The point being that either 'speed' is more than sufficient to whack the ball out of the ballpark which is a most interesting piece of the puzzle.
-----Original Message----- From: [email protected] [mailto:[email protected]] Sent: Friday, January 20, 2017 3:04 PM To: [email protected] Subject: Re: [Vo]:New paper from Holmlid. In reply to Axil Axil's message of Fri, 20 Jan 2017 16:44:39 -0500: Hi, [snip] I was hoping someone here would show me the error of my ways, before I made a complete fool of myself in public. (yes, I know I have already done that.) :) I think I may have found either my mistake or Einstein's. ;) Looking at https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation it looks like the Pythagorean relationship, which would seem to imply that we are looking at perpendicular vectors. However both energy and energy squared are scalars, so something is wrong. Note also that all the terms have the dimension of energy squared, implying that the "vectors" have the dimension of energy. Note that if Einstein is correct, then Ek is not p*c, which would be where I made my mistake here below (in my previous email). Consider this:- Et = Ek + Ep (where Et is total energy, Ek is kinetic energy, Ep is potential energy) => Et^2 = (Ek + Ep)^2 = Ek^2 + 2EpEk + Ep^2 Now compare this to https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation It looks very similar except that the term "2EpEk" is missing. That missing term is exactly what would happen if Ep and Ek were perpendicular vectors, forming the sides of a right triangle, and Et was the hypotenuse. In which case according to Pythagoras we get Et^2 = Ek^2 + Ep^2. Now this looks just like https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation The only problem is that Ek & Ep are not vectors, they are scalars....unless there is some dimension I am ignoring in which energy is the magnitude of a vector quantity?? >The is a comment section in the PLOS/1 format where a reader can submit >corrections as required for evaluation by the author. Why not submit >this proposed correction through this comment method. > >On Fri, Jan 20, 2017 at 4:21 PM, <[email protected]> wrote: > >> In reply to Axil Axil's message of Thu, 19 Jan 2017 11:58:00 -0500: >> Hi, >> [snip] >> >http://journals.plos.org/plosone/article?id=10.1371/journal.pone.016 >> >9895 >> >> I think Holmlid made a mistake in his velocity calculation. (Either >> that, or I did). >> >> He equates 500*MeV/u to 0.75 c. >> >> I think this derives from the formula:- >> >> (sqrt(500*MeV/u))/c = 0.733 which is close to 0.75 C. >> >> where u is the standard atomic mass unit. (i.e. mass of Carbon12 / 12). >> >> However I think the formula is incorrect, see the following derivation. >> >> From Einstein we get:- >> >> Kinetic energy (Ek) = p*c (where p is the momentum). >> >> p = m * v where m is relativistic mass, and v is the velocity. >> >> => Ek = m*v*c >> => Ek/m = v*c >> => Ek/(mc) = v >> >> 500 Mev /amu has the dimension of energy/unit mass, i.e. Ek/m, >> >> So >> >> 500 MeV / u = Ek/m = v*c >> >> => v = 500 MeV / (u*c) = 1.609E8 m/s or, as a fraction of c, >> >> 500 MeV / (u*c^2) = 0.537 (not 0.733) >> >> Regards, >> >> Robin van Spaandonk >> >> http://rvanspaa.freehostia.com/project.html >> >> Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
