In nuclear weapons, why does levitating a pit improve compression?

Question

Levitated pits were introduced after after solid pits. In this design the tamper is separated from the fissile with an airgap. From the Nuclear Weapon Archive:

The original Fat Man pit design used a Christy solid plutonium core, surrounded by a close fitting natural uranium tamper. The Sandstone devices all replaced the contiguous tamper-core approach with a "levitated core" in which the core was suspended within a larger hollow space within the tamper so that a gap existed between them. The collision between the tamper and core would create more efficient compression of the core than the explosive-driven shock in the wartime design.

Other vague descriptions I found:

The first improvement on the Fat Man design was to put an air space between the tamper and the pit to create a hammer-on-nail impact.

Efficiency of the implosion can be increased by leaving an empty space between the tamper and the pit, causing a rapid acceleration of the shock wave before it impacts the pit.

I recently started studying shockwaves and I don't understand this. How does leaving an empty space increase schokwave acceleration?

If I remove the airgap, the mass of the core won't change but now I can use the extra volume for more explosive. Inside the pit, compression of the fissile is the result of work performed by the explosive charge, so how is it possible that I can increase compression by reducing explosive volume ??

Answer 1

There is a Youtube video that explains this levitating pit quite well (Nuclear $101$: How Nuclear Bombs Work Part $1/2$ from the chanel of Belfer Center (see $t = 1579$ ff.)):

In the video we briefly say that this air space between the explosive and the fission material serves to give the outer shell, which is accelerated when the explosive explodes, more time to accelerate (so that the shell becomes faster and in the end more pressure is exerted on the core), since this air offers less resistance to the imploding shell than the core. Before the imploding shell reaches the core, it just needs to move as fast as possible, this air gap takes care of that.

Hence the analogy with the hammer and the nail: If you want to hit a nail with a hammer, where do you put the hammer / where do you get the momentum from? Right on the nail or do you pull it back/up/... from the nail? That's right, you pull the hammer back a little, accelerate it more because of the large distance, which gives it more momentum, and then this force acts on the nail to drive the nail in. Here the hammer stands for the explosive and the shell and the nail for the core.

So you might be leaving minimal explosives for room to accelerate. But you get that little bit back with the extra space to accelerate (get faster).

Answer 2

This is a fun and good question.

The size of the gap will not be big enough such that using that volume for more explosives will be of substantial difference. That is, the total energy of conventional explosives and the total inward momentum will be roughly the same between the case of having a gap, and filling up the gap.

To get a great shockwave, given a roughly constant total E and total momentum, we are thus trying to increase the forces inside the system. There is not much leeway to change the size of the system, i.e. $\mathrm dx$ is going to be difficult to change. But by adding a gap, we can change the $\mathrm dt$ quite a lot.

If there is no gap, then as the conventional explosive is going off, the inner material will always push back outwards. This means that the material always feels force $F-f$, where the explosive gives you $F$ and the inner material pushes back with $f$. By having a gap, $f$ is missing as the explosion $F$ pushes over the gap distance, and the time traversing the gap.

Then, suddenly, as the gap distance is traversed, the inward momentum is slammed shut, and over a tiny time $\mathrm dt$, a tremendous force is generated. That is how to vastly increase the strength of the shockwave given a roughly constant total energy E and total momentum.