Potentials for Radial and Central Fields

Radial Fields and Their Potentials

A radial field points along the line from a fixed center to the point where the field is evaluated. Its magnitude may vary with distance, but it has no separate dependence on direction. This structure makes it possible to find a potential using a one-variable antiderivative.

The key is to distinguish the scalar radial magnitude from the vector field. Integrating the magnitude gives a scalar function of distance; taking the gradient of that function recovers the full vector field, including its direction.

Definition and Setup

Work in two or three dimensions, and fix a center $\mathbf{c}$. For a point $\mathbf{x}\ne\mathbf{c}$, define the distance from the center by $r=\|\mathbf{x}-\mathbf{c}\|$. The associated radial unit vector points outward from $\mathbf{c}$.

\widehat{\mathbf{r}} = \frac{\mathbf{x}-\mathbf{c}}{\|\mathbf{x}-\mathbf{c}\|} = \frac{\mathbf{x}-\mathbf{c}}{r}.

A field with radial magnitude $f(r)$ is therefore written as $f(r)\widehat{\mathbf{r}}$. The sign of $f(r)$ determines whether the field points outward or inward: a negative value reverses the direction of the outward radial unit vector.

Definition: A scalar potential for a vector field $\mathbf{F}$ on a domain $D$ is a differentiable scalar function $\Phi:D\to\mathbb{R}$ such that $\nabla\Phi=\mathbf{F}$ throughout $D$.

For radial fields, the relevant domain must avoid the center, where $r=0$ and $\widehat{\mathbf{r}}$ is undefined. More generally, suppose $f$ is continuous on an interval $I\subset(0,\infty)$, and the field is defined at points whose distances from $\mathbf{c}$ lie in $I$.

Why Integrating the Radial Magnitude Works

Let $G$ be an antiderivative of $f$ on $I$, so that $G'(r)=f(r)$. Define the scalar function $\Phi(\mathbf{x})=G(\|\mathbf{x}-\mathbf{c}\|)$. The chain rule shows that its gradient points in the radial direction and has the required magnitude.

\frac{\partial r}{\partial x_i} = \frac{x_i-c_i}{r} \qquad\text{for each coordinate }x_i.

Applying the chain rule to $\Phi=G(r)$ gives the following calculation in every coordinate:

\frac{\partial\Phi}{\partial x_i} = G'(r)\frac{\partial r}{\partial x_i} = f(r)\frac{x_i-c_i}{r}.

Collecting these partial derivatives into a vector proves the formula.

Theorem: Suppose $f$ is continuous on an interval $I\subset(0,\infty)$, and let $G$ satisfy $G'=f$ on $I$. On any domain of points with $r=\|\mathbf{x}-\mathbf{c}\|\in I$, the radial field $f(r)\widehat{\mathbf{r}}$ has the potential $$ \Phi(\mathbf{x})=G(r), \qquad \nabla\Phi(\mathbf{x})=f(r)\widehat{\mathbf{r}}. $$

This derivation explains why the one-variable integration is sufficient: distance changes in the direction $\widehat{\mathbf{r}}$, and the chain rule supplies precisely that direction when the gradient is taken. The result is a potential on every such domain, including a domain that is not connected, provided the same radial formula applies throughout it. A constant may be added to a potential without changing its gradient. On separate connected components, those constants may be chosen independently.

Inverse-Power Radial Magnitudes

Consider a radial magnitude $f(r)=A r^{-p}$, where $A$ and $p$ are constants and $r>0$. The exponent $p=1$ is a special case because the antiderivative is logarithmic. All other exponents use the power rule.

G(r)= \begin{cases} \dfrac{A}{1-p}r^{1-p}, & p\ne 1,\\[6pt] A\ln r, & p=1. \end{cases}

Indeed, when $p\ne1$, differentiating the first expression gives $\frac{A}{1-p}(1-p)r^{-p}=A r^{-p}$. When $p=1$, the derivative of $A\ln r$ is $A/r$. If $r$ represents a dimensional quantity, the logarithm can instead be written $\ln(r/r_0)$ for a fixed reference distance $r_0>0$; this changes the potential only by a constant.

Worked Example: An Inverse-Cube Radial Field

Find a potential for $\mathbf{F}(\mathbf{x})=\frac{8}{r^3}\widehat{\mathbf{r}}$ on the domain $r>0$, and verify it.

Here $f(r)=8r^{-3}$, so $A=8$ and $p=3$. Since $p\ne1$, an antiderivative is

G(r)=\frac{8}{1-3}r^{1-3} =-4r^{-2} =-\frac{4}{r^2}.

Thus $\Phi(\mathbf{x})=-4/r^2$. To check it, differentiate with respect to $r$:

G'(r)=(-4)(-2)r^{-3} =8r^{-3} =\frac{8}{r^3}.

The gradient formula now gives

\nabla\Phi = G'(r)\widehat{\mathbf{r}} = \frac{8}{r^3}\widehat{\mathbf{r}} = \mathbf{F}.

The potential is valid throughout the domain $r>0$, where the field and the radial unit vector are defined.

Worked Example: An Inverse-Square Field Pointing Inward

Let $\mathbf{F}(\mathbf{x})=-\frac{6}{r^2}\widehat{\mathbf{r}}$ for $r>0$. Find a potential and check its sign.

The radial magnitude is $f(r)=-6r^{-2}$. Integrating with respect to $r$ gives

G(r) = \int -6r^{-2}\,dr = -6\left(\frac{r^{-1}}{-1}\right) = 6r^{-1} = \frac{6}{r}.

So take $\Phi(\mathbf{x})=6/r$. Differentiation verifies the radial magnitude:

G'(r) = 6(-1)r^{-2} = -\frac{6}{r^2}.

Therefore

\nabla\Phi = -\frac{6}{r^2}\widehat{\mathbf{r}} = \mathbf{F}.

The negative sign in the field is essential: it indicates that the field points toward the center, even though the potential $6/r$ is positive.

Polynomial Radial Magnitudes

If the radial magnitude is a polynomial, integrate each term as a power of $r$. Suppose $f(r)=\sum_{j=0}^{n}a_jr^j$, where the $a_j$ are constants. Each exponent $j$ is a nonnegative integer, so $j+1\ne0$.

G(r) = \sum_{j=0}^{n}\frac{a_j}{j+1}r^{j+1}.

To verify this expression, differentiate term by term: the derivative of $\frac{a_j}{j+1}r^{j+1}$ is $a_jr^j$, and the sum of those derivatives is $f(r)$. The theorem then turns this one-variable antiderivative into a potential for the vector field.

Worked Example: A Quadratic Polynomial Radial Magnitude

Find a potential for $\mathbf{F}(\mathbf{x})=(3r^2-4r+2)\widehat{\mathbf{r}}$ when $r>0$, and verify it in three dimensions with center $\mathbf{c}=\mathbf{0}$.

Integrate the radial magnitude one term at a time:

\begin{aligned} G(r) &=\int(3r^2-4r+2)\,dr\\ &=3\frac{r^3}{3}-4\frac{r^2}{2}+2r\\ &=r^3-2r^2+2r. \end{aligned}

Thus $\Phi(\mathbf{x})=r^3-2r^2+2r$, where $r=\sqrt{x^2+y^2+z^2}$. First differentiate with respect to $r$:

$$ G'(r)=3r^2-4r+2. $$

Since $\nabla r=(x,y,z)/r=\widehat{\mathbf{r}}$ for $r>0$, the gradient is

\nabla\Phi = (3r^2-4r+2)\frac{(x,y,z)}{r} = (3r^2-4r+2)\widehat{\mathbf{r}} = \mathbf{F}.

This check confirms both the radial magnitude and the direction of the field.

A Sign Error to Avoid

For an inward inverse-square field, it is easy to carry the negative sign incorrectly through the antiderivative. For example, given $\mathbf{F}=-6r^{-2}\widehat{\mathbf{r}}$, the incorrect choice $\Phi=-6/r$ gives

\frac{d}{dr}\left(-\frac{6}{r}\right) = 6r^{-2} = \frac{6}{r^2},

which produces an outward field, not the stated inward one. The correction is to integrate the radial magnitude itself: $\int -6r^{-2}\,dr=6/r+C$. Its derivative is $-6/r^2$, as required. Always differentiate the proposed potential before concluding that it matches the field.

When This Construction Applies

The construction depends on the field having the stated radial form on the domain under consideration. The magnitude must be a function of distance alone, and a differentiable antiderivative must exist on the relevant range of radii. For continuous $f$ on an interval, such an antiderivative exists there.

The center must be excluded if the field uses $\widehat{\mathbf{r}}$, since that unit vector is undefined at $r=0$. If $f$ also has singularities at particular positive radii, those radii must be excluded from the domain or handled on intervals that do not cross them. The potential formula applies on each such interval. Whether a field defined by some other expression can be extended to the center is a separate question; the radial-unit-vector formula itself does not define a field there.

Finding a potential matters because it gives a compact scalar description of a vector field and confirms that the field is a gradient on the specified domain. In this radial setting, the construction is especially direct: integrate the signed radial magnitude, keep track of the domain, and verify by differentiating.

Check Your Understanding

For each question, assume the domain excludes the center and any radii where the field is undefined.

For $\mathbf{F}=5r^{-2}\widehat{\mathbf{r}}$, find a potential and verify its derivative with respect to $r$.
What potential corresponds to the radial magnitude $f(r)=4/r$?
Find a potential for $\mathbf{F}=(2r+3)\widehat{\mathbf{r}}$.
Why does the formula $\nabla G(r)=G'(r)\widehat{\mathbf{r}}$ require $r>0$?
For $\mathbf{F}=-2r^{-3}\widehat{\mathbf{r}}$, determine whether $\Phi=1/r^2$ is a potential, and justify your answer by differentiation.