Supernovas, zinc mines, and North Korean nuclear blasts

At first glance, there is no obvious connection between zinc mining, neutron star formation, and nutty North Korean dictators. But a closer looks suggests that they may have something in common. And that something is the smallest object known to man.

The best way to determine whether or not the North Koreans actually detonated a nuclear weapon this week would be to detect a nuclear particle which is impossible to shield. One such particle is the neutrino. The problem with neutrinos that they are difficult to detect. But due to a stroke of luck, the Kamioka neutrino observatory, buried deep in an abandoned zinc mine in the Japanese Alps, just so happens to be located on the opposite side of the Sea of Japan as the North Korean test site.

If I was a qualitative blogger, I’d pat myself on the back for being so clever now, and go do something useful with my life. But since this is a science blog, I should at least try, in the roughest possible terms, to quantify this hypothesis. Is there any way to determine whether or not the Kamioka detector should be expected to pick up a nearby nuclear blast?

Addressing this problem could be tricky. On the one hand, I know jack-skittles about nuclear physics. On the other hand, I can look stuff up on the internet. So, taking the only two things I know about the Japanese neutrino detector, is it possible to estimate the sensitivity to a nuclear test.

First, the disclosure of those two things. I know that a recent paper used the neutrino detector to look at neutrinos produced by U and Th decay in the Earth’s interior. I also know that it detected the 1987 supernova in the Magellanic cloud.

Basically, this is a detection efficiency issue. If I can figure out the detection efficiency for a supernova, and then rescale that to the test, I can see how the number of expected neutrinos compares to the yield from the supernova.

The simplest test is this: to get a comparable number of counts, how much Pu do I need to fission?

Constant counts means that production/distance squared= constant for both the test and the supernova, assuming perfect transmission. If I can get the distances to both events, and the production for the supernova, then I should be able to calculate how much Pu I need to fission in North Korea.

So I need the following quantities
-distance from Kamioka to test site. (there’s a national geographic atlas on my shelf)
-distance from Kamioka to large Magellanic cloud (wikipedia)
-number of neutrinos produced in a supernova.
-number of neutrinos produced per unit mass of Pu fission.

I thought I would be able to simply look up the number of neutrinos in a supernova in wikipedia, but my searches of the obvious entries were fruitless. So if I want to know it, I will have to calculate it from first principles. Is that possible for an isotope chemist who knows squat about physics?

Well, wiki does tell me that the neutrinos are produced by the electron capture of protons in the Fe core of a giant star, as it transforms into a neutron star. How big a star? Rigel, in Orion, is given as an example of a blue supergiant. It is 20 solar masses. So if we simple assume that 20 solar masses of ⁵⁶Fe transforms all of its protons into neutrons, the number of neutrinos produces is simply the starting number of protons. 20 solar masses of ⁵⁶Fe worth of protons. A large, but tractable number.

The total number of nucleons in 20 solar masses is basically avagadro’s number times the mass in grams. In ⁵⁶Fe, 46% of those are protons. So, using Wiki or google again, we have 6.0221415x10²³ x 1.98x10³⁰kg x 1000 g/kg x 20= ~2.4x10⁵⁸ nucleons.

Holy leptons! I’ve never seen a number that big applied to physical science before. Astronomers are nuts. Of course, only 46% of them are protons- the rest are already neutrons, and thus don’t need to transform, but the total number of neutrinos emitted is still 1.1x10⁵⁸.

Distance to supernova: 168,000 light years x 9.46x10¹² km/lightyear = 1.59x10¹⁸ km. And I get to square it.

Distance to Kamioka from test site ~900 km.

Neutrinos generated at test site necessary to equal supernova flux in Kamioka: 7.68x10²⁷

That’s big, but not outrageous. Avagadro’s number is working on our side now, so to get back to kilos (not grams) we can divide by 6.03x10²⁶. In fact, we only need 5.8 kilos of nucleons to undergo beta decay in North Korea to produce an equivalent neutrino flux to the magellanic supernova. Depending on the bomb geometry, the total Pu needed for a bomb is in the 5-10 kg range.

So what fraction of Pu neutrons actually undergo beta decay? This is not a simple question to answer, but a rough estimate would suggest that it between 2 and 5%. For example, the net reaction ²³⁹Pu+n -> ⁹⁰Zr + ¹⁴⁷Sm + 3n requires eight beta decays. The neutrons will undergo beta decay in a vacuum, but in the real world, they are more likely to react with a nearby nucleus. The reaction product could either be stable or undergo further beta decay.

So in order to produce the neutrino flux equivalent to supernova 1987a, the North Koreans would have to fission between 110kg and 290 kg of Pu. This is about 2 orders of magnitude higher than what one would expect from a simple device, and 3 orders of magnitude larger than a dud.

Of course, the Kamiokande II detector used in 1987 has since been replaced by a bigger, better detector. And other factors, such as neutrino oscillation, neutrino energy, detection efficiencies between neutrinos and anti-neutrinos, and stuff I’ve never heard of have been ignored. But since the value is only a couple orders of magnitude short, it might be worth considering by someone less ignorant than I. So if I was a Japanese neutrino researcher, I’d lay down the line on the new nationalist PM. With a more sensitive detector, we might be able to call the North Koreans’ bluff.

Anyone with the tiniest smidgeon of knowledge in any of these subjects is welcome to tell me how wrong I am, and why.