# Interpreting path independent line integrals in terms of work done

I understand that integrating a force $\boldsymbol{F}$ along a curve $C$ represents the work done by that force. I am, however, struggling to interpret this in terms of path independent line integrals.

Say my force (conservative vector field) moves my particle from $(0,0)$ to $(1,1)$ along the curve $y=x$, for example. Say the same force moves my particle from $(0,0)$ to $(0,5)$ along $x=0$ then from $(0,5)$ to $(1,5)$ along $y=1$ then from $(1,5)$ to $(1,1)$ along $x=1$. We know that the result of the work done calculation will be the same since the vector field is conservative. This seems odd to me as the particle has 'moved' a greater distance in the second example.

Could anyone help to explain this?

I think I understand how a closed loop integral relates to a double integral over the area enclosed by the curve (Green's theorem), but I might yet have (potentially stupid) questions about this as well.

Any help would be great - thanks!

## 2 Answers

I always think of the example where the conservative field is the earth's gravitational field. Yes, the particle moved a greater distance, but in a certain part of its path it didn't have to exert energy to move, it actually gained energy (as it was "falling" down) so to speak, and then mathematically (actually physically as well), this comes down to the same amount of work for both paths, since the work was positive for some parts of the path, but negative for other parts, which then equal out to get the same amount of work no matter which path you choose.

This of course relates to the fact that for a conservative field you can find a potential for the field, so that you can invoke the fundamental theorem of calculus for gradients to see your result.

See, this is the magic of conservative fields. They do not depend on the path travelled, in this case what is "distance".They depend on the initial and final positions only.

For a prominent result, we can say that any conservative field $F$ can be written as $$\vec F=\nabla \phi$$

Now if you move a particle from the point $P_1$ to $P_2$, the work done is as follows: $$W=\oint_C \vec F \cdot d\vec r$$ $$=\int_{P_1}^{P_2} \nabla \phi \cdot d\vec r$$ $$=\int_{P_1}^{P_2} \left(\frac{\partial \phi}{\partial x}\hat i+\frac{\partial \phi}{\partial y}\hat j+\frac{\partial \phi}{\partial z}\hat k\right) \cdot \left(\hat i dx+\hat j dy+\hat k dz\right)$$ $$=\int_{P_1}^{P_2} \left(\frac{\partial \phi}{\partial x}dx+\frac{\partial \phi}{\partial y}dy+\frac{\partial \phi}{\partial z}dz \right)$$ $$=\int_{P_1}^{P_2} d\phi$$ $$=\phi(P_2)-\phi(P_1)$$

So the line integral is independent of what $C$ is, it only depends on $P_1$ and $P_2$.