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

[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.

 

[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, <br /> \Delta S = \int \frac{dQ}{T}<br /> 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, <br /> \Delta S = \int \frac{dQ}{T}<br /> 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 <br /> \frac{Q_1}{Q_2} = \frac{T_1}{T_2}<br /> 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.