Recent content by mborn

Thanks ansgar, I also could find there "transverse energy" which was my second question. ~mborn

Hi all, Just as the title says: What is transverse mass? Why do experimentalists use it instead of the usual mass (if it is something different and not just a jargon)? mborn

Thank you Rajini, So, I am OK now with normalization! Still I hope someone here will explain reweighting for us. As far as I know, reweighting is to divide the number of events in each bin of a histogram by the total number of events. If this is true, then why do we need to do reweighting? ~mborn

To make things eaiser, let's say you are in a a particle physics experiment that is trying to discover a particle. Now, for normalization, how can we make the area under the histogram equal one? Can you elaborate more on reweighting? What is the difference between reweighting (to Monte Carlo)...

Hi all, I am about to begin my studies as an experimentalist and I keep hearing about these terms when someone represents his data as histograms. Can some one here, please, give me a clear explanation about their meanings. My background is theory and you can use as much mathematics as you...

www.satirewire.com/news/may02/images/fermi.jpg I think he made the mistake again?

1eV=1.602 \times 10^{-19} J m_e=9.11 \times 10^{-31} Kg M B

Thanks all, may be he is talking about an INCLOSED charge, and in this case it is what you all said. But, if this charge is ON the spherical shell, then I have a right to confuse a little! :smile:

same as my answer without rounding. Thanks M B

Thanks! M B

that is what is said on my book? Charges tend to accumulate at the points at which radius of curvature is the smallest, that is at sharp points. M B

why didn't my [tex] appear? M B

\sqrt{3} x - x = \sqrt {3} then take x as a common factor M B

yes, it is negative on the upper surface, but I say there should be no charge on the inner surface

At two different points of an irregularly shaped conductor, the field had the following two values 5.6*10^4 and 2.8*10^4 respectively. Find the local surface charge density at; 1- the point with the greatest radius of curvature, 2- the point with the smallest raduis of curvature. I know...