Gradient of Newtonian Potential

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Theorem

Let $R$ be a region of space.

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

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

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:

$\grad S = -\dfrac {k \mathbf r} {r^3} = -\dfrac {k \mathbf {\hat r} } {r^2}$

where:

$\grad$ denotes the gradient operator
$\mathbf {\hat r}$ denotes the unit vector in the direction of $\mathbf r$.


The fact that the gradient of $S$ is negative indicates that direction of the vector quantities that compose the vector field that is $\grad S$ all point towards the origin.


Proof

From the geometry of the sphere, the equal surfaces of $S$ are concentric spheres whose centers are at the origin.

As the origin is approached, the scalar potential is unbounded above.

We have:

\(\ds \grad S\) \(=\) \(\ds \map \grad {\dfrac k r}\) Definition of $S$
\(\ds \) \(=\) \(\ds \map \nabla {\dfrac k r}\) Definition of Gradient Operator
\(\ds \) \(=\) \(\ds k \map {\paren {\mathbf i \dfrac \partial {\partial x} + \mathbf j \dfrac \partial {\partial y} + \mathbf k \dfrac \partial {\partial z} } } {\dfrac 1 {\paren {x^2 + y^2 + z^2}^{1/2} } }\) Definition of Del Operator
\(\ds \) \(=\) \(\ds -k \paren {\dfrac {x \mathbf i + y \mathbf j + z \mathbf k} {\paren {x^2 + y^2 + z^2}^{3/2} } }\) Power Rule for Derivatives
\(\ds \) \(=\) \(\ds -\dfrac {k \mathbf r} {r^3}\) Definition of $\mathbf r$ and simplifying
\(\ds \) \(=\) \(\ds -\dfrac {k \mathbf {\hat r} } {r^2}\) as $\mathbf r = r \mathbf {\hat r}$

$\blacksquare$


Sources

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