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I was thinking about a completely unrelated problem (Quantum Field Theory Peskin & Schroeder kind of unrelated!) when the diagram below sprang into my mind for no apparent reason. After some thinking, I can't figure out why it wouldn't work, other than the theoretical reason that it systematically decreases entropy in a closed system:

Entropy Decreasing Machine?

We have a thick, insulating barrier between two "ventricles" of a closed-off system. The only opening in this barrier is made by a solid shaft snugly (without permitting the transfer of heat between the two compartments) but frictionlessly. On the left ventricle we have a paddle attached, and on the right we have a coiled wire, as illustrated in the diagram. There are fixed magnets surrounding the coil of wire, providing a constant magnetic field through it.

Let's say the whole contraption is so small that a single air molecule hitting the paddle will contribute a small but not immaterial amount of angular momentum to the shaft (again, I never said this machine was practical!). Therefore, random chance will make air molecules hit the paddle so that it will start turning in Brownian-motion style. This rotation is damped by the energy dissipated when the coil turns the current induced by the magnets moving relative to the turning reference frame of the coil into heat through a resistor in the coil. Thus, the air molecules contribute a small portion of their kinetic energy to the paddle, which is then expended as heat on the other side of the border, making the air molecules on the left colder, while air molecules on the right heat up.

Doesn't this mean a decrease in entropy? (To see that it can't be an increase in entropy, take away the barrier, and note that the molecules go back to thermal equilibrium naturally, meaning that entropy increases naturally when undoing our actions).

To further show this mythical contraption to be an impossibility, we could create a barrier with two pieces of equipment forming a passage between the two ventricles. One port-hole of energy would be the paddle already envisioned, the other would be a Carnot engine taking energy from the hot right to the cold left. The paddle would take energy from the left to the right effortlessly for a period of time, and then the Carnot engine would move the heat back the other way, gaining energy that came from nowhere in the process!!

Where has my logic gone wrong? Clearly entropy must not decrease, and energy cannot be created by the fundamental axioms of physics. Why does this paddle fail to transfer energy from one ventricle to the other? An explanation of what has gone wrong with my reasoning would be greatly appreciated!

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    $\begingroup$ In thermal equilibrium, the paddle will continuously be wobbling back and forth, so it will be equally likely to speed up or slow down the particles hitting it. $\endgroup$
    – knzhou
    Sep 3, 2015 at 3:12
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    $\begingroup$ @Kevin Zhou Sure. Still, the energy will over the long run be taken out of the left and put in the right. Correct??? I'm assuming that the paddle is so small that individual hits are uneven, driving the paddle to wobble slightly, and each time a tiny amount of time is given between hits, the paddle slows down a tad bit by the tug of the coil in the magnetic field and the loss of energy due to radiation on the right. Correct? $\endgroup$
    – QuantumFool
    Sep 3, 2015 at 3:14
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    $\begingroup$ Nope! Basically, your machine is made of a lot of parts which exchange thermal energy with each other. If the whole thing is a perpetual motion machine, then some piece of it must be giving/receiving thermal energy asymmetrically, and we would be able to build a perpetual motion machine using only that one piece. $\endgroup$
    – knzhou
    Sep 3, 2015 at 3:26
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    $\begingroup$ With that in mind, we can always just keep zooming in on that piece, and the pieces of that piece. Inevitably we arrive at individual particles, which we know are symmetric, so the machine falls apart. $\endgroup$
    – knzhou
    Sep 3, 2015 at 3:26
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    $\begingroup$ For a similar idea, look up 'Feynman's ratchet'. $\endgroup$
    – knzhou
    Sep 3, 2015 at 3:27

3 Answers 3

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Thus, the air molecules contribute a small portion of their kinetic energy to the paddle, which is then expended as heat on the other side of the border, making the air molecules on the left colder, while air molecules on the right heat up. Doesn't this mean a decrease in entropy?

Yes it does.

However, we need to take the thermal noise of the resistor into account.

Hot resistors make noise

As discovered by John B. Johnson in 1928 and theoretically explained by Harry Nyquist, a resistor at temperature $T$ exhibits a non-zero open circuit voltage. This voltage is stochastic and characterized by a (single sided) spectral density

$$S_V(f) = 4 k_b T R \frac{h f / k_b T}{\exp \left(h f / k_b T \right) - 1} \, . \tag{1}$$

At room temperature we find $k_b T / h = 6 \times 10^{12} \, \text{Hz}$, which is a ridiculously high frequency for electrical systems. Therefore, for the loop of wire and resistor circuit in the device under consideration, we can roughly assume that

$$\exp(h f / k_b T) \approx 1 + h f /k_b T$$

so that

$$S_V(f) \approx 4 k_b T R \tag{2}$$

which we traditionally call the "Johnson noise" formula. If we short circuit the resistor as in the diagram where its ends are connected by a simple wire, then the current noise spectral density is (just divide by $R^2$)

$$S_I(f) = 4 k_b T / R \, .\tag{3}$$

Another way to think about this is that the resistor generates random current which is Gaussian distributed with standard deviation $\sigma_I = \sqrt{4 k_b T B / R}$ where $B$ is the bandwidth of whatever circuit is connected to the resistor.

Johnson noise keeps the system in equilibrium

Anyway, the point is that the little resistor in the machine actually generates random currents in the wire! These little currents cause the rod to twist back and forth for exactly the same reason that the twists in the rod induced by air molecules crashing into the paddles caused currents in the resistor (i.e. Faraday's law). Therefore, the thermal noise of the resistor shakes the paddles and heats up the air.

So, while heat travels from the air on the left side to the resistor on the right, precisely the opposite process also occurs: heat travels from the resistor on the right to the air on the left. The heat flow is always occurring in both directions. By definition, in equilibrium the left-to-right flow has the same magnitude as the right-to-left flow and both sides just sit at equal temperature; no entropy flows from one side to the other.

Fluctuation-dissipation

Note that the resistor is both dissipative and noisy. The resistance $R$ means that the resistor turns current/voltage into heat; the power dissipated by a resistor is

$$P = I^2 R = V^2 / R \, . \tag{4}$$

The noise is characterized by a spectral density given in Eq. (1). Note the conspicuous appearance of the dissipation parameter $R$ in the spectral density. This is no accident. There is a profound link between dissipation and noise in all physical systems. Using thermodynamics (or actually even quantum mechanics!) one can prove that any physical system which acts as a dissipator of energy must also be noisy. The link between noisy fluctuations and dissipation is described by the fluctuation-dissipation theorem, which is one of the most interesting laws in all of physics.

The machine originally looked like it moved entropy from the left to the right because we assumed the resistor was dissipative without being noisy, but as explained via the fluctuation-dissipation theorem this is entirely impossible; all dissipative systems exhibit noisy fluctuations.

P.S. I really, really like this question.

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    $\begingroup$ Dang, I was grasping at the fluctuation-dissipation theorem in my comments but I never knew it was a thing. This is super informative! $\endgroup$
    – knzhou
    Sep 3, 2015 at 8:04
  • $\begingroup$ a similar problem in at least two versions were discussed by Paul Penfield and several commenters following his letter in the Proceeedings of the IEEE "Unresolved Paradox in Circuit Theory", 1966, pp.1200-1201, and 1967 pp474-477, 2073-2076, 2173, 1968 p1225, etc. $\endgroup$
    – hyportnex
    Sep 3, 2015 at 12:37
  • $\begingroup$ @KevinZhou The quantum version is even more fun. Check it out some time. $\endgroup$
    – DanielSank
    Sep 3, 2015 at 15:37
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    $\begingroup$ @Michael actually not much changes. Now the Johnson noise (and the air molecules hitting the paddle) are driving a system not unlike a Brownian ratchet. $\endgroup$
    – DanielSank
    Sep 4, 2015 at 19:58
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    $\begingroup$ So is a good way to think of this to consider that a generator can also operate as a motor, and vice versa (e.g. you can crank a motor and get electric current out, and by passing current to a generator it will spin), and as the generator heats up, it starts acting more like a motor, and once it reaches equilibrium temperature it acts as a motor as much as a generator, and thus energy bounces back and forth uselessly, just as the ratchet where the pawl fails intermittently due to heating up and bouncing around? $\endgroup$
    – The_Sympathizer
    Mar 22, 2016 at 4:56
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The reason the "machine,"as designed, will not work, is that the paddle is not being hit by only one particle, on one quarter of its surface! Since it is being hit by several particles, "equally" on all 4 surfaces (two top and two bottom), there will be no net rotation of the shaft, thus no current generated. If the length of the paddle is made equal or less than the diameter of the particle (to exclude other particles), no current will be generated because the particles will be hitting the shaft, thus no net shaft rotation. If you control the direction of the particles, then it will work, by aiming the particles at one of the 4 paddle surfaces.

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    $\begingroup$ This is incorrect. While the mean angular velocity of the shaft is zero as stated, it still fluctuates about that mean. As it fluctuates, the wire loop's angular fluctuations induce fluctuating current and heat the resistor. You're mistaking a zero average for a constant zero. They're not the same thing. The power dissipated by the resistor is proportional to the absolute value squared of the instantaneous flux change in the loop. The fluctuations are tiny if the paddles are big, but not zero. $\endgroup$
    – DanielSank
    Sep 9, 2015 at 21:34
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This (ideal) device is a small variation of Maxwell's demon.

It is believed that Maxwell's demon violates the second law by decreasing the entropy of the universe (isolated system). However, that is not true; Maxwell's demon violates the first law of thermodynamics. See my article http://vixra.org/abs/1310.0181.

Your machine leads to violation of the principle of 'Perpetual motion of the first kind'.

What I mean by the above is this: Energy can be extracted in the form of mechanical work form a system when there exist between two points in the system, differences in any intensive property of the system. However, if we could extract mechanical work from a system which has uniform values for intensive properties, then we would violate the principle of 'Perpetual motion of the first kind'. This is impossible.

Your machine (device) makes it possible to extract mechanical work from an isolated (you call it closed) system which is at a uniform temperature through out, thereby leading to violation of the 'principle of 'Perpetual motion of the first kind'.This is the reason your machine does not work.

Radhakrishnamurty Padyala

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    $\begingroup$ "Where is the flaw in this machine that violates the laws of thermodynamics?" "Well the flaw is that it violates the laws of thermodynamics" is not helpful. I think the asker knows that the machine cannot violate the laws of thermodynamics, and they are asking why not. $\endgroup$
    – user253751
    Dec 30, 2019 at 14:14

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