Independence of Path Theorem

# Independence of Path Theorem

Recall from the Independence of Path page that if $\mathbf{F}$ is a continuous vector field on the domain $D$, then the line integral $\int_C \mathbf{F} \cdot \: dr$ is independent of path if for any two curves $C_1$ and $C_2$ in $D$ whose initial and terminal points coincide, we have that $\int_{C_1} \mathbf{F} \cdot \: dr = \int_{C_2} \mathbf{F} \cdot \: dr$.

The following theorem is critically important and relates independence of path to conservative vector fields.

Theorem 1: Suppose that $D$ is an open and connected domain, and suppose that $\mathbf{F}$ is a smooth vector field on $D$. Then the following statements are equivalent:a) The line integral $\int_C \mathbf{F} \cdot d \vec{r}$ has the same value for all piecewise smooth curves $C$ in $D$ whose initial point is $P_0$ and whose terminal point is $P_1$ for every $P_0, P_1 \in D$.b) $\mathbf{F}$ is a conservative vector field on $D$c) For every piecewise smooth closed curve $C$ in $D$ we have $\oint_C \mathbf{F} \cdot d \vec{r} = 0$. |

We will prove Theorem 1 in the case where $\mathbf{F}$ is a vector field on $\mathbb{R}^3$.

**Proof $a) \implies b)$**Suppose that the line integral $\int_C \mathbf{F} \cdot d \vec{r}$ has the same value for all piecewise smooth curves $C$ in $D$ whose initial point is $P_0$ and whose terminal point is $P_1$ for every $P_0, P_1 \in D$.

- Let $\mathbf{F} = P(x, y, z) \vec{i} + Q(x, y, z) \vec{j} + R(x, y, z) \vec{k}$. Let $P_0$ be the fixed point $(x_0, y_0, z_0)$ in $D$ and let $P_1$ be a point $(x, y, z)$ in $D$, and for a piecewise smooth curve $C$ in $D$ whose initial point is $P_0$ and terminal point is $P_1$, define the function $\phi$ as:

\begin{align} \quad \phi (x, y, z) = \int_C \mathbf{F} \cdot d \vec{r} \end{align}

- We want to show that $\mathbf{F} = \nabla \phi$ which by definition would mean that $\mathbf{F}$ is a conservative vector field (i.e, $P(x, y, z) = \frac{\partial \phi}{\partial x}$, $Q(x, y, z) = \frac{\partial \phi}{\partial y}$, and $R(x, y, z) = \frac{\partial \phi}{\partial z}$). We will only show that $P(x, y, z) = \frac{\partial \phi}{\partial x}$ and the other equalities are proven similarly

- Since $D$ is an open domain, there exists a ball $B$ centered at $P$ and contained in $D$. Select a point $P'$ with coordinates $(x_1, y, z)$ in $B$ such that $x_1 < x$. The line from $P'$ to $P$ is parallel to the $x$-axis (since $P'(x_1, y, z)$ and $P(x, y, z)$ has the same $y$ and $z$ coordinates).

- Now since $\int_C \mathbf{F} \cdot d \vec{r}$ has the same value for all piecewise smooth curves in $D$ whose initial and terminal points are the same, then choose the curve $C$ from $P_0$ to $P'$ to $P_1$. Then we have that

\begin{align} \quad \phi (x, y, z) = \int_{C_1} \mathbf{F} \cdot d \vec{r} + \int_{C_2} \mathbf{F} \cdot d \vec{r} \end{align}

- Now we take the partial derivative of both sides of this equation with respect to $x$. The first integral does not depend on the variable $x$ since $C_1$ is the path from $P_0(x_0, y_0, z_0)$ to $P'(x_1, y, z)$ and so partial differentiating this line integral with respect to $x$ is zero.

\begin{align} \quad \frac{\partial \phi}{\partial x} = \frac{\partial}{\partial x} \left ( \int_{C_1} \mathbf{F} \cdot d \vec{r} + \int_{C_2} \mathbf{F} \cdot d \vec{r} \right ) \\ \quad \frac{\partial \phi}{\partial x} = \underbrace{\frac{\partial}{\partial x} \left ( \int_{C_1} \mathbf{F} \cdot d \vec{r} \right)}_{=0} + \frac{\partial}{\partial x} \left ( \int_{C_2} \mathbf{F} \cdot d \vec{r} \right ) \\ \quad \frac{\partial \phi}{\partial x} = \frac{\partial}{\partial x} \left ( \int_{C_2} \mathbf{F} \cdot d \vec{r} \right ) \end{align}

- Now the straight line $C_2$ from $P'(x_1, y, z)$ to $P_1(x, y, z)$ can be parameterized as $\vec{r}(t) = (x(t), y(t), z(t)) = (t, y, z)$ for $x_1 ≤ t ≤ x$ and $\frac{d \vec{r}}{dt} = \vec{i}$ so $d \vec{r} = \vec{i} \: dt$ and so by the Fundamental Theorem of Calculus we have that:

\begin{align} \quad \frac{\partial \phi}{\partial x} = \frac{\partial}{\partial x} \left ( \int_{C_2} \mathbf{F} \cdot d \vec{r} \right ) = \frac{\partial}{\partial x} \int_{x_1}^x \left ( P(t, y, z), Q(t, y, z), R(t, y, z) \right) \cdot \vec{i} \: dt = \frac{\partial}{\partial x} \int_{x_1}^x P(t, y, z) \: dt = P(x, y, z) \end{align}

- Therefore $P(x, y, z) = \frac{\partial \phi}{\partial x}$, and if we repeated this in process, we'd see that $Q(x, y, z) = \frac{\partial \phi}{\partial y}$ and $R(x, y, z) = \frac{\partial \phi}{\partial z}$, so $\mathbf{F} = \nabla \phi$ and $\mathbf{F}$ is a conservative vector field.

**Proof $b) \implies c)$:**Suppose that $\mathbf{F}$ is a conservative vector field in $D$. Then $\mathbf{F} = \nabla \phi$ for some potential function $\phi$. Thus we have that:

\begin{align} \quad \mathbf{F} \cdot d \vec{r} = \nabla \phi \cdot d \vec{r} = \left ( \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z} \right ) \cdot (dx \vec{i} , dy \vec{j} , dz \vec{k}) \\ \quad \mathbf{F} \cdot d \vec{r} = \frac{\partial \phi}{\partial x} dx + \frac{\partial \phi}{\partial y} dy + \frac{\partial \phi}{\partial z} dz = d \phi \end{align}

- Let $C$ be a piecewise smooth closed curve in $D$ and let $C$ be parameterized by $\vec{r}(t)$ for $a ≤ t ≤ b$. Since $C$ is closed, we have that $\vec{r}(a) = \vec{r}(b)$ and so:

\begin{align} \quad \int_C \mathbf{F} \cdot d \vec{r} = \int_a^b \frac{d}{dt} \left ( \phi(\vec{r}(t)) \right ) \: dt = \phi(\vec{r}(b)) - \phi(\vec{r}(a)) = 0 \end{align}

**Proof $c) \implies a)$:**Suppose that for every piecewise smooth closed curve $C$ we have that $\oint_C \mathbf{F} \cdot d \vec{r} = 0$. Let $P_0$ and $P_1$ be two points in $D$ and let the curves $C_1$ and $C_2$ be in $D$, both of whose initial point is $P_0$ and whose terminal point is $P_1$. Define $C = C_1 - C_2$ to be the curve from $P_0$ to $P_1$ along $C_1$ and then from $P_1$ to $P_0$ along $C_2$. Clearly this curve $C$ is in $D$ and is piecewise smooth (since $C_1$ and $C_2$ are piecewise smooth) and closed (since its initial point is the same as its terminal point). Thus:

\begin{align} \quad 0 = \oint_C \mathbf{F} \cdot d \vec{r} = \int_{C_1} \mathbf{F} \cdot d \vec{r} - \int_{C_2} \mathbf{F} \cdot d \vec{r} \\ \int_{C_1} \mathbf{F} \cdot d \vec{r} = \int_{C_2} \mathbf{F} \cdot d \vec{r} \end{align}

- Thus our proof is complete. $\blacksquare$