Time Independent Form of Klein Gordon Eqn.: How to Reach (gδ3(x))

[Mohanraj S]

If equation of motion(K-G Eqn.,) follows,
∂_μ∂^μΦ+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.

 

[DuckAmuck]

Remember the Minkowski metric is involved here:

$$\partial_\mu \partial^\mu = \partial_t^2 - \partial_x^2 - \partial_y^2 - \partial_z^2$$

There's no time dependence so:
$$\partial_t \Phi = 0$$

Then to get the source term:
$$\rho = -g\delta(x)^3$$

 

[Mohanraj S]

Thank you. Also my query is how $$\rho = -g\delta(x)^3$$

 

[DuckAmuck]

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.