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Groups > sci.physics.relativity > #368577
| From | Thomas 'PointedEars' Lahn <PointedEars@web.de> |
|---|---|
| Newsgroups | sci.physics.relativity |
| Subject | Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 |
| Date | 2015-10-30 18:31 +0100 |
| Organization | PointedEars Software (PES) |
| Message-ID | <2895113.jQvESdZhOY@PointedEars.de> (permalink) |
| References | <5cf2af61-4464-4be0-b3a0-65878fdd881c@googlegroups.com> <be20fd36-5c65-4bcd-a0fe-2fb48a3eee46@googlegroups.com> <8c29812f-4705-4a7f-be4c-3b58280f4b28@googlegroups.com> <e2a78e57-a4c3-4e58-837e-29f59b31b7ff@googlegroups.com> |
fuller.david@hotmail.com wrote: > How much do the protons weigh in the LHC at 7Tev? > > The energy of a proton is 7 TeV. No, it is not. That is the *collision* energy coming from at least *two* colliding protons. You want to read page 3 and following of the 2009 edition of the LHC Guide: <http://cds.cern.ch/record/1165534/files/CERN-Brochure-2009-003-Eng.pdf#page-3> > Via E = mc2 the mass is simply 7 TeV/c2 - > and these are the units usually used. ["mc2" and "c2" are not proper notations. If you cannot use Unicode -- "mc²" and "c²", respectively -- you have to find an ASCII-compatible way to express powers. The usual way is to use the circumflex/caret character, so it would be "mc^2" and "c^2", respectively.] As has been explained here many times before (including several times by me), E = m c² (E = m c^2) is only true for the energy at relative rest, E = E(v = 0) = E(p = 0) =: E₀. The full equation for energy, and thereby the mass–energy equivalence, that can be derived from the norm of the four-momentum is E = √((m c²)² + (p c)²) (1) [E = sqrt((m c^2)^2 + (p c)^2)], with momentum p = γ m v (2) [p = gamma m v], and the Lorentz factor γ = 1∕√(1 – (v∕c)²) (3) [gamma = 1/sqrt(1 - (v/c)^2)], therefore, applying (3) to (2), p = m v∕√(1 – (v∕c)²) (4) [p = m v/sqrt(1 - (v/c)^2)], and finally, applying (4) to (1), E = m c³ √(1/(c² − v²)) [E = m c^3 sqrt(1/(c^2 - v^2))], assuming m > 0, c > 0, and v ≥ 0 (v >= 0). As you can read in the LHC Guide, the protons in the LHC, when they reach their greatest relative velocity, move with 99.9999991 % of the speed of light in vacuum, or 0.999999991 c. So it is very important not to ignore the part of their energy that comes from their relative motion. In fact, that is why we need particle *accelerators* for achieving high particle energies, therefore the chance for producing heavy particles like the discovered Higgs boson, in the first place. > 7 TeV/c^2 divided by the rest mass 0.938272029 GeV/c^2 gives us 7460.52 > times the rest mass Ex falso quodlibet. PointedEars -- Q: What happens when electrons lose their energy? A: They get Bohr'ed. (from: WolframAlpha)
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(((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-26 20:34 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-27 08:33 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-27 08:44 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-27 08:46 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-27 08:53 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-28 07:34 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-28 07:46 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-28 09:42 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-28 10:17 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-28 10:25 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-28 10:51 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-28 10:54 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-28 19:29 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-28 19:59 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-29 10:43 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-29 10:44 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-29 11:03 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 kefischer <emoneyjoe@iglou.com> - 2015-10-28 17:29 -0400
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-28 15:37 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-28 09:51 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2015-10-30 18:31 +0100
Something related to the Collision of Protons Bobby Sifuentes <bobsif@academicinstitute.org> - 2015-10-30 17:44 +0000
Re: Something related to the Collision of Protons Odd Bodkin <bodkinodd@gmail.com> - 2015-10-30 14:11 -0500
Re: Something related to the Collision of Protons Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2015-10-30 21:06 +0100
Re: Something related to the Collision of Protons Bobby Sifuentes <bobsif@academicinstitute.org> - 2015-10-30 20:47 +0000
Re: Something related to the Collision of Protons Bobby Sifuentes <bobsif@academicinstitute.org> - 2015-10-30 20:49 +0000
Re: Something related to the Collision of Protons Bobby Sifuentes <bobsif@academicinstitute.org> - 2015-10-30 20:54 +0000
Re: Something related to the Collision of Protons Bobby Sifuentes <bobsif@academicinstitute.org> - 2015-10-30 21:52 +0000
Re: Something related to the Collision of Protons Bobby Sifuentes <bobsif@academicinstitute.org> - 2015-10-30 21:59 +0000
Re: Something related to the Collision of Protons fuller.david@hotmail.com - 2015-10-30 22:20 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 John Heath <heathjohn2@gmail.com> - 2015-10-30 16:10 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-30 21:34 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-30 21:32 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2015-10-31 06:50 +0100
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 Jack Weidner <jackwd@webportal.au> - 2015-10-31 11:20 +0000
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-31 05:48 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2015-10-31 16:16 +0100
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-31 08:32 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-31 08:35 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2015-10-31 17:18 +0100
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 Hyperbolic.cc@outlook.com - 2015-10-31 09:30 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 Hyperbolic.cc@outlook.com - 2015-10-31 09:38 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 kefischer <emoneyjoe@iglou.com> - 2015-10-31 13:42 -0400
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 Hyperbolic.cc@outlook.com - 2015-10-31 09:58 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-31 12:15 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-11-04 07:58 -0800
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 Hyperbolic.cc@outlook.com - 2015-11-05 08:49 -0800
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 kefischer <emoneyjoe@iglou.com> - 2015-11-05 12:08 -0500
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 Poutnik <Poutnik4NNTP@gmail.com> - 2015-11-05 19:23 +0100
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-31 12:11 -0700
Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 Hyperbolic.cc@outlook.com - 2015-10-31 09:59 -0700
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