Potentials on Disconnected Domains

Why Disconnected Domains Need Care

A potential function describes a vector field through its gradient. On a connected domain, two potentials for the same field can differ only by one constant. If the domain is disconnected, there may be a separate constant on each component.

This is more than a bookkeeping detail. The endpoint formula for a line integral applies along a curve that lies where the potential is defined. A continuous curve contained in a disconnected domain cannot travel from one component to another. Thus the formula compares the endpoints of an actual curve within one component; it does not create a comparison between unrelated points in different components.

Key idea: A vector field can have a potential on each connected component, with an independent additive constant on each. The endpoint theorem applies to a curve contained in a component where the corresponding potential is defined.

Potentials on a Disconnected Open Set

Let $D$ be an open subset of $\mathbb{R}^n$. A connected component of $D$ is a maximal connected subset of $D$. The components are disjoint, and together they make up $D$. For open subsets of Euclidean space, each component is itself open.

A scalar function $\phi$ is a potential for a vector field $\mathbf{F}$ on an open set if $\phi$ is continuously differentiable there and its gradient equals the field at every point. On a disconnected set, this condition is checked on every component. The function $\phi$ may be specified by a different formula on each one.

Definition: A potential for $\mathbf{F}$ on $D$ is a function $\phi:D\to\mathbb{R}$ that is continuously differentiable on $D$ and satisfies $\nabla\phi=\mathbf{F}$ at every point of $D$. Equivalently, for each connected component $D_\alpha$, the restriction of $\phi$ to $D_\alpha$ is a potential for the restriction of $\mathbf{F}$ to $D_\alpha$.

“Continuously differentiable on $D$” is a local condition: it requires continuous first derivatives near each point of $D$. It does not require points in different components to be joined by paths, or values of $\phi$ on those components to fit together across a gap outside $D$.

In particular, specifying a potential on each component gives a potential on all of $D$. The additive constant chosen on one component has no effect on the gradient on another component.

Why the Constants Are Independent

On a connected open set, if $\phi$ and $\psi$ are both potentials for the same field, then $\nabla(\phi-\psi)=\mathbf{0}$. A connected open subset of Euclidean space is path-connected; along any piecewise smooth path in that set, the chain rule shows that $\phi-\psi$ does not change. It is therefore constant on that component.

For a disconnected set, this argument applies separately on each component. It gives no reason for the constants on different components to agree.

\nabla(\phi-\psi)=\mathbf{0} \quad\Longrightarrow\quad \phi-\psi=C_\alpha \text{ on each component }D_\alpha.

The constants $C_\alpha$ may differ from one component to another. Conversely, adding a constant separately on each component leaves the gradient unchanged everywhere in $D$. This explains precisely what “independent additive constants” means.

Worked Example: One Field, Two Independent Constants

Let $D=\{(x,y):x<0\}\cup\{(x,y):x>0\}$, the union of the left and right open half-planes. Define $\mathbf{F}(x,y)=(2x,2y)$ throughout $D$. Consider the function

\phi(x,y)= \begin{cases} x^2+y^2+2, & x<0,\\ x^2+y^2-3, & x>0. \end{cases}

On the left component, $\nabla\phi=(2x,2y)=\mathbf{F}$. On the right component, the derivative of the constant $-3$ is zero, so again $\nabla\phi=(2x,2y)=\mathbf{F}$. Therefore $\phi$ is a potential on $D$.

For a direct endpoint check, take the curve $\mathbf{r}(t)=(1+t,t)$, $0\le t\le1$. It lies in the right half-plane because its $x$-coordinate is at least $1$. Its derivative is $\mathbf{r}'(t)=(1,1)$, and the field along it is $\mathbf{F}(\mathbf{r}(t))=(2+2t,2t)$. Thus

\int_C \mathbf{F}\cdot d\mathbf{r} =\int_0^1 (2+2t,2t)\cdot(1,1)\,dt =\int_0^1 (2+4t)\,dt =[2t+2t^2]_0^1 =4.

The starting point is $(1,0)$, where $\phi(1,0)=1-3=-2$. The endpoint is $(2,1)$, where $\phi(2,1)=4+1-3=2$. The endpoint difference is $2-(-2)=4$, which agrees with the direct integral. The constant $-3$ cancels because both endpoints are in the same component.

The Endpoint Theorem on a Component

The fundamental theorem for line integrals says that if $\mathbf{F}=\nabla\phi$ on an open set containing a piecewise smooth curve $C$, then the line integral depends only on the endpoints of the curve. This statement does not require the entire domain to be connected. It requires a potential along the curve.

Theorem: Let $D\subseteq\mathbb{R}^n$ be open, let $\mathbf{F}=\nabla\phi$ on $D$, and let $C$ be a piecewise smooth curve in $D$ from $\mathbf{a}$ to $\mathbf{b}$. Then $$ \int_C \mathbf{F}\cdot d\mathbf{r}=\phi(\mathbf{b})-\phi(\mathbf{a}). $$ In particular, the theorem applies when $C$ lies within any one component of $D$ on which a potential is defined.

Here is the reason the theorem holds. On a smooth segment parameterized by $\mathbf{r}(t)$, the chain rule gives $\frac{d}{dt}\phi(\mathbf{r}(t))=\nabla\phi(\mathbf{r}(t))\cdot\mathbf{r}'(t)$. Since $\nabla\phi=\mathbf{F}$, integrating this identity gives the potential’s change along that segment. For a piecewise smooth curve, apply the identity to each smooth segment and add; the intermediate endpoint values cancel.

Parameterize a smooth segment.
Write it as $\mathbf{r}(t)$, for $a\le t\le b$.

Use the chain rule.
$\frac{d}{dt}\phi(\mathbf{r}(t))=\nabla\phi(\mathbf{r}(t))\cdot\mathbf{r}'(t)=\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)$.

Integrate and add segments.
The integral on each segment is the change in $\phi$; summing the changes gives $\phi(\mathbf{b})-\phi(\mathbf{a})$.

Every continuous curve has connected image. A connected subset of $D$ must lie within a single connected component. Consequently, a curve in $D$ cannot have one endpoint in one component and the other endpoint in another. This fact explains both the scope and the limit of the endpoint theorem in a disconnected domain.

Worked Example: Endpoint Calculations in Two Components

Let $D$ be the union of the open disks of radius $1$ centered at $(-2,0)$ and $(2,0)$. Define the field by $\mathbf{F}(x,y)=(2x,2y)$ in the left disk and $\mathbf{F}(x,y)=(2x,2y+3)$ in the right disk. Consider

\phi(x,y)= \begin{cases} x^2+y^2+11, & (x,y)\text{ in the left disk},\\ x^2+y^2+3y-4, & (x,y)\text{ in the right disk}. \end{cases}

On the left, $\nabla\phi=(2x,2y)$; on the right, $\nabla\phi=(2x,2y+3)$. Thus the displayed function is a potential for the stated field on both components.

First take the vertical curve $\mathbf{r}_L(t)=(-2,t)$, $0\le t\le\frac12$, in the left disk. It stays inside the disk because its squared distance from the center is $t^2\le\frac14<1$. Here $\mathbf{r}'_L(t)=(0,1)$, so

\int_{C_L}\mathbf{F}\cdot d\mathbf{r} =\int_0^{1/2}( -4,2t)\cdot(0,1)\,dt =\int_0^{1/2}2t\,dt =[t^2]_0^{1/2} =\frac14.

The potential values are $\phi(-2,0)=4+11=15$ and $\phi(-2,\frac12)=4+\frac14+11=\frac{61}{4}$. Their difference is $\frac{61}{4}-15=\frac14$, agreeing with the integral.

Now take $\mathbf{r}_R(t)=(2,t)$, $0\le t\le\frac12$, in the right disk. It also stays inside its disk. Since $\mathbf{F}(\mathbf{r}_R(t))=(4,2t+3)$,

\int_{C_R}\mathbf{F}\cdot d\mathbf{r} =\int_0^{1/2}(4,2t+3)\cdot(0,1)\,dt =\int_0^{1/2}(2t+3)\,dt =[t^2+3t]_0^{1/2} =\frac14+\frac32 =\frac74.

The endpoint values are $\phi(2,0)=4-4=0$ and $\phi(2,\frac12)=4+\frac14+\frac32-4=\frac74$. Their difference is $\frac74$, again matching the integral. Each calculation uses the potential on the component containing its curve.

What Cannot Be Inferred Across Components

If two points lie in different components, there is no curve in $D$ joining them. Therefore there is no line integral along a curve in $D$ for which those points are the endpoints. One must not substitute them into a potential and interpret the difference as a line integral in the domain.

Worked Example: A Cross-Component Comparison Has No Path

Take $D=\{(x,y):x<0\}\cup\{(x,y):x>0\}$ and let $\mathbf{F}=\mathbf{0}$ on $D$. Define

\phi(x,y)= \begin{cases} 2, & x<0,\\ -3, & x>0. \end{cases}

Both pieces have zero gradient, so $\nabla\phi=\mathbf{0}=\mathbf{F}$ throughout $D$. For example, the potential values at $(-1,0)$ and $(1,0)$ differ by $-3-2=-5$. This does not contradict the endpoint theorem: there is no continuous curve in $D$ from $(-1,0)$ to $(1,0)$. Any continuous path between them in the plane would have to pass through a point with $x=0$, which is excluded from $D$. The value difference is simply a comparison of two independently chosen component constants, not the integral along a path in $D$.

If instead the potential were defined as $0$ on both components, its gradient would still be $\mathbf{F}$. The cross-component difference would then be $0$. This confirms that such a comparison is not determined by the field through a path in $D$.

A Common Mistake

A tempting but incorrect shortcut is to assume that potentials on a disconnected domain must differ by one global constant. The zero-field example shows why this fails: the two valid potentials $0$ on both components and $2$ on the left and $-3$ on the right have different offsets on different components. Their gradients agree everywhere on $D$, even though their difference is not one constant on all of $D$.

The correction is to compare potentials component by component. On any one connected component, two potentials for the same field differ by a constant. Across distinct components, those constants are independent. Similarly, use the endpoint formula only when the curve lies in a component where the potential is defined.

Why This Distinction Matters

Component-wise reasoning prevents two errors: claiming a single additive constant when the domain does not support that conclusion, and applying an endpoint formula to points that cannot be joined by a curve in the domain. It also keeps the line integral calculation practical. Once a potential is known on the component containing a curve, the integral can be evaluated from endpoint values there, without needing to compare formulas on other components.

Takeaway: A potential on a disconnected domain is a potential on each component, assembled into one function on the whole domain. The endpoint theorem applies to every piecewise smooth curve in the domain, and each such curve stays in one component.

Check Your Understanding

Use the definitions and component-wise reasoning above to answer these questions.

Can the additive constants for two potentials of the same field differ on separate connected components? Explain.
Why must a continuous curve in a disconnected domain lie entirely within one connected component?
What hypotheses allow the endpoint formula to be used for a piecewise smooth curve?
For the zero field on the union of the two open half-planes, is the function equal to $4$ on the left and $9$ on the right a potential?
Why does a potential-value difference between points in different components not, by itself, give a line integral in the domain?