The h Bar

user218912

00:08

I asked it as a question so I can save the derivation

user218912

0

Q: Maxwell's equations in covariant form

Maxwell's equations of electrodynamics in vector calculus form are $$\nabla \times \mathbf{B} - \partial_t \mathbf{E} = \mathbf{J}$$ $$\nabla \cdot \mathbf{E} = \rho$$ $$\nabla \times \mathbf{E} + \partial_t \mathbf{B} = 0$$ $$\nabla \cdot \mathbf{B} = 0$$ Where $\mathbf{E}$ and $\mathbf{B}$ a...

user218912

answer if you want :)

user54412

00:31

Quote of the day:

user54412

Textbooks are for trivial beings. To write a textbook is to admit one used to be a trivial being. — Insulin69 yesterday

0celo7

01:42

@3750 no, indices do not match

user218912

01:55

@0celo7 answer my question please?

0celo7

@3750 I should VTC as homework like...

It literally is an exercise...

user218912

oh man

user218912

but I showed my work.

0celo7

well, take it up with @ACuriousMind @dmckee @DavidZ

unless it's a conceptual question, close

user218912

@0celo7 can I divide by $\epsilon^{ijk}$ and replace the indices by $(i, 0, j) = (\mu, \nu, \lambda)$ to get the bianchi identity?

user218912

02:04

or does that not make sense to do.

user218912

it doesn't really work nvm.

0celo7

well

I'm going to look in the book for a certain equation

if I find it, I'm voting to close your question because that means you didn't read

user218912

at least also tell me the equation.

user218912

and I know how to derive it using the $A_\mu$ definition.

user218912

I'm trying to do it without using that.

0celo7

02:09

It does not appear to be in there...

user218912

what is the formula?

user218912

D:

0celo7

Hmm, not even in the exercises

oh, it's in the comments.

user218912

comments?

user218912

oh

user218912

02:11

but he's working backwards

0celo7

no

there is a way to "divide off" a Levi Civita symbol

user218912

how

0celo7

did you read the wiki article

user218912

yeah

user218912

do you mean to

user218912

02:15

multiply by the inverse?

user218912

what I don't understand is basically

user218912

I have 4 indices in my equation

user218912

but the final answer has 3

user218912

different ones

dmckee

@3750 HI. Physics Stack Exchange is neither the same as either other physics related internet site you may be familiar with nor the same as other Stack Exchange site.

We're much stricter about homework-like questions that Mathematics, for instance.

That's because each stack negotiates it's own rules during the boot-up process.

And the users of Physics have consistently voted to not be a problem working site.

user218912

02:18

I don't use any other physics related website and I've been here for a year and the site is full of homework like questions like these, which have been answered.

dmckee

Your question is more advanced than many, and no one seems to be voting to close it, but a lot of our users don't like to just work problems for people.

You're more likely to get a few hints and be left to your own devices at first. Maybe someone will be bored or will take pity on you later.

user218912

yeah.

0celo7

@dmckee I like how you give him the beginner's speech

He's been around for a while

dmckee

Sorry. Didn't notice.

Have I mentioned that I have so-so people skills?

0celo7

savage

@dmckee Yes, unlike you I remember things :P

user218912

02:20

It's alright.

0celo7

@3750 It's not possible to do without working backwards a bit.

user218912

really?

dmckee

@3750 I did the work in your question once. In grad school. Haven't thought about it since and those notes are in a box in an attic in another state if they still exist at all.

user218912

well

user218912

I can't find a full derivation anywhere on the internet.

user218912

02:22

They all just say

dmckee

Egads! We did E&M out of Jackson, except where the professor thought that L&L was 'better'(apparently meaning more mysterious).

0celo7

Egads?

user218912

"obviously maxwell's equations can be written in terms of the field strength tensor as [...]"

dmckee

Ah. They consulted Jackson, then.

That means it's obvious to Jackson and the rest of us can simply consider ourselves stupid. Or something like that.

0celo7

@dmckee You people complaining about Jackson haven't read Sachs & Wu

Where the Geroch theorem is an EXERCISE

@3750 Ok, let's put it this way

In analysis, when proving that a function has a certain limit, you actually work backwards to find the proof

And then you write the proof down in one shot in the forward direction

Here it's similar, you need to work backwards to find all of the parts of the proof but you can write it down in the forward direction

Basically, you show that different combinations of $\mu,\nu,\lambda$ give zero

Then you write down all of the stuff that gives zero and claim it can be summed up in a fancy tensor equation

user218912

02:27

does 0 count as an extra index?

0celo7

hmm?

Well also note that cylic permuations are equivalent

so the index combination $00\lambda$ is the same as $0\lambda0$

for instance

It's not enlightening at all

and it's not as trivial as the other one, I don't think.

user218912

ok I'll work backwards and write an answer.

user218912

though is my equation correct?

user218912

the indices and such.

user54412

@dmckee Not your fault really, since the only clue that the persona has been around for a while is noticing that there seems to always be exactly 1 very blue avatar hanging around this chat room ;)

user218912

02:30

$$\epsilon^{ijk}\partial_k F_{0i} + \epsilon^{ijk}\partial_0 F_{jk} + \epsilon^{ijk}\partial_i F_{jk} =0$$

user218912

@ChrisWhite xD

0celo7

@3750 I'm assuming it because it is equivalent to part of the general equation

user218912

ok

user218912

I figured it out btw

0celo7

In particular the case $0jk$

$00\mu$ is trivial

Do you see why?

user218912

02:34

no.

0celo7

you're taking the antisymmetric part

so $00\mu=-00\mu$ in an obvious notation

which implies $00\mu=0$

so now just check the case $ijk$

which is the fourth Maxwell equation added to itself three times

I think...

0celo7

02:48

@3750 No, your equation is incorrect.

The indices are messed up

In one term there's a free $i$ not present in the others

user218912

03:02

yeah

0celo7

03:15

aha, there is a trick @3750

Hint: $\dim\Lambda^3(\Bbb R^3)=1$.

user218912

what does that mean

user218912

is that a lorentz transformation?

user218912

03:42

I just wrote a really long answer and deleted it.

vzn

05:19

No Escape From Black Holes? Stephen Hawking Points to a Possible Exit / Overbye, NYT

John Rennie

07:01

@vzn this is more on their idea that BMS modes on the horizon may contain all the information. The jury is still out.

John Rennie

10:21

@dmckee didn't you ask that a while ago?

@0celo7 do Americans say egad? It's one of those traditional Britishisms that no-one has said since Victorian times, and probably no-one ever actually said outside of a period novel. It's a euphemism for by God and dates back to the times when you were not supposed to take the Lord's name in vain.

Slereah

They only say "Egad!" when they want to sound quaint british

The same way they would say "Zounds!" or "Cor!"

John Rennie

"Zounds" :-) I bet no-one ever said "Zounds".

Slereah

Also only Gumbies say COR BLIMEY

John Rennie

I have heard "cor" used, though these days it's mostly used in an ironic way.

Slereah

Most religious minced oath are pretty archaic in a lot of languages

French has BOUDIOU

Which is one for bon dieu (good god)

So I'm trying to find facts about the geodesic equation for Krasnikov spacetime with two tubes

Just bounds

That shit is complicated

It has 0 Killing vector

John Rennie

10:37

While you're around, do you know anything about the mechanism of gravitational wave generation in a binary black hole merger?

Slereah

as far as I know it's mostly done numerically, so I don't know too much about it

There's one exact solution for a system of two black holes, but it's static

John Rennie

For real black holes, i.e. with only apparent horizons, does the radiated energy come from the gravitational potential energy?

Slereah

Mostly I would guess, yes

Although with real black holes, there's always a whole mess of matter to muddy the issue

John Rennie

The LIGO measurements recorded several solar masses being radiated away. If this was just PE being radiated away that makes sense. However if it was actual matter being converted to GW energy that's a bit less clear as I can't see what the mechanism would be.

Slereah

I don't think there is such a mechanism, classically at least

It would probably violate continuity equations of the matter

Although, you know

Violation of the conservation of energy in non-static spacetimes and all

Not sure it could take energy from whatever mass, though

Is Komar mass conserved?

Oh wait Komar is stationary

What's the non-stationary one

ADM?

apparently not conserved

So I dunno

Slereah

11:39

3

The essence of programming

0celo7

12:04

@JohnRennie maybe old ones do

Slereah

Nah

When old americans want a minced oath they will say darn

I'LL LOOK LIKE A GEE DEE IDIOT

GOSH DANGIT

GADZOOKS

BY GUM

LANDS SAKES ALIVE

HOLY TOLEDO

JUMPING JEOSAPHAT

0celo7

Stop smoking crack please.

Slereah

These are the fine words of your country, @0celo7

Do not be ashamed of them

0celo7

13:10

@Slereah I've heard one of those

@ACuriousMind Please explain

Slereah

This seems to be some sort of maths

user116211

Did PSE ever apply the blackboard theme?

user116211

Also, the witty @dmckee ;)

user116211

@Jonathan: At ::random land grant university:: the answer is the physics building was put up shortly after Sputnik and the boards haven't been replaced since then. Besides, half the inhabitants were hired shortly after Sputnik and they're a little set in their ways. (Though actually we're losing the real post-sputnik generation pretty fast, now.) — dmckee ♦ Feb 20 '11 at 17:00

David Z

13:17

@MAFIA36790 well not quite like that, but the very first design the site had was blackboard-based. It didn't work out very well.

0celo7

@Slereah well yes

but why does skew/(skew skew) equal that thing

user116211

ohh.

Slereah

According to Some Site, this is how you do a krasnikov tube

I'm 90% sure that the spaceship to go there looks like this

ACuriousMind

@Slereah wtf

@0celo7 The skew/(skew skew) thing is supposed to be a very brief illustration of what you do when you divide out that subalgebra from $\mathfrak{so}(n)$.

0celo7

@ACuriousMind yes, clearly

but how does the result follow?

ACuriousMind

13:31

By, uh, carrying out the quotient?

0celo7

hmm?

ACuriousMind

You do know how a quotient space is defined, right?

0celo7

You ask me this every time a quotient pops up

ACuriousMind

Because you act every time as if you haven't seen a quotient before! :P

0celo7

@ACuriousMind ok I clearly do not understand it in this case

ACuriousMind

13:36

You can write a general $\mathfrak{so}(n)$ matrix as $\begin{pmatrix} A & B \\ -B^T & C\end{pmatrix}$ where $A\in\mathfrak{so}(k),B\in\mathrm{Mat}(k,n-k),C\in\mathfrak{so}(n-k)$, agreed?

0celo7

No, else the result would be trivial and I would not have asked you.

ACuriousMind

Well, but it is just applying that $\mathfrak{so}$ are the antisymmetric matrices

I don't know what I could explain about that.

Slereah

Hm

Thinking about it

Krasnikov with one tube is causal

But it's not stably causal

You can have arbitrarily close points in a causal curve

0celo7

@ACuriousMind what is an antisymmetric matrix?

ACuriousMind

@0celo7 ???

0celo7

13:40

I clearly do not understand the concept.

ACuriousMind

It's one where $A^T = -A$

0celo7

What is $T$?

@ACuriousMind yes

what then

ACuriousMind

@0celo7 Taking the transpose

0celo7

@ACuriousMind clearly

I'm asking what it is, not what it is.

ACuriousMind

...what?

0celo7

13:43

I know what a transpose is

But I cannot feel what it is

@ACuriousMind Ok, let's say I accept your result

what then

I do not know what a quotient is, clearly.

ACuriousMind

@0celo7 You should take either more or less drugs, I'm not sure which :P

2

@0celo7 I thought:

9 mins ago, by 0celo7

No, else the result would be trivial and I would not have asked you.

0celo7

i changed my mind

ACuriousMind

14:01

@0celo7 Well, you just quotient out the matrices of the form with $B=0$ (that's what $\mathfrak{so}(n-k)\times\mathfrak{so}(k)$ is.

0celo7

@ACuriousMind what does that mean

user147690

14:21

He just wants to know what it isss.

0celo7

I want to know

What's the intuition @ACuriousMind

Hari Prasad

Hi @Slereah

Slereah

Hello

0celo7

@ACuriousMind I really am confused

user929304

@ACuriousMind Hi, any chance we can talk about that question? (the one I posted as a comment earlier) if you have some time.

ACuriousMind

14:43

@0celo7 That's apparent, but I don't really know what is confusing here

@user929304 Yes (you might want to ask here directly instead of using comments on unrelated posts, btw): The "propagation" of the wave is determined by the Hamiltonian/time evolution operator and will, for a free particle, be given by the momentum, not the speed of light. QM is non-relativistic, there is no reason for the speed of light to enter anywhere

Anubhav Goel

@ACuriousMind Did you earlier meant photon experience time or not?

ACuriousMind

@AnubhavGoel What?

A photon does not "experience" anything.

Anubhav Goel

Ok

user929304

@ACuriousMind oh I see, let me take an example to clarify my confusion: in a michelson interference scheme, for a single photon, the photon itself is not split (after BS), rather its wavefunction is split and propagates down the possible paths (of reflection and transmission), then it is known that even for one photon we observe interfenrece between the returning (off of the mirrors) probability waves. Question is, what is the propagation speed of this wave? can it be infinite in principle?

Anubhav Goel

If our frame moves with speed of light, can we say it photons rest frame, where photon seems to be at rest?

Slereah

14:52

@ACuriousMind Check your mass priviledge

user929304

@AnubhavGoel there's no rest frame for photons

Slereah

Rejecting the experience of massless particles

0celo7

@ACuriousMind I do not see why that matrix is a representant of the equivalence class of the quotient

0

Q: Why are boundary terms important in Chern-Simons theory?

I am learning about Chern-Simons theory. I work in Euclidean space. The action is given by $$ I=\int_{\mathbb{R}^4}d\omega=\int_{\partial\mathbb{R}^4}\omega $$ where $\omega$ is the usual Chern-Simons form, and I have used stokes' theorem. My first question is, what is the boundary of for dimensi...

chern-simons-theory

@ACuriousMind Am I insane or isn't $\partial\Bbb R^4=\emptyset$

ACuriousMind

@AnubhavGoel There are no frames moving at the speed of light. There is no Lorentz transformation into such a frame.

Anubhav Goel

@user929304 Why is it so please?

ACuriousMind

14:56

@0celo7 It's probably the good ol' "sphere at infinity". Or the user doesn't know what they're doing :P

@user929304 That would be the speed of light since it is, well, light that is moving there.

But non-relativistic QM indeed doesn't impose any restriction on that speed.

Anubhav Goel

@ACuriousMind That's absurd, I thought we could choose any frame of reference.

ACuriousMind

You need relativistic QM or better yet QFT to discuss causality issues.

@AnubhavGoel Yes, you can choose any frame that exists.

But there simply is no frame associated to "moving with the speed of light".

You can't choose what does not exist.

Anubhav Goel

Why no such frame exists? GW move at that speed too

Secret

@ACuriousMind Q3. p.223-226 Susskind seemed to run through a proof akin to the no communication theorem. While from that proof due to how the evolution operator of bob's $U_{bb'}$ is unitary, the density matrix of Alice is unaffected as $\sum_{b,b',b"} \psi^*(a'b")U_{b"b}U_{bb'} \psi(ab')=\sum_{b',b"} \psi^*(a'b")\delta_{b"b'} \psi(ab')=\sum_{b'} \psi^*(a'b')\psi(ab')$.

I am wondering then how can one show that the density matrix of Alice's be changed when Bob's subsystems evolve and the two subsystems are |spatially close enough that they can interact under the confines of the speed of li…

0celo7

@ACuriousMind Can you please explain this thing I'm helplessly confused

Emilio Pisanty

15:02

@dmckee Much perplexed (if only temporarily) by this comment

physics.stackexchange.com/questions/264708/…

(10k+)

ACuriousMind

@0celo7 Hmm...maybe what's confusing you is that the quotient is taken as a quotient of vector spaces?

0celo7

@ACuriousMind ...wrt addition?

ACuriousMind

@Secret I'm sorry, I don't really understand what you just said.

@0celo7 Yes

0celo7

@ACuriousMind sigh...thank you

ACuriousMind

If you were trying to take it as a quotient of rings or something then I finally know what was confusing you! :D

user929304

15:06

@ACuriousMind oh that would simplify many things for me. So it means the speed of propagation of a quantum system's wavefunction (which is not a physical wave), depends on the physical system itself, (e.g. if a photon, then bounded by speed of light). right?

0celo7

@ACuriousMind I was.

ACuriousMind

@user929304 Yes

0celo7

I know that algebras are vector spaces with multiplication, and I was trying to quotient wrt. multiplication

Slereah

What is the difference between the path integration of the action $\int d^4x \partial \varphi \partial \varphi$ and \int d\lambda \dot x^2$

ACuriousMind

@0celo7 Ah, that is the horrible thing about Lie terminology: Lie algebras are not algebras!

Slereah

15:07

With $\dot x$ a 4bector

0celo7

@ACuriousMind ...huh

ACuriousMind

They do not have a multiplication, they just have the Lie bracket

0celo7

is the Lie bracket not multiplication tho

ACuriousMind

They embed into their universal enveloping algebra where they get a multiplication, but the Lie algebra itself doesn't have multiplication.

0celo7

I don't see how the bracket is not multiplication.

what even is multiplication

ACuriousMind

15:09

@0celo7 It's neither commutative nor associative

user929304

@ACuriousMind awesome! thanks so much for your patience with me (as always)

0celo7

@ACuriousMind I don't require multiplication to be commutative

or associative, really

all I really want from my multiplication is that $x\star(cy)=c(x\star y)$, etc.

ACuriousMind

Well, then you can call it "multiplication" of course - but to me, a "multiplication" is the second operation of a ring, and therefore at least associative.

0celo7

@ACuriousMind well according to wiki its an algebra

Secret

@Acuriousmind Ok let me try again. p. 223-226 of susskind showed how entangled states cannot be used to transmit information by showing how because the time evolution operator of e.g. bob $U_{bb"}$ is unitary, Alice's density matrix $\rho_{aa'}=\sum_{b,b',b"} \psi^*(a'b")U^{\dagger}_{b"b}U_{bb'}\psi(ab')$ will be unchanged by $U_{bb"}$.

I am wondering how will $U$ be like if Alice's and bob's subsystem are placed next to each other such that they can interact directly, since if the two subsystems are placed next to each other, then intuitively, in principle they can interact and Alice's de…

Anubhav Goel

15:13

Can someone check my answer here physics.stackexchange.com/questions/230440/… It's on potential at centre of sphere due to external charge.

ACuriousMind

@0celo7 I guess my notion of algebra is narrower than that of Wiki

0celo7

@ACuriousMind clearly.

let us check the Standard References

ACuriousMind

To me, an algebra over a ring $R$ is just another ring $S$ with a map $R\to S$ that allows for "scalar multiplication".

0celo7

I hate crying children

@ACuriousMind No, the bracket is clearly mutiplication

Because this is consistent with the "abelian" terminology

ACuriousMind

@0celo7 But then what do you call the operation in the enveloping algebra that you need to define e.g. the Casimirs?

0celo7

15:17

@ACuriousMind I don't need that stuff as a geometer.

What you algebrists do in your spare time is your business

ACuriousMind

@0celo7 The "Abelian" terminology comes from thinking of the bracket as a commutator.

And, in fact, it does become the commutator in the enveloping algebra

0celo7

What is crying even supposed to do

ACuriousMind

BTW, the German Skyrim mod is out

0celo7

@ACuriousMind random

link?

ACuriousMind

@0celo7 dl.enderal.com

@0celo7 Get attention, I guess?

0celo7

15:19

@ACuriousMind well this child is crying for his mother who cannot hear him

so really it's doing nothing

@ACuriousMind you getting it?

yo it says end of July

you liar

ACuriousMind

@0celo7 I have started it just now.

0celo7

@ACuriousMind Will it wipe my Skyrim clean?

ACuriousMind

@0celo7 It will save your current Skyrim installation to a backup folder.

Secret

@ACuriousMind
Correction (and hopefully simplified enough version): p.223-226 of susskind showed a no communication theorem like proof on why Bob cannot transmit information to Alice by an entangled state by showing that $\rho_{aa'}=\sum_{b,b',b"} \psi^*(a'b")U^{\dagger}_{b"b}U_{bb'}\psi(ab')=\sum_{b',b"} \psi^*(a'b")\psi(ab')$

But what form will U take if the two subsystems are placed close enough such that they can directly interact as in such a scenario we expect both Alice and Bob's density matrix be affected by the state evolved by U?

0celo7

@ACuriousMind Well it won't let me dl the thing

it's racist against Americans

@ACuriousMind When I moved to Germany I was scared about going to German school because I thought the kids would be salty over the whole WW2 thing

user147690

15:31

^Lie algebras are the standard thing one thinks of when they think of a non-associative algebra

user147690

(else the octonions)

user147690

But I guess one could say it is an anticommuative algebra with the Jacobi identity
¯\_(ツ)_/¯

ACuriousMind

@0celo7 Wait, are you seeing a German or an English site?

It might autodetect your location - and the English version is indeed only due at the end of July

@Secret I still don't understand the question.

0celo7

@ACuriousMind English.

Why would you play the game in German?

ACuriousMind

@0celo7 Because it's made by Germans? Why would I choose the translated version when I can play the original?

0celo7

15:36

@ACuriousMind I could play the original too, but I won't

You forget that I speakth your tongue too

Wait, will the translated one be in the King's English

Secret

Suppose alice and bob's subsystems are placed next to each other (thus allowing them to interact with each other, how will the time evolution operator U that evolve the composite system look like. We knew it cannot be $U_{aa'}$ nor $U_{b'b"}$ as p.223-226 of susskind showed that these possibilities will not change Alice (and respectively, bob's) density matrices

Will U that describe the evolution of the composite system due to the subsystems interacting with each other (as they will be placed right next to each other, thus allowing them to physically interact) be $U_{aa'b'b"}$ and we will …

0celo7

Because German is preferable to that slop

ACuriousMind

@0celo7 I don't know

0celo7

@ACuriousMind let me know if it's good

ACuriousMind

@Secret How the two systems interact depends on how they interact. "Putting them ext to each other" doesn't actually induce any kind of interaction, unless you stipulate they're charged particles and attract each other or something like that. I still don't understand the question.

0celo7

15:44

> How the two systems interact depends on how they interact.

Wise words by Fut. Dr. Bajoran.

Anubhav Goel

Can someone follow thread of comments in physics.stackexchange.com/questions/259287/… this question between me and Sanya

Secret

Ok, in that case, since in susskind's book the entangled states he explores are qubits, suppose there are two entangled qubits (one from alice and one from bob) that are placed next to each other such that they can feel each other's magnetic moment caused by the spin. Will the time evolution operator that describes such interaction be $U_{aa'b'b"}$ when wrote in index form as intuitively we will expect the density matrices of alice and bob be changed by such interaction?

0celo7

16:12

@JohnRennie hello

John Rennie

Afternoon / morning

0celo7

it is 12

so...noon

what is noon anyway

that sounds like a made up word

Slereah

Hello

John Rennie

@0celo7 IIRC it's from an old English word

Presumably that would have come from the German since they invaded us in 500 AD or whatever. What's the German for "noon"?

Oh hang on, it's mittag isn't it?

0celo7

Yes

John Rennie

16:15

Which is midday. Is there another word for "noon" in Grerman?

David Z

BTW NASA has a press conference about the Juno spacecraft's approach to Jupiter

Slereah

From Old English nōn, from a Germanic borrowing of classical Latin nōna ‎(“ninth hour”) (short for nōna hōra), feminine of nōnus ‎(“ninth”).

David Z

ustream.tv/NASAJPL2

0celo7

@JohnRennie beats me how $tag\mapsto day$.

$mit\mapsto mid$ makes sense.

Slereah

From Middle English day, from Old English dæġ ‎(“day”)

0celo7

16:17

English is such a strange language.

@ACuriousMind Maybe German is better.

I'm willing to hear arguments for either side.

John Rennie

@DavidZ Cool :-) No, cooooooooooooooooooool! :-)

That's a brave move, livestreaming the insertion maneuver. If it goes wrong there'll be egg on faces ...

Slereah

heheheh

"Insertion maneuver"

David Z

@JohnRennie Are they really? I thought this was just a pre-approach press conference

then again I haven't been paying that much attention to it

John Rennie

@DavidZ yes, on closer examination the insertion maneuver isn't for some hours yet

@0celo7 the great thing about English is that it's a bastard mongrel of a language so there are usually several different ways to say the same thing but they all have slightly different meanings.

For someone like me who got English beaten into them courtesy of the English public school system you can have endless fun playing with the language.

0celo7

@JohnRennie can you give an example

maybe British English is a bastard mongrel...but not Glorious Murrican English

David Z

16:26

You misspelled 'MURICAN

0celo7

@DavidZ ...says the expat.

John Rennie

Well, midday and noon for example. Both are used in UK English.

One came from Latin and one from German.

Americans casually invent words as needed. That's not a criticism, after all, that's exactly what Shakespeare did, but the results aren't always edifying.

0celo7

example?

John Rennie

"deplane"

0celo7

is that not a word?

John Rennie

16:30

We native speakers would say disembark, which is a much nicer word.

0celo7

I would say both

David Z

"Deplane" sounds normal enough to me

Then again I'm used to hearing "alight" which seems to be popular in several non-English-speaking foreign countries

0celo7

@DavidZ Deplane is strange because plane is not a verb.

David Z

Yeah, just saying, it's common enough not to sound weird

John Rennie

I'm not criticising - I think people who get precious about language really need to get out more, and a language that never changes is a dead language.

0celo7

16:37

@JohnRennie Can you give another word?

*example.

Sir Cumference

Amazing

2

The Hubble telescope caught bright auroras on Jupiter

dmckee

@MAFIA36790 I seem to recall that the black background chalk-board theme was active for about a week. Certainly for long enough that someone wrote a greesemonkey script to override it (and I developed the habit of putting the site in its own window so I wouldn't run across it while flipping between tabs).

0celo7

16:57

@SirCumference not really.

What are the applications?

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