The h Bar

Secret

03:17

Last night dream, deals with a strange spacetime structure which is basically like a warp bubble except the flat spacetime in the middle is twisted into a tube and somehow punctured so that it continues with the relatively flat spacetime outside with no time dilation. The time coordinate, however has a displacement such that when passing through the tunnel, you will end up in some future time

This result in a very strange structure where locally there is no time dilation everywhere in the tunnel, but the overall time difference between the entrance and exit is nonzero

danielunderwood

03:46

Guys I've discovered a tremendous fact

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

Oops they're both $n-1$

Slereah

07:14

Hey

bolbteppa

07:29

@Slereah sup - think I got the point of brst finally

Slereah

So what was the secret

also which BRST

path integral BRST or canonical BRST

Path integral isn't too hard but canonical is the harsh one

bolbteppa

Maxwell/Yang-Mills is a theory derived based off the assumption of gauge invariance, well gauge invariance is a local symmetry so Maxwell/Yang-Mills are theories based off the assumption of the existence of a symmetry, it's just a local symmetry. The particular local symmetry known as gauge invariance results in a singular theory, and to measure things, we need a non-singular theory, so by fixing a gauge we arrive at a non-singular theory, great...

Slereah

that is indeed gauge theory

bolbteppa

However not only does fixing a gauge spoil the inherent symmetry of our theory, even worse, in fixing a gauge just by hand (which apparently people did in the early days before Faddeev-Popov), while giving measurable quantities, it apparently spoils the unitarity of the whole theory... A partial way to save unitarity was found by randomly inserting, again by hand, the things we now call ghosts into the theory.

Faddeev-Popov found a way to fix a gauge theoretically, i.e. not just by hand, and magically the ghosts naturally fell out of it (via the determinant)... Since we naturally fixed a gauge, spoiling symmetry, in an inherently symmetrical theory, we would expect that symmetry to still exist in our theory. Sure enough, a global symmetry on the ghosts is that symmetry, the BRST symmetry.

Thus, it seems that any theory with local symmetry can be re-formulated as a theory with global symmetry by re-expressing things in terms of ghosts. Previous points made here about BRST, e.g. using them as a way to show measurable quantities are independent of the gauge parameter, miss the main point of BRST, wahoo

Maybe that was obvious, but definitely not to me until now

Slereah

07:44

any good ressource for all that?

bolbteppa

So it makes sense to take the gauge-fixed FP Yang-Mills action, Peskin 16.34, and then look for some new symmetry which encodes the local symmetry you fixed (at the expense of new ghost variables), it's a bit crazy to look for fermionic symmetries unless you see the FP procedure on Yang-Mills and see how the fermionic ghosts arise and how you can re-write the action so that they act as a Lagrange multiplier (i.e. going from 16.34 to 16.44 in Peskin),

this way already hints at a first-principles way of introducing BRST from a given action by assuming fermionic symmetries and fermionic Lagrange multipliers

No good resource :p piecing it all together from loads of them :\

Slereah

well you know what you have to do

write one

one thing I should do rly is try to work out things for like

a toy model

Like the quantum theory for $L = \frac{1}{2} (\dot{x} - y)^2$

nice and short toy model

bolbteppa

I want to get BRST for the relativistic point particle, kept putting it off until I made sense of wtf you were even doing it all for

Slereah

It's not the most fun example to start with rly

because it's reparametrization invariance which is like

fancy

also there's like 4 different types of constraints in constraint theory so it's a bit hard to keep track

and they have similar names D:

bolbteppa

Yeah, I don't think it's that bad though

Slereah

07:59

could be worse yeah

bolbteppa

Think in this example you don't have to worry about those constraint cases other than the first one, still need to get that down too

Slereah

Also apparently the reason that you can't really have RQM in general is because you can't have a lorentz invariant theory of a point particle if the EM fields aren't zero

which is good to know

bolbteppa

Not sure about that, for one rel particle anyway I'm thinking the fact you don't have a non-trivial Schrodinger equation is because the Hamiltonian is zero (reparametrization invariance), and the Hamiltonian being zero is the KG equation, so KG holds immediately, but you unavoidably need to do the Dirac bracket stuff in the Heisenberg picture

Slereah

the whole reason why RQM doesn't work gets tricky once you realize you can still do BRST quantization of a relativisitic point particle

bolbteppa

and BRST sync's up with what you get from the Dirac stuff on the point particle

Slereah

08:04

it's fairly poorly explained in general

I think because between the era of RQM and the string theory era, people didn't care that much about making point particles rigorous

bolbteppa

Yeah it's pretty hard to make sense of it all

Basically, $S = - m \int ds$ has a zero Hamiltonian because of reparametrization invariance. You can derive that the Schrodinger equation from the quasi-classical approximation simply has to be $i \frac{\partial}{\partial \tau} \Psi = \hat{H} \Psi$ but $\hat{H}$ is zero. If you do the Dirac bracket stuff the constrained $\hat{H}$ is just a multiple of the constraint that $p^2 + m^2 = 0$, which in operator form is the KG operator,

but this still acts on $\Psi$ and it's not like the multiplier changes anything, so you get the KG equation from a point particle

Slereah

Well I understand the basics, just not like

The process to go from the constraint to the quantum theory

bolbteppa

I think that logic is right anyway, not 100% sure yet

Slereah

Like what justifies going from this classical theory to this quantum theory

though I'm guessing maybe nothing does

Since I doubt that there's a theorem guaranteeing a quantum theory that works in general

I dunno

bolbteppa

I think the idea is, in normal QM, both rel and non-rel, saying that $\Psi \sim e^{iS/\hbar}$ holds as you approach the classical limit is a way to state the quasi-classical approximation, and you literally derive Schrodinger from this by simply differentiating and using $\frac{i}{\hbar} \frac{\partial S}{\partial t} = - \frac{i}{\hbar} H$ so that $\frac{i}\hbar \frac{\partial }{\partial t} \Psi = \hat{H} \Psi$.

Clearly if $S$ is the relativistic action, we differentiate w.r.t. $\tau$ and find $H = 0$ so that $\hat{H} = 0$. That would all blindly seem to be 100% perfect, if it wasn't for the fact that we know reparametrization invariance forces $H = 0$ which forces a constraint on our system, so all we need to do is eliminate the inherent constraint and everything's fine.

The constraint and multipler is $\lambda(\tau) (p^2 + m^2)$, by the Dirac bracket stuff this turns out to be our constrained Hamiltonian. Now, repeating the process, $\hat{H} \Psi = \lambda(\tau) (\hat{p}^2 + m^2) \Psi$ holds, but the constraint $p^2 + m^2 = 0$ hasn't gone away, so the $\hat{H} \Psi = 0$ holds, the KG equation!

I think that's the claim of the theory

Kind of shocking you get KG from the relativistic point particle action for a single particle this way fundamentally, if it holds up

and BRST is supposed to repeat this and end up with the same claims apparently

Slereah

08:21

You can do reparametrization invariance in non-relativistic QM, too

which helps because the end quantum theory is the same

bolbteppa

Yeah, here reparametrization invariance arises because the choice of action was $S = - m \int ds$ to try get relativistic quantum mechanics for a single particle

Slereah

You get like $t = t(\tau)$, $$L = \frac{1}{2} m \frac{\dot{x}^2(t)}{\dot{t}^2}$$

bolbteppa

Without this picture, it's a bit random to say promote $p^2 + m^2 = 0$ to operators and apply it to a wave function, or 'lets look for a relativistically invariant action/eom' (the latter makes more sense from a Heisenberg QFT perspective I think)

Slereah

or something

really it's weird that it took all this time for constraint theory to appear

you'd think it would have come up in the 19th century

bolbteppa

$S = - m \int ds = - m \int \sqrt{\dot{x}^2} d \tau$, think your thing is the einbein form of it

Slereah

08:24

although I guess that back then people didn't use Lagrangian mechanics for EM a lot

bolbteppa

Yeah I'm wondering why it's not all just Lagrange multipliers

I was hoping to avoid the whole 'insert 1' stuff you do in the FP method on the YM action because it's so seemingly magic, was hoping to 'do' something to the action and have the result come from it, but that seems to be like 'fixing a gauge by hand' and thus risking the whole unitarity problem, I guess the virtue of FP is how magic it is

Have to do FP on the point particle action to set up brst and face the Dirac bracket stuff in an easy case, so it's a great example to fight through :p

Slereah

plus it's finite DoF

which is nice

bolbteppa

Working backwards from a Lagrange-multiplier version of the YM action, which looks natural, and getting to the FP action with ghosts by basically doing the normal steps backwards is already throwing away the inherent symmetry you wanted to preserve at every step by starting from a gauge-fixed action and then ending up with a theory with a fermionic global symmetry, it's like switching the symmetry off then back on, the normal way keeps it on the whole time

Slereah

08:39

the $(\dot{x} - y)^2$ example is nice, tho

Fairly simple model

and it's easy to see the redundancy

Since the EoM are $\ddot x - \dot y = 0$ and $\dot x - y = 0$

bolbteppa

6

Q: Can auxiliary fields be thought of as Lagrange multipliers?

In the BRST formalism of gauge theories, the Lautrup-Nakanishi field $B^a(x)$ appears as an auxiliary variable $$\mathcal{L}_\text{BRST}=-\frac{1}{4}F_{\mu\nu}^a F^{a\,\mu\nu}+\frac{1}{2}\xi B^a B^a + B^a\partial_\mu A^{a\,\mu}+\partial_\mu\bar\eta^a(D^\mu\eta)^a,$$ and in the superfield formalis...

Slereah

the spooky ghosts don't even rly matter for the classical theory

bolbteppa

@Slereah did you go through much of those YT constraint videos

Slereah

I wouldn't expect them to be Lagrange multipliers

@bolbteppa Some!

It's hard because the sound is pretty bad

But I'm on holiday next week

Should have some free time

bolbteppa

Did you take notes from what you went through, even like headlines of the topics he does in the first one or two?

Slereah

08:43

Well I did a few computations by hand to check

bolbteppa

Like, does he flesh out the primary vs. secondary, first class vs. second class stuff

The little Dirac book is just a bit too brief

Slereah

I didn't watch enough to find out yet

bolbteppa

Basically, the crazy thing is, KG is the Schrodinger equation, or rather the Hamiltonian

Slereah

I'm at that part of the work project where I can't just look at physics all day anymore :p

bolbteppa

You're supposed to differentiate w.r.t. $\tau$ not $t$, $t$ is in $H$, and $H = 0$ where $H$ is KG, nuts

Slereah

08:48

that is the trick indeed

and you can get rid of that by fixing $t = \tau $

bolbteppa

I am just stunned at that

Slereah

but then it's the awful hamiltonian

With square roots

bolbteppa

Wait no it's not that

Like, the literal equation $\hat{H} \Psi = (\partial_t^2 - \nabla^2 - m^2) \Psi = 0$

Should say, $i\frac{\partial }{\partial \tau} \Psi = \hat{H} \Psi = - (\partial_t^2 - \nabla^2 + m^2) \Psi = 0$, normally one would try to say $i\frac{\partial}{\partial t} \Psi = \sqrt{\mathbf{p}^2 + m^2} \Psi$ with $t$ instead of $\tau$, man why did that never register

Slereah

Well you can get that

using the action $$S = \int \sqrt{1 - \dot{x}^2} dt$$

or something

there's too many ways to write down point particles

bolbteppa

Hamiltonian of that is zero!

Or it should be if you re-insert $\tau$

Slereah

08:55

well why would you reinsert that

you silly man

the point of that hamiltonian is explicit time!

although I guess it's not Lorentz invariant

bolbteppa

Yeah it's relativity, time bends and light goes backwards and all that jazz, craaazyyy

I skimmed that EM interaction paper ages ago, no idea how it explains why RQM doesn't hold

Slereah

arxiv.org/pdf/gr-qc/0511088.pdf

bolbteppa

Interestingly, if $\frac{\partial}{\partial \tau} \Psi = \hat{H} \Psi = (KG) \Psi = 0$ is your Schrodinger equation, the probability current you get from this is still the usual KG current with the same RQM issues

Slereah

Yeah there's a Thing to do about it

it's described in the paper but I didn't get to it yet

something about only keeping the positive energy part of the Hilbert space

which still forms a Hilbert space

bolbteppa

09:15

When in doubt, throw it out

Slereah

hey, still a valid quantum theory!

you do the same thing in QFT, really

You only keep the forward light cone stuff

bolbteppa

09:26

If you throw them out, you keep positive probabilities in position space also

Novalium Company

10:26

Hey guys, a linear span is basically when you multiply a vector by a number, when you graph all of the possible multiplications, the span is basically infinite isn't it? Since you can multiply it with any number?

Is my understanding of linear span correct?

The span of a set of vectors is the linear combination of them, is that correct?

Anonymous

11:50

@NovaliumCompany Yes, that's correct

Novalium Company

Oh hi @Blue. I've been trying to understand linear combinations and span for the past few hours, making progress. How are you doing? Uni hard?

Is it worth is to try to visualize linear combinations. I mean, if you have 2 vectors for example, you can basically multiply them with specific numbers to get any vector you want. What's even the point of combination?

If you are busy, tell me :-)

Anonymous

@NovaliumCompany Yes, you can get any vector which lies on the plane in which those two vectors lie, by linearly combining them

Anonymous

Of course I'm speaking of real physics-y vectors a.k.a "arrows" here

user1732

@NovaliumCompany were they not drawn out that way here?

2 days ago, by user1732

Linear combinations and span | Vectors and spaces | Linear Algebra | Khan Academy

►

Anonymous

I don't know what type of answer you're seeking when you ask "what's the point of combination?"

Novalium Company

12:00

I mean, a linear combination is a particular way of combining things, is that it?

Anonymous

@NovaliumCompany Sure

Novalium Company

Then the span would always be infinite?

Anonymous

And it's the most natural way of combining too....linear combinations show up all over physics and mathematics

Anonymous

@NovaliumCompany Of a vector? Yup

Anonymous

Span is basically : "the region of space which I can cover by stretching or squishing my vector"

Novalium Company

12:03

Yep, that's my language :D

@Blue did you rated my game 4 stars :D?

Anonymous

@NovaliumCompany Yeah :P

user1732

3.999...

Novalium Company

Well, fair enough :D

@Blue How are your papers going?

Also I see span of 2 vectors for example, like span {v1, v2}. What does that mean? Is this the span of the two vectors added?

Anonymous

@NovaliumCompany Not bad. I'm avoiding sharing it here until I'm convinced people won't typesetting-shame me :P I'm a bit new to this LaTeX stuff and things tend to go horribly wrong at times

Anonymous

@NovaliumCompany It's the plane which those two vectors can cover. Basically by adding those two vectors like $a v_1+ b v_2$ you can get any vector in the plane

John Rennie

12:11

@Blue is the span of a set of vectors just the space of all linear combinations of the vectors, or is there more to it than that?

Novalium Company

@Blue But they are just two vectors, how can they create a plane?

user1732

think about two perpendicular vectors

i, j

Novalium Company

@user1732 Yes?

user1732

write their linear combination please

Novalium Company

ai + bj?

user1732

12:17

correct, now what does that equal?

Novalium Company

Depends on a, i, b and j? But simply, infinity?

user1732

infinity is not a number

Novalium Company

Ok, It can be equal to any point on the graph. (including the negative sides)

user1732

yes

Anonymous

@JohnRennie There are two equivalent definitions of span afaik: Given a vector space $V$ over a field $\Bbb F$, 1) the span of a set $S$ of vectors in a vector space $V$ is the smallest subspace of $V$ containing all the vectors in the set i.e. the intersection of all the subspaces containing that set 2) $\text{span}(S) = \sum_i a_i v_i$ where $a_i\in \Bbb F$ and $v_i \in S$

Anonymous

12:20

"is the span of a set of vectors just the space of all linear combinations of the vectors" - so yeah, it's essentially that...the keyword being "smallest" (in the first definition)!

user1732

does that make sense? @NovaliumCompany

Novalium Company

So the span of any 2 vectors is basically the whole graph plane?

user1732

13 mins ago, by Blue

@NovaliumCompany It's the plane which those two vectors can cover. Basically by adding those two vectors like $a v_1+ b v_2$ you can get any vector in the plane

Anonymous

Things become a little complicated if $S$ is infinite though.

Anonymous

Linear combination

In mathematics, a linear combination is an expression constructed from a set of terms by multiplying each term by a constant and adding the results (e.g. a linear combination of x and y would be any expression of the form ax + by, where a and b are constants). The concept of linear combinations is central to linear algebra and related fields of mathematics. Most of this article deals with linear combinations in the context of a vector space over a field, with some generalizations given at the end of the article. == Definition == Suppose that K is a field (for example, the real numbers) and V is...

Novalium Company

12:26

@user1732 I understand that the span of one vector is basically the line which you get when you stretch it. But when having 2 vectors, you say that it's the plane that they create, how?

Anonymous

Tbh I don't know how to deal with infinite sums. Balarka might know better

user1732

27 mins ago, by user1732

2 days ago, by user1732

Linear combinations and span | Vectors and spaces | Linear Algebra | Khan Academy

►

Novalium Company

@user1732 Ok, thanks. I suppose I'll watch it again, because last time I didn't pay much attention tbh. Thanks for the help and time spent :--)

Anonymous

Topological vector space

In mathematics, a topological vector space (also called a linear topological space) is one of the basic structures investigated in functional analysis. As the name suggests the space blends a topological structure (a uniform structure to be precise) with the algebraic concept of a vector space. The elements of topological vector spaces are typically functions or linear operators acting on topological vector spaces, and the topology is often defined so as to capture a particular notion of convergence of sequences of functions. Hilbert spaces and Banach spaces are well-known examples. Unless stated...

user1732

@NovaliumCompany np pal :-)

Anonymous

12:28

So, hmm, in topological vector spaces they claim that it makes sense. Interesting

user1732

thank you for your honesty

have fun

Novalium Company

@user1732 I always do :D

user1732

:D

Fawad

13:52

What does “common velocity of system of balls during collision” mean?

when two particles are colliding elastically

enumaris

15:54

oompah

ACuriousMind

loompah

Semiclassical

doopity

enumaris

doo

Secret

poo

enumaris

looks like my fishing expedition worked

Secret

16:04

::get hooked::

enumaris

the expression makes more sense in Chinese tbf

Secret

上釣

enumaris

The Chinese word for lurking is literally translated as "diving"

so to fish out the lurkers

you can go fishing :D

Secret

lol

ACuriousMind

@enumaris ...clever!

enumaris

16:08

:D

John Rennie

oompah loompah doopity poo actually has quite a nice rhythm to it

Secret

I lurk alot because my wavefunction is delocalised over at least 10 chat rooms

John Rennie

@Secret careful you don't collapse

enumaris

You should talk to the QC guys if you've figured out macroscopic coherence

heather

macroscopic coherence!? quick, write a pap...oh. =P

Semiclassical

16:11

I'm observed to be localized here and in the math chat. not sure if i'm symmetrized/antisymmetrized tho

Secret

Had I figured that out, the top priority is to not evolve into a momentum eigenstate, otherwise, my fate will be worst than ghost in antiman vs the wasp

Semiclassical

I guess maybe I'm neither, since I oscillate back and forth between them

Secret

Semi: Well there's an easy way to find out: Find someone who is also localised in here and math chat, and then see if bose condensation happened

having said that, I don't know what a chat condense is like

heather

@Secret the use of the word "quantum" in that movie was just...wince-worthy. at least they joked about it ("do you guys just add the word quantum to everything?")

Secret

@heather it does, MCU needs a lot of handwaving, but if you try very hard to try to comprehend what molecular disequlibrium is, the closest thing is that the person somehow slowly became a momentum eigenstate

thus in a sense, we should view ourselves lucky we are struggling so much to get coherence macroscopic,

else...

Jul 9 at 17:24, by Semiclassical

Everybody's gone to the wave existence

Semiclassical

16:17

(fun fact, I was watching a let's play of Xenogears when i wrote that. and 'the wave existence' is an actual plot point there. but then, Xenogears is a whole mess of weird appropriated terminology.)

Secret

Well, japanese are often very good to make science words sounds like mythology stuff for some reason

Anime love to abuse the word "timelines" and "dimensions" for example

Semiclassical

yeah

though I think talk of 'many worlds' is ripe for appropriation/abuse

Secret

Japanese culture, similar to Germans, also have a strong focus on the notion of time in their culture. You can sort of glance that by seeing how organised their time tables are

This time culture also make Japanese love to meddle with concepts of time in many of their works, it also explains why their time travel stories are often very wild

enumaris

which one is xenogears

is that the first one

and xenosaga is the second?

Semiclassical

yeah

xenogears was PS1 era

enumaris

16:24

ok, so that was the one I played

Semiclassical

Xenosaga was PS2 I imagine

enumaris

I got like 75% through

then I couldn't get past this jump obstacle

and I gave up

Semiclassical

and more recently there's Xenoblade

enumaris

lol

Semiclassical

lol, from that let's play I think I know the spot

I thought disk one was alright

enumaris

16:25

I recall a lot of cliffs

Semiclassical

then they started disk two and hoo boy

enumaris

I kept messing up a jump and falling all the way back down

was inside the xenogear at the time

that's all I can remember

kinda like the scene in The Dark Knight Rises...except even with no rope I can't make it to the top :P

Semiclassical

my understanding is that disk one of Xenogears took so much time/effort that they had to rush the second disk if they were to get it done at all

enumaris

very possible lol

Semiclassical

(with my impression being that it would've been better had they stuck with disk one and hoped it was successful enough for a sequel)

ACuriousMind

16:27

That's a very common thing to happen to games

Semiclassical

yeah

Secret

0

Q: Creating negative phase audio "voices" inside ambient sound

Is it possible to use a sample recording of a voice, then using the negate (180 phase) of ambient or white noise as a "brush" of sorts to create the negative of the voice as a cutout of the ambient sound and actually create a pattern of silence that the human ear would perceive as a "voice"?

acoustics

hmm...

Hearing an "antivoice" within a voice by dampening down the voice in a way such that it looks like the absence of soundwaves

I am not sure if we register patterns of silence as voice though

enumaris

16:45

interesting idea

kinda like if you paint an object by painting only the background

John Rennie

-4

Q: What if I throw a ball downward with velocity greater than but near to escape velocity of earth

As escape velocity does not depend on the angle of projection what if I throw a ball downward with velocity greater than but near to escape velocity of earth, will escape earth gravitational field.

escape-velocity

Presumbly there is a loud bang ...

Semiclassical

i was gonna say

the more serious answer, I suppose, is that the escape velocity applies under the assumption that we're in a 1/r^2 gravitational field, and that's the only force the object is subject to

enumaris

"I wanna know that, as it has more energy than Earth's binding energy so will it be repelled away from earth or will it feel weightlessness. "

mind=blown

Semiclassical

but, as JR's remark indicates, assuming that the only force acting on the particle is gravity is a rather poor assumption once the ball reaches the earth :P

that said, here's a more interesting version of that: Suppose you bore a hole through the globe, and you shot an object into it at just above its escape velocity

would it pass through the earth and escape on the other side?

Secret

But I think ultimately it will not work as well because silence technically has only one "pitch". you either let things very loud, or not at all

Semiclassical

16:51

(since the earth acts like a harmonic potential inside, I think the answer would be yes)

Secret

whereas soundwaves have both pitch and loudness

John Rennie

@Semiclassical yes, because the total energy (PE + KE) would be greater than zero.

Semiclassical

good call

enumaris

@Secret but a "not painted part of a picture" has only one color - the color of the canvas. So perhaps it's possible to pick out some outlines? O.O

energy methods - gotta love them

Semiclassical

plus, even if you don't bore a hole through the earth, if the object rebounds elastically then it'll still escape

I'm not sure how many rocket ships would rebound elastically upon collision with the earth tho :P

Secret

16:54

Well, I just run a test on a monotone and then volume modulate it in a way so that the curve of that modulation look like a speech wave. The result can sound like speech of some kind

John Rennie

@Semiclassical The dinosaurs tried that experiment. It didn't end well.

Secret

but I have yet to get to the point to make it sounds like words

Semiclassical

"in the cretaceous period, asteroid experiment upon you"

enumaris

my game theory book arrived :D

got something to read in my spare time now, fun beans

danielunderwood

17:12

Is this week beans week?

And I feel like I've reached some sort of ML enlightenment working on this trackml challenge. I actually have to sit down and think about the data instead of just throwing stuff into a model

ACuriousMind

@danielunderwood Every week is beans week for @enumaris

danielunderwood

I dunno. Last week was "hmmm" week lol

Emilio Pisanty

ah, great, hbar is crowded again, I needed to gripe

man, the RGB color space sucks

enumaris

:D

You're griping about the physical reality that is color?

Emilio Pisanty

@enumaris no

I'm griping about all the colours that are out there that I don't get to use in my figures because 99.99% of displays have terrible colour coverage

enumaris

17:25

You're advocating substractive CMYK methods of printing rather than additive RGB methods?

Emilio Pisanty

I mean, what the hell is this shit

enumaris

I see

Emilio Pisanty

look at all the colours that are out of that triangle that I can't use

@enumaris no, CMYK is equally bad

so, I was trying to design a nice cyclic color map

enumaris

Get a display that can display "a trillion colors"

or something

Emilio Pisanty

following the stuff here

N. Maneesh

17:26

Find the field inside and outside a sphere of radius
$R$, which carries a uniform volume charge density $\rho$.

[![enter image description here][1]][1]

[1]: https://i.sstatic.net/qfThu.png

We have $$\frac{dq}{dV}=\rho $$. Then by symmetry, $$d\vec{E}=\frac{1}{4\pi \epsilon_0}\frac{\rho (z-r\cos \theta) dV}{(r^2+z^2-2rz\cos \theta)^{3/2}}\hat{z}$$
$$\vec{E}=\frac{1}{4\pi \epsilon_0}\int_{r=0}^{R}\int_{\theta=0}^{\pi}\int_{\phi=0}^{2\pi}\frac{\rho (z-r\cos \theta) dV}{(r^2+z^2-2rz\cos \theta)^{3/2}}\hat{z}$$

Emilio Pisanty

3

Q: What are good periodic color scales for density plots of angle-like function values?

I am writing several papers that require density plots of complex-valued functions of two variables f:R^2→C, where I'm primarily interested in the complex argument arg(f(x,y)) as a function of the input variables, but I wish to emphasize only the regions where the modulus |f(x,y)| is high. Becau...

publications graphics

enumaris

I have no idea how they do the counting

but there you go

N. Maneesh

Am I correct?

Emilio Pisanty

and I was getting some pretty terrible color coverage

danielunderwood

Hey at least you don't have to have grayscale figures

enumaris

17:28

My PRD papers have color figures that are impossible to read without color

and yet I told them not to print it with color cus that's too expensive

Emilio Pisanty

like, Mathematica reckons that this color scale is perceptually uniform

enumaris

:P

Emilio Pisanty

but apparently it's just that I'm working in the CIE LAB color space and that ugly blue bit just steps out of RGB territory ¬¬

@enumaris that's what everyone does, right? I mean, I would not recommend reading anything I've published in PRA on the print version of the journal.

Loong

Reminds me that I should recalibrate my monitor.

danielunderwood

Hmmm no wonder something I just read said try not to use color as part of figure understanding if you don't have to

Emilio Pisanty

17:31

@danielunderwood well, in this instance I have to, so ¯\ _(ツ)_/¯

enumaris

@EmilioPisanty basically yeah

danielunderwood

I mean just look at that. All the answers to physics

I just couldn't use color

Loong

ah, found my colorimeter

danielunderwood

What's the consensus on non-continuous color maps? The only time I've had to use color in figures was for a numerical analysis class and I always used those. Stuff like this

Though I would imagine it's kind of bad to have a single color for a range of values

enumaris

17:46

uhhhh

danielunderwood

That's about what I thought the response would be. I don't really like it, but it seems a lot easier for me to get an idea of change than continuous color. Though I suppose a contour map may have done just as well

Come to think of it, I don't know that anyone has ever told me what's good and isn't in a figure aside from brief things I've read

enumaris

I can't view the image so I don't even know what you're talking about lol

danielunderwood

18:05

Oh right. In that case, it's the best surface plot ever made lol

enumaris

XD

Cows

18:30

I'm amazed by the fact that humans can ponder about how they ponder and remember what they pondered about how they pondered, while pondering.

Tweeted by KNjokom on August 6, 2018 at 6:29 PM

Semiclassical

18:44

What a ponderous statement

danielunderwood

I like to ponder $\textrm{(that I can ponder)}^n$

enumaris

19:35

ponderponder

danielunderwood

mathematicians may call it superpondering

enumaris

prolly not

mathematicians are boring, prolly just name it after the guy who thought it up

jokomian pondering

throws more shade

danielunderwood

Wait is super more of a physics thing than math thing?

enumaris

Superconductivity

Superfluidity

SuperHamiltonian

danielunderwood

superphysics

Hey at least the mathematicians didn't name quaternions Hamiltonians

even though they're $\mathbb{H}$ from what I've seen

enumaris

19:46

They knew better

John Reenie

19:58

hello there. hope you enjoy my new username :)

Semiclassical

@enumaris Ponderous media

danielunderwood

Do all SE sites have moderator elections or just when they need them? I just got a notification for Math

enumaris

dunno

Anonymous

Oh, cool. Nice to see Asaf Karagila in there.

danielunderwood

20:31

I didn't recognize any of the people in there. But I also don't really go to mse

enumaris

20:55

hmmm

Cows

21:22

I might have gone a bit too deep with the pondering lol

bolbteppa

21:54

@Slereah first lecture on constrained systems youtube.com/watch?v=vCk5N1F-Q00 was simply incredible

Sound is bad but blaring it is worth it, sets up constraints, primary constraints, gauge transformations in constrained systems, weakly-imposed constraints (weak-equalities), singular and non-singular systems for both $L = \frac{1}{2}(\dot{x}^2+\dot{y}^2)$ and $L = \frac{1}{2}(\dot{x}-y)^2$

Yellow

22:27

facebook.com/…

I beg an explanation :D

David Z

@danielunderwood All SE sites have moderator elections when they need them

(non-beta sites)

enumaris

23:07

this game theory book is pretty good mmm..tho I'm only 11 pages in lol

danielunderwood

23:20

What's the book?

enumaris

Game Theory

Drew Fundenberg and Jean Tirole

I bought it cus it's a textbook but paperback so easy to carry around :D

Cows

23:36

Very interesting :D

It looks like a cool text

enumaris

pretty good so far lol

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