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00:27

0

Q: Are questions about the field of physics off topic?

I often have questions about the field of physics itself (as opposed to physics questions). For example questions about the current state of a sub-field of physics, or questions about the implications of experiment results. The latter is especially important to me because my knowledge of physics ...

discussion

0celo7

02:40

@ACuriousMind Apparently it's a thing to measure mass in pounds

user116211

Just giving the definition of the topic doesn't count as a tag wiki, is it so?

0celo7

03:06

@Slereah Did you know that $S^1\times S^1\times S^1$ does not carry a metric of negative curvature

3075

03:20

@BernardMeurer so what finally happened to waterloo?

Bernard Meurer

@3075 RIP

3075

:(

Bernard Meurer

Yeah I was pretty sad

3075

what did they say?

Bernard Meurer

Like someone sent me an email about a week ago saying they needed an outline of my maths course

I said I didn't have that in hand and:
a. asked what would qualify as that and if a signed letter from my principal would do
b. why was it needed since CS didn't ask nor did any other school

they just replied a couple days ago "Hai no openings, kthxbai"

0celo7

03:21

RIP me

3075

damn.

0celo7

intro statics problem is kicking my ass

dmckee

@0celo7 Yes. If you're serious about doing engineering you need to be prepared to deal with pounds-mass versus pounds-weight (and in extreme cases kilograms-mass versus kilograms-weight; argh!).

3075

@0celo7 electrostatics?

0celo7

@3075 no, statics

Bernard Meurer

03:22

@3075 Yeah, sigh

0celo7

did I say electrostatics

3075

wow sorry.

i thought you finished first year mechanics.

0celo7

@dmckee kilogram weight???

@3075 no

dmckee

@0celo7 I've seen it. But the metric world has moved much farther along the road to making a clear distinction twixt weight and mass.

0celo7

twixt?

@3075 the picture has an acute angle but I got 270 degrees

3075

03:25

nice.

dmckee

@0celo7 Read you Shakespeare, young man. And get off my lawn.

0celo7

So I don't know if the drawing is supposed to show me where I'm supposed to measure the angle from

dmckee

Short for 'betwixt' which means 'between'.

0celo7

Or if it's actually supposed to be acute

@dmckee Owning land is a crime against humanity, old man

Bernard Meurer

@3075 Yeah, but my beer is good though

so it's alright I guess

3075

03:27

xD

Bernard Meurer

Anyway, I'll catch you guys later, hope you're recovering alright @3075

3075

thanks, I'm doing ok.

bye.

0celo7

who wants to calc a cross product for me

user54412

04:27

LIGO just made a sudden announcement to hold a press conference as part of the American Astronomical Society meeting today/tomorrow, 2016-06-15, 10:15 PDT (17:15 UTC). It will be broadcast here.

0celo7

"JK"

user116211

something special?

user54412

not sure -- there's excitement in the air

user54412

certainly there have been rumors of other signals in the pipeline

0celo7

ALIENS

user116211

04:31

ah!

John Rennie

06:04

@ChrisWhite A chat session. We need another special chat session for this. It has to be more mergers. @DavidZ @dmckee can we set up a chat session for this?

Secret

07:04

http://gmwgroup.harvard.edu/research/index.php?page=13

Studying dissipative systems by building them artificially

gmwgroup.harvard.edu/research/index.php?page=14

yuggib

07:38

here you go...another question on the continuity/differentiability of the wavefunction...

3

Q: Can a physical wavefunction be non-smooth (its first derivative is discontinuous)?

Here's an argument that might support the statement that such a non-smooth wavefunction is not physical: You cannot add a finite number of smooth functions to get a non-smooth function. By fourier expansion theorem, any function can be expressed as a sum of plane waves (which are smooth with res...

quantum-mechanics fourier-transform

this time I couldn't refrain from answering

how did the ama go yesterday?

(it was too late for me here in Japan)

user116211

08:53

@yuggib nice to the extent that you can be our next guinea pig ;P

David Z

@JohnRennie You can set up a chat session, and ask a mod to post an announcement

John Rennie

@DavidZ is there any setup needed? I assumed a chat session would just mean us all meeting here at 17:15 UTC.

IIRC that's how it worked for the first LIGO announcement ...

user116211

@JohnRennie all in the main page would know about that then if you do that.

yuggib

@JohnRennie I had a sneak preview of this new announcement... ;-P

John Rennie

@yuggib Spill,spill!

If an NDA stops you just tell us if we NEED to watch the live stream!!!

user116211

08:57

@yuggib don't make suspense ;P

user116211

4 hours ago, by Chris White

certainly there have been rumors of other signals in the pipeline

yuggib

there was at least another one

of course it was not confirmed, but I have been told to wait for such an announcement

John Rennie

Only one :-(
Still that's enough to make the live stream compulsory

yuggib

well, I know of one

maybe there were more

if they're announcing it probably means they've already another paper accepted

John Rennie

@DavidZ: assuming we just meet here at 17:15 UTC, can you make the announcement?

yuggib

09:02

@MAFIA36790 :-D

Feeds

Chris White has added an event to this room's schedule.

yuggib

@MAFIA36790 is there a date already? should I prepare a meta post?

John Rennie

@StackExchange Thanks Chris. It never occurred to me to see if I could add the event to the schedule myself.

user54412

Hopefully it's something interesting. Given that it's a press conference, and in fact the first press conference since the original, I'm hopeful.

yuggib

I bet on $\geq 2$ new signals

user116211

09:07

@yuggib 1st one: no; 2nd one: really don't know; after seeing last night's AMA, I'm sure we can expect further AMAs but of course, no official announcement either. And moreover, it would be too early, i suppose, to write a meta post. I'm waiting for @vzn for further details on that - when the next AMA would happen or so....

user54412

Also, I could have been at that meeting, in San Diego no less. Instead I'm working on a paper. I'm not sure if I'm doing grad school quite right.

user116211

:(

FrancescoS

10:15

Is there someone who knows how to use MadGraph to compute partonic cross sections???

Nobody on the internet explains you how to do it without saying a lot of useless details

Slereah

10:45

Are there any exact solutions to string theory?

Like to the Polyakov action

ACuriousMind

@Slereah Uh, it's just the string propagating in a straight line.

Slereah

Hm

What's the gravitational action like in string theory?

ACuriousMind

The "classical" string is a wholly uninteresting object.

Slereah

How do I get pants diagrams

I want a cornucopia of pants

ACuriousMind

@Slereah String theory is an S-matrix theory, not a Lagrangian theory. There is no action.

Slereah

10:54

Oh

How do I get the gravitational S Matrix, then

ACuriousMind

You do get various 10D SUGRA theories as the effective QFTs that produce the tree-level ST results, though

@Slereah You "just" insert the appropriate vertex operator into the string perturbation series.

Slereah

String theory is p. lame

Is there any non-perturbative definition

Also Polchinski defines the string everywhere in his book with a Lagrangian?

ACuriousMind

@Slereah That's the holy grail - but no one has found it yet. Under "M-theory", "F-theory" and other names, various non-perturbative bits are known, but no one knows what the correct non-perturbative formulation of string theory is.

@Slereah You're mistaking the Polyakov action for "the action" of the theory. The defining object of the dynamics of string theory is the string perturbation series/S-matrx.

Slereah

Is there a toy model maybe for string theory?

I dunno, 2+1D strings or something

ACuriousMind

The string theory books are really abysmal at making clear that string theory really is not a QFT

Slereah

10:58

Or are there no non perturbative definition at all

ACuriousMind

@Slereah That's called the "non-critical string" and it's harder than the critical one because there you have the Weyl anomaly hence no clue how to properly quantize

Slereah

Why do people even bother with string theory

ACuriousMind

@Slereah No, we know various things about the non-perturbative formulation provided it exists, but no one knows its proper definition.

Slereah

Do we have clues what it is expected to look like?

ACuriousMind

@Slereah Because it "miraculously" produces many structures that you haven't put in.

@Slereah Yeah, that's what is known as "M-theory"

We know some parts of the spectrum - the BPS objects - for example

But what actual kind of physical theory is might be is probably still open

Slereah

11:03

Is it ghosts

ACuriousMind

Is what ghosts?

Slereah

The real theory

Spooks and haunts

👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻

ACuriousMind

Well, the "M" might stand for "magic" :P

Slereah

Man is there any decent quantum gravity theory

That is both well defined and not proven wrong on some level

ACuriousMind

I think string theory might really be your best bet. It's just very much work in progress, it's as if you had stumbled into QFT before we knew anything about partition functions or path integrals.

That it reproduces QFT at low energy and so generically gives no low energy predictions doesn't help, of course :P

Slereah

11:09

What's the status of Lorentzian gravity, these days

ACuriousMind

I have no idea...I have to go to my string theory class, now

Slereah

Well ask them what the fuck is up

Ask them to make a real theory and stop wasting my time

0celo7

11:52

good morning plebs

irapub.com/images/short_pdf/1457154342_Haradhan_2.pdf

> Haradhan Kumar Mohajan
Assistant Professor, Faculty of Business Studies, Premier University, Chittagong,
Bangladesh

GP paper written by a business prof

OK, seems legit

0celo7

12:26

I managed to reference my blog in a PSE answer

0celo7

12:42

@ACuriousMind I told my prof you said a bunch of the stuff I was doing in Milnor and GP can also be done using algebraic topology and he said "where's the fun in that"

Secret

13:30

Interesting, I knew of a martin who also worked with quantum mechanics

@Martin do you by any chance, studied in UNSW in 2014?

0celo7

::attempts to stalk profs on MO::

user116211

13:48

@0celo7 How could it be o.O

0celo7

I meant GR paper of course

user116211

yeah... but still... how could it be o.O

user116211

a business prof? hmm...

user116211

It's nothing but weird ;\

0celo7

@JohnRennie How would you answer this?

Please don't take this the wrong way, but I'd like to see a "low tech" argument

user116211

13:52

Have anyone saw Feynman's derivation of Lienard-Wiechard Potential?

0celo7

Never heard of it

user116211

It's really giving me a shit.

user116211

@0celo7 potentials of a moving charge

0celo7

yeah, that's not a phrase we use in English

@ACuriousMind Are you the second close vote on this?

user116211

He used an elementary idea of discrete summation over a point charge dispersed over a cube moving with a non-relativistic velocity.

ACuriousMind

13:55

@0celo7 Yes, I don't know what that question is attempting to do.

If he wants to find out whether a metric is flat, why not just compute the curvature?

user116211

But actually he overcounts the charge in the volume slices.

0celo7

@ACuriousMind I'm still not convinced that curvature 0 <==> Euclidean space (locally)

The proof uses the damn black box of black boxes

The Frobenius theorem (PDE version)

user116211

There is a PSE post on this... but actually the answer doesn't seem to point the mistake of Feynman.

user116211

5

Q: Feynman's proof for Liénard-Wiechert's potential of a moving charge

Feynman's proof utilizes a geometrical and fundamental integration argument. I like it, except this bit: What makes me unconfortable somehow is that in (c) we are counting in some of the charge we counted at (b). It seems to me that it is this extra counting which makes the potential to be lar...

relativity classical-electrodynamics lienard-wiechert

0celo7

@ACuriousMind You know a black box is bad when it's used to prove other black boxes!

ACuriousMind

13:57

::shrugs::

0celo7

@ACuriousMind what's that shrug for

do you know the proof?

ACuriousMind

@0celo7 The shrug indicates I don't care

0celo7

:/

How can you maintain such an attitude

Martin

@Secret: No, in fact, I've sadly never been to Australia, yet.

FrancescoS

Guys, does anyone here knows how to use MadGraph?

John Rennie

14:36

@0celo7 What does constant curvature mean in 4D Lorentian manifold. In a 2D Riemannian manifold I'd say it was a constant Gaussian curvature, but I'm unsure how this extends to spacetime.

0celo7

@JohnRennie In this case it means the curvature scalar is constant

John Rennie

@0celo7 The Ricci scalar?

0celo7

@JohnRennie yes

Same difference

John Rennie

Well I suppose I'd compute the Ricci scalar - actually I'd just Google for the result.

0celo7

@JohnRennie I also believe it follows from Schur's lemma.

you'd probably have to modify it for spacetime

@JohnRennie erm, you don't know the metric yet

this is part of the derivation of it

frick

the idea of sectional curvature doesn't work that well

I know Beem wrote a bunch of papers on it.

because you can have degenerate planes, i.e. planes generated by null vectors

@JohnRennie also do you know what this guy is asking for physics.stackexchange.com/questions/262711/…

Obliv

14:50

is $\mathcal{O}$ a set?

nvm it is

ACuriousMind

@Obliv what?

0celo7

what the hell is that

what kind of question is that

Obliv

it's an equivalence class

called an orbit i think

ACuriousMind

Just writing down a letter and asking if it is a set is not a meaningful question.

Never assume your notation is universally known.

Obliv

@acuriousmind I was going to explain but i realized the answer before i could

John Rennie

14:53

@0celo7 I can't see the point of that. Any form of the flat space metric is related to the Minkowksi metric by a coordinate transformation. So the most general form would be written as an arbitrary transformation applied to the Minkwski metric. This strikes me as a boring thing to do.

0celo7

@JohnRennie *locally minkowski

You could have some weird topology

but yes I agree

not sure what OP wants

Obliv

Hey lagrange's theorem : "if $G$ is a finite group and $H$ is a subgroup of $G$, then $|H|$ divides $|G|$" only applies when there is a bijective map $\varphi: H \to \mathcal{O}$? Also, I get that this means $\forall h \in H$, $x = hb$ ,for some $x,b\in G$ so that the elements of $H$ divide the elements in $G$. How does this indicate the cardinality of the subgroup divides the cardinality of the group though?

Sorry if i'm not being clear let me know where I don't make sense

ACuriousMind

15:09

@Obliv You still haven't told us what $\mathcal{O}$ is. Also, Lagrange's theorem always holds.

Slereah

let's try to learn inverse scattering transform again

I want some of that sine gordon

Obliv

$\mathcal{O}$ is the equivalence class of an element $x \in G$ under ~ where ~ is defined as $x$~$b$ if $x = hb$ for some $h \in H$ and $x,b \in G$, $H$ is acting on $G$.

Slereah

Is sine gordon even well defined as a QFT

It has arbitrarily high derivatives

Is it even causal

0celo7

@Obliv sounds like some coset stuff

ACuriousMind

@Obliv Not indicating of which element it is such an equivalence class is poor form (that is, at least write $\mathcal{O}_x$). But, apart from that, what exactly is your question?

0celo7

15:13

@Obliv the best notation is always $[x]$

For instance, I am now classifying the equivalence classes of codimension $p$ Pontryagin manifolds where ~ is framed cobordism

Fun fun

And I'm calling them $[x_i]$

Obliv

@acuriousmind well suppose you have a bijective map $H \to \mathcal{O}_x$ where $H$ is a subgroup of $G$ acting on $G$. $x \in G$. How can this information alone let us deduce that $|H|$ divides $|G|$? I thought the orbit just shows that the elements of the group being acted on is divisible by the group's elements acting on it.

@0celo7 i have no idea what that means. what's a coset?

0celo7

@Obliv I don't know either

ACuriousMind

@Obliv I need not suppose that, the map is simply given by $h\mapsto hx$. This means that all equivalence classes have the same size as $H$. Furthermore all equivalence classes are disjoint (because equivalence classes always are). If $\lvert G \rvert$ is finite, then it's a finite disjoint union (of say $n$ classes), so $\lvert G \rvert = n\lvert H \rvert$.

Obliv

@acuriousmind $H \to \mathcal{O}_x$ maps elements $h \in H$ to the relation $x$~$h$, correct? So this means $h \to x = hx$ so $h \to hx$ I get that part. Doesn't this just mean the equivalence class of $x \in G$ under ~ is the size of $H$? This is only one element of $G$. Is this supposed to be extended to all elements in $G$?

0celo7

@Slereah The worst is translating math into GR

John Rennie

15:23

@0celo7: maybe this is a silly question, but does the assumption of homogeneity mean there cannot be an inhomogeneous solution? For example the vacuum is homogeneous, and while we do indeed get a homogeneous solution (Minkowski) we also get inhomogeneous solutions (Schwarzschild, Kerr, etc).

Slereah

especially since GR abuses a lot of math terms

ACuriousMind

@Obliv What does "maps to the relation $x\sim h$" mean? Your second sentence doesn'T make any sense at all to me. But yes, from the existence of a bijection $H\to \mathcal{O}_x$ we conclude that the equivalence class $\mathcal{O}_x$ has size $\lvert H\rvert$, and since nothing was special about $x$, this holds for all elements in $G$, there is nothing to "extend".

Slereah

There's like ten homogeneous vacuum solutions

In 4D

In 2D it's like

Flat space

Cylinder

Other cylinder

Torus

Klein bottle

Other Klein bottle

Projective plane

I think, anyway

Might be a little more

John Rennie

@Slereah: is that a reponse to my post?

Obliv

@acuriousmind well the orbit $\mathcal{O}$ is an equivalence class under an equivalence relation. Since the set $\mathcal{O}$ is made of elements that satisfy this relation, I think of it as those elements, which are mapped, having to satisfy this relation. I just wrote it as being mapped to it to save time

Slereah

15:26

somewhat

Obliv

Alright thank you :)

Slereah

I think those are the only homogeneous flat solutions

John Rennie

@Slereah I don't see how that answers my question

ACuriousMind

@Obliv I have no idea what you just said.

Slereah

I dunno mang

What is your question exactly

I'm not quite sure

John Rennie

15:28

@Slereah There are vaccum solutions that don't have constant curvature even though the vacuum is isotropic and homogeneous. Yes?

Slereah

They have constant curvature

Obliv

@acuriousmind $\mathcal{O}_x = \{a,b,c,...~|~x$~$a\}$, right? So if $H \to \mathcal{O}_x$ then $h \in H$ must satisfy the condition in the set $\mathcal{O}_x$?

Slereah

Well, they are Ricci flat, anyway

Obliv

So I just saved time by saying $h \in H$ was being mapped to $x$~$h$

John Rennie

@Slereah The Kretschmann scalar is not constant e.g. in the Schwarzschild metric

Slereah

15:29

If you mean the Riemann tensor, then yes, you can have vacuum solutions that are not flat

Secret

People often said quantum states are vectors in hilbert space, but why there isn't any states that acts like the zero vector, i.e. when superimposed onto another state, does not change the state?

Slereah

That's because in >2+1D, the Riemann tensor does not depend exclusively on the Ricci tensor

The 0 vector doesn't have norm 1

Quantum states are rays

0 is not a ray

ACuriousMind

@Obliv "What?" to the first question. "If $H\to\mathcal{O}_x$" doesn't make any sense,$A\to B$ is not a statement, and no, if $h\sim x$ for some $h\in H$, then $\mathcal{O}_x$ is the equivalence class of the identity.

You have to be way more careful with the way you state things in math.

John Rennie

@Slereah So can we have solutions for an isotropic and homogeneous non-vacuum solution that do not have constant curvature? i.e. the various scalars are not constants.

Slereah

Sure

The De Sitter Schwarzschild solution, for instance

Wait, no, it's not homogeneous

Wait if it's isotropic and homogeneous and non vacuum

It's gonna be either dS or AdS

John Rennie

15:32

@Slereah why?

Slereah

Well you can check it by hand

Apply all the Killing vectors

John Rennie

The vacuum can be Minkwski, or Schwarzschild or Kerr or etc etc

Slereah

You'll find out that the Riemann tensor is entirely determined by a scalar

But Schwarzschild isn't homogeneous

John Rennie

Why if you add a small constant energy density is the FLRW metric the only solution?

Slereah

$r$ isn't a Killing vector at all

Obliv

15:34

@acuriousmind by definition if the set $H$ is being mapped to a set that imposes a condition, the elements in the set $H$ must satisfy this condition in order for this to be a bijection, right? Also $H \to \mathcal{O}_x$ is supposed to signify mapping of $H$ to $\mathcal{O}$. Lastly, $h$~$x$ must be true in order for the $h$ to be in the orbit of $x$ I thought

ACuriousMind

@Obliv nononononono. You can map the odd numbers to the even numbers by adding 1. That doesn't mean that odd numbers satisfy the condition of being even.

John Rennie

Suppose you make the stress-energy tensor diag(e,0,0,0) for some small constant e. In the limit of e=0 we get all the vacuum solutions. For non-zero but infinitesimally small e the only solution is FLRW. Is that correct?

ACuriousMind

And I know that $H\to\mathcal{O}$ is supposed to "signify a mapping", it's just that "If $A\to B$" is not a meaningful statement. "If $f:A\to B$ exists" or "If $f:A\to B$ is a bijection" works, but "If $A\to B$" just doesn't mean anything.

Slereah

Errrr maybe?

John Rennie

Insufficient effort?

0

Q: What is the volume of a 4d sphere

What is the volume of a 4d sphere? I've seen so many site that would have answered this question, but all of them have so many numbers that some people (including me) don't understand. So I was hoping somebody here would give me a nice simple explanation.

volume

Slereah

15:39

Constant dust solution would be FRLW I think yeah

ACuriousMind

@JohnRennie Not about physics, homework-like question, no effort, take your pick :P

Obliv

Wait.. so $f: H\to \mathcal{O}_x$ is a bijection isn't implying that $x$~$h$ for all $h \in H$? So there is no significance in $\mathcal{O}_x$'s relation being $x$~$h$ iff $x = hb$ as long as it's an equivalence relation and this map $f$ is a bijection, $|H|$ divides $|G|$? I feel stupid now @acuriousmind

ACuriousMind

@Obliv I'll state this once again properly: The bijection is given by $f : H\to\mathcal{O}_x, h \mapsto f(h)=hx$ (prove that this is a bijection!). This means that $\lvert \mathcal{O}_x\rvert = \lvert H \rvert$ for all $x\in G$. Since $G$ is the disjoint union of $n$ such equivalence classes, $\lvert G\rvert = n\lvert H \rvert$.

Don't try anything else until you have understood this argument.

John Rennie

@Slereah well yes. When I started thinking about it maybe was the best I could come up with as well.

GPhys

so a little bird told me I should watch the new gravity wave conference today

John Rennie

15:45

@GPhys Absolutely, more merger detections!! (Allegedly :-)

Though it doesn't start for 90 minutes yet.

Slereah

Pretty sure that all tensor fields related to the metric are identical for a homogeneous and isotropic spacetime, though

GPhys

fun

John Rennie

@Slereah and yet the vacuum is homogeneous and isotropic ...

Obliv

What I'm about to say might sound really stupid and you might get mad at me @acuriousmind but I must say it. It's a bijection because by definition $f(h) = (x,h) \in \mathcal{O}_x = x$~$h = hx$ and all of the domain $H$ is being mapped to the equivalence class so all $h\in H$ map to $hx$. What's wrong with this reasoning?

Slereah

@JohnRennie No

There are plenty of vacuum solutions which are not

John Rennie

15:51

You misunderstand me. In a vaccum the stress-energy tensor is isotropic and homogeneous, and in an FLRW spacetime the stress-energy tensor is isotropic and homogeneous. So why are there many vacuum solutions and only one FLRW metric?

Slereah

Oh, you mean the stress energy tensor

Hm

Also when you say "isotropic and homogeneous", you mean just the spatial part, right?

John Rennie

@Slereah yes

Slereah

I'd say there are more than one solution type for the same homogeneous and isotropic stress energy tensor

Symmetries of the SET are symmetries of the Ricci tensor

Those will only determine the Ricci tensor

The Weyl tensor is still free

KISHORE_ZE

hello

John Rennie

@KISHORE_ZE Hi. Come for the gravity wave chat?

KISHORE_ZE

15:56

I suppose

LIGO Gravity WAve lab press conf.

ACuriousMind

@Obliv I have no idea what $f(h)=(x,h)\in\mathcal{O}_x = x\sim h = hx$ is supposed to mean, and I don't understand your reasoning at all. What is $(x,h)$?! Why is it $f(h)$, or an element of $\mathcal{O}_x$.

KISHORE_ZE

Wave*

I'm new to stack physics

why dosen't the formatting show up for me?

0celo7

@JohnRennie the assumption of homogeneity is cosmology

Schwarzschild is not a cosmological model

John Rennie

@KISHORE_ZE Mathjax isn't supported in the chat by default.

0celo7

@ACuriousMind damn you're pretty harsh

KISHORE_ZE

15:58

Oh

Thanks

John Rennie

@KISHORE_ZE see meta.stackexchange.com/questions/97938/…

0celo7

@JohnRennie homogeneous basically means that you can find coordinates in which the metric components are constant

@JohnDuffield still does not understand this

Slereah

He's not talking about symmetries of the metric tho

Only of the SET

John Rennie

@Slereah Yes!!! (SET = stress-energy tensor)

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