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In mathematical physics, scalar potential describes the situation where the difference in the of an object in two different positions depends only on the positions, not upon the path taken by the object in traveling from one position to the other. It is a in : a directionless value (scalar) that depends only on its location. A familiar example is potential energy due to gravity.

A scalar is a fundamental concept in and (the adjective scalar is frequently omitted if there is no danger of confusion with ). The scalar potential is an example of a . Given a , the scalar potential is defined such that:

\mathbf{F} = -\nabla P = - \left( \frac{\partial P}{\partial x}, \frac{\partial P}{\partial y}, \frac{\partial P}{\partial z} \right),

where is the of and the second part of the equation is minus the gradient for a function of the Cartesian coordinates . In some cases, mathematicians may use a positive sign in front of the gradient to define the potential.See [1] for an example where the potential is defined without a negative. Other references such as avoid using the term potential when solving for a function from its gradient. Because of this definition of in terms of the gradient, the direction of at any point is the direction of the steepest decrease of at that point, its magnitude is the rate of that decrease per unit length.

In order for to be described in terms of a scalar potential only, any of the following equivalent statements have to be true:

  1. -\int_a^b \mathbf{F}\cdot d\mathbf{l} = P(\mathbf{b})-P(\mathbf{a}), where the integration is over a passing from location to location and is evaluated at location .
  2. \oint \mathbf{F}\cdot d\mathbf{l} = 0, where the integral is over any simple closed path, otherwise known as a .
  3. {\nabla}\times{\mathbf{F}} =0.
The first of these conditions represents the and is true for any vector field that is a gradient of a differentiable single valued scalar field . The second condition is a requirement of so that it can be expressed as the gradient of a scalar function. The third condition re-expresses the second condition in terms of the curl of using the fundamental theorem of the curl. A vector field that satisfies these conditions is said to be irrotational (conservative).

Scalar potentials play a prominent role in many areas of physics and engineering. The gravity potential is the scalar potential associated with the force of gravity per unit mass, or equivalently, the due to the field, as a function of position. The gravity potential is the gravitational potential energy per unit mass. In the electric potential is the scalar potential associated with the , i.e., with the electrostatic force per unit . The electric potential is in this case the electrostatic potential energy per unit charge. In , irrotational have a scalar potential only in the special case when it is a . Certain aspects of the can be described by a . The potential play a prominent role in the Lagrangian and Hamiltonian formulations of classical mechanics. Further, the scalar potential is the fundamental quantity in quantum mechanics.

Not every vector field has a scalar potential. Those that do are called conservative, corresponding to the notion of conservative force in physics. Examples of non-conservative forces include frictional forces, magnetic forces, and in fluid mechanics a velocity field. By the Helmholtz decomposition theorem however, all vector fields can be describable in terms of a scalar potential and corresponding . In electrodynamics, the electromagnetic scalar and vector potentials are known together as the electromagnetic four-potential.


Integrability conditions
If is a conservative vector field (also called irrotational, curl-free, or potential), and its components have continuous partial derivatives, the potential of with respect to a reference point is defined in terms of the :

V(\mathbf r) = -\int_C \mathbf{F}(\mathbf{r})\cdot\,d\mathbf{r} = -\int_a^b \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt,

where is a parametrized path from to ,

\mathbf{r}(t), a\leq t\leq b, \mathbf{r}(a)=\mathbf{r_0}, \mathbf{r}(b)=\mathbf{r}.

The fact that the line integral depends on the path only through its terminal points and is, in essence, the path independence property of a conservative vector field. The implies that if is defined in this way, then , so that is a scalar potential of the conservative vector field . 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 is defined in terms of the line integral, the ambiguity of reflects the freedom in the choice of the reference point .


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 is the gravitational potential energy and is the height above the surface. This means that gravitational potential energy on a 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 always points straight downwards in the direction of gravity; . However, a ball rolling down a hill cannot move directly downwards due to the 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:

\mathbf F_\mathrm{S} = - m g \ \sin \theta

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

\mathbf F_\mathrm{P} = - m g \ \sin \theta \ \cos \theta = - {1 \over 2} m g \sin 2 \theta.

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

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

\theta = \tan^{-1}\frac{\Delta h}{\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 is inversely proportional to , which is not similar to force : 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 , 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 is the negative gradient of :

\mathbf{f_B} = - \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 (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:

F_B = - \oint_S \nabla p \cdot \, d\mathbf{S}.


Scalar potential in Euclidean space
In 3-dimensional Euclidean space , the scalar potential of an irrotational vector field is given by

\Phi(\mathbf{r}) = \frac{1}{4\pi} \int_{\mathbb R^3} \frac{\operatorname{div} \mathbf{E}(\mathbf{r}')}{\left\| \mathbf{r} - \mathbf{r}' \right\|} \, dV(\mathbf r')

where is an infinitesimal with respect to . Then

\mathbf{E} = -\mathbf{\nabla} \Phi = - \frac{1}{4\pi} \mathbf{\nabla} \int_{\mathbb R^3} \frac{\operatorname{div} \mathbf{E}(\mathbf{r}')}{\left\| \mathbf{r} - \mathbf{r}' \right\|} \, dV(\mathbf r')

This holds provided is continuous and vanishes asymptotically to zero towards infinity, decaying faster than and if the of likewise vanishes towards infinity, decaying faster than .

Written another way, let

\Gamma(\mathbf{r}) = \frac{1}{4\pi} \frac{1}{\|\mathbf r\|}

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

\nabla^2\Gamma(\mathbf r) + \delta(\mathbf r) = 0.

Then the scalar potential is the divergence of the of with :

\Phi = \operatorname{div}(\mathbf E * \Gamma).

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

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

Hence

\nabla\operatorname{div}(\mathbf E * \Gamma) = \nabla^2(\mathbf E * \Gamma) = \mathbf E

  • \nabla^2\Gamma = -\mathbf E * \delta = -\mathbf E

as required.

More generally, the formula

\Phi = \operatorname{div}(\mathbf E * \Gamma)

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

\Gamma(\mathbf r) = \frac{1}{n(n-2)\omega_n \|\mathbf r\|^{n-2}}

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

\Phi(\mathbf r) = -\frac{1}{n\omega_n} \int_{\mathbb R^n} \frac{\mathbf E(\mathbf r')\cdot (\mathbf r-\mathbf r')}{\|\mathbf r-\mathbf r'\|^n} \,dV(\mathbf r').


See also


Notes

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