Scalar potential

Altitude as gravitational potential energy

An example is the (nearly) uniform gravitational field near the Earth's surface. It has a potential energy $${\displaystyle U=mgh}$$ where U is the gravitational potential energy and h is the height above the surface. This means that gravitational potential energy on a contour map is proportional to altitude. On a contour map, the two-dimensional negative gradient of the altitude is a two-dimensional vector field, whose vectors are always perpendicular to the contours and also perpendicular to the direction of gravity. But on the hilly region represented by the contour map, the three-dimensional negative gradient of U always points straight downwards in the direction of gravity; F. However, a ball rolling down a hill cannot move directly downwards due to the normal force of the hill's surface, which cancels out the component of gravity perpendicular to the hill's surface. The component of gravity that remains to move the ball is parallel to the surface:

$${\displaystyle {\mathbf{F}}_{\mathrm{S}}=-mg\sin \theta }$$

where θ is the angle of inclination, and the component of F_S perpendicular to gravity is

$${\displaystyle {\mathbf{F}}_{\mathrm{P}}=-mg\sin \theta \cos \theta =-\frac{1}{2}mg\sin 2\theta .}$$

This force F_P, parallel to the ground, is greatest when θ is 45 degrees.

Let Δh be the uniform interval of altitude between contours on the contour map, and let Δx be the distance between two contours. Then

$${\displaystyle \theta ={\tan }^{-1}\frac{\mathrm{\Delta }h}{\mathrm{\Delta }x}}$$

so that

$${\displaystyle F_P=-mg\frac{\mathrm{\Delta }x\mathrm{\Delta }h}{\mathrm{\Delta }{x}^2+\mathrm{\Delta }{h}^2}.}$$

However, on a contour map, the gradient is inversely proportional to Δx, which is not similar to force F_P: altitude on a contour map is not exactly a two-dimensional potential field. The magnitudes of forces are different, but the directions of the forces are the same on a contour map as well as on the hilly region of the Earth's surface represented by the contour map.

Pressure as buoyant potential

In fluid mechanics, a fluid in equilibrium, but in the presence of a uniform gravitational field is permeated by a uniform buoyant force that cancels out the gravitational force: that is how the fluid maintains its equilibrium. This buoyant force is the negative gradient of pressure:

$${\displaystyle {\mathbf{f}}_{\mathbf{B}}=-\mathrm{\nabla }p.}$$

Since buoyant force points upwards, in the direction opposite to gravity, then pressure in the fluid increases downwards. Pressure in a static body of water increases proportionally to the depth below the surface of the water. The surfaces of constant pressure are planes parallel to the surface, which can be characterized as the plane of zero pressure.

If the liquid has a vertical vortex (whose axis of rotation is perpendicular to the surface), then the vortex causes a depression in the pressure field. The surface of the liquid inside the vortex is pulled downwards as are any surfaces of equal pressure, which still remain parallel to the liquids surface. The effect is strongest inside the vortex and decreases rapidly with the distance from the vortex axis.

The buoyant force due to a fluid on a solid object immersed and surrounded by that fluid can be obtained by integrating the negative pressure gradient along the surface of the object:

$${\displaystyle F_B=-{\oint }_S\mathrm{\nabla }p\cdot d\mathbf{S}.}$$

Scalar potential in Euclidean space

In 3-dimensional Euclidean space ⁠${\displaystyle {\mathbb{R}}^3}$⁠, the scalar potential of an irrotational vector field E is given by

$${\displaystyle \mathrm{\Phi }(\mathbf{r})=\frac{1}{4\pi }{\int }_{{\mathbb{R}}^3}\frac{\mathrm{div}\mathbf{E}({\mathbf{r}}^{\prime })}{\Vert \mathbf{r}-{\mathbf{r}}^{\prime }\Vert }dV({\mathbf{r}}^{\prime })}$$

where dV(r') is an infinitesimal volume element with respect to r'. Then

$${\displaystyle \mathbf{E}=-\mathrm{\nabla }\mathrm{\Phi }=-\frac{1}{4\pi }\mathrm{\nabla }{\int }_{{\mathbb{R}}^3}\frac{\mathrm{div}\mathbf{E}({\mathbf{r}}^{\prime })}{\Vert \mathbf{r}-{\mathbf{r}}^{\prime }\Vert }dV({\mathbf{r}}^{\prime })}$$

This holds provided E is continuous and vanishes asymptotically to zero towards infinity, decaying faster than 1/r and if the divergence of E likewise vanishes towards infinity, decaying faster than 1/r ².

Written another way, let

$${\displaystyle \mathrm{\Gamma }(\mathbf{r})=\frac{1}{4\pi }\frac{1}{\Vert \mathbf{r}\Vert }}$$

be the Newtonian potential. This is the fundamental solution of the Laplace equation, meaning that the Laplacian of Γ is equal to the negative of the Dirac delta function:

$${\displaystyle {\mathrm{\nabla }}^2\mathrm{\Gamma }(\mathbf{r})+\delta (\mathbf{r})=0.}$$

Then the scalar potential is the divergence of the convolution of E with Γ:

$${\displaystyle \mathrm{\Phi }=\mathrm{div}(\mathbf{E}\ast \mathrm{\Gamma }).}$$

Indeed, convolution of an irrotational vector field with a rotationally invariant potential is also irrotational. For an irrotational vector field G, it can be shown that

$${\displaystyle {\mathrm{\nabla }}^2\mathbf{G}=\mathrm{\nabla }(\mathrm{\nabla }\cdot \mathbf{G}).}$$

Hence

$${\displaystyle \mathrm{\nabla }\mathrm{div}(\mathbf{E}\ast \mathrm{\Gamma })={\mathrm{\nabla }}^2(\mathbf{E}\ast \mathrm{\Gamma })=\mathbf{E}\ast {\mathrm{\nabla }}^2\mathrm{\Gamma }=-\mathbf{E}\ast \delta =-\mathbf{E}}$$

as required.

More generally, the formula

$${\displaystyle \mathrm{\Phi }=\mathrm{div}(\mathbf{E}\ast \mathrm{\Gamma })}$$

holds in n-dimensional Euclidean space (n > 2) with the Newtonian potential given then by

$${\displaystyle \mathrm{\Gamma }(\mathbf{r})=\frac{1}{n(n-2){\omega }_n\Vert \mathbf{r}{\Vert }^{n-2}}}$$

where ω_n is the volume of the unit n-ball. The proof is identical. Alternatively, integration by parts (or, more rigorously, the properties of convolution) gives

$${\displaystyle \mathrm{\Phi }(\mathbf{r})=-\frac{1}{n{\omega }_n}{\int }_{{\mathbb{R}}^n}\frac{\mathbf{E}({\mathbf{r}}^{\prime })\cdot (\mathbf{r}-{\mathbf{r}}^{\prime })}{\Vert \mathbf{r}-{\mathbf{r}}^{\prime }{\Vert }^n}dV({\mathbf{r}}^{\prime }).}$$