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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.


Mar
31
awarded  Nice Question
Mar
31
comment The Lagrangian as a metric
When calculating distance from a metric tensor, the integrand depends upon not only the position of each point along a curve, but also upon the direction of the curve at that point. The Lagrangian has the same property -- if you plot $\mathbf{q}$ against $t$ it's easy to see that $\mathbf{\dot q}$ tells you the direction of the curve. It's because of this similarity that I think $L$ might formally be a metric in $\mathbf{r}$ space, if expressed in the right way.
Mar
31
comment The Lagrangian as a metric
That should be $\mathbb{R}\times\mathbb{R}^n$, or $\mathbb{R}\times\mathbb{R}^{3m}$ for an $m$-particle system. I think the other person who answered was confused by that as well, so I've edited it.
Mar
31
revised The Lagrangian as a metric
my weird cross-product notation was confusing people
Mar
31
comment Equivalence between Hamiltonian and Lagrangian Mechanics
I see, thanks. (I would write it $\left.\frac{\partial L(q,\dot q, t)}{\partial \dot q}\right|_{\dot Q(p,q,t)}$, but then I often find myself notating things differently from the conventions physicists use.)
Mar
31
comment Equivalence between Hamiltonian and Lagrangian Mechanics
What does your notation $\frac{\partial L}{\partial \dot q} (q, \dot Q(p,q,t), t)$ mean? The way it's written it looks like $L$ doesn't depend on $\dot q$, so that partial derivative should be zero.
Mar
31
comment The Lagrangian as a metric
By the way, I don't get notified when you update your answer, so if you make any changes you'd like me to see, it's a good idea to leave a comment to let me know.
Mar
31
comment The Lagrangian as a metric
Thanks for your second example, but it seems to be showing something different. The metric you describe is only for the spatial coordinates, and doesn't include time. Still, it's again suggestive of a "deep" relationship between Lagrangian mechanics and geodesics (at least in the case where there are no forces besides the constraints), and so I appreciate it.
Mar
31
comment The Lagrangian as a metric
The vector space I'm talking about is not space-time. For a single-particle system it has the same dimensionality but a different metric. (The Lagrangian not only has dimensions if energy but also depends on the particle's mass.) For an $n$-particle system it has $3n+1$ dimensions.
Mar
30
comment The Lagrangian as a metric
I suppose another way to put the question is, can we always write the Lagrangian in this form, even if $x$ refers to the "generalised space-time coordinates" for many particles (as defined in the question), and $\mu$ and $\nu$ range over a different number of dimensions than the dimensionality of space-time?
Mar
30
comment The Lagrangian as a metric
I'm interested in whether this carries over to the case of a general Lagrangian - multiple particles interacting under multiple forces, bound by multiple constraints, and not necessarily in a relativistic context. The fact that it works for a single relativistic particle moving inertially is suggestive, but by itself it isn't enough to demonstrate that the general case will hold. (See the final paragraph of my question.) But many thanks for the answer!
Mar
30
asked The Lagrangian as a metric
Mar
24
comment How clean will first generation fusion reactors be compared to fission reactors?
(The reason it's not obvious is that fusion power does produce radioactive waste, due to the reactor walls being bombarded by neutrons and becoming radioactive as a result. It would be pretty nice to know the rate at which this waste would be produced in comparison to a fission reactor producing similar power output, and how it compares to fission waste products in terms of safety and environmental concerns.)
Mar
24
comment How clean will first generation fusion reactors be compared to fission reactors?
The question in the title is a good one, but the rest of the text seems to assume the answer, so I can't vote for it.
Mar
24
answered Name of unknown effect where liquid moves when placed on a jagged surface
Mar
23
revised Does a current-carrying wire running through the centre of a solenoid experience force?
No need to be impolite. I think the geometry is quite clear from the question, but if you disagree, a request for clarification in a comment would be more appropriate.
Mar
22
answered Two suns, one moon, and one planet?
Mar
20
comment Are the Hamiltonian and Lagrangian always convex functions?
Thanks, I can mostly make sense of that. I wider if it might be the case that for any Hamiltonian, there exists a canonical transformation that makes it (positively or negatively) convex in the momenta. That would guarantee that a combination of canonical transformation and Legendre transformation could always put it into Lagrangian form.
Mar
19
comment Derivation of Noether's theorem - A problem with physical significance
@JessRiedel yeah, but the question as I understand it is, why is it the action that's preserved by symmetry transformations in this formalism, as opposed to, say, the energy, or just the equations if motion. After all, it seems easy to imagine that the equations of motion could stay the same while the action changes. I think it's a good question!
Mar
19
comment The sound when boiling water
Possible duplicates: physics.stackexchange.com/q/28069 physics.stackexchange.com/q/9333 physics.stackexchange.com/q/102032