Point Sources and Singular Flux

Why a Zero Divergence Can Accompany Nonzero Flux

A vector field can have zero divergence at every point where it is defined and still carry nonzero flux out of a region. The apparent contradiction occurs when the field is singular inside that region: the divergence theorem cannot be applied across a point where the field is not defined and continuously differentiable.

For a point source at the origin, the natural model is an inverse-square radial field. Its magnitude decreases with the square of the distance, while its direction points away from the source. Removing a small ball around the origin makes the field smooth on the remaining region. The flux through the new inner boundary then accounts for the flux through the outer boundary.

The Inverse-Square Point-Source Field

Write \(\mathbf{x}=(x,y,z)\), and let \(r=\|\mathbf{x}\|=\sqrt{x^2+y^2+z^2}\) be the distance from the origin. For a constant \(k\), define the field

Equivalently, \(\mathbf{F}=k\mathbf{\hat r}/r^2\), where \(\mathbf{\hat r}=\mathbf{x}/r\) is the outward radial unit vector. The field is undefined at \(r=0\). If \(k>0\), it points outward; if \(k<0\), it points inward.

This normalization follows from the flux calculation below. In particular, the coefficient \(k\) itself is not the total source strength; the factor \(4\pi\) comes from the area of the unit sphere.

Divergence Away from the Source

For \(r>0\), the components of the field are \(F_1=kx/r^3\), \(F_2=ky/r^3\), and \(F_3=kz/r^3\). Differentiating each component gives

Combining the terms and using \(x^2+y^2+z^2=r^2\), we obtain

This calculation establishes zero divergence only where \(r>0\). It says nothing about the origin, where the field is undefined and its derivatives cannot be evaluated.

Flux Through a Sphere Centered at the Source

Let \(S_R\) be the sphere of radius \(R>0\) centered at the origin, with outward unit normal \(\mathbf{\hat r}\). On this sphere, \(r=R\), so \(\mathbf{F}=k\mathbf{\hat r}/R^2\). The outward flux is therefore

The flux is independent of the sphere’s radius: doubling \(R\) reduces the field’s magnitude by a factor of four, but increases the sphere’s area by a factor of four. Their effects cancel.

Excision and the Divergence Theorem

The divergence theorem relates the flux across the entire boundary of a solid to the integral of divergence inside it. Its hypotheses require, in particular, that the vector field be continuously differentiable throughout the solid. Because \(\mathbf{F}\) is singular at the origin, the theorem cannot be applied directly to a solid containing that point.

Instead, let \(D\) be a bounded solid whose boundary is smooth or piecewise smooth, with the origin in its interior. Choose \(\varepsilon>0\) small enough that the ball \(B_\varepsilon\) is contained inside \(D\). Remove that ball and call the remaining region \(D_\varepsilon=D\setminus B_\varepsilon\). The field is smooth on \(D_\varepsilon\), so the divergence theorem applies there.

The boundary of \(D_\varepsilon\) consists of the original outer boundary \(\partial D\) and the small inner sphere \(S_\varepsilon\). Crucially, the outward normal of the punctured region on the inner sphere points toward the origin. It is \(-\mathbf{\hat r}\), not \(+\mathbf{\hat r}\). Since the divergence vanishes throughout \(D_\varepsilon\),

On the inner sphere, \(\mathbf{F}=k\mathbf{\hat r}/\varepsilon^2\). Its flux with the punctured region’s outward normal is

Substitution into the divergence-theorem equation yields the outer flux:

The radius \(\varepsilon\) does not affect the inner flux. As the excised ball shrinks, its surface area tends to zero, but the field’s magnitude grows like \(1/\varepsilon^2\). The product remains \(-4\pi k\), balancing the positive outer flux.

What If the Surface Does Not Enclose the Source?

When the origin lies outside a closed solid, the inverse-square field is smooth throughout that solid and has zero divergence there. The divergence theorem then gives zero net outward flux. This does not mean that the field vanishes on the boundary; it means the total outward flux, including any inward contributions, cancels.

Concentrated Source Strength

For a solid containing the origin, the pointwise divergence is zero everywhere except at one point where the field is undefined. Treating that divergence as an ordinary function and integrating it would miss the nonzero boundary flux. A more precise description treats the source as concentrated at the origin, with total strength \(4\pi k\).

This can be expressed using the Dirac delta distribution \(\delta_0\), which represents a unit point concentration at the origin: its action on a smooth test function \(\varphi\) is \(\varphi(0)\). In this distributional sense,

To see why the coefficient and sign are correct, take a smooth test function \(\varphi\) that vanishes outside a bounded region containing the origin. Integrate over that region with a ball of radius \(\varepsilon\) removed. The divergence of \(\varphi\mathbf{F}\) is \(\nabla\varphi\cdot\mathbf{F}\), because \(\nabla\cdot\mathbf{F}=0\) away from the origin. The outer boundary contributes zero since \(\varphi\) vanishes there. The inner normal points inward, so the divergence theorem gives

As \(\varepsilon\) tends to zero, \(\varphi\) approaches \(\varphi(0)\) on the shrinking sphere, whose area is \(4\pi\varepsilon^2\). Hence the right-hand side tends to \(-4\pi k\varphi(0)\). By the definition of distributional divergence, \(\langle\nabla\cdot\mathbf{F},\varphi\rangle=-\int\mathbf{F}\cdot\nabla\varphi\,dV\). Therefore \(\langle\nabla\cdot\mathbf{F},\varphi\rangle=4\pi k\varphi(0)\), which is exactly the action of \(4\pi k\delta_0\).

Common Pitfall: Applying the Theorem Across the Singularity

A tempting but invalid argument is: “The divergence is zero, so the flux through any closed surface is zero.” For a sphere containing the origin, this reasoning would give

The error is applying the divergence theorem to \(B_R\) as if the field were smooth throughout it. The field is undefined at the origin, so the theorem’s hypotheses fail. For example, when \(k=3\), direct computation gives the sphere flux \(12\pi\), not zero. Excision corrects the argument: the inner boundary contributes \(-12\pi\), so the total flux on the punctured region is zero while the original outer flux is \(12\pi\).

Always check both the regularity of the field throughout the region and the orientation of each boundary component. On an inner boundary of an excised region, the outward normal points into the removed hole.

Why Point Sources Matter

The excision method makes a useful distinction between local behavior away from a source and the net effect of the source itself. Away from the origin, the field has no local divergence. Yet any enclosing closed surface measures the same net outward flux, \(4\pi k\). That invariant flux is what identifies the source strength, independently of the size or shape of the enclosing surface.

The same reasoning also provides a diagnostic: a closed surface enclosing no singularity has zero net flux for this field, while a surface enclosing the source has flux \(4\pi k\). The key is not simply whether divergence vanishes at ordinary points; it is whether the field is regular on the entire enclosed region.