Perpetual motion?

http://neil.fraser.name/news/2008/10/19/

41 47 16 years ago

neil.fraser.name

_pius3• 16 years ago

The problem is: "Assume perfect efficiency."

I don't think so. In analysing designs for perpetual potion machines you're allowed to assume perfect efficiency, unless you can show it to be physically impossible. Here we only need to assume the conversion is sufficiently efficient. That's within the laws of thermodynamics, and the machine still works.

The problem lies elsewhere.

EDIT: In response to a down-vote, let me explain further.

By the laws of thermodynamics, even a perfectly efficient machine cannot result in a net production of energy. In the analysis of an engine the assumption of perfect efficiency is allowed, because really one is saying "assume sufficient efficiency that the production of energy will compensate."

In the case here we're assuming that we can convert mass to energy and vice versa and getting net production of energy. That shouldn't be able to happen, and the falw isn't in the assumption of perfect efficiency. The flaw is elsewhere, and the challenge is to find it.

I do know where it is, but it took me a long time to find it. I'm sure many people here will find it much more quickly than I did.

But the flaw is not in the assumption of perfect efficiency in the mass/energy conversion(s).

_pius3• 16 years ago

you're allowed to assume perfect efficiency, unless you can show it to be physically impossible

Even then, his specification is at too high a level of abstraction for us to prove that it's physically impossible. That seems unfair.

If the author wants people to ignore the abstractions that quickly prove something impossible, then he or she should provide enough detail to analyze it from a practical perspective.

nazgulnarsil4• 16 years ago

"It violates the laws of thermodynamics." This dismissive statement allows one to know that the machine won't work -- regardless of its details. But it is an unsatisfactory answer because it doesn't tell you why it won't work. Essentially it is a statement of faith that somehow the universe won't allow it.

Essentially it is a statement of faith that somehow the universe won't allow it.

statement of faith

GTFO of my science.

mkyc• 16 years ago

I get the feeling that you have no idea how science works, or what he's saying.

He's not saying that the 3rd law is faith, he's saying that you take it on faith. A statement of faith is one that rests on no proof or evidence. Someone who has either wouldn't be giving a dickish answer like "um, 3rd law" or "GTFO", they'd be busy explaining.

The most important part of science is questioning and attempting to falsify current theories. He has a legitimate question, answer it or be quiet.

amix• 16 years ago

Assuming that you can break the laws of thermodynamics is like assuming you can divide by 0 when doing real number arithmetic. If you allow this you would end with a highly inconsistent system/world.

nazgulnarsil4• 16 years ago

lol, it's called thermodynamics and if you didn't take it in college you can get a textbook pretty easily and check the math yourself. you can also do simple experiments at home to confirm the basic laws of thermodynamics. there is no "Science" that only men in important looking buildings wearing lab coats do. that's what distinguishes it from faith.

camccann3• 16 years ago

You know, I sympathize with your argument, but... seriously, the second law of thermodynamics? If there's anything that qualifies as the cornerstone of sanity in physics, it's that.

Of course it's illuminating to analyze the question and figure out where, exactly it's gone wrong, much like it's fun to puzzle out subtly flawed mathematical proofs that 0 = 1.

But in the end an argument that concludes with something equivalent to "therefore, causality is broken" or "therefore, time is meaningless" doesn't really warrant serious consideration unless it's backed by some very serious evidence.

Tichy• 16 years ago

I guess it costs energy to send that energy up the tower. Why wouldn't it?

chrisa• 16 years ago

I think that's it. A recent discussion on reddit[1] mentioned that energy actually weighs something (because of e=mc^2), so the energy required to lift "1kg" of energy would be exactly equal to the amount of energy that dropping it would produce.

[1] http://www.reddit.com/r/AskReddit/comments/awi5k/what_is_the...

lmkg• 16 years ago

This is my guess as well, although to really address his point you have to answer the question "why would it."

He seems to assume that energy doesn't have any mass. It does. "Converting" energy to mass is a misnomer--you're really just converting one form of energy to another, one of which may be stable enough last as "matter" but both of which obey the same laws of gravitation. Transferring the energy up the tower requires energy because the energy has (is) mass.

Note that the E=MC^2 equation is for rest-mass rather than total mass, while a mass in motion has something more like E^2=M^2*sqrt(C^4+V^4). In a different inertial frame of reference objects will appear to have different mass, although I forget if it's rest-mass or total mass. But the point is, in order to get up to the top of the tower, some of the rest-mass will have to be kinetic energy, which gets converted to potential energy as it climbs and back to kinetic energy as it falls.

If you try to beam the energy up as light, this manifests as a Doppler shift in the light frequency (I think this has been observed experimentally). If you send it up with electricity, I think you have gravitational drag on the electrons. In the case of a drive shaft whose axis is parallel with gravity... well, that one actually stumps me. I guess you'll have to work through the relativistic effects. It's probably something really subtle, like that the (slightly) different gravity at the different elevations slows time by different amounts, resulting in a different rate of rotation and therefore less energy out of the top than was put in at the bottom.

jacquesm2• 16 years ago

There is a basic flaw in this article that confuses 'perpetual motion' with 'over unity'.

The difference is subtle: A perpetual motion machine could in theory be constructed, and will continue to move as long as it's components don't wear out due to particle decay or some kind of impact. It simply means that a machine will continue to move.

The cleverest pseudo perpetual motion machines are very hard to debunk, some of the more intricate ones rob the earth of a little bit of momentum to function.

A 'real' perpetual motion machine (one without external energy input to overcome the inevitable engineering issues) has never been constructed but can't be entirely ruled out.

An 'over unity' machine produces more energy than goes in to it, and is an engineering impossibility.

For years I've had this up:

http://web.archive.org/web/20050113055344/http://www.greenbi...

But sadly no takers :)

dutchflyboy3• 16 years ago

Well, that's too bad. I would happily pay 100'000 for a machine that "produces" energy. I mean, you build a few, connect them to the grid, sell the electricity and you've got you're money back in no time.

pyre5• 16 years ago

I'm not sure whether you're trying to be sarcastic or not.

wwalker32• 16 years ago

I'm not a physicist, but here's why I think the energy appears to be less at the top of the tower than at the bottom.

If you send energy up the tower as a light beam, gravitational redshift reduces the energy of each photon as it travels up out of the gravity well.

If you send energy up the tower as electron current in a wire, the electrons spread out from each other slightly as they go up, due to the lower value of the electric constant (epsilon sub zero) in lower gravity, which makes the voltage appear to be lower.

If you send energy up the tower by rotating a shaft, with a motor at the bottom driving a generator at the top, then the energy produced by the generator at the top will appear to be less than that put into the motor at the bottom, again due to the lower value of the electric constant in lower gravity.

In all three cases, if you send the energy back down the gravity well (in whatever form), it'll appear to increase back to its original value, so you can't make a perpetual motion machine this way. And of course I'm assuming perfect conversion efficiency, lack of friction, perfectly rigid materials, et cetera in all three cases.

carbocation• 16 years ago

Absolutely. Thanks to wave-particle duality, gravity affects particles such as photons. We're not used to thinking of photons losing energy as they fight gravity, but gravitational redshift, as mentioned by wwalker3, is quite real.

Thanks to mass-energy equivalence, your choices for visualizing this are many. Perhaps the two easiest are: Physically lifting the 1kg matter+antimatter to elevation h, which is equivalent to Gravitational redshifting of the 1kg-worth (converted to energy) of photons between ground level and elevation h

As an aside, if momentum is conserved, shouldn't each photon be paired and emitted in opposite directions? The author charitably attributes the inability for such a system to be energy neutral to "engineering problems," but I think it's pretty clear that there are also physical laws that make this impossible.

steveitis• 16 years ago

"The author charitably attributes the inability for such a system to be energy neutral to "engineering problems," but I think it's pretty clear that there are also physical laws that make this impossible."

Why not just use a electromagnetic system? Electons with mass down, and em-waves without mass up? Then it's not impossible, just very small forces and superconductors.

Failing that why not use quantum teleportation?

There are several ways to move a subatomic particle in a massless way, and then return it to having mass at a distance.

EDIT My point is to say that the specifics of the enginerring aren't the real question. The real question is "Why doesn't the essential CONCEPT work"?

ewjordan• 16 years ago

Typically we don't consider epsilon naught to change as we move around in a gravitational field, we consider spacetime itself to change and leave the Maxwell equations unaltered. So that doesn't explain away the current in a wire or rotating shaft versions.

Really, the resolution of all of these problems is left to general relativity. Redshift is a consequence of gravitational time dilation, or the fact that if you bring a clock high up in the air and then down again, it will tick off more time than a clock you leave on the ground. If we consider each tick to be a spent pulse of energy, then it's pretty easy to see that the clock that went up and came back down spends more energy than the one on the ground.

Physically speaking, this would mean that if the shaft was, indeed, rigid, then it would actually be spinning at a lower rate if you went to the top and timed it with a stopwatch than it would be if you timed it at the bottom (your stopwatch is running "faster" higher up in a gravitational field, relative to the bottom). Similarly, the rate of electron arrival would be measured to be lower at the top than the bottom, etc.

NB: these rates I'm talking about are the locally measured rates, i.e. hz as measured at the spot with a properly functioning stopwatch. The shaft doesn't need to twist or anything for these rates to differ, in fact, our assumption is specifically that it does not twist.

So when the frequency-based energy is obtained at the top and converted to mass-based energy (which doesn't undergo redshift, but instead exchanges kinetic and potential energies), the "exchange rate" between the two is based on the local measurements of frequency, and we end up with less mass created at the top than we put in at the bottom.

Problem solved, as long as you trust the claim that any such energy transmission that doesn't suffer the kinetic energy penalty that normal projectiles suffer in a gravitational field has to be frequency based and suffer the time dilation effect. Briefly, energy is the time component of a four vector, so it has to be affected like this, no matter how you try to transmit it, but I'll leave that for another rant...

wwalker32• 16 years ago

That's a good point -- to avoid confusing people, I should mention that my explanations above are sort of the mirror image of how most physicists would think about this situation. I was trying to be careful to say "appears to be" instead of "is" to leave this interpretive door open :)

As you say, the most common interpretation is that time itself slows down as you enter a gravity well, leaving all the physical constants the same (since their locally measured values will appear the same to you).

But if you look at things from an imaginary "absolute" viewpoint outside of spacetime, you could think of spacetime itself as a sort of fluid that's denser in regions of high gravity. Light travels more slowly in this denser fluid, and since epsilon zero is inversely proportional to the square of the speed of light, you could interpret it as being larger where gravity is higher.

I guess you can tell that fluid dynamics was always easier for me than differential geometry -- whenever I see a system of nonlinear partial differential equations (like general relativity), it's just easier for me to think of them this way.

teamonkey2• 16 years ago

Redshift, certainly, but there's also the issue of recoil. A photon goes out of the laser at one end and momentum in the laser system must be conserved. Although we're talking about impossible "perfect" devices, a laser must also necessarily create heat.

artagnon• 16 years ago

Exactly. The trick is in realizing that gravity affects everything, including light, and not just physical objects as intuition might suggest. See also: The Museum of Unworkable Devices [ http://www.lhup.edu/~dsimanek/museum/unwork.htm ] for exercises.

cool-RR3• 16 years ago

I believe I have the answer.

There is no way to convert between energy and matter. Thinking that E=mc^2 is an "exchange rate" between energy and matter is a common misconception.

The meaning of E=mc^2 is not that you can exchange E amount of energy for mc^2 amount of mass. The meaning of E=mc^2 is that in every system, the total kinetic energy is equal to the the total mass times the speed of light squared.

The common belief is: When you cause nuclear fission in an atom, that very small mass gets converted into energy, and because the exchange ratio is c^2 (which is enormous), the lost mass turns into a huge amount of energy, which is the nuclear explosion.

This is wrong. The correct explanation is: Because of E=mc^2, we know that even in a small atom there is a huge amount of energy. That energy then gets converted using fission into a more useful form (the explosion.) The mass of the whole system stays constant through the whole process.

dutchflyboy3• 16 years ago

Well, no, that's not the answer. It's not important how the mass is transformed into energy, nor how much. He's just saying that if it's possible to transform x kg of something into y energy and back again (the second part being the real challenge), could I make a machine that "creates" energy?

cool-RR3• 16 years ago

He's just saying that if it's possible to transform x kg of something into y energy

It's not possible. That's what I'm saying. That's the wrong assumption.

ericd2• 16 years ago

From Wikipedia: "The total rest masses of the fission products (Mp) from a single reaction is less than the mass of the original fuel nucleus (M). The excess mass Δm = M – Mp is the invariant mass of the energy that is released as photons (gamma rays) and kinetic energy of the fission fragments, according to the mass-energy equivalence formula E = mc²."

So the energy released by the reaction would be Δmc², right?

cool-RR3• 16 years ago

It says "the total rest masses." This is talking about the rest mass, which is different than the relativistic mass. I believe that the relativistic mass is the one that actually matters.

To your question, I agree that the energy released would be Δmc², but there would be no conversion between mass and energy in the process. The total mass and the total kinetic energy would remain the same after the fission as they were before it.

notaddicted• 16 years ago

If you had a way to transmit and store energy with no loss, you could just transmit it back and forth, and that would be perpetual motion.

I don't know of any indication that the motion of http://en.wikipedia.org/wiki/Voyager_1 will ever stop.

blhack• 16 years ago

"Perpetual Motion" is a bit of a misnomer. The term should be "perpetual force".

If it were "perpetual motion", then everything would constantly be in relative perpetual motion until death of the universe. For example: despite me sitting motionless relative to my computer, I am moving relative to the moon, or the sun, or pretty much everything else in the universe.

teamonkey2• 16 years ago

Voyager 1 is passing through InterStellar Matter which would slow it down relative to the speed that the ISM is travelling.

mynameishere• 16 years ago

A perpetual motion machine is pretty useless until someone hitches a harness to it. That would probably kill it before anything tricky.

Perpetual motion?

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