6 ms·

Perpetual motion?

_pius 17y ago
The problem is: "Assume perfect efficiency."
- RiderOfGiraffes 17y ago
  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).
  - _pius 17y 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.
    - RiderOfGiraffes 17y ago
      So you're worried about the "perfect efficiency" and not concerned about the "convert the mass to energy" bit? It's a thought experiment.
      - _pius 17y ago
        It's a thought experiment. Fair enough. :)
nazgulnarsil 17y 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 17y 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 17y 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.
    - RiderOfGiraffes 17y ago
      Yes. So in which step are the laws broken?
  - nazgulnarsil 17y 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.
Tichy 17y ago
I guess it costs energy to send that energy up the tower. Why wouldn't it?
- chrisa 17y 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_craziest_fact_you_know_that_most/c0jrlxs http://www.reddit.com/r/AskReddit/comments/awi5k/what_is_the...
- lmkg 17y 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.
jacquesm 17y 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.greenbits.com/prize.html http://web.archive.org/web/20050113055344/http://www.greenbi... But sadly no takers :)
- dutchflyboy 17y 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.
  - pyre 17y ago
    I'm not sure whether you're trying to be sarcastic or not.
    - jacquesm 17y ago
      Obviously GP is 100% dead serious ;)
    - dutchflyboy 17y ago
      I'm not, I mean seriously, if you make a machine that "produces" energy, you have to be the worlds biggest idiot to sell it for 10'000$. It's like selling a chicken that lays golden eggs, you just don't. You keep it and make profits.
wwalker3 17y 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 17y 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 17y 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"?
cool-RR 17y 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.
- dutchflyboy 17y 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-RR 17y 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.
- ericd 17y 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?
  - 17y ago
notaddicted 17y 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 http://en.wikipedia.org/wiki/Voyager_1 will ever stop.
- blhack 17y 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.
- teamonkey 17y ago
  Voyager 1 is passing through InterStellar Matter which would slow it down relative to the speed that the ISM is travelling.
deleted 17y ago
[deleted]
mynameishere 17y ago
A perpetual motion machine is pretty useless until someone hitches a harness to it. That would probably kill it before anything tricky.

_pius 17y ago

The problem is: "Assume perfect efficiency."

RiderOfGiraffes 17y ago
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).
- _pius 17y 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.
  - RiderOfGiraffes 17y ago
    So you're worried about the "perfect efficiency" and not concerned about the "convert the mass to energy" bit? It's a thought experiment.
    - _pius 17y ago
      It's a thought experiment. Fair enough. :)

"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 17y 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 17y 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.
  - RiderOfGiraffes 17y ago
    Yes. So in which step are the laws broken?
- nazgulnarsil 17y 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.

Tichy 17y ago

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

chrisa 17y 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_craziest_fact_you_know_that_most/c0jrlxs http://www.reddit.com/r/AskReddit/comments/awi5k/what_is_the...
lmkg 17y 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.

jacquesm 17y 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.greenbits.com/prize.html http://web.archive.org/web/20050113055344/http://www.greenbi... But sadly no takers :)

dutchflyboy 17y 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.
- pyre 17y ago
  I'm not sure whether you're trying to be sarcastic or not.
  - jacquesm 17y ago
    Obviously GP is 100% dead serious ;)
  - dutchflyboy 17y ago
    I'm not, I mean seriously, if you make a machine that "produces" energy, you have to be the worlds biggest idiot to sell it for 10'000$. It's like selling a chicken that lays golden eggs, you just don't. You keep it and make profits.

wwalker3 17y 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 17y 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 17y 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"?

cool-RR 17y 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.

dutchflyboy 17y 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-RR 17y 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.
ericd 17y 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?
- 17y ago

notaddicted 17y 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 http://en.wikipedia.org/wiki/Voyager_1 will ever stop.

blhack 17y 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.
teamonkey 17y ago
Voyager 1 is passing through InterStellar Matter which would slow it down relative to the speed that the ISM is travelling.

deleted 17y ago

[deleted]

mynameishere 17y ago

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