Can pressure be negative?

[Michael Mozina]

Can you spot the problem in what you wrote, Michael? I've helpfully bolded two words.

Feel free to separate the concepts if you like, but use your formula of choice to demonstrate anything achieved a greater than infinite temp please.

Particles don't have a temperature, Michael. Temperature is a property of ensembles, not of individual particle.

The "ensembles" you're discussing are evidently finite collections of atoms and/or subatomic particles are they not?

 

[sol invictus]

Feel free to separate the concepts if you like, but use your formula of choice to demonstrate anything achieved a greater than infinite temp please.

That's already been done. There's no point in my repeating what others have already explained very clearly. And let me reiterate that this can be found in any text on thermal physics. Are they really all wrong, Michael?

The "ensembles" you're discussing are evidently finite collections of atoms and/or subatomic particles are they not?

Usually, yes. But bear in mind that particles and atoms have more than one kind of energy - kinetic energy is one type, but (for example) spin/magnetic dipole interactions are another.

 

[Michael Mozina]

That's already been done. There's no point in my repeating what others have already explained very clearly. And let me reiterate that this can be found in any text on thermal physics. Are they really all wrong, Michael?



Usually, yes. But bear in mind that particles and atoms have more than one kind of energy - kinetic energy is one type, but (for example) spin/magnetic dipole interactions are another.

The point you all seem to be overlooking is that the experiment you cited has a 'finite' set of atoms, and a finite amount of energy that has been applied to the spins of the finite numbers of atoms. The fact the number of atoms is finite, and the amount of energy used to modify the spin arrangement is finite, precludes you from claiming that anything, let along *everything* achieved 'infinite' temperatures.

What they did is change spin entropy by a "finite" amount, not an "infinite" amount. If the process was infinitely heating some physical thing (or all physical things in the system), we could add even more EM energy to change the spin configurations to any energy configuration we chose, but that is simply not the case. The spin alignment process itself is limited by the number of spins that can change inside a "finite" number of atoms. The most that you can hope for is a fully aligned "finite" number of spins. Since the spin potential is limited in number, the process itself is limited as well. There no "infinite" state achieved, just a "finite" state that is directly related to a finite number of spin positions inside a finite collection of atoms. Since adding more energy does not increase/decrease the entropy of the spin beyond a specified number, the process cannot be considered "infinite". The "temperature" isn't even reaching infinity (let alone greater than infinite temps) because you can't increase the EM energy infinitely and achieve higher/lower spin states beyond the number of spins present in a specific atom.

 

[GeeMack]

The point you all seem to be overlooking is that the experiment you cited has a 'finite' set of atoms, and a finite amount of energy that has been applied to the spins of the finite numbers of atoms. The fact the number of atoms is finite, and the amount of energy used to modify the spin arrangement is finite, precludes you from claiming that anything, let along *everything* achieved 'infinite' temperatures.

Define temperature.

What they did is change spin entropy by a "finite" amount, not an "infinite" amount. If the process was infinitely heating some physical thing (or all physical things in the system), we could add even more EM energy to change the spin configurations to any energy configuration we chose, but that is simply not the case. The spin alignment process itself is limited by the number of spins that can change inside a "finite" number of atoms. The most that you can hope for is a fully aligned "finite" number of spins. Since the spin potential is limited in number, the process itself is limited as well. There no "infinite" state achieved, just a "finite" state that is directly related to a finite number of spin positions inside a finite collection of atoms. Since adding more energy does not increase/decrease the entropy of the spin beyond a specified number, the process cannot be considered "infinite". The "temperature" isn't even reaching infinity (let alone greater than infinite temps) because you can't increase the EM energy infinitely and achieve higher/lower spin states beyond the number of spins present in a specific atom.

Define temperature.

 

[Tubbythin]

The point you all seem to be overlooking is that the experiment you cited has a 'finite' set of atoms, and a finite amount of energy that has been applied to the spins of the finite numbers of atoms. The fact the number of atoms is finite, and the amount of energy used to modify the spin arrangement is finite, precludes you from claiming that anything, let along *everything* achieved 'infinite' temperatures.

Errm. What part of the definition of temperature requires that the temperature remains finite when the number of atoms and the amount of energy remain finite? Please be specific. Otherwise you're just asserting blindly. And that aint science.

 

[W.D.Clinger]

Michael Mozina got something right

What they did is change spin entropy by a "finite" amount, not an "infinite" amount.

Yes, zero is finite.

The change in energy is finite also.

Now all we need is a definition that connects temperature to a small finite change in energy accompanied by zero change in entropy.

 

[Ziggurat]

I guess I need to see you get to a "greater than infinite" temperature with a very "finite' amount of EM energy (a few photons). I'll need to you identify the atom, or subatomic particle in question that you believe achieves an 'infinite+" energy state.

I already told you. It's in that paper I linked to. A paramagnet in a magnetic field that is magnetized in the opposite direction of the field is at negative temperature. A paramagnet in a magnetic field with zero magnetization is at infinite temperature. At no point is energy ever infinite, but temperature is, because temperature is not synonymous with energy. Which you'd know, if you knew what temperature is. But you obviously don't.

I think before I try to quibble over the formula itself, I'd like to see you or sol or Zig use your own definition to achieve an infinite+ temperature and I'll need you to identify the particle you believe achieves that temp.

The standard definition has been given multiple times in this thread. Now give us YOUR definition.

Your repeated failure has escaped no one's notice. It's hard to avoid the conclusion that you refuse to provide one because you don't WANT to find out the truth, because you might find out that you're wrong.

 

[Michael Mozina]

One quick question.....

If any or all of the atoms/physical things in that chamber are of 'infinite temperature', how come they don't heat up rapidly during the time the EM field is applied?

 

[Ziggurat]

One quick question.....

If any or all of the atoms/physical things in that chamber are of 'infinite temperature', how come they don't heat up rapidly during the time the EM field is applied?

When the applied magnetic field is reversed, the paramagnet DOES heat up.

Now what's your definition of temperature?

 

[Michael Mozina]

When the applied magnetic field is reversed, the paramagnet DOES heat up.

I'm not talking about while it's reversed, I'm talking about holding it in "infinite temperature" mode. Why doesn't the experiment simply explode?

Now what's your definition of temperature?

I've yet to see you present your whole definition in terms of that particular experiment, let alone demonstrate that anything ever achieves a greater than infinite temperature.

If all you're doing is moving the internal spins back and forth to heat the apparatus, you would have to keep doing it forever to achieve "infinite" temperatures!

 

[ben m]

I'm not talking about while it's reversed, I'm talking about holding it in "infinite temperature" mode. Why doesn't the experiment simply explode?

Temperature doesn't make things explode. (Pressure does.) Temperature is not the same thing as pressure.

I've yet to see you present your whole definition in terms of that particular experiment, let alone demonstrate that anything ever achieves a greater than infinite temperature.

It's very easy.

T = dS/dE.

For a system of N distinguishable spins of which U are "up" and N-U down, each with energy μ, the quantity E = µ(U - N/2) is the energy. The number of available states is Ω = "N choose U" = N! / (U! (N-U)!). The quantity S, the entropy, is S = k ln (Ω). That is a complete description of E and S in a two-state spin system. You have everything you need to figure out the derivative dS/dE.

(ETA: or, indeed, any other thermodynamic quantity you want---specific heat, for example.)

(ETA2: notice that I have not written "T = E/k" or T = E^2" or anything that suggests that high temperatures follow from high energies.)

If all you're doing is moving the internal spins back and forth to heat the apparatus, you would have to keep doing it forever to achieve "infinite" temperatures!

No, that's how you get to infinite energy. We're talking about temperature. They're different.

 

[Ziggurat]

I'm not talking about while it's reversed, I'm talking about holding it in "infinite temperature" mode. Why doesn't the experiment simply explode?

You can't hold it in infinite temperature mode because it's not stable. It passes through infinite temperature as it cools from negative temperature.

And it doesn't explode because it never has infinite energy, because temperature and energy are not the same thing.

Now what's your definition of temperature?

I've yet to see you present your whole definition in terms of that particular experiment

The definition of temperature doesn't depend on the details of that experiment. If it did, it wouldn't be a definition. If you want an equation of state for the temperature as it relates to other state variables, I can give that to you, but an equation of state is not a definition.

 

[sol invictus]

The point you all seem to be overlooking is that the experiment you cited has a 'finite' set of atoms, and a finite amount of energy that has been applied to the spins of the finite numbers of atoms. The fact the number of atoms is finite, and the amount of energy used to modify the spin arrangement is finite, precludes you from claiming that anything, let along *everything* achieved 'infinite' temperatures.

What they did is change spin entropy by a "finite" amount, not an "infinite" amount. If the process was infinitely heating some physical thing (or all physical things in the system), we could add even more EM energy ...

etc.

Do you see that the things colored differently are different quantities, Michael?

 

[Perpetual Student]

I would appreciate a little more help.

T = dE/dS

means that temperature is a rate of change: change in energy per change in entropy, in some units.

I have been struggling to get some kind of intuition about this definition.

Let's say some object is at some specific temperature, say 100 in some scale. So, temperature is defined as the rate of change in dE/dS of that object. Is it not true that nothing can stay at the same temperature without some input of energy (presumably it would otherwise naturally radiate and lose energy)? Now, for the temperature to remain constant and if the energy input equals the energy output doesn't dE = 0, which would make T = 0, which seems to be contradictory?
Am I making any sense here?

 

[ben m]

I have been struggling to get some kind of intuition about this definition.

Let's say some object is at some specific temperature, say 100 in some scale. So, temperature is defined as the rate of change in dE/dS of that object. Is it not true that nothing can stay at the same temperature without some input of energy (presumably it would otherwise naturally radiate and lose energy)? Now, for the temperature to remain constant and if the energy input equals the energy output doesn't dE = 0, which would make T = 0, which seems to be contradictory?
Am I making any sense here?

Hi PS, you seem to be viewing the derivative strictly in terms of ... well, in terms of the system taking an *actual* energy step dE, noticing that S changed by dS, and figuring out what its temperature is from those actual steps.

Try some analogies on other quantities for which the derivatives make sense. Tension in an elastic rope, for example, is F = dE/dx. Does that mean a rope that's not *actually in the process of stretching* doesn't have a tension ("there's no dx")? The slope of a mountainside is dy/dx. Does that mean the mountain-slope is only really there when you're climbing it, and ceases to be a slope when you stop? No.

A better way to think about it is from Zig's explanation. A real, macroscopic system, internally in thermal equilibrium, will have some stockpile of energy. That energy is almost-but-not-quite uniformly spread---the distribution jitters around a bit due to statistical fluctuations---and those statistical fluctuations correspond to entropy changes. So both of the quantities in the derivative---the dE and the dS---are being "explored" by various parts of the system, even if it's not losing net energy.

If you bring in another system and put it contact with the first one, you'll see that the statistical sloshing-around of energy can include both systems. But it's not obvious what that means---does the new system steal energy from the old one? Or donate it? Or do they basically keep (aside from temporary fluctuations) their initial energy stores?

This is a case where you can see an actual dE and an actual dS. Obviously the combined system will seek out a lower-entropy state; it'll try to increase the total entropy. You can convince yourself (try it!) that:

a) in a case where the new system has a larger dS/dE, the new system will have to donate energy to the old one in order to increase total entropy.

b) in a case where the new system has a smaller dS/dE, the new system will have to steal energy from the old one in order to increase total entropy.

So that's a nice way to see that this quantity "dS/dE" has the same behavior that we expect of a temperature---it's the thing that determines which direction heat flows.

 

[Ziggurat]

I would appreciate a little more help.

T = dE/dS

means that temperature is a rate of change: change in energy per change in entropy, in some units.

I have been struggling to get some kind of intuition about this definition.

Let's say some object is at some specific temperature, say 100 in some scale. So, temperature is defined as the rate of change in dE/dS of that object. Is it not true that nothing can stay at the same temperature without some input of energy (presumably it would otherwise naturally radiate and lose energy)? Now, for the temperature to remain constant and if the energy input equals the energy output doesn't dE = 0, which would make T = 0, which seems to be contradictory?
Am I making any sense here?

If dE = 0 identically, then dS = 0, and dE/dS is undefined.

But there's two reasons that this doesn't matter. The first is that we're just looking at the slope of the tangent to that function at that point. It doesn't matter if we're actually moving along the function, it's still got a slope at that point, and that slope is what we really mean by the derivative. The second reason is that dE != 0 identically, because you have small thermal fluctuations about equilibrium.

 

[Perpetual Student]

The analogy above provided by ben m may help:

"Tension in an elastic rope, for example, is F = dE/dx. Does that mean a rope that's not *actually in the process of stretching* doesn't have a tension ("there's no dx")?"

Is it not true that E = f(x), in this case so that tension, F = dE/dx = [latex] \lim_{\Delta x\to0} \dfrac{\Delta E}{\Delta x}[/latex] ?

If there is no movement dx, indeed, what does this mean? Are we saying that if we did have a dx the inevitable dE would define tension even though currently no dx may be happening; in other words, it's hypothetical, the slope of the curve if some Δx were to occur? How would this definition apply if the tension were applied to a rigid object (a steel rod)? Perhaps if I understood this definition of tension, temperature might become clearer.

 

[69dodge]

Is it not true that E = f(x), in this case so that tension, F = dE/dx = [latex] \lim_{\Delta x\to0} \dfrac{\Delta E}{\Delta x}[/latex] ?

If there is no movement dx, indeed, what does this mean? Are we saying that if we did have a dx the inevitable dE would define tension even though currently no dx may be happening; in other words, it's hypothetical, the slope of the curve if some Δx were to occur?

Yes, that's right.

How would this definition apply if the tension were applied to a rigid object (a steel rod)?

Even a steel rod can stretch very slightly.

But let's consider a theoretical perfectly rigid object. It can't stretch at all. (And it can't store any energy.) So, yeah, that definition won't work. If something can't stretch, it doesn't make sense to ask, "If it did stretch, how much more energy would it have?"

 

[Perpetual Student]

A brief search for tension = dE/dx has yielded no results. It does not seem to be part of Hooke's law or the definition of tension shown in the wikipedia article. Can anyone link a source showing the derivation of tension as dE/dx?

 

[ben m]

A brief search for tension = dE/dx has yielded no results. It does not seem to be part of Hooke's law or the definition of tension shown in the wikipedia article. Can anyone link a source showing the derivation of tension as dE/dx?

try F = -dE/dx. This is the standard definition of force in one dimension. I cited it as tension (a specific example of a force that's easy to picture in 1D) just for pedagogical purposes. Anyway, if you fiddle with F=Ma and E= 1/2 mv^2 you can derive this for yourself.