In case you’re not persuaded that ∇^2(1/r ) = −4πδ^3(r) (Eq. 1.102 with r' = 0 for simplicity), try replacing r by √ r ^2 + ε^2, and watching what happens as ε→ 0.^16 Specifically, let D(r, ε) ≡ −1/4π ∇^2 1/√r ^2 + ε^2 16This problem was suggested by Frederick Strauch.To demonstrate that this goes

Question

In case you’re not persuaded that [latex]∇^2(1/r ) = −4πδ^3(r)[/latex] (Eq. 1.102 with r' = 0 for simplicity), try replacing r by [latex]\sqrt{r^2+\varepsilon ^2} [/latex] , and watching what happens as [latex]\varepsilon \rightarrow 0.^{16} [/latex] Specifically, let

[latex]∇^2\frac{1}{\eta } = −4\pi \delta ^3(\eta )[/latex] (1.102)

[latex]D(r, \varepsilon ) ≡ −\frac{1}{4\pi }∇^2\frac{1}{\sqrt{r^2+\varepsilon ^2} } [/latex] .

[latex]^{16}[/latex] This problem was suggested by Frederick Strauch.

To demonstrate that this goes to [latex]\delta ^3(r) as \varepsilon \rightarrow 0[/latex] :

(a) Show that D [latex](r,\varepsilon )=\left(3\varepsilon ^2/4\pi \right)(r^2+\varepsilon ^2)^{-5/2} [/latex] .

(b) Check that D [latex](0,\varepsilon )\rightarrow \infty , as \varepsilon \rightarrow 0[/latex] .

(c) Check that D [latex](r,\varepsilon )\rightarrow 0, as\varepsilon \rightarrow 0,[/latex] for all r ≠ 0.

(d) Check that the integral of D(r, ε) over all space is 1.

Accepted Answer

(a) Since the argument is not a function of angle, Eq. 1.73 says

[latex]∇^2T=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial T}{\partial r} ight) +\frac{1}{r^2\sin heta } \frac{\partial}{\partial heta }\left(\sin heta \frac{\partial T}{\partial heta } ight)+\frac{1}{r^2\sin ^2 heta } \frac{\partial^2 T}{\partial \phi ^2}[/latex] (1.73)

[latex]D=-\frac{1}{4\pi }\frac{1}{r^2}\frac{d}{dr} \left[r^2\left(-\frac{1}{2} ight)\frac{2r}{\left(r^2+\varepsilon ^2 ight)^{3/2} } ight]=\frac{1}{4\pi r^2}\frac{d}{dr}\left[\frac{r^3}{\left(r^2+\varepsilon ^2 ight) ^{3/2} } ight] [/latex]

=[latex]\frac{1}{4\pi r^2}\left[\frac{3r^2}{\left(r^2+\varepsilon ^2 ight)^{3/2} }-\frac{3}{2} \frac{r^32r}{\left(r^2+\varepsilon ^2 ight)^{5/3} } ight] =\frac{1}{4\pi r^2}\frac{3r^2}{\left(r^2+\varepsilon ^2 ight)^{5/2} } \left(r^2 + \varepsilon^ 2 − r^2 ight)=\frac{3\varepsilon ^2}{4\pi \left(r^2+\varepsilon ^2 ight)^{5/2} } [/latex] .

(b) Setting r →0:

[latex]D(0, \varepsilon ) =\frac{3\varepsilon ^2}{4\pi \varepsilon ^5}=\frac{3}{4\pi \varepsilon ^3} [/latex],

which goes to infinity as ε→0.

(c) From (a) it is clear that D(r, 0) = 0 for r ≠ 0.

[latex]\int{} D(r, \varepsilon ) 4\pi r^2 dr = 3\varepsilon^2\int_{0}^{\infty }{} \frac{r^2}{\left(r^2+\varepsilon ^2 ight)^{5/2} }dr = 3\varepsilon ^2\left(\frac{1}{3\varepsilon ^2} ight) =1[/latex] .

(I looked up the integral.) Note that (b), (c), and (d) are the defining conditions for [latex]\delta ^3[/latex](r).

Question 1.64: In case you’re not persuaded that ∇^2(1/r ) = −4πδ^3(r) (Eq....

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