Could the Atomic Bomb Ignite the Atmosphere? The Math • The Science 360

Q: What was the LA-602 report?

LA-602, 'Ignition of the Atmosphere with Nuclear Bombs', is a Los Alamos report by Emil Konopinski, Cloyd Marvin and Edward Teller dated 14 August 1946, a year after Trinity. It wrote up the wartime analysis and concluded that energy losses to radiation always outweigh the gains from nitrogen fusion, while calling for better experimental data on the reaction rates.

Twenty-four kilotons of TNT is a lot of energy: 1 × 10¹⁴ joules, enough to flatten a city. It is also, almost exactly, the energy needed to heat a single cubic metre of air to the temperature at which nitrogen nuclei start to fuse at a useful rate. That one comparison is most of the answer to the question that hangs over the Trinity scene in Oppenheimer. The first atomic bomb had about enough energy to bring one cubic metre of the sky to fusion temperature, briefly, and that cubic metre would then have cooled in a few millionths of a second.

The film gets the history roughly right: Edward Teller raised the possibility in 1942, Oppenheimer took it to Arthur Compton, and Hans Bethe showed quickly that it was not going to happen. What the film and most articles on the subject leave out is why. "The calculations said no" is not an explanation. Below, we redo the core of the Los Alamos argument with modern nuclear data, and you will see that the margin was not a coin flip. It was enormous, and it does not depend on details nobody knew in 1945.

Quick answerNo. The fear was a fusion chain reaction in atmospheric nitrogen, not air burning. Hot ionised air radiates X-rays faster than nitrogen fusion can make energy at every temperature, by a factor of at least 12 even if fusion had no Coulomb barrier. Trinity's shocked air also stayed thousands of times too cool to try.

Infographic: Could Trinity ignite the atmosphere? A fireball sits on the desert floor with violet X-rays streaming out and two nitrogen nuclei colliding inside. Even with no Coulomb barrier, hot air radiates 12.4 times faster than nitrogen fusion heats it. With the real barrier the margin is 31,000 at 1 MeV, and Trinity's shocked air was about 6,000 times too cool. A log bar chart of loss divided by gain shows about 10 to the 24 at 0.1 MeV, 31,000 at 1 MeV, 35 at 10 MeV and a floor of 12.4, all above break-even. Tiles: 17.2 MeV per fusion, one cubic metre of 1 MeV air holds Trinity's 24 kilotons, and shocked air reached 0.17 keV. — Tap the image to view it full size. View the accessible, editable SVG.

What exactly were they afraid of?

Not fire. Nitrogen does not burn in the chemical sense, and nobody at Los Alamos thought the air would catch light like petrol. The worry was nuclear. Air is 78% nitrogen, and two nitrogen-14 nuclei that collide hard enough can fuse and rearrange into heavier pieces, releasing energy. The most energetic common outcome is magnesium-24 plus an alpha particle.

Energy released by ${}^{14}\mathrm{N} + {}^{14}\mathrm{N} \to {}^{24}\mathrm{Mg} + {}^{4}\mathrm{He}$ (NIST atomic masses)

\begin{aligned}\text{Mass before} &= 2 \times 14.003074\ \mathrm{u} \\ &= 28.006148\ \mathrm{u} \\ \text{Mass after} &= 23.985042\ \mathrm{u} \\ &\quad + 4.002603\ \mathrm{u} \\ &= 27.987645\ \mathrm{u} \\ \text{Mass lost} &= 0.018503\ \mathrm{u} \\ &\quad \times 931.494\ \mathrm{MeV/u} \\ &= \mathbf{17.2\ MeV}\ \textbf{per reaction}\end{aligned}

If the bomb heated a patch of air enough that these reactions happened often, each one would dump 17 MeV of fast particles into the surrounding gas, heating more air, triggering more reactions. That is ignition: a burn front that feeds itself and runs away through the atmosphere. Physics Today's 2024 account of the episode says Teller raised it among the theorists at a meeting in Berkeley in 1942, and Compton, the leading expert on how hot gases cool by scattering radiation, was consulted almost immediately.

Why the margin was large: losses beat gains at any density

For a self-sustaining burn, a hot parcel of air has to make energy from fusion faster than it loses it. The 1946 Los Alamos report LA-602, by Emil Konopinski, Cloyd Marvin and Teller, framed it exactly this way. Physics Today quotes its summary: "The energy losses to radiation always overcompensate the gains due to reactions."

Source note I could not open the LA-602 report itself (the public FAS copy was blocked when I tried). Its quoted sentences and the description of its three curves here come from the 2024 Physics Today article by Wiescher and Langanke, not from the original document.

The main loss is bremsstrahlung. At these temperatures the air is fully ionised, and free electrons swinging past nitrogen nuclei emit X-rays. Air is transparent enough to hard X-rays that they leave the hot region and take their energy with them. The key point, which most explanations miss, is how the two rates scale:

How gain and loss scale with density n

Fusion power per m³:

\begin{aligned}P_{\text{fus}} &= \tfrac12\, n_N^2\, \langle\sigma v\rangle\, Q \\ &\propto \mathbf{n^2}\end{aligned}

Bremsstrahlung per m³, also proportional to n²:

\begin{aligned}P_{\text{br}} &= 1.69 \times 10^{-32}\, n_e T_e^{1/2} \\ &\quad \times \sum Z^2 n_i\ \ \mathrm{W/cm^3} \\ &\propto \mathbf{n^2}\end{aligned}

Ratio of loss to gain: depends on temperature only.

Both processes need two particles to meet, so compressing the air speeds up both by the same factor. You cannot win by squeezing. Some popular accounts say fusion needs "great pressure" that the atmosphere lacks, but for the energy balance that is the wrong lever. What decides the question is temperature, through the term $\langle\sigma v\rangle$ : how often nitrogen nuclei actually fuse when they collide.

Log-log plot of power density in watts per cubic metre against air temperature from 10 eV to 30 MeV. The orange bremsstrahlung loss line rises gently across the whole range. A dashed purple line for fusion with no Coulomb barrier runs parallel to it, about 12 times lower. The blue line for fusion with the real Coulomb barrier appears only above about 500 keV, climbs steeply and approaches the dashed line near 30 MeV but never crosses the loss line. A shaded band at 10 to 200 eV marks Trinity's shocked air. — Figure 1. Fusion heating versus radiation cooling for sea-level air, computed per cubic metre. The blue gain curve uses a geometric cross-section times the quantum tunnelling probability through the Coulomb barrier, the same approach Physics Today describes for LA-602. The purple curve removes the barrier entirely. Neither reaches the orange loss line.

The worked calculation: an absurdly generous fusion rate still loses

Start with the most generous assumption possible. Suppose every collision in which two nitrogen nuclei touch causes fusion, with no Coulomb repulsion at all. The nuclear radius is about $1.2\ \mathrm{fm} \times A^{1/3}$ , so two nitrogen nuclei touch when their centres are 5.78 fm apart.

Step 1: a no-barrier cross-section

\begin{aligned}\sigma &= \pi\,(5.78 \times 10^{-15}\ \mathrm{m})^2 \\ &= 1.05 \times 10^{-28}\ \mathrm{m^2}\ (1.05\ \text{barn})\end{aligned}

Step 2: sea-level air (1.225 kg/m³)

\begin{aligned}n_N &= 3.98 \times 10^{25}\ \mathrm{m^{-3}}\ (\text{nitrogen}) \\ n_O &= 1.07 \times 10^{25}\ \mathrm{m^{-3}}\ (\text{oxygen}) \\ n_e &= 3.68 \times 10^{26}\ \mathrm{m^{-3}}\ (\text{all ionised})\end{aligned}

Step 3: at kT = 1 MeV

\begin{aligned}\langle\sigma v\rangle &= \sigma \times \text{mean relative speed} \\ &= 6.2 \times 10^{-22}\ \mathrm{m^3/s} \\ P_{\text{fus}} &= \tfrac12 \times (3.98 \times 10^{25})^2 \\ &\quad \times 6.2 \times 10^{-22} \times 17.2\ \mathrm{MeV} \\ &= \mathbf{1.4 \times 10^{18}\ W/m^3} \\ P_{\text{br}} &= \mathbf{1.7 \times 10^{19}\ W/m^3} \\ \text{Loss} \div \text{gain} &= \mathbf{12.4}\end{aligned}

Bremsstrahlung uses the NRL formula, with $\sum Z^2 n_i$ over N, O and Ar.

Because both rates grow as the square root of temperature in this no-barrier case, that factor of 12.4 is the same at 1 keV, 1 MeV or 10 MeV. Pure nitrogen would do a bit better, a ratio of 6.7, because oxygen and argon add radiating charge without adding fuel. But that is the ceiling. No temperature, no density and no bomb size makes a nitrogen plasma burn if it radiates freely, even in a world where the Coulomb barrier does not exist.

Now put the barrier back. Two nitrogen nuclei each carry a charge of +7, and their electric repulsion at touching distance is $49 \times 1.44\ \mathrm{MeV\,fm} \div 5.78\ \mathrm{fm} = 12.2\ \mathrm{MeV}$ . At a temperature of 1 MeV, only nuclei in the far tail of the thermal distribution, or those that tunnel through the barrier, ever fuse. Using the standard WKB tunnelling probability for a Coulomb barrier, averaged over a thermal distribution:

With the Coulomb barrier, sea-level air

Loss ÷ gain at each temperature kT:

\begin{aligned}0.1\ \mathrm{MeV}\ (1.2\ \text{billion K})&:\ \approx 10^{24} \\ 1\ \mathrm{MeV}\ (11.6\ \text{billion K})&:\ \approx \mathbf{31{,}000} \\ 10\ \mathrm{MeV}\ (116\ \text{billion K})&:\ \approx \mathbf{35}\end{aligned}

At very high temperature the ratio creeps down toward the no-barrier floor of 12.4, never below it.

Log plot of radiation loss divided by fusion gain against temperature from 0.1 to 30 MeV. The blue curve with the real Coulomb barrier falls from above a trillion at 0.3 MeV to 31,068 at 1 MeV and 35 at 10 MeV, flattening toward a dashed purple line at 12 labelled no Coulomb barrier at all. An orange break-even line sits at 1, well below both. — Figure 2. The "safety factor", radiation loss divided by fusion gain, as LA-602 defined it. With the Coulomb barrier, the margin is about 31,000 at 1 MeV. It shrinks as temperature rises, which is what worried people about bigger bombs, but it bottoms out near 12 and never reaches break-even at 1.

This is a simplified version of the Los Alamos approach, with three choices you should know about. It treats electrons and nuclei as having the same temperature. It uses the non-relativistic bremsstrahlung formula, which underestimates radiation above a few hundred keV, so the real losses are larger. And it ignores inverse Compton cooling, where hot electrons hand energy to the bomb's own X-rays, which again only adds loss. The last two choices make ignition look easier than it is. The equal-temperature assumption is the one that cuts the other way, since cooler electrons radiate less. But cooler electrons also drain heat from the nuclei faster, because Coulomb collisions hand energy to slow electrons more quickly. LA-602 calculated the electron temperature as a function of the nuclear temperature, which Physics Today notes comes out much lower, and the losses still won. The measured cross-sections from Oak Ridge in the early 1950s were also smaller than the geometric estimate LA-602 used, according to the Physics Today account.

What if the radiation can’t escape?

A fair objection: a big enough fireball is opaque, so the X-rays would not leave. Physics Today makes exactly this point about the real Trinity fireball, whose opacity kept its heat inside during the first few milliseconds. But trapping the radiation makes things worse for ignition, not better. Radiation trapped in a region comes into equilibrium with the matter, and at fusion temperatures the radiation field soaks up almost all the energy.

Energy to heat 1 m³ of sea-level air to kT = 1 MeV

Particles per m³ (nuclei + electrons): $4.2 \times 10^{26}$

\begin{aligned}E_{\text{th}} &= \tfrac32 \times 4.2 \times 10^{26} \\ &\quad \times 1.60 \times 10^{-13}\ \mathrm{J} \\ &= \mathbf{1.0 \times 10^{14}\ J \approx 24\ kt}\end{aligned}

Blackbody radiation filling the same 1 m³ at $T = 1.16 \times 10^{10}\ \mathrm{K}$

\begin{aligned}u &= aT^4 \\ &= 7.566 \times 10^{-16} \times (1.16 \times 10^{10})^4 \\ &= \mathbf{1.4 \times 10^{25}\ J} \\ &\approx \mathbf{3 \times 10^9}\ \textbf{megatons}\end{aligned}

That is my own estimate. Either the radiation escapes and carries energy away 12 or more times faster than fusion supplies it, or it is trapped and demands a hundred billion times more energy than the air itself holds. There is no third option where the air stays hot and the photons politely stay out of the way.

Horizontal bar chart on a log scale of energy in kilotons. Trinity yield 24.8 kt and the energy to heat one cubic metre of air to 1 MeV, 24 kt, are about the same length. Tsar Bomba is 50 megatons. Weaver and Wood's hypothetical ocean bomb is 20 million megatons. Blackbody radiation filling one cubic metre at 1 MeV is 3 billion megatons. — Figure 5. Energy budgets on a log scale. Trinity's entire yield matches the thermal energy of one cubic metre of air at 1 MeV. Keeping that cubic metre in equilibrium with its own radiation would take about a hundred billion times more.

And the clock is brutal. At 1 MeV, bremsstrahlung alone drains that cubic metre's thermal energy in about 6 microseconds. Fusion with the real barrier would take about 0.2 seconds to produce the same amount. A burn wave cannot outrun a cooling time 30,000 times shorter.

How hot did Trinity actually get the air?

Nowhere near 1 MeV. The device itself is hotter than its surroundings, but the air outside is heated mainly by a shock wave, and we can estimate its temperature from the most famous back-of-envelope calculation in physics. In 1950 G. I. Taylor published Julian Mack's time-stamped photographs of the Trinity fireball and showed that a point explosion of energy E in air of density ρ makes a shock of radius

\begin{aligned}R &= S(\gamma)\left(\frac{E t^2}{\rho}\right)^{1/5}\end{aligned}

with S = 1.03 for γ = 1.4 (Taylor computed $S^{-5} = 0.856$ )

Taylor's fit to the photographs (SI form, from Díaz's reanalysis):

(5/2)\log_{10} R - \log_{10} t = 6.915

\begin{aligned}R^5/t^2 &= 10^{13.83} \\ &= 6.8 \times 10^{13}\ \mathrm{m^5/s^2} \\ E &= 0.856 \times \rho \times R^5/t^2 \\ &= 0.856 \times 1.23 \times 6.8 \times 10^{13} \\ &= 7.1 \times 10^{13}\ \mathrm{J} = \mathbf{17\ kt}\end{aligned}

Taylor himself got 16.8 kt for γ = 1.4 and 22.9 kt for γ = 1.3. The official US figure was long 21 kt, and a 2021 Los Alamos reassessment raised it to 24.8 ± 2 kt. The method only needs the radius and the clock in each frame.

Log-log plot of shock radius in metres against time from 0.1 to 100 milliseconds. A dashed black line shows Taylor's fit to Mack's films, equivalent to about 17 kt. A blue line for Trinity at 24.8 kt runs just above it, reaching 120 metres at 16 milliseconds. An orange line for the 50 megaton Tsar Bomba runs about 4.6 times higher. — Figure 3. Shock radius versus time from Taylor's blast-wave law. Radius grows as $t^{2/5}$ and as $E^{1/5}$ , so even a bomb 2,000 times bigger makes a shock only 4.6 times larger at the same moment.

The shock speed is $dR/dt = 2R/(5t)$ , and a strong shock heats the gas it passes through in proportion to the square of that speed. For ideal air with γ = 1.4, $kT = 2(\gamma - 1)/(\gamma + 1)^2 \times m v^2$ , where m is the mean mass of an air molecule.

Shocked-air temperature, Trinity (24.8 kt), t = 0.1 ms

\begin{aligned}R &= \big(24.8 \times 4.184 \times 10^{12}\ \mathrm{J} \\ &\quad \times (10^{-4}\ \mathrm{s})^2 / (1.225 \times 0.856)\big)^{1/5} \\ &= 15.8\ \mathrm{m} \\ v &= 0.4 \times 15.8\ \mathrm{m} / 10^{-4}\ \mathrm{s} \\ &= 63\ \mathrm{km/s} \\ kT &= 0.139 \times (28.97 \times 1.66 \times 10^{-27}\ \mathrm{kg}) \\ &\quad \times (6.3 \times 10^4\ \mathrm{m/s})^2 \\ T &\approx \textbf{1.9 million K (0.17 keV)}\end{aligned}

Needed for the least hopeless case: about 1 MeV, or 11.6 billion K

That is an upper bound, because real air spends shock energy on breaking molecules apart and stripping electrons, so it ends up cooler than an ideal gas would. Physics Today's summary of the same history describes the heated air as reaching "a few million degrees", consistent with this. The gap to 1 MeV is a factor of about 6,000 in temperature, and the loss-to-gain ratio at the temperatures Trinity actually produced (the far left of Figure 1) is astronomically large.

Log-log plot of temperature just behind the shock in kelvin versus time from 10 microseconds to 100 milliseconds. The blue Trinity line falls from about 30 million K at 10 microseconds to a few hundred K at 100 ms. The orange Tsar Bomba line runs about 20 times higher, starting near 700 million K. Both stay well below a purple band at 1 to 10 MeV, about 10 to 100 billion K, where nitrogen fusion is least hopeless. — Figure 4. Ideal-gas strong-shock temperature behind the blast front. Even a 50 megaton weapon heats air to well under a tenth of an MeV equivalent once the shock has moved away from the device, and Figure 2 shows that losses win by a huge factor there.

The misconception: “they went ahead with a small chance of ending the world”

The story is usually told as a gamble: the scientists ran the numbers, found a tiny but real chance, and pushed the button anyway. Arthur Compton is often quoted as saying he would have stopped the project if the chance were more than about three in a million.

Source note The "three in a million" figure is widely attributed to Compton in a 1959 Pearl S. Buck article in American Weekly. Physics Today cites that article in connection with the ignition story, but I could not open it to confirm the exact wording or number.

Whatever Compton said, the physics is not a probability. The question is whether a hot parcel of air gains or loses energy, and that is a ratio of two rates. When the best case, with no Coulomb barrier, still loses by a factor of 12, and the real case at Trinity's temperatures loses by a factor beyond 10²⁰, there is no small chance to calculate. What the physicists were unsure about in 1945 was the nitrogen fusion cross-section, which nobody had measured. LA-602 handled that by assuming generous cross-sections, and still found the answer was no. Its authors were careful enough to add that "the absence of satisfactory experimental foundations makes further work on the subject highly desirable", as Alex Wellerstein quotes it.

What about hydrogen bombs and the oceans?

This is the part where the worry had more substance. Figure 2 shows the safety factor falling as temperature rises, and thermonuclear weapons are hotter and a thousand times bigger. Water also brings hydrogen and deuterium, which fuse far more easily than nitrogen. The Physics Today account says the concern followed the weapons programme through the megaton tests, and that an accelerator was built at Oak Ridge in the early 1950s partly to measure the nitrogen and oxygen fusion rates directly. The experimenters were not told why.

The definitive analysis is a 1979 Physical Review A paper by Thomas Weaver and Lowell Wood. Their abstract concludes that "neither the atmosphere nor oceans of the Earth may be made to undergo propagating nuclear detonation under any circumstances". For seawater, they found a self-sustaining detonation would need deuterium at about 1 part in 300, roughly 20 times its real concentration, and an initiating explosion of about 2 × 10⁷ megatons. That is 400,000 Tsar Bombas.

The Tsar Bomba itself, 50 megatons over Novaya Zemlya in 1961, settled the practical question for the atmosphere. If you want to compare that kind of number with other pop-culture energies, our posts on the energy to destroy a planet and on nuking an asteroid in Armageddon use the same kiloton bookkeeping.

The short version: gain and loss both scale with density squared, so only temperature matters. At any temperature, hot air radiates at least 12 times faster than nitrogen fusion can heat it, even with no Coulomb barrier. Trinity's shocked air reached roughly 0.0002 MeV, and even at 1 MeV, five thousand times hotter, the real loss still wins by 31,000 to 1.

FAQ

Could the atomic bomb have ignited the atmosphere?

No. The worry was that the bomb's heat would start nitrogen nuclei fusing in a self-sustaining chain. Hot, ionised air loses energy to X-ray radiation faster than nitrogen fusion can supply it at every temperature. Even if nitrogen nuclei fused on every contact with no Coulomb barrier, the loss would still be about 12 times the gain, and Trinity's air never got within a factor of thousands of the needed temperature.

What was the LA-602 report?

LA-602, "Ignition of the Atmosphere with Nuclear Bombs", is a Los Alamos report by Emil Konopinski, Cloyd Marvin and Edward Teller dated 14 August 1946, a year after Trinity. It wrote up the wartime analysis and concluded that energy losses to radiation always outweigh the gains from nitrogen fusion, while calling for better experimental data on the reaction rates.

Did Oppenheimer think there was a chance of igniting the atmosphere?

Edward Teller raised the possibility at a 1942 Berkeley meeting, and Oppenheimer took it to Arthur Compton. Hans Bethe's estimates quickly showed the temperatures were far too low. By the time of Trinity in July 1945 the scientists considered it settled, though Enrico Fermi reportedly joked by taking bets on it in the final minutes.

Could a hydrogen bomb ignite the atmosphere or the oceans?

No. The 50 megaton Tsar Bomba did not, and a 1979 analysis by Thomas Weaver and Lowell Wood concluded that neither the atmosphere nor the oceans can sustain a propagating nuclear detonation under any circumstances. Seawater would need about 20 times more deuterium than it has, and even then the trigger would need around 20 million megatons.

Sources

M. Wiescher and K. Langanke, "Manhattan Project astrophysics", Physics Today 77(3), 34 (2024), describing LA-602, its safety factor, the Oak Ridge cross-section measurements and the Trinity fireball: physicstoday.aip.org
E. J. Konopinski, C. Marvin and E. Teller, "Ignition of the Atmosphere with Nuclear Bombs", LA-602 (1946), FAS copy (not accessible when checked): sgp.fas.org
A. Wellerstein, Restricted Data blog, on the history of the ignition question, the Trinity bets and Weaver and Wood (1979): blog.nuclearsecrecy.com
J. S. Díaz, "Explosion analysis from images: Trinity and Beirut", arXiv:2009.05674 (Taylor's fit and $S(1.4)^{-5} = 0.856$ ): arxiv.org/abs/2009.05674
R. S. Baty and S. D. Ramsey, "On the Symmetry of Blast Waves", LA-UR-20-30042, arXiv:2103.05710 (Taylor's yields for different γ): arxiv.org/abs/2103.05710
J. I. Katz, "Fermi at Trinity", arXiv:2103.05784 (test date and the 2021 yield of about 25 ± 2 kt): arxiv.org/abs/2103.05784
NRL Plasma Formulary bremsstrahlung formula, as restated in "MCTrans++: A 0-D Model for Centrifugal Mirrors", arXiv:2304.01496: arxiv.org/abs/2304.01496
NIST, Atomic Weights and Isotopic Compositions (14N, 24Mg, 4He): physics.nist.gov

Method note: fusion rates use a geometric cross-section ( $r = 1.2\,A^{1/3}\ \mathrm{fm}$ ) times the WKB Coulomb-barrier transmission, averaged over a Maxwell distribution with reduced mass 7 u. Losses use the non-relativistic NRL bremsstrahlung formula for fully ionised sea-level air with electrons and ions at one temperature. Shock temperatures use the ideal strong-shock relation with γ = 1.4. The relativistic and inverse Compton corrections, both left out, would only increase the losses.

Header image: NASA (image S100E5498), regraded for this article.

What exactly were they afraid of?

Why the margin was large: losses beat gains at any density

The worked calculation: an absurdly generous fusion rate still loses

What if the radiation can’t escape?

How hot did Trinity actually get the air?

The misconception: “they went ahead with a small chance of ending the world”

What about hydrogen bombs and the oceans?

FAQ

Could the atomic bomb have ignited the atmosphere?

What was the LA-602 report?

Did Oppenheimer think there was a chance of igniting the atmosphere?

Could a hydrogen bomb ignite the atmosphere or the oceans?

Related on The Science 360

Sources