1
$\begingroup$

Consider a vector field, tangent to a curve. As a simple example, take the curve

$x(p)=\sin(p)$

$y(p)=\cos(p)$

$z(p)=p$

and its tangent vector field $\mathbf{A}(p)=\left(\begin{array}{c} \cos(p)\\ -\sin(p)\\ 1\\ \end{array}\right)$

Since this vector field lies on a spatial curve, does it have spatial derivatives? What is its curl?

What are the conditions for a vector field such as this one to have spatial derivatives? How does one express the spatial derivatives? Thank you.

$\endgroup$
6
  • $\begingroup$ For this case ${\bf r} = (x,y,z)$, ${\bf A} = (y, -x, 1)$, from this you can take the curl $\endgroup$
    – caverac
    Apr 12, 2017 at 18:06
  • $\begingroup$ But also $\mathbf{A}(p)=\left(\begin{array}{c} y(p)\\ -\sqrt{1-y^2(p)}\\ 1\\ \end{array}\right)$, or perhaps $\mathbf{A}(p)=\left(\begin{array}{c} \sqrt{1-x^2(p)}\\ -x(p)\\ 1\\ \end{array}\right)$, it depends how one writes $\mathbf{A}$ $\endgroup$
    – user142523
    Apr 12, 2017 at 18:47
  • $\begingroup$ The "tangent vector field" is a red herring. Let $f$ be a function whose domain is the $x$ axis in $\mathbb{R}^2$. What is $\partial_y f$? (Answer: you don't know, because there are many different ways to extend $f$ to a function on $\mathbb{R}^2$.) The situation is exactly the same here. $\endgroup$
    – Willie Wong
    Apr 12, 2017 at 18:49
  • $\begingroup$ Can the opposite be done? Can we find a curve, tangent to a given vector field $\mathbf{B}(x,y,z)$ $\endgroup$
    – user142523
    Apr 12, 2017 at 20:06
  • 1
    $\begingroup$ en.wikipedia.org/wiki/Integral_curve $\endgroup$
    – Willie Wong
    Apr 12, 2017 at 20:24

0

Reset to default

You must log in to answer this question.