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Charged particle in a box, variable m and e.

  1. Mar 25, 2012 #1
    Say we have one spin-less electron in a box in the ground state. Say we have dials on the outside of the box that can vary mass and charge of the spin-less electron. Suppose we start with one spin-less electron in a box with m and e set to as near zero as we please, but not zero. As m and e go to zero should the energy in the box go to zero?

    Now slowly raise the dial for mass of the spin-less electron in the box to m while keeping the charge near zero. What is the energy in the box, is it mc^2? Assume moving the dial requires a torque, now torque times the angular rotation of the dial gives an energy. Pretend that we transfer energy to the field of the spin-less electron in the box by moving the mass dial from near zero to m.

    Now slowly raise the dial for the charge of the spin-less electron while holding the mass dial. Assume again that any change in the field energy in the box comes from the energy to change the dials position. Does the energy in the box get bigger by making the charge larger? Is the extra energy stored in the electromagnetic field? As the charge dial is raised is there a torque on the mass dial?

    Now reverse the order, slowly raise the charge first and then the mass. Is the energy in the box the same for different paths?

    Thanks for any help!
    Last edited: Mar 25, 2012
  2. jcsd
  3. Mar 25, 2012 #2
    One more variation, vary the mass dial and charge dial together so that mass/charge = m/e. What is the energy in the box?

    Edit, what is the energy in the box for the mass dial at m and the charge dial at e?

    Thanks for any help!
    Last edited: Mar 25, 2012
  4. Mar 25, 2012 #3
    Can the answers to my questions be found in the lagrangian for this system?
  5. Mar 25, 2012 #4


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    Spinnor, For a particle of mass m in a box of side L the energy of the ground state is E = ħ2k2/2m where k = π/L. (Three times this for a three-dimensional box.) As you see, reducing m causes the energy to go not to zero but infinity.

    The reason being that a particle with an extremely small mass behaves more and more like a zero-mass particle, becoming relativistic. The 'infinite' energy is a signal to replace the Schrodinger equation with the Klein-Gordon equation and start over.

    Unfortunately the problem of a Klein-Gordon particle in a box has problems too. When m is so small that the Compton wavelength becomes comparable to the box size L, the walls leak despite their infinite height. Again this is a warning signal to start over, and take into account pair production.
  6. Mar 25, 2012 #5
    So if I pick some small nonzero mass m make sure the size of the box L is such that the ground state kinetic energy is small (pick L large enough) compared with mc^2.

    Does that get my problem back on track?

    Thanks for your help!
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