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11:05 AM
thanks for asking @Kaumudi.H I dont' live in NSW tho where the fires were I think
Oh, OK. They're still spreading, apparently...
Good to know that u're safe :-)
]I didn't even know about them
One of my nerdfighter friends lives there, which is how I came to know.
11:07 AM
oh cool
@YashasSamaga Google-able :-)
"Nerdfighters are people who, instead of being made up of cells and organs, are made up of pure awesome."
@YashasSamaga yeah they're tough blokes with a quest to destroy all nerds
they fight any nerd they come accross
@Kenshin On the contrary.
11:08 AM
Again, it's Google-able.
@YashasSamaga the light interacts with electrons in the glass and that slows it down. There have been several related questions on the site. I can dig out a couple if you're interested.
I have already dug out.
It seems to be having some QM stuff.
@YashasSamaga you can use either a QM or classical approach.
classical approach? How are you going to justify that the light ray which comes out is parallel to the incident ray?
The classical approach treats the oscillations in the dielectric electron density as forced oscillations so there is a phase lag. Then these lagged oscillations reradiate light out of phase with the original beam. Summing the two produces a wave travelling slower than $c$.
QM would describe the process as the formation of polaritons.
11:12 AM
I was reading the answers here
Q: How does light speed up after coming out of a glass slab?

Amey ShuklaAs I learned today in school, my teacher told me that when light enters a glass slab it slows down due to the change in density and it speeds up as it goes out of the glass slab. This causes a lateral shift and the light emerges out from the point different than that from where it should have act...

I don't think that's a good answer, even though it has 70 upvotes. It is a bit misleading.
The only classical answer for that question makes use of Huygens' construction.
If I ever use Huygens construction, I'll be punished at college.
@YashasSamaga really, why? Huygen's construction is a fundamental part of classical electrodynamics.
It does have problems, right?
11:16 AM
I thought it wasn't valid everywhere :|
Well, what Huygens said isn't really happening in reality, right?
There are situations where you can't ignore quantum effects, but aside from those special cases Huygen's construction is fine.
Do we have real wavefronts, secondary sources, etc?
Basically yes.
I am totally surprised now.
Wouldn't the existence of secondary sources violate conservation of energy?
Bear in mind that an EM wave is just an electromagnetic field i.e. at every point in space there is an electric and magnetic field.
11:18 AM
Ah, I can now make sense of it.
The propagation of energy through this field is described by Maxwell's equations, and simplifies to the Huygen construction for EM waves.
@YashasSamaga When a body is moving then also we can apply at any point
@help I am not sure what you mean by any point.
Most people use the C.O.M as the reference point on the rotating body.
@YashasSamaga not about centre of mass or axis of rotation
and convert it to the lab frame at the end
11:20 AM
@YashasSamaga how
there are different ways
$L = r \times mv + I\omega$
where m is the total mass
v is the c.o.m velocity
omega is the angular velocity in the rotating body's frame
The distribution of (up and down)votes in this post completely proved to me that most of the people don't vote based on reason.
You havewritten angular momentum , but I am taking of torque
The torque is just like force.
Nobody is going to write the torque equation for earth about the sun.
If you use a reference point other than the axis of rotation, you will unnecessarily complicate the problem.
@Mostafa to be fair that is a meta post so votes just express opinion and not a statement of rightness or wrongness.
11:25 AM
So at last we should write equation about COM or axis of rotation
When object is moving
@help in most cases COM is the axis of rotation
@YashasSamaga if both are different
for a sphere with uniform mass distribution, the axis of rotation passes through the C.O.M
use the axis of rotation
do you know what angular accleration means?
don't you measure it with respect to the axis of rotation?
11:27 AM
if you are measuring the angular accleration w.r.t to the axis of rotation, then there is no reason I can think of why you shouldn't write the torque equation about the axis of rotation.
Okay I got it
when you write the torque equation about the axis of rotation, you can use the following formula:
$\tau = I\alpha$
where $\alpha$ is the angular accleration
11:28 AM
if you take another point, you are going to convert it at some point unless you are a crazy guy who wants to solve the whole problem from an insane reference point.
Got it :-)
Thanks @YashasSamaga
@ACuriousMind operators form an algebra but distributions do not
Is that much of a problem for AQFT or am I thinking wrong
Since the operators stem from distributions
@JohnRennie What we're seeing there is the same question is upvoted here and downvoted there.
For example, DanielSank's question about mountain climbing, and my (and anonymous) question about her way of learning physics are downvoted but the same questions in the first (most upvoted) question aren't downvoted.
It doesn't follow any logic other than voting based on the current votes or the user.
People tend to upvote highly upvoted posts without reading it.
Yesterday (or the day before), I had quoted a research paper here which investigated how the people tend to upvote highly upvoted posts blindly.
11:40 AM
Sorry, I realise now what you mean.
deleted :)
@YashasSamaga That happens when a large number of people are voting. Here I think a very limited number of users that follow everything closely are involved.
Hi @Loong
@EmilioPisanty I'm not sure this is really the point:
A: Anisotropic electron orbitals in hydrogen

Emilio PisantyThe correct interpretation is the first one. A hydrogen atom in a pure $d$ state, like, say, the $l=2$, $m_{L_z}=0$ state, really does point in a specific direction (i.e. the quantization axis). There is nothing wrong with anisotropy, and particularly, there is nothing wrong with the existence o...

An isolated atom is spherical i.e. its $p$, $d$ etc electrons exist in a superposition of all possible orbitals.
Q: explain how Einstein explanation of photoelectric effect explain

Tshimangadzo Ace NesaneThe notes given to me claim that the intensity of electric current depends on the intensity of radiation flashed on the metal. Why is it so?

could that be a duplicate of
Q: Photoelectric effect: Why is the saturation current the same, as you vary the frequency of incident light but keep its intensity constant?

IanDsouza image courtesy of http://www.learncbse.in/ Why is the saturation current the same, as you vary the frequency of incident light but keep its intensity constant? If intensity is a measure of the energy of the incident beam, the product of the number of packets of light and the energy of each pa...

11:55 AM
@skillpatrol hi
is this on-topic?
Q: A gravity Experiment (Theorotical)

MathloverI imagined a theoretical experiment about gravity. First of all, I know that it is very hard to build such test tool with our current technology but it is just theoretical experiment to enlighten the general relativity. Let's imagine, We have a big perfect sphere balloon whose inner side of t...

I hope so seeing as I've just answered it.
1 hour later…
1:05 PM
@JohnRennie Consider also fixing the question title, maybe? :)
In option c how they can find KE using average speed
1:16 PM
@JohnRennie nope.
It is perfectly possible to prepare a hydrogen atom in a $p_z$ state, directionality included.
And, in fact, "a superposition of all possible orbitals" from a given representation (i.e. fixed $l$) will always vanish unless $l=0$. Try it yourself with $l=1$, for example.
@Slereah Does AQFT not require that you have a local net of algebras of observables? In particular, I think true AQFT does not have a primitive notion of "field" and therefore no distributions.
1:33 PM
But don't you have to construct said algebra that way?
I guess that abstractly it doesn't matter but when you construct it don't you need to do operator valued distributions with support in O
can neutrons go unaffected through the diameter of the earth?
Q: Can you shoot a neutron through earth?

maccoMy dad once told me that he read an article about some physicists wanting to shoot neutrons (or protons) through earth as a way of communication. The explanation to why it would work being that at the atomar level theres mostly just "free space", so it would't hit anything. This seems highly unl...

@Slereah The notion of a local net of observables makes no assumption about how that algebra is constructed. If you choose to construct it through operator-valued distributions, it's your duty to ensure that actually yields algebras.
I saw a wavefront at some point, i guess maybe you need the wavefront condition on your operators?
I'm afraid I've never delved into any detail so I can't tell you how one would actually construct such a net
For instance I'm guessing $a$ isn't in the algebra since $[a,a^\dagger]^2$ wouldn't make sense?
1:51 PM
@rob I think the OP wants the Q to be reopened.
Q: Can you shoot a neutrino through earth?

maccoMy dad once told me that he read an article about some physicists wanting to shoot neutrinos through earth as a way of communication. The explanation to why it would work being that at the atomar level theres mostly just "free space", so it would't hit anything. Is there any way this could work?

Thanks, xkcd, for making me waste a considerable amount of time trying to find out why one would introduce a UTC+8:45 timezone for a few houses in the middle of nowhere.
@YashasSamaga Uh, the OP already pinged rob in a comment, you don't need to ping him again here
@ACuriousMind I don't get it?
@ACuriousMind what is the answer?
@skillpatrol Look at Australia.
@ACuriousMind I did.
2:02 PM
And...the "+8:45 for a small area" doesn't strike you as odd?
To save others time, it's the region around a tiny settlement called Eucla, but no one seems to know why it has its own time zone.
In total, this is a unique and highly irregular time zone for around...200 people.
It also seems to not exist "officially", but is observed in the region nevertheless, whatever that means.
how is the reader suppose to know that?
Why do you think the reader is supposed to know that?
I said "thanks xkcd for making me waste my time" precisely because I went hunting for the explanation after seeing it there.
Ok that makes sense, I said "how" without further investigation...
...I don't normally expect to investigate cartoons.
I didn't know Western Australia does not use daylight saving time.
2:31 PM
@PhysicsMeta Documenting this here: previous report of that nature
Is that the right link? I get a completely different question about luminosity and stellar mass when I click that. — Tim Seguine Nov 25 '14 at 9:09
::flags as obsolete::
@EmilioPisanty so you think the question is obsolete?
@skillpatrol no, I think the comment is obsolete
True dat.
@JohnR: How ridiculously dirty is this keyboard:
2:46 PM
(Full disclosure: That is not my keyboard)
And no, the white spots are not a play of the light, they are, indeed, dirty marks.
The guy is a community ads machine. — hBy2Py Feb 10 at 12:01
@ACuriousMind =D
Your fame transcends the boundaries of physics.SE ;)
I do wonder - is there a physics.SE ad running on any other site?
2:54 PM
@ACuriousMind don't think so
is it really necessary?
No, I think we're large enough to not need to make other SE users aware of our existence
Was just curious because e.g. chem.SE has a pinned message in their chat with ads for their site in the other science ad threads.
@ACuriousMind that's...
I mean, advertising for beta sites is a Good Thing
but they can wean themselves off of that one, I should think
Hm, still got that $[\phi(x),\pi(y)] = \delta(x-y)$
So even if I just include phi and pi in the algebra
It will be nonsense
Why is all QFT a lie
What is the bloody algebra
2:58 PM
@Slereah Consider that the local net of observables does not associate operators to single points as far as I can tell, so objects like $\phi(x)$ do not exist in it.
Well yes bit I highly suspect it will also be nonsense with test functions
@Kaumudi.H I guess the dirty F1 key marks the user as someone who doesn't need much help with their computer.
I forget what the CCR with test functions is
also someone who doesn't use TeXMaker
Also, it's a gamer's keyboard, look at WASD and F5 (movement/quicksave) ;)
3:01 PM
that thing on the right is the scariest thing if you've got astigmatism
@EmilioPisanty why?
@ACuriousMind good spot
@ACuriousMind if you look at it without specs, some lines look blurry and other ones look sharp.
but if you tilt your head, the sharp ones get blurry and the blurry ones get sharp
So...it's the scariest thing if you're astigmatic and frightened by blurry lines? :)
having "how the world looks is" depend on which way your head is tilted is not something you normally experience in everyday life
3:04 PM
Actually, that's just one of the keyboards at a browsing center near my home.
@ACuriousMind There's a female character in MGSV who runs around in a bikini and has jiggle physics. They give a legitimate explanation for it too
@EmilioPisanty ---struck text--- for struck text
@ACuriousMind thanks
@Kaumudi.H someone has been playing free online games then
@EmilioPisanty I find that more fascinating than scary, actually
3:06 PM
@ACuriousMind fascinating too
@0celo7 I know
Grrr...why do some many papers talk about "rigid holomorphic curves" yet no one explains what the "rigid" is actually supposed to mean?
@0celo7 can you solve the mystery
@EmilioPisanty turns out the answer for my question yesterday was the Hellman-Feynman theorem
Of the algebra
One can compute the dipole moment of the electron by taking the derivative of the energy shift (caused by the electric field) wrt. the electric field
3:10 PM
Really I think the answer is in Haag
@Slereah what is the mystery?
But it's a lot to read for that
@ACuriousMind I'm assuming you googled?
You might consider asking Ted in the math chat.
The mystery of why QFT fields don't form an algebra since they are distributions
But AQFT needs an algebra
@Slereah Better get crackin', then :P
3:11 PM
Are you sure the algebra they need is based on fields?
@ACuriousMind How do I algebra? My algebraic topology homework wants me to compute presentations of free groups
What is the strategy?
(that is probably too vague)
It is, I don't know what it means to "compute" a presentation.
Well, the C* algebra is the abstraction of the operators on the Hilbert space
And for KG, that's gonna be built from field operators
@ACuriousMind find an isomorphism to some "common" group
So I'm assuming that the construction of the algebra will be based on the QFT fields
@0celo7 Ah. I think that problem (deciding whether two presentations are isomorphic) is provably NP-hard or something. There's no "good" algorithm, you have to use your creativity ;)
3:14 PM
But those do not form an algebra
@Slereah are you asking what $[\phi(x),\phi(y)]$ means?
Test functions do form an algebra
The famed Borscher algebra
@Slereah Sigh. Have you been reading Jost?
You should learn analysis first
Who has the time to learn anything
Didn't duffield teach you anything
No need for math baby
Has Duffield suffered here in vain
Hm, my computer is doing that thing again where arXiv downloads are named "document" instead of the identifier. Still haven't found out what causes that.
3:21 PM
Why do people downvote Duffield's posts?
@YashasSamaga I'd remind everyone that we don't encourage discussion of chat-banned users in chat, but the obvious answer is: Because they think the posts are wrong.
Oh, I never knew he was even banned from the chat.
@YashasSamaga We are all mourning
Let me join.
Although feel free to downvote duffield
3:26 PM
for unjustified reasons too?
He is always wrong
@Slereah Really?
Fish in a barrel
I don't think calling for unjustified downvotes is allowed.
Is it unjustified if he is always wrong
3:29 PM
@ACuriousMind Pls clarify
@Slereah We are doing Jordan measure in my measure theory class. It's not even a measure
Can you find an open set that has no defined area?
Up- or downvoting particular users and not their content as such is forbidden, and it's likewise dangerous ground to try to incite others to do so.
Are there open non measurable sets?
However, if a particular user has a consistent pattern of posting wrong or irrelevant content, it becomes very difficult to meaningfully distinguish user-voting and content-voting.
@ACuriousMind ;)
@Slereah isn't the Borel algebra meant to be precisely that generated by the open sets?
3:31 PM
@EmilioPisanty Yes.
But Jordan measure is not a measure.
It's a stupid name.
What is it then
Jordan measure is in many ways more intuitive than Lebesgue.
You shouldn't go looking for posts from a single user to downvote (and if you do you'll likely run into the serial voting detection script)...but if you downvote their posts for being wrong or irrelevant as you stumble upon them during your normal use of the site, it's fine.
@0celo7 ah, so non-Jordan-measurable open sets?
no idea
@EmilioPisanty Right. Jordan measure is the "measure" wrt. one defines the Riemann integral.
My measure theory prof decided we should do Riemann integration since that's what most people implicitly use, but never actually see the proofs.
3:33 PM
@0celo7 ok, so, remind me, what's a Jordan-measurable set?
@EmilioPisanty $A\subset\Bbb R^n$ bounded. Put $A$ in a rectange $R$
one where the upper limit of inscribed boxy sets coincides with the lower limit of superscribed boxy sets?
@EmilioPisanty yes
@0celo7 hmmmm, good question, then
my instinct would be no
i.e. open $\implies$ Jordan measurable
but then this is measure theory
Try the rationals in $(0,1)$.
3:35 PM
@ACuriousMind Nice open set you've got there
Doesn't the riemann integral fuck up for like indicator function of irrationals or something
@ACuriousMind the rationals is not an open set
Oh, open, nevermind me then :P
@EmilioPisanty Well, it's known that a set is Jordan measurable if and only if (might only be one direction) its boundary has Lebesgue measure zero.
There are open sets whose boundary have nonzero Lebesgue measure.
@0celo7 wait, what?
3:36 PM
So one of those would work. But those are probably terrible to construct
I guess... maybe?
@ACuriousMind That actually shows that Jordan measure is not a measure.
That set is not Jordan-measurable, but each point is.
So it can't be $\sigma$-additive.
@EmilioPisanty On the other hand, a Jordan measurable set is Lebesgue measurable.
@Kaumudi.H Ugh! :-) Though the dirt seems to be mainly dust. I've seen keyboards where the dirt was mostly finger grease!
@JohnRennie Oh no, that was not mostly dust, no Sir! I'm pretty sure that there were some forms of grease on it as well. I had to type on that!
3:41 PM
Did it though, flinching every few seconds.
Ugh! :-)
@JohnRennie I don't have a printer at home so have to resort to this :-(
@Kaumudi.H that was at a print shop?
Browsing center--u can print, scan, take Xerox copies also.
3:44 PM
Sigh... "You must award this bounty as soon as possible ", no, SE, I must not. Why this has been known as a bad formulation for more than four years and is still not fixed is a mystery to me.
You can cancel your bounty can't you ?
@ACuriousMind I'm considering extending that bounty for another week
@JohnRennie not that I'm aware of
I think they mean it as a moral imperative
@JohnRennie lol, yes, but that would be actual abuse of moderator power
If you don't you are a bad man
3:45 PM
The point is that I can decline to award it at all.
@EmilioPisanty mods can cancvel and refund bounties
@ACuriousMind when are you changing your avatar?
@JohnRennie in exceptional circumstances
btw, the message "you must award" also appears when there's no answer at all...which is terrible design
e.g. you want to migrate the question after having been pointed to a better site you were unaware of
3:46 PM
@ACuriousMind I take it that's the Majorana fermion question?
@JohnRennie oh, you mean "you" as in "you, ACM, specifically, as a mod"
@ACuriousMind actually mine is tue issue
It looks too much like the troll you once used
I keep thinking I'm you
What troll?
Thrall is an orc, dammit!
You once had a green troll face
trolls are not real
3:48 PM
@0celo7 That's...absurd (to say the least) :-|
@skillpatrol I take it you're new to the internet
@Kaumudi.H I love you too
x'D OK man, whatever u say.
Unless you are saying I'm just too handsome to confuse myself with ACM
I'll interpret it like that
@EmilioPisanty nah, just trolling ;)
3:49 PM
@JohnRennie Yup. I wish there was a button "I don't wish to award this bounty" since it's kinda pointless to have it hover in the grace period.
@0celo7 Lol, OK x'D
This guy seems to be taking the piss:
Q: why classical physics failed to explain the fact that atomic spectra are discrete

Tshimangadzo Ace NesaneI think it has something to do with the fact that they consist of separate lines

@JohnRennie what does that mean in English?
He is asking questions that a minutes Googling would answer,
Taking the piss is a Commonwealth term meaning to take liberties at the expense of others, or to be unreasonable. It is often used to mean (or confused with) taking the piss out of, which is an expression meaning to mock, tease, ridicule, or scoff. It is also not to be confused with "taking a piss", which refers to the act of urinating. Taking the Mickey (Mickey Bliss, Cockney rhyming slang) or taking the Michael is another term for making fun of someone. These terms are most widely used in the United Kingdom, Ireland, South Africa, New Zealand and Australia. == Usage == The term sometimes refers...
3:51 PM
@JohnRennie "Commonwealth term"?
^my thoughts exactly
does that just mean "everybody but the US"?
British Commonwealth
@JohnRennie Wow. My first thought was "being treated badly". Ya know, taking the piss and all.
3:52 PM
I never take the piss. Gross.
The Commonwealth of Nations (formerly the British Commonwealth[4]), also known as simply The Commonwealth, is an intergovernmental organisation of 52 member states that were mostly former territories of the British Empire. The Commonwealth operates by intergovernmental consensus of the member states, organised through the Commonwealth Secretariat and non-governmental organisations, organised through the Commonwealth Foundation. The Commonwealth dates back to the mid-20th century with the decolonisation of the British Empire through increased self-governance of its territories. It was formal...
@Kaumudi.H Ew.
@ACuriousMind maybe leave a vote here: meta.stackexchange.com/a/290612/271002
@JohnRennie obviously, but what are non-Commonwealth English-as-a-first-language countries?
Come on chaps. Keep up at the back.
3:52 PM
:-P Pardon me for being crude.
your pardon is under appeal :P
@JohnRennie also by OP:
Q: classical physics and photoelectric effect

Tshimangadzo Ace NesaneWhy did classical physics failed to explain the fact that in photoelectric effect, there is a threshold frequency value below which the effect does not occur? I not sure if my answer actually states why it failed but this is what i think: this is because different materials have different thre...

@Loong Thanks
Hi @BalarkaSen
3:54 PM
chillin. you?
@EmilioPisanty He asked at least two really basic PEE questions.
doing math, listening to music
@JohnRennie exactly three
Q: How did Einstein explain the photoelectric effect?

Tshimangadzo Ace NesaneThe notes given to me claim that the intensity of electric current depends on the intensity of radiation flashed on the metal. Why is it so?

(so far)
@0celo7 abbreviation for "Photoelectric Effect" used by people testing the boundaries of chat-appropriate language =P
3:56 PM
probably would also watch a movie today
There you go then, he is TAKING THE PISS
@BalarkaSen Nice! Which one?
@JohnRennie No, seriously, are there any non-Commonwealth countries with English as a first language other than the US?
@BalarkaSen Ooh, sounds...racy and also a bit disturbing.
3:58 PM
@EmilioPisanty Eire
@JohnRennie don't they speak Gaelic?
@Kaumudi.H It is disturbing. It's a big cult movie coming from the 70's counterculture
@BalarkaSen Hmm, I see...
But there are far disturbing stuff out there from that era of horror-genre films. Hellraiser et al

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