I get that a force dependent on path is non-conservative or a force which does non-zero work in completing a cycle. But how do you prove that mathematically.

Let's assume I have a force F= 2yi + x^2j. Now how do I conclude that it's work done is path dependent?

My attempt Work done=F.dr

(2yi + x^2j).(dxi + dyj)= 2y.dx + x^2dy

Now what do I integrate it to? Let's assume they take 2 paths to reach (10,10) 1.(0,10) to (10,10) 2. (10,0) to (10,10). Now if i get different work done, they are non-conservative. But again, how do I integrate it? Is there a way to tell it's non-conservative without all this?

  • $\begingroup$ Yes. Given that your field is continuously differentiable, if its curl is zero then it's conservative. $\endgroup$ – Omar Nagib Mar 24 '16 at 19:38

You need to use Stoke's theorem. $$\oint_{C} \mathbf{F} \cdot d\mathbf{r} = \int_S (\nabla \times \mathbf{F}) \cdot \mathbf{\hat{n}} dS$$ Where $\nabla \times \mathbf{F}$ is equal to $$ ({\partial F_z \over \partial y} - {\partial F_y \over \partial z}) \mathbf{\hat{x}} + ({\partial F_x \over \partial z} - {\partial F_z \over \partial x}) \mathbf{\hat{y}} + ({\partial F_y \over \partial x} - {\partial F_x \over \partial y}) \mathbf{\hat{z}}$$ For the line integral of the force to vanish on every closed path, its curl ($\nabla \times \mathbf{F}$) must be zero everywhere, too.

Calculate the curl for the force given. If it is zero everywhere, your force is a conservative one.

  • $\begingroup$ Ie there any other way? A year ago, our teacher asked us to integrate the force and further I don't remember nor do I have notes. Do you know about that method? Also how to find work done here? $\endgroup$ – jatin Mar 24 '16 at 20:24

If the force is generated by a potential, $F=-\nabla\Phi$, then its curl has to vanish, since $\nabla\times(\nabla\Phi)=0$. You can check the curl of your force field. Remember that the curl of a vector field at a point, according to its definition, is proportional to the line integral of the field along an infinitesimal loop around such point.

  • $\begingroup$ Can you explain by writing it down? $\endgroup$ – jatin Mar 24 '16 at 20:25
  • $\begingroup$ Consider a vector field $\mathbf{F}$ and an infinitesimal loop $C$ along which you want to integrate $F$. The loop $C$ encloses a region $S$ with area $A$ and unit normal $\hat{\mathbf{n}}$. This line integral is $\oint_C\mathbf{F}\cdot d\mathbf{s}$. As you shrink the loop, both the line integral and the area $A$ go to zero, however, the ratio does not. This ratio is precisely the curl, which is defined as $(\nabla\times\mathbf{F})\cdot\hat{\mathbf{n}}\equiv\frac{1}{A}\oint_C\mathbf{F}\cdot d\mathbf{s}$. This is what erenust just wrote. $\endgroup$ – Pedro Aguilar Mar 25 '16 at 19:21

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.