12:22 AM
hmm... what's this symbol that looks like a cursive r, in the first equation? detexify doesn't recognize it.
(it's in the denominator of the second fraction)
ah, nvm, found it:
12

The near-ubiquitous undergraduate reference on electrodynamics, "Introduction to Electrodynamics" by David Griffiths make extensive use of a script-r symbol, defined to be \vector{\scriptr} \equiv \vector{r} - \vector{r}^{\prime} and looks like: Although the question of how to produce such a s...

Nice work Sherlock :-)

ick, now I have to figure out how to format it in newcommand
so i take it back, my request for help still stands =P

1:11 AM
$\mathfrak{r}$
$\mathscr{r}$
$\mathcal{SUCKS}\,\,\mathcal{TO}\,\,\mathcal{BE}\,\,\mathcal{YOU}$
@heather I have a problem for you to think about.
A long-term goal.

0

WHY WHY WHY WHY WHY WHY WHY WHY WHY CAN'T I DO IT WHY why can't I do it that's stupid I want to do more

1:46 AM
@0celo7 okay, sure.

2:06 AM
Hey, would someone like to discuss a hypothetical universe ?
What would happen it the second law of thermodynamics was inverted
Energy would from more concentrated to less concentrated

@heather Ok, are you here?

@0celo7 yep

@heather Let $g:[0,1]\to\Bbb R$ be a fixed continuous function. Show that there exists a unique continuous function $f:[0,1]\to\Bbb R$ such that $$f(x)-\int_0^xf(x-t)e^{-x^2}\,dt=g(x).$$
$x\in[0,1]$ of course. @heather Does that make sense?

i'm still deciphering it =)
i think i'll need to learn a bit more to even understand the problem.

Hmm? What is confusing?

2:18 AM
well, i'm not sure what is meant by the notation $g:[0,1]$
I've seen it before, but I can't remember what it means.

$f:X\to Y$ is a function from the set $X$ to the set $Y$.

ah, okay
okay, so then "fixed continuous function" is a phrase I'm not familiar with.

Omit "fixed" then.
$g$ is given is what I'm trying to say.
@heather Do you not know what continuous means?

i think so...
there's no point where the function is undefined.
nothing like $\frac{1}{x}$ (because when x=0, the function is undefined) for example

wat

2:23 AM
okay, I'm wrong

$x\mapsto 1/x$ is not a function on $[0,1]$

no, no, I know it isn't
I'm saying 1/x on $\mathbb{R}$, sorry - I was just trying to give an example.

Hmm. I guess the best way to explain it is that a continuous function is one whose graph is "connected"
@heather It's not a function there either. $0\in\Bbb R$

wait, what?
I'm so confused then
I know 0 is in $\mathbb{R}$

$f(x)=1/x$ is not a function on $\Bbb R$
it is not defined at the origin

2:25 AM
yes, that's what I'm saying
it's a non-continuous function.
or is that not how continuous function is defined?

But it is continuous on its domain.
@heather It's one for which the graph is connected.

@0celo7 where there is no number you plug in for which the output is undefined?

brb

2:39 AM
sorry, my mom wanted me to do some stuff
@0celo7 okay, um...if you have a function $f:X\to Y$ and you input a $x\in X$, the output will be a $y\in Y$ (never a $y\notin Y$)

It's funny how a person can be blasé about one class of risks and get the heeby-jeebies about another one that is objectively no more substantial. This ...
0

I want to make small (3x2cm) pulsed (100kHz) laser with 905 nm and 1 uJ energy output per impulse(~10 ns). It can be done with laser driver + diode. But I found small laser drivers only up to 70kHz. The size is small so I dont think that it is can be made with resonator, am I right? May be the...

makes me shudder ... scary class 4 blindificator. Eeek!
But i have worked with potentially lethal (slowly and nastily lethal, at that) radioactive sources more than once and take very matter of fact approach to dealing with them.

i have to go for now. i will try to read more about your problem @0celo7

2:58 AM
@heather You will need Banach space theory to solve it.

2 hours later…
4:30 AM
Hi, everybody.

allo... Can I ask you an off-beat question about the site?
Maybe it's worth a question on meta...
you can tell me.

@ZeroTheHero Rule #1: Don't ask if you can ask. Just ask.

I just found a student in one of my courses posted a question... the question is not from my course but from the instructor next door.
... and it's a homework-type question.
so just wondering if there are guidelines for this type of situation.. couldn't find any... but maybe there are.

@ZeroTheHero Homework questions get closed pretty soon.

@anonymous yeah
the question isn't catching fire... but it's kinda very spooky.

4:45 AM
@ZeroTheHero Which question?

@anonymous I'd rather not say... it will die of its own death...

@ZeroTheHero Posting questions from homework assignments is totally acceptable here.

@DanielSank I realize this...

However, we have a so-called "homework policy" which essentially says that if you post asking for help with a specific problem, you have to 1) show effort, 2) identify a specific thing where you want help, and 3) ask something conceptual, not just "what did I do wrong".

@DanieS:

4:47 AM
@DanielSank which will lead to the question dying of natural death... I'm just suddenly confronted with an unexpected situation.

@ZeroTheHero What is so unexpected ? Isn't this the normal thing in every school/university ? :P
You can't stop a person from bypassing rules if they are bent on doing so. Just let them face the eventual consequences of their own actions.

@anonymous when the question was asked by your office neighbour, it's unexpected.
@anonymous yes I agree with this...
suddenly the homework policy takes on a very different materiality.

@ZeroTheHero Where do you teach? And what subject? :) (If you don't mind saying..)

@anonymous I teach physics... at least that's what I'm paid to do... the students might have a different opinion.

@ZeroTheHero LOL XD

4:56 AM
(as to whether or not I actually do any teaching)

5:06 AM
@anonymous out of curiosity: are you in the teaching business yourself?

@ZeroTheHero I'm just a school student

@ZeroTheHero Early in my teaching career I got a sheaf of solutions to a take-home part of an exam that were all wrong in the same, very strange way.
And one that was write in the same strange way because it used concepts more sophisticated than we'd gone over in class or in the text.

@dmckee you just have bright students!
inspired by a great teacher.

Sadly, no. Googling the question text brought up the place on Chegg where someone had answered my test question from a slick upper-division prospective.
One student had copied that down verbatium.
His friends has copied his work, only where he had written a rather florid and overwrought $v_2$, they read (and copied) $\sqrt{2}$.
And didn't notice that this had them equating $\sqrt{2} = 13 \,\mathrm{m/s}$.
I got five papers like that.

@dmckee 'been there done that...
my best ever will top that.

5:11 AM
@dmckee Isn't chegg paid? Or did he/she buy subscription just to copy the solutions ? :'D

It was the failure to notice the dimensional inconsitency that really annoyed me: we had drilled that all semester long.
@anonymous Chegg is paid. I don't know if the student bought in just for my class or made a habit of it.

many years ago... back in the pre-internet days... I was TAing and got a lab report where every other page was printed with a different printer. All the even pages with a daisy wheel, all the odd pages with a dot matrix.
it took all of 5 minutes to compare with the other TA in the lab... who had the dual version.

@ZeroTheHero Oh. My. The story behind that must be a riot.

the students had split the pages because they figured the lab report printed on the daisy wheel would get a better mark 'cuz it looked nicer.

@ZeroTheHero ::chuckles::

5:14 AM
@dmckee yeah... and you only catch the stupid ones.

Due to similar incidents our school stopped giving out graded homework assignments. Nowadays we get grades only based on test scores.

@ZeroTheHero Well, I always have an in-class component to exams, because I know that take-home is the same as open-resources (and I make that the explicit case with mine these days) but in class they have to do the work themselves. And under time-pressure, too.

@anonymous I am ambivalent about this... I always thought some longer more conceptual problems should be assigned.
@anonymous I recognize that this may not be practical anymore...

@ZeroTheHero Yes that is true. But it is not a good idea to grade them based on those assignments. IMHO.

We can obtain the momentum of a photon using $p = \frac{hf}{c}$
What is the $\Delta{x}$ there?
$h$ is known,$c$ is known, $f$ can be known as we control the source
so we know the exact momentum?
maybe we aren't sure if the source really produced a photon of frequency $f$?

5:21 AM
@YashasSamaga Try to find $\frac{dp}{dc}$.

@dmckee yes... we are encouraged to move more towards "in-class" type grades.

@YashasSamaga Error in c

for a change in the velocity of light?
Don't we know the exact value of $c$?

@YashasSamaga To answer that you have to start by understanding how the detector works.

5:22 AM
$c$ can't be accurately measured. It depends on medium.

Aha!
Let the medium be vacuum.
@dmckee I already know the information about the photon before it even enters the detector.

There is nothing called complete vacuum.
So letting it is an ideal situation...

@anonymous wait 'til you meet your first administrator.
... talk about complete vacuum.

Just use error analysis method to find error in $p$

@YashasSamaga AH, so you propose to determine this stuff at production. OK. Then we need to know how that works. With enough detail we can find the uncertainty.

5:25 AM
By taking log on both sides and differentiating..
@ZeroTheHero Who is an administrator ?

@anonymous Learning to understand the practical and intrinsic limits of measurement is a very useful thing, and it is not taught deeply enough in most programs due to lack of sufficient time. ::gets off experimentor soapbox::

@anonymous you will meet one in due time... and you will experience true vacuum.
@dmckee hear hear!

@dmckee I realize that :). I was saying just from a mathematical viewpoint.

@ZeroTheHero Oh please tell me you are another wrench and soldering-iron physicist. We need more experimenters on the site.

@dmckee regrettably no... It's not my fault but I'm merely a theorist.

5:30 AM
@ZeroTheHero You've made me curious. What is it? :P

@dmckee all the experimental skills of the family passed to an uncle, who incidentally is a master welder.

@anonymous Think of Dilbert.

@anonymous here's a taste: profron.net/fun/Administratium.html

@ZeroTheHero Ah. I'm not in that class—though I have wielded a welding torch once or twice—and I'm not actually that good with a soldering iron. But I am solid practical engineer, a hardware bit-twiddler of basic skill, and a wiz at just getting data around a big program.

If I try to produce a photon of frequency $f$ at the source, the machine needn't necessarily produce a photon of frequency $f$. There is uncertainty there.
What if a use a crystal source?
Aw, now there is uncertainty in measuring the frequency of the radiation emitted by that source.
I give up. Physics works!

5:33 AM
@YashasSamaga The physical dimension of the crystal will control how closely you can tune the resonance.
@YashasSamaga Yeah. And it can be fun to play "Hunt the source of uncertainty" with gedankenexperiments.
2
Any way, it's getting latish here and I have class tomorrow. Night, all.

bye tc :)

5:57 AM
Is there any easy way to divide numbers like 12431/12.091 accurately upto 2 decimal places, manually ? :/ These long divisions freak me out!

Vedic Mathematics

@YashasSamaga Which method ?

VM has all kinds of weird methods for each type of division.
There is even a trial and error method.

Vedic maths has hundreds of techniques.. most of which are difficult to remember

Yea.

5:59 AM
@YashasSamaga does it work for non-integer division? I only remember division by integers.

non integer division is good approximate of integer division
1.5/23 = 15/23 * 0.1

Why the hell JEE doesn't allow calculators!!! :P

because you don't need them
if you get horrible numbers, then your answer is wrong
or you are solving some FIITJEE paper

@YashasSamaga Look at JEE advanced 2016 modern physics question.
Even my teacher needed a calculator to solve it

@anonymous are you allowed a slide ruler?

6:01 AM
nope

Binding energy calculation
@ZeroTheHero No...nothing. Just a pen and pencil

1. The question tells you what unit to use
2. The answer choices aren't close enough
so approximations should work
paper 1 or paper 2?

There was binding energy problem
I forgot paper 1 or 2
I'll let you know

this paper?
aw that fail

See question 15 for example
I think the other question was in paper 2

6:06 AM
the paragraph question?

the integer type

it is just stupid
oh

Anyway...gotta go now :). Bye!

6:36 AM
@YashasSamaga only an infinite, NB infinite, plane wave has a precise frequency. Any wave packet of a finite length has a spread of frequencies.

Calculate $$\frac{3680}{3681}-\frac{1840}{1841}$$ manually upto 6 decimal places of accuracy (manually). This is from a Modern Physics problem. How would you do it? I would like to know @YashasSamaga
Is there any Vedic Maths method for this ?

I know how I'd do it...

@DavidZ How?

note that 1841 = 3682/2

@DavidZ Ok? Then..

6:42 AM
nvm lol

:35470365 What is that ? :'D
Ok
These problems freak me out

@anonymous what, you really want a complete walkthrough? I'm not here long enough for that :-P

@anonymous have you cleared KVPY?

@DavidZ Your method doesn't seem to provide any sort of shortcut
The denominators aren't related

Dumb question: what is the upper bound of the total surface area of the earth including all buildings and natural formations?

6:44 AM
@YashasSamaga I didn't appear for KVPY.

@anonymous If you use the fact I noted, they are related: they differ by one

Dumb question 2: why there are no world records for the largest surface area of earth covered by their travelling?

@DavidZ So how does that relation help in simplifying the problem? Am I missing something?

I don't have a complete solution written out or anything, but that probably lets you use identities like $$\frac{1}{n} - \frac{1}{n+1} = \frac{1}{n(n+1)} \frac{3680}{3681 \times 3682} 6:45 AM @DavidZ Hmm. Finding$$\frac{1}{n(n+1)} is a humongous task
By long division

cross 3680 and 3681
find 1/3700
uhm, you need 6 decimal places, right?
did you solve it correctly?
u dont get such Qs in IITJEE
that is a trick actually, you get bad numbers, your answer is wrong

@YashasSamaga Let's wait and see. After seeing last year's paper I don't trust them anymore :P
The JEE main paper had 4/5 mistakes. The advanced modern physics had huge calculations and that lens/mirror problem was wrong (that lens with tilted ray and axes).

Hi, everyone.

@YashasSamaga You gave IITJEE last year? You said you joined IISC...

why do people ask it again and again lol
I dropped out
I am made for engineering... science won't work for me

6:54 AM
@YashasSamaga I know that. I am asking whether you appeared for jee or not...

yes I did

What was your rank?
(Don't say if you mind)

:3 not good
in the early thousands
I wouldn't get CS in any IIT
except IIT Dharward which is new and it sucks

NIT ?

I made a mistake... should've joined NIT :P

6:56 AM
:)

there is no B Tech CS in IISc :(
there is MTech though
NITK is just few miles away from my home

Why can't an electron give away a part of its energy and change its frequency to excite an energy from a lower energy level to higher energy level? Why does its energy always have to exactly match the difference between any two energy levels?
Anyone here knows ? ^

give away the energy as what?

@YashasSamaga As KE ?

then it would be photoelectric effect
if you are talking in the relam of classical mechanics (Bohr model), then the K.E of an electron in a particular energy level is FIXED.

7:07 AM
@YashasSamaga Yeah, so why can't a photon give increase the KE of one electron and therby making it suitable for the higher energy level ?

When you talk about energy level of an electron, it is the sum of P.E. + K.E.
The photon must also have enough energy to help the electron overcome the electrostatic attraction.

@YashasSamaga KE and PE are interconvertible..

No.
It doesn't work that way with electrons.
You can derive the formule easily.

@YashasSamaga Why not?

You need to assume that the angular momentum is quantized, i.e: L = nh/2pi
K.E = 0.5 mv^2
mvr = nh/2pi
F = kZe^2/r^2

7:09 AM
@Kaumudi.H Morning :-)

mv^2/r = F

@YashasSamaga That is true only in a particular shell. A photon could lend its energy to provide KE and increase the energy of the electron.

once you get v, you get the kinetic energy
the kinetic energy of the electron in a given state is fixed
the electron cannot take up more K.E

@YashasSamaga hmm, then how does photoelectric effect work?

the photon used in photoelectric effect have way too much energy
more than the binding energy of the electron
in a ground state hydrogen atom, the electron's binding energy is 13.6 eV

7:10 AM
@YashasSamaga So when did I say the photons are of low energy...?

I am confused what you are asking now lol
anyway, if the photon has more than 13.6 eV, it can bring the electron out of the nucleus' influence.

I never said that the photons are of low energy. In most books it is written that photons can't cause an electron to jump to higher level if it doesn't have the exact energy difference between two levels. However, I feel it can if it has sufficiently large energy.
Say a photon has 20 eV energy
It could easily excite an electron to the next energy level in H atom.

You need around 10.2 eV to move an electron from 1 to 2 in the ground state hydrogen atom
suppose you give 10.4
where do you expect the 0.2eV to be stored?
to be honest, a 10.4eV photon can excite the atom

@YashasSamaga The photon can change its frequency to alter its energy to 0.2 eV. But most books say it can't. My question is :Why ?

you'll have another photon emitted
I think you can.

7:16 AM
@JohnRennie Could you give us some insight, please ? :)
2 mins ago, by Yashas Samaga
You need around 10.2 eV to move an electron from 1 to 2 in the ground state hydrogen atom
2 mins ago, by Yashas Samaga
suppose you give 10.4
2 mins ago, by Yashas Samaga
where do you expect the 0.2eV to be stored?
2 mins ago, by anonymous
@YashasSamaga The photon can change its frequency to alter its energy to 0.2 eV. But most books say it can't. My question is :Why ?

The photon can't change its frequency but it can transfer the extra energy to the whole atom.
I am turning pages in Feynman's book searching for it

@YashasSamaga Why can't photon change its frequency after transferring the required energy to excite electron to another energy level?

I don't know what you mean but I assume that you are saying it transfers the energy needed to excite the electron and goes back with the extra energy.
Well, there have been IITJEE problems where you get a photon with excess energy. We assume that the extra energy goes into atom's kinetic engery.

@YashasSamaga Those problems involved photons of energy greater than 13.6 eV
You can check.
In Compton effect photons do change their frequency upon collision
Why can't the same occur on hitting electrons?

Compton effect deals with free charged particles
and the charged particle can be an electron

7:24 AM
Hmm, so what prevents it in an atom?

If an atom could take in photons of all energies, we wouldn't have an absorption spectrum?
We would get some sort of continuous spectrum instead of a line spectrum.

@YashasSamaga I am not saying that the atom takes in photons of all energy. I am saying that the atom could take the required energy say 10.2 eV out of 10.4 eV and leave the photon with 0.2 eV.
Obviously it can't take in 10.4 eV completely

Your logic does not work. If the atom takes a photon of 15eV and releases a photon of 4eV, the 15eV dark line would be visible in the absorption spectrum.
but we don't observe it

@YashasSamaga Ah! So WHY don't we observe it?
Is there any logic ?

1

The electronic states of a molecule are eigenfunctions of the time independent Schrodinger equation (with a few approximations like the Born-Oppenheimer approximation). This means those states are time independent so a ground state will never rise to an excited state and an excited state will nev...

7:30 AM
Aha! "
We can calculate this probability using perturbation theory, and the equation is called Fermi's golden rule. Doing this calculation tells us how likely the photon is to promote the atom to the excited state, and we'll find that this depends on the photon energy and the probability is high only when the photon energy matches the energy difference between the states."

I expected such an answer :D We were talking about these things using the Bhor model.

Fermi's golden rule
Ok I need to see
^
Never heard of it
perturbation theory
I know nothing :/

You should have written KVPY.
If you had cleared KVPY, you could have attended the National Science Camp.

@YashasSamaga My father didn't get time to take me to India for KVPY...

You'd have lectures on basics of QM.
superconductors, superfluids, etc.
Only 600 students in India get that chance but it is worth it.

7:33 AM
He will take a leave for one month for my jee
Going to another country for giving exams is hectic.

Where are you now?

I usually stay in Kuwait, and on holidays in Assam or Dehradun or Durgapur
@YashasSamaga I've heard of it
I will get to learn those anyway in college

7:49 AM
Interestingly an electron can donate any part of its energy to an electron inside an atom upon collision.
@JohnRennie Thanks. That was a good answer. But I don't know many of terms used there. :)
Like Perturbation Theory
Hope to learn them soon in QM.