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In all of the below remarks, I am considering an object (a "mass") moving around above the surface of the earth.

What are some physical examples of situations that demonstrate path independence in a gravitational field (by that I mean: the condition where the work done by gravity on the object, as the object moves from point A to point B is independent of path)? We have all seen the randomly curvy paths leading from point A to point B... but is this physics or is this an artifact of multivariable calculus books? It seems to me the only true demonstrations of this would be projectile motion and free fall. So why do physics books have the super curvy paths when they introduce path independence? Am I missing something? Is there something physical to these strange paths?

Perhaps I misread the figures from the physics books, I don't know.

But the question remains: in a perfect book should only parabolas and straight lines be used for demonstrating "the geometry of path independence" for work done by a gravitational field?

What about roller coasters with no friction for making the demonstration of "the geometry of path independence" for work done by a gravitational field? For that matter, what about me holding a mass that I move around in the air so long as I promise to only provide the normal force like the track did for the roller coaster? Is the final word that, strictly speaking, constraints such as tracks are not allowed if you want to purely probe the interaction force between our mass of interest and the earth which is nothing other than the conservative force of gravity? So are roller coasters out [not because they exert a force normal to the movement which does no work, but, rather, because they introduce any force at all other that the one creating the conservative field... which necessarily takes us beyond the conditions we are considering]?

And finally, what if I want to move the mass from point A to point B with all different kinds of pushes and pulls with my hand... introducing tangential accelerations at will. Sometimes I move the mass from A to B very quickly and in a straight line. Sometimes I take a long path but with varying speeds. Once in a while I go slow at first and then accelerate to the fastest I can possibly go over a particularly long path. Do all of these paths have a place in the physics book when we do a picture of path independence? If yes, then obviously we are not constricting our focus to a situation where the only relevant force is the interaction force making the conservative field. So fine, we can still do the bookkeeping on the height-changes all the same... however this feels like a strange scenario for discussing a conservative field.

What is your take on the physicist's view on how one demonstrates, using a specific physical system, path independence in a conservative field.

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  • $\begingroup$ Do you mean the change in potential energy is independent of the path? $\endgroup$ – John Rennie Nov 28 '18 at 6:46
  • $\begingroup$ @JohnRennie I just edited the question and I think it is more clear now. Thanks for your input. $\endgroup$ – okcapp Nov 28 '18 at 7:06
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You can write the gravitational field like this $$ {\bf{g}}= - \nabla \phi \, \tag{1} $$ The work done along an arbitrary path $\gamma$ connecting two points $a$ and $b$ is proportional to

$$ \int_{\gamma(a,b)} {\bf{g} }\cdot {\bf{dx}} = \int_{\gamma(a,b)} (- \nabla \phi)\cdot {\bf{dx}}= -\phi(b) + \phi(a) \tag{2} $$ And as you can see It's independent from the path $\gamma$ you've chosen but it only depends on the initial and final point.

Notice that i didn't say who is performing the work, since both the work done by you against the field and the work done by the field are proportional to $(2)$.

Conclusion: the work done by or against gravitational field is path idepentent.

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Hopefully you will forgive me for posting an answer to my own question. But I have thought about it and I feel the answer could be (and I hope to receive feedback on this answer!):

In the perfect textbook several randomly curvy paths could be used as the figure for the discussion of path independence in the conservative gravitational field. But the caption should say:

"The work that gravity does on the object is the same for all paths from point A to point B. But if you are carrying the object along one of those paths from point A to point B then work will also be done on the object by you if you change the tangential speed of the object at any moment. That work need not be the same for all paths... it depends on how you choose to accelerate the object tangentially along each path. Similarly, if you don't neglect drag, then drag will do work on the object, even as you do work on the object to overcome drag. And here again work need not be the same for all paths... since the air pressure, wind, and velocity might change from path to path depending on where your path goes and how fast you are moving the object about."

I'm not looking for help with a figure for a textbook: I'm not writing a textbook... just attempting to have full clarity on the physics :)

Please let me know your thoughts.

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  • $\begingroup$ The work you do to against the gravitational field to move an object from a point that is at a certain height to a point at another height is path independent. That is it. The work done by other forces doesn't matter, we're talking about Gravity. There isn't air, and therefore there isn't air pressure or wind. The situation is idealised. $\endgroup$ – Run like hell Nov 28 '18 at 23:17

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