The h Bar

"You will also find in the literature that some people write `A' for the arrow $id_A$. This is a notation so ridiculous that it should be laughed at in the street."

GBeau

6:46 AM

what is "CME"

Shing

can anyone understand this lousy english sentence of mine ? ---> "now we take the observer who measures into a new system"

GBeau

as a physics graduate school topic to admit for

David Z

condensed matter experiment?

GBeau

ah, good guess

thanks

David Z

@Shing yeah, it's understandable, I think: there is an observer who measures something (perhaps as opposed to other observers who don't measure anything) and they are changing systems - presumably it's clear from the context what constitutes a system

JokelaTurbine

6:50 AM

La Gage's gravity problems were not asked properly before; physics.stackexchange.com/search?q=Le+sage%27s If there were some answers, they only referred "drag" physics.stackexchange.com/questions/211585/… or the level was very low like physics.stackexchange.com/questions/26692/…

or physics.stackexchange.com/questions/33977/…

So, I feel this question physics.stackexchange.com/questions/220050/… should be acceptable question, at least after my edit

Shing

@DavidZ Thanks for the help, I think I should add "without change anything" anyway to avoid confusion.

The Dark Side

Testing how to write something as a tag:

tag

Nope.

OK. Got it now.

David Z

There's also [meta-tag:tag]

The Dark Side

@DavidZ But that didn't work.

Anyways, I was wondering why do we have data-dump on Meta.Physics.SE?

David Z

Huh, maybe it's only in comments. But I thought something like faq was possible. edit: it is

The Dark Side

7:00 AM

meta.physics.stackexchange.com/questions/tagged/data-dump

David Z

[meta-tag:data-dump] -> data-dump

It's supposed to be for the periodic releases of the SE database, a.k.a. the "data dump"

It might be a default tag that exists on every meta site?

The Dark Side

Aha. But no guidance, no questions?

Absolutely nothing at all on our site.

David Z

Guess not. I don't think we have much of a need for it.

The Dark Side

So, we still have it only because it is there by default?

David Z

Presumably so

Though note that you can link to anything as a tag. e.g. mnomnop, it doesn't mean the tag actually exists

The Dark Side

7:04 AM

OK. That prevents a meta post :P Thanks.

John Rennie

@DavidZ: do we know who the Daniel Sank conspiracy nutcase is?

Incidentally he's just added aanother post you might want to delete

David Z

@JohnRennie we know some relationships between the accounts that have been posting

ROFL

got it

You tyrant

7:19 AM

I suppose that there are a good number of people that has nothing useful to do in their day... apart talking nonsense of course T__T

anyhow the electromagnetism tag is interesting...this is because DS controls minds and does harm by means of very potent EM emitters? superconducting magnets?

David Z

lol, who knows

oh, I think maybe we'd better keep these excerpts off the star list

DanielSank

@0celo7 That's not a cop out at all.

@ACuriousMind "restricting the domain of discourse from the beginning to consequential morals (as you seem to me to do) is therefore a mistake, in my opinion, because you first have to establish that morality is indeed decided by the actual consequences."

I think there is no issue of "establishing" that morality is decided by the actual consequences; this is, rather, a matter of how we decide to use the word "moral" (and derived words).

John Rennie

7:35 AM

@SirCumference-Pies it seems to me the answers to that question pretty much cover it. What extra info are you looking for? Something like:

21

A: Would the inside of a black hole be like a giant mirror?

Light cannot move outwards inside the event horizon. I would guess you're thinking that an outgoing light ray might leave you in the outgoing direction, then slow to a halt and return - hence you would see yourself. However this doesn't happen. The light leaving you moves inwards not outwards, bu...

DanielSank

I cannot think of a notion of "absolute" or "logical" morality which I think is worth wrapping up in a single, commonly used word.

If you can provide one (which doesn't degenerate upon inspection into religious doctrine), then I'll have a change in perspective.

My approach to morality and in particular what that word should even mean is influenced mostly by which notions are actually well formed.

I will not argue that my notion of morality is the better one as compared to Kant's because I think Kant's "isn't even wrong".

Slereah

9:09 AM

I'd better make a shortcut for $\textfrak{A}$ too

Too bloody long

Secret

9:27 AM

@Slereah I have an SR question
So basicaslly here in minkowski space I have observers B, O A

and I then plot the light cones for B and A at different points of their worldline

I then noticed that for B , the rays that intersect the t axis of O seemed to pile up (which seemed to correlate to how B is travelling very close to c)

and for C it occurs at a less extent

Suppose the observer measure a small time interval in his rest frame say for 5 proper seconds, then we should expect that interval would be 5 light seconds long and would basically be a vertical arrow of 5 units in the diagram

Slereah

That would be the Doppler effect

Secret

i then noticed the arrows made by the rays of obversers B and A are shorter than the 5 unit long vertical arrow that could have been obtained if O is measuring 5 light seconds of time

Is the ratio of the length of these red arrows with O give the time dilation between O and the corresponding observers? (and simialr the blu arrow give how much things are length contracted)

If the signal is light then is the "bunching" of these rays give the doppler effect?

yuggib

^ Doppler effect

Slereah

Nice zebra costume

The bunching is the doppler effect, yes

Secret

10:06 AM

Hmm, It seems my interpretaton that the red arrows measure time dilation is wrong, since here it gives the same arrwo length (the time interval between the receiving of the 1st and 4th light signal as measured by O)

In that case, better get back to the maths to see what this diagram mean

I can still see the doppler effect on the far right ,though

@Slereah I am guessing the horizontal spacing of the cones measures the wavelength of the light from E as observed by O (due to doppler effect), but what does the vertical spacing told us?

Slereah

the time between the reception of two wave packets

The talk page of wikipedia on AQFT : "The page is written in the language of category theory. I suggest that many people who would be interested in the Haag-Kastler axioms don't speak that language."

Who knew a theory entirely based on category theory would be written in category theory

yuggib

however I agree that who works on AQFT either do not know anything about physics, or do not know so much about category theory

Slereah

10:25 AM

wot

yuggib

I know two types of people that work on AQFT: mathematicians, and they do not know anything about physics (and essentially use AQFT as an excuse to do category theory); and (not so many) theoretical physicists who usually do not have a research-level knowledge of category theory

but that is just in my experience

Slereah

It ain't easy

yuggib

of course it isn't, but from a physical standpoint it did not yield many interesting results either

Slereah

10:42 AM

So

Which one is Haag

Apparently it's useful for QFT in GR but I still don't know why

John Rennie

Have we got any Mathematica users in the room? I want to ask what some Mathematica notation means ...

Slereah

I am, somewhat

John Rennie

Does this mean d/dr of the expression?

yuggib

maybe...but I think that we are far away from mathematically understanding QFT, and I have the feeling that AQFT, despite being interesting, is not the solution

John Rennie

i.e. d/dr(\sqrt{1-2M/r})

Slereah

10:54 AM

@JohnRennie I think so?

Although it's hard without more context

Could be $\partial_M$ for all I know

Nothing singles out $r$ as a variable out of context in Mathematica

Oh wait

I think the variable might be, without more context, directly $1 - \frac{2M}{r}$

John Rennie

I'm using Hartle's Christoffel.nb notebook to try and calculate the Christoffel symbols for the Schwarzschild interior metric. I get ...

Slereah

Yeah, I checked

Sqrt'[something] does the derivative wrt to the content of the brackets

So it should be $\frac{1}{2\sqrt{1 - \frac{2M}{r}}}$

John Rennie

@Slereah Thanks, that's what I guessed.

Slereah

11:18 AM

I've got a thing to compute GR tensor if you want

Probably not the best optimized code but it's fine

Danu

12:03 PM

I've also got a decent thingy for that stuff

@JohnRennie Yes it does mean that

yuggib

12:24 PM

scitation.aip.org/docserver/fulltext/aip/journal/jmp/56/11/…

I am still sad...

T__T

one day to refer and accept the paper...

and it is uninteresting, to say the least

and my two papers on the same journal took $\sim 1$year per paper on average to be accepted

damn peer review process!

innisfree

What do you guys think of Penrose's criticisms of inflation? They are basically ignored in the literature

Secret

I argue that color is one of the best way to illustrate superposition, because (as far I know) it and sound waves are the only things which the superposition is perceived as something all by itself instead of "simply a stronger or weaker intense thing" or "putting two different things together and still can tell there are two things being put together"

Secret

12:52 PM

What the hell is motion, really..?

0celo7

@Slereah did you really not sleep

user116211

1:46 PM

Sorry for disturbance but couldn't keep myself to share it. The funniest & one of the wittiest profile infos : "Well i am a school student from India whose science teacher can't explain a shit so i depend on physics stack exchange I am simply thankful to one who created this website" .

David Z

@JohnRennie assuming that's a prime, it means the derivative of sqrt. Just like how you'd write $f'(1 - 2M/r)$, where $f'(x) = df/dx$, except that here the name of the function is sqrt instead of f.

Note that the square root function in Mathematica is Sqrt with a capital S. One would hope that sqrt is defined with the same meaning though....

John Rennie

2:08 PM

@DavidZ Thanks. I'm attempting to use Mathematica to calculate Christoffel symbols, and since I have never used Mathematica before and my grasp of differential geometry isn't stellar it's proving an interesting experience :-)

Danu

@innisfree I think Penrose is not really taken too seriously anymore nowadays.

Mostly because of his weird opinions on non-strictly physics matters.

@JohnRennie Hey JR, I have a really simple thing to calculate GR tensors.

ACuriousMind

@Danu Don't you mean strictly non-physics matters? ;)

Danu

@ACuriousMind Not according to him ;)

@JohnRennie It's just a few lines of definitions that you actually paste into the .nb, (so no black box!) and it works like a charm to calculate the usual tensors.

0celo7

2:21 PM

@Danu Tsk. You're calculating the components.

Danu

@0celo7 You're "tsk"-ing me with regards to differential geometry? Tsk.

0celo7

Fight me.

Danu

@DavidZ I don't think sqrt is defined in Mathematica. Everything goes with caps.

@ACuriousMind Why does the statement that the conserved charge associated to a local gauge symmetry is gauge-dependent and hence unphysical lead to the statement that "gauge symmetry cannot be broken"?

I'm asking you because you're supposed to be our gauge theory dude here.

ACuriousMind

@Danu Breaking a symmetry means that the solution to the equation of motion (or "the vacuum state", or whatever you want to call it) is not invariant under the respective transformation. But the fundamental principle of gauge theory is that all physical states are gauge-invariant.

Danu

That's a decent statement.

But it does not directly tie in to the charge being unphysical. Is there a more direct argument?

I'm asking because my TA kept on saying this, but I don't really see how

David Z

2:38 PM

@Danu it's not a builtin function, but it is defined (it shows up in black in the screenshot)

ACuriousMind

@Danu Ah. See this answer, the conserved charge becomes non-conserved and is essentially the Goldstone boson of the symmetry breaking. If the charge is unphysical, you don't get the Goldstone boson, and the d.o.f. of the broken symmetry just vanished into nowhere.

Danu

@ACuriousMind Thanks a lot.

0celo7

3:25 PM

@ACuriousMind Back to the Lie group thing, how is $X(g)$ defined again?

i.e. given some vector field $v$ on $G$ how does one find the vector whose left translation to $g\in G$ is $v$ at $g$?

ACuriousMind

@0celo7 It's $v(e)$.

0celo7

@ACuriousMind lol

derp

$dL_gv(e)=v(g)$

thanks

John Rennie

5:22 PM

@Danu Cool, can you put the code somewhere I can find it? I've been using Hartle's package from:

wps.aw.com/aw_hartle_gravity_1/0,6533,512496-,00.html

and it seems to work for the Schwarzschild metric though I'm struggling to convince myself it's giving the correct answer for the interior metric.

user54412

6:00 PM

@ACuriousMind I've heard some bilingual philosophers choose an English translation, since the translator has had a chance to improve Kant's writing for the better. You have to have some faith in the accuracy of the translation, but on the other hand for German you would have to have faith that you aren't putting erroneous 21st century meanings to an 18th century text.

user54412

I suppose this is an issue that comes up with any old text.

0celo7

trying to manipulate Lie group things when there are three points and integrals and forms and derivatives involved is turning out to be not fun

user54412

Did Hartle really compress a small plaintext file? Some people shouldn't be allowed near computers.

3

0celo7

what's the issue?

Slereah

6:19 PM

@0celo7 Went to bed at noon in the end

0celo7

weakling

Slereah

I'm an old man :(

Hey @ACuriousMind

There is totally a set theoretic definition of categories

Apparently using the Groethiendieck axiom

⊢ ∃y(x ∈ y ∧ ∀z ∈ y (∀w(w ⊆ z → w ∈ y) ∧ ∃w ∈ y ∀v(v ⊆ z → v ∈ w)) ∧ ∀z(z ⊆ y → (z ≈ y ∨ z ∈ y)))

ACuriousMind

@Slereah I didn't say there wasn't, I said you shouldn't think about it that way and that most people just ignore the set-theoretic issues

Slereah

Yes, WEAK PEOPLE

also apparently the original Haag paper didn't use category theory at all

It was all Algebra

0celo7

@ACuriousMind so, now that I actually do this exercise

ACuriousMind

6:25 PM

@Slereah The proper category theorist founds set theory on category theory, cf fully formal ETCS ;P

0celo7

I don't see why the volume form has to be right invariant at all

Slereah

@ACuriousMind : Well I think all foundations of mathematics is expressible in terms of one another

I think that's a theorem of like

computation theory

0celo7

@ACuriousMind Left invariant of the new metric follows from left invariant of the old one...and right invariance follows from left invariance of the volume form math.stackexchange.com/questions/72333/…

Slereah

Set theory is expressible as $\mu$-recursive functions and vice versa

$\mu$-recursive functions are expressible as $\lambda$ functions

etc etc

⊢ Cat = {x ∣ ∃d∃c∃j∃r(x = <.<.d, c>., <.j, r>.>. ∧ ((<.<.d, c>., <.j, r>.>. ∈ Ded ∧ ∀f ∈ dom d∀g ∈ dom d∀h ∈ dom d(((d ‘h) = (c ‘g) ∧ (d ‘g) = (c ‘f)) → (hr(grf)) = ((hrg)rf))) ∧ (∀a ∈ dom j∀f ∈ dom d((c ‘f) = a → ((j ‘a)rf) = f) ∧ ∀a ∈ dom j∀f ∈ dom d((d ‘f) = a → (fr(j ‘a)) = f))))}

A category, apparently

ACuriousMind

@Slereah Tell me again how that notation is useful? :P

Slereah

6:28 PM

Isn't that much nicer than whatever you're using

Yeah that one gets a bit too long :p

ACuriousMind

@0celo7 Why are you telling me this?

Slereah

And that's not even exactly true

Because the guy had to decompose the definition

0celo7

@ACuriousMind because I need help?

Slereah

$Ded$ had to be defined first

Ded being

0celo7

@Slereah why would anyone do that

Slereah

6:29 PM

⊢ Ded = {x ∣ ∃d∃c∃j∃r(x = <.<.d, c>., <.j, r>.>. ∧ ((<.<.d, c>., <.j, r>.>. ∈ Alg ∧ ∀a ∈ dom j((d ‘(j ‘a)) = a ∧ (c ‘(j ‘a)) = a) ∧ ∀f ∈ dom d∀g ∈ dom d(<.g, f>. ∈ dom r ↔ (d ‘g) = (c ‘f))) ∧ (∀f ∈ dom d∀g ∈ dom d((d ‘g) = (c ‘f) → (d ‘(grf)) = (d ‘f)) ∧ ∀f ∈ dom d∀g ∈ dom d((d ‘g) = (c ‘f) → (c ‘(grf)) = (c ‘g)))))}

0celo7

you have got to be kidding

that's not real

Slereah

It totally is

There's a small community of crazy people that tries to prove math theorems entirely from syntactic proofs

ACuriousMind

@0celo7 And...I'm supposed to have your exercise from two days ago or so saved somewhere? You'll have to refresh my memory :P

0celo7

@ACuriousMind Yes.

I have proved left invariance. Right invariance is proved in the MSE post. Nowhere is part (a) used.

That's the proof of left invariance.

ACuriousMind

And you need help with what?

Slereah

6:36 PM

He needs help for everything

For he is just a baby

0celo7

@ACuriousMind Do you agree with my proof for left invariance?

@Slereah No

@Slereah Maybe?

I agreed with it until I completed the whole thing and I never used part (a).

Maybe I'm missing some subtlety?

Slereah

Hey @ACuriousMind

Why are the plane and hyperbolic space not conformal according to the uniformization theorem

They're both $\Bbb R^2$ and both of them have a metric $g = f(x) \delta$

ACuriousMind

@0celo7 The proof is correct, but you define $f$ but never use it, and you don't really show where you use that $\langle\dot{},\dot{}\rangle$ is left-invariant.

0celo7

@ACuriousMind I define $f$ for the proof of right invariance.

And the other objection: proof is in the book

ACuriousMind

@0celo7 The math.SE proof used a change of variable approach, which is basically equivalent to showing that left-invariant things are right-invariant things.

0celo7

6:44 PM

wait

ACuriousMind

The exercise just intends you to use right-invariance there and not apply the change of variable thing

0celo7

I use left invariance after "we see that"

@ACuriousMind o

ACuriousMind

@0celo7 Exactly, but you don't say it, that's why the "we see that" is a bit unclear to someone who doesn't know how the proof is supposed to work

0celo7

@ACuriousMind ah

ACuriousMind

@Slereah Because the hyperbolic space is conformally equivalent to the open unit disk?

0celo7

6:46 PM

I'll add in another step and explain

thanks

Slereah

Isn't the open unit disc the same topology as $\Bbb R^2$?

ACuriousMind

@Slereah So?

Slereah

what's the difference between the two metric?

Aren't they conformally related?

ACuriousMind

The uniformization theorem says there are three conformal equivalence classes: The open unit disk, the plane, the sphere.

Slereah

Or am I using a too physics definition of "conformal"

ACuriousMind

6:47 PM

Hyperbolic space is in the open unit disk, and the plane is, well, the plane

Slereah

What's the difference between the open disk and the plane outside of the geometry?

ACuriousMind

@Slereah I don't understand where your confusion lies at all - hyperbolic space is not the plane. There are several models of it, but none on the full plane

Slereah

I dunno

I guess it's because in GR, you usually consider manifolds to be the same if they are homeomorphic

0celo7

@Slereah *diffeo

Slereah

w/e

ACuriousMind

6:49 PM

Let's turn it around - what do you think is the actual map under which they are conformally equivalent?

0celo7

@ACuriousMind that's unfair

why did people star by dead bird comment

Slereah

$ds^2 = dz d\bar z \rightarrow ds^2 = \frac{dz d\bar z }{Im(z)}$

0celo7

wonder if they cleaned that up

Slereah

Or something

ACuriousMind

@0celo7 No, if someone asks me "Why isn't $A\cong B$?", it's a completely fair question to ask "Why do you think it should be?"

@Slereah That's not a definition of a map

Slereah

6:50 PM

Hm

0celo7

@ACuriousMind what?

ACuriousMind

You need to give an actual conformal map $f : H\to\mathbb{R}^2, z\mapsto f(z)$.

0celo7

that's the definition of a physics map

ACuriousMind

@0celo7 No, definitely not. Just writing down two metrics does not define a map at all

0celo7

@ACuriousMind wonder if I can find some book that disagrees

ACuriousMind

6:51 PM

A map is defined by mapping every point of one space to one point of the other.

@0celo7 People sometimes omit writing down the actual map if it is "obvious" and they are mostly interested in the metric

But that doesn't make writing down two metrics a definition of a map

0celo7

I'm not disagreeing with you for the record

Slereah

Isn't there always a map between a simply connected open set of $\Bbb R^n$ and $\Bbb R^n$

0celo7

@Slereah doesn't have to be a conformal isometry

ACuriousMind

@Slereah: Also, you should fix the model of hyperbolic space you're working with - upper half-plane, Poincare disk, projective space?

@Slereah No (at least not a homeomorphism, much less a conformal equivalence)

Slereah

Well it is usually given as the open disk

0celo7

6:54 PM

@ACuriousMind ._.

Slereah

$(D^2 , \frac{dzd\bar z}{Im(z)^2})$

I guess I have too much GR definitions in me brain

0celo7

@ACuriousMind I thought any simply connected open set of $R^n$ is homeomorphic to a ball...is that wrong?

ACuriousMind

@0celo7 Yes, we've been over this before.

0celo7

the fuck

where?

I would have bet my life on that

Slereah

"However, I must admit that his discussion of the "many worlds" in quantum mechanics, which Brian considers "inevitable", along with all of his thoroughly irrational attacks against the proper Copenhagen quantum mechanics, was driving me up the wall."

Motl really takes things too seriously

0celo7

6:58 PM

ah

the annulus

@Slereah No

Slereah

Well yes, you say that now you fake :V

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