So, there is a certain paragraph in Fenyman's book that I'm struggling with for quite some time. It says:

We imagine that there are two classes of machines, those that are not reversible, which includes all real machines, and those that are reversible, which of course are actually not attainable no matter how careful we may be in our design of bearings, levers, etc. We suppose, however, that there is such a thing—a reversible machine—which lowers one unit of weight (a pound or any other unit) by one unit of distance, and at the same time lifts a three-unit weight. Call this reversible machine, Machine A. Suppose this particular reversible machine lifts the three-unit weight a distance X. Then suppose we have another machine, Machine B, which is not necessarily reversible, which also lowers a unit weight a unit distance, but which lifts three units a distance Y. We can now prove that Y is not higher than X; that is, it is impossible to build a machine that will lift a weight any higher than it will be lifted by a reversible machine. Let us see why. Let us suppose that Y were higher than X. We take a one-unit weight and lower it one unit height with Machine B, and that lifts the three-unit weight up a distance Y. Then we could lower the weight from Y to X, obtaining free power, and use the reversible Machine A, running backwards, to lower the three-unit weight a distance X and lift the one-unit weight by one unit height. This will put the one-unit weight back where it was before, and leave both machines ready to be used again! We would therefore have perpetual motion if Y were higher than X, which we assumed was impossible. With those assumptions, we thus deduce that Y is not higher than X, so that of all machines that can be designed, the reversible machine is the best.

I've looked it up on Google, and I have found that somebody asked about this even here. But I still have a problem on drawing a picture. First let's assume that we place the two machines in the following order: on the left we plase machine B, and on the right the reversible machine A.

Initially, the machine B has the following state: the left pan has one unit of weight on it and the right one has those three units on it, which are lifted a distance Y, as Feynman states.

When we lower those three weights from Y to X, we can use the reversbile machine. So, we ease the machine B, because machine A takes those weights. Is that a right picture?

If so, when we have lowered the trhee weights a distance Y-X, the left pan which holds the one unit has been lowerd a certain distance, less than a unit of distance (I say less than a meter because Feynman stated that this irreversible machine can lift three units by lowering one unit of weight, a unit of distance).

So, when the weights are passed to the reversible machine, that one weight on the left pan of the irreversbile machine B lowers, lifting the right empty pan. Is that a right picture? If so, will the right (now) empty pan reach the distance Y again ?

Looking back to machine A: because it's reversible when those three weights touch its left pan, they are lowered by it, lifting the right pan that holds one unit of weight. And then, it goes back and forth, the one unit lowers, the three units are lifted up, and so on..

I can't see how comes that we do have perpetual motion. What 'free' energy have we obtained? Now, in the final states the machine B has its left pan lowered down, and its right pan held up with nothing on it because it passed its weights to machine A. And the machine A, goes on and on... So how comes that the machines are ready to be used again?

Hope I've managed to express my uncertainities right. If not so, please tell me, so I can review and correct my language, because I am not a native speaker.


3 Answers 3


When we lift those three weights from Y to X, we can use the reversbile machine. So, we ease the machine B, because machine A takes those weights. Is that a right picture?

There are two weights with mass $M$ and $3M$.

Initially, both weights are at the same height $h_0$ which we can freely set to zero: $h_0 = 0$.

Now, machine B lowers mass $M$ to height $h_M = - 1$ in order to raise mass $3M$ to height $h_{3M} = Y > X$.

At this point, with another machine, the $3M$ mass is lowered to height $h_{3M} = X$ thus extracting $W = 3Mg(Y - X)$ energy from the system.

Finally, machine A is used to lower the mass $3M$ from height $h_{3M} = X$ to height $h_{3M} = 0$ in order to raise the mass $M$ from height $h_M = -1$ to height $h_M = 0$.

So the system has returned to the initial state (with both masses at zero height) and thus has the same potential energy - but energy was extracted during the cycle. This cycle could be repeated with $W = 3Mg(Y - X)$ extracted on each cycle without end (in principle).

  • $\begingroup$ Could you explain a little more what dou you meancwhen you say the energy was 'extracted' . Was it extracted when the free pan of machine B was raised again? yes, i have coreected my question. I put 'lower' instead od 'lift'. $\endgroup$
    – Bardo
    Sep 20, 2014 at 19:14
  • $\begingroup$ @Bardo, once a mass has been raised, it can be attached to another machine, e.g., moved horizontally to another pan or attached to a different rope. Think of it this crude way: machine B lowers M and raises 3M where the weights are secured and machine B 'moved out of the way'. Then another machine is connected to 3M and gravity does work on this machine as the weight is lowered. 3M is secured and this machine is 'moved out of the way' and machine A is connected to allow the weights to come back to the zero height. $\endgroup$ Sep 20, 2014 at 19:33
  • $\begingroup$ So did you use three machines, right? stil don't get Feynamn's point, and I'm a little bit dissapointed with myself. I am still not able to see where is the 'free energy' everyone brags about...:( $\endgroup$
    – Bardo
    Sep 21, 2014 at 9:49
  • $\begingroup$ @Bardo don't beat yourself up. Many people on this site criticised chapter 4.2 as being unnecessarily complicated. My own take is that Feynman should have used more pictures - he's describing a dynamic system with lots of possible states but only uses a static figure to portray it all. $\endgroup$ Nov 26, 2020 at 21:41
  • $\begingroup$ @AlfredCentauri how did you deduce that Feynman was referring to using another machine to lower the 3𝑀 weight from height 𝑌 to 𝑋? The original text is vague: "Then we could lower the weight from Y to X, obtaining free power ..." $\endgroup$ Nov 26, 2020 at 21:45

here is what I think he is trying to explain.

You have two machines A and B. A is reversible B is not necessarily reversible. Both these machines are placed side by side. let both machines A and B be initially horizontal. (Left , right pans of machine A will hold 3 unit and 1 unit mass respectively. Similarly, the left and right pans of machine B will hold 3 unit and 1 unit mass respectively) Now we place a 3 unit mass on the left pan of machine B and 1 unit mass on the right pan of the same machine. Now let the machine do it job, that is to lower the 1 unit mass by 1 unit distance and lift the 3 unit mass by Y unit distance.

The situation now is the 3 unit mass is Y unit distance above the horizontal and 1 unit mass is 1 unit distance below the horizontal

(But if the machine A was to do the same job, 1 unit mass would be lowered by 1 unit distance and 3 unit mass would be raised but X unit distance above the Horizontal.)

Feynman’s assumes that Y > X.

If you read carefully you will see that machine a works backward, which means for machine A "right pan is already 1 unit below the horizontal".

We drop the 3 unit mass from height Y units to X units during which some amount of energy (potential energy) can be used to do other work. Simultaneously we move the 1 unit mass from the right pan of machine B to the right pan of machine A. Now we place the (dropped) 3 unit mass of the left pan of machine A. Since machine A is doing its job backwards. It will lower 3 unit from a height X unit distance to the horizontal (0 unit distance) and raise the 1 unit mass to the horizontal (from 1 unit distance below the horizontal).

Now we are back to our initial configuration of masses, i.e., initially all the masses were at the horizontal level and finally all the masses are also in the horizontal level. So change is potential energy is zero. So net work done by the system of masses is also zero.

Now here is where it gets interesting. When we dropped the 3 unit mass from a Y unit distance above the horizontal to a X unit distance above the horizontal there was some energy conversion which was used to do work(like generation of electricity in a dam.....) but we know that the net change is potential energy of masses is zero, i.e., the above energy(for generation of electricity) was obtained from nothing, i.e., energy was obtained from nothing which is not at all possible, that is we have created a perpetual machine which will run forever.

Anyone who has taken high school physics knows that perpetual machines are impossible, “which only means that our assumption Y>X is wrong”.

(Please note that system of machines must be completely isolated that is no other agent "other than earth" must do work on the system).

(Please also note Feynman is trying to derive the concept of gravitation potential energy through qualitative reasoning and Carnot’s theory, but I have used potential energy itself to explain Feynman argument).

I hope this helps, my is a little poor so please do forgive me for any mistakes in my English.


I also had hard time to picture what Feynman wanted to explain. It feels like my confusion arose from lack of proper definitions of system state, perpetual motion, and reversible machine. The examples he used are also not quite clear of the mechanical apparatus used, not sure if by design he wanted to abstract away that from the reader, but if so, I would suggest to not even bother to have those diagrams.

All that said, I would rephrase hypothesis and conclusions as follow:

  1. Assume there is no perpetual motion, i.e.: a system can't move between two states without external forces in both directions. Note that transition between states are performed by any arbitrary machinery.
  2. That's equivalent to say there is no actual reversible machine.

However, one can idealize an "almost reversible" machine that transition from state A to B without external force and from B to "nearly" A.

From (2), if the almost-reversible machine can be set to operate in reverse then this operation requires the use of external force.

This idealized machine is helpful to show that any non (almost)reversible machine can't take the system from A to a state C > B, otherwise it contradicts 1.

In other words, the best designed machine would require energy to keep the system in a cycle.

This invariant, is then called gravitational potential energy observing the conservation of energy law..

  • $\begingroup$ An almost reversible machine is one in which the energy required to tip the balance one way or the other is the limit to zero. $\endgroup$ Nov 26, 2020 at 22:08

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