Radial Fields and Flux Through Spheres

Why Radial Fields Simplify Spherical Flux

A vector field is radial about a point when its direction at every other point is along the line from that point to the location. On a sphere centered at that same point, the outward unit normal points in exactly the radial direction. This alignment makes flux especially direct to calculate: the field’s normal component is the same at every point on the sphere.

The field’s magnitude need not be constant as distance changes. It can increase, decrease, or even change sign with distance. For flux through one fixed sphere, however, its value is constant across that sphere because every point on it is the same distance from the center.

Key idea: For a radial field centered at the center of a sphere, the outward flux is the field’s signed radial magnitude at the sphere’s radius multiplied by the sphere’s area.

Radial Fields and Their Flux

Let the center of a sphere be the point with position vector $\mathbf{c}$. At a point $\mathbf{x}\ne\mathbf{c}$, define the distance from the center by $r=|\mathbf{x}-\mathbf{c}|$, and the outward radial unit vector by $\widehat{\mathbf{r}}=(\mathbf{x}-\mathbf{c})/r$. A radial field centered at $\mathbf{c}$ has the form $\mathbf{F}(\mathbf{x})=g(r)\widehat{\mathbf{r}}$, where $g$ is its signed radial magnitude.

Definition: The outward flux of a vector field through a sphere $S_R$ of radius $R$, centered at $\mathbf{c}$, is the surface integral $\iint_{S_R}\mathbf{F}\cdot\mathbf{n}\,dS$, where $\mathbf{n}$ is the outward unit normal.

At each point of this sphere, $r=R$ and $\mathbf{n}=\widehat{\mathbf{r}}$. Therefore $\mathbf{F}\cdot\mathbf{n}=g(R)$. Since this value is constant over the sphere and its area is $4\pi R^2$, the flux is $4\pi R^2g(R)$. A negative value of $g(R)$ means the field points inward there, so the outward flux is negative.

\iint_{S_R}\mathbf{F}\cdot\mathbf{n}\,dS = 4\pi R^2g(R).

This direct formula requires the sphere to be centered at the center of the radial field. If the sphere and field have different centers, the normal and the field direction generally do not agree pointwise, and this simplification does not apply.

Radial Divergence and the Divergence Theorem

The divergence theorem offers a second calculation for flux through a closed sphere. It states that if a vector field is continuously differentiable on an open region containing a solid $V$ and its boundary $S$, then the outward flux across $S$ equals the volume integral of the divergence over $V$. For a ball, the theorem can verify the direct radial-flux formula, provided the field is regular throughout the ball, including its center.

For $r>0$, the divergence of a radial field is $g'(r)+2g(r)/r$. The second term accounts for the way radial directions spread out as distance increases. The formula is evaluated away from the center; to use the divergence theorem over the entire ball, the field must also have a differentiable extension at the center.

Formula: For $\mathbf{F}(\mathbf{x})=g(r)\widehat{\mathbf{r}}$ and $r>0$, $$ \nabla\cdot\mathbf{F} = g'(r)+\frac{2g(r)}{r} = \frac{1}{r^2}\frac{d}{dr}\bigl(r^2g(r)\bigr). $$

Why the Radial Divergence Formula Works

Write the components of the field as $F_i=g(r)(x_i-c_i)/r$, for $i=1,2,3$. The divergence is the sum of the three partial derivatives of the corresponding components. The product rule separates the derivative of $g(r)$ from the derivative of the unit radial vector. Since $\partial r/\partial x_i=(x_i-c_i)/r$, the first contribution sums to $g'(r)$.

For the other contribution, write $\widehat r_i=(x_i-c_i)/r$. Its divergence is $2/r$: differentiating each component and summing gives $3/r-r^2/r^3=2/r$. Combining the two contributions gives the formula.

\nabla\cdot(g(r)\widehat{\mathbf{r}}) = g'(r)\sum_{i=1}^3\widehat r_i^2 + g(r)\sum_{i=1}^3\frac{\partial \widehat r_i}{\partial x_i} = g'(r)+\frac{2g(r)}{r}.

The equivalent derivative form follows by applying the product rule to $r^2g(r)$: its derivative is $2rg(r)+r^2g'(r)$, and division by $r^2$ gives $g'(r)+2g(r)/r$.

To see how this formula connects to sphere flux, integrate the divergence over the ball in spherical coordinates. The volume element contributes a factor $r^2$, and integrating over all directions contributes $4\pi$. Thus the volume integral is $4\pi\int_0^R (r^2g(r))'\,dr$. If the field is regular at the center, $r^2g(r)$ tends to zero as $r$ tends to zero. The integral is then $4\pi R^2g(R)$, exactly the direct surface calculation.

\iiint_{B_R}\nabla\cdot\mathbf{F}\,dV = 4\pi\int_0^R \frac{d}{dr}\bigl(r^2g(r)\bigr)\,dr = 4\pi R^2g(R).

This argument explains both the agreement between the methods and the condition at the center: the endpoint term at zero must vanish for the calculation over the full ball to give this result.

Worked Examples

Worked Example: A Radial Field Growing Linearly

Let $\mathbf{F}(\mathbf{x})=\mathbf{x}-\mathbf{c}$, and take the sphere centered at $\mathbf{c}$ with radius $R=2$. Find its outward flux directly and verify it using the divergence theorem.

At distance $r$, the field is $r\widehat{\mathbf{r}}$, so $g(r)=r$. On the sphere, the outward normal is $\widehat{\mathbf{r}}$, and $g(2)=2$. Its area is $4\pi(2)^2=16\pi$. Therefore the direct flux is $2(16\pi)=32\pi$.

\iint_{S_2}\mathbf{F}\cdot\mathbf{n}\,dS = 4\pi(2)^2(2) = 32\pi.

For the volume calculation, the radial divergence formula gives $g'(r)+2g(r)/r=1+2=3$ for $r>0$. The field is linear and differentiable at the center as well, so the divergence theorem applies. The ball’s volume is $\frac{4}{3}\pi(2)^3=\frac{32\pi}{3}$. Its divergence integral is $3\cdot\frac{32\pi}{3}=32\pi$, matching the direct flux.

Check: The outward normal agrees with the direction of the field, so the positive flux is consistent with the field pointing outward on the sphere.

Worked Example: A Radial Field with Quadratic Magnitude

Let $\mathbf{F}(\mathbf{x})=r(\mathbf{x}-\mathbf{c})$, where $r=|\mathbf{x}-\mathbf{c}|$, and use a sphere of radius $2$ centered at $\mathbf{c}$. Find and check its outward flux.

Since $\mathbf{x}-\mathbf{c}=r\widehat{\mathbf{r}}$, the field is $r^2\widehat{\mathbf{r}}$, so $g(r)=r^2$. On the sphere, the radial magnitude is $g(2)=4$; multiplying by the area $16\pi$ gives flux $64\pi$.

\iint_{S_2}\mathbf{F}\cdot\mathbf{n}\,dS = 4\pi(2)^2(2^2) = 64\pi.

For the divergence-theorem check, $g'(r)=2r$ and $2g(r)/r=2r$, so the divergence is $4r$. Using spherical coordinates, integrate over the ball of radius $2$: the angular factor is $4\pi$, and the volume element contributes $r^2\,dr$. The radial integral is $\int_0^2 4r^3\,dr=[r^4]_0^2=16$. The resulting divergence integral is $4\pi(16)=64\pi$.

\iiint_{B_2}\nabla\cdot\mathbf{F}\,dV = 4\pi\int_0^2 4r^3\,dr = 4\pi(16) = 64\pi.

Check: The volume calculation agrees with the direct surface flux. The field is differentiable at the center because it is $r(\mathbf{x}-\mathbf{c})$, whose components are $|\mathbf{x}-\mathbf{c}|(x_i-c_i)$; the radial divergence expression has the finite limit $0$ there.

Worked Example: A Magnitude That Levels Off with Distance

Consider $\mathbf{F}(\mathbf{x})=(\mathbf{x}-\mathbf{c})/(1+r^2)$, with $r=|\mathbf{x}-\mathbf{c}|$, and calculate outward flux through the sphere of radius $1$ centered at $\mathbf{c}$.

Rewriting the field as $g(r)\widehat{\mathbf{r}}$ gives $g(r)=r/(1+r^2)$. At $r=1$, this magnitude is $1/2$. The sphere has area $4\pi$, so the direct flux is $2\pi$.

\iint_{S_1}\mathbf{F}\cdot\mathbf{n}\,dS = 4\pi(1)^2\left(\frac{1}{2}\right) = 2\pi.

For an independent check, differentiate $g$: $g'(r)=(1-r^2)/(1+r^2)^2$. The divergence is therefore $(1-r^2)/(1+r^2)^2+2/(1+r^2)=(3+r^2)/(1+r^2)^2$. The field is continuously differentiable throughout the ball, so the divergence theorem applies. After including the spherical volume factor, the radial integrand is $r^2(3+r^2)/(1+r^2)^2$. This is the derivative of $r^3/(1+r^2)$. Integrating from zero to one gives $1/2$, and multiplying by $4\pi$ gives $2\pi$.

\iiint_{B_1}\nabla\cdot\mathbf{F}\,dV = 4\pi\int_0^1\frac{r^2(3+r^2)}{(1+r^2)^2}\,dr = 4\pi\left[\frac{r^3}{1+r^2}\right]_0^1 = 2\pi.

Check: The evaluated volume integral equals the direct flux, and its positive sign agrees with the outward-pointing field on the sphere.

A Common Divergence Error

A frequent mistake is to differentiate only the radial magnitude and omit the contribution from the changing radial direction. For example, for $\mathbf{F}(\mathbf{x})=\mathbf{x}-\mathbf{c}$, the radial magnitude is $g(r)=r$. The incorrect calculation $\nabla\cdot\mathbf{F}=g'(r)=1$ would predict, on a sphere of radius $2$, flux equal to the ball’s volume, $\frac{32\pi}{3}$. But the direct calculation gives $32\pi$, so the two values disagree.

The correction is to use both terms in the radial divergence formula: $g'(r)+2g(r)/r=1+2=3$. The divergence-theorem calculation is then $3\cdot\frac{32\pi}{3}=32\pi$, agreeing with the surface calculation. The extra term is essential; it reflects how the radial field’s direction varies from point to point.

Choosing and Checking a Method

For a sphere centered on the radial field’s center, direct integration is often the shortest route: evaluate the signed radial magnitude at the sphere’s radius and multiply by the sphere’s area. The divergence theorem is useful as a verification and shows how local divergence throughout the ball contributes to the total boundary flux.

Before using the divergence theorem, check that the field is continuously differentiable throughout the enclosed ball, not merely on its spherical boundary. When the field is defined by a radial formula, inspect its behavior at the center rather than assuming that a formula involving $1/r$ is valid there. If the field does not meet the theorem’s regularity conditions on the ball, this full-ball application is not justified.

Takeaway: On a sphere centered at a radial field’s center, outward flux is $4\pi R^2g(R)$. For a regular field on the full ball, the divergence theorem gives the same result because the radial divergence integrates to the boundary term $4\pi R^2g(R)$.

Check Your Understanding

Use the definitions and calculations in this tutorial to answer the following questions.

For a radial field with signed magnitude $g(r)$, why is the normal component constant on a sphere centered at the field’s center?
What is the outward flux through a sphere of radius $3$ if the radial magnitude there is $-2$?
For a radial field, what contribution is missing if its divergence is calculated using only $g'(r)$?
State the regularity condition needed to apply the divergence theorem to the entire ball bounded by a sphere.
For $g(r)=r^2$, calculate the divergence for $r>0$ and state the direct flux through a sphere of radius $R$.