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

[Drakkith]

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.

I understand perfectly what everyone is saying and nowhere will i say that you CANNOT have a negative temp. The negative temp is only applicable in a specific situation where you have a finite number of states.

Anyways, if you know the amount of energy that would be transferred from a negative temp, then why couldn't you convert that into a positive temperature in relation to the positive system? I definitely don't know the math behind all this, but I am fairly sure this wouldn't be impossible to do.

 

[Gerenuk]

Because temperature is defined by T=dE/dS. And all equations are based on this identity. You cannot arbitrarily introduce a minus sign there. Or at least it wouldn't make sense, because you had to rewrite all the other thermodynamics equation and thus going back to negative temperature.

 

[Drakkith]

Because temperature is defined by T=dE/dS. And all equations are based on this identity. You cannot arbitrarily introduce a minus sign there. Or at least it wouldn't make sense, because you had to rewrite all the other thermodynamics equation and thus going back to negative temperature.

In regards to the first post.

If -273 is Colder than -200, then you need to switch your T1 and T2 around.

For example. If t1=200 and T2 = 273, then its 200/273. The cooler temp is on the top.

Now, if T1 = -273 and T2 = -200, then its -273/-200. The energy in the system would flow from 2 to 1 because there is more energy in T2 than in T1, just like in the 1st example up top.

Is this correct?

 

[Drakkith]

Now, let's say that T1 = 200 and T2 = -200. If we know the amount of energy that -200 would transfer into T1, then we can say that T2 is actually equal to a positive temp that is greater than T1. For the purposes of this argument let's say that if the two temps were equalized, T2 would introduce as much energy into T1 as a temperature of 300k would.

So, instead of 200/-200, its actually 200/300.

Even though a negative temp has more heat than any positive temperature, it still has a finite amount of energy that it can transfer to a lower temperature.

Does this make sense?

Edit: I'm not changing the actual temperature of T2 here, but merely converting it to a positive number that will work in the equation. In this case, T still equals dE/dS.

 

[Dale]

Instead of ANYONE explaining to me why my explanations were not valid, all i have received is personal scorn and ridicule.

This is completely unsubstantiated. I have not once made any personal attack on you, have tried to teach the concepts involved, and have three times given different explanations of problems with your idea:

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.

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.

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.

In light of the above your assertion that you have not received any explanation and instead have received scorn is absurd. I have tried throughout this thread to both address the OP's question and to teach you and others about some interesting mainstream physics. I don't know what more I could possibly do, and your accusation here is both personally upsetting and factually unwarranted.

 

[Andy Resnick]

Hello Andy Resnick,

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


Thank you.

I said 'metastable'- a system assigned a negative temperature can only remain in that state if it does not interact with anything else, or energy is supplied to the system to maintain it in that state.

 

[Andy Resnick]

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?

Honestly, I don't recall- can you locate the thread? I'm not saying you are wrong.

 

[Drakkith]

Dale, you haven't understood AT ALL what any of my posts have been getting at have you? Granted I probably haven't used the 100% correct terminology, but its not that hard to understand.

In following with the title of this thread, a negative temperature would ONLY relate to the spin states of a system. Since the carnot cycle DOES NOT use spin states as a way to get work, then the negative temperature doesn't matter.

You guys decided to get off topic and when I stuck to it and tried to claim that the carnot cycle wouldn't work in this circumstance under real world physics then everyone has a freakin aneurysm. I thought we stuck to mainstream science here, not wishful thinking. Because that's about all this is right now. We "MIGHT" be able to get an engine using the carnot cycle to use the spin states at some point in the future, but not right now.

Edit: Also, $$\Delta S = \int \frac{dQ}{T}$$ must use a positive number for a temperature.

 

[Dale]

a negative temperature would ONLY relate to the spin states of a system.

This is incorrect and has already been addressed multiple times:

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.

Just any system where the maximum energy does not get infinite, can have negative temperature.

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.

Edit: Also, $$\Delta S = \int \frac{dQ}{T}$$ must use a positive number for a temperature.

No, if S decreases for an increase in Q then T must be negative. That is the whole point.

 

[Andy Resnick]

Careful here. This sounds like you (and Drakkith) are promoting personal theories.

I can't speak for Drakkith, but:

http://prola.aps.org/pdf/PR/v104/i3/p589_1

Has a very cogent discussion (with additional references) spelling out the needed condition to establish a system with negative temperatures. None of these criteria have been part of the OP's thought experiment, nor the subsequent "analysis".

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.

Sigh... The OP is proposing using *reservoirs* at negative temperatures to drive a heat engine. He did not propose the *existence* of a system at negative temperature. In fact, my objection was very clearly stated as such. Before claiming I am not making a rational objection, perhaps you could put forth a reference supporting your belief that a *thermal reservior* at negative temperatures is mainstream physics?

 

[Drakkith]

This is incorrect and has already been addressed multiple times:

No, if S decreases for an increase in Q then T must be negative. That is the whole point.

No, it isn't incorrect. I'm using real world science applicable to this thread. I don't know what your using. I know that negative temp is in any systems with a finite number of states. We don't have any systems like that other than spin states currently though.

Its obvious to me that you are unwilling or unable to understand my posts, and you really havnt even addressed them other than to point out some made up inaccuracies that you think i have in them because you don't understand what I am getting at. I'm done talking to a brick wall for the now. Goodnight all.

 

[Dale]

I can't speak for Drakkith, but:

http://prola.aps.org/pdf/PR/v104/i3/p589_1

Has a very cogent discussion (with additional references) spelling out the needed condition to establish a system with negative temperatures. None of these criteria have been part of the OP's thought experiment, nor the subsequent "analysis".

I can't access the link, I would be very interested.

Sigh... The OP is proposing using *reservoirs* at negative temperatures to drive a heat engine. He did not propose the *existence* of a system at negative temperature. In fact, my objection was very clearly stated as such. Before claiming I am not making a rational objection, perhaps you could put forth a reference supporting your belief that a *thermal reservior* at negative temperatures is mainstream physics?

AFAIK there is no real definition of what constitutes a "thermal reservoir". So I would have a hard time excluding a negative temperature system on that basis.

I don't think that the problem is in treating a negative temperature system as a reservoir, I think the problem is having a cycle in a negative temperature system. I don't know that such a system exists since I don't know enough negative temperature systems to know what other state variables they may have besides energy.

 

[Dale]

No, it isn't incorrect.

Yes, it is. Look at the formula.

If $$\Delta S$$ is negative for a positive $$dQ$$ then $$T$$ is negative according to the formula you supplied. I don't know how you can reach any other conclusion.

I don't know what your using.

Basic statistical mechanics definitions.

 

[nonequilibrium]

Phew, quite the discussion.

Andy Resnick: you might be confused if you thought I was suggesting an actual gas as the engine? I was assuming the system doing work itself could also have negative temperature. DaleSpam pointed out the problem might be in the actual finding of a system that could function as such a negative temperature engine, so we'd agree with you if you were to object that we can't just postulate an engine doing work on negative temperature. But your objections might lay elsewhere, but I'm not sure where: negative temperature seems to do its work just fine and has a logical definition, so I don't know why one would object. Of course the idea can be odd, since what we're saying is basically that if you'd throw an ideal paramagnet in the sun, the paramagnet would heat the sun up, but of course on that level our assumptions (that we can isolate certain modes) fall down, but this doesn't mean we can't explore it theoretically, and that if the theory tells us it's feasible, I'm sure we can find certain conditions where it can be actually construed. On a side-note: I'm not actually expecting to find a breach in the 2nd law, just finding the flaw in this situation.

DaleSpam, what about magnetic work! I got the idea of http://www.phys.uri.edu/~gerhard/PHY525/wtex4.pdf (but they seem to have specifically stuck to positive temperature)

 

[DrDu]

I don't think there is any principal flaw in the system considered and I am not even too surprised that it requires work to thermalize the two systems reversibly.
However, take in mind that the working medium of the Carnot machine has also to be a system capable of assuming negative temperatures, e.g. a collection of spins in a magnetic field. Then it should be possible to visualize the process in a simple model. Work can be done by the medium by shifting the magnet generating the field. At negative temperatures, the material is diamagnetic, i.e. the working substance and the magnet repell. The repulsion is higher the more negative the temperature. Shifting the magnet while in contact with one of the reservoirs would constitute the isothermal part of the process. Changing the magnetic field with the inversion hold fixed the adiabatic part.

 

[maimonides]

This question must have been settled somewhere. In the 5th volume of Landau/Lifgarbagez (a 1979 German edition) there is a chapter on negative temperatures.
IIRC there is another system with negative temperatures: the population inversion needed in lasers and masers is also described by negative temperatures.

 

[Andy Resnick]

I can't access the link, I would be very interested.

I believe the article is now public domain- I have the pdf, but I can't figure out how to stick it here.

AFAIK there is no real definition of what constitutes a "thermal reservoir".

You may claim these are "personal theories", but definitions can be found here:

Fermi, Thermodynamics, p.32
Sears and Salinger, Thermodynamics, Kinetic Theory, and Statistical thermodynamics, p.83
Halliday, Resnick,and Walker, Fundamentals of Physics, p438
Serway and Faughn, College Physics, p342
Tolman, Principles of Statistical Mechanics, p 556
Landau and Lifgarbagez, CTP, vol 5, pg 58

 

[Andy Resnick]

Phew, quite the discussion.

Andy Resnick: you might be confused if you thought I was suggesting an actual gas as the engine?

I am aware that you are trying to make a heat engine using Ramsey-type systems.

 

[Pythagorean]

Without reading Andy's resources, I was taught (in Thermo) that a heat reservoir was hot body such that

as you take sigmaQ (a small amount of heat compared to the total heat) from the reservoir to power your system, the reservoir's temperature stays the same.

 

[nonequilibrium]

Couldn't find a definition in Fermi or Tolman on mentioned pages.

 

[Andy Resnick]

Couldn't find a definition in Fermi or Tolman on mentioned pages.

Obstinance is not a virtue.

http://dictionary.reference.com/browse/reservoir : #5 is particularly relevant.


res·er·voir
   /ˈrɛzərˌvwɑr, -ˌvwɔr, -ˌvɔr, ˈrɛzə-/ Show Spelled[rez-er-vwahr, -vwawr, -vawr, rez-uh-] Show IPA
–noun
1. a natural or artificial place where water is collected and stored for use, esp. water for supplying a community, irrigating land, furnishing power, etc.
2. a receptacle or chamber for holding a liquid or fluid.
3. Geology . See under pool1 ( def. 6 ) .
4. Biology . a cavity or part that holds some fluid or secretion.
5. a place where anything is collected or accumulated in great amount.
6. a large or extra supply or stock; reserve: a reservoir of knowledge.

Origin:
1680–90; < F réservoir, equiv. to réserv ( er ) to reserve + -oir -ory2

 

[nonequilibrium]

Resnick, you're rude.

 

[Drakkith]

Yes, it is. Look at the formula.

If $$\Delta S$$ is negative for a positive $$dQ$$ then $$T$$ is negative according to the formula you supplied. I don't know how you can reach any other conclusion.

Basic statistical mechanics definitions.

Of course its negative if your using it for the negative temp. But to use these equations for a heat engine you can't have it negative, it must be positive. Otherwise weird things happen, like heat flowing from the cold to the hot. Or the engine doing work using less energy than was put in. Wasn't that the problem on the 1st page of this thread? The math wasn't working out correctly?

 

[nonequilibrium]

Drakkith - Well, if that is truly the reason why the math went wrong, then I'll agree that I won't like the concept of negative temperature anymore. But how can you know that the weird answer I got in the start is due to that? It might be, but if people throughout history just had gone "oh wrong answer let's delete the math" without even knowing it was the math, I think you'd agree that we wouldn't have gotten far. The concept of temperature is a well-defined thing in stat. mech, it's not something arbitrary; you can't just replace a minus-sign without harming the concept it has. If you don't understand what that concept really is, well that's okay, but then you have to go read a stat. mech. book and not just claim on forums that it can't make sense without having tried to understand it. I know you have repeatedly said that we don't just simply say where you are wrong, but I believe a few good people have tried but the fact is what you're basically asking us to do is to explain the whole of stat. mech., cause you can't just expect that you can understand what certain concepts are without having a course about it where you make exercises to gain a natural feeling of why that concept is useful and maybe even essential for such systems.

 

[Drakkith]

Drakkith - Well, if that is truly the reason why the math went wrong, then I'll agree that I won't like the concept of negative temperature anymore. But how can you know that the weird answer I got in the start is due to that? It might be, but if people throughout history just had gone "oh wrong answer let's delete the math" without even knowing it was the math, I think you'd agree that we wouldn't have gotten far. The concept of temperature is a well-defined thing in stat. mech, it's not something arbitrary; you can't just replace a minus-sign without harming the concept it has. If you don't understand what that concept really is, well that's okay, but then you have to go read a stat. mech. book and not just claim on forums that it can't make sense without having tried to understand it. I know you have repeatedly said that we don't just simply say where you are wrong, but I believe a few good people have tried but the fact is what you're basically asking us to do is to explain the whole of stat. mech., cause you can't just expect that you can understand what certain concepts are without having a course about it where you make exercises to gain a natural feeling of why that concept is useful and maybe even essential for such systems.

Vodka, here is the problem with the equation in your original post.

In $$\frac{Q_1}{Q_2} = \frac{T_1}{T_2}$$ you MUST put the system that the energy will flow into as temperature 1, and the system that the energy will flow out of and into the engine as temperature 2.

-273 is LESS temp than -200, but you used -273 as temperature 2. Thats the reason why it seemed like there was more heat being deposited on T2 than was coming out of T1.

If you converted both negative temperatures to a positive temperature, just like I've shown in previous posts, the equation would work perfectly as well.

Also, for $$\Delta S_{universe} = - \frac{Q_1}{T_1} + \frac{Q_2}{T_2} = 0$$ , where do you find the original equation? Why would you have to change the signs in it?

 

[xxChrisxx]

I'm not really going to make comment on the thread content. Being an engineer, I like to stick firmly in the realm of classical thermodynamics. This is frankly beyond me, and I have little interest in -ve temperatures being hotter than infinitely hot tempereatues or whatever... urgh just a recipie for melting my poor brain.

My one point would be, why is a statistical mechanics question posted on the classical physics board?
That's what could be causing comprehension issues.

 

[Drakkith]

I'm not really going to make comment on the thread content. Being an engineer, I like to stick firmly in the realm of classical thermodynamics. This is frankly beyond me, and I have little interest in -ve temperatures being hotter than infinitely hot tempereatues or whatever... urgh just a recipie for melting my poor brain.

My one point would be, why is a statistical mechanics question posted on the classical physics board?
That's what could be causing comprehension issues.

Could you just answer this for me? If -273K is less than -200k, then in the very original post, -273K should be T1 correct? (The math won't work, as its still negative, but it is still the state with less temperature)

 

[xxChrisxx]

Could you just answer this for me? If -273K is less than -200k, then in the very original post, -273K should be T1 correct? (The math won't work, as its still negative, but it is still the state with less temperature)

I don't know enough about statistical mechanics and how the temperature value range works to answer that.

 

[Drakkith]

I don't know enough about statistical mechanics and how the temperature value range works to answer that.

Fair enough.

 

[DrDu]

What he probably means is that a system with negative temperature cannot continuously be transfromed into a system with positive temperature going through T=0 but only going through T= infinity. Hence it would be better to consider beta=1/T instead of T.

 

[Drakkith]

Because temperature is defined by T=dE/dS. And all equations are based on this identity. You cannot arbitrarily introduce a minus sign there. Or at least it wouldn't make sense, because you had to rewrite all the other thermodynamics equation and thus going back to negative temperature.

You misunderstand what I'm saying. The temperature is still a negative temperature. However, you can calculate the energy that the system would deliver to another system and from that you could convert that negative temperature into a positive temperature for use in the math.

For example, if a 500k system delivered 50 joules of energy to a 200k system, and -100k system ALSO delivered 50 joules of energy to a 200k system, then for the purposes of math you could simply insert 500k into all the spots where you would have put -100k.

200k(T1)/500k(T2) = Q1/Q2.
200k(T1)/-100k(T2) = Q1/Q2.
You could also rewrite this as 200k/500k=200k/-100k. In terms of energy content, then both 200k/500k and 200k/-100k would be the same in this circumstance.

If QH is the energy transferred from the hot resevoir, and QC is the energy being transferred to the cold reservoir, then:
QH in the 200k/500k system would equal QH in the 200k/-100k system. Its the same for QC.

Note that in relation to value of 200k for T1, if i used 300k instead then -100k would still impart 50 joules into T1, so you would have to change the value of T2 to a higher value that would transfer 50 joules into 300k. If we say that 700k would do that, then:
300k/700k = 300k/-100k.

The total amount of thermal energy transferred between the hot reservoir and the system will be: QH = TH(SB - SA) If TH is a negative value, then QH, the amount of energy transferred to the system would be NEGATIVE according to this equation. Obviously that doesn't happen because the value for TH is NEVER a negative value, as you cannot have less energy than absolute zero at 0K.

I am NOT saying that negative temp doesn't exist. The values used in the math represent the energy content of systems at those temperatures. THAT is what is key here. THAT is why you cannot use a negative value to calculate the carnot cycle, because you cannot have absolute NEGATIVE energy in any system. Negative temperature does not imply negative energy, but if you use a negative value for T in your math, then that is implying that you have negative energy.Now, before anyone takes 1 little snippet out of my entire post and try to use that to tell me why I'm wrong like what's been happening, I request that you do the math first. Goodnight all.

 

[Dale]

But to use these equations for a heat engine you can't have it negative, it must be positive. Otherwise weird things happen, like heat flowing from the cold to the hot. Or the engine doing work using less energy than was put in. Wasn't that the problem on the 1st page of this thread? The math wasn't working out correctly?

I agree, but the solution isn't to mess with the definition of temperature but rather to derive an equation for a generalized Carnot cycle for a negative temperature system. That was my point way back in post 11.

Temperature and entropy are much more fundamental concepts, and the Carnot cycle equations are derived from them under certain assumptions. We are violating one of those assumptions (ideal gas -> positive temperatures) in this system, so we have to re-do the derivation, not redefine the fundamentals.

 

[Andy Resnick]

Temperature and entropy are much more fundamental concepts, and the Carnot cycle equations are derived from them under certain assumptions. We are violating one of those assumptions (ideal gas -> positive temperatures) in this system, so we have to re-do the derivation, not redefine the fundamentals.

The Carnot cycle (and thermodynamics generally) does not make any assumptions about ideal gases; it does not require the existence of an ideal gas (or atoms). The utility of Kelvin's (second) absolute temperature scale is that gas thermometers (using arbitrary gases) behave as ideal gas thermometers (i.e. identically) at low temperature. Thus, we can compare different thermometers by use of a material independent scale.

Assigning a unique temperature to a substance requires the substance to be at thermal equilibrium. Thus, applying temperature to a variety of systems is problematic- any system far from equilibrium, for example. Negative temperatures in physical systems arise when considering metastable systems- systems that can be held at steady-state conditions far from equilibrium, such as inverted atomic populations or spin states.

The situation with the proposed negative temperature heat engine is formally identical to that of a machine made of antimatter- as long as the machine does not interact with 'normal' matter, all is well.

 

[Dale]

The situation with the proposed negative temperature heat engine is formally identical to that of a machine made of antimatter- as long as the machine does not interact with 'normal' matter, all is well.

Yes, I like that analogy.

 

[Gerenuk]

The difference is that it is very easy to create negative temperature. As I said just take any system with finite energy capacity and almost fill it. You don't need metastability or any other anomaly. It can be an ordinary system as long as the maximum allow total energy is finite.

Of course it's a trivial matter to argue that if you put such a system in contact with gases, it will not keep it's temperature. But I wouldn't exclude negative temperature from consideration just because it cannot be in equilibrium with air.