Little Bang Posted February 15, 2011 Report Posted February 15, 2011 In a head on collision of two protons at the LHC. Has anyone seen any calculations trying to describe the electric fields as the two protons collide? Quote
Qfwfq Posted February 15, 2011 Report Posted February 15, 2011 For soft scattering events, a classical computation, from the radial density distributions, at least makes some sense. It makes less sense for hard events, where interactions between individual fermions dominate the picture, with the consequence that charge density can be approximated by superposition of the undisturbed distributions. Certainly the EM interaction isn't what matters the most in the events where they pass through each other. Quote
Little Bang Posted February 15, 2011 Author Report Posted February 15, 2011 As the two protons approach their electric fields must be tremendously compressed. Surely there must be a way to calculate the field density just before impact. Quote
Erasmus00 Posted February 17, 2011 Report Posted February 17, 2011 As the two protons approach their electric fields must be tremendously compressed. Surely there must be a way to calculate the field density just before impact. Why do you think the field is compressed? Electric fields are nice and linear- you can calculate the field of one proton by assuming the other isn't there at all, and then add the two fields. Its not quite the same as the simple inverse square law, because of relativistic effects, see http://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential for fields of relativistically moving point charges. Quote
Little Bang Posted February 20, 2011 Author Report Posted February 20, 2011 Do positive charges repel? I'm almost certain they do so in a very high velocity collision those fields would be compressed on the collision side.Those fields are no different from the magnetic field surrounding the Earth. Protons from the sun compress that field on the sun-ward side, I'm sure the same happens in a proton proton collision. Quote
Erasmus00 Posted February 21, 2011 Report Posted February 21, 2011 Do positive charges repel? I'm almost certain they do so in a very high velocity collision those fields would be compressed on the collision side.Those fields are no different from the magnetic field surrounding the Earth. Protons from the sun compress that field on the sun-ward side, I'm sure the same happens in a proton proton collision. Electromagnetic fields are linear. I'm not familiar with the compressing you are referring to, but its probably because the moving protons themselves constitute a plasma with a fairly strong magnetic field of their own. For the protons, up until the point where thinking of an electromagnetic field breaks down (and photons become important), the field is linear, and you can just add the fields of the individual protons. Quote
Rade Posted May 5, 2011 Report Posted May 5, 2011 I have a question that will expose my ignorance, but that is why I post, to learn from experts. Each proton {P} has an energy of 938.272 MeV/c^2. So, when two protons collide {P}+{P} there is only 1876.544 MeV total energy in the system, yet there are results of energy emerging from the collision in the 100-200 GeV range. So, what is the explanation for all this excess energy within the simple proton {P}--is it all from the momentum of each {P} just prior to the collision via E = Mc^2, or is there something "hidden" within each {P} that can be expressed when taken to such high momentum, or maybe a little of both are needed for a proper explanation ? Quote
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