Recent content by Anchovy

Mental arithmetic for definite, especially when three or more digits. I get the answer but it's not instant. I think part of the problem is I don't have one constant method of doing the calculation, e.g. sometimes I'll add the numbers together from say, units then tens then hundreds, sometimes...

Ah, well that's my question answered.

This is what made me come to this sub-forum. I get why the main sub-forums have to keep to non-speculative stuff, but why not have an exta bit where people can discuss stuff that would get deleted elsewhere? Does this exist already and I've not looked hard enough for it?

OK, thanks.

Ah, I see. Do things become much simpler if it's not actually two different experiments, but rather it's one Monte Carlo simulation, counting the same thing happening two different ways?

If the results of two separate experiments to measure the same quantity are stated in terms of upper and lower limits at the same confidence level, is it valid to say that the overall upper limit (at the same C.L.) is just the sum of the two individual upper limits? Or is something more...

OK, thanks.

OK, so what would cause such difference in a water Cerenkov? (I think in your example it'd be the other way round, I think Super K has set a higher lower-limit for the muon channel than a kaon channel). Do kaons produce worse-quality rings at the PMTs? Or would more muons be energetic enough...

If I look at the lower limits on the proton decay lifetime \tau set by, say, Super-Kamiokande, I'll see different lower limits depending on what the proton could decay into, eg. \tau_{min}(p \rightarrow K^{+} \overline{\nu}) < \tau_{min}(p \rightarrow \mu^{+} \pi^{0}) < \tau_{min}(p \rightarrow...

If we replace the nucleons (p, n) in the matter with antinucleons (pbar, nbar), there's nothing to say that they would have the same interactions, seeing as these two kaon doublets that are antiparticles of each other interact differently, so I'm going to say the asymmetry would flip signs?

Ah I deleted that CP violation comment earlier, didn't want to confuse things. I'm not 100% confident yet but let me see: (K+, K0) in one isospin doublet. (K-, K0bar) in another. These objects are each symmetric under isospin rotations? (ie. the strong interaction sees K+ and K0 as the same...

Is it that the charged kaon partners that are in each kaon isospin doublet are produced at different rates? Are we saying the doublet members, say (K+, K0), are interchangeable when a strong interaction occurs (and the same being true for (K-, K0bar) )? But if the K0 and K0bar don't interact at...

OK, so... p + S --> anything and n + S --> anything go at the same rate then. (ie. Nucleon is invariant under rotations in isospin space?) K+ + S --> anything and K0 + S --> anything go at the same rate (since (K+, K0) are in the same isospin doublet?) K0bar + S -> anything and K- + S...

Is it that their rates have to differ because p and n have different isospin 3rd components?

The K+ has isospin +1/2, neutron has -1/2, so a total of zero in the initial state so the final state should be zero overall too? The K0 has -1/2, proton has +1/2, so final state should be zero? The K- has +1/2, proton has +1/2, so final state should be +1 overall. The K0bar has -1/2, neutron...