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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 ?

Qmechanic
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  • https://en.wikipedia.org/wiki/Conservative_vector_field – G. Smith Oct 08 '20 at 04:21
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    @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. – Tob Ernack Oct 09 '20 at 20:19
  • First sentence: "we have $\int_C ; F_i(x) ,\mathrm{d}x^i$" ... is what? This phrase is incomplete. – Eric Towers Oct 09 '20 at 21:32
  • @EricTowers I copied this definition from here https://physics.stackexchange.com/a/31681/276626. –  Oct 09 '20 at 23:54
  • @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. –  Oct 10 '20 at 00:01
  • @TobErnack substitute $r^{n}$ for $r^2$ if you like; the proof still works. The assumption that needs to be made is that gravity is a central force. – ProfRob Oct 10 '20 at 07:32
  • @ROGNotes Rob Jeffries' answer says: "given...he work done by the gravitational field around a closed path is always zero." This is the definition of a "conservative force" used in mechanical engineering. If I understand his answer correctly, Rob Jeffries proves that both definitions are identical: He proves that gravity is conservative if gravity is "conservative" (according to the engineering definition of that term). So the next question should be: How do we know that gravity is "conservative" (in engineering terms). ... – Martin Rosenau Oct 10 '20 at 20:36
  • @ROGNotes ... And the simple answer is: This is a pure experimental result and we cannot be 100% sure that there will not be a sensational scientific discovery in the future that shows us that gravity is not conservative. – Martin Rosenau Oct 10 '20 at 20:39
  • @MartinRosenau if you propose some other definition of the gravitational force then of course it may not be conservative. I showed that gravity is conservative if it is a central force. – ProfRob Oct 10 '20 at 23:06
  • @RobJeffries The German-language Wikipedia explicitly mentions that there are also non-conservative central forces. So a force being a central force does not imply that the force is conservative. – Martin Rosenau Oct 11 '20 at 06:39
  • @MartinRosenau sure, a symmetric central force is required (which is often assumed to be part of the definition). – ProfRob Oct 11 '20 at 07:21
  • @MartinRosenau in fact the only reference in the article you refer to states (mathematically) that a central force is symmetric. – ProfRob Oct 11 '20 at 07:32
  • @RobJeffries In the introduction you'll find a sentence also found in the English language Wikipedia: "Not all central force fields are conservative or spherically symmetric. However, a central force is conservative if and only if it is spherically symmetric". However, this is not the point. I understood the original question in the way: "How do we know that gravity is a symmetric central force?" – Martin Rosenau Oct 11 '20 at 08:42
  • @RobJeffries ah yes you are right this works for any central force that has only radial dependence. Anyway my point was just to say that a mathematical proof of the conservative nature of gravity will still require some mathematical assumption such as symmetry that can't be proven experimentally, which seemed to be OP's original concern. – Tob Ernack Oct 13 '20 at 16:18
  • @TobErnack I have news for you - you can't prove any physical law experimentally. – ProfRob Oct 13 '20 at 18:21
  • I am not saying we can, and I am not sure if/why you think I am disagreeing with you. My comment was directed to the OP's last sentence before the bold question. It gave the impression that they weren't sure how a physical law could be verified if it theoretically requires an infinite number of measurements. They could have asked "What I cannot understand is How did you verify that at every point the gravitational field is a centrally symmetric force, I mean there are literally infinite numbers of points that we can have so How can we say that it is centrally symmetric ?" and the same happens. – Tob Ernack Oct 13 '20 at 18:34
  • Given that the OP is satisfied with the answer, clearly they consider the fact that the gravitational force is central symmetric to be more fundamental, which is reasonable but not the only possible way to build the theory. By the way by "centrally symmetric" I mean "central and spherically symmetric" to save some space. – Tob Ernack Oct 13 '20 at 18:51

5 Answers5

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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?

Oscar Bravo
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    I remember understanding this as one of my most eye opening moments in my student life! +1! – Deep Oct 09 '20 at 17:51
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    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". – ProfRob Oct 10 '20 at 07:38
  • @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 – Rasmus Damgaard Nielsen Oct 11 '20 at 10:11
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    @RasmusDamgaardNielsenn friction does not violate conservation of energy. Neither does work done by other non-conservative forces. – ProfRob Oct 11 '20 at 14:58
  • @RobJeffries At first I thought this was a crank comment - then I checked your credentials so I guess you're serious. Are you postulating a more complex field theory that collapses to the conventional inverse-square law in classical scales, but that has additional terms that give weird behaviour on galactic scales? In any case, how would you avoid that it gains energy with each loop and so becomes a free energy machine? – Oscar Bravo Oct 12 '20 at 09:23
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    @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. – ProfRob Oct 12 '20 at 10:27
  • @RobJeffries This must be new stuff... I thought that when we talk about a conservative field, the thing we are conserving is energy. So I confess to being baffled as to how you can have a non-conservative field that conserves energy (unless you're just talking about ordinary friction in the tyres, axles, air-resistance, etc.) – Oscar Bravo Oct 12 '20 at 12:39
  • @OscarBravo “when we talk about a conservative field, the thing we are conserving is energy” That’s a circular argument, or begging the question. The OP is asking why gravity is conservative in the first place. – user76284 Oct 28 '20 at 18:51
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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 and this is a defining property of a conservative field.

Note that the same argument applies to any central, symmetric force - for example if $g(r) = f(r)$ with no $\theta$ or $\phi$ dependence.

ProfRob
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  • Thanks @Rob Jeffries. –  Oct 10 '20 at 00:02
  • 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? – Steven Gubkin Oct 10 '20 at 19:06
  • 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? – Mithridates the Great Oct 10 '20 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

Luo Zeyuan
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    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. –  Oct 08 '20 at 07:38
  • Sure, I guess it will look something like the image I attached – Luo Zeyuan Oct 08 '20 at 07:48
  • Thanks @Luo Zeyuan. –  Oct 10 '20 at 00:02
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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.

anna v
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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.

Anthony Guillen
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Umaxo
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    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 $m$. – Zorawar Oct 11 '20 at 02:29
  • @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. – Umaxo Oct 11 '20 at 16:41
  • 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. – Zorawar Oct 11 '20 at 17:44