The h Bar

Slereah

12:20 AM

I have proven that an apple falls at 9.8 m/s² in Schwarzschild spacetime

2

Dawood ibn Kareem

3:12 AM

Just one apple, or all apples? And did you eat it afterwards?

peterh

3:36 AM

@DawoodibnKareem Wow, are you really from .ps?

Avantgarde

3:47 AM

zup

user228700

5:08 AM

@JohnrR: Morning! :-)

user228700

user228700

Yay! Ze flower!

John Rennie

Aha, you got your pictures of flowers transferred! :-)

That looks as if you've had a bit of rain ...

user228700

Nope. I'm using my mobile to upload it. The dumbass that I am, I didn't think of that yesterday.

user228700

@JohnRennie Yep, we most certainly have! In fact, it's drizzling right now!

John Rennie

5:12 AM

This week is forecast to be very wet in the UK. Nothing to do with the monsoon, just regular British summer weather :-)

user228700

I have mushrooms and cows and goats as well!

user228700

@JohnRennie Ah :-)

John Rennie

@Kaumudi.H All taken in Kerala?

user228700

@JohnRennie All taken here, yep!

John Rennie

5:17 AM

How did the train work out BTW?

user228700

Still in Kerala, Sir! Train's at 9:50 PM this evening.

John Rennie

Just you on the train?

user228700

5:28 AM

@JohnRennie Yep, why?

user228700

(Sorry about that; was devouring mangoes for breakfast)

John Rennie

I had mango on Saturday!

The local supermarket had them on offer.

user228700

@JohnRennie Nice! :-) These were freshly plucked just a few days ago!

John Rennie

Sadly the mangos from the supermarket are nowhere near as nice as freshly picked ones.

We used to get fresh mangos when I lived in the Sudan.

user228700

Right :-(

user228700

5:32 AM

@JohnRennie Mangoes in Sudan?! Cool!

John Rennie

The trouble is the mangos have to be picked unripe so they don't go bad while being shipped to the UK. They do ripen in transit, but they aren't as nice as mangos picked when they have ripened on the tree.

user228700

Ah, I see. Well, the ones I just had weren't picked ripe: since we're leaving today/tomorrow, my grandma had them picked a bit raw so that they would half ripen before we left.

John Rennie

@Kaumudi.H I don't think mangos are native to the Sudan because it's very dry and mango trees need a lot of water. But they are farmed there - or were.

user228700

They weren't sickly sweet but they tasted just fine!

user228700

@JohnRennie Right, right, which is why I was curious about that. I see...

user228700

5:36 AM

I'm excited about travelling alone tonight! Never done that before!

John Rennie

I suspect it will be exciting for a few minutes, then very boring!

Take a good book :-)

user228700

I agree :-P I might not be able to read; all the lights will be off :-/

user228700

I'll climb in and try to go to sleep!

John Rennie

I need to manually update the software on some servers. I've just run a query to check how many servers are affected, and there are 127 of them!!! That will keep me busy for a while.

user228700

Oh, sheesh, wow.

user228700

5:46 AM

OK, have fun!

John Rennie

Probably not that much fun :-) Oh well.

user228700

Ah :-/ I just got a shitton of money, BTW! Now would be a good time to order a harmonica and books!

John Rennie

6:16 AM

@Kaumudi.H In the UK second hand books are available from Amazon and eBay, and they can be very cheap. Is that the case in India as well?

Avantgarde

yeah

user228700

@JohnRennie Second hand? I'm not sure if they're second hand. The ones on Amazon anyway...

Avantgarde

Sorry for answering a question directed at you, @Kaumudi : P

user228700

And I'm not sure if I will find the specific books that I'm looking for on eBay...

user228700

@Avantgarde Lol, no problemo :-)

SBM

6:22 AM

school's over for today; just reached home.
Hello people

Avantgarde

hey SBM

user228700

Yello :-)

John Rennie

@Kaumudi.H it might be worth considering though. Although I prefer Amazon to eBay for buying second hand books.

user228700

Hmm. OK, I will look it up. Thanks :-)

SBM

@Kaumudi.H Yellow?

John Rennie

6:25 AM

Wow, it's just started raining cats and dogs in Wookey Hole :-(

Avantgarde

I just went through the amazon website of India to check for standard physics textbooks. There doesn't seem to be a section for selling used books

user228700

@SBM No, um, that was "Hello", disguised as a shortened version of the word "Yellow" because yellow is a happy color and...oh, never mind :-P

John Rennie

@Avantgarde yes, I just checked a couple of books that I know K has read and I couldn't find any second hand ones either.

user228700

@JohnRennie Not a fan of rain?

John Rennie

@Kaumudi.H no, rain just makes everything wet, muddy and dirty

Avantgarde

6:27 AM

Maybe not these particular books, but some others might be sold used (though I searched for the most standard references so it's already a bad start). Maybe there are different websites for this. Usual bookshops might have them. But it might be hard to find something elusive..

SBM

@Kaumudi.H Yellow reminds me of hufflepuff.

John Rennie

I guess I shouldn't complain since we need rain, but that doesn't mean I have to like it.

Avantgarde

I dislike rain too

user228700

@JohnRennie I agree with you 100%.

SBM

@JohnRennie True I hate summers but I also hate rain

user228700

6:28 AM

@SBM All the better for I am a ravenpuff!

Avantgarde

Aha! I found a used Weinberg's QFT Vol 1!

John Rennie

@Kaumudi.H Better than a Slytherclaw :-)

Avantgarde

at 25% of the new book's price. (Even after that it's not cheap at all lol)

SBM

I thought I was the only hufflepuff in here
Certain books are so costly.

John Rennie

Weinberg's QFT? That's a seriously tough book to read :-)

Avantgarde

6:30 AM

Oh yeah, I suppose.. It's definitely not the first (or second, or third..) book to pick for QFT

But it's a must-have. I presume the book is filled with nice personal insights and explanations not to be found anywhere else

kinda like the QFT analog of MTW's Gravitation. Though I've never read MTW.

user228700

@JohnRennie :-)

user228700

@SBM Nope, you have me for company! I'm a bit ambigious though, hence the "ravenpuff". I've been sorted into all four houses over the years but my values and passions align most with Ravenclaw and Hufflepuff in that order.

Avantgarde

@JohnRennie Have you read MTW? Or parts of it

John Rennie

@Avantgarde Parts of it.

Avantgarde

Did you like it?

John Rennie

6:34 AM

It is quite fun to read. It is written in a witty way.

SBM

$\large \ddot \smile$
MTW?

Avantgarde

I see. That's nice.

John Rennie

But it isn't for the beginner. The authors assume you're already pretty good at maths and it's written that way.

user228700

Damn, LaTeX still isn't working :-/

John Rennie

@SBM Misner, Thorne and Wheeler - Gravitation.

Avantgarde

6:35 AM

@SBM MTW = Misner, Thorne, Wheeler. The standard brick of a book on General Relativity

SBM

Okay two different things I see

@Kaumudi.H tinyurl.com/cfqcvpc

LaTeX rendering instructions

Avantgarde

It's more of a treatise than a book. It'd be a feat to actually read through the full book

user228700

Ooh, the updated one. Thanks! :-)

user228700

6:52 AM

@JohnR: Please ping me back when you're free(ish)!

John Rennie

@Kaumudi.H I'm freeish now. I'm doing bits and pieces at work but I'm interruptible :-)

user228700

Wokay! I was wondering about a book. Hang on, I'll message you on gchat.

Yashas

7:13 AM

Relativity 12a - gravitational waves I

►

At 4:03, does that really happen?

why are particles even going near the object

nvm

he acknowledges (at 4:44) that there are problems with that explanation

Emilio Pisanty

7:52 AM

Any mods around?

SBM

I'm afraid I don't know.

Shing

8:12 AM

I am doing some refreshing on Analytical mechanics,

almost all results in my book is based on an approxmation:

1st order approxmation

I wonder whether there is something interesting going on in higher order (say, strange result)

or the author just too lazy to give a complete proof.

Avantgarde

which book?

Shing

@Avantgarde David Morin's Introduction to Classical Mechanics

Avantgarde

Okay.. I've never heard of that one

Shing

@Avantgarde it has the best problem-set; however, the reading is a bit average.

and it has very good physics jokes

Avantgarde

haha

I did my learning from Goldstein

which was a hard book

but quite nice

Shing

8:32 AM

I am thinking my next step

should be Goldstein, or Landau?

(classical mechanics)

Avantgarde

I haven't read Landau

Goldstein is a good book but it is tough to read. The problems are hard too

Fortunately, my lectures were pretty good. I only read parts of Goldstein

Dawood ibn Kareem

9:42 AM

@peterh No, I'm just a wee bit political.

ACuriousMind

@EmilioPisanty What's up?

Emilio Pisanty

@ACuriousMind reckon this one's ready to be reopened

-2

Q: Why does this paper report potential energy in kHz, and what does it mean?

I am interested in studying theoretically the double well system shown in the figure below, which shows an experimental demonstration of an atomtronic battery. Caption: Figure 1. (a) Schematic of the double-well atomtronic battery system. The top panel shows the longitudinal potential e...

I've got an answer ready and would like to close the tab

Slereah

what's the superspace in superstring theory

Is it the 2D space or the target space

ACuriousMind

@EmilioPisanty I concur; done.

Emilio Pisanty

@ACuriousMind excellent

thanks

ACuriousMind

9:47 AM

@Slereah The target space, the worldsheet has no fermionic components. You can view the action as a "modified supervolume" for the embedding $\Sigma^{2\vert 0}\to \mathbb{R}^{10\vert \mathbf{N}}$, where the $\mathbf{N}$ controls which of the string theories you get, this view of string theory is called the Green-Schwartz action

Slereah

Alright

Is it fair to call a string a timelike 2D submanifold?

ACuriousMind

I don't think there's any requirement it be timelike.

Slereah

whaaat

Well i guess it could be null

Oh wait is that why some people say there are tachyons in string theory

are there strings that span a spacelike submanifold

ACuriousMind

@Slereah No; bosonic string theory has tachyons simply because the ground state has negative mass.

Slereah

oh, that kind of tachyon

Boring

ACuriousMind

9:55 AM

@Slereah I am not sure that is a well-defined question.

Given that the metric on the world-sheet has signature +-, I guess the classical answer is "no"

Slereah

Is the signature inherited from the target manifold?

or is it independant

ACuriousMind

That depends on your philosophical idea of how the string action works :P

I mean, you can view the action just as a 2d SCFT with a bunch of scalar and fermion field, you don't need to talk about the target spacetime at all if you're careful.

Slereah

well you have to at some point if you want to talk about the real world

ACuriousMind

If you take the traditional "worldsheet" stance, then you start with a worldsheet, i.e. some abstract thing that represents what a string traces out over time, and of course that already has a notion of time/space, so it starts with that signature, so it's not "inherited"

Slereah

Isn't the worldsheet embedded in Minkowski space?

Or at least some manifold with fixed geometry

ACuriousMind

10:00 AM

@Slereah Some people like to say "spacetime is not fundamental in string theory", and what they mean is that while you may assign one (or several!) target spacetimes to a particular string theory, there's no unique or "real" one.

Slereah

How do you perform measurements in string theory then?

ACuriousMind

@Slereah In the GS view with the modified supervolume you look at embeddings of an abstract worldsheet $\Sigma^{2\vert 0}$ into super-Minkowski space, so the abstract object obviously has to exist prior to embedding

Slereah

Abstractly, at least

ACuriousMind

@Slereah I do not understand the question :P

Slereah

@ACuriousMind Does it have to "match", though

ACuriousMind

10:02 AM

Match what?

Oh, I guess the embedding is "isometric"

Slereah

Does the restriction of the metric on the target superspace to the worldsheet have to be the sheet metric

ACuriousMind

Yes

Slereah

Ah yes

thanks

So I guess it is timelike

Danu

10:45 AM

o/ @ACuriousMind @Slereah

SBM

@Danu Hello

Danu

hi

Slereah

Wait

If you don't need the target space for string theory how do you differentiate superstring theory from string theory

Danu

@Slereah Fermionic fields on the worldsheet :p

Slereah

A superworldsheet?

the big problem with string theory is that there's a really big gap between GR + QFT and string theory

A lot of necessary things in between that are rarely done

ACuriousMind

11:02 AM

@Danu moin moin

@Slereah No, just a worldsheet with fermionic fields on it

Slereah

If strings are submanifolds of the target space with a conformal field defined on them, and that a Lorentz cobordism implies the formation of closed timelike curves, wouldn't the process of a closed string splitting in two closed string involve something weird?

ACuriousMind

Bosonic string theory is a 2d CFT, superstring theory a 2d SCFT (or rather, their actions are that; string theory is not the same as just CFT)

Slereah

The worldsheet would basically be CFT on the pair of pants spacetime

Which is singular and has CTCs

ACuriousMind

@Slereah No one claims that these processes "happen", I think

Slereah

yes, but still

ACuriousMind

11:04 AM

These diagrams where a string splits or whatever are exactly analogous to Feynman diagrams

Slereah

It's part of the "perturbative expansion" i guess

So it means that there's some possible causality violation in the perturbation

ACuriousMind

You should not take them as a literal representation of what happens/can happen

Slereah

which relates to my old question of the links between quantum gravity and CTCs

ACuriousMind

@Slereah Would you claim there's "causality violation" in Feynman diagrams?

Slereah

Well there are off-shell particles, certainly

SBM

11:05 AM

@ACuriousMind feynman diagrams?

ACuriousMind

The perturbation series is not claiming that these stringy Feynman diagrams are also embedded in the target spacetime, by the way

Slereah

Hm

but still

Asymptotically

ACuriousMind

It's just...abstractly summing over them, there's not really a trace of the embedding if you don't insist on it (yet another point where the target spacetime is rather inessential)

Slereah

You could have one closed string becoming two closed string

which means that the scattering matrix would be some Lorentz cobordism in between

I probably should read more string theory first tho

I'm still not up to string interactions

Even if the causality violation is "off shell" that would be interesting, if true

Danu

I'm stuck doing GR-style tensor computations :(

Slereah

11:10 AM

I mean you don't even need the target space to infer causality violations

SBM

@Danu GR?

Slereah

Any lorentz cobordism like that has causal weirdness

Danu

general relativity

SBM

oh

Slereah

I'm gonna take a guess and say that there's probably no exactly solvable string process

especially interacting ones

no string Sine Gordon

Is there a standard notation for the worldsheet space and target space?

Like if I want to write that $X : \Sigma \to \mathcal M$

With the worldsheet $\Sigma$ and target space $\mathcal M$

What's the usual notation

ACuriousMind

11:27 AM

@Slereah Define "exactly solvable"

Slereah

Yeah I guess that is also the issue

ACuriousMind

I have a feeling you're thinking far too "classical" about strings.

Slereah

Well gotta start somewhere

I am not quite ready to drop the target space quite yet :p

Also CFT on a curved spacetime is an odd definition of "classical"

ACuriousMind

I mean not the target space thingy, but thinking that the classical strings and their motion are in any way what string theory is about :P

Slereah

@ACuriousMind Still on the beginning of learning!

And the first chapters of string theory are always relativistic strings

ACuriousMind

11:30 AM

Sure, that's where one starts, but you seem to be trying to extrapolate all sorts of things about string theory from that description when no one really proposes that that classical part has much, if any, significance, I think

Slereah

Could be

ACuriousMind

You're gonna quantize that theory of the relativistic string and what comes out has very little to do with what you had as a classical picture, much like understanding a classical field theory doesn't really tell you much about its QFT

Slereah

Yeah but the fact that you have possible topology change on the 2D space seems very weird

Since independantly of any matter on it, that is causally very badly behaved

ACuriousMind

You don't, really. The string perturbation series that introduces these is purely quantum - no one claims the classical strings behave in that way.

Slereah

Sure, but you still have to perform calculations on it

How do you perform the "integral" on terms where the CFT might not even be well defined?

ACuriousMind

11:33 AM

Yes, but no one claims that is an "actual" worldsheet, so what does causality matter? It's just a computational tool.

@Slereah The CFT does not care one whit about GR or causality.

Slereah

Doesn't it?

ACuriousMind

Nope

Slereah

Weird

I should look up what the string "propagator" looks like, i suppose

But doesn't it depend on the worldsheet metric $\gamma$?

in the same way that CFT does on a 2D spacetime?

The CFT of a scalar field in 2D is gonna be like... $$G(\tau, \sigma) = \int \mathscr D\phi \exp{(i \int d\tau d\sigma \sqrt{-\gamma} \gamma^{\mu\nu} \partial_\mu \phi \partial_\nu \phi )}$$

Or something of the form

I'm not sure it's well defined on the trouser spacetime

Hence my confusion

Abcd

11:53 AM

Let vector A and Vector B be the two vectors of magnitude 10 unit each. If they are inclined to the X axis at 30 deg and 60 deg respectively, find the resultant.

The answer is 20 cos 15 deg

How do I solve this one?

Slereah

Inner product

$A \cdot B = |A| |B| \cos \theta$

Abcd

We want sum

Right?

We have to ad the vectors right? @Slereah

brb

Jim

Hear ye! Hear ye! Direct from the mouth of Jim, I am announcing officially that June 7 shall henceforth be Physics Pun Day here in the hbar. That's right, save up your best puns, research great ones, create new ones. On June 7 (this Wednesday), we shall unleash pun-ishment on all around. The creators of the best puns shall be immortalized, the worst puns will be mocked until I can't type any more. It's sure to be fun on the pun, so come out and bring a sense of humour

ACuriousMind

12:11 PM

@Slereah I don't know what you mean by "string propagator"

Slereah

Isn't there a propagator for string theory

Also that string book says that in 2D $R_{\mu\nu} = \frac 12 g_{\mu\nu} R$

SBM

String theory

Slereah

Isn't it $R_{\mu\nu} = g_{\mu\nu} R$?

cf

4

Q: the Ricci curvature in two dimension

In two dimensions, we know that the Ricci curvature can be written in terms of the Gauss curvature K as $Ric(g) = Kg$. Can anyone prove this? Sorry if the question is too trivial :).

differential-geometry riemannian-geometry curvature

ACuriousMind

@Slereah I'm gonna bet on "different normalizations" there :P

@Slereah See, I don't understand what you mean by that

Slereah

For a starting state of $|\mathfrak{string}_i \rangle$ and a finishing state of $|\mathfrak{string}_o \rangle$, how does one compute the transition amplitude between those states

I'm not sure this is related to different normalizations

I think it's part of the whole "2D GR isn't dynamic" rigamarole

Of which i have never seen a rigorous proof

ACuriousMind

12:21 PM

@Slereah Yeah, you should just read up on string interactions

Slereah

Will do!

I roughly understand conformal symmetry and supersymmetry so things should be easier now

I have a proof somewhere of the Gauss Bonnet theorem for 2D Lorentz surfaces

I should try to actually prove the whole 2D isn't dynamic thing

Not as easy as it sounds

because of the whole York Hawking Gibbon term

Danu

@Slereah This isn't so bad, is it? It's just that the Einstein tensor vanishes identically.

Slereah

But DOES IT?

I'm not 100% convinced that that is true

Danu

I think I have a proof written out.

It might not be rigorous.

Let me check.

Slereah

Since in 2D we have $R_{\mu\nu} = R g_{\mu\nu}$, that means the Einstein tensor is $\frac12 R g_{\mu\nu}$

Not quite identically vanishing

Danu

12:27 PM

Slereah

And the proof with the Gauss Bonnet theorem doesn't hold water because 1) the Gauss Bonnet theorem for Lorentz spacetimes is slightly different 2) it's valid only for compact sets so you have to work with the York term

Danu

You see if you think it's fine or not, I'm too lazy to check.

Slereah

I'll give it a look, thanks

I mean I'm sure it's true, but an actual proof would be nice

Danu

This purports to be an actual proof

I remember it not being so easy to see why those last expressions vanish, except in locally inertial coordinates

I may have overlooked some kind of subtlety, who knows.

Slereah

How does that mesh with $Ric = R g$ though

Can't be $Ric = Rg = Rg/2$, I don't think $R$ generically vanishes

Hm

I'll have to compare the two proofs

Sha Vuklia

12:36 PM

Guys, shouldn’t $\vec p$ point in the opposite direction? For we know that
$$
p=\sum_i q_ir_i=qr_+-qr_-=q(r_+-r_-)=qd,
$$
where $\vec d$ points from the positive charge to the negative charge. Since the electrons in the water molecules are attracted by the oxygen nucleus, it seems to me that $O$ is $\delta-$, and $H$ is $\delta +$, so the dipole moment should be pointing in the direction of $O$.

Danu

@Slereah My document also shows that it generically doesn't

Slereah

Ah wait

Apparently it's not $R$

It's $K$

And $K = R/2$

So it's all good

Phew

Damn Gaussian curvature

Sha Vuklia

oh never mind

apparently I made some error. what my book does is correct

I'll just have to reread it

Abcd

@Slereah Dont we have to add the vectors? Why are you finding the cross product?

Sha Vuklia

oh, my mistake was that i thought that $\vec r$ was the distance vector (between the charge and the observer), but it's just the position vector of the charges.

Sha Vuklia

1:18 PM

Guys, I’m confused by the minus sign that appears in $V$. We know that $\mathbf E$ points radially outward. We also know that $V=V_a-V_b$ (since $V_a$ is the potential on the positive shell with radius $a$, and $V_b$ on the negative shell), so I would think that we have
$$
V=-\int_b^a E\hat r\cdot d\vec l=-\int_b^a -Edr=\int_b^a Edr.
$$
Apparently I'm making some error with directions, but I wouldn't know where.

ACuriousMind

@ShaVuklia Where does the second minus sign come from?

Sha Vuklia

@ACurious Since $E$ points radially outward, and we integrate from $b$ to $a$, our infinitesimal vector $d\vec l$ points radially inward

ACuriousMind

@ShaVuklia How do you know $b$ is the outer sphere?

nothing in that solution you posted really declares it to be one or the other, so sign issues can be fixed by just adjusting what is $b$ and $a$

Sha Vuklia

It has to be, since $-\int_b^a Edl=V_a-V_b$, and we always subtract the potential on the positive shell (the inner sphere) from the potential on the negative shell

I see that since $a<b$, their potential $V$ is in the end positive

so they must be doing the right thing, since $V$ has to be positive @ACurious

though, if you think about it, $\int_b^a Edr$ seems like a positive quantity to me too

ACuriousMind

@ShaVuklia Not if $b > a$.

Sha Vuklia

1:30 PM

oh right

okay, well that's "good news", because then I'm doing something wrong.

ACuriousMind

So, I guess the issue here is the notation $\int_b^a$ for the line integral $E\cdot\mathrm{d}l$

Because that, actually, includes the orientation of the path "twice" as you're reading it: It has it once in the order of the boundaries, and once in your infinitesimal vector pointing radially inward

Which is clearly counting the orientation once too often, which is where your error with directions lies.

Sha Vuklia

oh

you mean $dl<0$

right?

ACuriousMind

The text you're reading cheats by including the orientation in the order of the boundaries, so it doesn't have to care about the orientation of the line element.

Sha Vuklia

like, within the integration, the $dl$ "adjusts" appropriately

ACuriousMind

@ShaVuklia I don't know what that is supposed to mean

My point is that writing a line integral as $\int_b^a E\cdot\mathrm{d}l$ is, strictly speaking, non-sensical

Line integrals are not over intervals $[b,a]$, but over oriented paths.

Sha Vuklia

1:33 PM

oh right

ACuriousMind

So that there is code for "integrate along the straight path from a point on the sphere at $b$ to the sphere at $a$".

Sha Vuklia

yea I know that. we could have taken any (continuous, I guess) path from $b$ to $a$

and in our case it made sense to choose a straight path

I'm guessing I should revise a bit on line integrals then

ACuriousMind

By the prescription for line integrals, one would then evaluate that as $\int_a^b E\hat{r}\cdot\mathrm{d}l$ (note the order of the integration boundary; the evaluation prescription for line integrals always has them in the order where the smaller bound is also really the lower bound), which is equal to $\int_b^a E\mathrm{d}r$ because $\hat{r}\cdot\mathrm{d}l = -1$, because it's pointing radially inward as you said.

Sha Vuklia

ohh, so you're saying they already switched $a$ and $b$, because they knew that $\hat r\cdot dl=-1$ in the case we went from $b$ to $a$?

ACuriousMind

Yes

Sha Vuklia

1:38 PM

that really is a dick move

SBM

Gauss theorem?

Sha Vuklia

(I still love Griffiths tho)

@SBM nah