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I'm a post-doctoral researcher with a wide range of interests. My career is in complex systems science (or maybe cybernetics) and the origins of life, but I also have research interests in

  • the foundations of statistical mechanics and its relationship to information theory
  • Earth systems science
  • non-equilibrium thermodynamics in general

I'm also generally interested in the foundations of quantum mechanics and in black holes, though I wouldn't say I'm an expert on those things.

It's probably worth noting that despite the fact that my research is in physics-related areas, all my degrees are in other subjects. If I occasionally seem to start talking in an alien language, this is probably why.


Oct
8
comment Definition of phase transitions in statistical mechanics
Many thanks for your helpful comments. I've updated my question with a part about the positive-$T$ transition case, although it's mostly just what you already wrote here. Could you say a little bit more about the relationship to "superselective particles"? I'm not familiar with those.
Oct
8
revised Definition of phase transitions in statistical mechanics
added addendum about a version with a positive transition temperature.
Oct
7
comment Definition of phase transitions in statistical mechanics
@user10001 as an aside though, I should mention that quantities like $Z$ can retain their thermodynamic interpretation even for small systems if you take care to interpret them in the right way. (You just can't get phase transitions.) A lot of the most celebrated recent results in stat mech (the fluctuation theorems, Jarzynski's equality etc.) rely on this.
Oct
7
comment Definition of phase transitions in statistical mechanics
@gatsu $\beta=1/(k_B T)$ is the inverse temperature (standard use in statistical mechanics), and $\lambda$ is a parameter that corresponds to the size of the system, e.g. its volume. Taking the limit of infinite system size is usual (and necessary) in showing that more sophisticated models have phase transitions. In this system it happens that the transition occurs at $\beta=0$, which does correspond to infinite $T$, though I don't see this as a problem for the point I'm making. As I mentioned to Jonas, there's a way to make the transition temperature finite, so I'll update the question later.
Oct
7
comment Definition of phase transitions in statistical mechanics
@user10001 that's certainly true. However, it's not clear that I can't do something trivial like take $10^{23}$ independent copies of this system, which would give me a system with a macroscopic number of states that exhibits the same behaviour. I need to think about this more though.
Oct
7
comment Definition of phase transitions in statistical mechanics
@Jonas though it's worth noting that in order for the transition temperature to be independent of the scale parameter, the degeneracy of the states has to increase exponentially with $\lambda$. This might actually provide some kind of clue about how this model relates to less trivial ones. (I'll edit these observations into the question later.)
Oct
7
comment Definition of phase transitions in statistical mechanics
@Jonas well spotted - but that can be changed by making the states degenerate. If there are several states with $E=0$ and fewer (or just one) with $E=\varepsilon\lambda$, then the transition temperature is positive. I didn't explain that in the question because I wanted to present the simplest example.
Oct
7
revised Definition of phase transitions in statistical mechanics
added 256 characters in body
Oct
7
asked Definition of phase transitions in statistical mechanics
Oct
7
answered Conservation of momentum when friction is present
Oct
6
reviewed Edit Do spin-spin interactions break time reversal symmetry?
Oct
6
revised Do spin-spin interactions break time reversal symmetry?
improved the format and added a tag
Oct
5
comment Are the perpetual machines in this video really perpetual?
you generate $mgh$ of energy for each weight you move from the top to the bottom, but you pay $mgh$ to get it back to the top again. So the total is $0$, always. It doesn't matter how they're connected or how they move, this will always be true. The impossibility of ever getting a non-zero amount of energy from such a machine is one of the most fundamental laws of physics there is.
Oct
5
comment Are the perpetual machines in this video really perpetual?
Hi @Adobe, sorry I forgot about your comment. The thing is, a force can't really be converted into energy, because they are different things, with different units. Gravitational potential (which is force times distance) can be converted into energy, but only by moving mass downwards. If you want to return the mass to where it was you have to move it up again, which uses up the same amount of energy as you gained by moving it down. This is why those devices won't really turn like that: the energy gained by moving mass down on one side cancels the energy lost by moving it back up on the other.
Oct
5
comment Deriving the condtion for spontaneity using gibbs free energy
@JohnRennie that assumption is pretty ubiquitous though, at least in the context of calculating the Gibbs energy change of a chemical reaction.
Oct
5
reviewed Approve Deriving the condtion for spontaneity using gibbs free energy
Oct
4
comment Action and reaction - not the same effect
I don't really get your point. Yes, you feel force on your hand, but your question was about why it hurts when you push into a rock. It hurts because the force is concentrated into small areas, that's all. In the case of pushing your palm into a surface the force is concentrated into smallish areas due to tense muscles, but not small enough to cause pain. Regarding free fall, I still don't get it. Are you asking whether it will hurt if both you and the rock are falling? If so, yes, I suppose it would, why wouldn't it? You will push yourself away from the rock in this case of course.
Oct
3
revised projectile that splits into two fragments of equal mass
homework tag applies even though it's not strictly homework
Oct
3
comment Are the perpetual machines in this video really perpetual?
Trust me, the ones in that video have hidden motors. (Apart from the perpetual train, where if you watch the weight rather than the slope it's on, you'll see it just rolls downhill for a bit then stops.) The thing that all those proposed machines have in common is that they don't work. Friction very rapidly stops them from turning.
Oct
3
comment Action and reaction - not the same effect
I just pressed my palm as hard as I could into a flat surface and I can confirm that it doesn't hurt at all. I'm not really sure what you mean by "the case of a free fall" - can you elaborate?