Divergence Calculation for $\bar f = \frac{1}{r^2}\hat r$

We are asked to find the divergence of the vector function $\bar f = \frac{1}{{{r^2}}}\hat r$ over a sphere of radius R that includes the origin. The vector function $\bar f$ represents a field directed radially outward from the origin, with magnitude decreasing as the inverse square of the distance $r$. Such fields are common in physics, like the electric field due to a point charge.

Divergence in Spherical Coordinates

To find the divergence of $\bar f$, we use the formula for divergence in spherical coordinates. For a vector field $\mathbf{F} = F_r \hat{\mathbf{r}} + F_\theta \hat{\boldsymbol{\theta}} + F_\phi \hat{\boldsymbol{\phi}}$, the divergence is:

For our function, $\bar f = \frac{1}{r^2}\hat r$, we have $F_r = \frac{1}{r^2}$, and $F_\theta = 0$, $F_\phi = 0$. Substituting these into the divergence formula:

Simplifying the expression:

This calculation shows that the divergence is zero for all points where $r \neq 0$. However, the vector field $\bar f$ has a singularity at the origin ($r=0$). Fields like $\frac{1}{r^2}\hat r$ originate from a point source. In mathematical terms, the divergence of such a field is related to the Dirac delta function centered at the origin:

The Dirac delta function $\delta(\mathbf{r})$ is zero everywhere except at the origin, where it is infinitely concentrated, and its integral over any volume containing the origin is 1.

Divergence Theorem Application

The question asks for the divergence "over a sphere". This typically implies calculating the integral of the divergence over the volume enclosed by the sphere. The Divergence Theorem (also known as Gauss's Theorem) provides a way to relate this volume integral to a surface integral (the flux) over the boundary of the volume:

Here, $V$ is the volume enclosed by the sphere $S$, and $\hat{\mathbf{n}}$ is the outward unit normal vector to the surface $S$. For a sphere of radius $R$, the outward normal is $\hat{\mathbf{n}} = \hat{\mathbf{r}}$.

Flux Calculation Through the Sphere

Let's calculate the flux of $\bar f$ through the sphere of radius R:

Flux $= \iint_S (\bar f \cdot \hat{\mathbf{n}}) \, dS$

On the surface of the sphere, the distance from the origin is constant, $r = R$. The unit radial vector is $\hat{\mathbf{r}}$. Therefore, $\hat{\mathbf{n}} = \hat{\mathbf{r}}$ and $\hat{\mathbf{r}} \cdot \hat{\mathbf{r}} = 1$. The vector field on the surface is $\bar f = \frac{1}{R^2}\hat{\mathbf{r}}$.

Flux $= \iint_S \left( \frac{1}{R^2} \hat{\mathbf{r}} \cdot \hat{\mathbf{r}} \right) \, dS = \iint_S \frac{1}{R^2} \, dS$

The surface area element $dS$ in spherical coordinates is $dS = R^2 \sin\theta \, d\theta \, d\phi$. The integral is over the entire surface of the sphere (from $\theta=0$ to $\pi$ and $\phi=0$ to $2\pi$):

Flux $= \int_0^{2\pi} \int_0^\pi \frac{1}{R^2} (R^2 \sin\theta \, d\theta \, d\phi)$

Cancel $R^2$ terms:

Flux $= \int_0^{2\pi} d\phi \int_0^\pi \sin\theta \, d\theta$

Evaluating the integrals:

Flux $= \left[ \phi \right]_0^{2\pi} \times \left[ -\cos\theta \right]_0^\pi$

Flux $= (2\pi - 0) \times (-\cos\pi - (-\cos0))$

Flux $= (2\pi) \times (-(-1) - (-1))$

Flux $= (2\pi) \times (1 + 1)$

Flux $= (2\pi) \times 2 = 4\pi$

Conclusion on Divergence Integral

According to the Divergence Theorem, the volume integral of the divergence over the sphere's volume is equal to the calculated flux:

This confirms that although the divergence is zero everywhere except at the origin, its integral over a volume containing the origin is $4\pi$, due to the source term represented by the Dirac delta function at $r=0$. Therefore, the divergence of the function over the sphere (interpreted as the volume integral of the divergence) is $4\pi$.