Can pressure be negative?

[ben m]

Not only did this experiment *not* achieve an "infinite" temperature inside the atom, it could not possibly do so because the energy input itself was "finite" to start with! The only thing that changed during the time in question is an internal spin arrangement inside the atom, not the actual "temperature" of the atom, or anything inside of the atom.

Energy is not the same as temperature. Telling me that the energy is finite tells me nothing about the temperature. When we talk about energy, we'll call it "energy" and often use the variable name E. Did anyone say that the energy was infinite, or zero?

"internal spin arrangement" is not the same thing as temperature---it's closer to being the entropy. Entropy is not the same thing as temperature. When we talk about entropy, we'll call it "entropy" and often use the variable name S. Did anyone say that the entropy was infinite, or zero?

Temperature is different than entropy, and it's different than energy. What does temperature mean, Michael? It's not what you think it is. You seem to think that "temperature" is just energy divided by (say) Boltzmann's Constant, and that you can convert from energy to temperature in the same way you'd convert joules to ergs.

 

[Skwinty]

But it's not a breakdown of physics, physics can handle infinite temperatures quite well without any problems.

In these instances, does renormalisation play a prominent role?

 

[Ziggurat]

In these instances, does renormalisation play a prominent role?

Nope. There's no need to renormalize anything, it's fine to leave temperature as an infinity. It doesn't cause any problems.

 

[Skwinty]

Nope. There's no need to renormalize anything, it's fine to leave temperature as an infinity. It doesn't cause any problems.

Why is that?

Is it because although T can be derived from the equations, you prefer beta and therefore the value of T is irrelevant?

 

[Ziggurat]

Why is that?

It doesn't cause problems because there are no problems for it to cause. If you tell me what problems you think there might be, I can tell you why they aren't problems, but otherwise I can't really provide a more specific answer than that.

Is it because although T can be derived from the equations, you prefer beta and therefore the value of T is irrelevant?

T isn't irrelevant, but one can always choose to use Beta instead (and vice versa), so if you're not comfortable with an infinite T, then you can simply use a zero Beta. But it's exactly the same thing.

 

[Skwinty]

Having read this http://en.wikipedia.org/wiki/Negative_temperature I think I understand the issue more.

My intuition, although irrelevant, tells me there is nothing magic in negative temperature.

After all, the macroscopic and microscopic realms are so different and if my $10 calculator can calculate tan (theta) then who am I to complain.

 

[dafydd]

Where did MM study,he seems to know more about physics than all the physicists in the world lumped together. Why doesn't he get a yearly Nobel prize?

 

[Calrid]

Breaches of Rules 0, 9, 10, and 12 removed.

Replying to this modbox in thread will be off topic  Posted By: Myriad

 

[Ziggurat]

Post edited to remove moderated content, and some responses to same.

Replying to this modbox in thread will be off topic  Posted By: Myriad

Edited by Myriad: 

edited to remove moderated content

Sol gave a very nice example of an easily measurable physical infinity. I have yet to see any objection to his example.

Edited by Myriad: 

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Hey, don't blame me, this is standard textbook physics. Go complain to Reif or Kittel.

Edited by Myriad: 

edited to remove moderated content

Why don't you invite them here so they can tell me themselves? I'd be more than happy to continue a discussion about the science of negative and infinite temperatures. Hopefully everyone in the field that you asked doesn't have similar verbal incontinence, though. That does sort of get in the way.

 

[Dr. Trintignant]

Nothing in that process requires the model or its parameters to conform to any kind of intuition or prior expectation - if it makes predictions that are consistent with experiment, it's a (potentially) valid theory. Since our intuition is mainly based on a certain limited subset of those experiments, it doesn't add anything.

Yup; that's essentially the attitude I take. Still, it take a bit of mental effort to get past billiard-ball models that correspond to macroscopic behavior. I might posit that this is the defining characteristic to modern (past century) physics: giving up the idea that our models are somehow just a poor approximation to reality; they may be approximate, but they aren't poor--they're the best we've got. Giving up the notions of the aether, or absolute time, or particles in a definite place, or what have you has essentially been the driver for 20-21st century physics.

Physicists have another definition of temperature - a definition based on mathematical models developed over centuries of study and experiment - that does allow infinities (and negative values) under certain circumstances. That's all there is to it.

Michael's position I find harder to understand. Temperature may correspond to conventional notions of hot and cold when it's applied to actual atomic motion, but there's no reason why it must only apply there. His reasoning is circular--his faulty personal definition of temperature (to the extent that it exists) leads him to believe that infinite temperature implies infinite energy, and he then uses this to conclude that models which have infinite temperature are false. Obviously not.

- Dr. Trintignant

 

[ben m]

Temperature may correspond to conventional notions of hot and cold when it's applied to actual atomic motion, but there's no reason why it must only apply there. His reasoning is circular--his faulty personal definition of temperature (to the extent that it exists) leads him to believe that infinite temperature implies infinite energy, and he then uses this to conclude that models which have infinite temperature are false. Obviously not.

That's a good summary of the situation.

It's almost identical to Michael's problem with pressure, the original topic of the thread.

Pressure may correspond to conventional notions of molecules crashing into walls when it's applied to actual atomic motion, but there's no reason why it must only apply there. His reasoning is circular--his faulty personal definition of pressure (to the extent that it exists) leads him to believe that negative pressure implies negative collision rates, and he then uses this to conclude that models which have negative pressure are false. Obviously not.

 

[Michael Mozina]

Yup; that's essentially the attitude I take. Still, it take a bit of mental effort to get past billiard-ball models that correspond to macroscopic behavior. I might posit that this is the defining characteristic to modern (past century) physics: giving up the idea that our models are somehow just a poor approximation to reality; they may be approximate, but they aren't poor--they're the best we've got. Giving up the notions of the aether, or absolute time, or particles in a definite place, or what have you has essentially been the driver for 20-21st century physics.

In this case it isn't the mathematical model that is "poor" in terms of it's predictive ability, it's Zig's "conceptual understanding" of what's really going on at the level of particle physics. At the level of particle physics, nothing ever does or ever could achieve "infinite temperatures', let alone a "higher than infinite temperature" (which was the term I actually took exception to by the way).

In this case a "finite" amount of energy is "added" to a 'near zero" system to "align" an internal spin arrangement. Once the magnetic field is removed, the spin arrangements return to "normal" but at no time did anything 'infinite' ever actually occur. The energy added was finite. The effect itself was finite, and the change over time was finite and measurable. To then claim it's an example of "infinity and beyond" in terms of temperature is patently absurd. It may "seem" that way if your concept of physics is limited to a math formula, and *ONLY* a math formula, but when you look at the physics itself, a "finite energy" injection that creates "finite" change in "near zero" atoms is not an example of "infinite temperatures".

Michael's position I find harder to understand. Temperature may correspond to conventional notions of hot and cold when it's applied to actual atomic motion, but there's no reason why it must only apply there.

Even if we attempt to apply the term to internal spin arrangements inside that atoms themselves, the net energy change is still "finite", not "infinite" and certainly not 'infinity and beyond'.

 

[ben m]

In this case it isn't the mathematical model that is "poor" in terms of it's predictive ability, it's Zig's "conceptual understanding" of what's really going on at the level of particle physics. At the level of particle physics, nothing ever does or ever could achieve "infinite temperatures', let alone a "higher than infinite temperature" (which was the term I actually took exception to by the way).

You are confused. The actual, physical definition of temperature is one in which "infinite" temperature makes perfectly good sense.

In this case a "finite" amount of energy is "added" to a 'near zero" system to "align" an internal spin arrangement. Once the magnetic field is removed, the spin arrangements return to "normal" but at no time did anything 'infinite' ever actually occur. The energy added was finite.

Finite ENERGY. Not finite TEMPERATURE. What definition of temperature are you using that makes you confuse these two things? It sounds like you think T = E/(kB). By that definition, Michael, there is no relationship between temperature and entropy, dQ != TdS, and the Mozina-laws of thermodynamics allow me to build perpetual motion machines. Great.

The effect itself was finite, and the change over time was finite and measurable.

Finite quantities sometimes have infinite derivatives. What's the slope of a vertical cliff, Michael? If I tell you I have a cliff whose slope (rise over run) is infinity, do you object because the cliff is of only finite height?

Even if we attempt to apply the term to internal spin arrangements inside that atoms themselves, the net energy change is still "finite", not "infinite" and certainly not 'infinity and beyond'.

Finite ENERGY. Do you have any understanding of this "temperature" quantity you keep mentioning? You seem to have it confused with energy.

 

[Ziggurat]

In this case it isn't the mathematical model that is "poor" in terms of it's predictive ability, it's Zig's "conceptual understanding" of what's really going on at the level of particle physics. At the level of particle physics, nothing ever does or ever could achieve "infinite temperatures', let alone a "higher than infinite temperature" (which was the term I actually took exception to by the way).

Again, how would you know if you don't know what temperature is? You continue to avoid the question. This is a pattern for you: you fail to understand the basic definition of a standard physics term (and yes, this is standard, undergraduate textbook thermodynamics), that leads you to wrong conclusions about that term, and when you are challenged, you refuse to define the term in question.

In this case a "finite" amount of energy is "added" to a 'near zero" system to "align" an internal spin arrangement.

Why all the quote marks? Is the amount of energy not really finite? Is it not really added to the system?

The effect itself was finite, and the change over time was finite and measurable. To then claim it's an example of "infinity and beyond" in terms of temperature is patently absurd.

Was Sol's example of an infinite slope patently absurd? No, it wasn't. There's no fundamental difference here.

It may "seem" that way if your concept of physics is limited to a math formula, and *ONLY* a math formula, but when you look at the physics itself, a "finite energy" injection that creates "finite" change in "near zero" atoms is not an example of "infinite temperatures".

It may seem that infinite temperature is impossible if your concept of physics is limited to your intuition. In other words, you can imagine limits which don't exist if you don't know what temperature is.

Even if we attempt to apply the term to internal spin arrangements inside that atoms themselves, the net energy change is still "finite", not "infinite" and certainly not 'infinity and beyond'.

I never claimed that the energy was infinite. I've always said that it's always finite. You think that infinite temperature requires infinite energy because you don't know what temperature is.

 

[Perpetual Student]

This is from wikipedia:

"For a system in thermal equilibrium at a constant volume, temperature is thermodynamically defined in terms of its energy (E) and entropy (S) as:

[latex] T \equiv (\dfrac{\partial S}{\partial E})^-^1 [/latex] "


For anyone willing to answer, my question is: If a system is in thermal equilibrium, why would T be equivalent to the partial derivatives shown in a situation where there should be no change (isn't that what thermal equilibrium means?)?

 

[Dr. Trintignant]

I don't know that I can add anything that Ben, Sol, and Zig have already, but...

Even if we attempt to apply the term to internal spin arrangements inside that atoms themselves, the net energy change is still "finite", not "infinite" and certainly not 'infinity and beyond'.

Fortunately, no one has claimed that there is ever an infinite amount of energy involved.

Temperature is defined as change in thermal energy divided by change in entropy. The change in energy is, of course, always finite. The change in entropy is always positive for atomic motion, but may be zero or negative for other systems. Temperature than thus be infinite or above.

You can try quibbling with the definition, but good luck creating a new one that:
1) Works the way we expect for normal hot and cold objects.
2) Doesn't break the laws of thermodynamics.

I don't think you can do it. Any formula must take both entropy and energy into account or it'll allow for the creation of perpetual motion machines. And given the units, there's only one way to combine them: T=dQ/dS.

Propose a different formula that meets the same basic constraints and we'll all think you very clever.

- Dr. Trintignant

 

[Dr. Trintignant]

For anyone willing to answer, my question is: If a system is in thermal equilibrium, why would T be equivalent to the partial derivatives shown in a situation where there should be no change (isn't that what thermal equilibrium means?)?

Well, we can speak of derivatives even if there is no actual change happening. So that equation just means that if you were to inject a small amount of energy dQ, the entropy dS would change by dQ/T. Just like a spring and weight in static equilibrium will displace by -F/k if you apply a small additional force.

- Dr. Trintignant

 

[Ziggurat]

For anyone willing to answer, my question is: If a system is in thermal equilibrium, why would T be equivalent to the partial derivatives shown in a situation where there should be no change (isn't that what thermal equilibrium means?)?

Thermal equilibrium is defined as the maximum of the total entropy. For two systems in thermal contact, if energy moves from one system to another, the entropy of each subsystem will likely change, but at thermal equilibrium, we're at a maximum of the combined entropy, so the derivative of the combined entropy with respect to energy flow between the subsystems will be zero. That defines our maximum: where the derivative of the total is zero with respect to energy flow within the system. But since the total entropy is simply the sum of the sub-entropies, this means that any entropy gained by one subsystem must be matched by an equal loss in the other subsystem. This will happen when their derivatives of S with respect to E are equal: I move a tiny bit of energy dE out of system 1, and decrease the entropy by (dS₁/dE)*dE, and that tiny bit of energy goes into system 2, increasing its total entropy by (dS₂/dE)*dE. So the derivatives must match. That means the temperatures (the inverse of the derivative) must also match at thermal equilibrium.

 

[Michael Mozina]

I don't know that I can add anything that Ben, Sol, and Zig have already, but...



Fortunately, no one has claimed that there is ever an infinite amount of energy involved.

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.

Temperature is defined as change in thermal energy divided by change in entropy. The change in energy is, of course, always finite. The change in entropy is always positive for atomic motion, but may be zero or negative for other systems. Temperature than thus be infinite or above.

You can try quibbling with the definition, but good luck creating a new one that:
1) Works the way we expect for normal hot and cold objects.
2) Doesn't break the laws of thermodynamics.

I don't think you can do it. Any formula must take both entropy and energy into account or it'll allow for the creation of perpetual motion machines. And given the units, there's only one way to combine them: T=dQ/dS.

Propose a different formula that meets the same basic constraints and we'll all think you very clever.

- Dr. Trintignant

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.

 

[sol invictus]

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.

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

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.

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