Can a system have a negative temperature based on its energy and entropy?

ak416
Messages
121
Reaction score
0
Ok i thought i had this under control, but now i come to think about it, i get more confused. I know d sigma/d U for any system must be positive because the temperature must be positive (sigma is the entropy and U is the energy). But what about this? Consider the model of the paramagnetic system where there's a constant N particles in a constant volume. The energy is proportional to how many spin down - spin up particles there is. So let's say for system 1, N= 30. 29 are spin down, 1 is spin up (Lets just say energy is 28). Let's say this system comes in thermal contact with another system (system 2). (the systems are allowed to exchange energy). So initially the entropy of the 1st system is 30 choose 1 = 30. Now let's say system 1 exchanges its spin up for spin down (Change in energy is +2). But now entropy is 30 choose 0 = 1. So change is entropy is -29. so change is entropy wrt to change in energy is negative? And temperature which is 1 over that is negative?
 
Last edited:
Physics news on Phys.org
Don't worry! As it turns out, negative temperature is a perfectly reasonable thing that has been realized in the laboratory in nuclear spin systems. Negative temperature is possible provided three conditions are met. First, the system must have bounded energy. Second, the system must be able to reach internal equilibrium quickly. Third, the system must be isolated (i.e. much longer relaxation time) from systems not obeying the first two conditions. Basically, systems with negative temperature are "hotter than infinity" since they will give up energy to any system with positive temperature. The standard example consists of a spin system where more than half the spins point anti-parallel to the applied field: as you increase the energy of the system, fewer and fewer microstates are available. Try googling for an article by Ramsey from the 50's where he talks about thermodynamics with negative temperature.

Hope this helps.
 
Last edited:
wow didnt know. I am going to have to read up about that, but since i have a test tomorrow, I am just trying to understand the more basic cases. Like for the example I gave, the multiplicity function is N choose how many up spins there is, and each up spins means negative energy. So doesn't this concept of negative temperature apply to anything that satisfies those properties (i.e, Multiplicity function g(N,k) N!/(k!(N-k!)) where k is proportional to -U? Or is the case i presented unrealistic because a system with so many down spins will never accept to switch for more down spins, it will instead switch a down spin to gain an upspin so that it can increase the joint multiplicity of the joint system(1+2)?

Edit: actually even if it switches its down spin to gain an upspin, Change in U will be negative and change in entropy will be positive, so its seems like its a general property of any system will the given type of multiplicity function regardless of how it switches its spins. Am i right?
 
Last edited:
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top