Given a quantum state, you can always tell me the entropy of that specific quantum state. It's 0. If that's the territory, then where is entropy in the territory?
Given a quantum state, you can always tell me the entropy of that specific quantum state. It's 0.
Only for pure states. Any system you have will be mixed.
these reduce bleeding but are the same as regular floss in removing plaque. If you have no problem with gums bleeding during regular flossing the benefits are only in whether you find it more convenient.
Some people report that it's easier to remove tonsilloliths as well as a possible reduction in formation. If you don't get them, not a concern, of course.
Well, the simplest way to represent a real number is with infinite decimal expansion. Every rational number with a finite digit expansion has two infinite digit expansions. Every Natural has one corresponding string of digits in any positional number system with natural base.
I think at the time of writing I considered 'completeness' to be ill defined, since the real numbers don't have algebraic closure under the exponential operator with negative base and fractional exponent, while with ordinary arithmetic it is impossible to shoot outside of the Complex numbers.
(EDIT: I cant arithmetic field theory today) The best I can come up with is uniqueness of representation, since it implies infinite representations and thus loss of countability. (insofar as I remember my ZFC Sets correctly, a set of all infinite strings with a finite alphabet is uncountable and isomorphic at least to the interval [0,1] of the reals)
EDITED to fix elementary error.
Every rational number has two infinite decimal expansions.
No. Every terminating number has two infinite decimal expansions, one ending with all zeros, the other with all nines.
1/3, for instance is only representable as 0.333... , while 1/8th is representable as 0.124999... and 0.125.
I thought it might be, and if I'd read it elsewhere, I'd have been sure of it - but this is LessWrong, which is chock-full of hyperintelligent people whose abilities to do math, reason and visualize are close to superpowers from where I am. You people seriously intimidate me, you know. (Just because I feel you're so much out of my league, not for any other reason.)
It's a standard joke about mathematicians vs everybody else, and I intended it as such. I can do limited visualization in the 4th dimension (hypercubes and 5-cells (hypertetrahedra), not something as complicated as the 120-cell or even the 24-cell), but it's by extending from a 3-d visualization with math knowledge, rather than specializing n to 4.
Now, if only this let you rotate things through the 4th dimension.
Doing specific rotations by breaking it into steps is possible. Rotations by 90 degrees through the higher dimensions is doable with some effort -- it's just coordinate swapping after all. You can make checks that you got it right. Once you have this mastered, you can compose it with rotations that don't touch the higher dimensions. Then compose again with one of these 90 degree rotations, and you have an effective rotation through the higher dimensions.
(Understanding the commutation relations for rotation helps in this breakdown, of course. If you can then go on to understanding how the infinitesimal rotations work, you've got the whole thing down.)
Try visualizing four spacial dimensions.
Just visualize n dimensions, and then set n = 4.
Morals are axioms. They're ultimately arbitrary. Relying on arguments with logic and reason for deciding the axioms of your morals is silly, go with what feels right. Then use logic and reason to best actualize on those beliefs. Try to trace morality too far down and you'll realize it's all ultimately pointless, or at least there's no single truth to the matter.
Morals are modeled as axioms in certain formulations.
Without the what? That isn't grammatical.
Without the fnord, of course.
In contrast, a few simple desiderata for "logical reasoning under uncertainty" establish probability theory as the only consistent way to do so that satisfy those criteria. Sure, other criteria may suggest some other way of doing so, but no one has put forward any such reasonable way.
Could Dempster-Shafer theory count? I haven't seen anyone do a Cox-style derivation of it, but I would guess there's something analogous in Shafer's original book.
I would be quite interested in seeing such. Unfortunately I don't have any time to look for such in the foreseeable future.
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Helmholtz free energy (A, or F, or sometimes H) = E - TS in the thermodynamic limit, right? So A = E in the case of a known quantum state.
Wouldn't Gibbs free energy be more appropriate? pV should be available for work too.
I find myself slightly confused by that definition. Energy in straight quantum mechanics (or classical Newtonian mechanics) is a torsor. There is no preferred origin, and adding any constant to all the states changes the evolution not at all. It therefore must not change the extractable work. So the free energies are clearly incorrectly defined, and must instead be defined relative to the ground state. In which case, yes, you could extract all the energy above that, if you knew the precise state, and could manipulate the system finely enough.