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A force field $F_i(x)$ is conservative if for every curve $C$ from a point $y_1$ to a point $y_2$, we have $\int\limits_C F_i(x)\mathrm{d}x^i$, so that the energy difference between $y_1$ and $y_2$ is independent of the curve taken from one to the other. Equivalently, the integral around a closed curve must be zero, $\oint\limits_C F_i(x)\mathrm{d}x^i=0$ for every closed curve $C$.

This is the definition of conservative force. Okay I agree but What I cannot understand is How did you verify that between two point in a gravitational field the work done in moving a object from one point to another is independent of the path, I mean there are literally infinite numbers of path that we can have between those two point so How can we say that it is independent of path ?

How did you reach to the conclusion that gravity is a conservative force ?

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  • $\begingroup$ en.wikipedia.org/wiki/Conservative_vector_field $\endgroup$ – G. Smith Oct 8 at 4:21
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    $\begingroup$ @Rob Jeffries' answer using Stokes' theorem is great. However I would like to observe that it again relies on knowing that the gravitational force satisfies the inverse-square law at every point in the universe. Inevitably you will have to make a certain assumption somewhere that cannot be "proved". In fact, it is not clear from your question which facts you accept as true for the gravitational force, from which to deduce that it is conservative. $\endgroup$ – Tob Ernack Oct 9 at 20:19
  • $\begingroup$ First sentence: "we have $\int_C \; F_i(x) \,\mathrm{d}x^i$" ... is what? This phrase is incomplete. $\endgroup$ – Eric Towers Oct 9 at 21:32
  • $\begingroup$ @EricTowers I copied this definition from here physics.stackexchange.com/a/31681/276626. $\endgroup$ – ROG Notes Oct 9 at 23:54
  • $\begingroup$ @TobErnack, Rob Jeffries' answer gave me the mathematical proof (though I didn't understand it as it requires high level calculus, but I am satisfied to know that it exists and someday I will be able to understand it.) and Luo Zeyuan's answer gave me the intuition (which I was able to understand). So I am confused about which answer to accept, so I gave upvote to both the answers. $\endgroup$ – ROG Notes Oct 10 at 0:01
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Are you looking for a mathematical proof (which has been given by others), or an experimental demonstration?

If gravity is not conservative then that means there would two paths up a mountain that take different amounts of energy to ascend (friction excluded).

So if you started on a bike at the top, you could free-wheel down the high-energy path, then coast up the low-energy path and when you got back to the top, you'd still have some excess energy (you'd still be moving). You could go round again and get even faster. And again and again, gaining free energy all the time.

Can you see what other conservation law you're breaking here?

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    $\begingroup$ I remember understanding this as one of my most eye opening moments in my student life! +1! $\endgroup$ – Deep Oct 9 at 17:51
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    $\begingroup$ But this isn't the only possibility. It could be a force that dissipates energy and so you lose energy by completing a circuit, whichever way you go, and this would not violate conservation of energy. Like friction. Of course the force would have to have other properties, like a speed dependence to any "drag". $\endgroup$ – Rob Jeffries Oct 10 at 7:38
  • $\begingroup$ @Rob You would still violate energy conservation, just in the negative direction. But then again, with a non-conservative gravity, we cant even talk about a gravitational potential energy, so what energy conservation would really mean is anyones guess $\endgroup$ – Rasmus Damgaard Nielsen Oct 11 at 10:11
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    $\begingroup$ @RasmusDamgaardNielsenn friction does not violate conservation of energy. Neither does work done by other non-conservative forces. $\endgroup$ – Rob Jeffries Oct 11 at 14:58
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    $\begingroup$ @Oscar I am saying that your argument that gravity must be a conservative force because otherwise conservation of energy is violated is incorrect or at least incomplete. Non-conservative forces exist and they do not violate conservation of energy. If you assume that it is a symmetric central force depending only on position then of course it is conservative. $\endgroup$ – Rob Jeffries Oct 12 at 10:27
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Stokes' theorem tells us that for any vector field, the closed line integral of that field is equal to the surface integral of the curl of that field over any surface bounded by the closed loop. In this case, for a gravitational field $$\oint \vec{g}\cdot d\vec{l} = \int (\nabla \times \vec{g})\cdot d\vec{A}.$$

Clearly, the LHS of this equation would be the work done (per unit mass) in moving an object around a closed loop in a gravitational field.

But $$\vec{g} = \frac{GM}{r^2} \hat{r}$$ in spherical coordinates, where $\hat{r}$ is a unit vector in the radial direction. Taking the curl of this field in spherical coordinates, then because there are no $\theta$ or $\phi$ components, and $g_r \neq f(\theta, \phi$), $$\nabla \times \vec{g}=0.$$ Given that, then the RHS of Stokes' theorem is always zero and so the work done by the gravitational field around a closed path is always zero.

Note that the same argument applies to any central, symmetric force

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  • $\begingroup$ Thanks @Rob Jeffries. $\endgroup$ – ROG Notes Oct 10 at 0:02
  • $\begingroup$ We only have curl free vector fields must be conservative on simply connected domains. Do you assume the universe is simply connected in this answer? $\endgroup$ – Steven Gubkin Oct 10 at 19:06
  • $\begingroup$ Well how we found out experimentally that $\vec{g}$ does not have angular components and by the way radial component is also symmetrical? Is there any experimental data onthis? $\endgroup$ – Alone Programmer Oct 10 at 20:44
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The force field due to a small element of mass (which we can think of as a point mass) is spherically symmetric and central, which makes it a conservative field. For the case of field due to a point mass, consider resolving each tiny segment of your path into 2 components, one along the radial direction and one along the circular direction. Work is only done when you travel along the radial components of the path, because that's where the force is along the path. No work is done along the circular component of the path because force is perpendicular to the displacement.

Now you can take any path to get from point A to point B, but no matter what path you take the radial components of the displacement and their corresponding force along that displacement will be the same (because of the central and spherically symmetric nature of the field), and we can hence conclude that the change in potential energy does not depend on the path.

The gravitational field of a continuous object is just the vector sum of all of the fields due to the individual mass elements, and is therefore also conservative even though the two fields (of a point mass and of an extended body) may look quite different. enter image description here

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    $\begingroup$ Among all other answers your answers seems to me somewhat intuitive to me. It would be a great help if you can portray an example of such a path between two points in a diagram. $\endgroup$ – ROG Notes Oct 8 at 7:38
  • $\begingroup$ Sure, I guess it will look something like the image I attached $\endgroup$ – Luo Zeyuan Oct 8 at 7:48
  • $\begingroup$ Thanks @Luo Zeyuan. $\endgroup$ – ROG Notes Oct 10 at 0:02
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The definition of a conservative force came after the observation of conservation of energy and the accumulation of data for the gravitational field.

How did you reach to the conclusion that gravity is a conservative force ?

First one accepts conservation of energy, and the proof that in a gravitational field the potential energy of an object is fixed by its position. Look at hydroelectricity. No matter which path the water has taken to enter the dam, the energy it can give is fixed by the dam height to the hydroelectric plant.

One sees this by mathematically modeling all the different paths an object can take to be found at a height h with the fixed potential energy.

A conservative force depends only on the position of the object. If a force is conservative, it is possible to assign a numerical value for the potential at any point and conversely, when an object moves from one location to another, the force changes the potential energy of the object by an amount that does not depend on the path taken, contributing to the mechanical energy and the overall conservation of energy. If the force is not conservative, then defining a scalar potential is not possible, because taking different paths would lead to conflicting potential differences between the start and end points.

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Just show that $F_idx^i$ is total derivative, i.e. $F_idx^i=dW$. Then the path integral reduces to $\int_C dW$, which is independent of the path and depends only on the initial and final point.

So: $$F_idx^i=-G\frac{mM}{r^3}\left(xdx+ydy+zdz\right)=-G\frac{mM}{2r^3}d\left(x^2+y^2+z^2\right)=-G\frac{mM}{r^2}dr.$$ As this is of the form $f(r)dr$, it is indeed a total derivative.

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  • $\begingroup$ This is the best answer, in my opinion. Perhaps you skipped too many steps. So, to redress the balance, $\vec{F}(\vec r) = -\frac{GMm}{r^3}\vec{r}$, and thus $F_i = -\frac{GMm}{r^3} x^i$. And at the end $-\frac{G M m}{r^2}dr = d(\frac{G M m}{r})$, so $\int_a^b F_i dx^i = \frac{G M m}{r} |_a^b$, and the quantity $GMm/r$ will be recognised as the potential energy of the mass $m$. $\endgroup$ – Zorawar Oct 11 at 2:29
  • $\begingroup$ @Zorawar. Thanks for these points. The reason I left with the form $f(r)dr$ and did not compute the potential energy $W$ was to leave with the point, that any force of the form $\vec{F}=g(r)\vec{r}$ will be conservative no matter the function $g(r).$ The rest I just assumed was obvious enaugh. $\endgroup$ – Umaxo Oct 11 at 16:41
  • $\begingroup$ Yeah, I would have thought it was enough too, but otherwise I can't explain how this answer doesn't have more upvotes. The question specifically asked how can you prove the statement for every path, and this answer shows how. Essentially, the interior points of the path cancel because of the integral, leaving just the boundary points. $\endgroup$ – Zorawar Oct 11 at 17:44

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