How to prove that a force is conservative?

[dyn]

I realized there's a better way to demonstrate the issue here. A field is conservative if the energy used to move something is independent of the path.

They have only told you a single path that a particle is taken. How can you possibly know what will happen if a particle takes a different path. Let's not worry about how people normally confirm something is conservative, you only have one path, how can you know what the result would be if you moved in a different path?

If a spring force is given by F = -kx is that force conservative ? We only know the force in one direction.

 

[Steve4Physics]

If a spring force is given by F = -kx is that force conservative ? We only know the force in one direction.

Hi. Hope no one minds if I chip in.

Yes. 'F=-kx' describes a conservative force.

The simple test is to show that no net work is done moving from some arbitrary extension, x=a, to a different arbitrary extension, x= b, and then returning from x=b to x=a. I.e. show the work done going round any arbitrary 'loop' is zero.

Edted: In physical terms this means (for b>a say) that the work done stretching an ideal spring equals the increase in its (elastic) potential energy. And when restored, this same amount of energy is 'returned' by the spring doing work.

If you really wanted to use ‘curl’ then you could take $\vec F = -kx\hat{i} + 0\hat{j} + 0\hat{k}$ and work out $curl(\vec F)$ in the usual way. (But then you don't get the bonus of deriving the expression for potential energy.)

 

[jbriggs444]

If a spring force is given by F = -kx is that force conservative ? We only know the force in one direction.

If we are working in a one-dimensional space where the spring is aligned then, since the force is aligned in that direction, there would seem to be no problem.

We know the force everywhere within the [sub-]space and the force falls within the [sub-]space. It is unambiguously conservative.

By contrast, for a craft in an orbit subject to a radial force, the force does not fall within the [sub-]space and we have a problem. Do we expand the sub-space to handle the force and cry "Not Necessarily Conservative"? Or do we project the force into the sub-space, wind up with a zero force and cry "A Zero Force is always Conservative".

 

[Office_Shredder]

If a spring force is given by F = -kx is that force conservative ? We only know the force in one direction.

As a serious question, do you know what the definition of a path is for the purposes of something being path independent? Because this is not really a counterpoint to what I said.

 

[dyn]

I have been told that force in #1 is not conservative because we only know the force on the circle which can be considered 1-dimensional and not anywhere else.
Yet i am also told that the spring force is conservative despite only knowing the force in 1-dimension and not anywhere else.
I fail to see the difference between the 2 cases

 

[jbriggs444]

I have been told that force in #1 is not conservative because we only know the force on the circle which can be considered 1-dimensional and not anywhere else.
Yet i am also told that the spring force is conservative despite only knowing the force in 1-dimension and not anywhere else.
I fail to see the difference between the 2 cases

The difference that I see is whether the force falls within the same sub-space as the motion. How can you have an inner product space if you can't figure out what space you are talking about?

 

[Orodruin]

I have been told that force in #1 is not conservative because we only know the force on the circle which can be considered 1-dimensional and not anywhere else.
Yet i am also told that the spring force is conservative despite only knowing the force in 1-dimension and not anywhere else.
I fail to see the difference between the 2 cases

You have been told the difference repeatedly. However, you seem to be ignoring this.

In the first case you are looking at the entire force field in three dimensions and a single possible solution to the motion. There is nothing a priori telling you that the motion is one-dimensional. Any particle in three dimensions will by definition move along what is a one-dimensional path.

In the second case, all possible solutions are one-dimensional and typically constrained by other forces to move in one dimension only. (Note: In the constrained case, curl is not an applicable concept.)

 

[dyn]

The university exam paper that i took the initial case from regarding the motion in a circle says that the force is conservative. If i accept that then i can also understand why the spring force is conservative,
Thank you to everyone who replied

 

[Orodruin]

The university exam paper that i took the initial case from regarding the motion in a circle says that the force is conservative. If i accept that then i can also understand why the spring force is conservative,
Thank you to everyone who replied

Again, the original question is ill posed. You should not pay too much attention to it. Being in a university exam paper is no guarantee that it is correct.

 

[Orodruin]

The model solution writes the components of F such as Fx = -mω2x.

To be more specific, the solution quotes the force field $\vec F_1 = - mω^2 \vec x$. However, all we can conclude from the problem statement is that this force field coincides with the actual force field along a circle of radius $R$. The actual force field may have any other behaviour away from this circle. We just don’t know. An example would be
$$
\vec F_2 = - mω^2 \vec x + (R^2 - x^2 - y^2) \vec k \times \vec x
$$
where $\vec k$ is a constant vector. The curl of this field is non-zero. We now have two fields, $\vec F_1$ and $\vec F_2$ that would both have the given circular path as a possible solution for the particle motion. One of them is conservative and the other not. It is therefore clear that it cannot be concluded with the given information whether the field is conservative or not.

 

[Office_Shredder]

I have been told that force in #1 is not conservative because we only know the force on the circle which can be considered 1-dimensional and not anywhere else.

You don't know the force on the circle. You know the force on the circle conditional on the particle moving in the way it moves. That's what I mean by you only know one path. If another particle moves another way (e.g. at a different speed around the circle), then you don't know what the force is going to be.

If the force is being generated by gravity, then the particle will move off the path of the circle. If the force is being generated by normal force from walls that constrain you to be inside the circle, then the particle may continue to be in the circle.