1
$\begingroup$

I need to check if a this vector field is irrotational and conservative:

$F=\langle e^{x \cos y} \cos y,-x e^{x \cos y}\sin{y} \rangle$

curl of F should be => $\dfrac{dQ}{dx} - \dfrac{dP}{y}=0$ (where 'd' = partial derivative).

With some calculation, I found out that the partials are equal :

$-e^{x \cos y}\sin y[1+x \cos y]$

But here is my confusion, does that mean that the vector field is irratational? or conservative? My doubt arises from the fact that the definition of curl I know is only for 3 dimensional vector field.

$\endgroup$
  • $\begingroup$ Your field is the gradient of $e^{x \cos y}$, confirming it is conservative $\endgroup$ – kmm Jan 14 at 14:26
  • $\begingroup$ yes, but how do I check if it is irrotational ? $\endgroup$ – NPLS Jan 14 at 14:27
1
$\begingroup$

You are right that the usual notion of curl is only defined for 3 dimensions. To prove a two-dimensional vector field is conservative, you can use Green's theorem which says that the following identity holds for a region $D$ bounded by a curve $C$

$$ \oint (P\,dx+Q\,dy)=\iint _{D}\left({\frac {\partial Q}{\partial x}}-{\frac {\partial P}{\partial y}}\right)\,dx\,dy$$

A conservative vector field is a vector field which can be expressed as the gradient of a scalar function. Remember that a conservative vector field has the property that the result of a line integral is path independent, implying the line integral over a simple closed path is always zero (there are some subtleties involving non-simply connected spaces). Hence the left integral will be zero if and only if the vector field $\vec{F} = P \hat{e}_x + Q \hat{e}_y $ is conservative. So to check if a field is conservative, we can simply check if the integrand of integral on the right hand side is zero. This expression is the two-dimensional analog of curl.

$$ curl \vec{F} = \frac {\partial Q}{\partial x}-\frac {\partial P}{\partial y} $$

This is the formula you wrote down. In two-dimensions, the curl is a scalar, not a vector field itself. To get a completely general description of the concept of curl, you need differential forms, but this a relatively complex matter.

Another way to see it is that your expression is the z-component of the 3-dimensional curl for a vector field which does not depend on the z-direction.

$\endgroup$
  • $\begingroup$ So what I can say about the irrotationality of my vector field? Does that mean that conservativeness implies irrotationality? (in the domain of definition) $\endgroup$ – NPLS Jan 15 at 8:18
  • $\begingroup$ It's irrotational if this two-dimensional analog of the curl vanishes. A conservative field will always be irrotational. To see this, consider a gradient $\vec{F} = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})$ . The 2D curl will be $ \frac{\partial }{\partial x}\frac{\partial f}{\partial y} - \frac{\partial }{\partial y}\frac{\partial f}{\partial x} $ which will always be zero as partial derivatives commute. So conservativeness implies irrotationality $\endgroup$ – kmm Jan 15 at 23:10

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.