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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.

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    $\begingroup$ Does this answer your question? Why can't conservative forces depend on velocity? $\endgroup$
    – Tobias Fünke
    Commented Jan 29 at 10:02
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    $\begingroup$ There are many more questions here. Please use the search function, click on related links etc. $\endgroup$
    – Tobias Fünke
    Commented Jan 29 at 10:58
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    $\begingroup$ You should point out where all other threads have not addressed your query. I don't see that this is the case, tbh. Also, how do you define conservative, to start with? If you want a proof, you have to state precise definitions. $\endgroup$
    – Tobias Fünke
    Commented Jan 29 at 11:07
  • $\begingroup$ Possible duplicate: Why can't we ascribe a (possibly velocity dependent) potential to a dissipative force? $\endgroup$
    – Qmechanic ♦
    Commented Jan 29 at 11:08
  • $\begingroup$ If the force depends on the velocity, it is not a function of $x$ alone. It is rather a function(al) of a trajectory: $t\to x(t)$. $\endgroup$
    – Tobias Fünke
    Commented Jan 29 at 11:09

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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!).

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    $\begingroup$ 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? $\endgroup$
    – IncredibleSimon
    Commented Jan 29 at 10:42
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    $\begingroup$ 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. $\endgroup$
    – Léo Vacher
    Commented Jan 29 at 10:52
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    $\begingroup$ If you want I can try to sketch (or find) a proof of the second type. $\endgroup$
    – Léo Vacher
    Commented Jan 29 at 10:52
  • $\begingroup$ 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. $\endgroup$
    – IncredibleSimon
    Commented Jan 29 at 11:06
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    $\begingroup$ Of course no problem! First time in classical mechanics is always hard. Let me know if I can help later. $\endgroup$
    – Léo Vacher
    Commented Jan 29 at 11:07

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