Contradiction/breach 2nd law? Simply with carnot engine + negative temperature

[Dale]

It doesn't matter what kind of mode the energy is in. As long as you have a system with some limited number of possible modes then you can get negative temperatures. The spin states is just one nice example.

 

[Pythagorean]

Wow, my mind was blown like two or three times in this thread

 

[nonequilibrium]

As mr. vodka already mentioned, the temperature of systems with a negative temperature is higher than that of any system with a positive temperature. The discovery of such systems did lead to a slight modification of the second law of thermodynamics. Google "Kelvin-Planck-Ramsey statement".

Hm, I don't really seem to get the statement. It just basically seems to say "you can't do what you just did" without saying why I can't: can you see what part is not allowed? Is it that $$\Delta S = \int \frac{dQ}{T}$$ does not apply to negative temperature systems for some reason?

 

[Drakkith]

It doesn't matter what kind of mode the energy is in. As long as you have a system with some limited number of possible modes then you can get negative temperatures. The spin states is just one nice example.

True, but as there is currently no upper limit on the number of translational, vibrational, rotational, electronic, and nuclear modes, then you cannot, in reality, get an actual negative temperature for the "whole" system. If you can devise a machine to get work out of ONLY spin states, then kudos. But the carnot cycle takes energy from the whole system, not just 1 part of it. The whole system wouldn't have a negative temperature, only the spin states (or whatever mode that has a finite number of states).

I guess what I am getting at, is that i don't see negative temperature as something that, in the view of the whole system, is an actual negative temperature. In this case, the spin states will transfer energy to balance themselves back out and that energy will be transferred to the other modes, increasing the overall temperature and entropy of the material. In the unique case of finite modes, yes you can have a negative temp. But not in the bigger picture.

Hows that sound? Does it make sense?

 

[JaredJames]

True, but as there is currently no upper limit on the number of translational, vibrational, rotational, electronic, and nuclear modes, then you cannot, in reality, get an actual negative temperature for the "whole" system. If you can devise a machine to get work out of ONLY spin states, then kudos. But the carnot cycle takes energy from the whole system, not just 1 part of it. The whole system wouldn't have a negative temperature, only the spin states (or whatever mode that has a finite number of states).

I guess what I am getting at, is that i don't see negative temperature as something that, in the view of the whole system, is an actual negative temperature. In this case, the spin states will transfer energy to balance themselves back out and that energy will be transferred to the other modes, increasing the overall temperature and entropy of the material. In the unique case of finite modes, yes you can have a negative temp. But not in the bigger picture.

Hows that sound? Does it make sense?

That sir, is exactly what I was trying to say in my posts, but failed miserably.

Although the maths may show a negative temperature occurs, in reality, it isn't so due to all the other factors.

 

[nonequilibrium]

So the argument of both of you is simply put "You can't isolate certain modes." but the fact you can't find a way to do that doesn't mean it can't be done... It might be practically hard, but I'm sure it can be done to a scale that's good enough for the theory to be sufficient. Anyway, it hardly seems to be of the point... No offence, but as you both knew nothing about statistical mechanics more than a day ago, I can't give you guys a lot of credit.

 

[Drakkith]

That sir, is exactly what I was trying to say in my posts, but failed miserably.

Although the maths may show a negative temperature occurs, in reality, it isn't so due to all the other factors.

Haha, yeah.

After a bit more thought, i realized that if you knew how much energy the negative system would transfer to balance itself out, then you could put that into a positive temperature for the purposes of calculating heat. In other words, if the negative temp system had enough energy to raise the temp of the other system by 50k, then one could say that instead of the temperature being -200k (or whatever the temp), it would actually be 50k MORE than the positive temp system. The actual math might not be right, but i think you can understand what I am getting at. In this case, you wouldn't plug a negative value into the carnot cycle, but a positive one, with one temp being 50k more than the other.

 

[Drakkith]

So the argument of both of you is simply put "You can't isolate certain modes." but the fact you can't find a way to do that doesn't mean it can't be done... It might be practically hard, but I'm sure it can be done to a scale that's good enough for the theory to be sufficient. Anyway, it hardly seems to be of the point...

I'm sorry if i seemed like i was trying to disprove the theory or something, i was not.

For the purposes of this thread, the negative temp wouldn't violate the 2nd law of thermodynamics, because you arent actually using a negative temp for a value in your equations of the carnot cycle.

 

[nonequilibrium]

Because you say so?

Sigh this thread is getting swamped :/ The chance of somebody bothering to read up and reply now are slim to none

 

[Drakkith]

Because you say so?

Sigh this thread is getting swamped :/ The chance of somebody bothering to read up and reply now are slim to none

Do you agree or disagree with what i was saying above?
Edit: I actually want to know. I just came up with all this over the last 30 minutes of even reading up on negative temperature, so if I've made a mistake somewhere or don't know something please tell me.

 

[nonequilibrium]

I'll PM you as to not get off-topic.

 

[Gerenuk]

I didn't read the last discussion, but back to the topic:

Now if you've followed closely, the net result of going from A to B: heat has been taken out of 2 and work has been delivered without any other effect in the universe.

There is a flaw here. If you allow for spontaneous heat transition, then you leave the shell of constant entropy. Basically this step is not reversible and leaves the two reservoirs in a completely new state, which cannot be brought back to the very initial state by reversible processes.

So does this have a basis in real life? Is it useful or just some piece of maths that shows A but we can only achieve B due to the laws of physics?

Just any system where the maximum energy does not get infinite, can have negative temperature. It's just that plain gases are mostly used. They of course can have arbitrarily fast molecules and thus cannot have negative temperature.

 

[Pythagorean]

so could you somehow exploit this effect (negative temperature systems) in the magnetocaloric effect to get higher gains from your system?

 

[nonequilibrium]

I didn't read the last discussion, but back to the topic:

Thank you!

There is a flaw here. If you allow for spontaneous heat transition, then you leave the shell of constant entropy. Basically this step is not reversible and leaves the two reservoirs in a completely new state, which cannot be brought back to the very initial state by reversible processes.

Well the spontaneous heat transition was not even essential, so you can leave it out. Leaving it out, you have an ever weirder situation:

heat has gone out of 2 and partially delivered work and partially gone into a reservoir WITH HIGHER TEMPERATURE! So it's like breaking the 2nd law twice :P

NOTE TO READERS: we're discussing a make-up situation I made up to try and show Gerenuk that if you have a (reversible) engine using reservoir 1 with T_1 = -200K as hot reservoir and reservoir 2 with T_2 = -273K as cold reservoir, that the engine will DELIVER work and not use it up.

But anyway, Gerenuk, is it not obvious that the engine will give work? If there is a spontaneous heat flow from 1 to 2 if allowed, then surely an engine can use this tendency to get work out, that's exactly the principle of an engine. The error must surely be somewhere else.

Pyth - I don't think so... but I can't find a flaw in this very elementary derivation, so I don't know what to answer yet.

 

[Drakkith]

Ok. Let me try to explain this again. =)

If you know the amount of energy transferred from one system to the other, then you can calculate your temperatures as a positive value instead of a negative. I think that is the key here. The equation for the carnot cycle will NOT work with a negative value for its temperature due to the math. One much use a positive value, which is achievable.

Thus, if your -273k system is having heat transferred to it from the -200k system, you could calculate the amount of energy transferred and how much that raised the temperature of the system, and by knowing that, could convert those temperatures into a positive value and plug them into the equation, getting the correct values from it.

Does this look right, or have a screwed up somewhere?

 

[JaredJames]

Ok. Let me try to explain this again. =)

If you know the amount of energy transferred from one system to the other, then you can calculate your temperatures as a positive value instead of a negative. I think that is the key here. The equation for the carnot cycle will NOT work with a negative value for its temperature due to the math. One much use a positive value, which is achievable.

Thus, if your -273k system is having heat transferred to it from the -200k system, you could calculate the amount of energy transferred and how much that raised the temperature of the system, and by knowing that, could convert those temperatures into a positive value and plug them into the equation, getting the correct values from it.

Does this look right, or have a screwed up somewhere?

Again, that is how I'm seeing it.

The negative values are simply another way of looking at the numbers.

The transfer from system A to B is 100K or from system B to A is -100K

Using the negative values just returns an inverted result so it looks as if B is working on A (ish, I think that's what I'm getting at).

 

[nonequilibrium]

Drakkith & jared, please stop spamming, there's just too little sense in those posts. I don't want to scare off people that actually know what they're talking about.

 

[Dale]

True, but as there is currently no upper limit on the number of translational, vibrational, rotational, electronic, and nuclear modes, then you cannot, in reality, get an actual negative temperature for the "whole" system. If you can devise a machine to get work out of ONLY spin states, then kudos. But the carnot cycle takes energy from the whole system, not just 1 part of it. The whole system wouldn't have a negative temperature, only the spin states (or whatever mode that has a finite number of states).

I guess what I am getting at, is that i don't see negative temperature as something that, in the view of the whole system, is an actual negative temperature. In this case, the spin states will transfer energy to balance themselves back out and that energy will be transferred to the other modes, increasing the overall temperature and entropy of the material. In the unique case of finite modes, yes you can have a negative temp. But not in the bigger picture.

Hows that sound? Does it make sense?

I don't know what you mean by "the whole system". You can arbitrarily define your system boundaries depending on what behavior you are interested in studying. There is no requirement that you compose your system such that there is always an infinite number of modes. And, in fact, in many QM systems you do have a limited number of modes. Statistical mechanics and negative temperature can become important for analyzing such systems. You cannot just wave this away in this manner.

Suppose that you are not building an engine to drive an automobile, but rather an engine on an atomic scale for some sort of "lab on a chip" application. QM effects and quantized states could become important depending on the scale, and could become the dominant source of energy. The OP is asking a rather general question that I don't think can be dismissed so easily.

 

[Pythagorean]

Wouldn't the 'hotter' system be -200, since -273 is less than -200?

 

[Drakkith]

Again, that is how I'm seeing it.

The negative values are simply another way of looking at the numbers.

The transfer from system A to B is 100K or from system B to A is -100K

Using the negative values just returns an inverted result so it looks as if B is working on A (ish, I think that's what I'm getting at).

Exactly.

The carnot cycle will produce work when "energy" is transferred between one system and another. Temperature is simply the means of getting and storing that energy if that makes sense. Since a negative temperature will transfer energy to another system with a positive temperature, its is the same as if that negative temp system was actually a positive temperature that was +X kelvin more than the other system.

 

[Drakkith]

Wouldn't the 'hotter' system be -200, since -273 is less than -200?

I was using the numbers from Mr Vodkas post from the hot and cold reservoirs.

I don't know what you mean by "the whole system". You can arbitrarily define your system boundaries depending on what behavior you are interested in studying. There is no requirement that you compose your system such that there is always an infinite number of modes. And, in fact, in many QM systems you do have a limited number of modes. Statistical mechanics and negative temperature can become important for analyzing such systems. You cannot just wave this away in this manner.

Suppose that you are not building an engine to drive an automobile, but rather an engine on an atomic scale for some sort of "lab on a chip" application. QM effects and quantized states could become important depending on the scale, and could become the dominant source of energy. The OP is asking a rather general question that I don't think can be dismissed so easily.

True, but in the context of this thread, which has to do with the carnot cycle and a heat engine, I believe it makes sense.

 

[Dale]

Mr. vodka, I think that this point is probably the key.

Basically this step is not reversible and leaves the two reservoirs in a completely new state, which cannot be brought back to the very initial state by reversible processes.

He is correct, the energy transfer from the hot reservoir to the piston is not reversible, and the piston cannot transfer energy to the cold reservoir. To make such a system work you would need to have your "piston" also be at a negative temperature, meaning that it could not use an ideal gas.

 

[Dale]

I have been thinking about this problem this afternoon and come to a conclusion. The reason that an ideal gas can be used in an engine is because the entropy is not solely a function of the energy, but also of the volume:
http://hyperphysics.phy-astr.gsu.edu/hbase/therm/entropgas.html

If entropy were solely a function of energy then a piston of ideal gas would occupy some 1D curve on a TS diagram and would therefore be unable to do any net work. But instead a piston can be anywhere on the TS diagram, with some specific volume at each point.

To do a negative temperature cycle you need a system with a maximum energy and an equation for entropy which is a function of energy and something similar to volume. If you have that, then you should be able to make a thermodynamic cycle purely in negative temperatures and define some efficiency for the cycle.

 

[nonequilibrium]

DaleSpam, you're right! A paramagnet cannot deliver work the way a gas can

Hm, this is kind of weird... This would imply that the 2nd law states you can't have a system with bound energy and the ability to do work? Surely there must exist an example (well not if my previous sentence is right, but I doubt that)

Nice insight, thank you :)

 

[Dale]

I agree that there probably is an example, but I don't know it. We just need some system where

S=f(U,V)

Umin<U<Umax and V is some other state variable. Any such system should be able to make a thermodynamic cycle in negative temperatures.

 

[Andy Resnick]

The main problem that I see is that a system at negative temperature is not in an equilibrium state- one can't use such a metastable state as a thermal reservior. Energy must constantly be supplied to keep the system in its metastable state.

Heat engines can operate via latent heat rather than specific heat (i.e. phase changes), but I don't see how to apply metastable systems as reservoirs.

Of course, the whole notion of allowing a thermodynamic negative temperature (as opposed to a statistical notion of 'effective temperature') is flawed.

 

[Drakkith]

Of course, the whole notion of allowing a thermodynamic negative temperature (as opposed to a statistical notion of 'effective temperature') is flawed.

Agreed.

 

[JaredJames]

Of course, the whole notion of allowing a thermodynamic negative temperature (as opposed to a statistical notion of 'effective temperature') is flawed.

Agreed also.

I believe it's what myself and Drakkith have been getting at from the start!

 

[Drakkith]

Agreed also.

I believe it's what myself and Drakkith have been getting at from the start!

That, and i was trying to explain how i thought the error with the equation could be resolved by converting the negative temp to a positive number.

 

[nonequilibrium]

Hello Andy Resnick,

Hm, in what way do you mean "a semistable state"? I don't know what you mean.

And as for the negative temperature: is you calling it "flawed" expression of a personal preference or something objective?

Thank you.

 

[JaredJames]

That, and i was trying to explain how i thought the error with the equation could be resolved by converting the negative temp to a positive number.

Of course, again I agree.

 

[Gerenuk]

Of course, the whole notion of allowing a thermodynamic negative temperature (as opposed to a statistical notion of 'effective temperature') is flawed.

Some time ago, in a long discussion people showed you that negative temperature is a perfectly valid equilibirum state and you agreed in the end.
All you need is a system or two whose state energy is bound. Then you put in so much energy as to almost saturate the energy capacity. In such a case the temperature of both systems is negative. Right?

 

[Dale]

The main problem that I see is that a system at negative temperature is not in an equilibrium state- one can't use such a metastable state as a thermal reservior. Energy must constantly be supplied to keep the system in its metastable state.

This is not correct. A system at negative temperature is not in equilibrium with any system at positive temperature, but it can certainly be in equilibrium with another system at negative temperature. In such a situation energy does not need to be constantly supplied.

Of course, the whole notion of allowing a thermodynamic negative temperature (as opposed to a statistical notion of 'effective temperature') is flawed.

Careful here. This sounds like you (and Drakkith) are promoting personal theories. The idea of negative temperatures is a result of the standard statistical mechanics definition of temperature. I don't know what your "effective temperature" definition is, but Drakkith's idea is not only a personal theory but an untenable one. Temperature is a state variable, and by definition state variables depend only on the state of the system, not on the state of other systems.

Andy, Drakkith, and jarednjames, your objections to studying this are not relevant. The OP's question is based on mainstream physics, simply applied to an unusual but physically realizable system. Should we also not discuss Bose-Einstein condensates simply because they are also unusual? None of your objections are rational objections based on mainstream physics, only personal discomfort with the standard idea of negative temperature.

 

[Drakkith]

This is not correct. A system at negative temperature is not in equilibrium with any system at positive temperature, but it can certainly be in equilibrium with another system at negative temperature. In such a situation energy does not need to be constantly supplied.

Careful here. This sounds like you (and Drakkith) are promoting personal theories. The idea of negative temperatures is a result of the standard statistical mechanics definition of temperature. I don't know what your "effective temperature" definition is, but Drakkith's idea is not only a personal theory but an untenable one. Temperature is a state variable, and by definition state variables depend only on the state of the system, not on the state of other systems.

Andy, Drakkith, and jarednjames, your objections to studying this are not relevant. The OP's question is based on mainstream physics, simply applied to an unusual but physically realizable system. Should we also not discuss Bose-Einstein condensates simply because they are also unusual? None of your objections are rational objections based on mainstream physics, only personal discomfort with the standard idea of negative temperature.

My statements are as relevant as trying to use a negative temperature in the equations of a carnot cycle. Not only that, i was trying to explain a way to resolve the problems that others were running into with that equation. Instead of ANYONE explaining to me why my explanations were not valid, all i have received is personal scorn and ridicule. I readily asked for feedback telling me why i was incorrect, yet i have still not received any other than "Your wrong".

Nowhere in this thread have I tried to come up with personal theories or told anyone that they shouldn't study this effect. I see the whole thread as a big misunderstanding of physics and have attempted to explain that. I'm only sorry so few of you can see that.

Now, i ask again, someone EXPLAIN how i was wrong anywhere in this thread in my thinking.

 

[Gerenuk]

Instead of ANYONE explaining to me why my explanations were not valid, all i have received is personal scorn and ridicule. I readily asked for feedback telling me why i was incorrect, yet i have still not received any other than "Your wrong".

People have given examples where a negative temperature can occur. It's not our fault if you skip reading these paragraphs.

EDIT: I myself actually skipped one argument you gave. Here is the answer: you cannot simple convert negativ temperature to positive, because temperature is defined by T=dE/dS. The signs of dE and dS are well defined.