# Feynman Vol.I 4–2 "Gravitational potential energy": a little confusion in the elegant proof

I'm new here and I've been reading through Feynman's great lectures from the very beginning.

In Feynman Vol.I 4-2 "Gravitational potential energy", Feynman uses an elegant proof to show the quantity $$X$$ must be $$\dfrac{1}{3}$$ and every $$Y$$ is strictly less than $$X$$ in the real world.

But after reading this I'm a little confused by some point and here I'm trying to explain that clearly as much as I can. Although I've known a little bit about basic physics, now I have to pretend to know nothing at first. Then how could I naturally think of "reversible" and "irreversible" and there is an $$X$$?

And in the elegant proof, if there is actually not a reversible machine to help us finish the process (though in our imagination) to derive a perpetual motion as a contradiction, why could we use this reversible machine? So I got stucked at some point in logic (maybe because I couldn't completely understand Feynman's thought).

I think it's similar to this situation in mathematics: if we at first know $$Y$$ has an upper bound $$X$$ we can derive $$X$$ must be $$\dfrac{1}{3}$$, but if we at first don't know $$Y$$ has an upper bound, how should we continue?

Any help will be appreciated.

• I understand your confusion, and I also have some questions about this: Since the book says that this thing does not exist, why are we able to use a non-existent thing to prove a real-world problem? Or does this mean that it's equivalent to assuming that we indeed have something like a minimal upper bound? I look forward to more explanations. Commented Apr 18 at 12:47

Feynman is probably going a little too fast and slick.

He is trying to point out two things at once. Namely, they are that

1. The best machines we can hope for are reversible, ideal, machines.
2. There cannot be any machine better than that, i.e. perpetual motion machines.

By considering a small enough mass and a small enough initial height, a reversible machine must have an upper bound $$X$$. After all, if you can have unbounded lifting of stuff under such conditions, then we can simply split stuff into small chunks like that and lift all of them with no upper bound. That is physically nonsense, so such an upper bound must exist. (Note that Feynman is so smart as to choose for the weights to be lifted are heavier than the weight being dropped; this means that $$X<1$$, making the bound even more obvious.)

Once we establish the limits of ideal reversible machines, i.e. bound $$X$$, then we can argue separately that there can be no better machine, no perpetual motion machines, by the argument that Feynman brought in. This limits $$Y\leqslant X$$

The rest of the argument then follows.

There is a tremendous difference between perpetual motion machines and ideal reversible machines, that you also have to be clear about. Although we all agree that ideal reversible machines are impossible in practice, the fact of the matter is that, for any tolerance factor, we can engineer a machine that is so efficient, that it lies within the tolerance factor of this ideal reversible machine.

Another way of phrasing it, is that an ideal reversible machine is an idealisation, a limit-taking of experimentally achievable machines. The important part is that these things are approximated well by experimentally achievable machines, to as good of an approximation as you want, except engineering difficulties to actually achieve stricter and stricter tolerances. They exist as theoretical constructs, but are physically allowed.

Perpetual motion machines are not even allowed as theoretical constructs, because we can prove that their existence breaks physics. That's what Feynman is also trying to argue. We can define them, we can reason with them, and immediately prove that they cannot exist. Square circles and all that.