Heat into work?

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Heat into work?

Postby ExplodingHat » Wed Jan 26, 2011 7:44 am UTC

http://home.sandiego.edu/~jhwright/Maxw ... ct2002.pdf

The basic idea of the article is:
A p-n junction results in a potential difference between the two ends of the diode (the diffusion of charges being a thermal process). If this diode is wrapped into an incomplete loop, the voltage difference at the ends creates a strong electric field (from which, theoretically, mechanical work can be extracted).

I'm no expert in solid-state, but from what I do know, the reasoning looks disconcertingly... solid. 8)
Anyone in the field with any insights into what else might be at play here?
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Re: Heat into work?

Postby ++$_ » Wed Jan 26, 2011 8:50 am UTC

We can make a similar device with a semipermeable membrane (J-I) and some salts. We'll put the membrane in the middle of a tube full of water, add sodium chloride on one side and potassium iodide on the other, and bend the thing into a U shape. If the membrane is permeable only to sodium and iodide, the sodium will flow from the sodium side to the potassium side, and the iodide will flow from the the potassium side to the sodium side, creating a membrane potential. Now, let's close the U shape with another identical semipermeable membrane (this is analogous to the way he closes J-II). What happens?

Well, nothing happens, obviously. The concentrations are at the equilibrium, with the osmotic pressure opposing the electric field and resulting in no net flow of ions. Whoops.

I know nothing about solid state physics, but it seems to me that the same thing happens with the diode. As electrons and holes diffuse across the p-n junction, a voltage is produced, but so is a concentration gradient. Eventually the concentrations will reach equilibrium, and at this point closing J-II does nothing.

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Re: Heat into work?

Postby Tass » Wed Jan 26, 2011 10:52 am UTC

Yes. If you have a temperature difference between the two junctions then you can get work out, that is well known and used.

If the second junction does not touch, then you do get a field in the air gap, but not a field that can be used for continuous work.

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Re: Heat into work?

Postby quantropy » Wed Jan 26, 2011 11:30 am UTC

Isn't it just a diode in the end? And hasn't everyone devised ways of breaking the second law of thermodynamics using a diode? The trouble is that hot diodes don't work properly. The effect of thermal noise always allows some reverse flow, and this cancels out the effect. Google 'Smoluchowski trapdoor' for more info.

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Re: Heat into work?

Postby ExplodingHat » Wed Jan 26, 2011 4:30 pm UTC

I looked into it a little more, and found this diagram (same guy):
http://www.sciencedirect.com/science?_o ... 246a85c514

It seems the idea is that the gap electric field provides an attractive force, pulling the loop closed. When it's closed the electric field effectively goes to zero, allowing the spring to open it up again, causing the now-open loop to charge up again. The hard part I'm guessing is matching the time constants of the diode's charging and the spring.
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Re: Heat into work?

Postby Antimony-120 » Wed Jan 26, 2011 4:51 pm UTC

Huh...I don't see a flaw in the logic at first glance, but I'd have to actually go through and check the formulae. I'm hesitant to say I expect interesting results from an experiment, partly because if something sounds this good in thermodynamics it's usually wrong. Partly because the wording sounds a little too optimistic, and that is often a bad sign in science papers.

And they accounted for thermal noise actually. It's thermal noise that's driving the thing (in a rather clever manner).

It'd be worth seeing if there are papers that cite this one, and see what they have to say on the issue.
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Re: Heat into work?

Postby ++$_ » Wed Jan 26, 2011 5:32 pm UTC

ExplodingHat wrote:It seems the idea is that the gap electric field provides an attractive force, pulling the loop closed. When it's closed the electric field effectively goes to zero, allowing the spring to open it up again, causing the now-open loop to charge up again. The hard part I'm guessing is matching the time constants of the diode's charging and the spring.
Like I said before, the electric field does not go to zero when you close the gap. When you close the gap, absolutely nothing happens.

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Re: Heat into work?

Postby ExplodingHat » Thu Jan 27, 2011 1:19 am UTC

++$_ wrote:When you close the gap, absolutely nothing happens.


Except that closing the gap creates another P-N junction. When they come into contact they would form another depletion region (which does not have the high electric field characteristic of the open configuration).
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Re: Heat into work?

Postby ++$_ » Thu Jan 27, 2011 5:52 am UTC

The electric field strength at a particular point is not relevant at all. Only the potential matters.

Let me explain. The potential difference tells me how much energy I can obtain by moving a charge from A to B. If I am trying to extract energy by moving charges from A to B, then I need to know the potential difference. I don't care what the electric field is -- in fact, that is the whole point of potential! Let's say the potential difference is 1 volt. If A and B are 1000 m apart, then the electric field is 10-3 V/m. If A and B are 1 mm apart, then the electric field is 103 V/m. However, in either case I obtain 1 J by moving 1 coulomb from A to B.

Now, Sheenan et al. believe at least one of the following two things. I can't tell for sure which one they believe, but they are both false.

Thing 1: They believe that it is possible to extract energy from the electric field without moving charges from one end to the other.
Thing 2: They believe that (positive) charges will flow from the positive end of their apparatus to the negative end without input of energy.

Whoa -- why is Thing 2 false? Simply, because positive charges do not like to flow into p-type semiconductors. There is a force that prevents them from doing so. This force probably has a name, but I'm just going to abuse terminology and call it osmotic pressure, because it is thermal in origin and arises from the fact that the concentration of positive charges inside the p-type semiconductor is higher than the concentration of positive charges outside of it.

So a neutral p-type semiconductor has a high osmotic potential which is going exclude any positive charges. In order to get the p-type semiconductor to accept more positive charges, you have to give it a strongly negative electric potential, so much so that the sum of electric and osmotic potential is negative. At that point it will accept positive charges, but only until the electric potential and osmotic potential are equal.

The setup of Sheenan et al. uses the initially high osmotic potential of a neutral p-type semiconductor in order to move positive charges into the n-type semiconductor (this is while J-II is open). This will equalize the electric and osmotic potentials, resulting in 0 total potential. Thus, when they close J-II, no charges will flow, since there is no potential gradient. Of course, there is an electric potential gradient (also known as an "electric field"), but only the overall potential gradient matters.

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Re: Heat into work?

Postby ExplodingHat » Thu Jan 27, 2011 8:00 am UTC

++$_ wrote:The electric field strength at a particular point is not relevant at all.


From what I understand the work on the cantilever device is extracted via electrostatic pressure, which is dependent upon field strength.
Thing1:
Wikipedia, re:'Energy density' wrote:A pressure gradient has a potential to perform work on the surroundings by converting enthalpy until equilibrium is reached.

Thing2:
Wikipedia, re:'Depletion region' wrote:A depletion region forms instantaneously across a P-N junction.

When J-II is closed, is it not a P-N junction?
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Re: Heat into work?

Postby ++$_ » Thu Jan 27, 2011 9:47 am UTC

From what I understand the work on the cantilever device is extracted via electrostatic pressure, which is dependent upon field strength.
The pressure (or force) is dependent on the magnitude of the field (i.e. the energy density). The amount of work that one can extract isn't.

Again, for a given potential difference, if you increase the separation by a factor of 10, you will decrease the force/field/pressure by a factor of 10 (using the simplifying assumption that we are talking about infinite charged sheets), but it will extend over a distance 10 times greater. Since [imath]W = \int F\,dr[/imath], you will get the same amount of work either way. This is what it means to say that any electric field is conservative.

While the size of a force is important for practical considerations (it needs to overcome friction, for example), we are concerned here with proving that this device does not even work in theory. Hence, we do not need to worry about frictional losses, because the device does not even work in the best case scenario where there are no frictional losses. So we don't care about the magnitude of the field.
A pressure gradient has a potential to perform work on the surroundings by converting enthalpy until equilibrium is reached.
Yes, this is correct, and the amount of work you can get out of the field is measured by the potential difference times the amount of charge you move. If you move no charges, you will not get any energy out. That is why I am saying that to extract energy from electric fields, you have to move charge from an area of higher potential to an area of lower potential.
A depletion region forms instantaneously across a P-N junction.

When J-II is closed, is it not a P-N junction?
Yes, and that's the whole problem. What Wikipedia is saying (correctly, as I understand it) is that as soon as a p-n junction forms, it will develop a potential difference of V (where V depends on the materials used and on the ambient temperature). In all of these setups, when J-II becomes closed, the newly formed p-n junction already has a potential difference of V across it. This means that nothing will happen when it becomes closed. In particular, no current will flow, so the hammer and anvil (c.f. the bottom part of the diagram you linked earlier) will not spontaneously separate, so the device does not oscillate.

Note that this is not true of the equivalent device driven by a battery (the diagram in the upper part of the image). The negative terminal of the battery is willing to accept the positive charge carriers, unlike the negatively charged (p-type) end of the semiconductor. This happens because the battery contains a low-energy sink for positive charge carriers. As a result, as soon as the hammer and anvil touch, a current will flow and eliminate the electric field, allowing the spring to separate the plates again. So the battery-powered device will oscillate until the sink has filled up. (Of course, once the sink is full, the battery is dead and it stops oscillating.)

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Re: Heat into work?

Postby Moose Hole » Thu Jan 27, 2011 3:39 pm UTC

Looks a lot like a Brownian ratchet.

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Re: Heat into work?

Postby ExplodingHat » Sat Jan 29, 2011 5:58 am UTC

++$_ wrote:Yes, this is correct, and the amount of work you can get out of the field is measured by the potential difference times the amount of charge you move. If you move no charges, you will not get any energy out.

When the two surfaces attract, are not the charges in the hammer moving? Isn't that how the battery-based model works?
++$_ wrote:In all of these setups, when J-II becomes closed, the newly formed p-n junction already has a potential difference of V across it. This means that nothing will happen when it becomes closed. In particular, no current will flow, so the hammer and anvil will not spontaneously separate, so the device does not oscillate.

I would imagine that in the case of the closed loop, the more favorable state would be to have 2 depletion regions (at J-I and J-II) than to have a depletion region at J-I and some sort of weird charge buildup thing at J-II. If you know of such a phenomenon, however, do tell; it sounds interesting. I would say the main issue of them not separating would be due more to van der Waals forces than anything, although presumably a small enough contact area could relieve this.
Moose Hole wrote:Looks a lot like a Brownian ratchet.

Aside from the proposed second-law contradiction, I don't see much resemblance here. (I.e: no scaling down of macroscopic phenomena beyond their realm of applicability)
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Re: Heat into work?

Postby ++$_ » Sat Jan 29, 2011 8:12 am UTC

ExplodingHat wrote:
++$_ wrote:Yes, this is correct, and the amount of work you can get out of the field is measured by the potential difference times the amount of charge you move. If you move no charges, you will not get any energy out.

When the two surfaces attract, are not the charges in the hammer moving? Isn't that how the battery-based model works?
Yes -- maybe I wasn't clear enough. It is possible to extract energy from bringing a p-type semiconductor into contact with an n-type semiconductor, but only a certain amount of energy. (This is true in the same way that if you bring water into contact with sulfuric acid, you will receive a certain amount of energy and no more.) In the case of the hammer/anvil device, the first strike of the hammer extracts all the energy that the device is capable of extracting. After that, the device reaches equilibrium no more energy is extracted, because no more charge moves through the electric field. If they think that any more energy can be extracted, they are wrong, because no charge is flowing!
I would imagine that in the case of the closed loop, the more favorable state would be to have 2 depletion regions (at J-I and J-II) than to have a depletion region at J-I and some sort of weird charge buildup thing at J-II. If you know of such a phenomenon, however, do tell; it sounds interesting. I would say the main issue of them not separating would be due more to van der Waals forces than anything, although presumably a small enough contact area could relieve this.
A depletion region is a charge buildup (or, more precisely, a charge separation -- negative charge builds up on the p-side and positive charge on the n-side).

The details of exactly where the charge goes are irrelevant in this case, though, because the error of Sheenan et al. doesn't depend on those details. They got the wrong sign* -- in order for the hammer to separate from the anvil after the first strike (other than through bouncing, which will eventually die out anyway), somehow the charge difference between the hammer and anvil must decrease. That is the only way to decrease the attractive force between hammer and anvil and allow the spring to separate them.

But in fact, the charge difference is going to increase or stay the same. When a p-n junction is formed, there is a specific potential difference that arises, depending on the materials and the temperature. Let's say the n-side gets a potential of +V and the p-side gets a charge of -V. (If you prefer to use charge density instead of potential, that's fine, and it makes no difference.) Then in the scenario given, the hammer is going to have a potential of at most +V. There is no way that it will have a greater potential than V, because the only mechanism charging it up is the potential difference across J-I. Meanwhile, the anvil will have a potential of magnitude at most -V, for exactly the same reason. Therefore, when they come into contact, a new p-n junction is formed, and this junction will immediately attempt to charge the n-side to +V and the p-side to -V To accomplish this, positive charge will flow from the anvil to the hammer. In no event will positive charge flow from the hammer to the anvil -- that is the wrong way, because the hammer wants to be charged more positively than it already is! Like I said, the details of the exact flow of charges don't matter at all -- the sign is still wrong even if the hammer is not charged to the full +V, or the anvil is not charged to the full -V, or both.

Van der Waals forces have nothing whatsoever to do with it.

Now, you might be able to make an engine using this technique, as long as you are willing to keep J-I and J-II at different temperatures. However, then you have built a heat engine, which is not a violation of the 2nd law.

*Yes, I am asserting that in this peer-reviewed and (miraculously) published paper, the researchers misunderstand the idea of "equilibrium" so badly that they can't even get the sign of the current right, and that I have somehow gotten it right despite knowing nothing about solid state physics.
Aside from the proposed second-law contradiction, I don't see much resemblance here. (I.e: no scaling down of macroscopic phenomena beyond their realm of applicability)
Yeah, I don't exactly see the connection to the Brownian ratchet either.

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Re: Heat into work?

Postby sikyon » Sat Jan 29, 2011 6:39 pm UTC

Would it be possible for someone to briefly explain what is supposed to happen in this device? I don't have the stomach to read 31 pages of the paper.

Let me try to say what I think the authors are trying to do, and if someone could correct me that would be great.

1. We have a loop of metal. Half of it is p-type and half of it is n-type.
2. There is a depletion region where n-type meets p-type. In this situation, electrons from the n-type and holes from the p-type diffuse into each other. What remains is positive charges left on the n-type and negative charges left on the p-type.
3. There is now a charge difference between the n-type and p-type regions. This leads to a charge imbalance between the two regions.
4. Now we take the back part of the device and say, "oh look there's a force from the 2 charged regions. The gap closes."
5. Since the gap closes, and there's an n-region touching a p-region, charge must flow from each region to the other one.
6. This neutralization in charge causes there to be no more force, and the device opens up again.
7. Process repeats, gain energy from gap closing/opening.

If I am correct, this is what happens: (+ is positive charges, 0 is no charge, - is negative charges)

Before contact:
n: +++++000000000 000000000------- :p

After contact:
n: ++++++++++++++--------------------:p

I don't understand how this device is purported to work.

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Re: Heat into work?

Postby ExplodingHat » Sat Jan 29, 2011 10:11 pm UTC

++$_ wrote:In no event will positive charge flow from the hammer to the anvil -- that is the wrong way, because the hammer wants to be charged more positively than it already is!

Could not the depletion region at J-I thin out to accommodate such a flow, though?
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Re: Heat into work?

Postby ++$_ » Sun Jan 30, 2011 4:00 am UTC

sikyon wrote:1. We have a loop of metal. Half of it is p-type and half of it is n-type.
2. There is a depletion region where n-type meets p-type. In this situation, electrons from the n-type and holes from the p-type diffuse into each other. What remains is positive charges left on the n-type and negative charges left on the p-type.
3. There is now a charge difference between the n-type and p-type regions. This leads to a charge imbalance between the two regions.
4. Now we take the back part of the device and say, "oh look there's a force from the 2 charged regions. The gap closes."
5. Since the gap closes, and there's an n-region touching a p-region, charge must flow from each region to the other one.
6. This neutralization in charge causes there to be no more force, and the device opens up again.
7. Process repeats, gain energy from gap closing/opening.
Yeah, that's how it's supposed to work.
ExplodingHat wrote:Could not the depletion region at J-I thin out to accommodate such a flow, though?
I don't understand what you mean. It doesn't matter how far the depletion region penetrates into the semiconductor, because there is still not going to be a greater potential difference than 2V.

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Re: Heat into work?

Postby sikyon » Sun Jan 30, 2011 3:02 pm UTC

++$_ wrote:
sikyon wrote:1. We have a loop of metal. Half of it is p-type and half of it is n-type.
2. There is a depletion region where n-type meets p-type. In this situation, electrons from the n-type and holes from the p-type diffuse into each other. What remains is positive charges left on the n-type and negative charges left on the p-type.
3. There is now a charge difference between the n-type and p-type regions. This leads to a charge imbalance between the two regions.
4. Now we take the back part of the device and say, "oh look there's a force from the 2 charged regions. The gap closes."
5. Since the gap closes, and there's an n-region touching a p-region, charge must flow from each region to the other one.
6. This neutralization in charge causes there to be no more force, and the device opens up again.
7. Process repeats, gain energy from gap closing/opening.
Yeah, that's how it's supposed to work.


OK, thanks. I actually have a fair bit of knowledge in solid state electronic device physics, so maybe I can provide my view on how it doesn't work?

Before contact we have:

n: +++++++0000000 0000000-------- :p

After contact we have a new depeltion region formed

n: ++++++++++---------------:p

As we can see, the new depletion region does not in any way oppose the old depletion region - the attractive force becomes stronger.

There is another case, however:

Before contact:

n: +++++++++++++ -----------------:p

So in this case we assume that the materials is doped lightly enough that the depletion region is the entire device.

After contact:

n:+++++++++++---------------:p

Note that the charges do not dissipate into each other. The structure formed is not:
n:+++++++++000000000000----------:p --> n:++++++++0000000 0000000000---------:p ----> n:+++++++++++ -----------:p

Why? Simple. The charges in the depletion region (positive charges left in the n-type region and negative charges left in the p-type region) are NOT mobile. They cannot annihilate each other. They are due to doping ions that are embedded into the structure of the semiconductor. This is the reason the depletion region exists in the first place.

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Re: Heat into work?

Postby ExplodingHat » Sun Jan 30, 2011 7:45 pm UTC

Sikyon, I'm not sure I quite understand your notation in 1-D. (I keep reading the :p side as an emoticon. >_<) In any case, the paper explicitly states that above a certain dopant concentration, the open configuration no longer has more energy than the closed configuration.

I think their reasoning is something along the lines of:

___0000000000000000
___0______________0
___+______________+
J-I +______________+ J-II
___-______________-
___-______________-
___0______________0
___0000000000000000
(Ignore the underscores.)

If I were to make a loop-shaped semiconductor, with half of the loop N-doped and the other P-doped, there is no argument that this should not be be the equilibrium configuration. (A depletion region at each junction.)
Now if I were to "cut" that loop at J-II, would this not be the new equilibrium state?

___0000000000000000
___+______________0
___+______________0
___+______________ <==Electric field.
___-______________
___-______________0
___-______________0
___0000000000000000

Sheehan et. al wrote:When J-II is open, there are about 330 free electronic charges on each gap face (calculable from Gauss' law); when it is switched closed, most of these disperse through and recombine in the J-II bulk. This net flow of charges is due to particle diffusion powered by concentration gradients and to particle drift powered by the large capacitive electric field energy of the open J-II vacuum gap. Thermodynamically, this energy release may be viewed as simply the relaxation of the system from a higher to a lower energy equilibrium state.

So it seems they believe that J-II should behave like any other P-N junction.
Edit: I.e: it will oscillate between the above 2 states.
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Re: Heat into work?

Postby ++$_ » Sun Jan 30, 2011 8:01 pm UTC

ExplodingHat wrote:Now if I were to "cut" that loop at J-II, would this not be the new equilibrium state?

___0000000000000000
___+______________0
___+______________0
___+______________ <==Electric field.
___-______________
___-______________0
___-______________0
___0000000000000000
No, this is not correct. There will not be an electric field between two neutral objects, as you have drawn there.

In any case, where did the charges go? You started with positive and negative charges separated across J-II. Afterwards, apparently both sides are neutral. How is this supposed to have happened?

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Re: Heat into work?

Postby ExplodingHat » Sun Jan 30, 2011 8:08 pm UTC

I guess my notation was confusing, I was using the ++-- thing to represent a depletion region.
Edit:
___0000000000000000
___+______________0
___+______________+
___+______________ <==Electric field.
___-______________
___-______________-
___-______________0
___0000000000000000
In terms of charge, I guess this would be what I meant to say; the field holds the charges at the J-II gap faces.
Last edited by ExplodingHat on Sun Jan 30, 2011 8:16 pm UTC, edited 1 time in total.
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Re: Heat into work?

Postby ++$_ » Sun Jan 30, 2011 8:11 pm UTC

Maybe you could do the diagrams instead using + and - to represent charges?

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Re: Heat into work?

Postby ExplodingHat » Sun Jan 30, 2011 8:33 pm UTC

The hammer-anvil design is actually not addressed in the originally linked paper, but is treated in more detail here, in case you were interested.
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Re: Heat into work?

Postby sikyon » Sun Jan 30, 2011 9:57 pm UTC

In any case, where did the charges go? You started with positive and negative charges separated across J-II. Afterwards, apparently both sides are neutral. How is this supposed to have happened?


What happens is that charges from the opposite region should drift around to the other side, thereby neutralizing the ions in the depletion region.

So ok, we have a closed loop and we cut it open and it goes to neutral state. Why should it close again? And for that matter, why should it open in the first place?

If the authors are arguing that at a neutral state there are still carriers on the gap faces, can you point me to their math? This does not make intuitive sense to me. When you dope a semiconductor with, for example, a n-type dopant, you have excess electrons, yes, but only relative to the states available in the valence band. However, you do not have excess charge because every extra electron is balanced by an extra proton. Putting an n-type and a p-type semiconductor in proximity does not make the 2 devices want to come together.

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Re: Heat into work?

Postby ++$_ » Sun Jan 30, 2011 10:08 pm UTC

ExplodingHat wrote:The hammer-anvil design is actually not addressed in the originally linked paper, but is treated in more detail here, in case you were interested.
Look -- I know exactly why this thing doesn't work. I've been trying to explain it to you, because you asked if anyone knew what was "at play here." And I do. I can't describe it in intimate submolecular detail, because I don't know anything about the details on that level, but I do know that they are totally irrelevant to the overall fact that the device does not represent a violation of the 2nd law. If you do want to understand the intimate submolecular detail, you had better ask sikyon, because he knows how it all works. But it is unnecessary.

What is "at play here" is that, as I've described, the device does not work as described by the authors; instead, it reaches an equilibrium and remains there forever. It doesn't matter how you connect the two sides of the device -- with a hammer/anvil design, or with a linear electrostatic motor design. It's still going to reach an equilibrium and remain there forever. I've given explanations of why this happens, and why what the authors are wrong about what will happen.

Any paper overlooking this fact, including the paper linked in the original post or the one you linked above, is a crock of shit. It's not helpful to cite the papers back at me, because they are just plain wrong. If you continue to accept the authors' conclusions at face value, you will never understand why the device doesn't work, because many of the authors' claims (e.g. that the device will oscillate) are false. Don't be mislead by the fact that the papers are full of equations and physicsy words like "van der Waals force"," "depletion region," and "metastable". They are like the claims of a person who declares that he has a method allowing people to walk on water, and discusses in detail how surface tension, viscosity, drag, wave motion, and so on must be taken into account in his scheme, but completely neglects to consider the force of gravity. Similarly, Sheehan et al. take it completely for granted that the potential difference at J-II will be neutralized when the gap is closed, and so they don't bother to consider whether this is true. They do consider mechanical damping, electrical resistance, van der Waals forces, and other such things, and conclude that they can all be accounted for with the right construction of the device, but they completely failed to examine their implicit assumption about the potential difference being neutralized. Needless to say, it is this one essential yet unexamined assumption that is false.

I guess what I'm trying to say is: no, I am not interested in the authors' defense of the hammer/anvil design. Nevertheless, I read the paper, and needless to say it is as wrong as the 2002 paper, although more readable. In fact, the clearer writing makes it even more obvious how they are wrong.

By the way, please don't ask me to account for the fact that these papers, containing completely elementary mistakes, were produced by someone who is apparently a tenured professor at a decent research university. I can't explain how someone who is apparently intelligent enough to get tenure could be so profoundly stupid and unreflective as to lend his name to this paper. Incidentally, I would give him the benefit of the doubt, except that he has also authored several other credulous papers on the subject of second-law violations. For example, I examined his paper of 2000 with Glick and Means, published in Foundations of Physics 30, and it also contained a purported second-law-violating device, discussed uncritically by the authors. This device, too, fails for trivial reasons (it is a version of the Brownian ratchet). The appendix B to this paper, which contains many straw-man objections to the supposed paradox, along with the authors' responses, is a particularly amusing read.

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Re: Heat into work?

Postby sikyon » Mon Jan 31, 2011 9:45 pm UTC

Just to add a thought:

One of the major reasons papers like this get published is because of the review process. Reviewers are not gods of knowledge or ultra-experts. Papers are basically the cutting edge, and it is very difficult to get reviewers who are experts in obscure fields. How many people do you think make it their life goals to analyze devices which violate the 2nd law?

The fact is that the more obscure and difficult the topic is, the easier it is to get it published much of the time.

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Re: Heat into work?

Postby Moose Hole » Mon Jan 31, 2011 10:07 pm UTC

sikyon wrote:Just to add a thought:

One of the major reasons papers like this get published is because of the review process. Reviewers are not gods of knowledge or ultra-experts. Papers are basically the cutting edge, and it is very difficult to get reviewers who are experts in obscure fields. How many people do you think make it their life goals to analyze devices which violate the 2nd law?

The fact is that the more obscure and difficult the topic is, the easier it is to get it published much of the time.
I'm not sold on the OP, but this doesn't seem like a good argument. You're saying that the reviewers let this one go because they don't know enough about physics, and yet the random posters on this board know enough to discount the paper.

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Re: Heat into work?

Postby Antimony-120 » Mon Jan 31, 2011 10:20 pm UTC

Moose Hole wrote:
sikyon wrote:Just to add a thought:

One of the major reasons papers like this get published is because of the review process. Reviewers are not gods of knowledge or ultra-experts. Papers are basically the cutting edge, and it is very difficult to get reviewers who are experts in obscure fields. How many people do you think make it their life goals to analyze devices which violate the 2nd law?

The fact is that the more obscure and difficult the topic is, the easier it is to get it published much of the time.
I'm not sold on the OP, but this doesn't seem like a good argument. You're saying that the reviewers let this one go because they don't know enough about physics, and yet the random posters on this board know enough to discount the paper.


This board is a little odd in that random posters often have physics degrees. Also, peer-review is notorious for having a group of busy scientists give a paper a pass (particularly if it's a minor paper in an obscure area) and assume that if there's a major problem somebody else will do it.

In fact that exact thing happened earlier on these fora, when I first posted in this thread and commented that I didn't see an immediate problem but that I didn't trust it. If I was a reviewer I might well have passed it on the merit of not seeing something on a five second glance. It scores remarkably low on the crackpot index for a Second Law Violating device (but I become more convinced it is wrong after a second glance. Still haven't gone through and done the physics though).
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Re: Heat into work?

Postby ExplodingHat » Tue Feb 01, 2011 3:31 am UTC

++$_ wrote:This device, too, fails for trivial reasons (it is a version of the Brownian ratchet).

Doesn't Feynman's Ratchet fail because the pawl is subject to Brownian motion? I don't see the analog for a moon-sized object.
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Re: Heat into work?

Postby ++$_ » Tue Feb 01, 2011 5:30 am UTC

According to them, their device should move towards the "sticky" side of the gravitator. They say this is because molecules hitting the "shiny" side collide elastically, whereas molecules hitting the sticky side are absorbed temporarily and then re-emitted at the temperature of the gravitator. For simplicity, let's assume that the atmosphere consists of only one molecule; to understand the behavior of an actual gas, we'll simply sum this approximation a large number of times. (Sure, this disregards collisions within the gas, but their argument does not even make mention of such collisions.)

Now, let's begin by assuming that the single molecule starts out with a higher temperature than that of the gravitator. (We'll soon analyze the case where its temperature is equal.) The molecule will strike the gravitator on the shiny side or the sticky side with equal probability. If it strikes the shiny side, it produces a large force towards the sticky side, causing the gravitator to move. If it strikes the sticky side, it produces a small force towards the shiny side, BUT it also loses energy (its temperature decreases to that of the gravitator). Therefore, the expected net force on the gravitator is towards the sticky side. However, there is nothing remarkable about this, because the force has been produced at the expense of the flow of heat from a hot object (the gas) to a cool object (the gravitator). So there is no second law violation when the gas is hotter than the gravitator. Therefore, this case is not of interest to us. Similarly, the case where the single molecule has a lower temperature than the gravitator is also not of interest, since the flow of heat from the gravitator to the gas is allowed to produce some work (and in fact it does, although the work is in the opposite direction of the previous case).

Now, let's say the gas is equal in temperature to the gravitator. This time, when the particle strikes on the sticky side, no energy is lost. However, now the force exerted by a particle striking the sticky side is equal to that of a particle striking the shiny side! Even though the particle striking the sticky side is temporarily absorbed, when it is later re-emitted, it has exactly the same speed it had before. Now, there is a subtlety, because it may not be re-emitted at the same angle. Assuming the re-emission angle is random, though, when we sum over many collisions, the emission angles on the sticky side have the same distribution as the absorption angles on the sticky side. This is also true on the shiny side, so there is no net force on the gravitator. Therefore, we extract no energy!

Sheehan et al. are not total fools, however. They obviously noticed this. This is why they specify that the object is a gravitator. That way, if we have a particle 100,000 miles away from the gravitator with temperature equal to that of the gravitator, by the time it comes into contact with the gravitator, it will have more energy (i.e. the impact will be "suprathermal").

This doesn't matter. After the particle has struck the sticky side for the first time, it comes away with some amount E of kinetic energy, which is just enough to make its temperature the same as that of the gravitator. As it drifts away from the gravitator, some of this kinetic energy is converted into gravitational potential energy (we measure this as a cooling of the gas). As it returns to the vicinity of the gravitator, the potential energy is converted back into kinetic energy. Regardless, when it returns to the gravitator it will once again have kinetic energy E, because gravity is conservative. So gravity does not affect the analysis of the device, except that we must make a note of the fact that the gas will reach thermal equilibrium with the gravitator when the temperatures are equal as measured at the surface of the gravitator.

There is one other thing that Sheehan et al. mention, which is a heat bath at temperature T. As far as I can tell, the heat bath is supposed to surround the entire cavity, thereby being in contact with the gas at the cavity wall. In this case, if the heat bath is hotter than (or cooler than) the gas at the cavity wall, we have a temperature gradient which is allowed to do work, so this case is uninteresting. Therefore, we assume that the heat bath is in thermal equilibrium with the gas at the cavity wall. Here, when equilibrium is reached, the gas inside the chamber has temperature T at the boundary of the chamber, while the gravitator and the gas in contact with it has temperature equal to T + t, where t is the change in temperature experienced by a particle falling under gravity from the cavity wall to the gravitator. There is still no energy extracted, and hence no contradiction.

I say this is like the Brownian ratchet because, like the Brownian ratchet, an asymmetry exists only when the gas is not in thermal equilibrium with the body it acts upon. As soon as the gas reaches thermal equilibrium with the body, the forces become symmetrical and no more work can be extracted.

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Re: Heat into work?

Postby ExplodingHat » Tue Feb 01, 2011 7:44 am UTC

++$_ wrote:I say this is like the Brownian ratchet because, like the Brownian ratchet, an asymmetry exists only when the gas is not in thermal equilibrium with the body it acts upon.

I would argue that the Feynman ratchet analogue is only valid in the case of a non-gravitational object, in that only thermal kinetic energy is considered. One feature seemingly characteristic of Sheehan's paradoxes, however, is that they involve a second type of energy that acts as a catalyst of sorts. (I.e: electrical, gravitational, chemical, etc.)

++$_ wrote:Sheehan et al. are not total fools, however. They obviously noticed this. This is why they specify that the object is a gravitator. That way, if we have a particle 100,000 miles away from the gravitator with temperature equal to that of the gravitator, by the time it comes into contact with the gravitator, it will have more energy (i.e. the impact will be "suprathermal").

From what I understand, the idea is that the temperature of the bath, cavity, and small-moon(space station? :wink: )-sized gravitator start out the same. Gas particles (of mass M) are gravitationally accelerated from their thermal velocity(V) to a "suprathermal"(V+) one by the object. Objects impinging on the shiny hemisphere of the Death Star hit the surface at this suprathermal velocity, just like the ones that hit the sticky side. The particles on the shiny side are reflected at the same speed they came with, while the sticky-side particles are re-emitted at the thermal velocity.The difference in momenta is imparted to the Death Star. Since, as you mentioned, gravitational energy is conserved, the energy driving this net motion must be thermal. Decreasing the overall kinetic energy of the gas would, of course, cool it down. Therefore, this system starting at a single temperature produces a temperature gradient, violating the Clausius definition of the Second Law.

*Edited for clarity.
Last edited by ExplodingHat on Tue Feb 01, 2011 6:17 pm UTC, edited 1 time in total.
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Re: Heat into work?

Postby ++$_ » Tue Feb 01, 2011 8:31 am UTC

ExplodingHat wrote:Therefore, this system starting at a single temperature produces a temperature gradient, violating the Clausius definition of the Second Law.
That is not the Clausius statement of the second law. The Clausius statement is (according to Wikipedia) "No process is possible whose sole result is the transfer of heat from a body of lower temperature to a body of higher temperature." In this case, the gas is moved relative to the gravitational field, so the sole result of the process is not the transfer of heat from a low-temperature body to a high-temperature body. Therefore, there is no violation.

For a simple counterexample to your version of the Second Law, if I have two giant space rocks at the same temperature out in the middle of nowhere, and I wait for them to crash into each other due to gravity, there will soon be a temperature gradient (because the rocks hit each other and the impacting side of each heats up). This is not a violation of the second law because there are other results (namely, the movement of the rocks relative to the gravitational field). For it to be a violation of the second law, the rocks would then have to return to their original positions while keeping the temperature gradient.

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Re: Heat into work?

Postby quantropy » Tue Feb 01, 2011 2:53 pm UTC

sikyon wrote:Just to add a thought:

One of the major reasons papers like this get published is because of the review process. Reviewers are not gods of knowledge or ultra-experts. Papers are basically the cutting edge, and it is very difficult to get reviewers who are experts in obscure fields. How many people do you think make it their life goals to analyze devices which violate the 2nd law?

The fact is that the more obscure and difficult the topic is, the easier it is to get it published much of the time.

But isn't the whole idea of peer review to find reviewers who do know enough about the subject matter to be able to judge it?

Also I think that there are probably more people than you think who know enough about the Second Law of Thermodynamics to say why devices designed to defy it won't work. When I was at school and invented a few such devices my physics teachers couldn't tell me why they wouldn't work, but when I got to university, some of the smart physics undergraduates there knew enough to tell me what the problem was.

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Re: Heat into work?

Postby ExplodingHat » Tue Feb 01, 2011 6:27 pm UTC

++$_ wrote:For a simple counterexample to your version of the Second Law, if I have two giant space rocks at the same temperature out in the middle of nowhere, and I wait for them to crash into each other due to gravity, there will soon be a temperature gradient (because the rocks hit each other and the impacting side of each heats up). This is not a violation of the second law...

...because the heat will then distribute throughout the rocks, reaching equilibrium. I meant to imply that the temperature gradient continues to widen after established. Sorry about that.
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