3

Let C denote the path from $\alpha$ to $\beta$. If $\textbf{F}$ is a gradient vector, that is, there exists a differentiable function $f$ such that $$\nabla f=F,$$ then \begin{eqnarray*} \int_{C}\textbf{F}\; ds &=& \int_{\alpha}^{\beta} \textbf{F}(\vec{c}(t)).\vec{c'}(t)\; dt \\ &=&\int_{\alpha}^{\beta} \nabla f(\vec{c}(t)).\vec{c'}(t)\; dt \\ &=&\int_{\alpha}^{\beta} \frac{\partial f(\vec{c}(t))}{\partial t}\; dt \\ &=&f(\vec{c}(\beta))- f(\vec{c}( \alpha)) \end{eqnarray*} That is, the integral of $\textbf{F}$ over $C$ depends on the values of the end points $c(\beta)$ and $c(\alpha)$ and is thus independent of the path between them.

This proof is true if and only if $\textbf{F}$ is a gradient vector, what if not ?

Eman Sammour
  • 47
  • Are you asking why line integrals do not depend on parametrization even in the non-gradient case, or are you asking why line integrals of gradient fields are independent of path? What you ask in the title does not agree with what you ask in the body. – Santiago Canez May 07 '16 at 19:56
  • My ask in the title is why line integrals of gradient fields are independent of path.. and I think it has two cases one of them as i mentioned (If F is a gradient vector) and the other if not. @Santiago canez – Eman Sammour May 07 '16 at 20:02
  • No, what you ask in the tittle says nothing about path independence, but rather it asks about independence of parametrization, which is a different concept. Please edit your title to match your actual question. – Santiago Canez May 07 '16 at 20:04
  • you are right, sorry about that @Santiago Canez – Eman Sammour May 07 '16 at 20:12

2 Answers2

0

If F is not a gradient vector field, there is no potential function to evaluate at the end points and the penultimate and last steps fail. The proof that path is irrelevant when there is a potential function relies on the fundamental theorem for line integrals, which no longer applies if there is no primitive.

For the proof, refer to my answer here: Is path length somehow irrelevent in this question? Line integrals Calc 3

Community
  • 1
operatorerror
  • 29,103
0

In general, for a smooth vector field $\vec F(\vec r)$, Helmholtz's Theorem guarantess that there exists a scalar field $\Phi(\vec r)$ and a vector field $\vec A(\vec r)$ such that

$$\vec F(\vec r)=\nabla \Phi(\vec r)+\nabla \times \vec A(\vec r)$$

Then, forming the line integral along a path $C$, from $\vec r_1$ to $\vec r_2$, we see that

$$\begin{align} \int_C \vec F(\vec r)\cdot \,d\vec \ell&=\int_C\left(\nabla \Phi(\vec r)+\nabla \times \vec A(\vec r)\right)\cdot \,d\vec \ell\\\\ &=\Phi(\vec r_2)-\Phi(\vec r_1)+\int_C \nabla \times \vec A(\vec r) \cdot \,d\vec \ell \tag 1\\\\ \end{align}$$

Inasmuch as the integral on the right-hand side of $(1)$ is, in general, path dependent, then the path integral of $\vec F(\vec r)$ is also, in general, path dependent.

Mark Viola
  • 179,405