# Interpretation of temperature for negative ones

1. Feb 23, 2013

### ShayanJ

In statistical mechanics and kinetic theory,it is shown that temperature can be interpreted as a measure of the energy of particles.But such an interpretation seems to need refinement because of introduction of negative absolute temperatures.
Can we still interpret temperature as a quantity showing the jiggling of particles?If not,how should we change that picture to take into account negative temperatures too?
Thanks

2. Feb 23, 2013

### Graniar

What is the meaning of negative temperatures do you mean?
It doesn't have physical sense.

The only one I know, is about population inversion in a gain medium of the laser,
reflecting the fact, that number of excited atoms at higher level is bigger, than of those at lower level.

3. Feb 23, 2013

### ShayanJ

Yes,That's exactly what I mean.

4. Feb 23, 2013

### Graniar

Than why to extra complicate the temperature?
It is the statistical concept for the wide class of systems.

Laser's gain medium is a bit out of that class.
May be there is a reason to percieve it as a mix of systems with different temperatures.
Something like nonequilibrium plasma.

5. Feb 23, 2013

### Jorriss

Are you familiar with the statistical mechanic definition of temperature?

6. Feb 23, 2013

### Mute

The thing about temperature is that there are actually lots of different ways one can introduce some notion of "temperature". The random jiggling of particles is just one of the more familiar notions.

From a "fundamental" standpoint, we define temperature as $T^{-1} = \frac{\partial S}{\partial E}$. In classical thermodynamics of particles, this gives rise to the equipartition theorem, which allows us to relate quadratic degrees of freedom to this statistical definition of temperature. Because a quadratic term only has one sign, the temperature will only have one sign. Since free particles' degrees of freedom are just $\frac{1}{2}mv^2$, we can relate the mean motion of particles to temperature, and this is where our familiar definition of entropy comes from.

However, there are more than just quadratic degrees of freedom. When we consider such non-quadratic degrees of freedom, such as certain spin systems, we no longer can use the equipartition theorem, but we still have the statistical definition of temperature based on energy and entropy. As a result, we have no guarantee that temperature is restricted to being non-negative, and indeed, certain systems can be set up in which temperature can indeed be negative.

The problem is that we are so used to thinking about temperature in terms of particle motion that we have a hard time wrapping our heads around negative temperatures. Negative temperatures are, in fact, not colder than absolute zero, but are in some sense "hotter" than infinite positive temperature. The reason that they are "hotter" is simply that energy will also flow from a negative temperature system to a positive temperature system.

7. Feb 23, 2013

### ShayanJ

In kinetic theory,temperature is defined via the following formula:
$T=\large{\frac{m \overline{v^2}}{3 k_B}}$
In which it is obvious that temperature can't be negative.

But can we still define kelvin temperature as the ratio of heats transferred between adiabatic surfaces in an isothermal process for negative temperatures?

Thanks all guys

8. Feb 23, 2013

### Graniar

And this, I believe, is not good.

The "temperature" term is not some sacred word given by heavens.
I think it's better to distinguish these concepts.

Then let us continue thinking this way.

Then let us use another word for that sense.

The temperature.2 of laser's gain medium, if one would like to calculate it's brightness,
differs from the temperature.1, if he designes a cooling system for it.

9. Feb 23, 2013

### ShayanJ

I almost understand what you're trying to tell Graniar.But the decision that we should make between extending the interpretation of temperature and defining a new quantity,is a little conventional here.So if we can extend the interpretation with little effort and problem,then why not?
I think the formula $T^{-1}=\frac{\partial S}{\partial E}$ does that for us.

10. Feb 23, 2013

### Staff: Mentor

Should this thread not be moved into "Classical"? Not a lot of QM going on here....

11. Feb 23, 2013

### ShayanJ

Negative absolute temperatures is a completely quantum phenomena impossible in classical physics!I guess we will have some discussions about QM.

12. Feb 23, 2013

### Jorriss

That is not a good statistical definition of temperature though. The reason one uses the statistical definition is it is the natural quantity two systems have in common at thermal equilibrium. It fits our intuitive understanding of temperature and it shows up naturally in stat mech. It's a better temperature definition than 'atoms jiggling.'

13. Feb 23, 2013

### Graniar

May be it has a sense. But not so simple, I think.
One should take care of the different substances.
Like for the nonequilibrium plasma, where ions and electrons forms two, with their different temperatures.

What <unified temperature> does have such a plasma?

The special thing about gain media, nonequilibrium for certain spectra levels.
Thus, all atoms excited by pumping light, forms such a substance for a while.

14. Feb 23, 2013

### Mute

Temperature is an equilibrium concept. What does it mean to talk about temperatures of ions and electrons in a non-equilibrium plasma? You have to be very careful with your definitions here.

Also note that many systems have "effective temperatures", which are measures that are analogous to temperature in certain non-thermal systems. For example, in statistical mechanical studies of granular systems, often an effective temperature is introduced to characterize fluctuations in the grain packings, which have nothing to do with heat flow. (The wikipedia article "effective temperature" talks about stars, which is a different notion that what I'm referring to here).

Strictly speaking, to get a negative temperature you just need a system with a maximum energy level. Classical spin models can still have negative temperatures. It's just very hard to engineer a real classical system with a maximum energy level.

15. Feb 23, 2013

### ShayanJ

Could you suggest some articles about the "effective temperature" which you mean?
Also the atoms regulated in an optical lattice used for reaching to negative temperatures,seems to me sth like a granular system.is it correct?
Thanks

Last edited: Feb 23, 2013
16. Feb 23, 2013

### Mute

Here are some slides from a talk which cover the concept: http://www.brandeis.edu/departments/physics/chakraborty/images/entropy-temperature.pdf [Broken]

The slides gives some references to some articles you might look into. I would find a couple good articles myself, but I'm not quite in this field, so I don't know any good articles off the top of my head, and it's late so I don't feel like logging into my university's vpn right now to go hunting for articles. I'll try to remember to find a couple decent-looking ones tomorrow or Monday maybe.

Last edited by a moderator: May 6, 2017
17. Feb 24, 2013

### stevendaryl

Staff Emeritus
A simple, exactly solvable problem that exhibits negative temperatures is the following:

Suppose we have a lattice of N magnetic dipoles each of which can either point up or down. We impose a magnetic field, and the resulting energy of the system is given by:

$E = E_0 + K j$

where $E_0$ is the minimum energy, when all the dipoles point down, and $j$ is the number of dipoles that point up, and $K$ is some constant.

Let's shift the energy so that $E_0 = \frac{K N}{2}$. So the energy is zero when half the dipoles are pointing up, and half are pointing down. Then the energy just counts the "excess" of spin-up dipoles: $E = (j - \frac{N}{2}) K$. The number of possible states with energy $E$ is combinatorially given by

$W(E) = \dfrac{N!}{(\frac{N}{2} + \frac{E}{K})! (\frac{N}{2} - \frac{E}{K})!}$

The entropy $S(E)$ is just given by:

$S(E) = k\ log(W)$

where $k$ is Boltzmann's constant.

We can use Stirling's approximation to compute the logarithms:

$log(n!) \approx n log(n)$

Using this, we compute:
$S(E) = -k\ [(\frac{N}{2} + \frac{E}{K})log(\frac{1}{2} + \frac{E}{NK}) + (\frac{N}{2} - \frac{E}{K})log(\frac{1}{2} - \frac{E}{NK})]$

Now, we compute $\dfrac{1}{T} = \dfrac{\partial S}{\partial E}$

$\dfrac{\partial S}{\partial E} = \dfrac{1}{K} \dfrac{\partial S}{\partial j}$
$= \dfrac{k}{K}\ log(\dfrac{\frac{N}{2} - \frac{E}{K}}{\frac{N}{2} + \frac{E}{K}})$

In terms of temperature, we can see:

• When $E \rightarrow -\frac{NK}{2}$, $\dfrac{\partial S}{\partial E} \rightarrow +\infty$, so $T \rightarrow 0+$.
• When $E \rightarrow 0$, $\dfrac{\partial S}{\partial E} \rightarrow 0$, so $T \rightarrow \infty$.
• When $E \rightarrow +\frac{NK}{2}$, $\dfrac{\partial S}{\partial E} \rightarrow -\infty$, so $T \rightarrow 0-$.

So things look kindof normal for small values of $E$. When $E$ is a minimum, $T \rightarrow$absolute zero. As the energy rises, so does $T$ until the point $E = \frac{NK}{2}$. At this point, the system is infinitely hot. For $E > \frac{NK}{2}$, the temperature is negative, meaning in a sense that it is hotter than infinitely hot.

The behavior of $T$ is actually more sensible if you look at the factor $e^{-\frac{1}{kT}}$. This quantity rises from $0$ at $E = -\frac{NK}{2}$ to 1 at $E = 0$ to $+\infty$ at $E = +\frac{NK}{2}$. This factor in statistical mechanics tells the extent to which higher energy levels are occupied, and the larger it is, the "hotter".

18. Feb 24, 2013

### ShayanJ

Thanks stevendaryl,your post is extraordinarily useful.
But there is a point.
why the energy is $E_0+Kj$?Why do you say $E_0$ is the minimum energy?since if we change one particle's spin from down to up,its energy won't be inside $E_0$ any more.
To me,its natural that it should be $n_u E_u+n_d E_d$ where subscripts u and d stand for up and down for spins and n is the number of particles having the indicated spin and the same thing for E which is energy.

19. Feb 24, 2013

### stevendaryl

Staff Emeritus
That's actually a mistake on my part. The way I defined it, the minimum energy is $E_0 - \frac{KN}{2}$, not $E_0$.

You're right. Except that I chose the zero of energy so that $E_u = K$, $E_d = -K$, and $j = (n_u - n_d)$.