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Klein-Gordon for a massless particle

  1. May 20, 2012 #1
    So I'm trying to find a solution of the Klein-Gordon equation for a massless particle. I reached the Klein-Gordon from the total energy-momentum equation. Then for a massless particle i get to this equation:
    $${ \partial^2 \psi \over \partial t^2 } = c^2 \nabla^2 \psi$$How do I solve for psi? I was thinking about trying the Frobenius method, but I'm not sure how to do that. Any help would be appreciated.

    Also how do I make my typed latex display in the latex format on this forum?
     
    Last edited by a moderator: May 20, 2012
  2. jcsd
  3. May 20, 2012 #2

    vela

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  4. May 20, 2012 #3
    so for a one dimensional wave equation the equation is:

    [tex] { \partial^2 \psi \over \partial t^2 } = { c^2 \nabla^2 \psi} = c^2 {\partial^2 \psi \over \partial x^2} [/tex]

    Right?

    I don't see how I could use separation of variables. If it were something like this:

    [tex] { \partial x \over \partial t } = { x } [/tex]

    Then I could say:

    [tex] {\frac{1}{x} \partial x } = {\partial t} [/tex]

    And then integrate but I dont see what you mean. What do you mean?

    The only thing i could see doing is this:

    [tex] {\partial^2 \psi \partial^2 x} = c^2 {\partial^2 \psi \partial^2 t} [/tex]

    And integrating twice? But that doesnt really make sense to me.
    I didnt think you could separate variables to solve a differential equation when there were double derivatives in the equation.
     
  5. May 20, 2012 #4

    vela

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  6. May 20, 2012 #5
    Cool, thank you.
     
  7. May 22, 2012 #6
    this equation looks pretty much like the wave-equation, for a wave moving with a speed v=c.
    So I guess you can think for solutions like cos or sin, or better exponential.

    If for example you say:

    Ψ(r,t)= ei(-kr+ωt) which is a spherical wave, you can see how this thing behaves in your equation.
     
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