The h Bar

0celo7

3:05 PM

@JohnRennie (removed)

Obliv

@0celo7 you know how groups have multiple presentations?

John Rennie

@Obliv presentations or representations?

Obliv

presentation, not sure what representations are. @johnR but it's not a problem I figured it out. Was wondering what determined if two presentations were equivalent. (i had a different presentation for a group than my book)

ACuriousMind

The problem of deciding whether two presentations are of the same group is actually undecidable (and equivalent to the halting problem) in general

Obliv

well then asking for a presentation for a group is a bad exercise problem, then @acuriousmind

ACuriousMind

3:20 PM

@Obliv Well, I'd consider it a boring exercise, but it's not a bad one - for almost all group you will ever encounter and all presentations you can write down for them a human will sooner or later figure out whether they are the same or not.

That the problem in general is undecidable just means that one can't give a single terminating algorithm that always works for that.

Obliv

oh you're saying you can check it yourself though?

to see if they produce the same group?

Danu

@ACuriousMind A bold statement.

ACuriousMind

@Obliv I'm saying that although there is no general algorithm, it's not actually that hard to determine whether groups are isomorphic or not in most cases.

@Danu The crucial part is sooner or later. I'd probably not give it as an exercise on an exam :P

John Rennie

@ACuriousMind Can't you use the first presentation to generate a group A, use the other presentation to generate a group B, then check to see if B is a representation of A or vice versa?

ACuriousMind

@JohnRennie What do you mean?

A representation is usually on a vector space

John Rennie

3:33 PM

Well a presentation is a compact way of writing down a group (I just looked it up on Wikipedia)

So ifyou start with the presentation you can generate arepresentation of the group.

ACuriousMind

If the groups are finite, you can of course just generate the group tables and check if any permutation of the rows/columns (or relabeling of the elements) makes them the same.

But if the groups are infinite, you can't give a terminating algorithm for the general case

John Rennie

Ah, yes, the key word is finite

I was forgetting about the infinite groups.

ACuriousMind

@JohnRennie A representation of a group is a group homomorphism from the group $G$ to the group of linear operators on a vector space.

Danu

@ACuriousMind Even some really simple groups like the fundamental group arising from excising a thickened torus knot from $S^3$ can give problems.

My TA (a postdoc working on geometry/topology) got stuck for half an hour trying to prove it was isomorphic to something simple, but didn't manage.

ACuriousMind

"simple group"..."excising a thickened torus knot from $S^3$"

Doesn't sound so simple to me because I have no idea what that means! ;P

Danu

3:36 PM

@ACuriousMind but the presentation is very simple. Something like $\langle a,b \mid a^{-2}b^3\rangle$ for the trefoil knot.

John Rennie

@ACuriousMind according to Ash and Gross: A group representation is nothing more nor less than a morphism from one group to another group.

Danu

The "another group" being $GL(V)$, yes.

John Rennie

That's a linear representation

ACuriousMind

@JohnRennie That's...an unusually broad definition of a representation.

Emilio Pisanty

Hey guys, check it out

doai.io

John Rennie

3:38 PM

Hey, I'm just copying and pasting :-)

Emilio Pisanty

> DOAI (Digital Open Access Identifier) is an alternate DOI (Digital Object Identifier) resolver that takes you to a free version of the requested article, when available.

> To use it, replace dx.doi.org by doai.io in any DOI link.

6

Danu

...excellent.

FrancescoS

If you know.. are there any formulations for Clebsch-Gordan decomposition and coefficients of ISO(2)?

Emilio Pisanty

@Danu Yeah, it's really good. I might make a community ad for it, come to think of it.

Danu

You want to decompose tensor representations of ISO(2)?

FrancescoS

3:43 PM

@Danu yes

Obliv

are homomorphisms kind of like 'local' isomorphisms?

Danu

@Obliv No.

They are structure-preserving maps.

ACuriousMind

@FrancescoS Have you looked at what dimension an irrep of ISO(2) has?

Danu

Note that (for groups), the trivial map into any other group is a homomorphism.

user116211

@EmilioPisanty Is it like Sci Hub?

Obliv

3:44 PM

what is the trivial map? the identity map? @danu

Danu

No, the constant map to the identity.

Obliv

uhm

ACuriousMind

@FrancescoS My point is supposed to be that the irreps of ISO(2) are one-dimensional. Tensoring 1D reps with each other just gives other 1D reps, there are no coefficients to compute.

FrancescoS

@ACuriousMind I am interested in 1/2 massless particles. So, I don't know which dimension has this irrep.

Obliv

what I mean is, homomorphisms are structure-preserving maps and isomorphisms are bijective homomorphisms, correct? So I feel like homomorphisms don't cover the whole group whereas the isomorphism does. That's what I meant by local. Is this line of reasoning correct @danu

FrancescoS

3:45 PM

@ACuriousMind can you give me a good reference? I am not really good with this argument

Danu

@Obliv No, because homomorphisms can fail to be injective, too.

What you're talking about is called an embedding (=injective homomorphism)

Obliv

I suppose if the homomorphism was surjective and not injective, then this description wouldn't work

ACuriousMind

@FrancescoS The construction of all the representations of relevant to the fundamental particles is done in Weinberg's QFT I

Slereah

@ACuriousMind no it's not

He lists them, he doesn't construct them!

ACuriousMind

He does? I must have imagined all those pages full of horrible computations about the little groups and so on then.

Slereah

3:51 PM

Mb

vzn

We want information / economist covers new paper by Hawking on black hole info paradox, via reddit

↑ & was someone just talking about the hairy ball thm? :P

FrancescoS

@ACuriousMind I'll check there. thanks

John Rennie

@vzn this has been around for a while hasn't it? Somewhere there's a video of a presentation by Malcolm Perry explaining the idea.

This one I think. It's an enjoyable talk.

05. Malcolm Perry - Black Hole Memory

►

Danu

That meta question by Chris is interesting, by the way.

vzn

@JohnRennie seems to be a new paper by coauthor Strominger re "soft particles". cant figure out which hawking paper theyre talking about yet. anyway what do you expect from an economics pub? :P

John Rennie

4:03 PM

@Danu I have to say I disagree. It's going over the same ground as the previous question on the subject.

Obliv

is a classification theorem just a theorem that predicts when groups are of the same isomorphism type based on some properties shared between them?

Danu

@JohnRennie But the answers it's getting are interesting. I also think some people (e.g. CuriousOne) have been a bit too aggressive, lately.

John Rennie

@vzn journals.aps.org/prl/abstract/10.1103/PhysRevLett.116.231301

vzn

> The wrinkle the new paper introduces is that the behaviour of the now-real Hawking-radiation particle is affected by the soft particles it encounters when it materialises. This causes the modulation that preserves causal determinism.

John Rennie

@Danu really? The top question is me doing my usual bleating about how ungrateful the site members are, and I didn't think the other two said anything especially incisive.

@Danu but I agree with your assessment on the contribution to the site made by CuriousOne's comments ...

Emilio Pisanty

4:08 PM

@MAFIA36790 No. It looks to see if there are legal copies, and if it doesn't it will send you to the paywall.

user116211

ohh... yeh you should make it a community ad @EmilioPisanty

Emilio Pisanty

@JohnRennie Just for that you get flagged as British, I'm afraid.

(If we didn't already know.)

John Rennie

@EmilioPisanty it's my upbringing - I struggle say what I really think of people. Fortunately my views are usually easy enough to guess :-)

Emilio Pisanty

@JohnRennie Yeah, it's the Euphemism Country thing.

In the same spirit as this (from my recent facebook posts)

There's some guys playing loud music on their phone in the train; I'm considering whether to use the nuclear option and clear my throat on top of looking at them with an annoyed look.
Man, Britain damaged me.

John Rennie

@EmilioPisanty :-)

Emilio Pisanty

4:14 PM

@JohnRennie Luckily I managed to escape before the damage was too bad. I'm now on my way to potentially being brusque with a stranger at some point in the future.

John Rennie

4:27 PM

Does taylor expansion of covariant derivative mean anything?

Mark Mitchison

I don't think so

It's a bizarre question. But a Taylor expansion is a property of a function, not of an operator on the space of functions.

I mean that the OP's post is bizarre. @JohnRennie your question is only as bizarre as the one which inspired it :) I voted to close since it appears to be a nonsense question.

Emilio Pisanty

@Mark look what the cat dragged in

;-)

Mark Mitchison

Best place to procrastinate bro

Or "a" place, anyway

How's the write-up going @EmilioPisanty?

John Rennie

@MarkMitchison I didn't think so either, but I'm not confident enough to VTC without checking.

Emilio Pisanty

@MarkMitchison Slow

Or rather, it's been on hold for a couple of months

I'm gearing it back up now

Or I'm meant to, anyway

Mark Mitchison

4:33 PM

@EmilioPisanty Fair enough. I'm starting to feel a bit panicky to be honest. We only have until September and it's quite a big job

But of course we'll manage. I'm glad there's other people in the same boat anyway.

Emilio Pisanty

@MarkMitchison I think I'll manage

Or at least I'm not panicking yet

Mark Mitchison

Wise

Emilio Pisanty

If you're worried then I might start getting worried too, though

Obliv

for a general group $G$ and an element in $G$ ,say $x$, is $x^0 = e$ in general? Raising to the 0 power is the same as $xx^{-1}$?

vzn

@MarkMitchison hi thx for your interest in the guest speaker series... would you consider it yourself sometime?

Mark Mitchison

4:34 PM

@EmilioPisanty Nah, don't worry. My biggest problem is that there is no overall theme to my research. I suspect yours will be a bit easier to sell as a coherent package.

Emilio Pisanty

@MarkMitchison Yeah, sort of.

We gave up on trying the coherent package and now I've got an explicit Part 1 and Part 2

but it's sort of coherent, yes

Mark Mitchison

@vzn Not sure what you mean about my interest. Are you talking about the AMA sessions? I literally learned what that stands for within the last hour

vzn

@MarkMitchison lol (yes AMA). you commented "great idea" in meta awhile back. despite widespread enthusiasm its proving not exactly easy to line up guests :|

Emilio Pisanty

@vzn and the surprise is...?

As I said at some point, you need to be really proactive at finding speakers (because it won't be easy to get people to agree to the spotlight)

Mark Mitchison

@EmilioPisanty Haha yeah the two part strategy seems to be a popular one. I'm doing the same. The problem is I have no idea what to write for the overall introduction. The common theme is "cold, trapped single atoms". How the hell am I supposed to summarise the current status of research on "cold, trapped single atoms" and place my research in context!? The very idea seems laughable.

vzn

4:38 PM

@EmilioPisanty hey the site is so utterly compelling, one would think that very high profile guests would be banging down the door to get in right? (dont recall you saying itd be not easy)

Emilio Pisanty

@MarkMitchison did cause me to laugh out loud just now

Mark Mitchison

Yeah

I can tighten it to cold and trapped single atoms, but still

vzn

@MarkMitchison not following, why is it laughable? is it too big a field? there are surely surveys etc

Emilio Pisanty

@MarkMitchison Just set a limit on wordcount and on number of references

Pick the $N$ best papers from your bigger list

Mark Mitchison

@vzn It's just very, very broad. And not something I have sufficient expertise to comment on. I haven't been alive for long enough, for a start...

Emilio Pisanty

4:40 PM

and write about those

vzn

@MarkMitchison the field is not that old. less than a few decades right? agreed it has exploded in "short time" though

Mark Mitchison

@EmilioPisanty You have a list? I'm just making it up as I go along...

Emilio Pisanty

@MarkMitchison On the whole thing? god, no

Mark Mitchison

@EmilioPisanty OK that's a relief

@vzn Well I'm only a few decades old ;) But I've only been doing research for about 5 of those years

vzn

@MarkMitchison gotta start somewhere right? guest speaker on internet would be cutting edge & will give you more confidence =D

Mark Mitchison

4:43 PM

@vzn But the point is more that it's ridiculously broad. My main problem is that I have two separate research themes whose connecting thread is tenuous as hell.

vzn

@MarkMitchison which 2 threads? there are a lot of "tenuous connecting threads" in QM/ research :|

Mark Mitchison

@vzn Ha, yeah OK. I don't remember ever making that comment, but I would be up for helping as long as it doesn't take too much time.

What would it involve?

Emilio Pisanty

@Mark On the other hand, the putting-the-thesis-on-hold thing definitely paid off

vzn

@MarkMitchison cool/ thx =D its really up to the guest to decide how much to put into it, we are keeping it very informal to start, think the fmt is intrinsically not amenable to a lot of preparation. Daniel Sank/ Google QM computing has committed. the talk/ session is only 1 hr & think it will likely go by fast.

Mark Mitchison

@EmilioPisanty Congrats. I look forward to not understanding it :P

Emilio Pisanty

4:46 PM

@MarkMitchison =D

@MarkMitchison I think you'll be able to get close to most of it.

Mark Mitchison

@vzn OK, well then I'd be happy to do it. The idea is basically just to hold a chat session where I try to answer questions, right?

@EmilioPisanty Strong field physics, I assume?

vzn

@MarkMitchison exactly. thx for the flexibility :)

Emilio Pisanty

@MarkMitchison Yeah, HHG. But it's a four-pager (nudge nudge), so if you're confused by it then the text isn't clear and accessible enough.

Mark Mitchison

ha

Emilio Pisanty

Anyways

vzn

4:49 PM

@Mark we can discuss it more in detail over coming wks. we have speakers for next few mos or so. slereah will be the 1st & yuggib is next, then DS. you can check out their sessions, hope you join, & we are planning an evaluation with DZ afterwards. & will probably write up more in a blog at some pt etc. also, there is quite a bit of meta content already.

Emilio Pisanty

Can I blatantly solicit votes for this one?

0

A: Community Promotion Ads - 2016

3

Mark Mitchison

@vzn OK, sounds good.

vzn

@EmilioPisanty am now going to put lots of hostile comments on that :P

← j/k, a softie, +1

Emilio Pisanty

@vzn yeah, obviously not a problem

No mysterious downvotes will appear on your account in the coming days

Or actually

That's probably more than anyone can promise

0celo7

@Obliv I'd never even heard of a group presentation

vzn

4:55 PM

@EmilioPisanty lol ofc all the copious downvotes will be un mysterious

0celo7

@ACuriousMind Would you accept $F\mid M$ for $F$ restricted to $M$?

or is only $F|_M$ acceptable

user116211

@EmilioPisanty voted; just 1 to go.

vzn

@0celo7 must be wrong then right? :P

0celo7

@vzn It wouldn't be the first mistake I've found in the book.

But HE is like that, you can tell when a part was written by one of them

Some parts are definitely less clear than others

vzn

@0celo7 lol, that was a joke. you think you can find mistakes in GR after studying it ~1 yr? and you think hawking/ ellis made one... because nobody else has anything like it? and how many ppl have called you a troll on here? :P

0celo7

5:02 PM

shrug

Whatever

I never said it was a mistake

vzn

yeah whatever

← read a ----load of 0celo7 and never seen anyone approach it like that and after a year have no clue what it's about

John Rennie

me neither - I had to google it.
And now ACM tells me I don't know what a *representation* is either :-)

0celo7

@JohnRennie a representation is a homomorphism $\rho:G\to\mathrm{GL}(V)$ for some vector space $V$

John Rennie

Isn't that a linear representation?

0celo7

there are others?

John Rennie

5:07 PM

Where linear means to do with matrices

0celo7

I work with Lie groups

everything is linear

and smooth

John Rennie

The word linear doesn't mean anything to do with lines. It's an adjective mathematicians use when thet mean to do with matrices

0celo7

@JohnRennie ...I know

...are you seriously telling me linear doesn't mean lines?

John Rennie

What about a non-faithful representation? Does that still have to be a mapping to a general linear group?

0celo7

beats me

I'm not an algebrist

never claimed to be one

John Rennie

5:11 PM

Ash and Gross say, and I quote:
A word about terminology: It gets boring using the word matrix both as a noun, naming the entity we are soon to define, and as an adjective. So we have a synonym for the adjectival use: linear. A linear representation is just a matrix representation, and a linear group is just a group of matrices. BEWARE: The adjective “linear” has many other meanings, such as “in the shape of a straight line,” and it has other uses in mathematics and even in representation theory. We hope these multiple meanings will not be a problem in the explanations to come. We will usua…

Emilio Pisanty

@MAFIA36790 Oh wow, that's the fastest zero-to-live dash I've seen a community ad do, like, ever

0celo7

@JohnRennie Well, that terminology is because the linear maps $\mathrm{End}\,\Bbb R^n$ are in a bijective correspondence with the matrices $\mathrm{Mat}(n)$

Actually the linear maps $\Bbb R^n\to \Bbb R^k$ are in a bijective correspondence with the $n\times k$ (or $k\times n$, can't remember which) matrices

I think it's $k\times n$

John Rennie

I can see there are dangers in reading books on number theory

0celo7

@ACuriousMind Wow Milnor is really getting interesting

@JohnRennie I could have told you that

vzn

@EmilioPisanty aka teamwork (in chat) =D

0celo7

5:16 PM

read topology with me :)

ACuriousMind

@JohnRennie Non-faithful just means that the representation is not injective, i.e. that two or more group elements are sent to the same target. "Representation" is usually (except when qualified as a projective representation) indeed reserved for group homomorphisms $G\to\mathrm{GL}(V)$ for some vector space $V$. If $V$ is finite-dimensional, this is indeed "matricial", but it is not for general vector spaces.

John Rennie

I suspect I am drawing too many conclusions from reading too few maths books. I should probably shut up until I've read some more.

ACuriousMind

I wouldn't call it a mistake of your book to define a more general notion of representation, but it will confuse you when you read other sources who don't bother prefacing the representation with "linear".

0celo7

@ACuriousMind I still have zero clue what a projective rep is

what the the hell does $O(x^k)$ mean again?

ACuriousMind

Use the Google.

0celo7

5:30 PM

is $f(x)=O(x^k)$ if $\lim_{x\to\infty} f(x)/x^k=0$?

or is it a constant

not zero

@ACuriousMind ...what do I google

or maybe when $x\to 0$

help

ACuriousMind

@0celo7 "Big O notation"

0celo7

...it has a name?

ACuriousMind

Also called "Landau notation/Landau symbols"

Bernard Meurer

5:49 PM

We use big O for complexity notation in CS

Emilio Pisanty

6:07 PM

oh, bejeesus

> the apparatus, which we name MAZEL-TOV for MAch-ZEhnder-Less for Threefold Optical Virginia spiderwort

This has to stop

3

Mike Miller

@ACuriousMind Did you figure out $\delta^2$?

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

@MikeMiller o/

ACuriousMind

@MikeMiller Not yet...I'll ask the lecturer tomorrow what exactly that was supposed to mean

Mike Miller

Let me know when you do, I'm pretty curious

ACuriousMind

@MikeMiller Will do

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

6:20 PM

@ACuriousMind have you changed your opinion about the validity of the abbreviations +ve and -ve :P

ACuriousMind

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ What? I still don't understand why anyone would put the "ve" there, if that's what you mean

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

I guess I do it because I'm lazy @ACuriousMind

ACuriousMind

But why not just "+", then?

That's even lazier

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

But I want to abbreviate the "word" positive, not the symbol for plus.

Emilio Pisanty

@ACuriousMind Don't be so -ve about it.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

6:29 PM

Ex. Positive numbers = +ve numbers

+ numbers would mean plus numbers to me

ACuriousMind

@EmilioPisanty ::twitch::

Emilio Pisanty

bingo

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

:D

Wouter

Anyone know a good (application-oriented) optics book at an advanced level?

(Hello everyone, by the way :P )

Physics Guy

6:48 PM

@JohnRennie Yes, I want. It seemed noone has understood my question.

Slereah

7:03 PM

Hello

So Visser defines $$A(s) = P\{ \exp[\int_\gamma (\vec \nabla \otimes \vec t) ds]\}$$

With $P$ the path ordering operator

the fuck is path ordering

ACuriousMind

@Slereah Essentially a Dyson series: $P\exp(\int_\gamma A) = 1 + \int_0^1 A(t_1)\mathrm{d}t_1 + \int_0^1\int_0^{t_1} A(t_1)A(t_2)\mathrm{d}t_2\mathrm{d}t_1 +\dots$

It's exactly like time-ordering for the usual Dyson series, just the "time" is the parameter of the path $\gamma$ instead

Slereah

Aight

errr

Physics Guy

What are you talking about ?

Slereah

shouldn't the third term be $\int_{t_1}^1 \int_0^{t_1}$

ACuriousMind

@Slereah Nope.

Slereah

7:15 PM

Also Visser references HE

And I left mine at home :V

0celo7

@ACuriousMind How on earth does one orient $[0,1]\times M$?

given an orientation on $M$

$M$ is boundaryless

Slereah

Take the orientation form, multiply it by a 1-form of $[0,1]$?

0celo7

I'm assuming one orients $[0,1]$ simply by a right-pointing vector

@Slereah not via forms

ACuriousMind

@0celo7 By using the general orientation of a product?

0celo7

@ACuriousMind Milnor probably forgot to include that

He says "orient $[0,1]\times M$ as a product" but never explains what that means

@ACuriousMind Basically, I need to show that $\partial([0,1]\times M)$ is two copies of $M$, one with the correct orientation and one with the incorrect orientation

this is kinda clear when I draw a picture

I think the issue is that he defined the boundary orientation heuristically

John Rennie

7:31 PM

@PhysicsGuy Hi, are you still around?

0celo7

I don't think "points outward" is rigorous

Physics Guy

Yes

0celo7

Unless you do it like in my stokes theorem answer.

John Rennie

@PhysicsGuy OK. I think what you're asking is the relationship between strings and particles i.e. how strings manage to look like particles. Is that correct?

Physics Guy

I was wondering what a particle is in context of string theory, yes.

John Rennie

7:33 PM

@PhysicsGuy OK. First you need to know what a particle is in the context of quantum field theory, because that's what string theiry reproduces.

When we quantise a field we write it as a sum of quantised simple harmonic oscillators.

We end up with a vacuum state and an (infinite) array of creation operators.

Physics Guy

I (think I) know that. the unification of field theory and classical quantum theory, by second quantisation.

John Rennie

Each creation operator acts on the field to produce a particle of a certain frequency. So the particles are in a sense just excitations of the quantised field produced by a creation operator.

Physics Guy

Ok. Related to vacuum fluctuations ?

ACuriousMind

@JohnRennie It acts on the vacuum to produce particles, not on the field. And it would be better to say "momentum" instead of "frequency" because there is no actual frequency involved.

John Rennie

@PhysicsGuy No, it has nothing to do with vacuum fluctuations. And yes ACM is correct and I should really talk about momentum rather thn frequency.

But the point is that we can think of these operators as being arranged along a single axis.

Physics Guy

7:38 PM

I just knew that it is a specific solution of Schrödinger-equation.

John Rennie

In the positive direction we get particles of increasing momentum while in the negative direction we get antiparticles of increasing momentum.

Now, we do a similar sort of thing when quantising a string.

Again we get a vacuum state and some creation operators that act on the vacuum to produce states of the string.

Incidentally I'm going to talk about the bosonic string because I understand supersymmetric strings even less than I understand bosonic strings.

ACuriousMind

@JohnRennie Hmmmmmmm

John Rennie

@ACuriousMind go with me on this one :-)

Physics Guy

It is dificult to understand, because I am a foreigner, but please go on.

John Rennie

But when we look at the creation operators for a bosonic string we find they naturally fall into a different pattern.

The operators can be arranged as a 2D array not just along a line.

ACuriousMind

7:42 PM

@JohnRennie How did you get to that picture? The existence of distinct antiparticles is when you get two sets of different creation/annihilation operators when expanding the field (due to it being complex or spinorial) Are you trying to appropriate the Dirac sea picture here?

John Rennie

The width of this array is the number of spacetime dimensions - 2, which for a bosonic string is 24. The height is infinite, just like in the particle case.

And this structure is equivalent to the product of 24 particle like states.

So we find that the creation operators for the string look like a combination of the creation operators for lots of different particles.

And that's how we get particles from strings.

Slereah

But aren't string particles all bosonic :O

and you need SUSY shit to get actual particles

John Rennie

It's because there are states of the string that behave like the states we get in QFT by quantising a field.

QED

At this point the audience are stunned into silent disbelief

DanielSank

@EmilioPisanty What's that from?

Slereah

You know what's great?

Misner space.

DanielSank

7:47 PM

@JohnRennie ::says nothing because stunned::

Slereah

Misner space is so bloody great for calculations

John Rennie

I skimmed Zwiebach last night. This string theory stuff is a piece of cake. I don't know what all the fuss is about :-)

Physics Guy

@JohnRennie Ok, thank you for your description. I think I have understood it more. I just was not clear about a few things. Greetings from Germany.

John Rennie

You might want to wait for ACM's opinion before deciding what I've said actually makes sense :-)

ACuriousMind

@JohnRennie I'd say it's not wrong, but I'm undecided whether it conveys useful information :P

That said, I'm also undecided whether actual string theory conveys useful information ;)

John Rennie

7:52 PM

@ACuriousMind actually I'm relieved to have even got that far :-)

Anyhow this has been a flying visit. I'm at work in eight hours and seven minutes so I'm off to bed. Tomorrow I shall explain the answer to life, the universe and everything.

Slereah

Hm

John Rennie

@ACuriousMind admit it, that was quite an explanation from someone who knows bugger all about string theory and almost bugger all about QFT :-)

Slereah

To show chronology protection, Visser uses the Hadamard form, to show that the contribution from the CTC geodesics fuck up the divergence

But Robert Low's spacetime has no CTC geodesics

Which means...

I'll have to do the real calculations to check if it's divergent or not

blah

Physics Guy

@JohnRennie@ACuriousMind For me, it does, because I was just confused about some definitions, that helped me a lot to "get back on the right train".

ACuriousMind

@JohnRennie You are uniquely gifted :)

vzn

7:57 PM

@JohnRennie have long wondered, does string theory have any explanation/ model for (std theory) particle masses?

John Rennie

@vzn no. All the observed particles belong to the lowest state of the string and all those modes are massless. We need a higgs mechanism to get the masses.

I suppose in principle string theory should be able to tell us the Higgs couplings ...

vzn

@JohnRennie ok, does higgs have any explanation for different particle masses?

Physics Guy

Spontaneus symmetry breaking. Coupling with Higgs-field. that stuff.

John Rennie

@vzn no. The Higgs couplings are arbitrary parameters in the standard model

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