/sci/ - Science & Math

[Anonymous No.2729201]

what gives mathfags?

[Anonymous No.2729234]

>mfw i am learning that shit tomorrow in calc 3

[Anonymous No.2729319]

Would you care to write better, and actually say what everything in your notation is?
I really don't understand your notation.

[Anonymous No.2729330]

>>2729319
fuck greek symbols, use cyrillic for your variables!

[Anonymous No.2729334]

[Anonymous No.2729338]

>>2729330
they are not symbols they are letters from our fucking alphabet

[Anonymous No.2729347]

>>2729338
>letter
>symbol
>glyph
>character

same fucking shit

[Anonymous No.2729362]

>MFW you use <span class="math">\int[/spoiler] instead of <span class="math">\oint[/spoiler] for closed integrals

[Anonymous No.2729379]

>>2729362

What makes closed integrals special? I've seen some examples of them but I don't get what's done differently. Does it have something to do with line integrals?

[Anonymous No.2729411]

>>2729379
They ARE line integrals, but closed.

[Anonymous No.2731048]

OP here, was sleeping.

>>2729319

I am using physicists' differential calc notation. Let's restrict to Reimannian manifolds, then the first equation says that the integral of the divergence of a vector field over the whole manifold is zero (Stokes Thm); the second equation says for any scalar function f on the manifold we can find a vector field v satisfying div(v)=f. The third equation says, for all f, the integral of f over the whole manifold is zero.

>>2729362

btw, I see what goes wrong in 1 dimension (S1), namely that you can't solve eqn 3 for all periodic functions to get another periodic function (eg if f=sin(theta)^2, and dv/dtheta = f, then v is not periodic, so it isn't a vector field on the 1-sphere.

My notation is the usual one for differential geometry, the M under the integral sign means integrate over the whole manifold

[Anonymous No.2731062]

OP here, was sleeping.

>>2729319

I am using physicists' differential geometry notation. Let's restrict to Reimannian manifolds, then the first equation says that the integral of the divergence of a vector field over the whole manifold is zero (Stokes' Theorem for compact without boundary manifolds); the second equation says for any scalar function f on the manifold we can find a vector field v satisfying div(v)=f. The third equation says, for all f, the integral of f over the whole manifold is zero.

>>2729362

My notation is the usual one for differential geometry, the M under the integral sign means integrate over the whole manifold


btw, I see what goes wrong in 1 dimension (S1), namely that you can't solve equationn 3 for all periodic functions to get another periodic function (eg if f=sin(theta)^2, and dv/dtheta = f, then v is not periodic, so it isn't a vector field on the 1-sphere.) But I'm not sure what the general argument for any dimension and any compact boundary-less manifold is.

[Anonymous No.2731141]

In general, not writing the measures, i.e. dx, dV and so on is a bad idea.
Before I say anything,
are you familiar with the notion of differential forms?

[Anonymous No.2731145]

>>2731141

I was using the notation of a fixed volume form which was omitted (i've seen this done a lot). Yes, technically f should be a n-form and v should be an n-1 form

[Anonymous No.2731165]

>>2731145
I don't have a definitive global answer but the implication

okay then first of all, your statement can be written

dv=f·dVol

can be solved for all f such that v is a smooth vector field is a bit over the top. Let f be constant for example, which would violate inf_M f=0, then v would be a vector field with only one entry which is a coordinate function which is certainly nonperiodic

[Anonymous No.2731171]

-"okay then first of all, your statement can be written"

[Josef !!nUf2NflSAyw No.2731185]

Stokes doesn't say anything with covariant derivatives.

[Anonymous No.2731193]

mhm, you got me confused about the conditions for stokes theorem now..

let T^2 be the flat torus of length 1cm·1cm, then the metric is g=diag(1,1) and thus the volume form is det(g)dx^dy=dx^dy=d(x·dy)=dVol

clearly int_M d(x·dy)=1 as is has to be since the Torus has an area of 1cm^2.

I conclude that it can't be stated that

Int_M dv = 0

on a compact manifold.

There must be some conditions (maybe they are so that there are just so few possible admissible vector fields)

[Anonymous No.2731197]

>>2731185
ein thread über stokes theorem
muss ja ein gefundenes fressen für dich sein :D
Und, irgendwelche ideen was die conditions in dem fall angeht

und das \del wird wohl ein konventioneller "grad" sein

[Anonymous No.2731209]

I guess to give a general answer we need some kind of "hairy ball theorem" which says admissable (periodic) vector fields are already closed such that dv=0 and
dv=f·dVol
can never be solved

[Josef !!nUf2NflSAyw No.2731232]

>>2731197
Stokes hat kein del oder nabla drin sondern ein d und nix anderes.

[Anonymous No.2731280]

Emma Stone is the best thing about this thread.

MOAR

[Anonymous No.2731286]

>>2731232
ich würde nie auf die idee kommen was anderes als d zu schreiben ;)

>>2731280
I agree

[Josef !!nUf2NflSAyw No.2731319]

... überhaupt das Nabla "Del" zu nennen geht mir echt aufn Sack. Scheiß Erstimathe.

[Anonymous No.2731365]

>>2731319
mhm, ich mag die überschneidung auch nicht, aber ich würde nicht sagen, dass es die kovariante Ableitung zwingen mehr verdient hat

[Josef !!nUf2NflSAyw No.2731446]

>>2731365
Naja, einfach nur "Ableitung einer Form" reicht ja nicht, da gibt's ja äußere, Lie, kovariante, partielle, ...

[Anonymous No.2731459]

>>2731446
Das sind alles Lie Ableitungen :)

[Josef !!nUf2NflSAyw No.2731460]

Oh, and I might want to add that OP's question is wrong. Ignoring the retarded notation, the integral over an exact form f is zero on a manifold with boundary.

[Anonymous No.2731464]

Pawn to E5

[Anonymous No.2731513]

OP here

talked to a math guy. On a compact manifold without boundary, one cannot always find a v satisfying D.v=f, as I certainly found on S1.

[Anonymous No.2731547]

wow, so much notation fail in here...since when does one use a nabla for divergence?! what the fuck is this shit?!! yesimad

[Anonymous No.2731562]

So much foreignfag ITT.

Still, the Emma Stone pics make up for it.

[Anonymous No.2731570]

>>2731562
there's just two germanfags in here, one of them spouting major nonsense...

[Josef !!nUf2NflSAyw No.2731644]

>>2731570
Only one Germanfag.

[Anonymous No.2731657]

>>2731644
and what's with your friend? Austrian or something?

[Anonymous No.2731712]

>>2731547

in physicists' differential geometry notation, nabla usually denotes the metric compatible derivative operator. why's everyone so bitchy about notation, it don'r matter.

[Josef !!nUf2NflSAyw No.2731773]

>>2731712 in physicists' differential geometry notation
There's still no measure in your integral.

[Anonymous No.2731790]

>>2731712
>metric compatible derivative operator
uh, why isn't the Levi-Civita connection (i.e. covariant derivative) "metric compatible"? I don't quite see your point here...

[Anonymous No.2731796]

>>2731657

[Anonymous No.2731807]

>>2731712
>why's everyone so bitchy about notation
Because you're using non-standard notation without defining it properly in your OP. Either you use something that can be found in all mainstream math literature or you carefully define it for us. But blindly assuming that everyone is gonna use your retarded (and for fucks sake it's retarded, where's your measure?!) notation is retarded. Retarded.

[Anonymous No.2731823]

∫ ∂ fag_in_here > ∞

(I ran out of Emma)

[Anonymous No.2731826]

>>2731773

I'm using a common shorthand. For example the integrand of the left hand side of the first equation is really shorthand for (nabla_[a1) (V^b)_|b|a2...an] where n is the dimension of the manifold, [ ] denotes anti-symmetrization, | | denotes that there is no anti-symmetrization over the contracted indices, and I'm using abstract index notation. So we're integrating an n-form over an n-dimensional manifold, which is perfectly well-defined.

[Anonymous No.2731831]

>>2731790

it is, that's the derivative I'm using

[Anonymous No.2731845]

>>2731807

sorry, it's standard notation in physics texts. I guess it's not in math circles. anyway, i wouldn't think there are that many ways to misinterperet what I wrote.

[Josef !!nUf2NflSAyw No.2731850]

>>2731826 (nabla_[a1) (V^b)_|b|a2...an]
Get out.

[Anonymous No.2731869]

>>2731831
wtf are you on mate? first you say it's Stokes' theorem and implying your nabla stands for divergence and suddenly is the Levi-Civita connection?! Stokes' theorem has nothing to do with the Levi-Civita connection!!

[Anonymous No.2731891]

>>2731869
Stokes' theorem: The theorem that says that the integral of df over a region is equal to the integral of f over the boundary, where f is a differential form and d is the exterior derivative.

This can be written in terms of any derivative operator, including the Levi-Civita one.

[Anonymous No.2731896]

>>2731869
http://en.wikipedia.org/wiki/Stokes%27_theorem

[Anonymous No.2731908]

>>2731891

Actually, Stokes theorem is much more fundamental, though the special case you´ve mentioned is often seen in physics, especially electrodynamics.

[Anonymous No.2731984]

>>2731908

I think the version I wrote, in terms of differential forms on smooth manifolds, is pretty damn general.

[Anonymous No.2732032]

>>2731891
>>2731896
what?! no for fuck' sake. if the nabla in OP's pic denotes the covariant derivative then Stokes' theorem cannot be applied here. period.

[Anonymous No.2732042]

>>2731984

lrn2references

[Anonymous No.2732119]

can someone tell an englishfag who's only an amateur differential geometer what just happened itt

[Anonymous No.2732145]

>>2732119
physicist uses retarded notation without defining it for anyone and gets confused himself trying to apply Stokes' theorem where it isn't applicable. Also how is one an "amateur differential geometer"? You just reading books for yourself or are you in a introductory lecture to differential geometry and thus consider yourself a novice ergo "amateur"?

[Anonymous No.2732682]

>>2732032
>>2732145

fuck you. The generalized Stokes Theorem includes the divergence theorem ("Gauss' Theorem") as a special case, and as long as the levi civita volume element (sqrt(g)dx1dx2...dxn) is assumed (what the fuck other measure do you think it would be, from the context), this theorem says that the first equation in the OP picture holds on a compact manifold without boundary. Suck it.

[Anonymous No.2733583]

>>2732682
>levi civita volume element
nigawut

[Anonymous No.2733753]

>>2733583

http://en.wikipedia.org/wiki/Levi-Civita_symbol#Ordinary_tensor

[Anonymous No.2733790]

>>2733583
I wrote what the fuck I meant, sqrt(g)dx1dx2...dxn. If you have a fetish for precise definitions, read appendix B of Wald's relativity book; equation B.2.9 defines the volume element I was talking about, and the rest of the appendix explains integration on manifolds. Stop being such a pussywad.

[Anonymous No.2733823]

lol at OP and thread

it is that "for all f" that is the problem/troll

when the picture is just about functions are del v

[Anonymous No.2733842]

>>2733823
many red faces

[Anonymous No.2733904]

>>2733823

yes, we already figured out what the problem was, see >>2731513 for example. Thanks for your fucking input.

[Anonymous No.2733938]

(OP btw) so i ask i question whose answer wasn't obvious to me, eventually figure out the answer thanks to help from some helpful people, and everyone else bitches that I didn't know the answer to begin with. If I knew what was wrong with my question in the fist place, I wouldn't have asked it. Guess I couldn't have expected any different; it's 4chan after all. gonna go fap on /s/ now.

[Anonymous No.2733957]

ITT: OP doesn't know what the Levi-Civita connection is and thinks it's divergence. Then cries when everyone points out what an uninformed moron he is.

[Anonymous No.2734066]

>>2733957

lol. the contraction of the covariant derivative of a vector field is the generalization of divergence to curved spaces, and in fact it is commonly called the divergence of the vector field (at least in physics circles).

[Anonymous No.2734078]

[Anonymous No.2734095]

[Anonymous No.2734108]