The h Bar

dmckee

00:09

Mostly a blessing, really, but also the last resort of snowflakes too precious and delicate to subject themselves the same work that their peers do.

dmckee

00:23

Their job is to help students resolve or cope with non-school issues that are interfering with their studies, and to detect signs of possible academic trouble early and to guide the student toward solutions to that to.

That is an important job at a less selective school, where many of the students have heavy external obligations and may have a less than stellar backgrounds.

But some people see them as a way to get special consideration.

Danu

Interesting talk that gets to frontiers of complexity theory but starts at introductory level by a guy who just came up with a major breakthrough (which is what his talk is about): people.cs.uchicago.edu/~laci/2015-11-10talk.mp4

An "almost polynomial" algorithm for checking graph isomorphism!

user54412

@Danu Your manifold. It's so... vibrant.

Danu

@ChrisWhite It's the Hopf fibration of $S^1$ over $S^2$ :)

@ChrisWhite Or rather its stereographic projection onto $\mathbb R^3$ I think...

Qmechanic

00:43

0

Q: Nonconservative forces such as velocity-Dependent forces, friction/drag/etc. Do they really overall defy Newtons' Laws

(Note: I have limited physics background, please keep it fairly simple if possible) On this site I've read about nonconservative forces regarding Newton's laws. Such as velocity-Dependent forces, friction/drag/etc. Can someone suggest, or say for certain, whether events occur in reality which d...

forces

Unclear what's been asked?

Danu

@Qmechanic Certainly!

dmckee

00:59

ARGH!

Comment to not post on ..

-6

Q: The first plasma?

What the German Engineers did two days ago, is it really the first time that we create artificial plasma? Source: https://t.co/lWk1ykaisA

plasma-physics

> Go down to Spencer's Gifts or other knick-knacks store. Ask them for "plasma sphere" or "plasma generator". Pay money. Take purchase home and make your very own plasma. -- dmckee

user54412

I was going to suggest lighting a match...

dmckee

@ChrisWhite That would allow you to point the OP at our questions(s) on the degree to which flames are ionized (only a little mostly), so that's something.

Emilio Pisanty

Man, undergrads

just... just... no

Danu

01:18

@EmilioPisanty Hahhahaha

Let $x=p$. Then clearly $[x,x]=i\hbar$

0celo7

01:46

@dmckee I think you meant to use the backspace key.

@Danu My god, I didn't recognize you

ACuriousMind

@0celo7 Yes and no, see Wiki

dmckee

@EmilioPisanty Is that some random substitution ansatz? Ouch.

0celo7

@ACuriousMind We've been over this.

@EmilioPisanty fum8

Emilio Pisanty

@dmckee I suspect Health and Safety will be glad to know the office furniture is working correctly =D

@dmckee Yeah, just a change of variables to simplify the integral. But $x$ for a momentum... why, just why did he think that was a good idea?

ACuriousMind

Perhaps messing with you intentionally? ;)

0celo7

01:51

$x$ is a momentum ratio in proton structure stuff

Emilio Pisanty

@0celo7 Yeah, this is the momentum space wavefunction of the simple harmonic oscillator

and they've explicitly been told to transform from the position representation

0celo7

@EmilioPisanty no need to slam undergrads

Emilio Pisanty

@0celo7 Who's slamming undergrads?

0celo7

47 mins ago, by Emilio Pisanty

Man, undergrads

ACuriousMind

That's clearly a statement of affection!

0celo7

01:56

^no

Emilio Pisanty

I'm glad you can read my mind then. Seriously, chill out.

0celo7

ACuriousMind

Maybe you need a Snickers :D

0celo7

@ACuriousMind I didn't take you for a commercial watcher

ACuriousMind

It's been years since I saw one, but that ad is quite old

0celo7

02:01

@ACuriousMind Einstein no longer sits with me in my room :'(

ACuriousMind

:'(

0celo7

::abducts cat and places on chair::

he needs to be retaught

dmckee

@0celo7 Nope. Emilio is right about this. Sure, (s)he might have done that because they always use $x$ by default, but they should have seen the trouble as they proceeded and fixed it.

0celo7

@dmckee Right about what?

dmckee

By the time they've gotten through the sophomore sequence these students need to be making an intentional effort to begin writing like grown-up physicists.

@0celo7 That the variable selected for the substitution he showed is just atrocious.

0celo7

02:08

@dmckee Did I disagree with that?

I would have picked $k$.

Danu

You objected against him expressing that sentiment.

0celo7

@Danu The sentiment that being an undergrad has anything to do with shitty notation.

ACuriousMind

$p'$ is obviously the way to go. No need to pick a new letter until you've stacked at least a prime, a tilde and a bar on that letter ;)

0celo7

If we want bad notation, let's pick some Weinberg book...

Danu

$\tilde{\bar{p'}}$

confirmed

0celo7

02:09

what about ${}'p$

Danu

^this guy

ACuriousMind

@0celo7 That's some out-of-the-box thinking right there

0celo7

${}'\bar{\underline{\tilde p}}'$

is \underline not a thing?

Danu

it is, but might need a package

0celo7

well you guys get the idea

can we stack tildes? $\tilde{\tilde p}$

how high can we go? $\tilde{\tilde{\tilde{\tilde{\tilde{\tilde{\tilde{\tilde{\tilde p}}}}}}}}$

Danu

02:12

@ACuriousMind When substituting "length-type variables" I do like to go $x\to y \to z$ (in one-dimensional problems)

0celo7

Great, now the cat is just staring at me...

@ACuriousMind It's like in ME2 when you're dead for 2 years and then Liara is kinda freaked out when you come back.

Exactly like that.

Einstein: "I mourned you, human."

L33ter

06:26

hi

Huy

08:14

@0celo7 pls

how should I play GTA with you when I can't even finish simple calculations and you won't help me @0celo7

Illustionist

@Huy

Huy

sup Karim

Illustionist

do you want to discuss something in physics

Huy

like what

I'm no good at physics

Illustionist

so, I am having a problem

suppose the particle is released from rest in the $xz$ plane so it is constrained to move on the parabola $z = ax^2$. Use the method of lagrange multipliers to find the equations of motion the particle and the force of surface reaction on the particle.

so I did the lagrangian etc

now I got complicated expression for the multiplier

$\lambda$

Huy

08:30

idk man

Illustionist

its ok

Physics Meta

09:16

0

Q: what is this "cite" in the same line as edit and close

I must have missed something. What is this "cite" in the same line as "edit" and "close" in a question, and "edit" and "flag" in an answer. Cite what and where?

discussion

Balarka Sen

11:04

@Danu You're a circle bundle!

John Rennie

@Danu: re the GR Mathematica notebook you gave me. Is there an easy way to calculate the Ricci and Kretschmann scalars? The Ricci scalar I could do by hand if pushed since your notebook does give me the Ricci tensor. The Kretschmann scalar I could in principle do by hand but it's a loooooong calculation.

Jacobadtr

11:52

Hi,

when referring to the total spin of a pair of electrons, should I use $\vec S = 1$ even though i'm not specifying a direction, or should I give it as a scalar quantity?

I want to mention LS coupling where it makes sense to express the total angular momentum as the sum of vectors and don't want to be inconsistent.

I haven't done any formal QM yet if that influences your opinions (this is for a lab report on Mercury spectroscopy)

David Z

12:19

@Jacobadtr Strictly speaking, you should only put the same kind of quantity on either side of an equality. So $\vec{S} = 1$ is inconsistent because it has a vector on the left and a scalar on the right. If you want to say that the system has spin 1, I'd go with $S = 1$.

That's a general rule, nothing specific to QM or spin.

Jacobadtr

Absolutely I agree - It just feels strange talking of how spin is a vector quantity but expressing it as a scalar.

David Z

Well, "spin" can mean any of several things - the spin angular momentum vector, the magnitude of that vector, or the spin quantum number

Actually, come to think of it, maybe you should be writing $s = 1$, not $S = 1$, because $s$ is the quantum number (which takes half-integer values) whereas $S$ is the magnitude of the angular momentum vector, which takes values of $\hbar\sqrt{s(s+1)}$

If I remember correctly. I haven't done this in a while.

Jacobadtr

I have discussed the half integer values of s, and need to mention that they can either add together or 'cancel' each other.

My lab script speaks of $S = 1 or 0$ so I think i'm safe in following it's lead

it refers to $S$ as 'total spin'

David Z

@Jacobadtr unfortunately that's ambiguous

Anyway you would be far from the first person to not be pedantic about the difference between $s$ and $S$. As long as your notation is clear to you and to whoever has to read your report, go for it.

Jacobadtr

Ok, thanks for your help @DavidZ

0celo7

12:44

@JohnRennie Those calculations build character.

Huy

@0celo7: PLS

0celo7

@Huy Those calculations build character.

Huy

@0celo7: PLS JUST ONE STEP

0celo7

oh ffs

I don't even have paper

Huy

._.'

0celo7

12:56

Oh, @Huy , I'm going to see SW7 with the girl and she promised the SJW will not try to tag along.

Huy

<3

0celo7

I shall find some paper...

Tell me again what I need to calculate?

And our progress so far

you're lucky I have GR books at my house

Huy

@0celo7: ultimately, that this is coordinate-independent, but if you cba doing such a long calculation, just a short step like this one would probably be helpful already

0celo7

I'll just do the whole thing

or fail miserably

aw hell I need the derivative of the inverse of a matrix

$-M^{-1}\partial MM^{-1}$?

Huy

I don't know ._.

0celo7

13:06

you're a mathematican wtf

Huy

so what

I don't differentiate matrices

like ever

only when I do physics

0celo7

you @ss this is not physcis

you think Hawking or Wald do this shit

Huy

@

yes all day long exclusively

0celo7

ok so I need to do $\partial_j' (K \sqrt{g(y)} (K^{-1})^i{}_k X^k' )$

f u chatjax

@Huy can you read that

Huy

no

0celo7

13:10

$$\partial_j' (K \sqrt{g(y)} (K^{-1})^i{}_k X^k' )$$

mother

Huy

no

0celo7

@Qmechanic Do you have any idea why that won't parse?

damn, there's a weird transpose going on now

I need to use the full power of physics notation

@Huy this is not looking good

Qmechanic

@0celo7 : ::experimenting:: $$\partial_j^{\prime} (K \sqrt{g(y)} (K^{-1})^i{}_k X^{k\prime} ).$$

0celo7

@Huy how far did you get in this calculation?

Qmechanic

@0celo7 : The primes. I have seen this happen before on the main site.

Huy

13:14

@0celo7: didn't even know how to do the first part of the product rule

0celo7

@Qmechanic Ah, thanks.

@Huy How are you a math student?

Huy

@0celo7: proving != computing

0celo7

holy crap I have to do the derivative of the square root of a determinant

Huy

yes

0celo7

time to look it up in a GR book

Huy

13:15

I was scared of that one

0celo7

you have to do this derivative to derive the Einstein equations

Huy

I didn't derive them

0celo7

ffs this is terrible

Huy

then why did you laugh at me like it was some easy shit

0celo7

$$\partial\sqrt{\det M}=\frac{1}{2}\sqrt{\det M}\partial\ln\det M$$

Huy

13:17

lol

0celo7

@Huy : Good luck.

Huy

with what

0celo7

doing this on your own

Huy

I thought you show me the full power of physics notation now

0celo7

I did say that

Huy

13:18

yes

I am waiting for that to happen

0celo7

if I'm not done in 40 minutes, I give up

Huy

ok

0celo7

can I give up now

Huy

pls no

0celo7

hmm maybe I need a different approach

I'll do it from the other side as well

and try to meet in the middle

Huy

13:20

okidoki

0celo7

is that acceptable?

Huy

sure

0celo7

hmm this won't work

I still need $\partial\log\det M=\partial \tr \log M=\tr M^{-1}\partial M$

:(

@Huy wow these low level GR books are really helpful for these calculations

Huy

lol

why

many formulae?

0celo7

yes

I need to figure out what $\tr (g^{-1}\partial g)$ is

$$g^{ij}\partial_kg_{ji}$$

maybe...

oh there's four product rules btw

Huy

13:28

-_-

0celo7

I thought you said 3

I did not sign up for 4

Huy

._.'

0celo7

can you show, without my help, that $\partial_j^\prime KX^{j\prime}+K_i{}^j\partial_j^\prime(K^{-1})^i{}_kX^{k\prime}=0$

Huy

probably not, but if I see your calculation maybe I then also know how to properly compute these kind of things

0celo7

ok I'll do this

or I'll give it a try

crap now I'm getting into trouble....there's a transpose floating around that I did not keep track of

@Huy there's a determinant floating around and it should be gone

Huy

13:41

what do you mean by that

0celo7

I think everything cancels but one term has an extra determinant

Huy

ok

you can send me what you have and I'll look at it and try to find why the determinant is there

0celo7

13:54

@Huy hmm, I think I should have paid closer attention to the tranposes...there are some indices that are backwards

I'm terrible at this

i.imgur.com/3HvMUW7.jpg?1

Huy

thx

is that like a flashcard

0celo7

a big one

Huy

ok

very cute

0celo7

the new link is the rotation

oh that should be $\operatorname{div}X$

dunno why I used $V$

@ACuriousMind can you please check my work?

ACuriousMind

No

0celo7

14:06

wow...

@Huy to get the derivative of the inverse of a matrix, take the derivative of $MM^{-1}=I$

to get the derivative of the determinant, take the derivative of $\det M=\exp \operatorname{tr}\log M$

Secret

https://books.google.com.au/books?id=I3W6oSaRlMsC&redir_esc=y
FINALLY, finsihed reading this book from cover to cover, after almost 3 weeks!
Now I have more idea on how to read spectroscopy papers which is needed for my honours stuff

---
On a more related note to the discussion here...
http://www.math.uwaterloo.ca/~hwolkowi/matrixcookbook.pdf

Secret

A wave of questions will be coming in a few days after I have digested all the matrials and came up with the conceptual questions I want to clarify

0celo7

on?

Secret

the math uwaterloo link contains derivatives for various matrix expressions

0celo7

no what questions will you have

Secret

14:23

After reading the spectroscopy book, and reflect on what I learnt in the spring school, I think there's a lot of similarities between the maths of particle physics and molecular states

---

@0celo7 I don't know the exact numbers yet, but they are related to integrals that describe transition between states, as well some group theory stuff. The only question that I can currently think of atm is about internal conversion (IC), on whether it is irreversible or can go both ways, but I will see whether I can clear that up tmr at uni when I ask my peers. (This is because the example of IC given in the book the start and end states have the same energy)

0celo7

you should take a look at the calculation I did above

@Huy have you completed it?

I must have made an error somewhere

@Huy cancel that determinant in the last line

it goes away, I made a mistake

Secret

@0celo7
$$\bf {MM^{-1}}=\bf{I}$$
$$\frac{\partial}{\partial \bf {x}}(\bf {MM^{-1}})=\frac{\partial}{\partial \bf {x}} \bf {I}$$
$$\frac{\partial \bf {M}}{\partial \bf {x}}\bf {M^{-1}}+\bf {M}\frac{\partial \bf {M^{-1}}}{\partial \bf {x}}=\bf {O}$$
$$\bf {M}\frac{\partial \bf {M^{-1}}}{\partial \bf {x}}=-\frac{\partial \bf {M}}{\partial \bf {x}}\bf {M^{-1}}$$
$$\frac{\partial \bf {M^{-1}}}{\partial \bf {x}}=-\bf {M^{-1}}\frac{\partial \bf {M}}{\partial \bf {x}}\bf {M^{-1}}$$

ok, didn't saw the last comment, sorry

0celo7

14:53

@Secret uh, what's that for

Huy

@0celo7: no, don't get it yet. I'll probably need a few hours.

0celo7

@Huy :(

are you confused by a step

Danu

15:16

@JohnRennie I think it's not too hard to modify...

no_choice99

15:34

hi guys I've a question very similar to physics.stackexchange.com/questions/80349/… I'm not sure whether to seek help here, create a new post, start a bounty on that question or comment it seeking more help

basically if I place an object very close to the eye (closer than the focal length), I want to show or see that the image formed on the retina (it's not a real image, not sure if virtual. in any case it's blurry) is inverted

I've done a sketch. and I can see that 2 rays starting from teh top of a vertical object won't focus on a single point on the retina but almost cover all the retina

and a ray starting at the bottom of the object will hit the retina in its middle

so it seems to me that a part of the image should be inverted and the other part should be upward, due to the blurriness of the image on the retina. but obviously in real life this doesn't happen, so I know something's wrong... but I can't figure out what

no_choice99

15:52

ok I've added a comment to that post

0celo7

@ChrisWhite an index expert!

@ChrisWhite you wanna prove, explicitly, that $\mathrm{div}X$ is coordinate invariant?

user54412

uh, that the divergence of a vector field is invariant?

user54412

because it's a scalar, so....

0celo7

proof?

i.imgur.com/3HvMUW7.jpg?1

my attempt at helping @Huy

ACuriousMind

@ChrisWhite Nice tautology ;)

David Z

16:09

@no_choice99 for future reference: if the new question you would ask goes beyond the existing question in some way, then you can post a new one - otherwise, comment or put a bounty.

0celo7

@ACuriousMind proof of the tautology?

ACuriousMind

@0celo7 Wat? Saying something is "invariant" under a transformation exactly means it's a scalar under that transformation, hence saying "It's invariant because it's a scalar" is a tautology.

0celo7

I still don't know why @Huy won't accept $L_X\sqrt{g}\mathrm{d}x^1\wedge\dotsc\wedge \mathrm{d}x^n=\partial_i(\sqrt{g}X^i)\mathrm{d}x^1\wedge\dotsc\wedge\mathrm{d} x^n$

@ACuriousMind he needs to prove that it's a scalar

Huy

Lie derivative too advanced

0celo7

wtf?

It doesn't even need a connection

how could you be less advanced than flows

ACuriousMind

16:17

@0celo7 The Lie derivative is...less known than the books you've read may suggest

(Likewise the concept of flows)

0celo7

It's not in do Carmo

but it's in Lee, Jost and every GR book

@ACuriousMind also less known ~= too advanced

ACuriousMind

@0celo7 I'm pretty certain that when @Huy says "too advanced", he really means "I don't know that" :P

Huy

we did Lee and I also somewhat studied a bit of GR so I know it

0celo7

o.o is this true @Huy

@ACuriousMind liar

Huy

but Rosenberg's book has an exercise that tells me to show the expression is coordinate invariant

the exercise looks a lot like "in a way without fancy Lie derivatives etc"

ACuriousMind

16:21

@0celo7 I said "I'm pretty certain" - that's not a lie, even if what I think is false :P

Huy

I actually thought it would be a very simple computation until I tried it ._.

0celo7

have you checked my work yet

I'll continue the calculation after I get back from errands

Huy

yeah I'm a bit confused by it because I feel like some $g$s are magically vanishing etc

0celo7

or you could try to convince the German to do it...it would take him 5 minutes tops

Huy

I can point out explicitly what I mean after dinner

0celo7

16:24

ok

Huy

who's the German

too many Germans here

0celo7

@ACuriousMind <

user54412

Huy

no Christoffel

0celo7

^

Huy

16:26

just transforming this here: chat.stackexchange.com/transcript/message/26112840#26112840

user54412

you want to prove something about a covariant derivative without acknowledging a connection? uh... have fun then

Huy

??

0celo7

@ChrisWhite that' what I told him

and then he guilted me into trying it

Huy

I don't even use the words "covariant derivative" or connection

I don't see why I should

I just want to define a Laplacian

ACuriousMind

@Huy Because your expression there is the covariant divergence.

Huy

16:28

how would I know

ACuriousMind

Slereah said it right after you wrote that!

Huy

and I said right away that I don't want to use all that fancy stuff

ACuriousMind

It's not fancy, why on earth would you define the Laplacian like that if not because you know it's a covariant derivative and hence plays nicely with coordinate changes?

I.e. why would you look at that expression at all if you didn't know about covariant derivatives?

Huy

I know how to define a gradient

and I know the divergence operator is its adjoint

from the formal adjoint condition you get that local expression for divergence

ACuriousMind

And that's not fancy? :P

Huy

16:31

no because I'm coming from an Analysis background

not from a GR background or whatever

ACuriousMind

But anyway, if you don't like the Christoffel, just replace it by its definition in terms of derivatives of the metric, and then @ChrisWhite's calculation still works to show what you want to show.

Huy

it's like some kid in MSE asking how to prove that the sum of even numbers is again even and people starting to talk about prime fields or whatever

0celo7

Doesn't that follow from distributivity in reverse

Aka factoring out a 2

Huy

bravo

0celo7