Gravitational Potential Energy: Feynman's "3 Balls" Example

[physicsYum]

Feynman's lecture on physics: chapter 4 derives equation for gravitational potential energy by a "lifting 3 balls" example. The book notes:

"But the strange thing is that, in a certain way of speaking, we have not lifted two of them at all because, after all, there were balls on shelves 2 and 3 before. The resulting effect has been to lift one ball to a distance of 3X."

How is this argument true?

I am a physics student. And I have no rights to question Feynman's point of reasoning. But, this line of reasoning seems very strange to me.

Just because we can not see that the other two balls have also moved, does not mean that they have not been moved. If seeing is everything we could have used 3 different colored balls.

What am I missing here?

 

[TheShrike]

I believe he is saying that, in terms of energy, moving 3 balls by a distance X each is equivalent to moving 1 ball a distance 3X. This holds if the balls are identical, and the gravitational field is uniform.

 

[physicsYum]

That's what he is trying to derive, without any assumptions. Because he then goes on to proving X <= 1 ft. Right?

Intuitively, if we know " moving 3 balls by a distance X each is equivalent to moving 1 ball a distance 3X", we are not deriving the equation, we are using it, right?

 

[Michael C]

Look at it like this. We perform one of two actions on the set of balls.

Action A: we move each ball up a distance X. We end up with one ball on shelf 2, one on shelf 3 and one on shelf 4.
Action B: we take the bottom ball and move it up by a distance of 3X. We also end up with one ball on shelf 2, one on shelf 3 and one on shelf 4.

Of course if we colour the balls differently the two outcomes won't look the same, but since the balls all have the same weight, it's clear that the gravitational potential energy level of the whole system is the same after performing action A as it is after performing action B. Whether we move one ball up a distance of 3X, or three balls up each a distance of X, we have produced the same total change in potential energy.

 

[physicsYum]

Thanks Michael. That sounds convincing.

 

[codelieb]

Look at it like this. We perform one of two actions on the set of balls.

Action A: we move each ball up a distance X. We end up with one ball on shelf 2, one on shelf 3 and one on shelf 4.
Action B: we take the bottom ball and move it up by a distance of 3X. We also end up with one ball on shelf 2, one on shelf 3 and one on shelf 4.

Of course if we colour the balls differently the two outcomes won't look the same, but since the balls all have the same weight, it's clear that the gravitational potential energy level of the whole system is the same after performing action A as it is after performing action B. Whether we move one ball up a distance of 3X, or three balls up each a distance of X, we have produced the same total change in potential energy.

Michael C,

I think your explanation is very good. I would add only the following (largely superfluous) comment: Never mind about gravitational potential; Just assume that all the balls are identical. Then, if I go out of the room, and you perform action A or action B, when I return, there is no experiment I can perform that will distinguish which action you performed. The end-result is the same physical state. That's the main point. From this it follows that Action A or Action B result in a state with the same physical attributes, including gravitational potential.

Mike Gottlieb
Editor, The Feynman Lectures on Physics
www.feynmanlectures.info

 

[Fastman99]

If they were identical quantum particles that were indistinguishable, then yes you wouldn't be able to tell Action A and B would be indistinguishable. With large objects like these balls, you could theoretically be able to tell based on studying the balls closely enough before hand, detecting subtle differences between the three balls.

 

[TheShrike]

If they were identical quantum particles that were indistinguishable, then yes you wouldn't be able to tell Action A and B would be indistinguishable. With large objects like these balls, you could theoretically be able to tell based on studying the balls closely enough before hand, detecting subtle differences between the three balls.

By hypothesis, all the balls are identical. There are no subtle differences between them. But I see where you're coming from.

 

[codelieb]

If they were identical quantum particles that were indistinguishable, then yes you wouldn't be able to tell Action A and B would be indistinguishable. With large objects like these balls, you could theoretically be able to tell based on studying the balls closely enough before hand, detecting subtle differences between the three balls.

Earlier you asked "What am I missing here?"

What it seems to me you are missing is the fact that in this discussion Feynman is not talking about real balls in a real machine. He is talking about ideal (identical) balls in an ideal machine (that operates in a perfectly uniform gravitational field, with perfect precision and repeatability, without any friction or other energy losses, etc.) There are no such machines. It's a thought experiment. Of course a real machine could be made to work the same way, within a given precision. And in fact, there is such a real machine! It was built for this lecture, and you can see it http://www.basicfeynman.com/images/chalkboard/4_07.jpg.

Such abstractions are necessary in order to achieve a proper understanding of physics. Without them one is led to fallacies such as Aristotle's, who believed that in order to keep an object in motion, you have to constantly apply a force to it. He drew his conclusion from what he observed in the real world. And that's what people believed for almost 2 millennia before Galileo realized that it was wrong. To reach his conclusion that "A body moving on a level surface will continue in the same direction at a constant speed unless disturbed" Galileo had to make an abstraction (perfectly level surface, zero net gravitational force, no friction or other energy losses, etc.) in character similar to the abstraction Feynman makes in his argument.

You can overcome this kind of confusion by working on physics problems, because in order to solve them, you will have to make similar abstractions. Essentially, it's a matter of learning what things to pay attention to and what things to ignore in a given physical situation, in order to be able to make some analysis of it using the abstract mathematical laws of physics. In this particular case, the color of the balls or quantum mechanical differences between them are irrelevant and should be assumed not to exist, or simply ignored.

Mike Gottlieb
Editor, The Feynman Lectures on Physics
www.feynmanlectures.info

P.S. I will confess that when I first read this argument in The Feynman Lectures on Physics, something about it bothered me too; I was bothered by the statement "(a) First we roll the balls horizontally from the rack to the shelves, (b), and we suppose that this takes no energy because we do not change the height." This bothered me because, clearly, to move the balls at all we have to impart to them some kinetic energy, so how could that take no energy?! The answer is two-fold: (1) it takes no minimal amount of energy to get the balls moving (though, of course, the less energy we impart, the slower they move), and (2) however much energy is imparted to the balls to get them moving, if we are clever in our design we can recapture that energy when they stop moving! For example, a latched compressed spring could be released to push a ball one way, and another spring could be compressed by the energy of the ball, stopping the ball and latching that spring. Then that spring could be released to push the ball back the other way, etc. Of course, this cycle could be repeated indefinitely only in an ideal machine where there are no energy losses, but that is what is being discussed.