The h Bar

1:02 AM

@StanShunpike I finally bought some new music!

It passed my patented test with flying colors: If the album has at least (album price in dollars)x2 songs that I instantly like and do not already have, it's a purchase.

I think a lot of hardstylers are experimenting, I'm hearing a lot of new sounds on this album.

There's only so many ways you can spin a 150 bpm with hard house synths.

Icosahedron

1:25 AM

@0celo7 You think I'm that bad?

0celo7

@Icosahedron No. $[\xi_i,\xi_j]=f_{ij}{}^k\xi_k$.

Icosahedron

I'm not, really

0celo7

What if I'm being sarcastic about being sarcastic?

Icosahedron

The syntax should include two separate equations.

0celo7

@ACuriousMind Fun fact: 32% of times someone said your name in chat, it was me. 50% of times someone said my name (spelt correctly), it was you.

@StanShunpike I'm convinced iPhones were never meant to have this much music on them. Currently performing maintenance on my iTunes library. (Note the tiny sliver of free space!)

ACuriousMind

1:43 AM

@0celo7 That might have something to do with us rather consistently using the reply feature even when in longer conversations.

...or with us just spending too much time here :P

0celo7

@ACuriousMind A bit of both ;)

@StanShunpike I just deleted a whole bunch of apps I haven't used in forever, and when I synched the phone, iTunes added them all back!

GG Apple, nice alphabetical sorting there.

ACuriousMind

@0celo7 I'm not sure if I'm more disturbed that there is S between A and A or that it decided to file that under S...

0celo7

@ACuriousMind The actual name of the album is Scantraxx 131 - EP, so that explains the S. Not sure why it displays it with the wrong title and puts it in with the A's.

Note that there are 96 songs in a compilation album with only 81.

user54412

1:58 AM

@0celo7 Because music metadata databases were filled in by drunk monkeys with typewriters, vainly trying to recreate the works of Shakespeare.

0celo7

@ChrisWhite Interesting theory.

I can't say for sure if that is true or not.

ACuriousMind

...why do the monkeys have to be drunk? Does that actually affect their performance?

Or is it just for their enjoyment?

0celo7

Combinatorially speaking, would they have an easier time typing all of Shakespeare on typewriters, or Einstein's seminal paper on GR, if we extend the keyboard and do everything in LaTeX?

@ACuriousMind Both, of course.

You can't expect fully sober monkeys to actually sit there and type for that long.

There are 884,421 total words in Shakespeare's 43 works.

I expect Einstein's paper to be much shorter.

How do we determine the number of words in a PDF?

Icosahedron

Depends on the reader.

ACuriousMind

16

Q: Count the number of words in a PDF file

How can I get the word count of a PDF file? I think that most pdf files for which I want to get total word count have text layer embedded, so I need no OCR. The task was arisen from searching for some scientific papers of known size, e.g. 15000 words. Most moders papers are published in pdf format

0celo7

2:09 AM

We'd also need to determine the number of equations, both in-line and not.

ACuriousMind

That...might be impossible

0celo7

Impossible? Nay. We just need a dedicated graduate student.

ACuriousMind

@0celo7 Why not use drunk monkeys for this also?

0celo7

@ACuriousMind TIL German drunk monkeys can count.

Icosahedron

Do you still need the HE text pdf?

0celo7

2:16 AM

Need is a strong word.

Like, yes.

Icosahedron

Then no.

0celo7

That's a shame.

ACuriousMind

@0celo7 Our beer has magical qualities ;)

2

user54412

2:29 AM

@0celo7 You have much to learn about copyright violation ;) In particular, there is a magical land of searchable pdf's of every book ever printed. This land is called China, and its search engine is called baidu.

0celo7

@ChrisWhite I'm not going to search for it.

user54412

2:44 AM

Also, wow the (result insightfulness)/(proof detail) ratio is quite low for Prop 4.5.1.

Icosahedron

@0celo7 Answer more questions.

0celo7

@ChrisWhite Unless I'm dumb, you're saying the result doesn't deserve as long of a proof as they give?

Small numerator?

user54412

Yeah. Or at least, the reward for understanding the whole proof is pretty low.

0celo7

The result is really obvious.

The rigorous proof is really dumb though.

4.5.3 is the one I'm really having trouble with, however.

@ChrisWhite Have textbook companies ever gone after anyone for getting a digital copy if they had previously purchased a physical copy?

If I were to obtain such a digital copy, this would be my situation.

user54412

Everything imaginable has happened to someone I'm sure, but it seems rare. Especially if you're not a distributor.

0celo7

2:58 AM

@ChrisWhite Do you mind taking a look at 4.5.3? My biggest issues are: I can't equate the $\sigma$ in this proof to the one in Lemma 4.5.2, and I can't prove $\rho=\sqrt{-\sigma}$.

user54412

I'm reading it now, because sometimes GR is a nice distraction from... GR.

0celo7

The issue with the first one is that I can prove that, assuming $g$ is constant within the CNN.

Let $v_0$ be the initial tangent to the geodesic $\gamma(t)$. Then $\exp_q^{-1}p=v_0$.

Thus the $\sigma$ from 4.5.2 is $\sigma=|\exp_q^{-1}p|^2=|v_0|^2$.

Using normal coordinates within the CNN, we may give the geodesic the coordinates $tv_0^i$. Then the components of the tangent are simply constant, $v_0^i$. Thus, the $\sigma$ in 4.5.3 is $$\sigma=\int_0^1 g_{ij}v_0^i v_0^j\,\mathrm{d}t$$

So $g_{ij}$ must be constant in the CNN for that result to hold. Is this true?

user54412

I think $g_{ij} = \delta_{ij} + \mathcal{O}(x^2)$, where $x$ is the distance from $p$. Maybe?

0celo7

I think that's correct.

user54412

Where I reserve the right to replace all $\delta$'s by $\eta$'s in the event we move in a timelike direction.

0celo7

3:05 AM

But they don't say anything about approximations, do they?

@ChrisWhite Of course.

user54412

@0celo7 No, they don't. Then again, we're doing variational calculus...

0celo7

@ChrisWhite Note that under the corollary on page 105 they say we first study the case where the points are close together.

I'm not sure if that's a hint.

user54412

I would think "close together" means "in a CNN," not "infinitesimally close."

0celo7

Do you agree with my analysis of the two $\sigma$'s?

user54412

The first one certainly.

0celo7

3:12 AM

Do you agree that the components of the tangent to the geodesic are constant in the CNN coordinates?

user54412

Okay I agree with your analysis.

0celo7

We're both equally confused then?

user54412

3:35 AM

Yeah, I don't know what to say.

user54412

I guess whatever magic makes them think the $\sigma$'s are equal probably also allows interchanging integration and square rooting.

0celo7

@ChrisWhite Wald cites this proof in HE for one of his propositions.

@ChrisWhite Are the geodesics perhaps normalized so that $g(\partial_t,\partial_t)=-1$?

user54412

I thought they were parameterized to have endpoints 0 and 1

0celo7

Does that prevent that normalization?

user54412

Does the first definition of $\sigma$ actually get used anywhere?

0celo7

3:45 AM

I'm really lost.

Maybe.

This whole section has been very vague.

user54412

I don't quite see where it enters. Like, if you redefined the first $\sigma$ with an integral, would anything explode?

0celo7

In the figure, it seems as though $\sigma=-s^2$.

$s$ is used.

Not sure if redefining $\sigma$ does anything.

user54412

$\sigma = -s^2$, sure, but that's just a roundabout way of defining $s$ to be the coordinate such that $\partial_s$ is in the direction of the geodesic from $q$ to $p$.

user54412

I think all the pullbacks through exp() in Lemma 4.5.2 just rely on this property

0celo7

And since they call $\sigma$ the "length" in the diagram, this implies some integral, right?

On a unrelated note, I now have 199 Dwarven Metal Ingots in my inventory.

1990 Dwarven Bolts coming up!

@ChrisWhite Ok, let's suppose now that in Lemma 4.5.2 we define $\sigma=\int_0^s \sigma_\text{original}\,\mathrm{d}t$.

Is that the upper bound?

What confuses me in these proofs is the difference between $s$ and $t$.

Like they say that $\gamma(t)$ is a geodesic with curve parameter $t$, but it seems like $t$ is the variational parameter too.

In the picture, it looks like $s$ is the curve parameter.

Stan Shunpike

3:59 AM

@0celo7 dude, your music. omg that's crazy. how many gigs do you have?

0celo7

@StanShunpike 42? Maybe more.

Stan Shunpike

huh, i thought they came in 32 and 64

still a hell of a lot of music

0celo7

I have the 64.

It's 42 gigs of music + other stuff + OS

Stan Shunpike

wow, that's a ton. what genres?

0celo7

So much hardstyle

A lot of it is double because I uses to do this thing where I bought compilations, took them apart and gave the songs the proper album art.

Over the summer, I might clear out everything that's double.

It's still a shitload of music.

I also deleted 20+ gigs of mixes.

Stan Shunpike

4:02 AM

wow, have you listened to most of it

?

0celo7

I think so.

I buy a lot of hard dance compilations from festivals, and I don't like all of the genres. There's some stuff that I haven't listened to and never will.

I'm forcing myself to listen to jumpstyle right now to see if I like it.

@StanShunpike I suspect that when I get my new laptop for college, I'll go through my hard drive and do a massive clean-up, including all duplicate songs, so my new setup is cleaner.

@ChrisWhite Redefining $\sigma$ does not help with the relation to $\rho$, does it?

user54412

4:39 AM

@0celo7 Does it? Define $\rho$ to be the length, $\rho = (-\sigma)^{1/2}$. Then the final equation before the end of the proof on p. 106 is essentially $\int (-\lVert \partial_t \rVert)^{1/2} \leq (\int (-\lVert \partial_t \rVert))^{1/2}$, and isn't this just Jensen's inequality?

user54412

Anyway, time to leave the office.

0celo7

Is Jensen a special form of Hölder?

Google tells me that it is not, but I think that inequality follows from Hölder as well.

0celo7

5:03 AM

(It doesn't. There'd be a factor of 2 I think.)

user54412

5:22 AM

Let's just agree to cite the general Jensen-Hölder-Cauchy-Bunyakovsky-Schwarz-Minkowski inequality.

6

0celo7

@ChrisWhite Jensen is the correct one in this case.

My problem is that they say "let $\rho$ be the length of the geodesic", but on the previous page they defined the length functional as $(\int -||\partial_t||)^{1/2}$, which is not $\sqrt{-\sigma}$. There has to be a typo somewhere.

Maybe the best solution here is not to try to decipher their shitty proof but rather write a new one.

user54412

@0celo7 $\rho$ matches the second definition of sigma, and we're agreeing to never again mention the first definition, right?

0celo7

@ChrisWhite I don't think $\rho$ matches either $\sqrt{-\sigma}$.

We are not mentioning the first definition, agreed.

Ok, let's ignore the first part of the proof.

user54412

Oh I see. If we take "length of the curve" at face value, $\rho = L$.

0celo7

Let's work on "It now remains"

That's the important part, anyway.

user54412

5:34 AM

whereas $\sigma$(2) gives some sort of (squared, negated) length of the preimage of the curve under exp() back in the tangent space, or something

0celo7

@ChrisWhite Any idea what $f(t)$ is?

user54412

The thing just above the second offset equation on p. 106

user54412

$\lambda(t) = \alpha (f(t), t)$

0celo7

In that, what is $f(t)$?

user54412

it's the s-coordinate of the geodesic as a function of the affine parameter?

0celo7

5:40 AM

But isn't $t$ the variation parameter of the initial condition?

In Fig. 11, $t$ clearly plays a different role.

For the record, Straumann proves the following on page 636: Let $v$ be a vector. Then $\exp_q(tv)=\gamma_v(t)$, where $\gamma_v(0)=q$ and $\dot\gamma_v(0)=v$.

user54412

well, whatever it is, it's the parameter for $\lambda$

0celo7

So here is how I look at $\alpha$: Fix $t=t_0$. Then $\alpha(s,t_0)=\exp_q(sX(t_0))=\gamma_{X(t_0)}(s)$

This seems to be supported in the proof of Lemma 4.5.2

Actually, it's pretty much said word for word there.

So $\alpha(f(t),t)=\gamma_{X(t)}(f(t))$?

user54412

sure

0celo7

So how does the next equation follow?

This is so vague.

Obviously the tangent of the curve decomposes into a part tangent and perpendicular to the surface.

user54412

that's just the chain rule basically

0celo7

5:51 AM

I thought that, but I don't follow.

user54412

perpendicularity of t and s doesn't enter yet I think (that's the following equation)

0celo7

Yeah, I don't see the chain rule.

user54412

this is just saying that the point $\lambda(t)$ has coordinates $s = f(t)$ and $t = t$, so the curve $lambda$ has tangent $(\dot{s},\dot{t}) = (f'(t),1)$ at $t$.

0celo7

I still don't understand what $t$ is.

Is it the parameter of $\lambda$ or the variation parameter?

I seriously think there are two different $t$s.

user54412

We went from one parameter $t$ of a not-necessarily-geodesic (but rather just timelike) curve $\lambda$ to two parameters $(s,t)$. We could have reparameterized the curve with any number of parameters with any number of interrelations.

0celo7

5:58 AM

Ok, I get the next equation and inequality.

The vector $(\partial_t)_\alpha$ is spacelike.

Why is $(\partial_t)_\alpha=0$ the condition for a geodesic?

user54412

Because there is a unique geodesic from $q$ passing through a given surface of constant $s$ orthogonal to that surface, and we constructed things such that $t$ measures the shift away from that point of intersection.

0celo7

Wtf, I thought $t$ is the curve parameter!

I just said this!

user54412

Yeah, I mean the t-direction on the 2-surface $\alpha$ when parameterized by s and t

0celo7

So $t$ plays a different role for $\lambda$ than for $\alpha$?

For $\lambda$ it is the curve parameter and for $\alpha$ it measures the deviation of $\lambda$ from geodesity?

user54412

I suppose that's not inaccurate.

0celo7

6:05 AM

@ChrisWhite Are you referring to both statements?

If yes, I think I finally get this.

user54412

I think so.

0celo7

Ok, hang on to your seat:

We define $f(t)$ such that $\int f'(t)=\rho$, right?

user54412

I feel like you need to define one of $\lambda$, $f$, and $X$ in terms of the other two

user54412

Or if $f$ is defined as you want, then I guess that forces a choice of $X$ on us.

0celo7

Then we'd have to show such an $X$ exists.

user54412

6:13 AM

It was never really clear whether $X$ was free to be whatever we wanted, or handed down from on high, or what.

0celo7

"handed down from high" lol

@ChrisWhite We need to see if the curve constructed as $\lambda(t)=\alpha(f(t),t)$ fits the definition of a curve joining two points. We have $\lambda(0)=\alpha(f(0),0)=\exp_q(f(0)X(0))$. Choosing $f(0)=0$ we get $\lambda(0)=q$. For the other endpoint we have $\lambda(1)=\exp_q(f(1)X(1))$. By definition, $f(1)=\rho$. Not sure how to continue.

@ChrisWhite I wonder if this is circular: can we pick $X(1)$ such that $\exp_q(\rho X(1))=p$?

user54412

6:31 AM

sigh -- it's far too late to be chasing circular definitions

0celo7

Are we going to continue this or are you convinced?

user54412

umm, I'm not the one who ever needed convincing :P

0celo7

I wish you'd just said "no it's not circular, you can go to bed now"

user54412

I'm perfectly content to let sleeping proofs lie

2

0celo7

If you tell me you're convinced, I'll be convinced.

user54412

6:32 AM

@0celo7 Ok I'm convinced.

0celo7

@ChrisWhite Nice try. Now I don't believe you.

@ChrisWhite Thanks for your help. I'll sleep on this at take a look at it with fresh eyes tomorrow.

user54412

Back to magnetic fields in Kerr spacetimes for me.

0celo7

@ChrisWhite How bad is the derivation of Kerr-Newman? I skimmed Straumann's derivation of plain Kerr and it was intense.

Stan Shunpike

7:55 AM

LHC smashes collision energy record bbc.co.uk/news/science-environment-32809636

Danu

8:29 AM

smashes, hehe

alarge

12:25 PM

@Hippalectryon As you probably noticed there was a typo in my equations. Not that it really had much on an effect, and the result you're going to get is still different from the paper. Now it can be that there are several roots to the equation. If you just numerically differentiate the solution you get, you'll find that it does satisfy the ODE and the constraints, so.

The Dark Side

1:37 PM

-4

Q: Are professors designated to wear the same cloth every single day?

Do professors sign a contract with a list of clothing and fashion styles at the beginning of their tenure that's why their outfit NEVER changes? Is this to prevent sexual harassment?

professors

^ Troll-o-LOL !!!

0celo7

2:56 PM

@ACuriousMind Disturbing revelation last night: Einstein approves of QFT in curved spacetime.

ACuriousMind

@0celo7 Bah, Einstein never liked proper quantum theories, anyway :P

0celo7

@ACuriousMind Did you see the discussion between Chris and me about that "proof"? It was more like a giant hint.

ACuriousMind

@0celo7 Saw it, was glad I don't do GR ;)

0celo7

@ACuriousMind Lol, that's not the lesson to be gotten from that!

I might type up my own proof in TeX, see if I can make it make sense.

Icosahedron

The lesson is that HE is a great introduction to GR for beginners.

0celo7

3:10 PM

@Icosahedron Definitely.

Are we eschewing sarcasm equations?

Qmechanic

@0celo7 : Note that 'proof by lack of sleep' is not the same as proof by exhaustion :)

14

0celo7

Didn't see that.

Icosahedron

I should post a picture of my cat.

0celo7

Can he measure a great dead theorist's satisfaction with modern physics?

Icosahedron

Define what is meant by "can he measure".

0celo7

3:21 PM

Is your cat's satisfaction guaranteed to represent the dead theorist's?

Icosahedron

Yes, you need only to look at his face.

0celo7

What's his name?

Icosahedron

Planck.

0celo7

No one cares about him.

Icosahedron

He doesn't care about anyone either.

I only need to figure out how to do this without a phone.

0celo7

3:32 PM

@Icosahedron I use my iPad to take pictures too.

Danu

So I'm trying to read this old paper and I've encountered some notation that makes me go ?!?!?!

Majumdar (1947) - A class of exact solutions of Einstein's field equations

ACuriousMind

Christoffels

Danu

$\{ij,a\}:=\Gamma^{a}_{ij}$ ?

ACuriousMind

See the alternative notations as given on Wiki

Danu

Ah, very nice. Thanks.

ACuriousMind

3:39 PM

No problem. I have no idea why I know that, actually :D

Icosahedron

Aren't christoffel symbols very common? I knew about them before I even did differential geometry. They're in the first few pages of most books on the topic.

ACuriousMind

Probably remembered it from the Wiki article...

0celo7

@ACuriousMind Probably, that's the only place I've seen that notation too.

ACuriousMind

@Icosahedron The issue is not what Christoffels are, it is denoting them by $\{ij, k\}$ or $[ij, k]$ instead of $\Gamma^k_{ij}$, as is modern.

Danu

@Icosahedron lol that'd be embarrassing, if I didn't know what the Christoffel symbols were while trying to read GR papers :P

Glen The Udderboat

3:52 PM

I'd like to build a database from scratch. I haven't even got the software yet. What's top-notch/standard? Is it MS Access???

Danu

A database...?

Glen The Udderboat

@Danu Yeah, you know, a relational database.

Danu

I do not know :P

Glen The Udderboat

I thought you (all of you) were whizzkids. Oracle?

ACuriousMind

lol, Danu knows nothing about computers :P

Glen The Udderboat

3:56 PM

@ACuriousMind OK. I'll take this elsewhere... Hmm. which chat is recommended?

0celo7

@GlenTheUdderboat We have come computational physicists here, but most of us know nothing about it.

Danu

@ACuriousMind I'll have you know that I tex all my problem sets!

ACuriousMind

@Danu ...and?

Physics Meta

4:22 PM

0

Q: Why is question-migration limited?

When a question is asked on the wrong site, it can be migrated to another site via the flagging function. Why is this migration limited to only a few sites? It should rather be a list of all sites to choose between. Specifically, from my point of view a specific question is better off on the Eng...

bug flagging migration

Icosahedron

4:39 PM

@0celo7 I need assistance on the tensor exercises, when do you have time?

(I can't do it now though, I have to go return a book to the library)

Danu

Am I cray-cray or is the $\phi\phi$ component of the standard metric on the sphere equal to $r^2\sin^2\theta$?

ah, wait. In curved spacetime, do we have $F^{ij}=\epsilon^{ijk}B_k$ where $\epsilon$ is now the actual tensor rather than the Levi-Civita symbol (i.e. is it multiplied by $\sqrt{-g}$)?

ACuriousMind

@Danu Yes, pretty sure it is.

Danu

@ACuriousMind Yikes :P

I can't find an explicit reference for it in textbooks, but I feel it should be true as well

ACuriousMind

@Danu The spatial part of the electromagnetic field strength is the Hodge dual of the magnetic field, and the Hodge dual involves the metric in curved space.

That you can't find an explicit reference for this is because they are all index fetishists :P

Danu

4:54 PM

@ACuriousMind I've got an implicit one though, now :)

does anyone know of any "inverse \coloneqq" command in latex?

i.e. =: in a single symbol

user54412

5:42 PM

@Danu ctan.org/pkg/comprehensive?lang=en

user54412

In particular, see p. 41 of this pdf for \equalscolon from the colonequals package

Danu

@ChrisWhite Problem already solved, but thanks anyways'

mathtools also has a solution: \eqqcolon

Danu

7:17 PM

I'm trying to get access to Nordström's 1918 paper containing the derivation of the metric named after him and Reissner. Can anyone help me out? It was published in an obscure journal...

0celo7

@Danu Why all the interest in historical sources?

Danu

Why not?

0celo7

You're just looking them up because you want to read them?

Danu

In this particular case, I'm preparing a talk on the topic. But in general, I do look up and try to read a lot of old papers

I have a large list of unread stuff on my computer

Things like Dirac's work, Feynman's original path integral paper, Dyson's most famous stuff, etc

0celo7

I see.

Danu

7:20 PM

A lot of biographical stuff on mathematicians, too

0celo7

I rarely have access to historical sources.

@Icosahedron Yeah, I can see how these exercises require fluency in indices. I'm available whenever.

@Danu If you're looking for a slick-ish derivation, Straumann does the RN metric in the tetrad formalism.

Danu

@0celo7 I'm looking for the original derivation :P

0celo7

@Danu Just letting you know :)

Danu

Full reference for anyone willing to help: G Nordström, 1918: On the Energy of the Gravitation field in Einstein's Theory. Verhandl. Koninkl. Ned. Akad. Wetenschap., Afdel. Natuurk., Amsterdam 26: 1201–1208. (adsabs.harvard.edu/abs/1918KNAB...20.1238N)

Danu

9:20 PM

ALright, done for the day

Stan Shunpike

lol sayonara

or do you mean you are done with work?