the fuel burns. Find the fuel radius, mass and 30 %-burn yield for a uniform D-T sphere at (a) cryogenic solid density, 0.25 g cm⁻³, and (b) after a thousandfold compression to 250 g cm⁻³. Take 3.375 × 10¹¹ J g⁻¹ and 1 kt TNT = 4.184 × 10¹² J.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- At fixed areal density the mass is M = (4π/3)ρR³ = (4π/3)(ρR)³/ρ²; with (ρR)³ = 27 g³ cm⁻⁶ the numerator is (4π/3) × 27 = 113.1 g³ cm⁻⁶, and M falls as 1/ρ².
- (a) ρ = 0.25 g cm⁻³: R = ρR/ρ = 3.0/0.25 = 12 cm, and M = 113.1/0.0625 = 1810 g = 1.81 kg.
- (a) Yield at 30 % burn: 0.30 × 1810 g × 3.375 × 10¹¹ J g⁻¹ = 1.83 × 10¹⁴ J = 43.8 kt of TNT. That is a weapon; no chamber survives it.
- (b) ρ = 250 g cm⁻³: R = 3.0/250 = 0.012 cm = 120 μm, and M = 113.1/62 500 = 1.81 × 10⁻³ g = 1.81 mg — a million times less, because (10³)² = 10⁶.
- (b) Yield: 0.30 × 1.81 × 10⁻³ g × 3.375 × 10¹¹ J g⁻¹ = 1.83 × 10⁸ J = 183 MJ, about 44 kg of TNT; at 10 shots per second that is 1.8 GW of fusion power.
Answer. (a) R = 12 cm, M = 1.8 kg, Y = 1.8 × 10¹⁴ J ≈ 44 kt; (b) R = 120 μm, M = 1.8 mg, Y = 183 MJ. Same ρR, 10⁶ less mass — the entire case for compression.
HardExample 3
A capsule of radius R = 1.0 mm is driven at g = 1.0 × 10¹⁴ m s⁻² for 4.0 ns with ablation velocity vₐ = 3.0 μm ns⁻¹. Using Takabe's γ = 0.9√(kg) − 3kvₐ with k = ℓ/R: (a) find the cutoff mode ℓc and the fastest-growing mode; (b) find its linear growth factor and what a 10 nm seed becomes, against a 30 μm in-flight shell; (c) find the vₐ that would hold the worst growth to e⁶.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- (a) Cutoff where γ = 0: 0.9√(kc g) = 3kc vₐ gives kc = 0.09 g/vₐ² = 0.09 × 10¹⁴/(3.0 × 10³)² = 1.0 × 10⁶ m⁻¹, so ℓc = kc R = 1.0 × 10⁶ × 10⁻³ = 1000. Ripples finer than that are eaten by ablation.
- (a) Fastest mode from dγ/dk = 0.45√(g/k) − 3vₐ = 0: k* = 0.0225 g/vₐ² = kc/4 = 2.5 × 10⁵ m⁻¹, so ℓ* = 250, and γₘₐₓ = 0.9√(2.5 × 10⁵ × 10¹⁴) − 3 × 2.5 × 10⁵ × 3.0 × 10³ = 4.5 × 10⁹ − 2.25 × 10⁹ = 2.25 × 10⁹ s⁻¹ (equivalently 0.0675 g/vₐ).
- (b) e-folds over the acceleration: γₘₐₓ t = 2.25 × 10⁹ × 4.0 × 10⁻⁹ = 9.0, a growth factor e⁹ = 8.1 × 10³. A 10 nm seed at ℓ = 250 reaches 81 μm, larger than the 30 μm shell, so linear theory says that mode alone breaks the shell.
- (b) In practice a finite density-gradient scale length lowers γ below Takabe's fit and the amplitude saturates near λ/10 = (2π × 1 mm/250)/10 = 2.5 μm — but the mode number that sets the surface-finish specification is now known.
- (c) Require γₘₐₓ t = 6: γₘₐₓ = 1.5 × 10⁹ s⁻¹ = 0.0675 g/vₐ, so vₐ = 0.0675 × 10¹⁴/1.5 × 10⁹ = 4.5 × 10³ m s⁻¹ = 4.5 μm ns⁻¹. The cutoff drops to ℓc = 0.09 × 10¹⁴/(4.5 × 10³)² × 10⁻³ = 444 and the worst mode to ℓ ≈ 111 — a thicker or doped ablator buys stability at the price of drive energy.
Answer. ℓc = 1000, worst mode ℓ ≈ 250 at γₘₐₓ = 2.25 × 10⁹ s⁻¹; nine e-folds, ×8100, so 10 nm → 81 μm; holding it to e⁶ needs vₐ = 4.5 μm ns⁻¹.
Quick check
Select an answer and explain it before checking.
Review & continue
Inertial confinement fusion compresses a millimetre capsule of deuterium-tritium to hundreds of g cm⁻³ so that its own inertia holds it together for the ~100 ps needed to burn, with the areal density ρR setting the burn fraction and alpha self-heating igniting a central hot spot.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.