Newtonian Potential satisfies Laplace's Equation

From ProofWiki
Jump to navigation Jump to search


Let $R$ be a region of space.

Let $S$ be a Newtonian potential over $R$ defined as:

$\forall \mathbf r = x \mathbf i + y \mathbf j + z \mathbf k \in R: \map S {\mathbf r} = \dfrac k r$


$\tuple {\mathbf i, \mathbf j, \mathbf k}$ is the standard ordered basis on $R$
$\mathbf r = x \mathbf i + y \mathbf j + z \mathbf k$ is the position vector of an arbitrary point in $R$ with respect to the origin
$r = \norm {\mathbf r}$ is the magnitude of $\mathbf r$
$k$ is some predetermined constant.

Then $S$ satisfies Laplace's equation:

$\nabla^2 S = 0$

where $\nabla^2$ denotes the Laplacian.


From Gradient of Newtonian Potential:

$\grad S = -\dfrac {k \mathbf r} {r^3}$

where $\grad$ denotes the gradient operator.


\(\ds \operatorname {div} \grad S\) \(=\) \(\ds \map {\operatorname {div} } {-\dfrac {k \mathbf r} {r^3} }\) where $\operatorname {div}$ denotes divergence
\(\ds \) \(=\) \(\ds \nabla \cdot \paren {-\dfrac {k \mathbf r} {r^3} }\) Divergence Operator on Vector Space is Dot Product of Del Operator
\(\ds \) \(=\) \(\ds -k \paren {\mathbf i \dfrac \partial {\partial x} + \mathbf j \dfrac \partial {\partial y} + \mathbf k \dfrac \partial {\partial z} } \cdot {\dfrac {x \mathbf i + y \mathbf j + z \mathbf k} {\paren {x^2 + y^2 + z^2}^{3/2} } }\) Definition of Del Operator
\(\ds \) \(=\) \(\ds 0\) Power Rule for Derivatives and tedious messy algebra
\(\ds \leadsto \ \ \) \(\ds \nabla^2 S\) \(=\) \(\ds 0\) Definition of Laplacian on Scalar Field