Newton's law of cooling for negative Kelvin temperatures

Question

Newton's law of cooling says that $$\frac{\mathrm{d}T}{\mathrm{d}t}=-k(T-T_a)$$ where $T$ is the temperature of the substance, $T_a$ is the ambient temperature, and $t$ is the time. Usually, we say that if $T>T_a$, the object is cooling down, which makes sense intuitively as well as mathematically, because the time derivative of temperature is negative.

However, let's imagine that $T_a=3\,\text{K}$ and $T=-3\,\text{K}$ (random numbers, but the signs are important). I think the object will 'cool down': $$T=\frac{\mathrm{d}Q}{\mathrm{d}S}$$ suggests that 'cooling down' in this case can be thought of as a decrease in entropy.

But when you put the signs in Newton's law of cooling, you predict an increase in temperature, since $\mathrm{d}T/\mathrm{d}t$ becomes positive.

Is Newton's law of cooling applicable to negative temperatures? If yes, what's the mistake in the aforementioned reasoning?

If it helps, this page suggests that Newton's law of cooling would have correct implications in the discussion of engine efficiencies at negative temperatures. But that's a conclusion I drew, and I would like to see some publications/theories which actually say something directly.

Answer 1

To answer this question, we need to understand a couple of things about negative temperatures.

First, rather than defining $$T = \frac{\mathrm{d}Q}{\mathrm{d}S}$$ as you have done, it's more typical to define the reciprocal of temperature, $$\frac{1}{T} = \frac{\mathrm{d}S}{\mathrm{d}E}.$$ This may seem like a trivial change but it helps to explain how negative temperatures arise, and from there how we might reason about them.

With this definition in hand, we can ask, "Under what circumstances is $T$ negative?" The answer is that $T$ will be negative if increasing the energy of the system decreases its entropy, which only happens in systems with a maximum total energy, and then only when the energy in the system is more than half of the maximum.

Let's get specific. Assume a system of $N$ spin-1/2 particles in a magnetic field. (You can do this with arbitrary magnetic moments, but it's easier to keep track of everything if each particle can only take on two energy values.) The particles can either be aligned with the field (low energy) or aligned opposite the field (high energy). Thus the energy of the system is quantized, with $$E(n) = 2\mu_B B n,$$ where $\mu_B$ is the magnetic moment of each particle and $n$ is the number of particles in the high-energy configuration. (The factor of 2 comes from setting the zero point of our energy to correspond to having all of the particles in the low-energy configuration. You don't have to do this, but it simplifies the energy expression a little bit.)

The last preliminary bit of information we need is the entropy as a function of energy. In this case, for an energy $E(n)$, we have $n$ particles in the high energy configuration. There are $$\Omega = \frac{N!}{(N-n)!}$$ ways to accomplish this, which gives an entropy of $S = k_B\ln\Omega = k_B(\ln N! - \ln (N-n)!).$

Note that $S$ will be a maximum when $n=N/2$; it reaches a minimum value of $0$ when $n$ is either $0$ or $N$. This leads to the statement I made above: $T$ is negative when $n>N/2$. There are two other consequences, though, which are a little counterintuitive:

1. Negative temperatures are hotter (have more energy) than positive temperatures.
2. As you put more energy into a system with negative temperature, the temperature gets closer to zero.

So, in order of increasing energy, the temperature goes $$T = +0 \rightarrow T = \text{positive} \rightarrow T=+\infty \rightarrow T=-\infty \rightarrow T=\text{negative} \rightarrow T=-0.$$

Now, back to Newton's law of cooling. As you've written it, $$\frac{\mathrm{d}T}{\mathrm{d}t} = -k(T - T_a)$$ has three related assumptions buried in it which are relevant to this discussion:

1. Temperature increases as energy increases. (The equation is written in terms of temperature instead of energy.)
2. Higher temperatures correspond to higher energies. (The minus sign on the right-hand-side.)
3. The relationship between temperature and energy is linear. (The constant $k$ on the right hand side. Strictly speaking, $k$ accounts for the thermal coupling between the two systems, but making it temperature dependent could also deal with a nonlinear relationship between $T$ and $E$.)

Except for the pathological point at $n=N/2$, the first of these is satisfied regardless of the sign of the temperature, but the second fails if the two temperatures have different signs. The third assumption breaks down if you get close enough to the transition between positive and negative temperatures.

So, in the situation you've outlined ($T=-3\,\text{K}, T_a = 3\,\text{K}$) Newton's law of cooling will not work:

• $(T - T_a)$ is negative,
• so the right hand side is positive, which would mean that
• $\mathrm{d}T/\mathrm{d}t$ is positive, so
• the temperature would become less negative (move toward negative zero) over time, which would mean
• energy would flow into the system from its environment.
• In terms of entropy, the system has a negative temperature, so increasing its energy decreases its entropy, and
• the environment has a positive temperature, so decreasing its energy decreases its entropy, so
• this process would decrease the overall entropy of the universe, and
• this is a violation of the 2nd law of thermodynamics.