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A Time Independent form of Klein Gordon equation. How?

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  1. Oct 4, 2016 #1
    If equation of motion(K-G Eqn.,) follows,
    μμΦ+m2Φ=ρ
    where 'ρ' is point source at origin.

    How time independent form of above will become,
    (∇2-m2)Φ(x)=gδ3(x)
    where g is the coupling constant,
    δ3(x) is three dimensional dirac delta function.
     
    Last edited: Oct 4, 2016
  2. jcsd
  3. Oct 4, 2016 #2
    Remember the Minkowski metric is involved here:

    [tex] \partial_\mu \partial^\mu = \partial_t^2 - \partial_x^2 - \partial_y^2 - \partial_z^2 [/tex]

    There's no time dependence so:
    [tex] \partial_t \Phi = 0 [/tex]

    Then to get the source term:
    [tex] \rho = -g\delta(x)^3 [/tex]
     
  4. Oct 4, 2016 #3
     
  5. Oct 5, 2016 #4
    That's what rho needs to be to get the equation you have. To have a point source, you use a dirac delta with some kind of charge or coupling.
     
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