The h Bar

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

00:03

:P

Slereah

So I'm watching the Flash series

It's painful to see Holliwood trying to do science

dmckee

@Danu Mmmm ... yes. I got the wrong name. I mean Goodstein's "States of Matter"

I'll see if I can run down the quote.

Ah. Posted in this bar before..

Physics Meta

01:00

0

Q: The practice of answering a question in its comments area

I've noticed that some members provide a perfectly good answer to a perfectly good question in the comments area under the question rather than posting their response as a standalone answer. As this practice often occurs among members of high standing in the community, it seems to be acceptable...

discussion

Slereah

I wonder what fields are the least represented in mad science

Metrology perhaps

HE HAS MEASURED WHAT MAN WASN'T MEANT TO MEASURE

3

yuggib

01:38

aeroports are boring

flights are even more boring

at least on long distance you can watch movies

yuggib

02:50

america-based regulars are rather silent today...they should still be awake

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

@yuggib o/

0celo7

@yuggib I'm on a trip

yuggib

\o

you came as summoned...good

;-P

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

:D

yuggib

@0celo7 me too...a long one

and boring

5 hours into it, and I've not yet completed a fourth of it

T__T

I have to board my plane...see ya in roughly 12 hours

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

03:32

cya later

vzn

04:42

@Slereah by "mad science" are you referring to hollywood depictions?

Danu

05:57

@DanielSank sleep

Danu

06:18

0

A: Gauge redundancies and global symmetries

Answer posted by Lubos Motl in the comments; I reproduce most of it here. This answer was posted in order to remove this question from the "unanswered" list. Some (sketches of) answers to your questions, one by one: Physical states have to be invariant under gauge symmetries, so all of them ar...

^let's get this off the "unanswered" list.

user116211

08:01

17

Q: Gauge redundancies and global symmetries

It is often said that local (gauge) transformation is only redundancy of description of spin one massless particles, to make the number degrees of freedom from three to two. It is often said that these are not really symmetries because it means that there are only apparently different points in c...

quantum-field-theory symmetry gauge-theory conservation-laws noethers-theorem

user116211

Is it 'too broad'?

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

09:27

@vzn mad science needs a mad scientist :P

Loong

10:26

@JohnRennie Braking a train from 300 km/h to zero in about 2.8 km braking distance:

Schnellbremsung aus 300 km/h ICE 3

►

You can also see the three different braking systems (electric motors, eddy current brakes, and disc brakes).

Danu

11:42

@MAFIA36790 it's asking 8 questions.

user116211

Ah! got the answer, @Danu.

0celo7

12:48

@ACuriousMind bist du around?

ACuriousMind

@0celo7 sort of

Secret

https://www.amazon.com/General-Relativity-Introduction-Theory-Gravitational/dp/0521379415

Is this the stephani book that slereah uses?

0celo7

No

Secret

what is the name of that stephani book?

0celo7

@ACuriousMind I'm confused by something on page 56 of Milnor. What does he mean by "locally a diffeomorphism" and why must therefore the segments of the graph end on the boundary?

Actually I'm also not sure why that composition exists on the entirety of the intervals

Maybe it doesn't

I just feel like it should only be defined on some open set of I

Secret

13:01

http://www.goodreads.com/author/show/250289.Hans_Stephani

there are at least 4 books by stepani, which one corresponds to

Aug 16 '15 at 18:10, by Slereah

It's in the Stephani book

@0celo7 ?

ACuriousMind

@0celo7 It doesn't, it exists only on (the respective preimages of) $f(I)\cap g(J)$.

and you can omit the "locally" there, it's just a diffeomorphism.

0celo7

@ACuriousMind but why does it have to extend to the boundary

And actually why is it made of at most two segments, why can't it be many smaller segments with that slope

@Secret check my GR book list

Secret

ok

book found, thanks

ACuriousMind

@0celo7 If we write $I=[a,b]$, $J = [c,d]$, then "extend to the boundary" means that at least two of the four points $f(a),f(b),g(c),g(d)$ lie in $f(I)\cap g(J)$.

@0celo7 That he already explains - because $g^{-1}\circ f$ is one-to-one; $\Gamma$ having more than two components would mean it hits one of the boundaries more than once, contradicting the bijectiveness of $g^{-1}\circ f$.

0celo7

13:19

@ACuriousMind but why

@ACuriousMind uh, no?

In the picture, slice one interval in half

Why isn't this allowed?

ACuriousMind

@0celo7 Okay, let's do this in small steps:

1. $\Gamma$ is diffeomorphic to $f(I)\cap g(J)$, yes?

0celo7

Sure

(Taking a shower, brb)

ACuriousMind

2. Let $X:= f(I)\cap g(J)$. $g^{-1}\circ f : f^{-1}(X)\to g^{-1}(X)$ is a diffeomorphism, mapping a relatively open subset $K:=f^{-1}(X)\subset I$ to a relatively open subset $L:=g^{-1}(X)\subset J$.

ACuriousMind

13:43

Ugh, I see what the argument is, but I have a hard time verbalizing it any more precise than Milnor did

Ah: If one connected component of $\Gamma$ ended somewhere before the boundary, that would mean there is an $x\in I^o$ such that $x\in K$ and $x+\epsilon\neq K$ for all $\epsilon > 0$.

0celo7

What is $I^o$ @ACuriousMind

ACuriousMind

But that would mean that one connected component of $K$ looked like $(...,x]$ for $x\in I^o$, so it could not be relatively open in $I$

@0celo7 interior of $I$ (relative to $\mathbb{R}$, I just mean to say it's a point not in the boundary)

So all connected components of $\Gamma$ have to end on the boundary, and then bijectivity of $g^{-1}\circ f$ means that there are at most two.

Not sure if that's more understandable than what Milnor wrote :P

0celo7

@ACuriousMind Oh, you mean $I^\circ$

ACuriousMind

@0celo7 ::squints very hard:: Yeah.

0celo7

@ACuriousMind to me $I^o$ means the identity connected component

I've been reading too much about Lie groups...

@ACuriousMind I'm not sure what this means

Why is "relatively open subset" important

Milnor says that too

@ACuriousMind I still don't get it

ACuriousMind

13:59

@0celo7 We don't know whether $I$ is closed or open, so we don't know whether there are relatively open subsets of the form $[a,b)$ where $a\in \partial I$.

But we do know that no subset of that form can be open if $a\in I^\circ$, regardless of the form of $I$

@0celo7 Why all connected components have to end on the boundary, or why that means there are at most two?

0celo7

@ACuriousMind the first part

@ACuriousMind hmm, what?

oh, relatively open wrt. $\Bbb R$

@ACuriousMind ok I understand this

@ACuriousMind what

I don't understand what the local diff property of $g^{-1}\circ f$ has to do with this

@ACuriousMind and what does this have to do with anything?

sorry for being dumb :/

ACuriousMind

As I said, I have trouble verbalizing the argument any better than Milnor did :P

0celo7

Milnor doesn't verbalize it well at all

I have no clue what he's talking about

ACuriousMind

@0celo7 I do not dispute that :D

0celo7

@ACuriousMind well you're able to get it

somehow

ACuriousMind

14:07

Let me try on last time. Each connected component of $\Gamma$ must map to an open subset of $I$ as well as $J$.

0celo7

Yes.

(maybe)

Ahhhh, why?

ACuriousMind

The connected components of $\Gamma$ look, a priori, like $\{ (x,y) \mid x\in |i_0,i_1 |\subset I, y\in |j_0,j_1|\subset J\}$ where by $|i_0,i_1|$ I mean that we are as of yet agnostic whether these are open or closed intervals.

@0celo7 Because $\Gamma$ is diffeomorphic to $f(I) \cap g(J)$, which as an open usbset of $M$ maps to open subsets $I$ and $J$ under $f^{-1}$ and $g^{-1}$, respectively.

Now you must convince yourself that a closed subset of $I\times J$ consisting of straight lines must have the $\lvert i_0,i_1\rvert$ up there as $[i_0,i_1]$ for its connected component unless $i_0$ or $i_1$ is on the boundary (then $($ instead of $[$ i permitted if $I$ resp. $J$ are open intervals).

0celo7

@ACuriousMind I've never heard agnostic used that way.

@ACuriousMind I don't know how to convince myself of that...

ACuriousMind

And I fucked up, we must add $f(x)=g(y)$ to the condition ont he connected components up there

@0celo7 I have convinced myself using the sequential characterization of closedness/openness (as well as continuity of $f$ and $g$)

0celo7

Sequential characterization?

like $A$ is closed iff it contains its limit points?

ACuriousMind

14:14

Yes, that one

0celo7

dunno what the sequential characterization for openness is

ACuriousMind

That the complement is closed ;P

0celo7

lol

ok I'll think about it

Slereah

Does the fundamental theorem of calculus hold for path integrals?

Does $$\int \frac{\delta f[\phi]}{\delta \phi} d\mu[\phi] = f[\phi_b] - f[\phi_a]$$

ACuriousMind

@Slereah That doesn't make sense. $\delta f[\phi]/(\delta\phi)$ is not a functional itself, it is a function.

You can't integrate it against $\mathrm{d}\mu[\phi]$.

0celo7

14:24

@Slereah Pretty sure that's used to derive Ward-Takahashi

Secret

sciencedirect.com/science/article/pii/S0007449704000107

0celo7

or something similar at least

Secret

Actually, given some contnuity assumptions, the fundemental theorem of calculus generalise to path integrals

(Theorem 7 in this open access paper)

ACuriousMind

@0celo7 Not in the derivation I know

I've never seen any physicist write boundary conditions for a path integral over fields at all :P

Slereah

Well then obviously you haven't read Weinberg

Or Rovelli

ACuriousMind

14:28

That's largely correct

I have read a bit of Weinberg until I decided I don't have the patience to decode that notation of his :P

Slereah

yeah Weinberg is a bit of a slog to go through

And obviously you haven't read the question on SE where I asked about it

A kind soul answered it

(It was me)

ACuriousMind

@Slereah I know that you've been going on about boundaries of path integrals for a while ("any physicist...at all" was hyperbole :P)

Slereah

I just want to compute quantities with it, is it so much to ask

and apparently you totally can for the free field

ACuriousMind

But the fact remains that the functional derivative is itself not a functional, you cannot integrate it against a measure on functionals, so the l.h.s. of your proposed theorem is really not defined

Slereah

I suppose so yeah

Wondering if there is some kind of equivalent, though

Secret

14:35

Actually, I don't know what is time slicing approximation, thus I am not sure if my answer is actually responding to the correct context

vzn

14:54

anyons/ qm computing error correction decoduku/ via reddit

0celo7

@ACuriousMind no I mean like

$$\int\frac{\delta S}{\delta X}\,DX=0$$

I've seen that

guaranteed

@ACuriousMind can you please explain this

this argument is just too confusing

Slereah

@0celo7 : That's in the proof for the Dyson equation

ACuriousMind

@0celo7 Ah, that does...kind of appear. What you actually derive is $\int \int \left(\frac{\delta S}{\delta \phi}(y)\Delta\phi(y)\right)\mathrm{d}^4 y \mathrm{e}^{\mathrm{i}S[\phi]} \mathcal{D}\phi$ and when you set $\Delta \phi(y) = \delta(x-y)$ you get Dyson-Schwinger

0celo7

I don't know what that is @ACuriousMind

"Dyson-Schwinger"

Never heard of it

Slereah

That is only the most important QFT equation

0celo7

15:09

is it in Weinberg?

Slereah

Yes

ACuriousMind

But Ward-Takahashi wasn't far off - choosing $\Delta\phi$ to be the change under a symmetry, you get the W-T identity for that symmetry

Slereah

yeah

0celo7

@ACuriousMind ok I don't care about physics, can you please elaborate on the Milnor proof :(

ACuriousMind

@0celo7 I said "convince yourself" because it's another argument that I see but don't really know how to clearly explain in words

But I'll try

0celo7

15:11

I don't see it...

Secret

After cross checking this with acuriousmind's, 0celo7 and the paper I found, in order to have some form of "fundemental theorem of calculus" defined, you need a functional in the integrand. The expression you showed above does not have a functional, thus Theorem 7 cannot be applied

(All the above based on (hopefully correctly understood) analysing these sources...)

0celo7

@ACuriousMind could you please sketch this?

draw a picture please :3

ACuriousMind

Since $\Gamma$ is closed as a subset of $I\times J$, we need that for every pair of sequences $x_n\to x,y_n\to y$ with $x_n,x\in I, y_n,y\in J$ and $f(x_n) = g(y_n)$ that $(x,y)\in I\times J$. If one of the intervals $\lvert i_0,i_1\rvert,\lvert j_0,j_1\rvert$ is open, say $(i_0,i_1]$ and $i_0$ is not in the boundary of $I$, then there is a sequence $x_n\to i_0$ from which we can form a sequence $(x_n,g^{-1}(f(x_n))$ that is in $\Gamma$ but whose limit point is not.

...that second $I\times J$ should've been a $\Gamma$.

0celo7

I understand the first sentence.

(the definition of closedness :P)

Just to check, it's closed because it's the preimage of $0$ under $(x,y)\mapsto f(x)-g(y)$, right?

ACuriousMind

Yes

0celo7

15:25

It took me a long time to realize that's why graphs are closed

ACuriousMind

@0celo7 Let me perhaps rephrase the second sentence then: If $i_0$ and $g^{-1}(f(i_0))$ both lie in the interiors of their respective intervals, then for $x_n\to i_0$ with $i_1 > x_n > i_0$, $(x_n,g^{-1}(f(i_0)))$ is a sequence in $\Gamma$ whose limit point is in $I\times J$ but not in $\Gamma$, so $\Gamma$ would not be closed.

0celo7

@ACuriousMind I did not read the second sentence yet :P

I was simply reporting my progress.

@ACuriousMind Mmm, this might make sense.

what intervals though

ACuriousMind

@0celo7 By "their respective intervals", I mean $I$ respectively $J$.

0celo7

wait wait what does this have to do with the boundary of $I\times J$

I really need a picture :<

ACuriousMind

@0celo7 Currently I'm just discussing how the connected components of $\Gamma$ must look. Right now I'm essentially saying the components not ending on a boundary must come from closed intervals $[i_0,i_1],[j_0,j_1]$ in $I$ resp. $J$.

0celo7

15:36

Good lord I don't know what's going on

What question are you answering

ACuriousMind

Explaining what I think is Milnors argument

This is one of the steps

0celo7

Which step??

ACuriousMind

More precisely, it's the first of two steps to explain why the components of $\Gamma$ must end on the boundary.

The next will be to explain why $\Gamma$ can't actually have components coming from such a closed interval.

Which is hopefully simpler as follows: This would mean that a connected component of $f(I)\cap g(J)$ is closed in $M$ - which cannot be since all connected components of that are open in $M$ and $M$ has no clopen subsets since it is connected.

0celo7

@ACuriousMind I still don't understand what step you proved

What exactly did you show

ACuriousMind

@0celo7 That all connected components of $\Gamma$ are of the form $\{(x,y)\mid x\in [i_0,i_1]\subset I, y\in [j_0,j_1]\subset J, f(x) = g(y)\}$ unless $i_0,i_1,j_0,j_1$ are on the boundary of $I,J$ in which case the intervals can be open if $I,J$ are open there.

And then I gave an argument that such components can't actually exist, so all components must belong to the exceptional case where at least one of $i_0,j_0$ and $i_1,j_1$ is on the boundary.

(Then what's left is to count how many components you can actually get by this without getting two $y$ with $f(x) = g(y_1) = g(y_2)$, and that's two components.)

0celo7

16:03

@ACuriousMind I'm sorry, I have to look at this tomorrow

And probably again Monday

And Tuesday morning

And then Tuesday afternoon tell my prof I don't get it

Secret

Possibly a huge chunk of info I am missing, including the details about $\Gamma$

therefore this pics is not even coherent

X(

NB do not ask me for a caption of this pics, because I am missing possibly the first half of the problem

thus it probably does not make sense

(Unless you guys, as usual, somehow can connect the dots as yoou are ver good at piecing my brain together by supplying the words I want to say)

Secret

17:16

@loser arxiv.org/abs/1506.05136

Mike Miller

The SW equations on 4-manifolds and the Bogolmony equations on 3-manifolds both have solutions that are called monopoles. What's the physical relationship between the two?

Physics Meta

17:39

0

Q: Starting from an expression of E(V) and P(V), is there a way of obtaining an expression for E(P)?

I have a question about my Physics Stack Exchange post: Starting from an expression of E(V) and P(V), is there a way of obtaining an expression for E(P)? Does anyone have an idea on how to solve this issue? Thanks

discussion

ACuriousMind

@MikeMiller Both are the equations of motion of a physical theory that has a notion of electric and magnetic charge and where the magnetic charge associated to the field $F$ is a topological invariant.

Also, the solutions "look like" the magnetic charge is concentrated at a single spot.

Mike Miller

If $(A,\phi)$ is the input data (one a connection, the other a section), what do we mean by the electric and magnetic charge?

Slereah

How can we determine that the Hilbert space of nonrelativistic QM is $L_2$

But QFT isn't

ACuriousMind

@MikeMiller The electric charge "within" a 2-cycle $C$ is $\int_C F$ and the magnetic charge within a $d-2$-cycle $D$ is $\int_D {\star} F$

Slereah

Is it related to the CCR

Mike Miller

17:43

And when you say they're the equations of motion, do you mean that if I put these equations in temporal gauge on $\Sigma \times \Bbb R$, where $\Sigma$ is of lower dimension, then the equation becomes time-evolution of a system on $\Sigma$?

Certainly this is true of SW, which is the time-evolution eqn of the Chern-Simons-Dirac functional, though I'm not sure if CSD is physically interesting.

Thanks for your patience, by the way.

Slereah

So this episode of the Flash has a physicist talking about time travel

And his blackboard is nuclear physics

How gauche

ACuriousMind

@MikeMiller Actually, they are the equations of motion in two different ways ;) The Bogolmolny equations are indeed as you say something where we've forgotten the temporal part - they are actually the static part of a 't Hooft-Polyakov monopole. I think the SW equations are already the full equation of motion for the massless degrees of freedom of some SYM theory

Mike Miller

I guess I'm not sure what it means for something to be an equation of motion, then, if it's not in temporal gauge on something of the form $\Sigma \times \Bbb R$. What does a solution represent (if it 'solves the equations of motion')? A stationary point of an appropriate functional?

ACuriousMind

@MikeMiller Yes, an equation of motion is the Euler-Lagrange equation for an appropriate action functional

Mike Miller

ehehe, my ignorance shows.

ACuriousMind

17:50

(well...you have areas in physics where not all equations of motion come from an action, but this isn't one of them)

Mike Miller

Anyway, thanks. I guess I have a little homework.

Danu

You never heard of the equations of motion @MikeMiller? :D

Mike Miller

I got a B in physics 1.

Of course, I've heard of them, but my personal equation of motion dictates I usually run in the other direction when I see them.

Danu

Nice

Minimize that action, boy!

ACuriousMind

19:02

@Slereah Well, the point is not that the Hilbert space "is $L^2$. All quantum mechanical Hilbert spaces are abstractly isometrically isomorphic to $L^2$ because they are all assumed separable

What matters is the representation of the algebra of observables, and it is the Stone-von Neumann theorem that tells you that there is only one isomorphism class of representing the CCR for finitely many degrees of freedom - of which the canonical representant is $L^2(\mathbb{R}^n,\mathrm{d}x)$ with $x$ as multiplication and $p$ as differentiation.

For infinitely many degrees of freedom, the SvN theorem fails, and there are uncountably many non-isomorphic representations of the CCR of fields

Danu

Small question: My lecturer said $H^n(T^n,T^n\setminus\text{pt})\cong H^n(I^n,\partial I^n)$ due to excision.

What do I excise?!

(first note that $\partial I^n$ is a defo. retract of $I^n\setminus\text{pt}$)

Makes no sense to me... $T^n$ is not a subspace of $I^n$...

ACuriousMind

$I^n$ the n-cube and $T^n$ the n-torus, or what are these things?

Danu

Yes

Mike Miller

Pick a small ball around the point and excise its complement. You get that the first thing has the same relative homology as $H^n(B^n, B^n - 0)$.

Now use the relative long exact sequence to show that $(B^n, \partial B^n) \to (B^n, B^n - 0)$ induces an isomorphism in (co)homology.

Danu

Sure

Thanks.

I was thinking in terms of $T^n$ arising from $I^n$, while I should've been thinking the other way around.

Mike Miller

19:11

Right, that isomorphism has nothing to do with the dimension of the cohomology group or the manifold used as input.

Any manifold has $H_n(M, M - p) \cong H_n(B^n, B^n-0) \cong H_n(S^n)$.

Danu

@MikeMiller Just defo retract is fine right

@MikeMiller Yes, I know that.

I just failed to remember it :)

Mike Miller

It's - it's in the book

Danu

Yes, I know that.

Mike Miller

@Danu Be careful. Can you say precisely the argument you have in mind? I can think of two ways it can be wrong.

Danu

@MikeMiller $B^n\setminus 0$ deformation retracts onto $S^{n-1}$

Mike Miller

19:14

So why are the relative homologies $H_n(B^n, B^n - 0)$ and $H_n(B^n, S^{n-1})$ the same? Give me some axiomatics or something.

Danu

@MikeMiller The deformation retraction induces a chain homotopy equivalence, no?

(also, if you're referring to axioms for homology, we didn't do those)

Mike Miller

Yes, between $C_*(B^n - 0)$ and $C_*(S^{n-1})$. Do you see why you're not answering my question?

Danu

Functoriality in the pair, something something?

Functorial dependence of relative homology on the pair, no?

Mike Miller

I'm unsatisfied, since I docked points on a similar argument recently. But perhaps the physicists are going to grow tired of our garbage soon.

Danu

@MikeMiller ?? But really though, isn't functoriality enough?

Mike Miller

19:19

Yes, there is indeed an induced map $H_k(B^n, S^{n-1}) \to H_k(B^n, B^n - 0)$. Why's it an isomorphism?

Danu

Use the homotopy inverse?

Mike Miller

Now we're at the crucial problem. The homotopy inverse of what?

Danu

Of the map of pairs, right?

So you need to show that the prism operator descends to a prism operator on pairs

(which we did in class)

Mike Miller

Ah, and the jig is up. What are the maps of pairs $(B^n, B^n - 0) \to (B^n, S^{n-1})$?

Danu

It's up... for me? :(

Mike Miller

19:23

The point is, if $f(B^n - 0) \subset S^{n-1}$, then $f(0) \in S^{n-1}$ by continuity. You cannot possibly write down a homotopy inverse of that map; those two pairs are not homotopy equivalent.

Danu

@MikeMiller Now comes the part where you enlighten me? :D

My prof. completely glossed over this, and I never thought about it.

Mike Miller

That was the enlightenment. Your argument does not work, because there is no good map of pairs in the opposite direction. Indeed use what I said above to prove that any map $(B^n, B^n - 0) \to (B^n, S^{n-1})$ induces zero on $H_n$.

However, if you have a map $(A,B) \to (X,Y)$ that induces an isomorphism $H_k(A) \to H_k(X)$ and $H_k(B) \to H_k(Y)$, (or on homotopy groups etc) it induces an isomorphism on homology of the pair $(A,B) \to (X,Y)$, by the relative LES and the five lemma. That's all you needed.

Danu

The relative LES for what pair?

Mike Miller

Both of them. The LES is functorial.

You get a diagram with two infinite rows.

Danu

Ah, a ladder diagram,

Of course, I should've known when you said 5 lemma.

Mike Miller

19:28

Uncouth.

Danu

I love the five lemma. You don't?

ACuriousMind

@MikeMiller Wait a moment, how can you first say that any map $(B^n,B^n-0)\to(B^n,S^{n-1})$ induces zero and then tell us that there are maps that induce isomorphisms on homology? oO

Danu

I really loved proving it.

@ACuriousMind Induces zero? What?

The 5 lemma thing obviously works.

Mike Miller

@ACuriousMind It's a map in the other direction :)

ACuriousMind

@MikeMiller Ahhhh

Mike Miller

19:30

It really upset me to realize this when I was grading that homework.

Danu

So... You also didn't know this before?

ACuriousMind

@Danu What "What?" - Mike said that "any map $(B^n,B^n-0)\to(B^n,S^{n-1})$ induces zero on homology", and this confused me.

Danu

That'd make me feel less terrible about myself.

Mike Miller

The map $(B^n, S^{n-1}) \to (B^n, B^n - 0)$ induces an isomorphism on all homotopy groups but is not a homotopy equivalence; relative Whitehead doesn't apply to the latter, since it's not a CW pair.

Danu

@ACuriousMind I just didn't read 'till the end of that message of his. Sorry :P

Mike Miller

19:32

@Danu I was not consciously aware of it, no. But I'm supposed to be careful when grading.

I'm sure Hatcher points this out at some point, since he always does.

Danu

@MikeMiller So what is the other way it could be wrong? ;)

(you said you could think of two ways)

Also thanks a lot for that little lesson.

Mike Miller

The less subtle mistake is thinking that if $A$ has the same homology as $X$ and $B$ has the same homology as $Y$, then $H_k(A,B) \cong H_k(X, Y)$.

Danu

Right.

Mike Miller

One needs a map inducing both the isomorphisms $H_k(A) \to H_k(X)$ and the same for $B, Y$ at the same time.

Danu

Really, I love the five lemma.

It was essential for me to understand the "spirit" of diagram chases.

Use surjective maps to go up/left/right/down, use injective maps to close squares :)

Mike Miller

19:39

I don't understand algebra, is all.

I'm told if you really want to understand diagram chases you should do them in abelian categories. I decided I didn't really want to understand them.

Danu

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I wish I knew...

Physics Meta

20:20

0

Q: On the infallibility of Wikipedia edit histories

This is not really a question about this site (though it does have some impact), but I'm not sure where exactly to raise it and I'd like to pose it to this community, which can probably / hopefully help. Recently I discovered that the Wikipedia data I used to produce an Astronomy Stack Exchange ...

discussion

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

20:41

How can anybody say this and not find it contradictory?

in Mathematics, 28 mins ago, by Monad

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ They are not the same, but they do not have a number in-between

Re: the difference between 0.999... and 1.

:-/

ACuriousMind

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ "In between" is a bad way to try to go about showing they are the same - you are implicitly using things like that the reals are totally ordered and that there is another real between any two distinct reals.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Shouldn't graphing the numbers on a line convince them? @ACuriousMind

ACuriousMind

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Graphs can convince you of many things that may or may not be true :P

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

:P

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

21:04

smbc-comics.com/index.php?id=3777

Anthony

21:37

Why is there only one H2? I'm not familiar with quantum chemistry, but is there some fundamental property of something that makes there only be one stable configuration in molecular bonds?

*covalent

ACuriousMind

@Anthony What do you mean "one H2"? What is your notion of "being the same" here?

Anthony

Isn't there only one thing really meant by H2? Like it doesn't come in different varieties does it?

Like why aren't there other stable ways to combine two hydrogen atoms?

ACuriousMind

@Anthony Again, what do you mean by "varieties"? There are, for instance, the spin isomers para- and orthohydrogen

Anthony

Well then maybe I'm just wrong. Are the atomic spacings constant?

ACuriousMind

I'm...not sure "atomic distance" is well-defined for something like H2, and if it is, I have no idea about its possible variation

But I'm still not sure what you're asking - the two H atoms just have one electron each - what are they supposed to do except share it with the other atom to form a bond?

In any case, this might also be more a question for the people from Chemistry

Anthony

21:49

Alright, maybe this is a better question. The two electrons in H2 have some electronic structure. Are those energy levels the same in every H2 molecule? I'm just confused because I've never seen this work through, I don't know what parameters there are, and I don't know if uniqueness of solutions for differential equations plays a role anywhere in this.

ACuriousMind

I would say that the energy levels in para- and orthohydrogen are different (but very close), similar to the (hyper) fine structure of atomic orbitals, but I'm really out of my depth here

One thing I can say, though: You've never seen anyone "work through this" because there is nothing to work through - we can't solve for any orbitals analytically except for the lone hydrogen atom. all others must be computed in approximations and numerically.

Steven Stewart-Gallus

22:44

The potential energy in a capacitor is cV^2/2. More generally the potential energy stored in a voltage difference is the integral of voltage with respect to charge. Likewise, the kinetic (I guess) energy stored in an inductor is LI^2/2 and the energy stored in a moving current is the integral of current with respect to the integral of voltage with respect to time. What does the integral of voltage with respect to time mean? It's like charge is for voltage but for current.

dmckee

22:56

@Anthony There are bonding and anti-bonding molecular orbitals available, but the ground state is to have both electrons in the bonding orbital (and two fit because of the spin degeneracy).

And @ACuriousMind is right bout the para- and ortho- but you need a pretty good spectrometer to discern the splitting.

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