Could negative kelvin be achieved?

What the title says I guess. I'm looking into ways to make a plausible exotic matter source for a story with an Alcubierre drive and I think recall that I saw that there was a possibility to use lasers to cool materials, and since "quantum gases" (I'm not sure what it means) have been observed to go below absolute zero I was wondering if a laser could be said to be used for cooling materials below absolute zero in order to create exotic matter required to bend space by achieving the negative density needed.

I'm just not clear on the science and all of this is things I've gathered from googling, but I am by no means a physicist.

I am wondering about the energy it would require too, would it be achievable with antimatter reactors or even nuclear fusion?

• Sounds like something better asked on Physics, although do check their help centre first.
– VLAZ
Dec 20 '20 at 14:43
• Well since it's theoretical and highly sci-fi for a story, I thought this one would be better Dec 20 '20 at 14:45
• Well actually, an Alcubierre warp requires negative mass/energy to create -- and negative mass has negative temperature by definition. See Negative mass: Runaway motion. But regular matter can't have negative absolute temperature. See e.g. news.mit.edu/2013/…
– dbc
Dec 21 '20 at 0:27
• ISTR in one of the Star Trek nitpickers guides there was a line of dialogue analyzed. A character reported a temperature of minus 300 degrees Celsius. That is less than 0 degrees Kelvin. I forget the exact temperature, or which episode. The point is, that it was lower than the theoretical minimum of the universe. No explanation was given. Most likely an error by the creators, but maybe you could invent some Handwavium concept of negative heat to explain it.
– Pete
Dec 21 '20 at 2:50
• @Pete sounds like someone rounded down the reading. Dec 21 '20 at 19:26

To use scientifically correct terms: forget about it.

First reason comes from the definition of temperature. To explain it in layman terms, the temperature measures how much the molecule of matter vibrate around their rest position. When the temperature is 0 K they don't vibrate, when it's above 0 K they vibrate. From this is follows that you can't have a negative temperature, because molecules either stand still or vibrate.

Second reason, dictated by that harsh mistress which is thermodynamic: one cannot reach 0 K because the energy taken from any temperature above 0 K has to be dumped into something colder than that. When everything around you is hotter than that, you cannot take that energy away in any way. Let alone go below 0 K.

This question and its answers on Physics.SE go more in details in the explanation

a substance with a negative temperature is not colder than absolute zero, but rather it is hotter than infinite temperature.

• Comments are not for extended discussion; this conversation has been moved to chat.
– L.Dutch
Dec 23 '20 at 5:26

The short answer is that it is indeed possible to create systems with negative temperatures. Unfortunately, that doesn't imply that the system in question has a negative energy density.

What is temperature?

I think you need to move away from the idea of temperature as simply a measure of the kinetic energy per particle in a system. Other folks have used it, but in thermodynamics, it can be extremely misleading - temperature encapsulates a much wider range of behaviors than just moving around. Instead, I like to use the statistical mechanics definition of temperature, and think about it as a way of measuring how the number of possible configurations (called "microstates") of a system changes as you increase or decrease the system's energy.

Let's say we have a system consisting of two particles, with a total energy of two "chunks" of energy.$$^{\dagger}$$ How many ways can we distribute those chunks of energy between the particles?

Energy of particle 1 Energy of particle 2
2 0
1 1
0 2

This gives us three possible ways we can distribute the energy; we then say that the system has three possible microstates. Now say we increase the energy of the system, and add a third chunk of energy. Now how many ways can the energy be distributed?

Energy of particle 1 Energy of particle 2
3 0
2 1
1 2
0 3

We can distribute the energy four ways, so the system now has four possible microstates. By increasing the system's energy, it turns out that we've increased its number of possible microstates.

The entropy of a system, $$S$$, is related to the number of microstates $$\Omega$$ by the relation $$S=k_B\ln\Omega$$ where $$k_B$$ is Boltzmann's constant. An increase in microstates means an increase in entropy. Here's where temperature comes in. The statistical mechanical way of defining temperature is $$\frac{1}{T}=\frac{\partial S}{\partial E}$$ where $$E$$ is the energy of the system; put in words, temperature describes how the entropy of a system changes in response to changes in the system's energy.

We can see that the toy system above has a positive temperature: An increase in the energy increases the entropy, and a decrease in the energy decreases the entropy. Therefore, for any change of energy, we always have the condition that $$\frac{\partial S}{\partial E}>0$$ and so the system has a positive temperature. By using the general properties of Einstein solids, you can show, in fact, that regardless of the number of particles or the total amount of energy, the temperature of Einstein solid is always positive.

Now, negative temperature systems do exist, as you've alluded to in comments (e.g. Braun et al. 2013, which you mentioned above). However, they have some strange properties. For instance, the Hamiltonian describing them must be bounded from above (feel free to skip this section if you're not familiar with Hamiltonians - you won't miss too much!). This means we can show easily that certain systems can never have negative temperatures.

For example, consider an ideal gas of $$N$$ identical particles, each with mass $$m$$. We assume that the particles don't interact, so there is no potential term in the Hamiltonian - just a bunch of kinetic terms: $$H=\sum_{i=1}^N\frac{\mathbf{p}_i^2}{2m}$$ where $$\mathbf{p}_i$$ is the momentum vector of particle $$i$$. As there is no limit to the momentum of a particle, the Hamiltonian is not bounded from above. Therefore, an ideal gas must have a positive temperature.

Now let's look at the Einstein solid case in more detail. Each "particle" is actually a three-dimensional quantum harmonic oscillator, with mass $$m$$, frequency $$\omega$$ and Hamiltonian $$\hat{H}=\sum_{i=1}^N\frac{\hat{\mathbf{p}}_i^2}{2m}+\frac{1}{2}m\omega^2|\hat{\mathbf{r}_i}|^2$$ Again, the momentum operator is unbounded, so the Einstein solid must have a positive temperature. In general, unless the kinetic energy is somehow bounded, the presence of a kinetic term implies a positive temperature (as I understand it). To create a negative temperature system, you would need the appropriate bounds on the Hamiltonian.

In negative-temperature systems, if you increase the total energy, you decrease the entropy - and by our statistical definition of temperature, the system must have a negative temperature. This doesn't mean that the average energy per particle in the system has a negative energy; therefore, it's not useful for creating something like an Alcubierre drive.

There are some interesting consequences to this. In the absence of performing work (as is the case in a refrigerator, where work is performed to cool your food), when two objects with positive temperatures are placed in contact, heat will flow from the one with the higher temperature to the one with the lower temperature. On the other hand, if one object has a negative temperature and the other has a positive temperature, heat will always flow from the negative-temperature object to the positive-temperature object. In this sense, an object with negative temperature is always hotter than an object with positive temperature.

A way of quantifying this is to define $$\beta\equiv\frac{1}{k_BT}=\frac{1}{k_B}\frac{\partial S}{\partial E}$$ and note that for any two objects with $$\beta_1$$ and $$\beta_2$$, heat will flow from object 1 to object 2 if $$\beta_1<\beta_2$$, and from object 2 to object 1 if $$\beta_2<\beta_1$$.

As I've stated above, we need to separate the notion of temperature from the notion of kinetic energy. In the classical case where we have, say, particles at a mean temperature $$T$$ moving around, then we can assign them each thermal energies $$f\cdot\frac{1}{2}k_BT$$, with $$f$$ the degrees of freedom of the particle. From this, you'd think that a negative temperature corresponds to a negative kinetic energy - but as the sort of negative-temperature systems we're talking about don't involve billiard ball-like particles moving around classically, it's not really correct to say that they have negative energies.

$$^{\dagger}$$This is effectively a simplistic Einstein solid model. In general, if there are $$q$$ chunks of energy and $$N$$ particles, there are $$\Omega(q,N)=\frac{(q+N-1)!}{q!(N-1)!}$$ microstates. You can check that the two cases I described above correspond to $$\Omega(2,2)=3$$ and $$\Omega(3,2)=4$$.

• I like this answer that explains the problem in plain, simple words - understandable by your average Arts Major. (just kidding, +1 because it is neat and my PhD in physics helps :))
– WoJ
Dec 21 '20 at 17:37
• To check that I understand correctly, does this mean that $\Omega$ of a system with one regular particle and one exotic particle has infinite entropy even at $E=0$, since the kintetic energy $\frac{1}{2}mv^2$ could be distributed (0, 0), (1, -1), (2, -2)...? Dec 21 '20 at 21:45
• @MartinvanIJcken What do you mean by an "exotic particle"? Also, what do you mean by $E=0$? If there's no energy, there's only one microstate. Dec 21 '20 at 21:48
• @HDE226868 as I understand, exotic matter is matter with negative mass, allowing it through $\frac{1}{2}mv^2$ to take on negative chunks of energy, with an exotic particle I then mean one of the particles that makes up exotic matter. Dec 21 '20 at 21:50
• @MartinvanIJcken I'd say that that's going beyond the bounds of my example - it has nothing to do with negative mass, or indeed the mass of the particles at all. It's just a pretty simple toy system as a way to explain the statistical definition of temperature - I wouldn't read too much into it. Dec 21 '20 at 21:53

Absorb energy by turning it into matter.

Warning: these foggy musings are strictly suitable for a science fiction endeavor!

Consider iron. https://en.wikipedia.org/wiki/Iron_peak In typical stars which are making new elements via fusion, iron is the heaviest element they make. Fusion of elements lighter than iron gives off energy and heat the stars. Fusion of iron and elements heavier absorbs energy and so cools the star, ultimately making it either fade or explode. Actually fusion of some lighter isotopes also absorbs energy as well which might be good for the story if iron fusion seems trite.

In any case: people in your world have figured out how to do muon catalyzed fusion, inducing fusion without extreme temperatures and pressures. That is where they get their energy. Induced fusion of heavy elements is endothermic, absorbing energy. If this spooky muon tech is used to cause fusion of iron nuclei but the need of this matter for energy is greater than what can be met in the surroundings, the resulting matter goes into debt - a temperature lower than absolute zero, because all ambient energy has been absorbed and additional input energy disappears into trying to complete formation of the heavy element product of the fusion.

What this indebted matter would look like, or be good for, is a fine subject for the story.

• Definitely not science-based Dec 21 '20 at 17:12

Uncertain

The problem is that we barely scratched the surface in these fields. They work with uncertainty in their principles even, allowing for things to basically teleport across barriers, or at the same time be, not to be and any combination thereof.

Incredibly simplistic: Temperature is basically a way to say how fast the molecules are moving around. Kelvin is measured from when molecules have stopped moving completely. It isn't possible to get lower than not moving. They did discover these negative Kelvin degrees in a quantum gas or similar, but how they derive it's lower than 0 Kelvin is lost on me.

Negative density is something else however. It is not just the absence of mass, but negative mass. I really don't understand either further than these definitions. Temperature is a property of mass and it seems unlikely to me it can create negative mass.

Perhaps you can think of the situation like this: in the essence of "laser cooling" its much like cancelling the motion/jittering of the molecules, which essentially lowers the temperature. A parrellel to this is like if a bowling ball is rolling at you, you can strike if halfway with another bowling ball, of equal and speed, and this effect will cancel the bowling ball comming at you and cancel the motion. Laser cooling takes this same effect, instead shooting waves of light with wavelengths that are almost matching the jitterering of the molecules the gas is hitting. So it cancels the motions and therefore reducing the temperature. But the laser can theoretically only cancel the motion to zero/nothing/no motion, so having negative temperature with this formulation would be something like having anti Or negative motion, which realistically may be hard to grasp!

It turns out, you can build a perpetual machine of second kind, thus violate the second law of thermodynamics, if you can extract energy from negative temperature systems. The proof of such assertion is as follows.

Kelvin's statement of second law of thermodynamics: It is impossible to build a machine whose sole result is the complete conversion of heat into work, as known as, there is no miraculous heat engine, that sits idly by and taking all the heat it could get from environment by itself, and dumping energy to you - we could have near unlimited infinite energy with such a machine! - YAY! LET'S BUILD ONE!.

Consider a carnot engine. A carnot engine is defined as a reversible cyclic engine [aka, it operates in a cycle, and, the cycle can be reversed]. It turns out that, if you consider two carnot engines operating in series, it is possible to prove that all carnot engines have the same efficiency, provided they operate between the same heat sources. Because of this, it can be shown that the efficiency of the carnot engine depends on the temperature of the cold source $$T_C$$ and the temperature of the hot source $$T_H$$, only:

$$1-\eta = \frac{Q_C}{Q_H} = \frac{T_C}{T_H}$$

In this, the standard definition of thermodynamic temperature was chosen, but any other temperature definition could have been chosen, and you would only change the efficiency by the addition of a functional form on the temperature. Furthermore, from the formula above it is clear that the temperature must be proportional to the actual heat.

Now, let's plug our carnot engine in between a positive temperature hot source $$T_H > 0$$, and a negative temperature cold source $$T_C < 0$$. This automatically implies that $$Q_C < 0$$ and $$Q_H > 0$$. From the drawing above, $$Q_H$$ is going into the machine, but now, because $$Q_C$$ is negative, the arrow reverses: the machine is no longer dumping the heat into $$T_C$$, it is actually taking heat from $$T_C$$.

So, we build a machine that takes heat from $$T_H$$ and $$T_C$$ and uses all the heat to produce work $$W$$. This is, by definition, a perpetual machine of the second kind, and we violated the second law of thermodynamics: all we needed to build such miraculous machine, was the existence of a system with negative temperature.

Since the second law of thermodynamics holds, we hereby proved that all thermodynamic temperatures (doesn't matter how you define them), must be positive.

=].

Just a note. Yes, negative temperature systems do exist in the real world, but, for them to exist, they need to be isolated from the everything else - they need to be adiabatic systems. And, you can't extract heat from adiabatic systems, by definition. If you were to build a negative temperature system [say, an idealized two level system, or a more realistic somewhat prepared meta-stable magnetic spin system, or something], and, after you done, if you remove its adiabatic barriers so you could extract energy from it, it turns out it would lose excess energy spontaneously to the environment and it would be bought to a positive temperature.

So, no violating the second law of thermodynamics today.