01:00 - 16:0016:00 - 00:00

1:00 AM
what is a dipole limit?
anyone here?

1:20 AM
@Paul777 it's a hard limit on the number of Polish twins that can exist in the universe at any given time.

lol

@0celo7 why are people calling Cartesian coordinates "rectangular coordinates" in my class

@Paul777 I think there's a good definition in the Wikipedia article

@BernardoMeurer it's an alternative name for them

@BernardoMeurer It's a perfectly good description when set next to "cylindrical coordinate" and "spherical (polar) coordinate" neither of which have a nominal appellation I am aware of.

1:30 AM
I see
I like Cartesian better, sounds fancier

@BernardoMeurer In my class(es) I prefer when possible to use names that have a plain-language connection to the underlying concept. Rectangular coordinates (or, even better IMO, "rectilinear") reference the notion that the coordinate system being described has linear axes oriented at right angles to each other. "Cartesian," on the other hand, connects only to the history of the concept, not to the concept itself.
(When no descriptive term exists I'm perfectly happy with name-based terminology. Khayyam quadrilaterals, for instance, abound in my modern geometries course.)

@nitsua60 I get that, and maybe it is indeed a better practice, but I've called it Cartesian my whole life, it's the only name in Portuguese, and I don't intend on changing now :P
I understand and agree with your point though

@BalarkaSen yo are you willing to read some highly experimental metric geometry with me
Gromov sanctioned

@BernardoMeurer It's no hill to die on, IMO. I.e. I'm never going to tell a colleague they're wrong to use "Cartesian" in any way. I will, however, point out that I think we should be exposing kids to the notions of oblique, curvilinear, and (hyper)spherical coordinates they'll be seeing later, and "rectangular" sits nicely in that scheme.

Yeah, that makes sense. I would never suggest forcing anyone to say "rectangular" instead of "Cartesian" but it's definitely important that people understand what "rectangular coordinates" means when they read or hear it.

2:24 AM

2 hours later…
4:21 AM
@user187604 weirdly phrased, but d)

@G.Bergeron a lot of people are saying that option but my book says option c. Now I am stumbled what is right!

@user187604 Electric potential decreases across resistors in the direction of current flow, in this case left to right. In this case, S is 10V + 20V lower than Q, which we posit to be at 0V (grounding). The answer is d). c) does not make any sense and is an error. Either the author wanted to write R instead of S or he uses a non-conventional definition of ground to mean +20V
@user187604 Most likely, it's just a typo.

4:40 AM
@G.Bergeron thanks.

5:24 AM
heavy rain makes my shoes and socks all wet and trousers partly wet.

5:45 AM
@JohnR: Hi! Are you around?

@KaumudiH Morning :-)

Morning! :-) Hangouts?

it has been afternoon.

1 hour later…
7:16 AM
mornin

Is there anything sacred about picking $i$ for the base vector of $U(1)$ in EM

7:48 AM
@JohnRennie I did it!
I have a working linux server!
With DNS and all
@BalarkaSen I can finally host you any files you want

Where is it? In the cloud somewhere?
Not that I'm trying to wee on your bonfire, but isn't it pretty easy to set up a Linux server?
Even I can do it :-)

@JohnRennie It can be a bit tricky!
Especially if you don't have a fixed IP

:: John looks sceptical ::

@JohnRennie Well ain't that the quaintest of British sayings!

7:53 AM
He'd hate to be ants at your picnic

Is that what the French say?

@DawoodibnKareem :-)

It is not

A pity. It would amuse the rest of us.

@JohnRennie With a dynamic IP not so much
I'll give out the address as soon as I get my firewall up

7:56 AM
There are Ways with a dynamic IP
but it's a bit annoying

DynDNS?

yeah

Hardly the leading edge of technical issues ...

So's knapping a flintstone, but I can't do it

@BernardoMeurer is that the server you were building on an RPi, or Arduino or something like that?
Or am I thinking of someone else?

7:58 AM
@BernardoMeurer I think it's a bit late for that. If you've got a server on the Internet with no firewall, then choosing the correct time to tell John Rennie your IP address is the least of your worries.

@JohnRennie Well, I did dyndns with just a script

This is one of the oldest meme I remember, btw

@DawoodibnKareem Nah, it's fine, the ports are all closed

To be fair setting up a LAMP server takes little more than choosing that option at install.
I don't know if the acronym is still used. If so presumably "P" refers to Python rather than PHP these days :-)

I don't use a server, I just use a telegraph button to send the requests manually

8:02 AM
I haven't heard it for a few years. What's wrong with PHP?

@DawoodibnKareem It's a POS?

PHP was always a bit of a hack, and I think it's regarded as a hopelessly ugly hack these days by the Web community. But this is just the impression I get so I wouldn't swaer to that under oath.

is there any language that isn't a hack
Except glorious LISP of course

It's alright
Although I'm not fond of the whole "Entirely object oriented" thing
I like having global variables and actual functions!

8:06 AM
@JohnRennie Can you test my server? Firewall should be up

also : mandatory garbage collection
Sometimes I like handling my memory!

@BernardoMeurer if you mean do a penetration test then I don't have the kit to do that.

IIRC you can do it in C# but it's quite a hassle

@JohnRennie I mean poke around :P

@Slereah But garbage is garbage. If it's of any possible use, it's not garbage.

8:07 AM
But if you give me the DNS name I can have a play. On Facebook if you don't want to post it here.

@DawoodibnKareem But garbage collection is more efficient if done when the object ceases to be useful

rpi.meurer.xyz @JohnRennie

rather than every few steps

@JohnRennie I fear nothing
@BalarkaSen Keep it up and you won't even get access to my movie and music

8:09 AM

@JohnRennie well that's quite an indictment of his server already

Hackerman indeed

@Slereah getting iptables right is something of a blackart ...

I am a wizard

I don't even remember how I set up my server

8:11 AM

And I'm gonna have to redo it once I move in the new flat

@BalarkaSen Did I tell you the kids at the high school here call me that, lol
@BalarkaSen Don't have one

@BernardoMeurer you can make servers disappear from the internet!!! :-)

My social security number is the 0000000000002

@JohnRennie Yes :P

8:12 AM
@BernardoMeurer Sounds legit

Hey @JohnRennie, you're an expert in VBA coding
Can you help me to do this
take a gun and shoot me

set gun = CreateObject("WScript.Handgun")

The one nice thing is that at least I don't have to update the VBA
I just have to convert it into a website

gun.Target = "idiot who wrote this stuff in the first place"

@JohnRennie You know who the culprit is

8:15 AM
@JohnRennie Not Nice

It's the company that didn't have any computer man to work on this project and just said "Let's do it in Office, it will be easy"

@Slereah Famous last words :-)
@BalarkaSen gun.Mode = GUN_INFLICTPAINFULBUTNONFATALFLESHWOUND

there's a bunch of hidden sheets that have no clear use
they're not even used in the code
@DawoodibnKareem But at what cost

@Slereah I sometimes use hidden sheets as staging areas. e.g. my reactor corrosion app has a function to copy the report to the clipboard. I write the data to a hidden sheet in a convenient layout, then copy it, then clear the sheet.

@Slereah Way less than the Java application they thought they wanted.

8:18 AM
@JohnRennie But I checked every bit of the VBA code, it's not even referenced once there!

@JohnRennie It's nowhere near Nice to threaten others, be the target of the threat a user of this chat or not, to inflict them with painful wounds, fatal or otherwise, jokingly or otherwise.

@DawoodibnKareem I meant the human cost
The cost of dignity

The culture of this room does not endorse such vile commentary

@BalarkaSen in principle I agree, but this is a theoretical and unspecified other and the statement is clearly a joke. I find myself disinclined to take seriously opinions that I'm being nasty by posting along those lines.

@Slereah The humans were given a choice afterwards. They picked it.
@BalarkaSen To be fair, Sam did say "take a gun and shoot me".

8:20 AM
@BalarkaSen ah OK, your statement was ironic :-)
I'm obviously too serious today.

@DawoodibnKareem Whether that's Nice or not depends on the country's assisted suicide laws
@JohnRennie Lol
You fell into my trapp
Use homeomorphism to escape

@BalarkaSen It will preserve my continuity but not my dignity

Only if you can differentiate the two.

@BalarkaSen If I wanted to be nasty my comment would have involved wooden stakes, techniques borrowed from Vlad Dracula and for extra effect a liberal application of chilli to the wooden stake before insertion :-)

John Rennie puts the extra "effect" into "extra effect".

8:23 AM
I don't think Europe had chilli back in the days of Vlad Tepes

the face of terror

@BalarkaSen the "a" means "son of". His father was Vlad Dracul.

LOOOL

8:24 AM
And his great grandson was Vlad Draculaaaa!

Is "Draculi" the plural of Dracula?
Maybe Draculae

I don't know medieval Romanian well enough to comment :-)

Dracula is the plural, the singular is draculum
draculus?

In Latin ...

8:26 AM
@JohnRennie Then what possible use are you?

Lel
My knowledge of medieval Romanian is limited to the infamous O-Zone song

@DawoodibnKareem people ask me that all the time. I have yet to find a convincing answer :-)
@BalarkaSen Nooooooooooooooooooooooooooooooooooooooooooooo
If a YouTube link to that song appears there will be trouble ...

what you gonna do

Where's that stake?
And chilli?

@Slereah The official music video

8:29 AM
@BalarkaSen This is more official than the official one

Absolutely
It's what made the meme blow up originally

I wonder if anyone has done a death metal version ...

probably

Oh god

I remember the original big meme, tho

8:31 AM

@JohnRennie It exists
We finally have a reason to end it all

Any experimental physicists (or even engineers!) around?
1

The representation of bits in different technological areas: Normal digital bits are mere abstractions of the underlying electric current through wires. Different standards, like CMOS or TTL, assign different thresholds to such signals: "if the voltage goes above this level, then the bit is 1;...

@BalarkaSen I quite like that - kill me now!

Also, for the more theoretically-minded: a couple of bounties ending in less than 1h, and which remain totally unanswered so far:
5

The context: We are in the solid state. After a photon absortion by a system with a singlet ground state, the system undergoes the spin-conserving fission of one spin singlet exciton into two spin triplet excitons (for context, see The entangled triplet pair state in acene and heteroacene materia...

5

The context: We are in the solid state. After a photon absortion by a system with a singlet ground state, the system undergoes the spin-conserving fission of one spin singlet exciton into two spin triplet excitons (for context, see The entangled triplet pair state in acene and heteroacene materia...

8:52 AM
@BernardoMeurer youtube.com/watch?v=3tZtish6RPU I don't get how this dude is not a legend. Do the rap community really hate verbosity that much?

2 hours later…
10:41 AM
So suppose you've a rod through the middle of which passes an axle, with some mass distribution. Now if you push at some point in the rod perpendicular to it, every particles receives a force perpendicular to the rod, right ? What happens in the molecular level to develpe a rough mental picture of how this force propagatoin works ?

1 hour later…
11:50 AM
Could someone explain how they arrive at $-\nabla V=-\frac{P}{3\epsilon_0}\hat z$? If I use the gradient in polar coordinates, I get:
$$\nabla V=\frac{P}{3\epsilon_0}\cos\theta\,\hat r-\frac{P}{3\epsilon_0}\sin\theta\,\hat\theta.$$
I’m guessing $\hat z$ is just the unit vector that points in the positive $z$ direction. So somehow we would need: $\cos\theta\,\hat r -\sin\theta\,\hat\theta=\hat z$.

12:04 PM
Take a look at the back cover of Griffiths’ @ShaVuklia
Among other things, he has all the conversions between basis vectors of the three main coordinate systems

why can't I just use the polar form of $\nabla$?
o wait
o jesus
it's literally there indeed
alright, well that solves it quickly

Sure

Right, so how do they know the sign of $p\cdot\hat r$? They want to write $\frac{P}{3\epsilon_0}\frac{R^3}{r^2}\cos\theta$ in vector form. I understand we need some vector $x$ and $\hat r$, but in this case they use $p$, so given the fact that we have $\cos\theta$, I would think then that $p$ points upwards. However, how do we know this in advance?
wait
I haven't given full context

I recognize the example

12:12 PM
Field of a uniformly polarized sphere

right, well I've uploaded it just in case

They should say at the start of the problem that the uniform polarization is assumed to be in the z-hat direction

ohh okay
that's a quick answer too then
o wait
they did actually

"We may as well choose the z axis to coincide with the direction of polarization"
well thanks then!

12:15 PM
np

1 hour later…
1:26 PM
Hm
The inner product on a Fock space involves the permanent of a Matrix
I've never heard of this
$$\operatorname{perm}(A)=\sum_{\sigma\in S_n}\prod_{i=1}^n a_{i,\sigma(i)}$$

greeting @Slereah and commies @EricSilva @G.Bergeron
@Slereah does that have a more normal expression

Why would that not be normal
Perfectly well defined in $M_{n\times n}$

does it have a fancy basis free definition

It's basically the determinant except the minuses are all pluses

oh

1:32 PM
So replace $\varepsilon$ in the determinant by its norm
both of which are versions of the immanent
wot
Immanant redirects here; it should not be confused with the philosophical immanent. In mathematics, the immanant of a matrix was defined by Dudley E. Littlewood and Archibald Read Richardson as a generalisation of the concepts of determinant and permanent. Let λ = ( λ 1 , λ 2 , … ) {\displaystyle \lambda =(\lambda _{1},\lambda _{2},\ldots )} be a partition of ...

now you're just making things up

it is apparently real

"The skin transform $\langle A\rangle$ detects inhomogeneities in the underlying space $H\setminus\Sigma$ and
gives us a means to uniformly reshape and unfold this delicate and wrinkled geometry."
@BalarkaSen you know you want to

>1934
it's old, too
But apparently, for the bosonic $n$-particle Hilbert space, you have $$\langle u_1 \circ ... \circ u_n, v_1 \circ ... \circ v_n \rangle = \operatorname{Per}(\langle u_i, v_j \rangle)$$
although I think there's probably a factor of $n!$ somewhere, too

2:04 PM
Equation 4.23 is $\oint D\cdot da=Q_{\text{f,encl}}$ and eq. 4.25 is $\nabla\times D=\nabla\times P$. I don’t really understand how we use them. Let’s start with eq. 4.23. I know how we obtain the discontinuity of the perpendicular component of the electric field. In that case, we consider a thin Gaussian pill box, and use the electric field of an ‘infinite’ plane. This yields the required discontinuity.
But here I’m not sure what to use. I know that $\sigma_f$ is the charge that causes the the electric field from the polarisation of the dielectric, but that’s not $D=\epsilon_0 E+P$ right? In any case, I'm not really sure how to show 4.26

2:19 PM
I basically have to argue that $D$ points in the same direction as $E$ caused by a sheet with $\sigma_f$ surface bound charge, and that $D=\epsilon_oE_{free}$

@ShaVuklia you don’t need to show that right. By symmetry, D will have the same component throughout the plane along the horizontal direction
So while applying gauss law for pillbox , that horizontal component gets cancelled
And you’re left only with D_perpendicular

right, but I still need that the perpendicular component of $D$ is the same as the perpendicular component of $E_{free}$ (the field caused by the free surface charge), or is that exactly what 4.23 tells me..
oh of course, because we also have $\oint E_{free}\cdot da=Q_{f,encl}$

Which equation are you trying to prove?

4.26
I know that $E_{free,above}^\perp-E_{free,below}^\perp=\sigma_f/\epsilon_0$
that's just the regular electrostatic boundary condition
so it seems that $E_{free}^\perp\epsilon_0=D^\perp$
this is close to $D=\epsilon E_{total}+P$. I guess if I get rid of $P$ by rewriting things, I will get $E_{free}$ in the right form, but I don't know how
oh.. I guess there is no $P$ at all, because we are looking slightly above and below the surface
so we have $D=\epsilon_0 E_{total}$, which should be equal to $\epsilon_0E_{free}$ (or at least, the perpendicular components)

3:11 PM
@ShaVuklia sorry, had to go for dinner. So 4.26 doesn’t involve the E at all, just the D. And while showing the similar relationship for E, we never assumed anything about the components of E other that the gauss law, which holds for D too. so the same relationship must hold for D too

@PrathyushPoduval We used more than just Gauss's law; we also used Coulomb's law (because we used the electric field of a surface charge)

Aren’t they equivalent?
(And where do we use columns law in the E version ?)

I don't know, I've only ever seen Coulomb $\implies$ Gauss, but I guess they could be equivalent
we use it because we use the electric field of a surface charge
I don't know how $D$ behaves given a surface free charge (at least not yet), but I do know how $E_{free}$ behaves close to the surface charge (it's just the electric field of an infinite plane), and that's how I know how to use the Gaussian pill box for $E_{free}$

The proof I know is to use a Gaussian pillbox. Make it very thin so the horizontal contribution of the field D is neglegible. The only contribution is now from the perpendicular component and then the equation follow
@ShaVuklia you don’t need to know how D behaves, just use the gauss law for D, like I said above

ohhhhh
I completely forgot that we could use $\oint D\cdot da$ (divergence theorem)

3:25 PM
Lol yeah

yea, then Gaussian pillbox works indeed
alright, that solves 4.26

Is it examseason for you now?

no not yet
in 4-5 weeks I think

Ah best of luck

thanks!
@PrathyushPoduval mind looking with me at 4.27?
I have a fair idea already, just not the end part

3:29 PM
Lemme see...

We know that $\int \nabla\times P\cdot da=\oint P\cdot dl$, by Stokes' theorem. But now I have to argue that 4.27 holds
so the total paths are the same, but apparently the difference between up and down is the same as well (thinking of the same argument as for $E$ parallel)

That expression with E above is 0

yea sorry
that was indeed zero, but not for $P$, because $\nabla\times P\neq 0$

So when you do curl of D, the E contribution is cancelled and you’re left only with P

yes

3:32 PM
So take a rectangle which is the cross section of the pillbox we took
The perpendicular contribution this time are negligible
So you’re left with only parallel component
A
and you get said equations

hm?
I don't see that
I had the same idea in mind with the rectangle and stuff (just copying the argument for the parallel electric field), but I don't see the connection with this equation

Write the line integral for D along the rectangle
Substitute $D=\epsilon E +P$

we lose $\epsilon E$ due to Stokes

E term of the line integral is 0, so you’re left with only P
And the perpendicular component contribution of P and D goes to 0, since the height of the rectangle is almost 0

yes, so we know that the total paths are the same
but what guarantees that the difference between up and down is the same
ohh
I know!
I can use a symmetry argument

3:37 PM
And you left with $(Dp2-Dp1)l=(Pp2-Pp1)l$
p stands for parallel here
@ShaVuklia how?
@ShaVuklia umm that’s what I’m proving here

wait I have to think
@PrathyushPoduval where exactly do you prove it?
@PrathyushPoduval I guess you're using this
but to me that's not evident, but let me look at it a little bit longer

Just write the line integral
Wait a minute

$\oint D\cdot dl=\int D_1dl+\int D_2 dl$

Oh you’re asking why one is a minus sign?
Why it’s the difference instead of a sum?

not exactly, but sort of. I don't see for instance why we have $D_{2}\cdot l$ for $\int D_2dl$
we are working with such a rectangle that $D_{p2}$ is considered constant on one part of the rectangle?

3:48 PM
Ah okay I didn’t explicitly mention that
Yes, you consider the horizon length of rectangle small so that D and P is approximately constant

right, because we only care about 'points' really (for each point, we consider such a small rectangle)
okay then that's solved, now I just have to think about the minus sign

@ShaVuklia that’s because of the direction you’re taking the line integral
We assume that D is positive along the same direction
So if in the top side, D and dl are in the same direction, in the bottom they’re in the opposite

is that trivial?

Remember, all the values here are magnitudes of components , which have a fixed sign Convention (positive along one direction, negative along opposite direction)
@ShaVuklia nope, it’s a perfectly valid question

let me reread the parallel argument for the electric field

3:54 PM
I need to go now, have an observation session now
Goodbye!

yea but my point is that if we mirror out set up, then we still have the same physical situation, but the direction of $D$ will then be reversed (the positive and negative directions are exchanged)