Update | James says I misrepesented his method, which I may well have done. He is going to post a comment that explains his method in better detail than I have apparently done.
Apologies that it took so long to post the companion piece on yield estimation to my Foreign Policy column, North Korea’s Big Bang.
The good news is that there are now several yield estimates that help illustrate the point I want to make: Early yield estimates are very rough approximations.
|Aster||Mb=4.05+0.75 log (W)||5.1||20||15||40|
You will undoubtedly notice the very large range. I still think the best thing is to say the event was “several kilotons” or “on the order of ten kilotons” (as opposed to 1 or 100.) But the most important information to convey is that these are approximate yields based on a well-understood but rough method.
Some of the difference is explained by differences in estimating the magnitude of the body wave generated by the explosion, which ranges from 5.0 to 5.2.
But a much greater discrepancy arises from the equation that is used to model the geology of the test site and the resulting relationship between the size of the explosion (yield) and body wave (Mb). This equation is normally given in the form of Mb = A + B Log (Yield).
This equation differs from test site to test site, and while a good fit can be achieved, it is far from perfect. Here, for example, is some data from the then-Nevada Test Site (now Nevada National Security Site) that is used in the foundational study by Nuttli (1986) to give us the canonical equation: Mb = 4.05 + 0.75 log (W), where W is yield.
|Reported Yield (W) (in kt)||Reported Mb||Expected Yield (from reported Mb)||Expected Mb (from reported W)|
Reported yield and reported Mb are the actual data. ”Expected yield” is what one would expect based on the equation given the reported Mb. “Expected Mb” is what one would expect based on the equation given the reported yield.
So, for example, the 4.8 Mb event had a reported yield of 35 kilotons — much greater than the 10 kilotons that would expect based on the equation. Several events with different yields have the same Mb. There is nothing wrong here, it’s just that the physical relationship is an approximate one. A usefully approximate one.
Now, Nevada is not a perfect example — seismic waves don’t propagate all that well at the Nevada National Security Site. As a result, the yields for a given Mb in Nevada are a lot higher than in North Korea, which may lead to some head-scratching. (Blame the terroir!) Also, the data is noisier than it would be if we could make the model out of reliably reported North Korean events.
The major take-away is that while it’s a good fit, the standard method is not all that precise. In the Nuttli paper, he had a standard deviation of 0.2 Mb units. That’s quite a lot. (Later work seems to suggest that corrections for bias can get that number down to about 0.1 Mb unit.)
It is very hard to know what the proper relationship is for the North Korean test site. James Acton did something rather clever, assuming that the 2006 and 2009 tests were at the same depth, which allows him to assume that B=1. Fitting a line to two data points does not inspire huge confidence, especially when we don’t know the yield, location, and depth of burial, but it is better than not doing it at all and probably more appropriate than using Nuttli’s equation for Nevada or Murphy’s equation for Semipalatinsk, Lop Nor, and Pokhran. The Chinese seismological community takes a similar approach, using Mb = 4.25 + 0.75 Log W for events of more than 1 kiloton — the Bowers et al equation for Novaya Zemlya — based on three chemical explosions detonated in 1998.
As James is careful to point out that, these are back-of-the-envelope calculations that are used to help us start sorting through what this might mean. So, for example, we have a general sense of whether the test was successful (seems so) and whether it was a thermonuclear device (seems not, with some qualifications).
Competent seismologists — i.e., not policy hacks like myself or lapsed physicists like James — will come along and make much more sophisticated estimates. This will take some time and, I am sorry to tell you, almost certainly fail to produce a definitive result. Debate still exists about the yield of the 2009 event. The relevant literature illustrates a number of more sophisticated ways to estimate yield, none of which agree. While the US intelligence community placed the size of the 2009 test at approximately 2 kilotons, open-source estimates — including work done at Livermore and Los Alamos — range from 2 kilotons using wave-form coda to more than 5.7 kilotons using hydrodynamic calculations and satellite images. Nice to see Livermore and Los Alamos can still disagree. I would write a literature review, but I am not a seismologist and I find this stuff very, very dense. Feel free, in the comments, to do so, and to add any corrections.
So, for the DPRK test, my advice is for everyone to take a deep breath. The explosion was clearly bigger than the last. There is no harm in making rough yield estimates, provided we are all as careful as James. But let’s not take any of this too seriously just yet.