Feynman's Brownian Ratchet analysis

[cianfa72]

TL;DR Summary

About the statistical thermodynamics analysis of the Feynman's brownian ratchet

Hi, as in a previous thread I would like to better understand the Feynman's analysis of brownian ratchet as described here:
https://www.feynmanlectures.caltech.edu/I_46.html
https://en.wikipedia.org/wiki/Brownian_ratchet

Consider the case in which the two boxes (i.e. heat baths) are at the same temperature $T$.
The probability to gain the energy $\epsilon$ by a molecule hitting the paddle wheel is $e^{-\epsilon/kT}$. What is at a given point in time the probability to turn forward the ratchet wheel?

The energy taken from the vane is ϵ+Lθ. The spring gets wound up with energy ϵ, then it goes clatter, clatter, bang, and this energy goes into heat. All the energy taken out goes to lift the weight and to drive the pawl, which then falls back and gives heat to the other side.

In the above quote from Feynman 46-2, at first glance, the energy transformed in heat on the paddle wheel side should be $\epsilon$ and not $\epsilon + L\theta$.

 

[Bystander]

Nobody?

"There is no such thing as a free lunch."

 

[hutchphd]

Feynman does a bettr job of explaining than I ever could......Why do you think he is wrong?
All the energy taken out goes to lift the weight and to drive the pawl, which then falls back and gives heat to the other side.
Try a second and third glance assuming he is right!

 

[cianfa72]

All the energy taken out goes to lift the weight and to drive the pawl, which then falls back and gives heat to the other side.

Sorry, when the pawl then falls back it releases only the energy stored in the spring (so the energy that goes into heat on the paddle wheel side does not include the energy used to lift the weight).

 

[cianfa72]

Any comment on my post#4 ? Thank you.

 

[DrClaude]

Sorry, when the pawl then falls back it releases only the energy stored in the spring (so the energy that goes into heat on the paddle wheel side does not include the energy used to lift the weight).

Which case are you considering: the ratchet going forwards or backwards?

 

[cianfa72]

Which case are you considering: the ratchet going forwards or backwards?

The ratchet going forwards.

 

[DrClaude]

The racket going forwards.

Then there is no heat going to the paddel side. ε of heat is added to the ratchet side.

 

[cianfa72]

Then there is no heat going to the paddel side. ε of heat is added to the ratchet side.

Yes, sorry my fault, it was the energy going into heat on the ratchet side. As you said such energy is only $\epsilon$ and not $\epsilon + L\theta$.

So the energy going into heat to the ratchet side is $\epsilon$ while the energy going into heat to the paddle wheel side is $\epsilon + L\theta$. Therefore there is an imbalance of energy transferred into heat on both the sides even though both temperatures are the same.

 

[DrClaude]

So the energy going into heat to the ratchet side is $\epsilon$ while the energy going into heat to the paddle wheel side is $\epsilon + L\theta$. Therefore there is an imbalance of energy transferred into heat on both the sides even though both temperatures are the same.

Yes, according to the scenario as presented up to this point. On which Feynman then reflects:

According to Carnot’s hypothesis, it is impossible. But if we just look at it, we see, prima facie, that it seems quite possible. So we must look more closely. Indeed, if we look at the ratchet and pawl, we see a number of complications.

 

[cianfa72]

Yes, according to the scenario as presented up to this point. On which Feynman then reflects:

Sorry, this statement from Feynman predates his analysis of the ratchet and pawl device as presented in section 46-2.

I believe the point is that, even though the value of energies transferred into heat are different on each side (paddle wheel side vs ratchet side) however, on each side separately, there is a balance between the energy transferred from heat into work and from work into heat.

On the paddle wheel side it is $\epsilon + L\theta$ while on the ratchet side is just $\epsilon$. In any case, since there is a balance on each side, no net work is done (on average no weight is lifted) and the same temperature $T$ on each side doesn't change.