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I am currently taking multivariable calculus and we have just gone over line integrals over conservative vector fields. A vector field F is called conservative if it may be generated from some scalar potential function $\phi$ by taking its gradient $$\nabla \phi$$ and satisfies $$\nabla \times \mathbf{F}=0$$ The algebraic proof of path independence in such a field which invokes the Fundamental Theorem and Clairaut's theorem makes perfect sense to me. However I am having difficulty visualizing this from a geometric standpoint. Textbooks on electromagnetism will usually justify this path independence by saying that the components of the work $\mathbf{F} \cdot d\mathbf{r}$ will cancel each other as we integrate along the closed path.

This is easy to visualize for a radially directed conservative vector field (like the electric field). But for a non-radial conservative vector field, is there some sort of graphical simulation or visualization which really allows one to see such path independence? Again, just to clarify, I am not confused by path independence, I just would like to see a graphical, geometric justification.

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The geometric idea is:

enter image description here (Source: Wikipedia)

where $F$ should be defined in the interior region of the loop.

For two paths $S1$ and $S2$, starting at $1$ and ending at $2$ but otherwise arbitrary, we see that they form a closed curve $C$, if we reverse the direction along one of the paths.

We also know from parametrization of the line integral, that integration along the reverse path of $S$, which we write as $-S$, we get the integral value along $S$, but with reversed sign. \begin{align} 0 &= \int\limits_C F(x) \cdot dx \\ &= \int\limits_{S1 + (-S2)} F(x) \cdot dx \\ &= \int\limits_{S1} F(x) \cdot dx + \int\limits_{-S2} F(x) \cdot dx \\ &= \int\limits_{S1} F(x) \cdot dx - \int\limits_{S2} F(x) \cdot dx \end{align} which is equivalent to $$ \int\limits_{S1} F(x) \cdot dx = \int\limits_{S2} F(x) \cdot dx $$ Update:

Note that by Stokes's theorem $$ \DeclareMathOperator{curl}{curl} \int\limits_S \curl F \cdot dS = \int\limits_{\partial S} F \cdot dx $$ so vanishing curl implies vanishing integrals over closed curves.

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    $\begingroup$ Thank you for your reply, but is there a method which does not take $\int_C \mathbf{F} \cdot d\mathbf{r}=0$ as an a priori? $\endgroup$ – nguzman Apr 29 '17 at 21:14
  • $\begingroup$ The vanishing circulations are the result of vanishing curl. $\endgroup$ – mvw Apr 29 '17 at 21:39

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