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

[JaredJames]

A negative temperature system is hot, not cold. Remember, when we have two systems which are exchanging energy through thermal contact the energy will be divided up among the two systems in the manner which maximizes the total entropy. If you have one system that has monotonically increasing entropy and one system that has some peak entropy and if your system with peak entropy has more energy than its peak entropy state then entropy is always increased by transferring that energy to the other system regardless of how hot the other system is. This means that it has a higher temperature than infinity. So thermal energy transfer will always be from a system of negative temperature to a system of positive temperature.

See now that I understand.

So what if you have two systems with an equal peak entropy? Where would they be transferring energy to? (Assuming a closed system). Or do you always need one entropy level higher than the other for this to occur?

 

[Dale]

If you had two systems with peak entropy then it is possible for them to come to equilibrium at some negative temperature if they had sufficient total energy.

 

[Drakkith]

After reading up a bit on this, I think i understand what it is all about.

Anyways, i don't think the carnot cycle is applicable under these circumstances as the negative temperature isn't based on the "thermal temperature" of the material, but on the entropy of the spin states. Unless you can run a heat engine on spin states, then i don't see how it applies. And even if you can, it wouldn't violate the 2nd law of thermodynamics, as the entropy of the negative temp system would increase until it reached equilibrium with the positive temp system. The energy required to push that material into the negative temperature would be transferred into the positive temperature material until both systems were at a positive temperature. Right?

Someone correct me if I'm wrong here.

Edit: Also, you are only counting the spin states as the temperature here. In reality, the actual thermal temperature of the negative temp system, the energy among the other modes would need to be included in this, right? If your temperature for your spin state system were -200, you could still have a temperature of 100k for the rest of the modes for that system, couldn't you?

Edit 2: Also, depending on the amount of energy transferred when the spin states flip, the negative temp system could still have more energy transferred into itself rather than to the other positive temp system if the temperature of the other modes of your negative temp system were lower than the temp of the positive system by enough.

Sorry if this is confusing, I am kind of getting this all out to get myself to understand it as well.

 

[Dale]

i don't think the carnot cycle is applicable under these circumstances as the negative temperature isn't based on the "thermal temperature" of the material, but on the entropy of the spin states.

I don't know what you mean by "thermal temperature of the material".

the entropy of the negative temp system would increase until it reached equilibrium with the positive temp system. The energy required to push that material into the negative temperature would be transferred into the positive temperature material until both systems were at a positive temperature. Right?

Yes.

 

[Drakkith]

I don't know what you mean by "thermal temperature of the material".

The energy in the other modes other than the spin states is what i mean. The translational, vibrational, rotational, electronic, and nuclear modes.

 

[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.