# Feynman Lectures Chapter 4.2: Understanding the obtainment of free power

Excerpt:

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.

I understand reductio ad absurdum in general, so that is not the issue. The "obtaining free power" is where I got lost. How is this free power? He kind of implies that there is an external force applied to the machine when he says "we could lower the weight from Y to X".

In addition, I struggle to understand his explanation visually. When he says to use the reversible Machine A that runs "backwards", does he mean that someone picks up the three-unit weight with their hands and places it on the balance pan on Machine A?

Note: This answer is probably cheating from a logical perspective, since I think that Feynman is trying to use his example to justify the principle I am using. However, I think it can be useful pedagogically to have the argument laid out in energy terms, and then go back to Feynman's exposition later to see exactly how he reaches the same conclusion without needing to directly use the concept of energy.

As a refresher, the work needed to raise a mass $$m$$ by height $$h$$ is $$mgh$$.

Let's say the "unit mass" has mass $$m$$, and "unit distance" is $$h$$. Then let's say that Machine A raises a mass $$3m$$ by distance $$X$$, and Machine B raises a mass $$3m$$ by distance $$Y$$.

We will use the following basic principle. The net change in internal energy of the system, is equal to the total work done on the system. $$\begin{equation} \Delta U = W \end{equation}$$ In fact, for this reversible machine, we will always have that the net work done on the system is $$0$$. (I can go into more detail about this if needed).

Let's consider Machine A. Initially, Machine A consists of a mass $$m$$ at height $$h$$, and a mass $$3m$$ at height zero. Therefore the initial energy is $$U_i=mgh$$. We then lower the little mass, and raise the larger mass to a height $$X$$. The net work that is done on the system is zero. The final energy of the system is $$U_f=3mgX$$. (Of course this implies that $$X=h/3$$, but let's pretend we don't know this since it would lead us to conclude that $$Y=X$$ "too early" to follow Feynman's argument.)

Similarly, if we consider Machine B, then if we use the little mass to raise the 3 masses, the initial energy is $$U_i=mgh$$ and $$U_f=3mgY$$.

Now suppose $$Y>X$$. Initially, the energy is $$mgh$$. Then we use Machine B to raise the three masses to height $$Y$$, so after the three masses are in the air, the energy of the system is $$3mgY$$. The final step is to use Machine A to lower the three masses. At the end of this process, the little mass $$m$$ will be raised to a height $$H$$. Now, you should expect that $$H=h$$ -- that is, the little mass $$m$$ should be at the same height at the end of the cycle, as when we started. What else could the answer be? But this cannot be the case. We have previously seen for Machine A that it only takes $$3mgX$$ worth of energy to raise $$m$$ to height $$h$$. On the other hand, we have also said that the energy in the system is $$3mgY > 3mgX$$. Therefore, Machine A must end up raising the mass $$m$$ to a height $$H>h$$, as we lower the three masses to the ground. However, this implies that the final energy is $$mgH>mgh$$. By going through a cycle we have ended up with more energy than we started with, despite doing no work on the system. This contradicts the basic principle $$\Delta U=W$$.

This fact can be exploited, since we can now extract work $$W=mg(H-h)$$ from the system repeatedly by performing this loop over and over. By simply raising and lowering a weight on the magic seesaw formed by joining Machines A and B, we can extract work. Say we attach a piston to the seesaw, then we can raise and lower the weights and spin a wheel, to power a motor, to turn on a light.

• Hey I deleted both of my previous comments because I realised I fooled myself into thinking I understand why the net work done on this reversible machine is always zero. I don't, although I suspect that the reason is that nobody applies an external force on the system (or if they do it is an infinitesimally small one). Sorry to pester you but can I kindly ask you expound on this? Nov 24, 2020 at 22:56
• I think this is where I get into trouble logically speaking, since I think this is more or less what Feynman is trying to prove. But here it goes. If we go through an entire cycle, and there were no losses, and we input work into the system, then we ca have returned to exactly the same state we started with. We have to have ended up in a state with more energy, since we input some work into the system. But the premise of this process combining Machines A and B is that it forms a closed loop and we have returned to exactly the same state we started. So we can't have input any work. Nov 25, 2020 at 0:33

When Feynman says

lower the weight from Y to X

he simply means

let the weight fall from height Y to the lower height X

The “free power” is simply saying that we could use the falling weight to do work - turn a shaft, raise another weight, run a clockwork mechanism etc.

We now have the three-unit weight at height X and the one-unit weight at height $$-1$$ unit - remember it went down as machine B lifted the three-unit weight. We can use machine A to lift the one-unit weight back to height $$0$$ and lower the three-unit weight to height $$0$$ at the same time. Because machine A is reversible it can either raise the three-unit weight by X units and lower the one-unit weight by $$1$$ unit or it can lower the three-unit weight by X units and raise the one-unit weight by $$1$$ unit. It just needs a small nudge to set it off in one direction one the other.

Yes, we need to detach the weights from machine B and attach them to machine A to do this, but Feynman is assuming moving weights around without changing their height requires no work.

Now both weights are back at height $$0$$ where they started, and we could run through the whole cycle again and again, producing as much “free power” as we like.

The whole explanation is complicated because Feynman does not want to use the word “energy” since he is working up to an explanation of what energy is and why it is conserved.

• Right so he's assuming that there is no external actor acting on the system except for Earth's gravity. That is the logic of "let the weight fall from height Y to to the lower height X". Nov 26, 2020 at 22:13
• Question about your 3rd paragraph though .. can you really assume that moving weights around without changing their heights requires not work? If it did require work, would his explanation still be valid? I'm new to physics and this assumption-laden thinking takes me a while to get used to it. Nov 26, 2020 at 22:22