Mathematics - 2018-08-06 (page 1 of 3)

Zest

00:43

Hello there

Ted Shifrin

hi Zest

Nick

Hi :D

@TRiG Interestingly, this was the first question I had about imaginary numbers as well.

Ted Shifrin

Hi Nick. What's up?

Daminark

Hey everyone!

Ted Shifrin

00:59

Heya Demonark

Daminark

How's it going?

Ted Shifrin

warmly, as everywhere

Daminark

This is true, had to turn on the fan today

Ted Shifrin

Well, we know you love to have fans.

Daminark

Heh, better than enemies in any event

Iza_lazet

01:40

Complex numbers aren't just vectors. They are multivectors (scalars + bivectors in particular), equipped with the geometric product.

I have made it my mission in life to call people out when they confuse vectors with bivectors. The magnetic field is a bivector field!

vzn

02:55

in theory salon, 1 min ago, by vzn

The dangerous disregard for fat tails in quantitative finance / SRSV

user2236

03:24

inverse.com/article/…

and

icm2018.org/wp/2018/08/04/…

Secret

03:52

in The h Bar, 6 mins ago, by danielunderwood

Guys I've discovered a tremendous fact

$$
{10^n \choose 2} = 4 \underbrace{9 \cdots 9}_{n + 1} 5 \underbrace{0 \cdots 0}_{n - 1}
$$

@user2236 These vanishing instances where you still have hope of humanity

Jacksoja

@LeakyNun are you good at statistics?

Leaky Nun

no

Jacksoja

okay thanks anyway

vzn

← wonders if the fields medal was stolen by anyone associated with a brazilian ~~slum~~ favela... seems maybe odds are good o_O telegraph.co.uk/travel/news/…

2

> “It’s often said also that drug gangs prefer to avoid conflict with tourists as this attracts unwanted attention to their territory.” lol

user2236

04:41

@Secret yeah, it was a nice gesture

Eric Silva

05:04

@vzn what does it mean to be "associated" with a favela, it's literally just a place where people live (and most of the people who live there are firmly middle class, but whatever you can also believe propaganda pieces written by people who dont actually understand the local politics).

MatheinBoulomenos

@rschwieb localizations at prime ideals, yeah

Secret

05:30

Hmmmm...

Let {1,2,3,4,5} be a set with discrete topolpgy

consider a permutation map:

Then given any open set, it's orbit will intersect all other open sets

zhk

Hello everyone!
I need help to come up with a weak form of a system of PDEs
Anyone please guide me
Thanks

Secret

Thus for any pair of open set, there exists some n such that $f^n(A) \cap B \neq \varnothing$

So the permutation map is topologically transitive and hence topologically mixing

So is that chaos?

[Random]
Self indulgent hyperbolic mixing

dspace.lafayette.edu/handle/10385/1154

Weak formulation

Weak formulations are important tools for the analysis of mathematical equations that permit the transfer of concepts of linear algebra to solve problems in other fields such as partial differential equations. In a weak formulation, an equation is no longer required to hold absolutely (and this is not even well defined) and has instead weak solutions only with respect to certain "test vectors" or "test functions". This is equivalent to formulating the problem to require a solution in the sense of a distribution. We introduce weak formulations by a few examples and present the main theorem for the...

hmm...

Leaky Nun

06:54

$$\lim_{s \to 1^+} (s-1) \sum_{n=1}^\infty n^{-s}$$

Leaky Nun

07:12

$$\sum_{n=1}^\infty n^{-s} \le 1 + \int_1^\infty x^{-s} \ \mathrm dx = 1 + \frac1{s-1}$$

$$\sum_{n=1}^\infty n^{-s} \ge \int_1^\infty x^{-s} \ \mathrm dx = \frac1 {s-1}$$

so the limit is $1$

s.harp

07:23

Hello!

As everybody knows, the measure of a point wrt Lebesgue measure is 0. This leads to a conundrum when one wants to "draw a sample from the uniform distribution on [0,1]"

Every point has measure zero, so it is impossible for any point to be returned as a sample, instead one can only sample a range, ie an array of yes/no statements of the form: The sample is in the set A_n

I was wondering whether anybody knows a nice exposition on this thing?

Alessandro Codenotti

08:07

You can also have statements of the form "the sample is the point 0.5", they just have negative answer with probability 1

You can find more in any book on probability (with the measure-theoretic approach)

s.harp

08:31

Mathematical probabilty theory texts dont talk about a sampling process

and that is what I am interested in, what I describe above is a dumbed down version of what von Neumann describes in his book on quantum mechanics, but I am interested in something more modern and explicitly addressing this

Shobhit

09:20

Hello, i am searching for material on $k$-th complete coefficient $C_{k}$ for a continued fraction. I have searched online for the same and looked at wiki but found nothing nice, if anyone knows anything please tell me.

pointguard0

09:37

interesting question: math.stackexchange.com/questions/2868922/…

Secret

10:24

Chaotic sphere

Contrast this with:

Secret

11:42

Some dynamical systems exhibit shearing ; that is, the differentials are transvections. Then they can stretch open sets, and wrap them around so as to get topological mixing, while still having zero Lyapunov exponents. An example would be the horocycle flow on a compact surface of negative curvature, which exhibits 2, but neither 1 nor 3. — D. Thomine 1 hour ago

Ah interesting...

So that means, chaos = positive Lyapunov coefficient + topological mixing + dense periodic orbits

ok the latter two is very easy to visualise, it's the former that is a headache

In summary:
1,2,3 = chaos

1 only: e.g. scaling

2 only: e.g. shearing map that topologically mixes

3 only: e.g. tent map + permutations

1+2: e.g. constant rotations of points on a circle consists of only irrational points

1+3: e.g phase space where there are orbits around every rational point, such that no two orbits intersect each other (think of them stacked like a 2D generalisation of ford circles, and there are orbits inscribed in each in a dense manner)

2+3 implies 1

So that means... once we saw signs of topological mixing and locate the dense orbits, we have chaos

So... the logistic map's chaotic region should be full of these cycles lurking somewhere in that seemly homogeneous mass

But can we do better: If we knew the full set of dense orbits, and their whole profile of interaction strengths on nearby trajectories, can we then figure out which trajectory we are on and hence predict chaos?

Chaos is deterministic, meaning everything is determined by the initial condition. Alternately, it means given a particle at an initial condition, the dynamics will cause it to trace out a trajectory. Even if this trajectory is almost impossible to know in principle, the particle will have zero chance of leaving that trajectory

Secret

12:03

So it goes like this:

Secret

12:14

Deterministic, nonrandom and locally predictable: Almost all Newtonian dynamics, where every future and past part of a trajectory is contained in the differential equations and can be computed using only information at a given point on the trajectory

Deterministic, nonrandom and predictable: Some long range dynamics, where more than one point can influence a trajectory. All information still contains in the differential equation systems

Deterministic, nonrandom and limited predictablity: Chaotic dynamics, where exact and error free information of the whole trajectory is needed to predict the dynamics given initial condition, but at small enough ranges, the probability of finding a point somewhere can be predicted

Deterministic, nonrandom and unpredictable: Systems which has more than one solutions, future evolution of a point cannot be predicted at those branching points

Deterministic, random and locally predictable: Brownian motions and other random walks

Deterministic, random and predictable: Stochastic systems with a lot of mixing, or long range interactions

Deterministic, random and limited predictability: Chaotic dynamics plus noise

sorry, the previous 3 should be nondeterministic

[s]I cannot think of any deterministic system that you only have pdfs as information[/s]

Actually, there is one...

Deterministic, random and limited predictability: Quantum systems

Deterministic, random and unpredictable: ??? Something that, when at equlibrium, gives quantum mechanics

Nondeterministic, random and unpredictable: Unknowns

Nondeterministic, nonrandom and unpredictable: Unknown unknowns

In pictures:

Semiclassical

13:03

@Secret if one takes the randomness to reflect our inability to know the initial conditions of the system, then I’d say dBB fits here

Secret

ah, so we cannot calculate the specific trajectories beforehand because we cannot know where the particle initially is?

Semiclassical

you cannot calculate the specific trajectory, singular

once you specify the initial wavefunction and the evolution equation of the system, you can deduce how the wavefunction evolves in time

that time-dependent wavefunction, in turn, will enforce a set of allowed trajectories for the particle

with the initial position dictating which one of these allowed trajectories is actually chosen

user2236

~~trajectory~~ possibilities

Semiclassical

trajectories

user2236

possible paths

Secret

13:11

Hmm.. so it is kinda a flipped around version of the narrative in standard QM as for that, you knew the initial condition, but there is no trajectory, whereas in dBB, you knew the set of trajectories but cannot know the initial condition, thus you can at best only said the probability of which trajectory the particle landed on

Semiclassical

It's worth keeping in mind that, in dBB, the 'complete' initial specification of a system is the initial wavefunction plus the initial position

However, one can learn about the wavefunction statistically---i.e. make many measurements of $x$ at time $t$ in order to deduce $\rho(x,t)$

But by that same token, one can only learn about the initial position in a statistical way

Secret

hmm...

Semiclassical

The basic problem is that the question of 'what trajectories are allowed to the electron' is a contextual one.

It depends on how the system is prepared, and on what interactions subsequently take place

You can measure the electron to learn its position at a given time, but to do so requires interacting with the electron

and once you've done that, the experimental context has changed. hence you lose any further opportunity to learn about the trajectory of the electron in the original context.

Now, you can characterize the original context in a statistical sense by repeatedly preparing said system and measuring it

which is to say, you can estimate $\rho(x,t)$ from actual measurements

@Secret the reason I distrust that way of putting it is that, even in usual QM, one doesn't speak of knowing the position of the electron at time t=0 to infinite precision

one rather says that you measure $x$ at $t=0$ with a certain uncertainty

Secret

right

Semiclassical

The point is more that, if you can only ever measure the position of an electron along a trajectory in a given context once, then what use is that concept of trajectory?

In classical Newtonian physics, the utility of a trajectory is precisely that we can measure the particle at different times along the path and see what is happening

in QM, you lose the ability to make sense of successive measurements as forming a path

or, at least, you lose the option of interpreting those successive measurements as revealing the path in a sense that's independent of the measurement device

Secret

13:25

yup

because unlike in classical, every measurement disrupt the system and thus changing its context

so after every measurement, we are technically dealing with a fresh system which we knew nothing about

Semiclassical

Not nothing, but not much

Take the example of two SG devices which are at 45 degree angles to one another

If an electron comes out as spin up in one of them, then the odds of coming out spin up/down in the other won't be fifty-fifty. (it would be if they were at 90 degrees to each other)

and that difference in probabilities means that we do know 'something' about our system. it's just that knowing 'something' isn't the same as being able to predict the outcome

Secret

Right, we knew the statistics to some extent, but statistics only give some information on what the outcome is like

Semiclassical

Yeah. Huzzah for conditional probabilities

Secret

(Unrelated) Currently applying the explosive generalisation operator onto the discussion we had above as I chat with WSE people about how to think of very very hard magic

Rudi_Birnbaum

hi guys!

what is dBB?

Secret

13:41

de Brogile Bohmian mechanics

Semiclassical

(a brief soapbox: to regard the concept of a trajectory as having no utility in QM is, to me, a perfectly legitimate reason for not talking about it. but to regard such trajectories as therefore 'nonscientific' and worthy of mockery is frankly irritating.)

Rudi_Birnbaum

You know Motl's take on it?

motls.blogspot.com/2016/04/…

Semiclassical

yes

but given his propensity for conflating disagreement with him as proof of intellectual (and moral) inferiority

I basically don't have any interest in what he has to say.

Rudi_Birnbaum

ok. don't know what to think about it. that the heat cap. would come out wrong would kill it. hes a foul mouth.

but politeness would not alter the argument i supp.

Secret

Best way to deal with Motl = tell him to walk from 0 to $\omega_1$

Rudi_Birnbaum

13:52

lol!

Semiclassical

Debating with a normal person is like a random walk on a sphere

Rudi_Birnbaum

Do you have exp. with him?

Semiclassical

you can credibly hope that if you keep going long enough, you'll eventually get where you need to go

debating with LM, however, seems more like a random walk in 3D

Rudi_Birnbaum

:-)

Anyway it seems dBB is not settled, or is it?

Semiclassical

Depends on what we're talking about

In terms of dBB unto itself, i.e. unmodified and restricted to a non-relativistic context

in that context, you've got three postualtes: waves evolve according to the Schrodinger equation, particles are guided by waves, and the initial positions of particles are distributed according to the Born rule

once you grant those, then you can show the equivalence to QM in all its predictions

Rudi_Birnbaum

13:58

Heat capacities of solids?

Semiclassical

is a prediction of QM and therefore is a prediction of dBB

Secret

well maybe except the relative intensities of the fringes, I read some 2013 paper saying that the intensities are slightly different between QM and dBB

Semiclassical

@Secret people have raised stuff like that, but as far as I'm aware it's not true

in order to get dBB to disagree with QM, you have to introduce additional premises/assumptions which modify dBB

but if you do that, then what you're critiquing isn't dBB anymore

Rudi_Birnbaum

@Semiclassical how would that work when you'd have the same phase space volume for a particle as in classical mechanics?

Semiclassical

You wouldn't. dBB isn't a Newtonian theory.

Rudi_Birnbaum

14:02

oh

Semiclassical

To the extent that you'd talk about volume, it'd be configuration space volume not phase space volume

For better or worse, dBB is a story on configuration space not on classical phase space

but this is indicative of a sleight of hand that people like to pull

What, exactly, is meant by 'classical mechanics'?

Rudi_Birnbaum

OK!

Semiclassical

If you're talking about phase space, I can only infer you're talking about some formulation of Newtonian mechanics

But in that sense, dBB is not a theory of classical mechanics

Rudi_Birnbaum

This is how imagined the Bohmian trajectories

Semiclassical

On the other hand, people also like to talk about any theory in which trajectories exist as 'classical mechanics'

Rudi_Birnbaum

14:05

I identify that with Newtonian. Without thinking about it ...

I mean in nature there are not that many possibilities (in principle ...)

Semiclassical

I think a very minimal account of 'classical mechanics' would be that Newton's laws hold

And, well, dBB doesn't even accept that much :)

suppose you have a single particle, starting at rest in a system with no external forces

Rudi_Birnbaum

I did not know that. So the Bohmian trajectories violate Newtons laws?

Semiclassical

in Newtonian physics, you'd conclude that the particle must remain at rest for all times

Rudi_Birnbaum

yes

Semiclassical

In dBB, though, you've also got the wavefunction

and that's what dictates the allowed trajectories of the particle

as such, it's trivial to come up with particles which will start at rest but will accelerate to a final velocity

Rudi_Birnbaum

14:09

But there is no resting particle

in QM

Semiclassical

my point basically being that particles in dBB can accelerate despite the absence of an external force

now, you can actually write the theory in such a way that the wavefunction contributes a term to the potential and therefore gives a 'quantum' contribution to the force

but the more basic lesson seems to be that the wavefunction dictates that the velocity is a function of position and time, which is not the case classically

Rudi_Birnbaum

OK.

Semiclassical

I guess the way I'd say it is that, in the standard QM story, you interpret the uncertainty principle as forcing you to give up on both configuration and phase space

whereas in dBB you only give up on phase space

Now, that choice isn't without it's consequences

And I think it's not unreasonable to conclude that this cure is worse than the disease

Rudi_Birnbaum

You understand by phase space the "position space" I suppose and only by "phase space" both pos. and mom.?

Semiclassical

configuration space = positions of particles, phase space = positions and momenta

Rudi_Birnbaum

14:18

yes.

Semiclassical

@Rudi_Birnbaum (you said phase space twice)

Rudi_Birnbaum

yes.

oh sorry I confused it

Semiclassical

this does mean, I should note, that dBB for better or worse does not treat momentum and position symmetrically

you do recover such a symmetry once you go to the classical limit, of course

but the interpretation does place position measurements as being privileged as compared with velocity measurements

Rudi_Birnbaum

As far as i know, that symmetric treatment in QM has very profound (practical) limits.

as well

Semiclassical

I think that privileging position isn't so unreasonable, given that the only way I know how to measure velocity is by making two position measurements

The problem, though, is that it seems to commit you to thinking of an N-particle wavefunction---a function on 3N-dimensional configuration space---as an objective entity

This is (at least one reason) why Heisenberg, for instance, couldn't stomach the idea of the N-particle wavefunction as being objective

Rudi_Birnbaum

14:23

Of course in pure QM theory that is not the case.

Semiclassical

An objective wave in 3D space is one thing. An objective wave in 3N-D space?

Rudi_Birnbaum

Sure.

Semiclassical

The fact that it's a wave on 3N-dimensional configuration space also commits one to some version of nonlocality

if you act on one of the particles, you change the entire wavefunction

Rudi_Birnbaum

there is no "one"

of the particles.

you have fermions or bosons

and nothing were a particle has an "identity"

Semiclassical

Well, there's two responses here

Rudi_Birnbaum

14:27

which makes a lot of sense for the world of the "small" in my opinion.

Semiclassical

one is that, in ordinary QM, there's also no notion of indistinguishability. You only get that by further imposing symmetry/antisymmetry on the wavefunctions

which, to be sure, one has good reasons to do

but it's something you impose upon from the formalism from without. it's not something that the formalism itself tells you to do

Rudi_Birnbaum

Well the wave function is not an observable.

its Physics, no formalism lives there on its own

Or what you mean by formalism?

Semiclassical

my point is that the definition of fermions/bosons isn't one which is told to you by either the Schrodinger equation or the measurement postulates

Rudi_Birnbaum

I think that the wavefunction is not an observable should be somehow deducible from that its no eigenvalue of an hermitian operator.

Only the density is-

and then you got only two choices left

Semiclassical

to say that the wavefunction is not an observable and therefore not physically real is already to assume the Copenhagen interpretation, though

in particular, one has to have already assumed the validity of the measurement postulates

in dBB, you don't do that.

Rudi_Birnbaum

14:31

don't know about "real" its just not observable. Since obsevables are eigenvalues of operators acting on the wavefunction. that is "the formalism", isn't it?

I see.

Semiclassical

in particular, there's no collapse postulate in dBB

you still have collapse for practical purposes, but it's not a postulate

Rudi_Birnbaum

What I wanted to say, earlier already. When dBB and QM are formally equivalent then I do not see why any one should have a problem with it.

Well, and then its a question about predictions as well. You like both to agree.

When they not agree both, something must by fishy ...

Semiclassical

well, there's two ways dBB can 'not agree' with QM

one is basically by not using dBB correctly :P

Rudi_Birnbaum

lets forget that

Semiclassical

I'd love to, but the examples in the literature of such are basically all of this kind as far as I know

the other way dBB can fail to agree with QM is if one consciously modifies it or violates its postulates

Rudi_Birnbaum

14:36

ok, but thats just some more wrong stuff in primary literature, its full of that ...

Semiclassical

in particular, one of the postulates is the so-called 'quantum equilibrium hypothesis'

Rudi_Birnbaum

then its great.

what is that?

Semiclassical

this postulate in particular has seemed rather artificial to people

more or less, it states that the born rule is fulfilled at some moment in time

once you assume that, then the other two postulates---that the wave evolves according to an appropriate Schrodinger equation, and the particles are guided by the wave in a specific way---ensure that the Born rule will be fulfilled at every other time

The natural response, of course, is: Why should we expect, a priori, the Born rule to be fulfilled at any particular instant?

If you're interpreting $\rho(x,t)$ as a probability density, as in the usual story, then the Born rule basically is a definition. but the dBB story doesn't have that option.

Now, the QEH is not quite as strict a requirement as it might seem. There's been work showing that, if QEH were not true at some time t=0, then nevertheless for times t>0 the system will rapidly 'relax' to equilibrium

just as one would have a system relaxing to thermal equilibrium at some rate in thermodynamics

The implication being that, even if quantum non-equilibrium is possible in dBB, it is by no means probable.

that said, one can consider the possibility of observing such quantum non-equilibrium

and such observations would violate the predictions of QM. (they'd also violate the postulates of dBB, but that's considered a plus for people who consider the QEH postulate as artificial.)

Rudi_Birnbaum

Well, so I see that we have two equivalent theories and one requires a slighty more "volumninous"/complicated formulation.

Semiclassical

Yep.

I mean, in some ways the construction is rather elegant

Rudi_Birnbaum

14:49

That means its a matter of taste or style which one to use at which instance.

Semiclassical

for instance, the guidance law for particles is simply that they move in the direction of their probability current

But I tend to agree with your framing of it.

Rudi_Birnbaum

How does that work for many-body wave functions?

I mean that guidance principle?

You have one current but many particles in the wave function.

Semiclassical

you have one current in 3N-dimensional configuration space, but that projects down to currents for each particle in 3-space

Rudi_Birnbaum

But I suppose they will be all the same?

Semiclassical

not necessarily. indeed, if you have two particles colliding head-to-head, the probability currents should not be the same

Rudi_Birnbaum

14:55

I am thinking about electrons in a static magnetic field.

Semiclassical

I think there you run into some complications due to having to worrying about spin, as there's an additional contribution to the probability current

Rudi_Birnbaum

That can be separated (at least non-relativistically) The wave function gets an imaginary component proportional to the field, hence a non-zero real current.

Semiclassical

hmm

Rudi_Birnbaum

In a non-time dependent problem you have only an imaginary current density.

without field

Semiclassical

no. the current density is always real-valued

has to be, otherwise the continuity equation wouldn't make sense

Rudi_Birnbaum

14:59

Well for a static problem, non-time dependent, the wavefunction is constant and real.

Semiclassical

only if we're talking about stationary states

Rudi_Birnbaum

Thus the current is imaginary.

Semiclassical

No. In that case, the current is zero.

Rudi_Birnbaum

yes (in Chemistry thats most relevant).

Semiclassical

Maybe we're just talking different definitions

Rudi_Birnbaum

15:00

No its a phase fluctuation.

Lest check it - a second.

Semiclassical

when I say probability current, I mean $\vec{j}=\frac{\hbar}{2im}(\Psi^*\nabla \Psi-\Psi\nabla \Psi^*)$

in which case $\partial_t \rho+\nabla \cdot \vec{j}=0$

Rudi_Birnbaum

Sorry you are right its 0!

imaginary zero ;-)

Semiclassical

lol

Rudi_Birnbaum

Anyway when you switch on a static B field you get an imaginary component to wave function.

the

a static imaginary component.

Semiclassical

you're almost right from my perspective, tho. you can rewrite that as $\vec{j} = \frac{\hbar}{m}\text{Im}[\Psi^* \nabla \Psi] = \frac{1}{m}\text{Re}(\Psi^* \hat{\vec{p}}\Psi)$

so $\vec{j}$ is the imaginary part of a complex-valued function

Rudi_Birnbaum

15:05

Yes,

The real part btw would be the density gradient

which continues me to puzzle ...

but thats a different story

Semiclassical

right

anyways, the guidance law would be that $\vec{v}=\vec{j}/\rho$

Rudi_Birnbaum

But now we can have of course a many particle wave function. Ah OK!

thats the velocity field!

Semiclassical

which, if you rearrange it a bit, is $m\vec{v}=\text{Re}\left[\dfrac{\Psi^* \hat{\vec{p}}\Psi}{\Psi^*\Psi}\right]$

Rudi_Birnbaum

"momentum field"?

Semiclassical

in that rewriting of it, yeah

I'm not sure whether to take velocity or momentum as more fundamental in this POV, though

Rudi_Birnbaum

15:09

So but how are now the "individual" electrons flowing?

Semiclassical

Well, in the equation I just wrote, that really is a one-particle equation

so the statement would be that, if an electron is at spacetime point (x,t), then you plug that into the RHS and so compute the momentum of the particle at that point

Rudi_Birnbaum

(Well the velocity field has the same form like the vector potential of the magnetic field ...) So it means it works for one particle based approximations, I suppose.

Semiclassical

which is to say, the wavefunction imposes a specific relation between the velocity and the position of a particle

I'm not sure this takes into account the coupling to the vector potential, though

Rudi_Birnbaum

SO in qBB the velocity field really is a velocity field ...

Semiclassical

Yep.

Rudi_Birnbaum

15:13

Thats nice you know.

I am doing lots of research in currents (mainly applied).

Semiclassical

If you go to the N-particle context without worrying about spin, then the only real differnence is that $\vec{v}$ is now a vector on 3N-dimensional configuration space

but of course if you go component by component you'll get a guidance equations for each of the N particles

this description, though, really only works if you take the particle to be spinless. the probability current gets modified if you include spin

Rudi_Birnbaum

Would continuity hold for those?

Usually we do "closed shell" so all spins are paired in all "orbitals"

Semiclassical

it should. You'd basically have $\vec{v}=(\vec{v}_1,\vec{v}_2,\cdots,\vec{v}_N)$

and $\hat{\vec{p}} = (\hat{\vec{p}}_1,\hat{\vec{p}}_2,\cdots ,\hat{\vec{p}}_N)$

where $\hat{\vec{p}}_N = -i\hbar \nabla_i$

so the 3N-dimensional guidance equation projects to N guidance equations on 3-space

Rudi_Birnbaum

I see (sorry I have to leave for an hour or so!).

Semiclassical

np

Rudi_Birnbaum

15:19

cu

Semiclassical

l8r

sP_

What is √-25*√-4? I suddenly got so confused. Is it -10 or 10? I feel both the answers are right :x

Semiclassical

@sP_ some discussion here: math.stackexchange.com/questions/437908/…

Secret

> Remark: Ordinary laws can give inconsistent results, giving lawyers opportunities to get rich. Mathematical laws have to meet a higher standard.

lol

Semiclassical

lol

Secret

15:23

I really hope all legal systems are mathematical, that way, no more stupid loopholes

Semiclassical

I think if you're hoping all existing legal systems to be like that, you're going to be disappointed

Secret

indeed, human nature is weird

Semiclassical

plus, even if you had multiple legal systems which were each consistent unto themselves

sP_

@Semiclassical ty! :)

Semiclassical

you could easily have scenarios where they each have jurisdiction and therefore issue competing demands

Reminds me of the following. You've got three people trying to decide on beverages. The first prefers beer to hard liquor to water (B>L>W), the second L>W>B, and the third W>B>L

Secret

15:27

that's classical rock paper scissors scenario

Semiclassical

yep

If they vote on beer versus hard liquor, then beer wins 2-to-1.
If they vote on hard liquor versus water, then hard liquor wins 2-to-1.
If they vote on water versus beer, then water wins 2-to-1.

So even though they individually hold transitive preferences, as a group they don't have pairwise transitive preferences

Hence why pairwise voting methods are rather problematic :)

(It also means you can basically get any outcome you want. If you want to have beer, then have them vote on liquor vs. water (liquor wins) and then beer vs. liquor (beer wins). Since beer beats liquor and liquor beats water, beer "must" have been the preferred outcome.)

Secret

cyclic orders often give me an impression that they have some underlying stability, but one can never get that if they look at things pairwise or partially

Semiclassical

there's an underlying symmetry, yeah

Donald Saari uses that perspective to advocate on behalf of the Borda count: If you've got three candidates, then your top-ranked choice gets 2 votes and your second-ranked choice gets 1 vote

in that case, the above cyclic set of voters would lead to all three choices getting 3 votes i.e. a tie

which is pretty much the only reasonable interpretation of the outcome

now, to be fair, that's also true if you just have people give one vote for their top ranked candidate (i.e. voting by plurality)

but the borda count is also symmetric between voters with A>B>C vs. C>B>A

a plurality vote would instead give A,C two votes and B no vote

Hence why plurality votes tend to give short shrift to candidates people like but don't love :P

Secret

sounds like something to be aware of as here in Australia, we use preferential voting

Semiclassical

ah, neat

saari's got a nice summary here: colorado.edu/education/DMP/voting_b.html

vzn

15:55

@EricSilva dont have an axe to grind but from misc sources brazilian favelas seem to be massive, crimeridden and some of the most dangerous places in the world & think many locals would not disagree with those sentiments. its assertions to the contrary that sound like propaganda to me, some of it to decrease tourism alarm. another harrowing/ (now) high profile story that surfaced lately, Christina Rickardsson, TED talk etc brasilwire.com/the-lottery-of-life-christina-rickardsson

Semiclassical

bet - call - raise

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