- #1

bladesong

- 30

- 0

## Homework Statement

By integrating (2-55), over a small volume containing the origin, substituting ψ = Ce

^{-jβr}/r, and letting r approach zero, show that C = 1/4π, thus proving (2-58).

## Homework Equations

(2-55): ∇

^{2}ψ + β

^{2}ψ = -δ(x)δ(y)δ(z)

(2-58): ψ = e

^{-jβr}/(4πr)

## The Attempt at a Solution

I am SO lost on this.

I integrated the RHS first. I would assume I have to do this in spherical coordinates. I know from the internet that to convert the delta function between coordinate systems you need to incorporate the jacobian, but because it's at the origin (i.e., x,y,z = 0) I just assumed that:

∫∫∫(-δ(x)δ(y)δ(z))r

^{2}sin(θ)dr dθ d∅

= -∫∫∫r

^{2}sin(θ)dr dθ d∅

= -(4π/3)r

_{0}

^{3}

Where I've integrated from r = 0 to r

_{0}, θ from 0 to pi, and ∅ from 0 to 2pi. Is this even correct?

For the next part, the LHS, I'm totally lost. I'm trying to follow the instructions chronologically, thus

∫∫∫(∇

^{2}+ β

^{2})ψ r

^{2}sin(θ)dr dθ d∅

Where I'm using the same integration limits as before. I'm really, really unsure how to even do this, but I assumed that psi is regarded as scalar (or otherwise does not have dependency on r, theta or phi). Thus, starting with the r component:

∫∫∫(1/r

^{2})(∂/∂r)[r

^{2}(∂/∂r)] (ψ r

^{2}sin(θ)dr dθ d∅)

= ∫∫∫(1/r

^{2})(∂/∂r)[r

^{2}(ψ 2r sin(θ)dr dθ d∅)]

= ∫∫∫(1/r

^{2})(∂/∂r)[2r

^{3}(ψ sin(θ)dr dθ d∅)]

= ∫∫∫(1/r

^{2})[6r

^{2}(ψ sin(θ)dr dθ d∅)]

= ∫∫∫[6 (ψ sin(θ)dr dθ d∅)]

= 4ψ∏r

_{0}

For the next part (θ component):

∫∫∫(1/[r

^{2}sin(θ)])(∂/∂θ)[sin(θ)(∂/∂θ)] (ψ r

^{2}sin(θ)dr dθ d∅)

=∫∫∫(1/[r

^{2}sin(θ)])(∂/∂θ)[sin(θ)cos(θ)] (ψ r

^{2}dr dθ d∅)

=∫∫∫(1/[r

^{2}sin(θ)])(∂/∂θ)[(1/2)sin(2θ)] (ψ r

^{2}dr dθ d∅)

=∫∫∫(1/[r

^{2}sin(θ)])[cos(2θ)] (ψ r

^{2}dr dθ d∅)

=∫∫∫(cos(2θ)/ sin(θ)]) (ψ dr dθ d∅)

But I can't integrate this because (cos(2θ)/ sin(θ)]) doesn't converge. I'm really lost on where to go from here and don't fully understand what I'm being asked to do.

ANY help is appreciated. Thank you. Please note that I'm an electrical eng so our notation may be different, and I may have some difficulty with more of the hardcore physics notation.