# Brownian Ratchet (exercise framed by Feynman's Lectures, l. 46)

• Hak
Hak
Homework Statement
In 1912, Smoluchowski proposed a Brownian ratchet device. In 1962, Feynman conducted its first quantitative analysis. This device consists of two parts: a small paddle immersed in an ideal gas at temperature ##T_1## and a ratchet immersed in an ideal gas at temperature ##T_2##, connected to the paddle by a linkage. A pulley shaft is fixed in the middle of the linkage, with an object suspended below it. The ratchet is asymmetric in shape and has a small spring piece called a pawl that catches on the ratchet teeth. All components of the system are lightweight. Smoluchowski pointed out that when liquid molecules undergo random motion and collide with the paddle wheel, they will set it in motion.
The pawl restricts the ratchet to rotate in only one direction. Therefore, even if ##T_1 = T_2##, the system will continue to rotate in the same direction, thus extracting mechanical work from the internal energy of a thermally equilibrated system, violating the second law of thermodynamics.

1. When ##T_1 = T_2##, how will the system rotate?
(a) Completely stationary
(b) Rotating clockwise
(c) Both clockwise and counterclockwise rotation, with a net clock-
wise rotation on average
(d) Both clockwise and counterclockwise rotation, with no net di-
rection

2. When ##T_1 > T_2##:

(a) Estimate the average angular velocity of the system. Assume that each tooth of the ratchet corresponds to a central angle of ##\theta##. The radius of the pulley shaft in the linkage is
##R##, and the mass of the suspended object is ##m##, with all other components being lightweight.
(b) Determine the energy ##E## transferred from the gas on one side of the small paddle wheel to the small paddle wheel during the rotation of one tooth of the ratchet.
Relevant Equations
/
1. (d) Both clockwise and counterclockwise rotation, with no net direction. Because the pawl, which is also at the same temperature as the paddle and the ratchet, will undergo Brownian motion and bounce up and down randomly. Sometimes, it will fail to catch the ratchet teeth and allow the ratchet to slip backward. The net effect of many such random collisions and failures will be that the system will move back and forth with no preferred direction.

2. When ##T_1 > T_2##:

(a)
I assume that the system is in a steady state, meaning that the angular velocity of the paddle, the ratchet, and the pulley shaft are equal. I also assumed that the system is in equilibrium, meaning that the net torque on the system is zero.
I would calculate the average force ##F## exerted by the gas molecules on the paddle by using the ideal gas law, which relates the pressure, volume, temperature, and number of molecules of an ideal gas; then, the viscous drag coefficient ##\gamma## of the paddle by using Stokes’ law for a sphere, which relates the drag force, viscosity, and radius of a spherical object moving in a fluid. Then, I would equate the torque due to ##F## and the torque due to ##\gamma##, which are opposite in direction, to find the angular velocity ##\omega## of the paddle. Then, I will multiply ##\omega## by the ratio of ##R## and ##r##, which are the radii of the pulley shaft and the ratchet, respectively, to find the angular velocity of the ratchet. Dividing the angular velocity of the ratchet by ##2\pi##, which is the angle of one full rotation, I'll find the number of rotations per unit time: multiplying the number of rotations per unit time by ##\theta##, which is the angle of one tooth of the ratchet, I'll find the average angular displacement per unit time. I get:$$\omega=\frac{F}{\gamma}\frac{R}{r}\frac{\theta}{2 \pi}$$

The average force ##F## can be approximated by using the ideal gas law:$$F=\frac{N_1k_BT_1}{A}−\frac{N_2k_BT_2}{A}$$where ##N_1## and ##N_2## are the number of gas molecules on each side of the paddle, ##k_B## is the Boltzmann constant, ##T_1## and ##T_2## are the temperatures of the gases, and ##A## is the area of the paddle.

The viscous drag coefficient ##\gamma## can be calculated by using Stokes’ law for a sphere:$$\gamma=6\pi \eta r_p$$,where ##\eta## is the viscosity of the gas and ##r_p## is the radius of the paddle.Substituting these expressions into the formula for ##\omega##, we get:$$\omega=\frac{\frac{(N_1T_1 -N_2T_2) k_B}{A}}{6\pi \eta r_p}\frac{R}{r}\frac{\theta}{2 \pi}$$(b) The energy ##E## transferred from the gas on one side of the small paddle wheel to the small paddle wheel during the rotation of one tooth of the ratchet can be obtained by multiplying the average torque ##\tau## by the angle ##\theta##:

$$E = \tau \theta$$The average torque ##\tau## can be found by multiplying the average force ##F## by the radius ##r_p## of the paddle:$$\tau=Fr_p$$Using the expression for F from part (a), I get:$$\tau=\frac{(N_1T_1−N_2T_2)k_B}{A}r_p$$Therefore,$$E = \frac{(N_1T_1−N_2T_2)k_B}{A}r_p \theta$$Where do I go wrong?

Does anyone have any ideas?

Nothing?

Hak said:
Homework Statement: In 1912, Smoluchowski proposed a Brownian ratchet device. In 1962, Feynman conducted its first quantitative analysis. This device consists of two parts: a small paddle immersed in an ideal gas at temperature ##T_1## and a ratchet immersed in an ideal gas at temperature ##T_2##, connected to the paddle by a linkage. A pulley shaft is fixed in the middle of the linkage, with an object suspended below it. The ratchet is asymmetric in shape and has a small spring piece called a pawl that catches on the ratchet teeth. All components of the system are lightweight. Smoluchowski pointed out that when liquid molecules undergo random motion and collide with the paddle wheel, they will set it in motion.
The pawl restricts the ratchet to rotate in only one direction. Therefore, even if ##T_1 = T_2##, the system will continue to rotate in the same direction, thus extracting mechanical work from the internal energy of a thermally equilibrated system, violating the second law of thermodynamics.

1. When ##T_1 = T_2##, how will the system rotate?
(a) Completely stationary
(b) Rotating clockwise
(c) Both clockwise and counterclockwise rotation, with a net clock-
wise rotation on average
(d) Both clockwise and counterclockwise rotation, with no net di-
rection

2. When ##T_1 > T_2##:

(a) Estimate the average angular velocity of the system. Assume that each tooth of the ratchet corresponds to a central angle of ##\theta##. The radius of the pulley shaft in the linkage is
##R##, and the mass of the suspended object is ##m##, with all other components being lightweight.
(b) Determine the energy ##E## transferred from the gas on one side of the small paddle wheel to the small paddle wheel during the rotation of one tooth of the ratchet.
Relevant Equations: /

1. (d) Both clockwise and counterclockwise rotation, with no net direction. Because the pawl, which is also at the same temperature as the paddle and the ratchet, will undergo Brownian motion and bounce up and down randomly. Sometimes, it will fail to catch the ratchet teeth and allow the ratchet to slip backward. The net effect of many such random collisions and failures will be that the system will move back and forth with no preferred direction.

2. When ##T_1 > T_2##:

(a)
I assume that the system is in a steady state, meaning that the angular velocity of the paddle, the ratchet, and the pulley shaft are equal. I also assumed that the system is in equilibrium, meaning that the net torque on the system is zero.
I would calculate the average force ##F## exerted by the gas molecules on the paddle by using the ideal gas law, which relates the pressure, volume, temperature, and number of molecules of an ideal gas; then, the viscous drag coefficient ##\gamma## of the paddle by using Stokes’ law for a sphere, which relates the drag force, viscosity, and radius of a spherical object moving in a fluid. Then, I would equate the torque due to ##F## and the torque due to ##\gamma##, which are opposite in direction, to find the angular velocity ##\omega## of the paddle. Then, I will multiply ##\omega## by the ratio of ##R## and ##r##, which are the radii of the pulley shaft and the ratchet, respectively, to find the angular velocity of the ratchet. Dividing the angular velocity of the ratchet by ##2\pi##, which is the angle of one full rotation, I'll find the number of rotations per unit time: multiplying the number of rotations per unit time by ##\theta##, which is the angle of one tooth of the ratchet, I'll find the average angular displacement per unit time. I get:$$\omega=\frac{F}{\gamma}\frac{R}{r}\frac{\theta}{2 \pi}$$

The average force ##F## can be approximated by using the ideal gas law:$$F=\frac{N_1k_BT_1}{A}−\frac{N_2k_BT_2}{A}$$where ##N_1## and ##N_2## are the number of gas molecules on each side of the paddle, ##k_B## is the Boltzmann constant, ##T_1## and ##T_2## are the temperatures of the gases, and ##A## is the area of the paddle.

The viscous drag coefficient ##\gamma## can be calculated by using Stokes’ law for a sphere:$$\gamma=6\pi \eta r_p$$,where ##\eta## is the viscosity of the gas and ##r_p## is the radius of the paddle.Substituting these expressions into the formula for ##\omega##, we get:$$\omega=\frac{\frac{(N_1T_1 -N_2T_2) k_B}{A}}{6\pi \eta r_p}\frac{R}{r}\frac{\theta}{2 \pi}$$(b) The energy ##E## transferred from the gas on one side of the small paddle wheel to the small paddle wheel during the rotation of one tooth of the ratchet can be obtained by multiplying the average torque ##\tau## by the angle ##\theta##:

$$E = \tau \theta$$The average torque ##\tau## can be found by multiplying the average force ##F## by the radius ##r_p## of the paddle:$$\tau=Fr_p$$Using the expression for F from part (a), I get:$$\tau=\frac{(N_1T_1−N_2T_2)k_B}{A}r_p$$Therefore,$$E = \frac{(N_1T_1−N_2T_2)k_B}{A}r_p \theta$$Where do I go wrong?
Can you please provide a schematic of the setup?

The problem doesn't mention viscous drag so why did you bring it in? First assume the system is as simple as possible.

Chestermiller said:
Can you please provide a schematic of the setup?
Yes. Here it is.

Last edited by a moderator:
bob012345 said:
The problem doesn't mention viscous drag so why did you bring it in? First assume the system is as simple as possible.
I wouldn't know how to do that. Do you have any advice?

Hak said:
I wouldn't know how to do that. Do you have any advice?
I think you are going to need to include viscous effects to solve this (as you already seem to be doing).

This problem should be framed by the following chapter of "The Feynman Lectures on Physics.": https://www.feynmanlectures.caltech.edu/I_46.html. Could you help me figure out how to solve it based on this newly reported paper? I will be very grateful if any of you could help me out. Thank you.

Hak said:
This problem should be framed by the following chapter of "The Feynman Lectures on Physics.": https://www.feynmanlectures.caltech.edu/I_46.html. Could you help me figure out how to solve it based on this newly reported paper? I will be very grateful if any of you could help me out. Thank you.
I see you found Feynman's discussion about it. I think the heart of it is the third paragraph of 46-2 where Feynman introduces his terminology. First see if you can follow how he sets up the problem. As I understand it, he is simplifying the problem to it's core basics and applying thermodynamic probabilities as though each event (such as the ratchet moving through a small angle ##\theta##) were a single particle and not like a macroscopic group of particles.

bob012345 said:
I see you found Feynman's discussion about it. I think the heart of it is the third paragraph of 46-2 where Feynman introduces his terminology. First see if you can follow how he sets up the problem. As I understand it, he is simplifying the problem to it's core basics and applying thermodynamic probabilities as though each event (such as the ratchet moving through a small angle ##\theta##) were a single particle and not like a macroscopic group of particles.
Thank you very much. I paid attention to the third paragraph of 46-2. In the problem reference is made to mass ##m##, whereas in Feynman's discussion within his Lectures no reference is made to mass, at least symbolically. I cannot find a way that is collimated with what Feynman says and with the data provided by the problem. Do you have any suggestions?

Hak said:
Thank you very much. I paid attention to the third paragraph of 46-2. In the problem reference is made to mass ##m##, whereas in Feynman's discussion within his Lectures no reference is made to mass, at least symbolically. I cannot find a way that is collimated with what Feynman says and with the data provided by the problem. Do you have any suggestions?
You are being asked to frame the problem slightly differently that Feynman. You have the angle and radius of the gear which can relate to the height ##h## the mass is raised against gravity. Feynman just calls the energy raising the mass ##\epsilon## which should be the same as ##mgh##.

bob012345 said:
You are being asked to frame the problem slightly differently that Feynman. You have the angle and radius of the gear which can relate to the height ##h## the mass is raised against gravity. Feynman just calls the energy raising the mass ##\epsilon## which should be the same as ##mgh##.
OK, thank you very much, but what about ##L## and ##\tau##? Furthermore, one note from the exercise was: a great use of thermodynamics; actual thinking is also involved, not just simple maths.

Hak said:
whereas in Feynman's discussion within his Lectures no reference is made to mass, at least symbolically
As I recall Feynman analyzes the effect of hanging a flea on the thread attached to the pulley. So ##m_{flea}##. I believe his solution is that for ##T_1\gt T_2## the machine will rotate according to your analysis but that the pall requires viscosity to keep from bouncing higher and higher with each impact (you do the mechanics of that if you wish) thus heating the fluid in its container until ##T_2=T_1##.
edit: pawl dangit

Last edited:
Lord Jestocost
hutchphd said:
As I recall Feynman analyzes the effect of hanging a flea on the thread attached to the pulley. So ##m_{flea}##. I believe his solution is that for ##T_1\gt T_2## the machine will rotate according to your analysis but that the pall requires viscosity to keep from bouncing higher and higher with each impact (you do the mechanics of that if you wish) thus heating the fluid in its container until ##T_2=T_1##.
OK, thank you very much. What would this change in the analysis? I originally included viscosity in my process, but it is probably not correct. How can this be changed? If you would give me some hint, I would try to draft a new procedure. Thank you.

Hak said:
OK, thank you very much, but what about ##L## and ##\tau##? Furthermore, one note from the exercise was: a great use of thermodynamics; actual thinking is also involved, not just simple maths.
Feynman uses ##L## as torque. You can figure that from the force of gravity, the angle and radius. ##\tau## is an assumed rate. Yes, this is a thermodynamics problem primarily, not just a mechanics problem but the mechanics is consistent with energy expended or generated. As I said and understand Feynman reduces the problem to thinking of individual particle equivalents which is why he uses the fundamental probability ~##e^{-kT}## for each action or event. It is all summarized in Table 46-1.

Viscosity is never free. All "frictional" forces just sweep energy to degrees of freedom we do not look at in detail. What is the "solution" to the problem that you desire ? I described the physics above. The details of the viscosity are very complicated, so I don't understand what a solution is in your mind.

Lord Jestocost
hutchphd said:
As I recall Feynman analyzes the effect of hanging a flea on the thread attached to the pulley. So ##m_{flea}##. I believe his solution is that for ##T_1\gt T_2## the machine will rotate according to your analysis but that the pall requires viscosity to keep from bouncing higher and higher with each impact (you do the mechanics of that if you wish) thus heating the fluid in its container until ##T_2=T_1##.
Feynman does not seem to use viscosity in this treatment, at least not in a direct and recognizable way. He reduces the problem to essentially individual molecules in effect. Please correct me if I am misunderstanding the argument. This is the chapter the problem corresponds to.

https://www.feynmanlectures.caltech.edu/I_46.html

hutchphd said:
Viscosity is never free. All "frictional" forces just sweep energy to degrees of freedom we do not look at in detail. What is the "solution" to the problem that you desire ? I described the physics above. The details of the viscosity are very complicated, so I don't understand what a solution is in your mind.

I agree with what you said. You rightly say that viscosity is needed to solve the problem, but how to apply it if it is so complex to deal with? I don't have any solution in mind, I just want to understand how to go about calculating the expression of angular velocity and energy. Thank you very much.

bob012345 said:
Feynman uses ##L## as torque. You can figure that from the force of gravity, the angle and radius. ##\tau## is an assumed rate. Yes, this is a thermodynamics problem primarily, not just a mechanics problem but the mechanics is consistent with energy expended or generated. As I said and understand Feynman reduces the problem to thinking of individual particle equivalents which is why he uses the fundamental probability ~##e^{-kT}## for each action or event. It is all summarized in Table 46-1.
Is it ##L = mgr \sin \theta##? Thanks.

The torque does not depend upon ##\theta## for a string unwinding from an axle

hutchphd said:
The torque does not depend upon ##\theta## for a string unwinding from an axle
Is it just ##L = mgr##?

hutchphd
Incidentally a similar analysis holds for the question of "why doesn't a diode in a circuit provide free power from fluctuations" This is an aside

Lord Jestocost
hutchphd said:
Incidentally a similar analysis holds for the question of "why doesn't a diode in a circuit provide free power from fluctuations" This is an aside
Interesting. However, I am waiting for a final decision on how to proceed (to apply viscosity or not, etc.), then I will sketch out a process. Thank you.

What particular question are you are trying to answer?

hutchphd said:
What particular question are you are trying to answer?
The second, (a) and (b).

Could you please restate one question completely and then we will do it without ambiguity. Then we can do that again for the second one, and so on.

Yes, thanks.

2. When ##T_1 > T_2##

(a) Estimate the average angular velocity of the system. Assume that each tooth of the ratchet corresponds to a central angle of ##\theta##. The radius of the pulley shaft in the linkage is ##R##, and the mass of the suspended object is ##m##, with all other components being lightweight.

hutchphd
Hak said:
Interesting. However, I am waiting for a final decision on how to proceed (to apply viscosity or not, etc.), then I will sketch out a process. Thank you.
I would suggest asking your teacher for guidance. Otherwise, try the simpler case first then the other.

bob012345 said:
I would suggest asking your teacher for guidance. Otherwise, try the simpler case first then the other.
Ok, thanks, I'll try the simpler case first. I will submit my result shortly.

hutchphd
I now have another new favorite Feynman chapter.......he does his version of the fluctuation-dissipation theorem in 46-2 of this lecture. I think it probably not too different from the original sources Einstein-Smoluchowslki.

bob012345
Hak said:
Ok, thanks, I'll try the simpler case first. I will submit my result shortly.
I'm looking forward to it.

bob012345 said:
I'm looking forward to it.
I am working on it, between tomorrow and the day after tomorrow I will publish my process. Thanks.

bob012345

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