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In Mathematical Methods of Classical Mechanics Arnold defines a central field as a vector field $\mathbf{F}\colon \mathbb{R}^n \to \mathbb{R}^n$ which is invariant under the group of galilean transformations which fix the origin (if I understand him correctly). From this he concludes that $\mathbf{F}(\mathbf{r}) = \Phi(r) \mathbf{e}_r$, where $r = \lVert \mathbf{r} \rVert,$ for some scalar function $\Phi\colon \mathbb{R}_{\geq 0} \to \mathbb{R}$.

This is fine. Next, he shows that any central field is conservative by showing that the line integral between two points $P_1$ and $P_2$ is independent of path. To this end he simply writes:

$$ \int_{P_1}^{P_2} \mathbf{F}(\mathbf{r}) \cdot \mathrm{d}\mathbf{S} = \int_{r(P_1)}^{r(P_2)} \Phi(r) \, \mathrm{d}r. $$

(Presumably $r(P)$ is the radial coordinate of the point $P$.) But I have a problem with this equality. Sure, it "follows" easily from calculating $\mathbf{F} \cdot \mathrm{d}\mathbf{S}$, but I have never seen any justification for this other than the usual physical one. I was wondering if there is any way to write out the definition of the line integral and show the above equality that way? I have tried choosing an arbitrary path and parametrisation, but I don't end up with anything constructive.

Incidentally, if there is any rigorous justification for simply calculating $\mathbf{F} \cdot \mathrm{d}\mathbf{S}$, I would love to hear that as well.

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We can prove that ${\bf F}$ is conservative by exhibiting a potential of ${\bf F}$. Assume $${\bf F}({\bf r})=\phi(r)\>{{\bf r}\over r}\ ,$$ choose an $R>0$, and put $$\Phi(r):=\int_R^r\phi(s)\>ds\ .$$ I claim that $$f({\bf r}):=\Phi\bigl(\|{\bf r}\|\bigr)\qquad({\bf r}\in\dot{\mathbb R}^n)$$ is a potential of ${\bf F}$.

Proof. We have to show that $\nabla f={\bf F}$. To this end compute $$\bigl(\nabla f({\bf r})\bigr)_k={\partial f\over\partial x_k}({\bf r})=\Phi'(r){\partial r\over\partial x_k}=\phi(r){x_k\over r}=\bigl({\bf F}({\bf r})\bigr)_k\ .$$

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Not sure if can get to the core of your question, but I'll place a few key points here:

For any vector field $f:\mathbb{R}^3 \rightarrow \mathbb{R}^3$, the integral over a curve $\gamma$ is denoted by: $$ \int_\gamma f(\gamma)d\gamma $$ Which is something I also dislike.

Assuming it is a Parametrized curve $\gamma:\mathbb{R}\rightarrow \mathbb{R}^3 $ , then: $$ \gamma=\gamma(t) $$ $$ d\gamma =\gamma'(t)dt $$

Such that the previous integral becomes: $$ \int_\gamma f(\gamma)d\gamma = \int_{\gamma (t_1)}^{\gamma (t_2)} f(\gamma(t)). \gamma'(t) dt $$

Where the dot product was made explicit. Noting that: $$ g(t) = f(\gamma(t)). \gamma'(t) $$

We have $g:\mathbb{R}\rightarrow \mathbb{R}$. Assume there exists $F:\mathbb{R}^3\rightarrow \mathbb{R}$ such that: $$ \nabla F(x,y,z) = f(x,y,z) $$ Then: $$ \frac{dF(\gamma(t))}{dt} = \nabla F(\gamma(t)) . \gamma '(t) = f(\gamma(t)) . \gamma '(t) $$

So if there exists $G(t)$ such that $G'(t)=g(t)$, then $G(t) = F(\gamma(t)) $:

$$ \int_\gamma f(\gamma)d\gamma = \int_{\gamma (t_1)}^{\gamma (t_2)} f(\gamma(t)). \gamma'(t) dt = G(t_2)-G(t_1) = F(\gamma(t_2))-F(\gamma(t_1)) $$

So the line integral only depends on the points $\gamma (t_1)$ and $\gamma (t_2)$, not on which curve $\gamma$ was chosen, hence the path independence.

In the case the book presents $\gamma(r)=r\textbf{e}_r$ such that $\gamma'(r)=r'\textbf{e}_r+r\textbf{e}'_r=\textbf{e}_r+r\textbf{e}'_r$ ($r'=1$), thus:

$$ F(\textbf{r}) d\textbf{S} = (\Phi(r)\textbf{e}_r).(\textbf{e}_r+r\textbf{e}'_r) dr $$

But since $\textbf{e}_r$ is a unit vector, $\textbf{e}_r . \textbf{e}'_r = 0$. Therefore:

$$ F(\textbf{r}) d\textbf{S} = \Phi(r) dr $$

Does this help?

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  • $\begingroup$ Not really, I don't think. First of all, how do you see that "the book presents $\gamma(r) = r \mathbf{e}_r$"? And we're trying to show path independence, so I don't see how we can choose a specific path. Secondly, manipulating differentials like you are doing is what I wanted to avoid, especially since I have no idea what a differential is (again, apart from the physical interpretation "very small change in"), so I am not comfortable with that. I can write down the line element in spherical coordinates and "derive" the result that way, but I don't know what a line element is either. $\endgroup$ – Danny Hansen Mar 28 '18 at 16:43
  • $\begingroup$ I claimed the the book presented $\gamma (r) = \textbf{e}_r$ from experience. I recall the physics books I've used to employ some sloppy notation tricks like these. I thought you are okay with path independance. I'll edit the aswer to prove that. $\endgroup$ – Mefitico Mar 28 '18 at 16:49
  • $\begingroup$ Ah sorry, I see how my post wasn't clear on that point. I've clarified it now. But I'm still not sure I follow. In assuming that there exists an $F$ such that $\nabla F = f$, aren't you already assuming that $f$ is conservative? Which is equivalent to path independence. $\endgroup$ – Danny Hansen Mar 28 '18 at 17:48
  • $\begingroup$ I didn't prove that the central force would be conservative, I've tried to explain the notation. But note that $\nabla (h°g)=h'(\nabla g)$. So it suffices to integrate $\Phi(r)$ to find the potential, as is done in the answer by @Christian Blatter. $\endgroup$ – Mefitico Mar 28 '18 at 17:58

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