# Negative Kelvin temperature? (Recent Science paper)

by xnwkac
Tags: kelvin, negative, paper, science, temperature
 P: 743 Entropy is positive, so it can't approach zero from the other side. Entropy tends to zero as you approach zero temperature from either side. If you plot entropy versus energy for a "finite system" (that is, a system with a maximum energy), the graph looks qualitatively like a parabola or a sine half period. See top plot of attachment. The maximum energy that the system can hold is 1. At low energy, the entropy is near 0, because with little energy, there are few ways to arrange the energy. (Think of it this way. You have n energy quanta and m bins to put them into. You have fewer options if n is small.) But at high energy, entropy is also near 0, because the system is filling up with energy, and there are fewer ways to arrange the holes. (Since this is a finite system, you can only stick at most, say, 2 quanta of energy into each bin. Then, when they are almost all filled up, you have few options of ways to arrange the vacancies.) The vacancies or holes behave more or less analogously to energy quanta. If you plot temperature versus energy, it looks like bottom plot of attachment. For this example, temperature doesn't ever reach 0. But for a larger system, temperature can get very close to 0, and entropy will get relatively close to 0. So, in a system where there is a maximum energy capacity, a low energy (positive temperature) and high energy system (negative temperature) can behave very similarly. But the positive temperature system deals with quanta of energy and the negative temperature system deals with quanta of holes. Attached Thumbnails
 P: 8 I can't understand the following from Baez: " We define the temperature, T, by 1/T = dS/dE, so that the equilibrium condition becomes the very simple T1 = T2. This statistical mechanical definition of temperature does in fact correspond to your intuitive notion of temperature for most systems. So long as dS/dE is always positive, T is always positive. For common situations, like a collection of free particles, or particles in a harmonic oscillator potential, adding energy always increases the number of available microstates, increasingly faster with increasing total energy. So temperature increases with increasing energy, from zero, asymptotically approaching positive infinity as the energy increases. " Isnt S(E) a concave function in "common" situations? Why does it say "faster with increasing total energy" ? If S increases faster and faster with increasing E, then the derivative dS/dE is larger and larger, then 1/T is larger and larger, than T is smaller and smaller . What am I missing? thankx Wentu
HW Helper
P: 6,164
 Quote by Wentu I can't understand the following from Baez: " We define the temperature, T, by 1/T = dS/dE, so that the equilibrium condition becomes the very simple T1 = T2. This statistical mechanical definition of temperature does in fact correspond to your intuitive notion of temperature for most systems. So long as dS/dE is always positive, T is always positive. For common situations, like a collection of free particles, or particles in a harmonic oscillator potential, adding energy always increases the number of available microstates, increasingly faster with increasing total energy. So temperature increases with increasing energy, from zero, asymptotically approaching positive infinity as the energy increases. " Isnt S(E) a concave function in "common" situations? Why does it say "faster with increasing total energy" ? If S increases faster and faster with increasing E, then the derivative dS/dE is larger and larger, then 1/T is larger and larger, than T is smaller and smaller . What am I missing? thankx Wentu
Normally S increases slower and slower with increasing E.
That is why T increases with increasing E.

However, for some substances, at some point S starts decreasing with increasing E.
This is when the sign of the temperature flips.

After that S starts decreasing quicker and quicker, making temperature (which is the inverse) approach zero.
 P: 8 Serena, this is exactly what I am asking. I understood it should be like you are saying, but Baez is saying that in COMMON systems Entropy increases FASTER with INCREASING energy and this means temperature rises... and this is contrary to what you said (and to what I understood so far) I am asking why Baez is saying something that seems contrary to what i have read elsewhere thankx W.
 Mentor P: 9,650 Baez is saying that the number of available microstates increases faster with energy. But entropy is related to the logarithm of the available microstates, and that increases slower.
P: 90
 Quote by Hurkyl $$(0^+) --- (1) --- (\infty) --- (-1) --- (0^-)$$
Can someone add a couple more values after the +infinite and -1 value?
 Mentor P: 9,650 ##(\infty) - (-10^{100}) - (-10^5) - (-10) - (-1) - (-0.1) - (0^-)## Like that? The first one is actually ##\pm \infty## and not a real number, as it corresponds to ##\frac{1}{T}=0##.
 P: 90 Yeah I get it now.
P: 82
 Quote by I like Serena However, for some substances, at some point S starts decreasing with increasing E. This is when the sign of the temperature flips. After that S starts decreasing quicker and quicker, making temperature (which is the inverse) approach zero.
I understand how in a situation where you get the usual entropy at a high energy, and the molecules sort of "peak" in their form or melt or destabilize further into a gas. Where I'm confused is on the loss of energy and S reaching equilibrium for the system... if that system is aided in cooling it to where S drops faster than E can balance, what's to stop it from becoming -K?

Also, if the motion of the intended atoms becomes negative K, from a relativity perspective, wouldn't they still have motion? And from a logic standpoint (and what I know about these things)... approaching absolute zero in an ideal system is really just continually slowing the entropy until you reach a "new" absolute zero... a new limit for how much you can slow something before the inverse energy (cold) catches up with the rest of the system and balances out like what would happen if you were adding energy and not taking it away.

P.S. - Think I sort of answered my own questions... tends to happen when I type thoughts out. Help, clarifications, corrections, a dunce hat-etc would be appreciated.
 HW Helper P: 6,164 At some point you can no longer extract energy from a system. When that happens its entropy approaches zero, which is basically what the third law of thermodynamics states.
 P: 82 What stops it from being possible? Edit: I mean, I would assume it's the limit of the system being finite, and time restrictions. Would increasing the system to a (insert degree of hugeness required) change anything? Or just allowing the system to be considered limitless?
 Mentor P: 9,650 If you find some source of materials with negative temperatures (you won't), maybe. That is as useful as a statement "I can use [magical box that always maintains 1000K at its outside] to extract as much energy as I want!". It is true, but that magical box does not exist.

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