I want to know
∇ 2 ( 1 ∣ r ⃗ − r ⃗ ′ ∣ ) = ∇ ⋅ ( ∇ ( 1 ∣ r ⃗ − r ⃗ ′ ∣ ) ) . \nabla^2 (\frac{1}{|\vec r - \vec r'|}) = \nabla \cdot \left( \nabla (\frac{1}{|\vec r - \vec r'|}) \right). ∇ 2 ( ∣ r − r ′ ∣ 1 ) = ∇ ⋅ ( ∇ ( ∣ r − r ′ ∣ 1 ) ) . Method 1 — Cartesian Coordinates Let’s first look at ∇ ( 1 ∣ r ⃗ − r ⃗ ′ ∣ ) \nabla (\frac{1}{|\vec r - \vec r'|}) ∇ ( ∣ r − r ′ ∣ 1 )
∣ r ⃗ − r ⃗ ′ ∣ = ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 , |\vec r - \vec r'| = \sqrt{(x-x')^2 + (y-y')^2 + (z-z')^2}, ∣ r − r ′ ∣ = ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 , and so
∂ ∂ x ( 1 ∣ r ⃗ − r ⃗ ′ ∣ ) = x ′ − x ∣ r ⃗ − r ⃗ ′ ∣ 3 , ∂ ∂ y ( 1 ∣ r ⃗ − r ⃗ ′ ∣ ) = y ′ − y ∣ r ⃗ − r ⃗ ′ ∣ 3 , ∂ ∂ z ( 1 ∣ r ⃗ − r ⃗ ′ ∣ ) = z ′ − z ∣ r ⃗ − r ⃗ ′ ∣ 3 . \frac{\partial}{\partial x}\left( \frac{1}{|\vec r - \vec r'|}\right) = \frac{x'-x}{|\vec r - \vec r'|^3}, \\ \frac{\partial}{\partial y}\left( \frac{1}{|\vec r - \vec r'|}\right) = \frac{y'-y}{|\vec r - \vec r'|^3}, \\ \frac{\partial}{\partial z}\left( \frac{1}{|\vec r - \vec r'|}\right) = \frac{z'-z}{|\vec r - \vec r'|^3}. ∂ x ∂ ( ∣ r − r ′ ∣ 1 ) = ∣ r − r ′ ∣ 3 x ′ − x , ∂ y ∂ ( ∣ r − r ′ ∣ 1 ) = ∣ r − r ′ ∣ 3 y ′ − y , ∂ z ∂ ( ∣ r − r ′ ∣ 1 ) = ∣ r − r ′ ∣ 3 z ′ − z . Thus,
∇ ( 1 ∣ r ⃗ − r ⃗ ′ ∣ ) = ( x ′ − x ) ı ^ + ( y ′ − y ) ȷ ^ + ( z ′ − z ) k ^ ∣ r ⃗ − r ⃗ ′ ∣ 3 = r ⃗ ′ − r ⃗ ∣ r ⃗ − r ⃗ ′ ∣ 3 . \nabla (\frac{1}{|\vec r - \vec r'|}) = \frac{(x'-x)\ihat+(y'-y)\jhat+(z'-z)\khat}{|\vec r - \vec r'|^3} = \frac{\vec r' - \vec r}{|\vec r - \vec r'|^3}. ∇ ( ∣ r − r ′ ∣ 1 ) = ∣ r − r ′ ∣ 3 ( x ′ − x ) ı ^ + ( y ′ − y ) ȷ ^ + ( z ′ − z ) k ^ = ∣ r − r ′ ∣ 3 r ′ − r . Then, we want ∇ 2 ( 1 ∣ r ⃗ − r ⃗ ′ ∣ ) = ∇ ⋅ ( r ⃗ ′ − r ⃗ ∣ r ⃗ − r ⃗ ′ ∣ 3 ) . \nabla^2 (\frac{1}{|\vec r - \vec r'|}) = \nabla \cdot \left( \frac{\vec r' - \vec r}{|\vec r - \vec r'|^3} \right). ∇ 2 ( ∣ r − r ′ ∣ 1 ) = ∇ ⋅ ( ∣ r − r ′ ∣ 3 r ′ − r ) .
∂ ∂ x x ′ − x ∣ r ⃗ − r ⃗ ′ ∣ 3 = − ∣ r ⃗ − r ⃗ ′ ∣ 3 + 3 ∣ r ⃗ − r ⃗ ′ ∣ ( x ′ − x ) 2 ∣ r ⃗ − r ⃗ ′ ∣ 6 , ∂ ∂ y y ′ − y ∣ r ⃗ − r ⃗ ′ ∣ 3 = − ∣ r ⃗ − r ⃗ ′ ∣ 3 + 3 ∣ r ⃗ − r ⃗ ′ ∣ ( y ′ − y ) 2 ∣ r ⃗ − r ⃗ ′ ∣ 6 , ∂ ∂ z z ′ − z ∣ r ⃗ − r ⃗ ′ ∣ 3 = − ∣ r ⃗ − r ⃗ ′ ∣ 3 + 3 ∣ r ⃗ − r ⃗ ′ ∣ ( z ′ − z ) 2 ∣ r ⃗ − r ⃗ ′ ∣ 6 . \frac{\partial}{\partial x} \frac{x'-x}{|\vec r - \vec r'|^3} = \frac{-|\vec r - \vec r'|^3 + 3|\vec r - \vec r'|(x'-x)^2}{|\vec r - \vec r'|^6},\\ \frac{\partial}{\partial y} \frac{y'-y}{|\vec r - \vec r'|^3} = \frac{-|\vec r - \vec r'|^3 + 3|\vec r - \vec r'|(y'-y)^2}{|\vec r - \vec r'|^6},\\ \frac{\partial}{\partial z} \frac{z'-z}{|\vec r - \vec r'|^3} = \frac{-|\vec r - \vec r'|^3 + 3|\vec r - \vec r'|(z'-z)^2}{|\vec r - \vec r'|^6}. ∂ x ∂ ∣ r − r ′ ∣ 3 x ′ − x = ∣ r − r ′ ∣ 6 − ∣ r − r ′ ∣ 3 + 3∣ r − r ′ ∣ ( x ′ − x ) 2 , ∂ y ∂ ∣ r − r ′ ∣ 3 y ′ − y = ∣ r − r ′ ∣ 6 − ∣ r − r ′ ∣ 3 + 3∣ r − r ′ ∣ ( y ′ − y ) 2 , ∂ z ∂ ∣ r − r ′ ∣ 3 z ′ − z = ∣ r − r ′ ∣ 6 − ∣ r − r ′ ∣ 3 + 3∣ r − r ′ ∣ ( z ′ − z ) 2 . The divergence is to adding them up, which is
− ∣ r ⃗ − r ⃗ ′ ∣ 3 + 3 ∣ r ⃗ − r ⃗ ′ ∣ ( ( x ′ − x ) 2 + ( y ′ − y ) 2 + ( z ′ − z ) 2 ) ∣ r ⃗ − r ⃗ ′ ∣ 6 = − ∣ r ⃗ − r ⃗ ′ ∣ 3 + 3 ∣ r ⃗ − r ⃗ ′ ∣ ⋅ ∣ r ⃗ − r ⃗ ′ ∣ 2 ∣ r ⃗ − r ⃗ ′ ∣ 6 = − ∣ r ⃗ − r ⃗ ′ ∣ 3 + 3 ∣ r ⃗ − r ⃗ ′ ∣ 3 ∣ r ⃗ − r ⃗ ′ ∣ 6 = 0. (For ∣ r ⃗ − r ⃗ ′ ∣ ≠ 0 ) \begin{align*} &\frac{-|\vec r - \vec r'|^3 + 3|\vec r - \vec r'|\left((x'-x)^2+(y'-y)^2+(z'-z)^2\right)}{|\vec r - \vec r'|^6} \\ &= \frac{-|\vec r - \vec r'|^3 + 3|\vec r - \vec r'|\cdot |\vec r - \vec r'|^2}{|\vec r - \vec r'|^6}\\ &= \frac{-|\vec r - \vec r'|^3 + 3|\vec r - \vec r'|^3}{|\vec r - \vec r'|^6} \\ &= 0. \quad \text{(For $|\vec r - \vec r'| \neq 0$)} \end{align*} ∣ r − r ′ ∣ 6 − ∣ r − r ′ ∣ 3 + 3∣ r − r ′ ∣ ( ( x ′ − x ) 2 + ( y ′ − y ) 2 + ( z ′ − z ) 2 ) = ∣ r − r ′ ∣ 6 − ∣ r − r ′ ∣ 3 + 3∣ r − r ′ ∣ ⋅ ∣ r − r ′ ∣ 2 = ∣ r − r ′ ∣ 6 − ∣ r − r ′ ∣ 3 + 3∣ r − r ′ ∣ 3 = 0. (For ∣ r − r ′ ∣ = 0) We thus get for ∣ r ⃗ − r ⃗ ′ ∣ ≠ 0 , |\vec r - \vec r'| \neq 0, ∣ r − r ′ ∣ = 0 ,
∇ 2 ( 1 ∣ r ⃗ − r ⃗ ′ ∣ ) = 0. \nabla^2 (\frac{1}{|\vec r - \vec r'|}) = 0. ∇ 2 ( ∣ r − r ′ ∣ 1 ) = 0. Method 2 — Spherical Coordinates It is best to calculate this in spherical coordinate system. Here’s the result for the laplacian operator in spherical coordinate system.
∇ 2 = 1 r 2 ∂ ∂ r ( r 2 ∂ ∂ r ) + 1 r 2 sin θ ∂ ∂ θ ( sin θ ∂ ∂ θ ) + 1 r 2 sin 2 θ ∂ 2 ∂ ϕ 2 . \nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial}{\partial r}\right) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta \frac{\partial}{\partial \theta}\right) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial \phi^2}. ∇ 2 = r 2 1 ∂ r ∂ ( r 2 ∂ r ∂ ) + r 2 sin θ 1 ∂ θ ∂ ( sin θ ∂ θ ∂ ) + r 2 sin 2 θ 1 ∂ ϕ 2 ∂ 2 . Our function has nothing to do with θ \theta θ ϕ \phi ϕ
∇ 2 1 ∣ r ⃗ ∣ = 1 r 2 ∂ ∂ r ( r 2 ∂ ∂ r ( 1 r ) ) = 0. \nabla^2 \frac{1}{|\vec r|} = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial}{\partial r}\left(\frac{1}{r}\right)\right) = 0. ∇ 2 ∣ r ∣ 1 = r 2 1 ∂ r ∂ ( r 2 ∂ r ∂ ( r 1 ) ) = 0. You just offset the thing and get the form in r ⃗ − r ⃗ ′ \vec r - \vec r' r − r ′
What Happens At The Origin? Because the divergence is zero everywhere except origin, by a consequence of Divergence Theorem, any shape around the origin will have the same flux. Let’s just pick V V V
∯ ∂ S r ⃗ ∣ r ⃗ ∣ 3 ⋅ d A ⃗ = ∯ ∂ S d A = 4 π = ∭ V ∇ ⋅ ( r ⃗ ∣ r ⃗ ∣ 3 ) d V ⟹ ∇ ⋅ ( r ⃗ ∣ r ⃗ ∣ 3 ) = 4 π δ 3 ( r ⃗ ) . \oiint_{\partial S} \frac{\vec r}{|\vec r|^3}\cdot d\vec A = \oiint_{\partial S} dA = 4\pi \\ = \iiint_V \nabla\cdot\left(\frac{\vec r}{|\vec r|^3}\right)\,dV \implies \nabla\cdot\left(\frac{\vec r}{|\vec r|^3}\right) = 4\pi\,\delta^3(\vec r). ∬ ∂ S ∣ r ∣ 3 r ⋅ d A = ∬ ∂ S d A = 4 π = ∭ V ∇ ⋅ ( ∣ r ∣ 3 r ) d V ⟹ ∇ ⋅ ( ∣ r ∣ 3 r ) = 4 π δ 3 ( r ) . Thus we conclude
∇ 2 1 ∣ r ⃗ ∣ = − 4 π δ 3 ( r ⃗ ) , \nabla^2 \frac{1}{|\vec r|} = -4\pi\,\delta^3(\vec r), ∇ 2 ∣ r ∣ 1 = − 4 π δ 3 ( r ) , And the offset version
∇ 2 1 ∣ r ⃗ − r ⃗ ′ ∣ = − 4 π δ 3 ( r ⃗ − r ⃗ ′ ) . \nabla^2 \frac{1}{|\vec r - \vec r'|} = -4\pi\,\delta^3(\vec r - \vec r'). ∇ 2 ∣ r − r ′ ∣ 1 = − 4 π δ 3 ( r − r ′ ) .
