The h Bar

Slereah

16:04

So what's a good paper that shows the construction of Hadamard states

0celo7

@Slereah what is that even

is that some QFT thing

why even bother with QFT

Slereah

Well it's a bit more general

It's like

Green functions in curved space

Wait

If I have 2D GR

with a cosmological constant

Will the EoM just try to minimize the spacetime volume

Since $S = -\int d^2x \sqrt{-g} \Lambda = -\Lambda V[g]$

ACuriousMind

16:22

@Slereah What's $V[g]$?

Slereah

The total volume

$V[g] = \int d^2x \sqrt{-g}$

ACuriousMind

Ah

@0celo7 Nah, I clicked accept, but since I never configured broadcasting, it probably doesn't do anything

@Slereah: Is it even meaningful to look at 2D GR+cosmo? Gravity in 2D is trivial because the E-H action is just the Euler characteristic, right?

Slereah

No

$\int R \sqrt{-g} dx = \chi$

But only that term

The cosmological term isn't a constant

Or any other term, if you decide to add them

If you put in like $\alpha R^2 + \beta R_{ab}R^{ab}$ or whatever

ACuriousMind

But those are not GR terms, they're just stuff you just made up :P

Slereah

They totally are GR terms!

Well not basic GR but still

They are part of Lovelock gravity, for a start

They also appear in semiclassical gravity renormalization

ACuriousMind

16:28

@Slereah That's a generalization of GR.

Slereah

Cosmological constant is part of GR I would say, though :p

ACuriousMind

Okay, so the action that is to be varied is $-\Lambda V[g]$. I don't think that has a minimum on the space of Lorentzian metrics $g$.

0celo7

@ACuriousMind aww

@ACuriousMind would that not maximize volume

if you try to minimize $-$volume

ACuriousMind

@0celo7 Technically, you're searching for stationary points, not minima/maxima

0celo7

@ACuriousMind I know

ACuriousMind

16:32

But you're right, a minimum of $f$ is a maximum of $-f$.

0celo7

but he said minimize

Slereah

Hm, I guess the variation would be $g_{ab}\Lambda = 0$

Indeed it seems a bit weird

ACuriousMind

It's just a bad action functional :P

0celo7

@Slereah using abstract indices

@ACuriousMind way to label, bigot

Slereah

Though it is used in simple examples of QG

ACuriousMind

16:33

The obvious and unique solution is $g=0$, but that isn't an admissible metric

So this system has no classical solution for its e.o.m.

0celo7

@ACuriousMind why not

Slereah

Well either that or $\Lambda = 0$ :p

So I guess the quantum version works out better

ACuriousMind

@0celo7 Stop trolling :P

0celo7

@ACuriousMind :(

all vectors are null in this spcaetime

I don't see the issue

ACuriousMind

@Slereah It's a bit worrisome to have a system which has no classical solution, though, right?

Slereah

16:34

Because the metric is degenerate

0celo7

@Slereah hmm

Slereah

@ACuriousMind True, but how often can you say "I have solved quantum gravity"

0celo7

that's true

@Slereah every morning

Slereah

Me I mostly surf the web in the morning

ACuriousMind

I mostly sleep in the morning.

Slereah

16:35

2D gravity is pretty trivial so it makes its quantization rather simple

0celo7

@ACuriousMind you should enable broadcasting

it's off by default

David Z

@ACuriousMind just when I might have thought seeing the word "trolling" meant there was actual trolling going on... :-/

0celo7

@DavidZ exactly

people here use that word and it's like crying wolf

one day the real trolls will invade and the mods will just twiddle their thumbs

46

Q: How exactly does gravity work?

The electromagnetic force and strong and weak forces require particles like photons and gluons. But in case of gravity there is no such particle found. Every mass bearing object creates a gravitational field around it, and whenever another mass bearing object enters its field the gravitational f...

ACuriousMind

@DavidZ "Pretending not to understand something" is trolling in my book. Just not necessarily malicious :P

Slereah

I guess it has no solution because there is no such thing as a maximal volume

0celo7

16:38

43

A: How exactly does gravity work?

You're quite right that the other fundamental forces of Nature possess mediator particles, e.g. the photon for the electromagnetic force. For gravity, a graviton particle has been postulated, and is included in the five standard string theories which are candidates for quantum gravity. From a qu...

David Z

@0celo7 either we actually won't, or we'll sit back and laugh because y'all brought this on yourselves :-P

Slereah

you can always pick a solution with a greater volume

0celo7

Am I the only one who thinks it's a bit pretentious for Jamal to say "since you're a high school student, here is the simple explanation"

Maybe not pretentious...

Something in that vein.

Slereah

Do you say that because you're a high school student

0celo7

@Slereah I'm not.

Slereah

16:39

Are you sure you're not

Do you need an explanation of what a high school is

John Duffield

I know exactly how gravity works.

0celo7

9-12

@JohnDuffield Inhomogeneous space, right?

Slereah

Oh, Duffield is back

Hadn't noticed with all that blocking :p

Adolfo

Seeing that you're chatting about GR and gravitation, could you take a look at this: physics.stackexchange.com/questions/219470/… ?

0celo7

we know thing about GR

sorry, this is a math chat

@ACuriousMind So why is $\mathrm{Ad}_g$ smooth in $g$?

yuggib

16:45

@ACuriousMind because it was not sufficiently good to be accepted immediately of course :-P

0celo7

@yuggib I still don't see the issue with the $\pi$ paper

yuggib

@ACuriousMind well, according to your definition of morning, that implies that you always sleep

ACuriousMind

@yuggib Yep, that's why I get nothing done

Slereah

Are there any $1+1D$ topologically trivial metrics with CTCs

yuggib

@0celo7 the $\pi$ paper is a masterpiece

it just made me feel like a giant crab who is shot in the face

user54412

16:48

@Slereah Give me a 0+1 spacetime with a CTC and I'll be impressed

John Duffield

@0celo7 : yes. A concentration of energy in the guise of a massive star "conditions" the surrounding space altering its metrical properties. The effect diminishes with distance and is modelled as curved spacetime. But note that light doesn't curve because spacetime is curved, it curves because die Ausbreitungs-geschwindigkeit des Lichtes mit dem Orte variiert. If you plotted this you'd plot a curve.

Slereah

@ChrisWhite : $S^1$

user54412

@Slereah Realized this as soon as I said it

Slereah

heheh

user54412

well, I'll still be impressed

Slereah

16:49

Well there's only two spacetimes in one dimension

0celo7

is $S^1$ even a spacetime

what is the metric

Slereah

$\mathbb{R}^1$ and $S^1$

0celo7

$-dt^2$?

Slereah

All 1D manifolds have the same metric

yes

because you can always diffeomorphic them back to flat space in 1D

0celo7

@Slereah did you hear

quantomorphism is a thing

Slereah

16:50

Is it

0celo7

yes, it is.

Slereah

$ds^2 = -f(t) dt^2$, then $dt' = \frac{1}{\sqrt{f(t)}} dt$

So $ds^2 = -dt'^2$

So only the topology matters, and there's only 2 connected 1D manifolds

0celo7

@JohnDuffield Zitierst du Einstein auf Deutsch jetzt?

David Z

By the way, general announcement: if interpersonal issues, arguments, insults, etc. disrupt the chatroom too severely, we (moderators) will do what we need to do to keep things under control, which may include chat suspensions or other measures. There's no requirement that these measures be triggered by a specific message, no requirement that we give a specific justification at the time, and no requirement that they be immediate. (We try, though)

2

This isn't prompted by anything specific, just saying

0celo7

@DavidZ the "no requirement that they be immediate" makes no sense to me

Slereah

16:52

Well, unless the manifold isn't Hausdorff, of course :p

Then there are infinitely many of them

0celo7

so does someone want to tell me why $\mathbb{R}\cup\mathbb{R}\ne \mathbb{R}^2$?

Slereah

@0celo7 : $\forall A,\ A \cup A = A$

0celo7

@Slereah proof?

Slereah

grid.ece.ntua.gr/sites/metamathsite/mpeuni/unidm.html

David Z

@0celo7 because mods aren't always here, it's possible that one of us comes in to the room, sees an argument from 18 hours ago, and issues a suspension because of it. That's what I'm getting at. The point is, if someone posts something and doesn't get immediately called out on it, that doesn't automatically make it acceptable.

John Duffield

16:54

@0celo7 : Ja. Because then the meaning isn't lost in translation. Such as to "A curvature of rays of light can only take place when the velocity of propagation of light varies with position."

David Z

Not that I really expect people here to think "I didn't get suspended within 10 seconds, it must be fine", but you never know.

Slereah

⊢ ((φ ∨ φ) ↔ φ), ⊢ ((x ∈ A ∨ x ∈ B) ↔ x ∈ C) equivalent to ⊢ (A ∪ B) = C

So ⊢ (A ∪ A) = A

0celo7

what the heck does any of that mean

Slereah

Are you not a math student

0celo7

also, we're talking about the disjoint union

@Slereah haven't taken PhD courses yet

Slereah

16:58

Do you mean "basic logic course"

Well then write the disjoint union :V

$A \sqcup A$

ACuriousMind

@Slereah: No mathematician except for logicians uses those notations, and even they try to avoid it when possible.

0celo7

Basic logic?

John Duffield

@0celo7 : it means light curves rather like sonar.

0celo7

Tfw you lose weight and clothes are loose

Slereah

Disjoint union is $A \sqcup B = \bigcup_{i \in A, j \in B} \{(i,1), (j,2)\}$

@ACuriousMind For shame

It's a gr8 notation

You want a bad notation try the Principia Mathematica

ACuriousMind

17:02

To obfuscate things, sure :P

0celo7

@ACuriousMind what is the German word for "German speaker"

Slereah

^a bad notation

I solved the first 12 chapters of this by hand so don't tell me about bad notation :V

ACuriousMind

@0celo7 You mean as in "He's a German speaker"? That'd just be "Er spricht deutsch", we don't use/have a nominal construction like that. There's Muttersprachler for "native speaker", but we don't extend that construction to "<language> speaker".

Slereah

Hm

I think it was that file

dl.dropboxusercontent.com/u/19940612/…

I rewrote it in modern notation because seriously

0celo7

@ACuriousMind ok, so sprachler is a word

I thought that was some Pfalzisch thing

ACuriousMind

17:08

@0celo7 No, it only occurs as part of Muttersprachler, it's not a word on its own.

0celo7

German doesn't make sense

John Duffield

Sounds like motherspeaker.

Slereah

@0celo7 : Basically disjoint union has cardinality $a + b$ while the cartesian product has cardinality $ab$

Disjoint unions is just the two sets together while the cartesian product is that, for every element of $A$, you join a full copy of $B$

0celo7

Wtf is cardinality

I'm not a PhD

Slereah

do you know finger counting

It is the number of elements in a set

you lunkhead

0celo7

17:23

Finger counting?

How do I count the real numbers

Slereah

You know

You extend one finger

and it is one

two fingers, that is two

0celo7

Those are integers

Slereah

Yes.

0celo7

I don't think there is a bijection from my fingers to the real numbers

Slereah

You can't exactly do the same for the reals but the analogy is the same

$R \sqcup R$ is two copies of $R$

$R \times R$ is a copy of $R$ for every point of $R$

$R\sqcup R$ is the manifold made of two disconnected copies of $R$

$R \times R$ is the plane

0celo7

17:36

@Slereah yes, that's what makes it clear

Secret

(Actually I need to read again the start of this discussion...)

0celo7

@ACuriousMind Is the statment "$\mathrm{Ad}_g$ is smooth in $g$" equivlent to "$\mathrm{d}_g^n(gxg^{-1})$ exists for all $n$"?

ACuriousMind

@0celo7 Yes

0celo7

also, how is differentiation defined here

is there any noncommutative lie group shenanigans

like can we move $\mathrm{d}_g$ past $g$ when doing the product rule?

ACuriousMind

@0celo7 Choose coordinates for $G$, then differentiate with respect to the coordinates for $g$

0celo7

17:39

@ACuriousMind so I can't just do the differentiation as I wrote it there

Slereah

Is there a classification of non-compact 2-manifolds without triangulation

ACuriousMind

@0celo7 Well, strictly speaking not, but it turns out that one can in the end just symbolically differentiate w.r.t. $g$, e.g. $\partial_g(gxg^{-1}) = xg^{-1} + gx\partial_g(g^{-1}) = xg^{-1} - gxg^{-2}$

0celo7

@ACuriousMind why

ACuriousMind

@0celo7 Tedious computation.

0celo7

@ACuriousMind what does one compute?

and how do you know one can do it

ACuriousMind

17:45

@0celo7 Write $g = g^{ab}$ in some coordinates (it's best to choose the coordinates in which the Lie group is a matrix group, hence the two indices), compute $\frac{\partial}{\partial g^{ab}}$ of that expression.

0celo7

@ACuriousMind oh god

proof by "I don't wanna do that"

ACuriousMind

Yeah, it gets ugly and very full of indices

Slereah

Indices are lovely

INDICE IS NICE

0celo7

but that only proves $C^1$ :P

well, that's all you need, really...

the idea is to show that it's continuous

ACuriousMind

@0celo7 Well, ofc you should just compute that $\partial_g g^n = n g^{n-1}$, then you can do symbolic calculus with $g$ after that.

No need to repeat this nonsense for every function of the group you encounter.

0celo7

17:49

@ACuriousMind oh god how does one do that

induction?

ACuriousMind

Hm, yes, I guess induction is possible

But you have to do it "twice", one starting from $n=-1$, and once from $n=1$.

0celo7

@ACuriousMind oh god

I like my TA's "it's a Lie group so it's likely smooth" better

your "easy way"

boards.4chan.org/v/thread/317363003

discussion about the endings of the game

OH MY GOD

wow 4chan just spoiled the game for me

@Slereah have you completed the main quest yet?

Slereah

nah

0celo7

I'm thinking I should make a new char and only do the MQ

Just so I can see the endings

Slereah

Either that or look at youtube videos of it

Hm, is the Gauss Bonnet theorem $\int R dx \propto \chi$ or $\int \sqrt{-g}R dx \propto \chi$

I guess it's with the determinant, but since they are math people it's implicit in the fact that's it's a surface integral

0celo7

18:11

Well only one of those is coordinate invariant...

Slereah

Also wait what the fuck is the Gauss Bonnet theorem, really

I keep seeing it written in different ways

Some of which don't even have the Euler characteristic

Oh wait, is it only valid for compact surfaces

Hm

Is the Hilbert action still a constant for a non-compact manifold

"Compactness of the surface is of crucial importance. Consider for instance the open unit disc, a non-compact Riemann surface without boundary, with curvature 0 and with Euler characteristic 1: the Gauss–Bonnet formula does not work."

:O

Cohn-Vossen's inequality

In differential geometry, Cohn-Vossen's inequality, named after Stephan Cohn-Vossen, relates the integral of Gaussian curvature of a non-compact surface to the Euler characteristic. It is akin to the Gauss–Bonnet theorem for a compact surface. A divergent path within a Riemannian manifold is a smooth curve in the manifold that is not contained within any compact subset of the manifold. A complete manifold is one in which every divergent path has infinite length with respect to the Riemannian metric on the manifold. Cohn-Vossen's inequality states that in every complete Riemannian 2-manifold S with...

Hm

that's only an inequality, though

Could still have a dynamic

0celo7

18:45

@Slereah D:

so what about 2D GR on a noncompact manifold

Slereah

I know right!

0

Q: Two dimensional spacetime and the Gauss Bonnet theorem

Generally two dimensional spacetimes are deemed to be static, as the Gauss Bonnet theorem implies that the Einstein Hilbert action would be a constant independent of $g$. But as far as I can tell, the Gauss Bonnet theorem only applies to compact manifolds, even in the version for Lorentzian man...

general-relativity differential-geometry dimensions topology

Keep up with the news here!

ACuriousMind

@Slereah: In 2D you have diffeomorphism invariance + Weyl invariance, and a Weyl rescaling allows you to locally (on an open set, not at a point) make the metric Minkowski

Therefore, locally, there cannot be dynamics, and I think this also means the "Gauss-Bonnet" integral is constant even for non-compact manifolds.

It's a purely topological expression, but classifying non-compact manifolds is annoying, see this MO answer

Slereah

Ah, good to know

Was worried for a bit!

But then the expression isn't gonna be $S_H \propto \chi$, I suppose

Still would be nice to find a paper with the proof that isn't just "see the Gauss Bonnet theorem"

carol of the old ones

►

0celo7

19:18

barracks complete

My adhesive farm is in full swing now

Slereah

How do you do an adhesive farm

0celo7

@Slereah corn, tato and mutfruit

Slereah

Oh

0celo7

the issue is I'm gonna run out of purified water soon...

and to get that you need...whiskey

and I don't have a source for that

Slereah

Ask ur mum

gonenc

19:25

What is the hodge dual of $$\partial_k(F^ig_{ij}) dx^j\wedge dx^k$$?

0celo7

Why do you need to know

gonenc

@0celo7 curiosity

I actually wanna know what $\star(dF^\flat)^\sharp$ is but for that I need this step

Slereah

In how many dimensions

$\varepsilon^{jkl} \partial_k(F^ig_{ij}$

gonenc

umm sorry 3

Slereah

$\star (dF^\flat)^\sharp$ should be like

$\varepsilon^{jkl} \partial_k(F^ig_{ij})$

@ACuriousMind is probably sad to see all those indices

gonenc

19:41

@Slereah that is exactly what I thought but somehow it seems like it isn't correct

user54412

If this is GR, what's with the $\partial$? If SR, why $g$?

0celo7

ACM doesn't hate indices

Slereah

@ChrisWhite : For forms, $\nabla = \partial$

0celo7

he hates when indices make things complicated

and unnatural

gonenc

basically $\nabla \times F$ should be that form

0celo7

19:41

@gonenc it is

Slereah

@gonenc Oh yeah, I see the problem

0celo7

so what @Slereah wrote is correct

user54412

@Slereah Is this standard notation I've somehow missed?

Slereah

You forgot to put in the antisymmetrisation

0celo7

@ChrisWhite what are you talking about...

Slereah

19:42

It should be $\partial_{[k} F_{j]}$

gonenc

@0celo7 then why the hell don't I get the correct thing in cylinder coord

0celo7

the exterior derivative is calculated using partials because the Christoffels die

Slereah

So sad :(

ACuriousMind

@gonenc You used the correct $\nabla$ for cylinder coordinates, right?

@Slereah wat

gonenc

@ACuriousMind I looked it up so it is gotta be correct

0celo7

19:43

@Slereah this is not true

Slereah

@ACuriousMind You know what I mean

You monster

ACuriousMind

@Slereah No, not really

0celo7

@Slereah no, I don't think he does

I don't either

user54412

Raise your hand if you understand anyone else right now

0celo7

I understand ACM

Slereah

19:44

$\nabla_{[a} T_{bcde...]} = \partial_{[a} T_{bcde...]} $

0celo7

::raises hand::

user54412

::ACM lowers your hand::

2

0celo7

@ChrisWhite ::I punch CW::

ACuriousMind

@Slereah Ah.

user54412

@0celo7 y u hate homotopy theory?

0celo7

19:46

my water purification system works

I now have unlimited adhesive

Slereah

what are you gonna do with them

Since the game is over

ACuriousMind

@gonenc If you're also using the correct $g$, then it remains to conclude that you can't calculate :P

0celo7

gonenc

@ACuriousMind possible

0celo7

this makes quite a lot of water

Slereah

19:49

@gonenc : Also did you remember to antisymmetrized everything and all

It's a form, it should be antisymmetric

gonenc

@ACuriousMind I only have a problem at the $\theta$ component

I have a factor of $1/r$, which shouldn't be there

user54412

As long as I'm being confused, what's up with these questions?

user54412

3

Q: Why doesn't a backward wave exist?

Huygens principle says every point of wavefront emit wavelet in all directions. Then why does a back ward wave not exist? Can any expert tell real answer? On different sites I get different and contradictory answer. I am asking question from my class 12 book. It says " Huygens argued that the am...

waves interference geometric-optics

user54412

0

Q: Is Wikipedia wrong about Huygens-Fresnel Principle?

Someone posted a question about why backward waves don't exist according to the Huygens-Fresnel principle: Why doesn't a backward wave exist? In following up on this question I read the Wikipedia article https://en.wikipedia.org/wiki/Huygens%E2%80%93Fresnel_principle, which makes a big deal abou...

visible-light diffraction

user54412

Why is there so much consternation about backward waves?

ACuriousMind

19:53

@ChrisWhite I have genuinely no idea what those questions are talking about

gonenc

@ACuriousMind me neither :D

ACuriousMind

I think the confusion is that, if every point of a wavefront emits a spherical wave, then why does the wave travel forward?

But:

16

Q: Why is Huygens' principle only valid in an odd number of spatial dimensions?

Apparently Huygens' principle is only valid in an odd number of spatial dimensions: http://mathoverflow.net/a/5396/21349 Huygen's principle in curved spacetimes Why is this? [EDIT] This is somewhat perplexing, since AFAIK it's pretty common to teach freshmen about double- and single-slit dif...

optics electromagnetic-radiation waves

user54412

It's like asking "How come when I drop a pencil and it falls to Earth, it wasn't falling from even higher before I dropped it, since the equations are time-symmetric?"

Slereah

Well why doesn't it???

ACuriousMind

This question, or rather Moretti's answer, shows that you can't think of Huygens principle as simple as "everything emits a spherical wave" and then it follows that there's onloy a forward wavefront

Rather, the fact that only the "forward" convex hull of the spherical waves is physical is a non-trivial statement

Which cannot be proven from the spherical wavefronts alone.

Hm...maybe people are also confused abot the difference between a "full", "retarded" and "advanced" solution

Slereah

20:03

you just went full retarded

ACuriousMind

Because the Huygens' principle implictly only searches for the advanced solution when it take the convex hull "forward"

The "backward" wave is the retarded solution, i.e. where the wave came from.

Hm, I should make that an answer, I think.

Slereah

I wonder what EM looks like if you allow the solution to be time symmetrical

I think Feynman did a thing on that

0celo7

found it

Slereah

Take it

It's breddy good

Also mod it to hell

You can kill a courser in two shots

0celo7

@Slereah uh

I'm the director of the institute

I can get a courser as a follower

!

JokelaTurbine

20:08

@ACuriousMind comment to your comment under my answer; Well, it's of course very simply according to the Newtons laws. This all is actually said in the question; "The point is that I don't understand how this tells the way we should compute A. I've read this some times now and I can't get the idea." ...and further. So the question is basically; How are these fancy formels connected to reality.

0celo7

what is this beauty

missing the arms .-.

Slereah

@0celo7 : Kill everyone

0celo7

@Slereah is this thing really good?

ACuriousMind

@JokelaTurbine No, the question is explicitly about how to calulcate the gauge potential $A$ in the question. Your answer does not provide an answer to that at all.

Slereah

@0celo7 It's no nukes, but it's pretty good and the ammo isn't too rare

user54412

20:11

@ACuriousMind I feel like this all boils down to people forgetting that t=0 (when I switch on the light or whatever) is a boundary condition, and a non-time-reversible one at that.

Slereah

Also it's kind of an axiom in EM that only retarded solutions are kept

Only solutions in the forward light cone, in general!

JokelaTurbine

@ACuriousMind The answer is in question; To cancel these forces and torques, we must correct the motion by subtracting a Stokes’ flow corresponding to a rigid displacement of the shape with the same leading behaviour at infinity as our trial solution. The result is the actual fluid motion. By our definition, this rigid displacement is 1×A(t)δt. This completes our outline of the method for calculating A.

Slereah

Another obviously fake name : Cutkoski

Cutkoski cutting rules?

0celo7

This is just getting silly

Slereah

Obviously someone russified "cut" to make the name

user54412

20:21

@gonenc Did you get the coefficient to work out? If your original form had coefficients $T_{jk} = \partial_k(F^i g_{ij})$, I calculate ${\star}T^\theta = \partial_r F_z - \partial_z F_r$, which is the $\theta$-component of $\nabla \times F$, isn't it?

gonenc

@ChrisWhite but you have the term $1/\sqrt{|g|}$ upfront, which comes from the levi civita tensor

I think I found what the problem is

user54412

Where am I supposed to have that? I used ${\star}T_l = \sqrt{\lvert g \rvert} [j\, k\, l] T^{jk}$ at some point.

gonenc

$\varepsilon^{jkl} \partial_k(F^ig_{ij})= \tilde \epsilon^{jkl}\partial_k(F^ig_{ij}) 1/\sqrt g$

where $\tilde\epsilon$ is the lc symbol

@ChrisWhite and from that I get $\frac 1 r$ of what you have

the problem arises from some normalization mistakes imo

cause in divergence I also get the $\theta$ term wrong

this time I miss a factor of $r$

0celo7

20:40

I need to raise gun nut, science and armored to 4 to make my character godly

0celo7

21:00

@Slereah have you uncovered the mayor's true identity yet?

gonenc

@ChrisWhite btw everything gets waay worse in spherical coords.

Slereah

No

0celo7

He's not who he says he is!!

ACuriousMind

@gonenc Just compute that the term is the curl in Cartesian coordinates and observe that both the curl and the $\star(\mathrm{d}F^\flat)^\sharp$ are covariant under coordinate transformations, no need to compute this ugly equality in spherical coordinates :P

However, I understand that now that you've begun that you want to find the error

gonenc

@ACuriousMind what I want is a simple neat formula for div or curl in funny coords

@ACuriousMind and yes it is fucking annoying me

I mean you should be able to "derive" the funny looking div and curl from the diff form right?

0celo7

21:41

the perfect hacking perspective

Slereah

21:59

too bad there's no keys on that keyboard

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