Radial Vector Fields from Magnitude Rules

From a Magnitude Rule to a Vector Field

A radial vector field is organized around a fixed center. At each point away from that center, its vector points along the line connecting the center to the point: either away from the center or toward it. The length of the vector can change with the point’s distance from the center.

This gives a direct way to build a field from two pieces of information: its center and a rule for how its magnitude depends on distance. The essential step is to convert the displacement from the center into a unit direction, then multiply by the required magnitude.

Radial Directions and Magnitudes

Work in two or three dimensions. Let $\mathbf{x}$ be a position vector and let $\mathbf{c}$ be a fixed center. The displacement from the center to the position is $\mathbf{x}-\mathbf{c}$, and its distance from the center is

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

When $r>0$, the unit vector pointing outward from the center is the displacement divided by its length. The inward unit vector points in the opposite direction.

\mathbf{e}_{\mathrm{out}}(\mathbf{x}) = \frac{\mathbf{x}-\mathbf{c}}{\|\mathbf{x}-\mathbf{c}\|} = \frac{\mathbf{x}-\mathbf{c}}{r}, \qquad \mathbf{e}_{\mathrm{in}}(\mathbf{x}) = -\frac{\mathbf{x}-\mathbf{c}}{r}.

Definition: A radial vector field centered at $\mathbf{c}$, with magnitude rule $m(r)$, has vectors parallel to $\mathbf{x}-\mathbf{c}$ at every point other than the center. Here $m(r)\geq 0$ specifies the vector length at distance $r$. The outward and inward formulas are $$ \mathbf{F}_{\mathrm{out}}(\mathbf{x}) = m(r)\frac{\mathbf{x}-\mathbf{c}}{r}, \qquad \mathbf{F}_{\mathrm{in}}(\mathbf{x}) = -m(r)\frac{\mathbf{x}-\mathbf{c}}{r}, \qquad r>0. $$

The condition $r>0$ matters: at the center, the expression $(\mathbf{x}-\mathbf{c})/r$ divides by zero, and there is no radial direction determined by the displacement. A field may be defined at the center separately if that makes sense for the chosen rule. For example, if $m(r)=kr$ with $k\geq0$, the formula away from the center simplifies to a linear expression, and assigning the zero vector at the center gives a well-defined field there.

Why the Formula Works

The construction follows from the meaning of a unit vector. For $r>0$, the outward unit vector has length one because

\left\|\frac{\mathbf{x}-\mathbf{c}}{r}\right\| = \frac{\|\mathbf{x}-\mathbf{c}\|}{r} = \frac{r}{r} = 1.

Multiplying this unit vector by the nonnegative number $m(r)$ preserves its direction and changes its length to $m(r)$. More explicitly, the length of the outward field is

\|\mathbf{F}_{\mathrm{out}}(\mathbf{x})\| = \left\|m(r)\frac{\mathbf{x}-\mathbf{c}}{r}\right\| = m(r)\frac{\|\mathbf{x}-\mathbf{c}\|}{r} = m(r).

The inward formula has the same length because multiplying a vector by $-1$ reverses its direction without changing its length. This verifies both requirements: the field points radially in the selected direction, and its magnitude follows the specified rule.

Worked Examples

Worked Example: An Outward Field with Magnitude Proportional to Distance

Find and evaluate the outward field centered at $\mathbf{c}=(1,-1)$ whose magnitude at distance $r$ is $m(r)=2r$.

Write the position as $\mathbf{x}=(x,y)$. The displacement and distance from the center are

\mathbf{x}-\mathbf{c}=(x-1,y+1), \qquad r=\sqrt{(x-1)^2+(y+1)^2}.

Insert the magnitude rule into the outward formula. For $r>0$,

\mathbf{F}_{\mathrm{out}}(x,y) = 2r\frac{(x-1,y+1)}{r} = (2(x-1),2(y+1)).

At the center, this simplified expression equals $(0,0)$, so define the field there by $\mathbf{F}_{\mathrm{out}}(1,-1)=(0,0)$. Now evaluate at $(4,3)$. The displacement is $(3,4)$, and $r=\sqrt{3^2+4^2}=5$. The magnitude is $m(5)=10$, and the outward unit direction is $(3/5,4/5)$. Thus

\mathbf{F}_{\mathrm{out}}(4,3) = 10\left(\frac35,\frac45\right) = (6,8).

Check: $\|(6,8)\|=\sqrt{36+64}=10$, the prescribed magnitude, and $(6,8)$ points in the same direction as the displacement $(3,4)$. At $(-1,-1)$, the displacement is $(-2,0)$, so $r=2$, $m(2)=4$, and the outward unit direction is $(-1,0)$. Therefore $\mathbf{F}_{\mathrm{out}}(-1,-1)=(-4,0)$, whose length is $4$ and whose direction is outward.

For a sketch, place the center at $(1,-1)$. Draw arrows directly away from it. Since the magnitude is $2r$, arrows farther from the center should be longer in direct proportion to their distance; the arrow at the center is zero.

Worked Example: An Outward Field with Inverse-Distance Magnitude

Find the outward field centered at the origin with magnitude $m(r)=6/r$, and evaluate it at $(3,4)$ and $(-1,0)$.

For $r=\sqrt{x^2+y^2}>0$, the radial formula gives

\mathbf{F}_{\mathrm{out}}(x,y) = \frac{6}{r}\frac{(x,y)}{r} = \frac{6}{r^2}(x,y) = \left(\frac{6x}{x^2+y^2},\frac{6y}{x^2+y^2}\right).

The field is not defined at the origin: its magnitude rule $6/r$ is itself undefined there. At $(3,4)$, $r=5$, the magnitude is $6/5$, and the outward unit direction is $(3/5,4/5)$. Hence

\mathbf{F}_{\mathrm{out}}(3,4) = \frac65\left(\frac35,\frac45\right) = \left(\frac{18}{25},\frac{24}{25}\right).

Check: the vector is $(6/25)(3,4)$, so it points along the displacement. Its length is $\sqrt{18^2+24^2}/25=\sqrt{900}/25=6/5$, as required. At $(-1,0)$, $r=1$, the magnitude is $6$, and the outward unit direction is $(-1,0)$. Therefore $\mathbf{F}_{\mathrm{out}}(-1,0)=(-6,0)$; its length is $6$ and it points away from the origin.

In a sketch, arrows point away from the origin but become shorter as distance increases, because $6/r$ decreases as $r$ increases. The formula does not assign an arrow at the origin.

Worked Example: An Inward Field in Three Dimensions

Find an inward field centered at $\mathbf{c}=(0,0,1)$ whose magnitude is $m(r)=r^2$. Evaluate it at $(2,1,1)$ and $(0,0,3)$.

For $\mathbf{x}=(x,y,z)$, the displacement is $(x,y,z-1)$ and $r=\sqrt{x^2+y^2+(z-1)^2}$. The inward formula is

\mathbf{F}_{\mathrm{in}}(\mathbf{x}) = -r^2\frac{\mathbf{x}-\mathbf{c}}{r} = -r(\mathbf{x}-\mathbf{c}), \qquad r>0.

At $(2,1,1)$, the displacement is $(2,1,0)$, so $r=\sqrt{2^2+1^2}=\sqrt5$ and the magnitude is $5$. The inward unit direction is $(-2/\sqrt5,-1/\sqrt5,0)$. Thus

\mathbf{F}_{\mathrm{in}}(2,1,1) = 5\left(-\frac{2}{\sqrt5},-\frac{1}{\sqrt5},0\right) = (-2\sqrt5,-\sqrt5,0).

Check: its length is $\sqrt{20+5}=5$, and it points opposite to the displacement $(2,1,0)$, toward the center. At $(0,0,3)$, the displacement is $(0,0,2)$, so $r=2$ and $m(2)=4$. The inward unit direction is $(0,0,-1)$, giving $\mathbf{F}_{\mathrm{in}}(0,0,3)=(0,0,-4)$. Its length is $4$, and its direction is toward $(0,0,1)$.

Since $m(r)=r^2$, the simplified formula $-r(\mathbf{x}-\mathbf{c})$ approaches the zero vector as $\mathbf{x}$ approaches the center. Define the field at the center to be zero. A sketch has arrows directed toward $(0,0,1)$, with lengths increasing as distance from the center increases.

A Reliable Method for Sketching

A radial sketch should communicate both where arrows point and how their lengths vary. Use the same process in two or three dimensions:

Mark the center.
Locate $\mathbf{c}$, the point from which distances are measured.

Choose sample points.
Use points in several directions and at different distances from the center.

Set each arrow’s direction.
Point away from the center for an outward field and toward it for an inward field.

Compare arrow lengths.
Compute $m(r)$ at each sample distance and use those values to represent relative lengths.

Check the center.
Do not draw an arrow there unless the field has been defined at the center.

For example, if $m(r)=kr$ with $k>0$, doubling the distance doubles the arrow length. If $m(r)=k/r$ with $k>0$, doubling the distance halves the arrow length. These comparisons follow from the stated magnitude rules; they do not change the radial direction.

Common Mistake: Forgetting to Normalize

The displacement $\mathbf{x}-\mathbf{c}$ points outward, but its length is $r$, not $1$. Multiplying it directly by the desired magnitude usually gives the wrong arrow length. For the inverse-distance example, a mistaken formula would be

\mathbf{G}(x,y)=\frac{6}{r}(x,y).

At $(3,4)$, where $r=5$, this gives $\mathbf{G}(3,4)=(18/5,24/5)$. Its length is $\sqrt{(18/5)^2+(24/5)^2}=6$, but the prescribed magnitude is $6/5$. The error is that $(3,4)$ has length $5$, so it is not a unit direction. Normalize first:

\frac{(3,4)}{5} = \left(\frac35,\frac45\right), \qquad \frac65\left(\frac35,\frac45\right) = \left(\frac{18}{25},\frac{24}{25}\right).

In formula form, the coefficient multiplying $\mathbf{x}-\mathbf{c}$ must be $m(r)/r$, not $m(r)$. The extra division by $r$ ensures that the displacement is first converted to a unit vector.

Why Radial Magnitude Rules Matter

These fields provide compact models for quantities that point directly toward or away from a distinguished location. The formula separates the geometric choice of direction from the rule governing strength: the center determines the radial direction, while $m(r)$ determines the arrow length at distance $r$. That separation makes it possible to create and sketch many different fields without changing the basic construction.

Always check three details before using a formula: the chosen inward or outward sign, the normalization by $r$, and whether the center is included in the domain. Together, they prevent the most common direction, magnitude, and undefined-point errors.

Check Your Understanding

For each question, identify the center, direction, magnitude rule, and any issue at the center where relevant.

Write the outward radial field centered at $(2,0)$ with magnitude $m(r)=3$, for points away from the center.
For an inward field with magnitude $m(r)=r$, what is the vector at distance $4$ in the direction from the center to the point?
Why must the displacement vector be divided by its length before multiplying by a prescribed magnitude?
For an outward field with magnitude $m(r)=5/r$, does the formula define a vector at the center? Explain.
For a field with $m(r)=2r$, how should arrow lengths at distances $2$ and $6$ compare?