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Electron-electron scattering

  1. Apr 1, 2009 #1
    My emphasis is on the solid-state/mesoscopic physics and I have the following question:

    Can the interaction of two electrons that scatter off one another due to Coulombic repulsion be considered elastic?

    I know that the phases of the electrons will be randomized and I also sense that this will not reduce the tranmission (although the momenta will be randomized, the lost momentum will be picked up by the other electron) but I don't know whether I can assume that they are going to travel with their initial wave vector ( frequency)?

    This is somewhat important for my research. Thanks for your responses.
     
  2. jcsd
  3. Apr 2, 2009 #2
    I'm going to try some dimensional reasoning on this and see where it goes. Correct me if I'm wrong....

    When you talk about inelasticity in electron-electron scattering, I'm assuming you mean the losses to radiation. Let's start by looking at the Hydrogen atom, with its 13.6 eV electrons and it 1 Angstrom radius. I'm going to think of it as a Bohr atom with the planetary orbits...I know this is wrong but it lets me ballpark some figures.

    The v-squared-over-r forces in the Hydrogen atom are similar to what you would have if you shot two 13.6-eV electrons at each other and collided them head-on. So the radiative losses for the scattering case should be similar to the radiation from one cycle of the Bohr atom. This comes to around one part in 10 000 000 of the total energy. (In e-m theory the reciprocal of this fraction is called the Q factor of the corresponding antenna.)

    As you increase the energy, the Q factor goes down by what I have reasoned out is the 3/2 power of the energy. For example: four times the energy gives 1/4 the radius gives twice the velocity gives one-eighth the interaction time gives sixteen times the accelleration gives 256 times the power taken over 1/8 the interaction time gives 32 times the energy which implies one-eighth the Q factor. (The Q factor, remember, is the ratio of the energy-loss-per-cycle versus the total energy. Eight is related to four by the 3/2 power.

    To go from a Q-factor of 10 000 000 down to unity, you therefore need an increase in energy by a factor of some 50 000 or so. That's the 3/2 power relation.

    If you start from the scale of the Hydrogen atom and keep increasing the energy until the Q factor is reduced to unity, then you have basically ALL the energy of the electrons being radiated away in a single pass. That's when the collision becomes totally inelastic. It happens at an energy of around 500 keV and a proximity of around 10^-15 meters. These are pretty well-known numbers.
     
  4. Apr 2, 2009 #3
    Thank you for the terrific response.
    I have surely benefited from it, hope others would find it useful, too.

    edit: (I think for my purposes the interaction could be considered fully elastic - since the energies are never greater than, say, 10 eV's in my field)
     
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