Gauss s integral theorem

When d is a spatial domain with the closed boundary dA and the respective unit vector field of the surface normals e is directed outwards, then Gauss s integral theorem states... 

The surface integral in Eq. (3.32b) can be transformed into a volume integral over the enclosed domain by using Gauss s integral theorem, Eq. (3.7). Equating with Eq. (3.32a) leads to the electrostatic equilibrium condition, which is known as one of Maxwell s equations in integral form ... 

Integrating (246) on the V volume of the current tube and applying the Gauss s divergence theorem... 

This integral relation is known as Gauss s theorem. The most familiar example is in electrostatics. 

The survival probability of a pair before it is formed,p( °° r0, t0), is zero and the integral can be simplified by using Gauss s theorem (see Appendix A.3) to give... 

Because 8 is arbitrary and small and independent of Gauss s theorem for the volume integrals may be used and... 

Using Gauss s theorem to reduce the volume integral and applying the inner boundary condition of eqn. (353c) together with eqn. (352) with p = exp (+ G(R, t)... 

The minus sign is included since, with n pointing outward, V n > 0 when system points flow out of the volume element and therefore cause a decrease in the number of system points in the volume element. It is inconvenient to have a volume and a surface integral in the same equation, so we convert the surface integral to a volume integral using Gauss s theorem ... 

The right-hand side may be converted to a surface integral using Gauss s theorem, Eq. (5.14), and we find... 

We introduce now a domain Vr, bounded by a sphere dVa of a radius / , with its center at the origin of some Cartesian coordinate system, x, y, z. Integrating both sides of equation (9.65) over the domain Vr, and applying Gauss s theorem, we find ... 

Thus, according to Gauss s theorem (13.100), we obtain the following integral... 

The surface integral in eqn (5.76) comes from the application of Gauss s theorem to the term involving V -(V,i/ 5 ). As before, all such surface integrals vanish except for r,- = fi because of the vanishing of on the boundaries at infinity. From this point on, the coordinate Tj and the volume element d-Tj will be set equal to r and dr, respectively, and Vj and to their corresponding unscripted quantities. 

Integration of the right-hand side of eqn (5.82) in the manner indicated in eqn (5.80) transforms it into an integral of the Laplacian of the charge density. A typical term in this integration can be transformed using Gauss s theorem to yield... 

The force may be equivalently expressed using Gauss s theorem as an integration of the force density — V r (r) over the basin of the atom,... 

Applying Gauss s theorem to convert the surface integrals to volume integrals on the right-hand sides of Eqs. (10.4.4) gives three equations such as Jyd3r Y = 0. Since the volume V is arbitrary, Y — 0. Thus we find the local conservation laws... 

Now consider a volume H in configuration space where the wave function i), satisfies the Schrodinger equation, i.e., (H - E) = 0 and we also assume that fl contains the interaction region. For the closed surface 5 surrounding fl the use of Gauss s theorem (integration by parts) yields a very important property of the solutions of the wave equation as... 

This equation shows the relationship between field values observed on various points of the surface S and can be interpreted from two points of view. If the charge e is known, eq. 1.34 can be considered to be an integral equation in an unknown variable the normal component of the field. In contrast, when the electric field is known, the use of the flux allows us to determine the sources of the field. If we wish to find the relationship between flux and source within an elementary volume, we can make use of Gauss s theorem ... 

Both the integration and differentiation operations carried out on the right-hand side of eq. 1.100 arc performed with respect to the same point p, so that one can apply Gauss s theorem, which results in ... 

This can be physically interpreted as the indication that magnetic charges do not exist and that magnetic flux lines are closed. Applying Gauss s theorem, we obtain the integral form of this equation ... 

Application of Gauss s law, or the divergence theorem, transforms the snrface integral on the right side of equation (d) to an integral over the entire control volume ... 

Equation (3) presupposes that there is no source of particles Xi within volume V. Transformation of the surface integral to a volume integral by the help of Gauss s theorem yields the relationship... 

Now integrate over all the infinitesimal volume elements. The left-hand side of Equation (17.23) must be integrated over the whole surface S, and the right-hand side over the whole volume V. Because the flux into one volume element equals the flux out of an adjacent element, the only component of the surface integral that is non-zero is the one that represents the outer surface. This integration gives Gauss s theorem ... 

Using Gauss s theorem, the first term can be converted into a surface integral. Since we assume the value of is time independent at the boundary, i.e. the boundary conditions are time independent, this surface integral vanishes. Using the relations... 

In order to show that the probability density is time independent, we now take its space integral and find by Gauss s theorem that... 

The last term in (4.9) is the divergence of a vector and, according to Gauss s theorem, its volume integral can be converted to a surface integral therefore it describes only contributions to the surface and not to the volume eneigies. 

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