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In this website, it was written that

If a force acting on an object is a function of position only, it is said to be a conservative force.

Why is this so? According to wikipedia, a force is conservative if it meets any of the following conditions:

  1. $\nabla \times \vec{F}=0$
  2. $\vec{F} = -\nabla V(\vec{r})$
  3. $W = \oint \vec{F} \cdot d\vec{s} = 0$

How can one derive that a force that only depends on position is conservative from these three conditions?

Community
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TaeNyFan
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2 Answers2

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That statement simply isn't true. It's easy to construct an example of a non-conservative force which depends only on position, e.g.

$$\mathbf F(x,y) = \pmatrix{-y\\x}$$

In one dimension, any force which depends only on position is automatically conservative, but for higher dimensions this is not true. One can see why by trying to construct a potential energy function explicitly.

Consider a 2D system subject to a force $$\mathbf F = \pmatrix{-x\\-y}$$ We're looking for some $U(x,y)$ such that $\mathbf F = -\nabla U$, i.e. $$\pmatrix{-x\\-y} = \pmatrix{-\frac{\partial U}{\partial x} \\ -\frac{\partial U}{\partial y}}$$ From the first component, we have that $\frac{\partial U}{\partial x} = x$; taking the antiderivative, it follows that $U(x,y) = \frac{1}{2}x^2 + C(y)$ where $C$ is an arbitrary function which may depend on $y$ but not $x$.

From the next component, we have that $$\frac{\partial U}{\partial y} = C'(y) = y$$ which implies that $C(y) = \frac{1}{2}y^2 + C_0$ where $C_0$ is a constant. Therefore, any potential energy function $U(x,y)= \frac{1}{2}x^2 + \frac{1}{2}y^2 + C$ produces the given force, and we can say that $\mathbf F$ is conservative.

The reason that this doesn't always work is that, while we can always find an antiderivative for the first component, we may run into inconsistencies in the others. The non-conservative force I wrote first is an example; if we try to apply the same procedure to that one, the first component will give us that $U(x,y) = xy + C(y)$, but the second will give us that $x + C'(y) = -x \implies C'(y)=-2x$. But since $C(y)$ cannot depend on $x$, this is inconsistent. There is no $U(x,y)$ such that $-\nabla U = \pmatrix{-y \\ x}$, and so the force is not conservative.

J. Murray
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I'm sure that is because in that website they are only considering one dimensional forces, so that if $\vec{F}=F(x)\hat{x}$ then $\nabla\times\vec{F}=0$. In general that is not true, for example the magnetic field created by stationary currents only depends on position, but it is not conservative!

Urb
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  • Hmm I think you may be wrong. $ \nabla $x $\vec{F} = 0$ is a general statement. For your case of magnetism, a stationary current does in fact produce a conservative force https://physics.stackexchange.com/questions/118498/is-magnetic-force-non-conservative. – baker_man May 19 '20 at 16:14
  • Also, an easy application of Stoke's theorem shows 1 implies three – baker_man May 19 '20 at 16:18
  • @Nick but the magnetic force depends on velocity and position. – JEB May 19 '20 at 16:21
  • @JEB have a read at the link. It shows that for the Lorentz force $ \nabla \times \vec{F} = 0$ if the magnetic field is non-varying. Can also think of it as the component of the Lorentz force from the magnetic field is perpendicular to the velocity, so does no work (hence conservative). – baker_man May 19 '20 at 16:37
  • @Nick Look that when I put the example of the magnetic field I am talking about the field itself. My point is that for a general field $\vec{F}$ (in my example $\vec{F}=\vec{B}$), dependent only on position, like the one created by a linear current in the $z$ direction, which goes as $1/\rho$, the field is not necessarily conservative. – Urb May 19 '20 at 17:42
  • Seems like a confusing thing to talk about considering she's talking about forces, not general vector fields. They're different things – baker_man May 19 '20 at 18:53
  • Isn't a force field a vector field like any other? – Urb May 19 '20 at 19:18
  • A conservative force is defined as the work done in moving a particle is independent of the path taken. How can you talk about work done if it's just a general vector field? – baker_man May 19 '20 at 19:31
  • A conservative vector field is something very well defined in mathematics and there is no need to talk in physical terms (a particle, work, etc). What is the work done by a force anyway? It's the circulation of the force along a given path (the integral of the tangential component to the path). A conservative vector field is one in which the circulation of the field along any closed path is $0$, or alternatively one in which the circulation between any two points depends only on the points and not on the path. – Urb May 19 '20 at 19:51
  • There is a need to talk about it in physical terms because the force generated by a field isn't the same as the field itself. For the magnetic field, the force is the Lorentz force. So if you want to talk about the magnetic field being a conservative force, what you need to talk about is the lorentz force field. If you want to talk about the magnetic field being a conservative field, then you can just talk about the mathematical definition you just gave. If you want to continue the discussion, I suggest moving to private message as it's gotten quite long – baker_man May 19 '20 at 20:03