The Los Alamos Primer - Part 2

• 11 February 2026
• physics

Simplest Estimate of Minimum Size of Bomb

In Part 1 I covered the first few sections of the Los Alamos Primer: energy release, cross-sections, and why ordinary uranium can't sustain a chain reaction. This post picks up where we left off - figuring out how big a lump of fissile material needs to be before it goes critical.

We will model neutron transport as a diffusion process and solve for the geometry where production exactly balances leakage. The original text skips a lot of steps and rushes through this section, so I'll try to fill in the gaps.

Please note that this is what the Primer calls the "simple" diffusion theory. We won't get quite the "proper" results from it. The Primer says as much:

Specifically, this bit:

In ordinary diffusion theory (which is valid only when all dimensions of boundaries are large compared to the mean free path of the diffusing particles - a condition not fulfilled in our case)

Still, I think going through this is valuable:

• we get an overestimated result but within the right ballpark
• we can attempt to correct some of the assumptions we make for a better result
• the general "diffusion process" modelling is very powerful

The Diffusion Equation

We start with neutron density $N$ (number of neutrons per unit volume). Two things affect how $N$ changes over time:

1. Production: Each fission produces $\nu$ neutrons on average, and fissions happen with mean time $\tau$ between successive fission events. We can estimate $\tau$ from the fission mean free path ${\lambda }_f$ and the neutron speed $v$: $\tau ={\lambda }_f/v$. For fast neutrons in U-235, $v\sim {10}^9$ cm/s, giving $\tau \approx {10}^{-8}$ seconds. The net production rate per neutron is $(\nu -1)/\tau$ - the "$-1$" accounts for the neutron consumed in the fission process (one neutron goes in, $\nu$ come out, so the net gain is $\nu -1$).

2. Diffusion: Neutrons scatter around and tend to flow from high-density regions to low-density regions. The current (flux of neutrons) is proportional to the gradient: $\mathbf{j}=-D\mathrm{\nabla }N$, where $D$ is the diffusion coefficient. Physically, $D$ captures how quickly neutrons spread through the material. It's related to the transport mean free path ${\lambda }_{tr}$ (which accounts for scattering angles) and neutron speed by $D={\lambda }_{tr}v/3$. The factor of $1/3$ comes from averaging over all directions in three dimensions.

Aside: Wait, what is a "Mean Free Path" actually?

If you're wondering why the Primer keeps swapping between $\lambda$ and $\sigma$ (cross-sections), here is the quick and dirty version.

Imagine running blindfolded through a forest.

• $N$ is the density of trees (trees per $m^3$).
• $\sigma$ is the "cross-section" of a tree trunk - literally how "fat" the target is (in $m^2$).

If you multiply them, you get $N\sigma$ (often called the Macroscopic Cross-section, $\mathrm{\Sigma }$). The units are $1/m^3\times m^2=1/m$. This represents the probability of smacking into a tree per meter you run.

If you have a probability of hitting something per meter, the average distance you travel before hitting it is just the reciprocal:

$$\lambda =\frac{1}{N\sigma }$$

That's it. It's just a probability inversion. High density ($N$) or fat targets ($\sigma$) mean a short path ($\lambda$).

In our case, we actually have two different mean free paths because the trees are weird:

• ${\lambda }_f=1/(N{\sigma }_{fission})$: How far you run before you hit a tree that explodes (fission).
• ${\lambda }_{tr}=1/(N{\sigma }_{transport})$: How far you run before you hit a tree and bounce off (scattering/transport).

Combining these gives the diffusion equation:

$$\dot{N}=D\mathrm{\Delta }N+\frac{\nu -1}{\tau }N$$

where $\mathrm{\Delta }={\mathrm{\nabla }}^2$ is the Laplacian. The first term describes neutrons spreading out; the second describes their multiplication.

Separation of Variables

We want to find steady-state behaviour, so assume the solution separates into spatial and temporal parts:

$$N=N_1(x,y,z)\cdot e^{{\nu }^{\prime }t/\tau }$$

Here ${\nu }^{\prime }$ is the "effective neutron number" - it captures the actual growth rate once leakage is accounted for. If ${\nu }^{\prime }>0$, the neutron population is growing exponentially and you have a supercritical system. If ${\nu }^{\prime }<0$, neutrons are leaking out faster than they're being produced and the reaction dies.

Taking the time derivative:

$$\dot{N}=\frac{{\nu }^{\prime }}{\tau }N$$

And substituting back into the diffusion equation:

$$\frac{{\nu }^{\prime }}{\tau }N=D\mathrm{\Delta }N+\frac{\nu -1}{\tau }N$$

The exponential factor cancels (since the Laplacian only acts on spatial coordinates), leaving:

$$\frac{{\nu }^{\prime }}{\tau }{N}_1=D\mathrm{\Delta }{N}_1+\frac{\nu -1}{\tau }{N}_1$$

Rearranging:

$$D\mathrm{\Delta }{N}_1=\frac{{\nu }^{\prime }-(\nu -1)}{\tau }{N}_1$$

$$\mathrm{\Delta }{N}_1+\frac{(\nu -1)-{\nu }^{\prime }}{D\tau }{N}_1=0$$

If we define $k^2=\frac{(\nu -1)-{\nu }^{\prime }}{D\tau }$, this becomes the Helmholtz equation:

$$\mathrm{\Delta }{N}_1+k^2{N}_1=0$$

Note that $k^2>0$ requires ${\nu }^{\prime }<\nu -1$, i.e. the effective neutron number is less than the bulk production rate. This is the physically relevant regime: leakage always reduces the effective multiplication below its theoretical maximum.

This is an eigenvalue problem. The geometry and boundary conditions determine $k$, and then we can solve for ${\nu }^{\prime }$.

Solving for a Sphere

A sphere is a natural starting point. For a sphere of radius $R$, we assume $N_1$ depends only on $r$. The Laplacian in spherical coordinates (for radially symmetric functions)^[1] is:

$$\mathrm{\Delta }{N}_1=\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d{N}_1}{dr}\right)=\frac{d^2{N}_1}{d{r}^2}+\frac{2}{r}\frac{d{N}_1}{dr}$$

So we need to solve:

$$\frac{d^2{N}_1}{d{r}^2}+\frac{2}{r}\frac{d{N}_1}{dr}+k^2{N}_1=0$$

There is a standard trick here. Substitute $N_1=u(r)/r$. The middle terms cancel beautifully, leaving:

$$u^{\prime \prime }+k^{2}u=0$$

This is just simple harmonic motion. The general solution is:

$$u(r)=A\sin (kr)+B\cos (kr)$$

So:

$$N_1(r)=\frac{A\sin (kr)+B\cos (kr)}{r}$$

Boundary Conditions

This is where we plug in our assumptions.

At the origin ($r=0$): $N_1$ must be finite. Since $\cos (kr)/r\to \mathrm{\infty }$ as $r\to 0$, we need $B=0$.

At the surface ($r=R$): Neutrons escape. The simplest model sets $N_1(R)=0$.

$$N_1(R)=A\frac{\sin (kR)}{R}=0$$

For non-trivial solutions, $\sin (kR)=0$, which means $kR=n\pi$ for integer $n$. The fundamental mode ($n=1$) gives:

$$k=\frac{\pi }{R}$$

The solution is:

$$N_1(r)=A\frac{\sin (\pi r/R)}{r}$$

The Critical Radius

Now we connect $k$ back to the physics. Recall:

$$k^2=\frac{(\nu -1)-{\nu }^{\prime }}{D\tau }$$

With $k=\pi /R$:

$$\frac{{\pi }^2}{R^2}=\frac{(\nu -1)-{\nu }^{\prime }}{D\tau }$$

The critical radius $R_c$ is where ${\nu }^{\prime }=0$ - the boundary between growth and decay:

$$0=(\nu -1)-\frac{{\pi }^{2}D\tau }{R_c^2}$$

$$R_c^2=\frac{{\pi }^{2}D\tau }{\nu -1}$$

This is the key result. Now, let's substitute our definitions for $D$ and $\tau$ back in to see what physical constants drive this.

• $D={\lambda }_{tr}v/3$
• $\tau ={\lambda }_f/v$

The velocity $v$ cancels out, leaving us with:

$$D\tau =\frac{{\lambda }_{tr}{\lambda }_f}{3}$$

And the critical radius formula becomes:

$$R_c=\pi \sqrt{\frac{{\lambda }_{tr}{\lambda }_f}{3(\nu -1)}}$$

This makes physical sense: the critical size scales with the geometric mean of the transport and fission mean free paths. Longer mean free paths mean neutrons travel further before interacting, so you need a bigger sphere to keep them contained. More neutrons per fission ($\nu$) means you can afford more leakage, so the critical size shrinks.

What About Other Shapes?

The sphere is optimal - it has the smallest surface-area-to-volume ratio, minimising neutron leakage. But what if your fissile material is cast into a different shape?

The Cube

We are presented with the following exercise:

Show that if the gadget has the shape of a cube, $0<x<a0<y<a0<z<a$ that the critical value of $a$ is given by $a=\sqrt{3}{R}_c$.

Let's consider a cube of side length $a$. Note that everything up until the Helmholtz equation was shape-agnostic: we then get different results depending on the Laplacian we plug in. Starting again with the Helmholtz equation:

$$\mathrm{\Delta }{N}_1+k^2{N}_1=0$$

Except now we use good old Cartesian coordinates. We also assume a separable solution:

$$N_1(x,y,z)=X(x)\cdot Y(y)\cdot Z(z)$$

Substituting and dividing by $XYZ$:

$$\frac{X^{\prime \prime }}{X}+\frac{Y^{\prime \prime }}{Y}+\frac{Z^{\prime \prime }}{Z}+k^2=0$$

Here is the key insight: each term depends on a different variable, but they sum to a constant. This is only possible if each term is individually constant^[2]:

$$\frac{X^{\prime \prime }}{X}=-k_x^2,\frac{Y^{\prime \prime }}{Y}=-k_y^2,\frac{Z^{\prime \prime }}{Z}=-k_z^2$$

with

$$k_x^2+k_y^2+k_z^2=k^2$$

Applying boundary conditions $X(0)=0$ and $X(a)=0$, we find $k_x=\pi /a$, and by symmetry $k_y=k_z=\pi /a$.

So:

$$k_{cube}^2=\frac{{\pi }^2}{a^2}+\frac{{\pi }^2}{a^2}+\frac{{\pi }^2}{a^2}=\frac{3{\pi }^2}{a^2}$$

Comparing Sphere and Cube

At criticality, both systems have the same $k^2$ (since it depends only on material properties). Setting $k_{cube}^2=k_{sphere}^2$:

$$\frac{3{\pi }^2}{a^2}=\frac{{\pi }^2}{R_c^2}$$

$$a^2=3R_c^2$$

$$a=\sqrt{3}{R}_c\approx 1.73R_c$$

A critical cube has side length about 1.73 times the critical radius of a sphere made from the same material. To put that in terms of material needed: the critical cube has volume $a^3=3\sqrt{3},R_c^3\approx 5.2,R_c^3$, while the critical sphere has volume $\frac{4}{3}\pi R_c^3\approx 4.19,R_c^3$. So the cube needs about 24% more fissile material.

Physical Reality

Now to tie this all together. The Primer uses the following values for U-235:

• ${\lambda }_f\approx 13$ cm (fission mean free path)
• ${\lambda }_{tr}\approx 5$ cm (transport mean free path)
• $\nu \approx 2.2$ (neutrons per fission)

Plugging into our formula:

$$R_c=\pi \sqrt{\frac{5\times 13}{3(2.2-1)}}=\pi \sqrt{\frac{65}{3.6}}$$

$$R_c=\pi \sqrt{18.05}\approx \pi \times 4.25\approx 13.35\text{ cm}$$

This is extremely close to the Primer's stated result of $13.5$ cm.

A Note on the Numbers

You might notice we got $13.35$ while the Primer gets exactly $13.5$. That's because the Primer doesn't actually use the "two rounded lambdas" method we just did. Instead, it calculates $D\tau$ directly from the cross-sections rather than from rounded mean free paths, getting a slightly different value. Working backwards from the Primer's result:

$$R_c^2=\frac{{\pi }^{2}D\tau }{\nu -1}\approx 183$$

$$\sqrt{183}\approx 13.5\text{ cm}$$

The discrepancy arises because we approximated ${\lambda }_{tr}$ and ${\lambda }_f$ as independent rounded integers (5 and 13). In reality, the transport cross-section ${\sigma }_{tr}$ technically includes the fission cross-section (it's the total collision probability weighted by angle), so the exact ratio isn't perfectly captured by simply multiplying our two rounded lambdas. But $13.35$ vs $13.5$ is close enough for government work.

Using $R_c=13.5$ cm, the critical mass is roughly $M_c=\frac{4}{3}\pi R_c^3\rho \approx 200$ kg for a bare sphere (using $\rho \approx 18.7$ g/cm³ for metallic uranium).

So is 200 kg the actual answer? Not quite.

Our Accuracy

The math checks out, but the physics is still an overestimate. The elementary diffusion theory gives us $\approx 13.5$ cm, but the correct critical radius is approximately 9 cm.

So where did we go wrong? It seems the main culprit is our boundary condition. We set $N_1(R)=0$ precisely at the surface, which assumes that any neutron touching the edge is instantly lost forever.

However, real neutron transport is apparently a bit messier. From what I understand of the "proper" transport theory (which gets complicated very quickly), the neutron density doesn't actually hit zero until a short distance outside the physical sphere. This is known as the "linear extrapolation distance" - essentially, the vacuum outside the sphere reflects the lack of incoming neutrons in a way that makes the sphere "feel" slightly larger than it is.

The standard correction involves pretending the sphere has a radius of $R+0.71{\lambda }_{tr}$​. When you apply that correction to the diffusion equation, the physical radius $R$ needed for criticality drops significantly.

With the corrected radius of $\approx 9$ cm, the critical mass drops to $\frac{4}{3}\pi (9{)}^3\times 18.7\approx 57$ kg - roughly a third of our estimate. That's a significant overestimate, but for a back-of-the-envelope diffusion model, landing within the same order of magnitude isn't too shabby.

TBC

Next time I'll cover the tamper and revisit this diffusion theory.

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