recently I have studied conservative forces and non-conservative forces in halliday book and while doing some exercise I saw some questions asking for proving if a force is conservative so after doing some research I found out that there's an mathematical operator called curl wich is used to prove conservativeness of a general force. but my math knowledge was not enough to use that so I just didn't solve those problems. but in some problems in my books I saw the solution was proving a force being conservative by comparing only to different paths. I wonder if that is a scientific solution for that? can't we find a counterexample?


2 Answers 2


$\int \nabla × F \cdot da= \int F \cdot dr$

Where dr is a closed curve and da is the area enclosed by that curve. This is called Stokes theorem.

If $\nabla × F$ is zero then the closed line integral about any path must be zero.

This proves that your field is Conservative as,

For a Conservative field by definition the line integral is only dependant on the ends of the path and not the path inbetween. we also know that the line integral from a to b is the negative line integral from b to a , so... the conclusion is, is that any path taken from a to b, and then back from b to a would be 0 ( as they are same magnitude but negative)

We can also say that If $F = \nabla V $then it is also Conservative as $ \nabla × ( \nabla V) $ = 0 and hence fits our Proof of the curl being zero

There is also a separate proof if it can be written as the gradient of a scalar function then its Conservative, by performing a line integral about a general path , the line integral reduces to v(b)-v(a) which is only dependant on the points a,b and not the path inbetween


Given $ F = \nabla V(x,y,z)$ is integrated about the path $$R(t) = x(t) I + y(t) j + z(t) k$$ $$R(t0)= a , R(t1)= b$$

Then its line integral from A to B is

$$\int_{a}^{b} \nabla V(x,y,z) \cdot dr$$

$$dr/dt = R'(t)$$ $$dr = R'(t)dt$$

$\int_{t0}^{t1} \nabla V(R(t)) \cdot R'(t)dt$

$V(R(t))$ as we want to change V to be evaluated along my path at some specific time t

Substituting the definition of $\nabla V$: $$\nabla V = (\partial V/ \partial x) i + (\partial V /\partial y) j + (\partial V /\partial z) k$$

and $R'(t) = x'(t) I + y'(t)j + z'(t) k$

into our integral, taking the dot product between these vectors gives

$$\int_{t0}^{t1}[ (\partial V/ \partial x) x'(t) + (\partial V /\partial y) y'(t) + (\partial V /\partial z) z'(t) ] dt$$

Notice now this is an integration between some complex expression and dt where the bounds are t

Recall: if I have a function $ f(x(t))$ then to differentiate this with respect to time I would use the chain rule$ (df/dx)(dx/dt)$

This is EXTREMELY similar to what I have above in my complex expression. Using the multivariable chain rule instead

$$(\partial V/ \partial x) x'(t) + (\partial V /\partial y) y'(t) + (\partial V /\partial z) z'(t) ]$$

Is just simply $dV/dt$ where V is a function of$ x(t),y(t),z(t)$

So, our integral is just

$$\int_{t0}^{t1} (dV/dt) dt$$

Which is just simply a nice integration, V(r(t)) Evaluated at T1 and t0

=$$ V(R(t1)) - V(R(t0))$$ $$V(b) - V(a)$$

What does this mean? Well, this proves that my field is Conservative as my line integral is only dependant on the starting and end positions, (a and b), and not the path inbetween

Thus if, $F = \nabla V$ Then my field is Conservative

$$ or \nabla × (\nabla V ) = 0$$ $$\nabla×F = 0$$

  • $\begingroup$ thanks. I've newly learned integrals. so I read about stokes theorem in a bunch of websites and also I read your answer several times. So this is what I concluded: if we can prove that W in two general paths then we can say that the force is conservative. is that right? $\endgroup$
    – infinite
    Nov 21, 2021 at 6:04
  • $\begingroup$ Let's say I want to compute a line integral from point A to point B, now... for a random vector field F, the line integral would depend on the path that I take. By definition a Conservative vector field is path independent l. meaning , for Conservative vector fields, any path I take from point A to point B the line integral is say "W" . now, consider that I now want to compute the line integral from B to A aka the reverse path. Clearly for a Conservative field , any lath I path from B to A should he the same. there is a property of line integrals that the line integral from A to B about some- $\endgroup$ Nov 21, 2021 at 12:06
  • $\begingroup$ specific path is the negative of the line integral from B to A. thus for Conservative fields, if I were to compute the line integral about ANY path from A to B, and then add on the line integral about ANY path from B to A then I would be doing "W +(-W)" = 0, thus for ANY closed loop line integral from (A ->b )+ (b-> A) = A->A would.be zero if my.field is Conservative, which is exactly the condition imposed via the curl = 0 forcing my vector field to have this property $\endgroup$ Nov 21, 2021 at 12:11
  • $\begingroup$ And in essence yes If you can prove that from A to B about ANY path, is the same as the negative from B to A about any pathm then your field is Conservative ( Note this lath isn't some specific path, you have to prove it using a generalised path) $\endgroup$ Nov 21, 2021 at 12:12
  • $\begingroup$ look at my edit, I prove it more rigourously $\endgroup$ Nov 21, 2021 at 12:42

A little more information.

In basic physics tests, a conservative force is defined as one where the work done by the force in moving from point a to b is independent of the path taken in going from a to b. For a conservative force, the change in potential energy is defined as the negative of the work done by the force. The use of potential energy instead of evaluating the work for the conservative force simplifies many problems. For example, for the gravitational force near the earth the change in potential energy is simply $mgh$ where $m$ is mass $g$ is the acceleration of gravity and $h$ is the change in height, regardless of how complicated the actual physical path is in changing the height. $mgh$ is easy to evaluate and equals the negative of the work done by gravity $-\int_{r_1}^{r_2} \vec F_g \cdot d\vec r$ which is not always easy to evaluate. $\vec F_g$ is the force of gravity and $\vec r$ describes the path taken in changing the height by $h$.

In classical mechanics texts it is shown that a conservative force has zero curl, as @jensen paull says. Also, it is shown that a potential energy function $V$ can be defined for a conservative force, and $\vec F = -\nabla V$. See the text Mechanics by Symon for details.


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