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I'm studying calculus, but since the example involved a physical concept. I will ask here:

This is how it goes:

This means that in a conservative force field, the amount of work required to move an object from point a to point b depends only on those points, not on the path taken between them.

But what does it mean? How does it not depend on the path?

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    $\begingroup$ Is there something the Wikipedia article doesn't cover about this? $\endgroup$ – ACuriousMind Jun 5 '15 at 19:39
  • $\begingroup$ Well, mathematically it's proven in my book, too, but somehow it didn't make an intuitive sense. But maybe math is all I need to rely on here. So then, in a conservative vector field, the object will always perform the maximum amount of work possible between 2 arbitrary points? Did I get that right? $\endgroup$ – Max Jun 5 '15 at 19:52
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    $\begingroup$ you do work, say to lift a weight, then, in a conservative field, say gravity with no friction, you get back, exactly, the work done in you having lifted the weight in the first place. not a real life common effect. the path in gravity only depends on vertical motion, horizontal motion does not come into it $\endgroup$ – user81619 Jun 5 '15 at 20:01
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If $\vec{F}$ is a conservative force field, then it satisfies the property $$ \tag{1} \vec{\nabla} \times \vec{F} = 0, $$ and it can be written as $$ \tag{2} \vec{F} = \vec{\nabla}V, $$ for a scalar function $V$ (which corresponds to potential function in physics).

Note that, when you put $(2)$ into $(1)$ it becomes a "curl of a gradient" and is automatically vanishes. You can derive this result by using simple mathematical knowledge.

In physics, work done on a particle by applying a force $\vec{F}$ along a path is defined as $$ \tag{3} W = \int_C \vec{F} \cdot d\vec{s}, $$ where $C$ is any path connecting two points in the space, call $A$ (initial point) and $B$ (end point). These are the points we start/finish applying the force. Regarding this, we can rewrite $(3)$ as $$ W = \int_A^B \vec{F} \cdot d\vec{s}. $$

Now, if $\vec{F}$ is conservative then we can also use property $(2)$, which gives \begin{align} W & = \int_A^B \vec{\nabla}V \cdot d\vec{s} \\ & = \int_A^B \frac{\partial}{\partial \vec{s}}V \cdot d\vec{s} \\ & = V(B) - V(A). \end{align}

Thus, with the given property that force field is conservative we find work done on a particle by exerting this force field only depends on the end points but not on the path we choose.


NOTE: Conventionally, in physics we write $(2)$ with a minus sign in front, $$ \vec{F} = -\vec{\nabla}V. $$ However, in the above text, I used it in the formal mathematical description without regarding any physical concerns.

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  • $\begingroup$ Thank you. Actually, the proof is given in my book, too. I guess I just didn't feel how could it be, but then...I should probably rely on math here. I'll choose your answer as the best, maybe it will help someone in the future. $\endgroup$ – Max Jun 5 '15 at 20:14
  • $\begingroup$ I just saw the comments above. As @Acid Jazz mentioned, teachers also usually give the same gravity example in lectures. If gravity pulls an object down from a height $h$, then independent of the path chosen, it will always exert the force $\vec{F} = m\vec{g}$ and as a result work done on the particle will always be $W = mgh$, again independent of the path. $\endgroup$ – sahin Jun 5 '15 at 20:23

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