| bio | website | nathanielvirgo.com |
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| location | ||
| age | ||
| visits | member for | 1 year, 8 months |
| seen | 20 mins ago | |
| stats | profile views | 859 |
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.
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1d |
revised |
Solving Klein-Gordon equation in the Rindler coordinates - the Unruh effect corrected spelling and capitalisation in title |
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2d |
comment |
on Brownian motors Sorry, I agree that talking about "adding a process" was confusing. I've changed it so that it talks about putting a small "device" in the system instead, and I've added another example that should help make it clearer. |
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2d |
revised |
on Brownian motors added 2426 characters in body |
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Jun 16 |
comment |
A box with cooler and heater on opposite faces It could happen if the gas is very diffuse. If you imagine the extreme case where there's only one particle, it's fairly obvious that momentum has to be transferred from the hot side to the cold one, because the particle is (on average) moving faster when it's moving in one direction than when it's moving the opposite direction. |
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Jun 16 |
comment |
A box with cooler and heater on opposite faces So conflicted... on the one hand I don't think this is correct - you're assuming that the cricket ball's momentum is the same when it hits the cold wall as it was when it hit the same wall, whereas in the case of the box of gas this isn't true, so I think there is a net transfer of momentum from the hot to the cold side. On the other hand, you based your answer on an episode of Doctor Who that was first shown in 1982, which makes me really want to give you a +1. |
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Jun 16 |
awarded | Custodian |
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Jun 16 |
reviewed | Reopen How is the energy of an electron-shell related to the speed of electrons in that shell? |
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Jun 16 |
reviewed | Leave Open Creating a stream of single file atoms |
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Jun 16 |
reviewed | Leave Open what makes time proceed |
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Jun 16 |
comment |
on Brownian motors I wouldn't say that thermodynamic equilibrium is free of fluxes "by definition." Rather, the absence of fluxes is a somewhat non-trivial consequence of the second law. |
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Jun 16 |
answered | on Brownian motors |
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Jun 16 |
comment |
on Brownian motors Also note that the statement you quote is from the very beginning of the introduction - I would expect that a detailed explanation of it will be found later in the paper. |
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Jun 16 |
comment |
on Brownian motors Note that a non-paywalled version of the report in question can be found here. |
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Jun 12 |
revised |
Why does this screwdriver roll in a curved trajectory? more descriptive title, and corrected "vertical" to "horizontal". |
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Jun 11 |
revised |
How many points are required to make a black box spelling |
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Jun 11 |
comment |
How does light behave within a black hole's event horizon? maybe not, but why wouldn't it? Consider a photon moving upwards in a constant gravitational field that has energy $E=hf$. As it moves upwards its gravitational potential energy increases, so its $hf$ energy has to decrease in order for the total to be conserved. If it moves far enough its $hf$ energy must reach zero - but what happens then? The photon can't just disappear, and it can't keep moving upwards without getting a negative energy, so the only option is for it to start moving downwards and increasing in frequency. That's my guess, anyway. |
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Jun 11 |
comment |
How does light behave within a black hole's event horizon? I don't know if this is the correct answer to your question, but light moving in the direction of increasing gravitational potential will be redshifted, thus losing energy without losing speed. I can imagine it being redshifted all the way through zero into "negative" frequencies, which (I guess) amounts to light propagating in the opposite direction. |
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Jun 11 |
comment |
How many points are required to make a black box Maybe there's some redundancy in those constraints. E.g. if you know the total mass, centre of gravity, and five of the inertia tensor's degrees of freedom, perhaps there's a way to deduce the sixth. If this is the case there would be only nine constraints and the answer would be four. However, if there is redundancy in those constraints, I haven't been able to see it yet. |
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Jun 10 |
comment |
Why the heat flux vector at a point must be perpendicular to the temperature isothermal surface? Is it a definition or a deduction? fffred has it exactly right, you can only get to the orthogonality condition through symmetry arguments. It's not true, for example, in the case of a material with a highly anisotropic thermal diffusivity. I don't know if such materials exist, but they're certainly not ruled out by thermodynamics. |
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Jun 10 |
comment |
Why the heat flux vector at a point must be perpendicular to the temperature isothermal surface? Is it a definition or a deduction? "[It's] a basic principle of thermodynamics that heat flows from hot to cold bodies. The direction of the heat flux vector is precisely that. Therefore, it should be obvious why this vector is orthogonal to isothermal surfaces once we accept that principle." - that doesn't follow at all, and in fact isn't true in general but only for a homogeneous material. You can get to it from symmetry arguments, but it simply isn't the case that it follows from thermodynamics. |