Conservative forces and Variation

Question

I am currently studying "Classical mechanics by Goldstein" and have just started. The book introduced something simple. For a conservative force, the work done in taking a mass from one position to other does not depend on path taken. I know it's a very obvious statement but I wanted to become a bad physics student and wanted to complicate the problem. I rather asked the question: Can I prove this statement with the help of Variation? By substituting $ \delta W =0$, I wanted to know the condition for it.
Here is how I proceeded, $$ \delta W_{12} =\delta \int_1^2 \vec{F}\cdot d\vec{s}. $$ I won't like to tell how it went (TBH it was something useless). I hope that you can help me out with it. It will be really helpful. (I know I should have shown my own math but it is just that I didn't wanted to make the question to long and mostly unreadable.)

Answer 1

mathematically rigorous:

If a differential form F is conservative, its differential vanisehs: Be $U,\ V$ vector spaces and $F:U\rightarrow V$ a differential form. The vectorfield F is conservative is equivalent to $$ \mathrm{d}F=0 $$ The integral of this differential form is interpreted pyhsically as energy: $$ W= \int F $$ independent of coordinates. If the differential form $F$ - thus $\mathrm{d}F=0$ - is conservative this translates to $$ \delta W=\int\mathrm{d}F=0 $$ This is interpreted in such a way that in case of a conservative force the energy only depends on the starting point and finishing point but not the path inbetween.

General idea:

A variation of the integral $$ W=\int_{a}^{b}\vec{F}\mathrm{d}\vec{s} $$ is obtained by taking the difference of the integral between two paths: $$ \delta W=\int_{\gamma(0)}^{\gamma(1)}\vec{F}\cdot\mathrm{d}\vec{s}-\int_{c(0)}^{c(1)}\vec{F}\cdot\mathrm{d}\vec{s} $$ If the work done between two points does not depend on the path between the two points, you would go on and calculate the difference between the energies of both paths. Then you would use that your force is conservative to show that this difference vanishes and thus the energy is path independent:

Chosing two appropriate paths:

So you take any two curves, say parametrised: $$ c: [0,1]\rightarrow[a,b]\ \mathrm{with }\ c(0)=a\ \mathrm{and}\ c(1)=b;\\ \gamma:[0,1]\rightarrow[a,b]\ \mathrm{with }\ \gamma(0)=a\ \mathrm{ and }\ \gamma(1)= b$$

$$\delta E= \delta \int_{a}^{b}\vec{F}\cdot\mathrm{d}\vec{s}\\ E_{c}-E_{\gamma}=\int_{c(0)}^{c(1)}\vec{F}\cdot\mathrm{d}\vec{s}-\int_{\gamma(0)}^{\gamma(1)}\vec{F}\cdot\mathrm{d}\vec{s}$$

Now you need to observe that the paths along the two integrals are different so in teh general case of an arbitrary force they would not vanish. However, as for the two curves $c(t)$ and $\gamma(t)$ the starting and finishing points are equal, the difference of the curves describes a closed loop.

Transforming into closed loop integral:

As you have stated that your force is conservative this will be useful. Another statement of conservation of a force is that its curl vanishes.

$$E_{c}-E_{\gamma}={\int_{a}^b}_{c} \vec{F}\cdot \mathrm{d}\vec{s}-{\int_{a}^{b}}_{\gamma}\vec{F}\cdot\mathrm{d}\vec{s}=\oint\vec{F}\cdot\mathrm{d}\vec{s}$$

Invoking Stokes theorem:

Now we can use Stokes theorem: The integral of a vectorfield along a closed perimeter of a surface equals the flux (the scalar product of the curl of the force with the surface normal) through said surface. $$ \oint_{\partial S}\vec{F}\mathrm{d}\vec{s}=\int_{S}curl(\vec{F})\mathrm{d}\vec{A} $$ where $\mathrm{d}\vec{A}$ describes the surface integral.

Obtaining the independence of energy for conservative force from chosen path:

As the curl vanishes because the force is conservative $$ \int_{S} curl(\vec{F})\mathrm{d}\vec{A}=\int_{S} 0\mathrm{d}\vec{A}=0 $$

you see that finally: $$ E_{c}-E_{\gamma}=0 $$ and thus the energy is independent of the path taken to calculate the energy as long as starting and finishing point are the same for all chosen paths.