The h Bar

2:34 PM

Who was the genius who came up with staples?

@Slereah Is there any way that "inégalités" means something else than inequality?

$${T^\mu}_\lambda &=& \frac i2 [\bar \gamma^\mu \psi D_\lambda \psi - (\overline{D_\lambda \psi}) \gamma^\mu \psi] - \frac{1}{4\pi} [{F^a}_{\mu\nu} {F_a}^{\mu\nu} - \frac 14 {\delta^\mu}_\lambda {F^a}_{\alpha\beta} {F_a}^{\alpha\beta}]\\
&+& \frac 12 [(D_\lambda \phi)^\dagger D^\mu \phi + (D^\mu\phi)^\dagger D_\lambda \phi - {\delta^\mu} \{ (D_\alpha \phi)^\dagger D^\alpha \phi - \mu^2 \phi^\dagger \phi - \frac{2\lambda}{4!} (\phi^\dagger \phi)^2 \} ]$$

Solve for the boundary condition of the universe

@0celouvskyopoulo7 unlikely

$${T^\mu}_\lambda = \frac i2 [\bar \gamma^\mu \psi D_\lambda \psi - (\overline{D_\lambda \psi}) \gamma^\mu \psi] - \frac{1}{4\pi} [{F^a}_{\mu\nu} {F_a}^{\mu\nu} - \frac 14 {\delta^\mu}_\lambda {F^a}_{\alpha\beta} {F_a}^{\alpha\beta}]\\ + \frac 12 [(D_\lambda \phi)^\dagger D^\mu \phi + (D^\mu\phi)^\dagger D_\lambda \phi - {\delta^\mu} \{ (D_\alpha \phi)^\dagger D^\alpha \phi - \mu^2 \phi^\dagger \phi - \frac{2\lambda}{4!} (\phi^\dagger \phi)^2 \} ]$$

Dang it

Of course the original morse paper is 50 pages long

you rboke it

this paper is oooold

He calls Betti numbers "connectivity numbers" and $\epsilon$ $e$

2:52 PM

to be fair connectivity number is a much better name

if longer

Sorry Enrico Betti

I think he calls them $R_n$ as well.

@JaimeGallego Yes, 9th Feb is my birthday though I'm not sure why sharing a birthday is a paradox. Unless of course we are both the same fermionic eigenstate of the birthday operator :-)

Birthday problem

In probability theory, the birthday problem or birthday paradox concerns the probability that, in a set of n {\displaystyle n} randomly chosen people, some pair of them will have the same birthday. By the pigeonhole principle, the probability reaches 100% when the number of people reaches 367 (since there are only 366 possible birthdays, including February 29). However, 99.9% probability is reached with just 70 people, and 50% probability with 23 people. These conclusions are based on the assumption that each day of the year (except February 29) is equally...

Yeah, it's often called "birthday paradox" but it is a bit of a misnomer :-)

3:01 PM

@Slereah What does cherche mean in this context?

Search does not make sense.

The global theorem we were looking for is obvious in the case where etc etc

Aha

Ah, I think I found the theorem I was looking for!

"If $M$ is a Hausdorff smooth manifold, we have a canonical $C^\infty(M)$-linear isomorphism $$\tau : \mathfrak X (M) \to \text{Der}(M)$$ where $\mathfrak X (M)$ is the $C^\infty(M)$-module of vector fields on $M$"

Apparently you do need the Hausdorff property to prove this

Because you require bump functions

Because you require the lemma that if $f$ vanishes on some open subset, then $D f(q) = 0$ for all $q$

Hm, what's a non-Hausdorff space where that wouldn't be true

you're off the deep end

I'd rather read that than Sobolev spaces

3:09 PM

what's wrong with a sobolev space?

"I think that the theorem relating derivations to vector fields still goes through. As I recall, all one needs is the existence of a single bump function around every point, and you still get this as it's true locally in $R^n$."

@ACuriousMind ahlp

what does "ao. Univ.-Prof. i.R. tit. Univ.-Prof. Dr." mean?

Getting closer to an answer!

I need to use all my bibliography skill

@0celouvskyopoulo7 ao = außerordentlich, i.R. = im Ruhestand, tit. = Titular (Titularprofessor is a Swiss/Austrian title)

The Germans do love their titles

3:22 PM

@ACuriousMind why does he have Univ.-Prof. twice?

perhaps he was twice professor'd

@0celouvskyopoulo7 I'd guess because "ao. Univ-Prof i.R." and "tit. Univ.-Prof." are two distinct titles conferred on him

@ACuriousMind Ah.

the love of formal title is what they share with the Japanese

That is why historically they shared ventures together

Hello mister Einstein

Hi

3:33 PM

So anyway @AlbertEinstein, why did you marry your cousin

That's pretty sketchy

I am tired of people citing Morse 1925

It's a 50 page paper

Can you at least give me a section?

quora.com/…

cat's eye

hey is it true that the first exam held was a physics exam?

Hello @JohnRennie

@AlbertEinstein Hi

3:39 PM

@JohnRennie, how are you?

@AlbertEinstein good thanks, and you?

The Wonderful World of Mathematics Constructing the 120 Cell

@JohnRennie, I am also good

@

Secret

►

@ACuriousMind: I know we are hesitant to delete posts, but this seems an obvious candidate for deletion.

@JohnRennie @rob just got to it

3:50 PM

It says removed for reasons of moderation.

@JohnRennie, A particle is moving in a circular path of radius r with uniform velocity v. The change in velocity when the angular displacement $\theta $ is

Aren't all things removed removed for reasons of moderation

@AlbertEinstein Yes?

@Slereah No, someone could just have deleted their own question for reasons unknown to us and unrelated to moderation

@Slereah This wasn't deleted by the moderators.

3:54 PM

@JohnRennie, it's complete

"removed for reasons of moderation" != "deleted by a moderator" - posts deleted by massed spam/abusive flagging are also shown as "removed for reasons of moderation" even if no moderator ever interacted with the post

@AlbertEinstein Remember that velocity is a vector. In circular motion the magnitude of the velocity vector is constant, but the direction is continuously changing.

rob

@JohnRennie That sort of gibberish is appropriate to flag as "abuse," since it's a waste of time trying to figure out whether there's any meaning to it. Somewhere there is a meta.SE post to that effect.

@AlbertEinstein I would guess that the question is asking you to give the velocity vector as a function of time.

Or as a function of the angle $theta$, but $theta$ is proportional to time anyway.

@JohnRennie, what is the way to approach this problem?

4:01 PM

Why do we need manifolds with corners

Wouldn't a corner be homeomorphic to a boundary

@AlbertEinstein Do you know how to write the position $(x,y)$ as a function of time for circular motion?

@Slereah Yes but not diffeomorphic.

@JohnRennie $e^{i\omega t}$

how many definitions of manifolds are there, anyway

There's manifolds

Manifolds with boudnaries

manifolds with corners

Conifolds

Orbifolds

are there any other

varifolds

@0celouvskyopoulo7 from Albert's question I doubt he is familiar with that notation ...

4:04 PM

Branched surface

In mathematics, a branched surface is a generalization of both surfaces and train tracks. == Definition == A surface is a space that looks topologically (i.e., up to homeomorphism) like ℝ². Consider, however, the space obtained by taking the quotient of two copies A,B of ℝ² under the identification of a closed half-space of each with a closed half-space of the other. This will be a surface except along a single line. Now, pick another copy C of ℝ and glue it and A together along halfspaces so that the singular line of this gluing is transverse in A to the previous singular line. Call this...

"In mathematics, a branched surface is a generalization of both surfaces and train tracks."

what

haha

@JohnRennie

Oh apparently that's a math term

@JohnRennie, no

Are you familiar with unit vectors?

4:05 PM

Branched manifold

In mathematics, a branched manifold is a generalization of a differentiable manifold which may have singularities of very restricted type and admits a well-defined tangent space at each point. A branched n-manifold is covered by n-dimensional "coordinate charts", each of which involves one or several "branches" homeomorphically projecting into the same differentiable n-disk in Rn. Branched manifolds first appeared in the dynamical systems theory, in connection with one-dimensional hyperbolic attractors constructed by Smale and were formalized by R. F. Williams in a series of papers on expanding...

Yeah @JaimeGallego

I think that's what you get if you identify the adjacent points of a non-Hausdorff manifold

Secret

@Albert How would you define the position function for the object?

"In physics and mathematics, supermanifolds are generalizations of the manifold concept based on ideas coming from supersymmetry."

what

4:07 PM

@JaimeGallego, no idea

Well, it's going around in a circle, so trigonometric functions are useful here

@JaimeGallego,.

@AlbertEinstein give me a moment and I'll draw you a diagram ...

Ok

Think: which trigonometric function determines the x position on the unit circle as a function of $\theta$?

4:13 PM

@JaimeGallego, is it cos?

Cosine, exactly. But that is for the unit circle. How would you adjust the function for a circle with radius $r$?

how does one prove that anyway

it's not clear why cosine gives the $x$ position

@JaimeGallego, r cos$\theta $

Yup. That's for the $x$ position, so you multiply it by $\hat{\textbf{i}}$

4:16 PM

@AlbertEinstein If it helps conceptually, imagine some machine called $\hat R$ that acts on a vector $\mathbf V$ and maps it to another vector $\mathbf V$ of the same magnitude. Imagine that it rotates it continuously around the origin in the plane x,y.

$\mathfrak B$

@AlbertEinstein Ah, I see you and the other member of our degenerate pair are already discussing this ...

@0celouvskyopoulo7 what do you mean about proving cos(x) represents the $x$

@JohnRennie, yeah.

@Phase starting from $\cos x=1-x^2/2+\cdots$, derive $\cos \theta$= adjacent/hypotenuse.

4:18 PM

@AlbertEinstein Hopefully the diagram makes it clear what is going on.

what letter is that supposed to be?

I would guess B

That's a $t$

$\mathfrak t$

$\mathfrak{P}$

Capital P.

4:20 PM

@0celouvskyopoulo7 I don't understand, the cos(x) is DEFINED to be opposite/hypotenuse

That's how it was first defined with trigonometry

AFAIK

Perhaps

But that's not a useful definition.

@AlbertEinstein: anyhow, can you now write down the expression for the position $(x,y)$ as a function of time?

@0celouvskyopoulo7 Defining it in terms of the Unit circle? Why not?

$\cos x$ is the unique solution to $f''=-f, f(0)=1, f'(0)=0$.

@JohnRennie, no

4:21 PM

@Phase Hard to do calculus

@AlbertEinstein Well, what is $x$? Hint, it's written on the diagram :-)

@0celouvskyopoulo7 i see your point

@JohnRennie, you mean rcos..

What about defining them all in terms of tan(x)?

wot

4:22 PM

using tan(x) = sin(x)/cos(x)

Well, then you have all the trigonometric functions and you could define them in terms of the unit circle, but then also relate the derivatives etc

Or, well, you could just define it as the real part of $e^{ix}$

@AlbertEinstein: I wonder if it would be a good idea for you to put your question on hold for a bit and read about circular motion. It feels as if you are missing the fundamentals.

Is @AlbertEinstein the same guy who was @LeonardEuler or @RobertFrost?

@Phase How does that help?

Well you're saying it's hard to do Calc without a good definition of cos(x) etc right?

@JohnRennie I agree.

4:26 PM

@SirCumference No one was Leonard Euler and Robert Frost

But Albert Einstein was Leonard Euler.

@ACuriousMind No idea what you mean

@SirCumference Edit history is visible, that doesn't work here :P

@ACuriousMind What?????

@ACuriousMind Your screen is glitching, or possessed

@AlbertEinstein do you know what matrices are?

4:27 PM

Euler was a Jew?

The dates don't line up @ACuriousMind

@ACuriousMind This conspiracy theory doesn't seem to hold up

@phase, do you mean the rectangular arrangement of elements in to rows and columns

@Phase That will complicate things a bunch :)

@0celouvskyopoulo7 is right

When am I not?

4:28 PM

I suppose, it was just the first way I properly learnt circular motion so I think it deserves a mention

@0celouvskyopoulo7 when you say astronomy is useless

@AlbertEinstein yep

;-;

Sep 28 '16 at 1:20, by 0celo7

I was wrong!

I think most things are useless

4:29 PM

@Phase, yeah I know

@ACuriousMind I can be wrong about algebra, that's fine.

Ok, well if you rotate a vector by 90 degrees anti-clockwise, and it was lying along the x axis to begin with, what do you get?

@AlbertEinstein

@0celouvskyopoulo7 I was playing through MGS TPP and I realised something and got confused

Why did Ocelot go from trying to kill big boss to helping him

@Phase he became obsessed with him after the 3rd one

And he was a CIA plant all along anyway

@Albert Still there?

@0celouvskyopoulo7 So him being a sadist was an act too? if so, why was he sadistic in 1 too?

4:43 PM

There was a scene in 3...when you jump off the waterfall, where he doesn't take the shot

@Phase no, he's a sadist

He's a sadist in all of the games

@0celouvskyopoulo7 huh? in TPP he was basically the voice of reason and Kazuhira was the Ocelot

Super Mario Brothers Super Show Credits - Do the Mario!

►

5:05 PM

@Phase Kaz is a little crazy

But Ocelot is still a sadist

He does all of the torturing

@0celouvskyopoulo7 eh.. I guess, I just didn't get any vibes like that from him, his VA in 3 was much better

So now the question is

If for non-Hausdorff manifold, there's no canonical mapping from vectors as curve tangents and vectors as derivatives

which one is best

I would guess curves

isn't seeing vectors as derivatives like better though, because at least you can do things with derivatives (like apply them to functions or something?) but what can you do with seeing vectors as curve tangents?

Non-Hausdorff manifolds all have covers by Hausdorff submanifolds on which you can define those derivatives, if necessary

@JoshuaLin you can visualize

5:19 PM

why are non-hausdorff manifolds interesting in the first place?

there's a few proposals to have spacetime having such a structure

daaang
spacetime would have to be pretty weird to be non-hausdorff right? wouldn't that mean that like... there are distinct events in spacetime that.. can't be separated..

Usually it's some attempt to have quantum mechanics and measurement be explained topologically

The different measurements corresponding to a "branching"

holy crap that's pretty neat

Well it's a nice idea, but I don't think anyone ever did much with it

They discuss the topology a lot, but I've never seen anyone actually try to do a simple QM model of it

Which would probably be fairly awful

5:26 PM

I have an idea to solve the measurement issue of QM, let's add "Dark Observers" I'm sure that'll work

having not enough observers is not the problem in QM

It's having too many

If there were only one observers things would be much simpler

The dark observers act as negative observers

Stop making me legitimise this satire

Could any experiment even decide it spacetime is hausdorff?

Probably not

Actually, I'm not even sure why non-Hausdorff would be shocking

5:33 PM

especially if the branching occurs on the light cone

The topology is not observable

and Hausdorff is a technical condition on the open sets

It might have some relevance to causality, though

If you have CTCs, you could conceivably go back to before a branching point and take a different branch

That may be true

although thinking about it, since you're always going inside the light cone, you wouldn't really "go back" to before the branching point, you'd just go to another branch

6:29 PM

@Jim Well, that time you answered his post and then continued the conversation.

Admittedly this guy (or at least this account) is new to the site, so you weren't to know.

I'm sure we all have our candidates for the "most annoyingly wrong user on the site" award, and this guy isn't even in my top ten yet.

my top ten is 0celo7 ten times

@AccidentalFourierTransform Hawh. He's annoying, but not annoyingly wrong.

@dmckee What?

fair enough

still in my top 10

What have I done to either of you

6:37 PM

My number one candidate like to to read exactly what he already believes from whatever he's told. You can cite experimental evidence that his notions are utterly broken and he'll pick out one phrase, generalize from that and claim you've just supported him.

And I'm not talking about J.D.

@0celouvskyopoulo7 Perhaps I should have put a smiley after than line.

I never remember until after I've hit "enter", however.

Hmm

Please accept by assurance that I hold you in high regard.

aww

shanks

Can someone help me with a question?

Maybe. What's up?

6:44 PM

How to calculate the resistance of this circuit?

When the switch is closed

when the switch is closed, the inner resistors are...

Treat the leg with the closed switch as having a very small but non-zero resistance (say $0.001 \,\mathrm{\Omega}$).

Then the parallel resistance will be similarly very small, and you can work from there.

I am not able to

Or just recognize that the equivalent resistance of a parallel set is always smaller than the smallest branch.

Arent 6 ohm and 3 ohm in parallel

and 10 ohm is in series

right?

6:47 PM

Are you unsure about how to do compound crciuts at all?

@dmckee rule out that possibilty

@dmckee I know how to calculate the equivalent resistance of "simple" grade 10 level circuits.

what is the resistance of this part?

you have three parallel resistors

You do compound circuits by fist doing the simple sets and then re-drawing the circuit with a single resistor where a set used to be.

@dmckee I am not able to read what you have written after resistance - It appears in the form of a code

A: Any chance of MathJax in chat?

19

As a workaround while this request is pending, there exist several client-side workarounds that can be used to enable LaTeX rendering in chat, including: ChatJax, a set of bookmarklets by robjohn to enable dynamic MathJax support in chat. Commonly used in the Mathematics chat room. An altern...

6:49 PM

@AccidentalFourierTransform how three?

@AccidentalFourierTransform It's 2 ohm considering 6 ohm and 3 ohm resistors

and one with 0 ohm :-)

@Abcd Or very close to zero Ohms when the switch is closed.

@AccidentalFourierTransform okay thanks

then?

@AccidentalFourierTransform But he can't proceed naively with the parallel formula if he uses exactly zero—all the math teachers will cry—so he shuld use a very small resistance instead.

but I aint no math teacher!

6:52 PM

@dmckee No need for that please

lol

ok, so you have three resistors, $R_1=6,R_2=3,R_3=\epsilon$

what is R_eq?

^ That.