Scalar potential

A scalar potential is a fundamental concept in vector analysis and physics (the adjective 'scalar' is frequently omitted if there is no danger of confusion with vector potential). Given a vector field F, its scalar potential "V" is a scalar field whose negative gradient is F,

:$mathbf\{F\}\; =\; -\; abla\; V$.

Conversely, given a function "V", this formula defines a vector field F with the scalar potential "V". Scalar potential is also frequently denoted by the Greek letter "Φ", for example, in electrodynamics.

The physical meaning of the scalar potential depends on the type of the field. For a velocity field of a fluid or gas flow, the definition of the scalar potential implies that the direction of the flow at any point coincides with the direction of the steepest decrease of the potential at that point, and for a force field the same is true of the acceleration at a point. The scalar potential of a force field is closely related to the field's potential energy.

Not every vector field has a scalar potential; those which do are called conservative, corresponding to the notion of conservative forcein physics. Among velocity fields, any lamellar field has a scalar potential, whereas a solenoidal field only has a scalar potential in the special case when it is a Laplacian field.

Integrability conditions

If F is a conservative vector field (also called "irrotational", "curl-free", or "potential"), and its components have continuous partial derivatives, the potential of F with respect to a reference point $mathbf\; r\_0$ is defined in terms of the line integral:

:$V(mathbf\; r)\; =\; -int\_C\; mathbf\{F\}(mathbf\{r\})cdot,dmathbf\{r\}\; =\; -int\_a^b\; mathbf\{F\}(mathbf\{r\}(t))cdotmathbf\{r\}\text{'}(t),dt.$

where "C" is a parametrized path from $mathbf\; r\_0$ to $mathbf\; r,$

: $mathbf\{r\}(t),\; aleq\; tleq\; b,\; mathbf\{r\}(a)=mathbf\{r\_0\},\; mathbf\{r\}(b)=mathbf\{r\}.$

The fact that the line integral depends on the path "C" only through its terminal points $mathbf\; r\_0$ and $mathbf\; r$ is, in essence, the path independence property of a conservative vector field. The Fundamental Theorem of Calculus for line integrals implies that if "V" is defined in this way, then $mathbf\{F\}=\; -\; abla\; V,$ so that "V" is a scalar potential of the conservative vector field F. Scalar potential is not determined by the vector field alone: indeed, the gradient of a function is unaffected if a constant is added to it. If "V" is defined in terms of the line integral, the ambiguity of "V" reflects the freedom in the choice of the reference point $mathbf\; r\_0.$

Altitude as gravitational potential energy

An example is the (nearly) uniform gravitational field near the Earth's surface. It has a potential energy:$U\; =\; m\; g\; h$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 which is perpendicular to the hill's surface. The component of gravity which remains to move the ball is parallel to the surface::$F\_S\; =\; -\; m\; g\; sin\; heta$where "θ" is the angle of inclination, and the component of "F_S" perpendicular to gravity is:$F\_P\; =\; -\; m\; g\; sin\; heta\; cos\; heta\; =\; -\; \{1\; over\; 2\}\; m\; g\; sin\; 2\; heta.$This force "F_P", parallel to the ground, will be 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 :$heta\; =\; an^\{-1\}\{Delta\; h\; over\; Delta\; x\}$ so that:$F\_P\; =\; -\; m\; g\; \{\; Delta\; x\; Delta\; h\; over\; Delta\; x^2\; +\; Delta\; h^2\; \}$.However, on a contour map, the gradient will be 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 will be permeated by a uniform buoyant force which will cancel out the gravitational force: that is how the fluid maintains its equilibrium. This buoyant force is the negative gradient of pressure::$mathbf\{f\_B\}\; =\; -\; abla\; p$.Since buoyant force points upwards, in the direction opposite to gravity, then pressure in the fluid will increase 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 which are parallel to the ground. The surface of the water can be characterized as a plane with zero pressure.

If the liquid has a vertical vortex (whose axis of rotation is perpendicular to the ground), then the vortex will cause a depression in the pressure field. The surfaces of constant pressure will be parallel to the ground far away from the vortex, but near and inside the vortex the surfaces of constant pressure will be pulled downwards, closer to the ground. This will also happen to the surface of zero pressure: therefore, inside the vortex, the top surface of the liquid is pulled downwards into a depression, or even into a tube (a solenoid).

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::$F\_B\; =\; -\; oint\_S\; abla\; p\; cdot\; ,\; dmathbf\{S\}.$

A moving airplane wing makes the air pressure above it decrease relative to the air pressure below it. This creates enough buoyant force to counteract gravity.

Calculating the scalar potential

Given a vector field E, its scalar potential "Φ" can be calculated to be:$phi(mathbf\{R\_0\})\; =\; \{1\; over\; 4\; pi\}\; int\_\; au\; \{\; abla\; cdot\; mathbf\{E\}(\; au)\; over\; |\; mathbf\{R\}(\; au)\; -\; mathbf\{R\_0\}\; ,\; d\; au$where "τ" is volume. Then, if E is irrotational (Conservative), :$mathbf\{E\}\; =\; -\; abla\; phi\; =\; -\; \{1\; over\; 4\; pi\}\; abla\; int\_\; au\; \{\; abla\; cdot\; mathbf\{E\}(\; au)\; over\; |\; mathbf\{R\}(\; au)\; -\; mathbf\{R\_0\}\; ,\; d\; au$.This formula is known to be correct if 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/"r²".

See also

* Electric potential
* Fundamental theorem of vector analysis
* Vector potential
* Yukawa potential