Searching Neutralino Decay in the Sun w/ Amanda

sam_021
Messages
38
Reaction score
0
Hi
so I am doing a presenation, on the indirect search for the neutralino from the sun at amanda
In short:
neutralinos are graviationally captured and settle int he sun.. they decay to produce neutrinos.. we detect the neutrinos at the amanda detector through muon productionSo what I am confused about is basically the following:
1) in the detection if u look at papers like the following slide show -page 8
http://docs.google.com/viewer?a=v&q...zFvLRG&sig=AHIEtbS-vyEGTsEvTGs7SRevo8kNyg1rTQ
it goes on about events and live time.. so am i right in saying the live time is how long detector is running and the number of events is the number of muon productions (as in any muon be it from the neutralino or cosmic rays) .. I don't think I am right but I am not sure ?

2) on the same page as above ..it talks about simulations, so these different simulations are to obtain theoretical results, decrease background and etc right ?
 
Physics news on Phys.org


yes :D live tiime is the time the detctor is on till :D
 
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}...
Back
Top