Newtonian Potential satisfies Laplace's Equation
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Theorem
Let $S$ be a Newtonian potential over $R$ defined as:
- $\forall \mathbf r = x \mathbf i + y \mathbf j + z \mathbf k \in R: \map S {\mathbf r} = \dfrac k r$
where:
- $\tuple {\mathbf i, \mathbf j, \mathbf k}$ 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.
Proof
From Gradient of Newtonian Potential:
- $\grad S = -\dfrac {k \mathbf r} {r^3}$
where $\grad$ denotes the gradient operator.
Then:
\(\ds \operatorname {div} \grad S\) | \(=\) | \(\ds \map {\operatorname {div} } {-\dfrac {k \mathbf r} {r^3} }\) | 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 |
$\blacksquare$
Sources
- 1951: B. Hague: An Introduction to Vector Analysis (5th ed.) ... (previous) ... (next): Chapter $\text {V}$: Further Applications of the Operator $\nabla$: $9$. The Vector Field $\map \grad {k / r}$: $(5.12)$