0
$\begingroup$

For the method of finding a potential function of a conservative vector field, usually I would let $\nabla \phi$ be equal to the vector field, and do some integration (indefinite) and differentiating whilst equating it to other components of $\nabla \phi$. However, in this youtube video (at 11:47), the technique was to do a definite integral twice, over some random path. Where could I learn this method and why does this work?

The question is to find the potential scalar field for the conservative vector field:
$$F = \begin{pmatrix} x \\ y \end{pmatrix} r^{n},\qquad \text{with $r = \sqrt{x^2 + y^2}$}$$ The method in the video found the potential vector field by evaluating the following integral:
$$\int_{1}^{y_1} (1+y_1 ^2)^{\frac{n}{2}} y_1\, \mathrm{d}y + \int_{1}^{x_1} (x_1 + y_1)^{\frac{n}{2}} x\, \mathrm{d}x.$$

$\endgroup$
  • 1
    $\begingroup$ Perhaps summarize the content of the video here, to make this question more self-contained. $\endgroup$ – Travis Willse Jun 1 '17 at 12:26
  • $\begingroup$ Edited into the OP. $\endgroup$ – Twenty-six colours Jun 1 '17 at 13:01
  • 1
    $\begingroup$ I assume one $y_1$ should be a $y$ (inside the integral). If you know that the vector field is conservative, the integration is path independent (for proper paths...). That means you can just choose the simplest possible paths. $\endgroup$ – Gesbesgue Jun 1 '17 at 13:18
  • $\begingroup$ Ah because it only depends on the endpoints? So them choosing that specific path was arbitrary (so we could've, say, picked a path that had four segments to it instead of this example where it had two)? $\endgroup$ – Twenty-six colours Jun 1 '17 at 13:54
  • $\begingroup$ The path isn’t so much “random” as it is chosen to make the integrals easy to compute. $\endgroup$ – amd Jun 1 '17 at 23:42
1
$\begingroup$

Glossing over some fine points, the underlying idea is as follows: Choose a point $\mathbf x_0$ in the domain of the conservative vector field $\mathbf F$ and define $\phi:\mathbf x\mapsto\int_{\Gamma(\mathbf x)}\mathbf F$, where $\Gamma(\mathbf x)$ is a piecewise smooth path from $\mathbf x_0$ to $\mathbf x$. The value of hits integral is independent of the choice of $\Gamma(\mathbf x)$, so $\phi$ is well-defined. It’s a fairly straightforward exercise to prove that $\mathbf F=\nabla\phi$. The scalar potential $\phi$ is determined up to a constant; choosing a different point for $\mathbf x_0$ amounts to changing this constant of integration.

In practice, one often chooses the origin for $\mathbf x_0$ and a straight line segment with the obvious parameterization for $\Gamma$ so that the integral is evaluated from $0$ to $1$, but you can choose any convenient path from the fixed starting point to $\mathbf x$. It would appear that in this video the integral is taken along the path that consists of the line segment from $(1,1)$ to $(1,y_1)$ and the line segment from there to $(x_1,y_1)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.