Answer on Question #51711 - Math - Vector Calculus
V ( 1 / r ) = ? , V 2 ( 1 / r ) = ? \mathcal {V} (1 / r) = ?, \qquad \mathcal {V} ^ {2} (1 / r) = ? V ( 1/ r ) = ? , V 2 ( 1/ r ) = ?
where
r ~ = x i ~ + y j ~ + z k ⃗ \tilde {r} = x \tilde {i} + y \tilde {j} + z \vec {k} r ~ = x i ~ + y j ~ + z k Solution
V ( 1 / r ) \mathcal{V}(1 / r) V ( 1/ r )
V ( 1 r ) = [ ∂ ∂ x ( 1 r ) ] i ~ + [ ∂ ∂ y ( 1 r ) ] j ~ + [ ∂ ∂ z ( 1 r ) ] k ⃗ \mathcal {V} \left(\frac {1}{r}\right) = \left[ \frac {\partial}{\partial x} \left(\frac {1}{r}\right) \right] \tilde {i} + \left[ \frac {\partial}{\partial y} \left(\frac {1}{r}\right) \right] \tilde {j} + \left[ \frac {\partial}{\partial z} \left(\frac {1}{r}\right) \right] \vec {k} V ( r 1 ) = [ ∂ x ∂ ( r 1 ) ] i ~ + [ ∂ y ∂ ( r 1 ) ] j ~ + [ ∂ z ∂ ( r 1 ) ] k
Since r = x 2 + y 2 + z 2 r = \sqrt{x^2 + y^2 + z^2} r = x 2 + y 2 + z 2
∂ ∂ x ( 1 r ) = ∂ ∂ x ( 1 x 2 + y 2 + z 2 ) = − ∂ ∂ x ( x 2 + y 2 + z 2 ) 2 ( x 2 + y 2 + z 2 ) 3 = = − 2 x 2 ( x 2 + y 2 + z 2 ) 3 = − x r 3 \begin{array}{l}
\frac {\partial}{\partial x} \left(\frac {1}{r}\right) = \frac {\partial}{\partial x} \left(\frac {1}{\sqrt {x ^ {2} + y ^ {2} + z ^ {2}}}\right) = - \frac {\frac {\partial}{\partial x} \left(x ^ {2} + y ^ {2} + z ^ {2}\right)}{2 \left(\sqrt {x ^ {2} + y ^ {2} + z ^ {2}}\right) ^ {3}} = \\
= - \frac {2 x}{2 \left(\sqrt {x ^ {2} + y ^ {2} + z ^ {2}}\right) ^ {3}} = - \frac {x}{r ^ {3}} \\
\end{array} ∂ x ∂ ( r 1 ) = ∂ x ∂ ( x 2 + y 2 + z 2 1 ) = − 2 ( x 2 + y 2 + z 2 ) 3 ∂ x ∂ ( x 2 + y 2 + z 2 ) = = − 2 ( x 2 + y 2 + z 2 ) 3 2 x = − r 3 x
Similarly
∂ ∂ y ( 1 r ) = − y r 3 \frac {\partial}{\partial y} \left(\frac {1}{r}\right) = - \frac {y}{r ^ {3}} ∂ y ∂ ( r 1 ) = − r 3 y ∂ ∂ z ( 1 r ) = − z r 3 \frac {\partial}{\partial z} \left(\frac {1}{r}\right) = - \frac {z}{r ^ {3}} ∂ z ∂ ( r 1 ) = − r 3 z
Substituting these into the first equation we obtain
V ( 1 r ) = [ ∂ ∂ x ( 1 r ) ] i ~ + [ ∂ ∂ y ( 1 r ) ] j ~ + [ ∂ ∂ z ( 1 r ) ] k ⃗ = − x r 3 i ~ − y r 3 j ~ − z r 3 k ⃗ = − r ⃗ r 3 \mathcal {V} \left(\frac {1}{r}\right) = \left[ \frac {\partial}{\partial x} \left(\frac {1}{r}\right) \right] \tilde {i} + \left[ \frac {\partial}{\partial y} \left(\frac {1}{r}\right) \right] \tilde {j} + \left[ \frac {\partial}{\partial z} \left(\frac {1}{r}\right) \right] \vec {k} = - \frac {x}{r ^ {3}} \tilde {i} - \frac {y}{r ^ {3}} \tilde {j} - \frac {z}{r ^ {3}} \vec {k} = - \frac {\vec {r}}{r ^ {3}} V ( r 1 ) = [ ∂ x ∂ ( r 1 ) ] i ~ + [ ∂ y ∂ ( r 1 ) ] j ~ + [ ∂ z ∂ ( r 1 ) ] k = − r 3 x i ~ − r 3 y j ~ − r 3 z k = − r 3 r V 2 ( 1 / r ) \mathcal{V}^2 (1 / r) V 2 ( 1/ r )
V 2 ( 1 r ) = ∂ 2 ∂ x 2 ( 1 r ) + ∂ 2 ∂ y 2 ( 1 r ) + ∂ 2 ∂ z 2 ( 1 r ) \mathcal {V} ^ {2} \left(\frac {1}{r}\right) = \frac {\partial^ {2}}{\partial x ^ {2}} \left(\frac {1}{r}\right) + \frac {\partial^ {2}}{\partial y ^ {2}} \left(\frac {1}{r}\right) + \frac {\partial^ {2}}{\partial z ^ {2}} \left(\frac {1}{r}\right) V 2 ( r 1 ) = ∂ x 2 ∂ 2 ( r 1 ) + ∂ y 2 ∂ 2 ( r 1 ) + ∂ z 2 ∂ 2 ( r 1 ) ∂ 2 ∂ x 2 ( 1 r ) = ∂ ∂ x ( − x r 3 ) = − ∂ ∂ x ( x ) r 3 − x ∂ ∂ x ( 1 r 3 ) = = − 1 r 3 − x ∂ ∂ x ( 1 ( x 2 + y 2 + z 2 ) 3 ) = \begin{array}{l}
\frac {\partial^ {2}}{\partial x ^ {2}} \left(\frac {1}{r}\right) = \frac {\partial}{\partial x} \left(- \frac {x}{r ^ {3}}\right) = - \frac {\frac {\partial}{\partial x} (x)}{r ^ {3}} - x \frac {\partial}{\partial x} \left(\frac {1}{r ^ {3}}\right) = \\
= - \frac {1}{r ^ {3}} - x \frac {\partial}{\partial x} \left(\frac {1}{\left(\sqrt {x ^ {2} + y ^ {2} + z ^ {2}}\right) ^ {3}}\right) = \\
\end{array} ∂ x 2 ∂ 2 ( r 1 ) = ∂ x ∂ ( − r 3 x ) = − r 3 ∂ x ∂ ( x ) − x ∂ x ∂ ( r 3 1 ) = = − r 3 1 − x ∂ x ∂ ( ( x 2 + y 2 + z 2 ) 3 1 ) = = − 1 r 3 + x 3 2 ∂ ∂ x ( x 2 + y 2 + z 2 ) ( x 2 + y 2 + z 2 ) 5 = − 1 r 3 + 3 x 2 ( x 2 + y 2 + z 2 ) 5 = − 1 r 3 + 3 x 2 r 5 = - \frac {1}{r ^ {3}} + x \frac {3}{2} \frac {\frac {\partial}{\partial x} \left(x ^ {2} + y ^ {2} + z ^ {2}\right)}{\left(\sqrt {x ^ {2} + y ^ {2} + z ^ {2}}\right) ^ {5}} = - \frac {1}{r ^ {3}} + \frac {3 x ^ {2}}{\left(\sqrt {x ^ {2} + y ^ {2} + z ^ {2}}\right) ^ {5}} = - \frac {1}{r ^ {3}} + \frac {3 x ^ {2}}{r ^ {5}} = − r 3 1 + x 2 3 ( x 2 + y 2 + z 2 ) 5 ∂ x ∂ ( x 2 + y 2 + z 2 ) = − r 3 1 + ( x 2 + y 2 + z 2 ) 5 3 x 2 = − r 3 1 + r 5 3 x 2
Similarly
∂ 2 ∂ y 2 ( 1 r ) = − 1 r 3 + 3 y 2 r 5 \frac {\partial^ {2}}{\partial y ^ {2}} \left(\frac {1}{r}\right) = - \frac {1}{r ^ {3}} + \frac {3 y ^ {2}}{r ^ {5}} ∂ y 2 ∂ 2 ( r 1 ) = − r 3 1 + r 5 3 y 2 ∂ 2 ∂ z 2 ( 1 r ) = − 1 r 3 + 3 z 2 r 5 \frac {\partial^ {2}}{\partial z ^ {2}} \left(\frac {1}{r}\right) = - \frac {1}{r ^ {3}} + \frac {3 z ^ {2}}{r ^ {5}} ∂ z 2 ∂ 2 ( r 1 ) = − r 3 1 + r 5 3 z 2
Therefore
∇ 2 ( 1 r ) = ∂ 2 ∂ x 2 ( 1 r ) + ∂ 2 ∂ y 2 ( 1 r ) + ∂ 2 ∂ z 2 ( 1 r ) = − 1 r 3 + 3 x 2 r 5 − 1 r 3 + 3 y 2 r 5 − 1 r 3 + 3 z 2 r 5 = = − 3 r 3 + 3 x 2 + y 2 + z 2 r 5 = − 3 r 3 + 3 r 2 r 5 = − 3 r 3 + 3 r 3 = 0 \begin{array}{l} \nabla^ {2} \left(\frac {1}{r}\right) = \frac {\partial^ {2}}{\partial x ^ {2}} \left(\frac {1}{r}\right) + \frac {\partial^ {2}}{\partial y ^ {2}} \left(\frac {1}{r}\right) + \frac {\partial^ {2}}{\partial z ^ {2}} \left(\frac {1}{r}\right) = - \frac {1}{r ^ {3}} + \frac {3 x ^ {2}}{r ^ {5}} - \frac {1}{r ^ {3}} + \frac {3 y ^ {2}}{r ^ {5}} - \frac {1}{r ^ {3}} + \frac {3 z ^ {2}}{r ^ {5}} = \\ = - \frac {3}{r ^ {3}} + 3 \frac {x ^ {2} + y ^ {2} + z ^ {2}}{r ^ {5}} = - \frac {3}{r ^ {3}} + 3 \frac {r ^ {2}}{r ^ {5}} = - \frac {3}{r ^ {3}} + \frac {3}{r ^ {3}} = 0 \\ \end{array} ∇ 2 ( r 1 ) = ∂ x 2 ∂ 2 ( r 1 ) + ∂ y 2 ∂ 2 ( r 1 ) + ∂ z 2 ∂ 2 ( r 1 ) = − r 3 1 + r 5 3 x 2 − r 3 1 + r 5 3 y 2 − r 3 1 + r 5 3 z 2 = = − r 3 3 + 3 r 5 x 2 + y 2 + z 2 = − r 3 3 + 3 r 5 r 2 = − r 3 3 + r 3 3 = 0
Answer: ∇ ( 1 r ) = − r ⃗ r 3 , ∇ 2 ( 1 r ) = 0. \nabla \left(\frac{1}{r}\right) = -\frac{\vec{r}}{r^3}, \quad \nabla^2\left(\frac{1}{r}\right) = 0. ∇ ( r 1 ) = − r 3 r , ∇ 2 ( r 1 ) = 0.
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#340153 on Dec 2023