Can we determine the NK nuclear detonation strength?

mkarger
Messages
56
Reaction score
0
Last edited by a moderator:
Physics news on Phys.org
I asked similar question back in nineties, when French did their last nuclear test. The short answer I got - from some seismologist - was "no". Too many variables.

I suppose some rough estimates on minimum and maximum value are possible. Wikipedia page on Richter scale gives something in the range of 0.5kT as an equivalent (in terms of amount of energy) of 5.0. But it is not even clear to me if the magnitude given is on the Richter scale, plus it is unlikely that all energy was used to shake the ground, so I doubt it is even a reasonable low estimate.
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

Similar threads

Back
Top