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The inverse-square field is divergence free, and its flux through the sphere bounding the translated unit ball vanishes

Example

On U=R3{0} let F(x,y,z)=(x,y,z)(x2+y2+z2)3/2. Then divF=0 on U. Consequently the outward flux of F through the sphere bounding the translated unit ball B={(x,y,z):(x2+y2+(z2)2)1} is 0.

Facts & Assumptions

Given: The field F on U=R3{0}, and the translated closed unit ball B={(x,y,z):(x2+y2+(z2)2)1}.

[L1]

For a finite gluing of elementary solid regions, a C1 field on an open set containing the union whose divergence vanishes there has zero outward boundary flux (A field with vanishing divergence has zero outward flux through the boundary of a glued elementary solid).

[L2]

The closed ball admits the octant presentation adapted in all three coordinate directions (The closed ball is an elementary solid region, presented by the eight spherical octants).

[F1]

The divergence of a field is the sum of its coordinate partial derivatives (Divergence and curl of a C1 vector field).

[L6]

For every real α, the function ssα is continuous and differentiable on (0,), with derivative αsα1 (Continuity and derivatives of positive-base real powers).

[F2]

The Jacobian matrix records the coordinate partial derivatives (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

[L5]

The divergence theorem is the identity EdivF=EF,n (The divergence theorem on an elementary solid region).

[F4]

Flux is computed against the oriented area vector of a patch (Unit normal fields, orientations, and flux through a regular surface patch).

[F5]

A subset of a metric space is open when every one of its points contains an open metric ball lying in the subset; the Euclidean metric on R3 is induced by 2 (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Open ball, closed ball and sphere in a metric space, Rn as the set of functions nR, and d1, d2, d are metrics on it).

Verification

technique · direct
1.1

Put s(x,y,z)=x2+y2+z2, which is positive and continuous on U by [L7]. The ith component of F is Fi=xis3/2. By the product and chain rules [L3, L4], the positive-base power rule [L6], and the coordinate interpretation of partial derivatives [F2], every coordinate partial derivative is jFi={s3/23xi2s5/2,j=i,3xixjs5/2,ji. The coordinate projections are continuous by [L7], so s is continuous; because s>0 on U, [L6] and [L7] make every function in the displayed formulas continuous there. Hence F is C1 on U.

L3L4L6L7F2F3given
1.2

Translating the octant presentation of the unit ball by (0,0,2) gives an elementary solid region presentation of B, because translation adds a constant to each patch and changes no derivative.

L2F4given
2.1

Summing the three diagonal formulas of step 1.1 gives divF=3s3/23ss5/2=0 on U by [F1].

step 1.1F1
2.2

Every point of B has distance at least 1 from the origin, so BU. The set U is open: if pU, then p2>0 and the ball B(p,p2/2) cannot contain the deleted origin, whose distance from p is p2. Step 1.1 proves that F and all nine coordinate partial derivatives are continuous throughout this open set, so F is C1 on an open set containing B.

step 1.1step 1.2F3F5given
3.1

Step 2.1 gives vanishing divergence and step 2.2 gives the required open neighbourhood hypothesis, so [L1] and [L5] give zero outward flux through B.

step 2.1step 2.2L1L5

Remarks

  • The translation in step 1.2 is not cosmetic. The origin is the singular point of the field, so moving the ball off it is exactly what makes the divergence theorem applicable.

Depends on

Used by

Dependency tree · two levels

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Sources