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I am currently an undergraduate taking a course on Newtonian mechanics. The lecturer defines a force to be conservative if there exists a scalar function (we call it potential function), say $V(x,y,z)$, such that the force $$\mathbf{F}=-\Big(\frac{\partial V}{\partial x}\mathbf{\hat i}+\frac{\partial V}{\partial y}\mathbf{\hat j}+\frac{\partial V}{\partial z}\mathbf{\hat k}\Big).$$ Then the lecturer told us that whenever you see an expression of $\mathbf{F}$ which depends on velocity $\mathbf{v}$, then this force must be non-conservative. Then I asked why, because I think in some scenario $\mathbf{v}$ can be expressed solely in terms of an object's position vector $\mathbf{r}$, but the response was not clear. Could anyone comment on the lecturer's statement and explain why?

I know there are similar questions on this community, but I has not found a rigorous proof to, say, how to show that $\mathbf{F}=A\mathbf{v}$, where $A$ is a non-zero constant, is not conservative? Thank you.

IncredibleSimon
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A force is not conservative if work depends on the path taken by the particles. Work is defined as

$$W = \int_A^B \vec{F}\cdot d\vec{\ell}$$

Now you see that if $\vec{F}$ depends on the velocity $\vec{v}$, you will not have the same integral for $W$ depending on the path you take.

You are stating that $\vec{v}$ can depend solely on $\vec{r}$. This is true, but the changes in velocity $\vec{v}$ of the particle moving along the path $A\to B$ is generated by the forces acting on it, such that it must be obtained from solving the equations of motions. For forces dependent in $\vec{v}$ it is thus impossible to express the velocity solely in term of $\vec{r}$ by solving the second law of dynamics ($m\dot{\vec{v}}=\vec{F}(\vec{v})$), such that you can never find a path independent expression for $W$. You can think of simple examples as motion with friction e.g. $\vec{F}=-k\vec{v}$. In that case, you will obtain $|\vec{v}| \propto e^{\frac{k}{m}t}$, such that $W$ will depend on the path (the longer the path, the greater the time $t$ and the greater the work!).

Léo Vacher
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    Thank you for your answer, but I still do not understand. Say $\mathbf{F}=A\mathbf{v}$ where $A$ is a non-zero constant, then how to show with rigor that $\mathbf{F}$ is not conservative? – IncredibleSimon Jan 29 '24 at 10:42
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    So I would say that you can take two approaches. Either you compute explicitly $W$ and you will obtain something like

    $$W = \int_A^B \vec{F}\cdot d \vec{\ell} = \int_A^B A \vec{v}\cdot d \vec{\ell} \propto \int_A^B A e^{\frac{k}{m}t} d\ell$$

    And here you showed that $W$ is proportional to the time to go from $A$ to $B$, proving that the force is non conservative because it depends on the path.

    Otherwise you can try to prove the impossibility to find a $V(x,y,z)$ such that $\vec{F}=-\vec{\nabla}V$. For sure examples of this can be found in textbooks.

     – Léo Vacher Jan 29 '24 at 10:52
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    If you want I can try to sketch (or find) a proof of the second type. – Léo Vacher Jan 29 '24 at 10:52
  • Thank you for the reply! I realize that I am lack of some knowledge of vector calculus, for which I only know some rules of differentiating dot and cross products, in order to understand your response. So, I am leaving this question for now, but I'll come back to it later. – IncredibleSimon Jan 29 '24 at 11:06
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    Of course no problem! First time in classical mechanics is always hard. Let me know if I can help later. – Léo Vacher Jan 29 '24 at 11:07
  • You are so kind. Thank you! I will start learning about vector calculus in two weeks. Hopefully, at the end of March, I would be able to understand your response. – IncredibleSimon Jan 29 '24 at 11:14