The h Bar

JokelaTurbine

00:44

....And a happy new year... physics.stackexchange.com/questions/240979/…

DanielSank

03:12

Enter Bernardo.

Bernardo Meurer

@DanielSank HAI

So

Here's my problem, there's this humongous question

DanielSank

k

Bernardo Meurer

and at one point I gotta find $g'(t)$

Ramanujan

@PhysicsMeta are you bot?

Bernardo Meurer

$$g(t) = \int_2^t (2-t)f(x)dx$$

DanielSank

03:14

ok

Bernardo Meurer

@Ramanujan Yes, he is

DanielSank

What is question?

Bernardo Meurer

@DanielSank How do I do this? I just did $(2-t)\int_2^t f(x)dx$ but then I got stuck

Should I just apply the fundamental theorem of calc on that integral?

Oh no

I gotta do the product rule for the derivative too

Sigh

DanielSank

Yep.

:D

Bernardo Meurer

Ged demmit

Gimme a sec

user228700

03:22

Hello, everyone :-)

DanielSank

Hello.

Tiramisu, eh?

I have been following your food journal.

user228700

:-) Yep. It was delicious.

user228700

I have a quick question about the rate law for a given reaction. My textbook says:

user228700

> "Generally rate law expressions aren't simple and these may differ for the same reaction depending on conditions under which the reaction is being carried out"

user228700

Is this really true? Can the same reaction have different expressions of the rate law depending upon the conditions?

DanielSank

03:25

I have no idea what your book means by "same reaction" under "different conditions".

That sounds like a contradiction in terms.

user228700

I think they mean the same reactants reacting to give the same set of products except, maybe the they add a catalyst or the temperature is increased, etc.

Bernardo Meurer

@DanielSank That derivative will be wrt $t$ and not $x$ right?

Ramanujan

@DanielSank different conditions means different temperature,pressure…

DanielSank

@BernardoMeurer What do you think?

Bernardo Meurer

so, for example $(2-t)' = -1$

DanielSank

03:28

@Ramanujan Ah.

Bernardo Meurer

@DanielSank yes

DanielSank

The $'$ notation is ambiguous, but yes.

Bernardo Meurer

I think it will be a derivative on $t$ sine $g$ is a function of $t$ and not of $x$

DanielSank

Right, $x$ is meaningless. It's just a way to tell you what's being integrated. It's a for-loop dummy index.

Bernardo Meurer

This prof loves dumb code

DanielSank

03:32

prof?

user228700

Never mind, thanks!

Bernardo Meurer

@DanielSank Applying L'Hopital 8 times in a row?

@DanielSank I'm a little stuck, this is what I got

$$-\int_2^t f(x)dx + (2-t)f(t)f'(x)-f(2)f'(x)$$

DanielSank

I dunno maybe. I don't have paper in front of me.

Bernardo Meurer

.-.

It's not right

DanielSank

ok just a sec

$$\frac{d}{dt} \int_2^t (2-t) f(x) dx $$

Bernardo Meurer

03:40

$$g'(t)=-\int_2^t f(x)dx + (2-t)f(t)$$ is the right one

DanielSank

$$\frac{d}{dt} \left( (2-t) \int_2^t f(x) dx \right)$$

$$(2-t) \frac{d}{dt} \int_2^t f(x) dx - \int_2^t f(x) dx$$

$$(2-t) f(t) - \int_2^t f(x) dx$$

So yeah, you got it.

Bernardo Meurer

Should that last one be $f(t)$?

So I got it right?

DanielSank

@BernardoMeurer I don't know what "the last one" means. Use math, not English.

@BernardoMeurer Pretty sure, yes.

Bernardo Meurer

@DanielSank How is mine right? It looks different

$$-\int_2^t f(x)dx + (2-t)f(t)f'(x)-f(2)f'(x)$$

DanielSank

@BernardoMeurer That one is right.

Bernardo Meurer

03:50

Hm

Well then

DanielSank

Where did you get $f(2)f'(x)$?

@ACuriousMind d00d, use a password manager.

Secret

newscientist.com/article/…

Tool using ants

Bernardo Meurer

@DanielSank I just applied the fundamental thm

DanielSank

@BernardoMeurer Can you please use math instead of english?

Bernardo Meurer

$$F(x) = \int_{a(x)}^{b(x)} \\F'(x) = f(b(x))b'(x)-f(a(x))a'(x)$$

DanielSank

03:54

@BernardoMeurer That's because you have $x$ in both limits.

Secret

NOW THAT's REALLY STRANGE:

phys.org/news/…

Looped trajectories in 3 slit experiments

Bernardo Meurer

@DanielSank How would I apply that here then?

Secret

Full paper here (is open access): nature.com/articles/ncomms13987

DanielSank

@BernardoMeurer $a'=0$ here, d00d.

Bernardo Meurer

AH

and t is constant wrt x

so it's just $f(t)$

Secret

03:58

0

Q: A visual perception of time

So I'm creating a character that is an extradimensional being exists outside of time. If we experience time as a linear event, how would time be perceived if this being has the ability to see time in a way that we look at how people move through three dimensions. I was initially going to base it...

physics

=>Refer to canonical answer

DanielSank

@BernardoMeurer I don't know what "it" means. Listen, please, use math to talk about math.

Please.

Bernardo Meurer

It's right, don't worry about it

DanielSank

04:09

ладно

Bernardo Meurer

No more slavic languages today

Sorry

Nika already cursed me in like all the different Yougoslav languages while playing darts

4

Bernardo Meurer

04:54

@DanielSank

$$\int_0^{2\ln(x+1)} e^{\cos t}dt = e^{\cos(2\ln(x+1))}(2\ln(x+1))' - e^{\cos(0)}(0)'$$

David Z

05:06

9

Q: Does anyone know what that blue thing is?

It looks like the bulb of the lamp beside it but how why what

visible-light geometric-optics

On topic? I'm a little skeptical

Bernardo Meurer

@DavidZ Do you know integrals?

Or rather, do you still remember how to integrate stuff?

user228700

05:25

Hi again. I have a "quick" question. I've been given that $rate=K1C/(1+K2C)$ and I've been asked to find the order of the reaction when $C$ s very very low. Since $C$ is very very low, I estimated that $1+K2C=1$

user228700

Is this an okay approximation to make? What my textbook has done is that they assumed $1+K2C=K′$. Hence, we get $rate=K1C/K′$. Either way, I find that the reaction is a first order reaction for very very small values of $C$. Which way is correct tho? Why does it make sense to assume that $1+K2C$ is just some constant?

Bernardo Meurer

05:41

@Kaumudi.H @0celo7 Wants to know why you are always talking about food

4

user228700

@BernardoMeurer x'D Firstly, tell him I say "Hi" and secondly, well, food is an excellent "simple pleasure" and I enjoy speaking about it to other enthusiasts, much like physics/math. However, I promise that there will be no more food-talk from my side from now till June.

Secret

0

Q: Are particles really in a superposition before you observe the particle

So if I understood correctly, Schrodinger's Cat is a thought experiment that puts a cat inside a box, and there's a mechanism that kills the cat with 50% probability based on a quantum process. The argument is that the cat now must be in a superposition of dead and alive. But, is the cat really i...

quantum-mechanics superposition schroedingers-cat

It is, otherwise entanglement will not work

But what about isolated paritcles?

Secret

06:00

Update: Ok, looped trajectories are not strange, but still kinda interesting. I wonder when we will want more looped trajectories?

anonymous

@BernardoMeurer Hi. Do you know how to integrate that function ? (Just curious as I could not figure out a direct way) What you did there seems more of differentiation using Leibniz rule...

Bernardo Meurer

@anonymous I do, yeah

Just follow from what I wrote

the right part will vanish

then it's handwaving

@JohnRennie I need help integrating :P

anonymous

@BernardoMeurer Well, this term $$e^{\cos(2\ln(x+1))}(2\ln(x+1))'$$ remains

So I use substitution ?

Umm nice

Bernardo Meurer

Yeah, so, solve the derivative for $2ln(x+1)$

and you'll be done

John Rennie

@BernardoMeurer These days I generally leave integrals to Mathematica I'm afraid. I'm badly out of practice with integrals.

Bernardo Meurer

06:07

you should get $$\frac{2}{x+1}e^{\cos(2\ln(x+1)}$$

anonymous

Ah great! But unfortunately it works only for this particular limit :P ...if we changed the bounds of integration then it won't be so simple :) BTW do you know of any direct method when we suppose solve the integral from 0 to $\pi$ ? @BernardoMeurer

Bernardo Meurer

IIRC, it's buried in my notes already

@JohnRennie physics.stackexchange.com/questions/303471/…

@anonymous Meh, I guess just use the fundamental theorem of analysis and handwave

@JohnRennie Have you seen that question yet?

anonymous

@BernardoMeurer $\int_0^{\pi}e^{\cos(x)}dx$. Nah, I don't think fundamental theorem helps here.

Bernardo Meurer

@anonymous Then just solve it normally, i.e. $\int_0^\pi e^{\cos(x)} = e^{\cos(\pi)} - e^{\cos(0)}$, no?

$=e^{-1} - e$

anonymous

@BernardoMeurer What??? You didn't even integrate the function... You just put in the bounds without integration

wolframalpha.com/input/?i=integrate+0+to+pi+e%5E(cos+x)

If you don't believe see Wolfram Alpha gives a totally different answer

Bernardo Meurer

06:15

it's 5am

I'm stupid

anonymous

LOL XD

Bernardo Meurer

Anyway, yeah, it seems like you can't solve for that one

anonymous

Seems so...I have to ask on Math SE

Bernardo Meurer

Not with Riemann integrals at least

Hmmm

anonymous

Contour Integration will work I guess

John Rennie

06:16

@BernardoMeurer That's waaaaaaaaaaay over my head!

Bernardo Meurer

Although

I guess the set of discontinuities is countable so it does suffice for Riemann integrability

Yeah, well, it is Riemann integrable I believe

Just hard af

anonymous

@BernardoMeurer e^(cos x) isn't discontinuous...

Bernardo Meurer

@anonymous Which is exactly equivalent to what I said

anonymous

@BernardoMeurer In a twisted fashion :P

anyway it was a nice problem

Bernardo Meurer

I used the stronger version of the statement :P

i.e. that the function doesn't have to be continuous, it just needs to have a countable set of discontinuities

anonymous

06:22

@BernardoMeurer True :)

Bernardo Meurer

(Proof provided by Lebesgue's Theorem)

user228700

@JohnR: Hello :-) Did u see the message I left for u last night?

Bernardo Meurer

@JohnRennie I might just get rekd in my calc exam

John Rennie

@Kaumudi.H Hi :-)

@Kaumudi.H No-one can work all the time. We all have to have some time off, and chatting with friends is a great way to relax. If you want to procrastinate some of the time I think that's an excellent idea. Just not all the time :-)

Balarka Sen

I procrastinate a lot

John Rennie

06:28

@BernardoMeurer <insert morale boosting message here> :-)

user228700

@JohnRennie No, it's not about procrastination, per se. I don't procrastinate that much, truth be told. It's just that when I start talking about music/food, I generally miss deadlines that I may have set for myself. Dyou understand what I'm saying?

Bernardo Meurer

@JohnRennie I much prefer if you could mail me your knowledge in a pill :P

2

John Rennie

@Kaumudi.H Yes, absolutely, and I used to do that as well. (I say used to because I no longer have deadlines not because I got better at meeting them :-)

user228700

Ah :-) Anyway, alright then, I will control urges.

John Rennie

@BernardoMeurer You guys are much better than me at the hard core stuff. I can't even understand 0celo7's question about metric, let alone answer it :-)

anonymous

06:31

You might find this article about procrastination interesting :) time.com/51883/procrastination-is-in-your-genes I don't want to believe it though :-P

John Rennie

@Kaumudi.H it's OK to miss some deadlines. If you don't miss some of your deadlines you're probably not setting them aggressively enough. The point is to get a balance.

Get ahead of your schedule some days and behind it on others.

Bernardo Meurer

@JohnRennie I doubt that Mr. PhD

user228700

@anonymous Me neither :-P Then again, it's not about procrastination itself.

user228700

@JohnRennie Hmm, I don't have that many days, you see.

John Rennie

There will be days when physics seems easy and other days when it just seems a faff. On the latter days talking about food and Coldplay is a good way to get your mojo back.

user228700

06:33

:-D

anonymous

@Kaumudi.H That's why I think regular exams are important. My speed of studying increases 10 times on the eve before the exam day :D

(Fear of failure :P)

user228700

Yeah, I agree :-)

Balarka Sen

Speaking of music I have been rehearing some of Bowie's stuff and it strikes me similar to the works of one of India's earliest rock bands. Funny, huh.

Communication and influence of art is one of the strangest things I'd never get.

anonymous

@Kaumudi.H Both methods look correct but your one is better. K' is nearly equal to 1 anyway as C tends to 0.

John Rennie

@Kaumudi.H I would use a binomial expansion for that ...

user228700

06:54

Hmm, it doesn't actually matter anyway, since they've only asked about the reaction order. Thank you :-)

Swapnil Das

Hello. Happy Galileo + Hawking's B'day!

anonymous

@SwapnilDas Galileo' s bday is on 15 th Feb I guess

(If you meant Galileo Galilei)

user228700

07:18

I've a quick-ish question. This is what my textbook says:

John Rennie

@Kaumudi.H Whenever you see $(1 + x)^n$, where $x$ is small, always think about a binomial expansion. We use that all the time in physics.

user228700

@JohnRennie Riight. Isn't the Taylor series for this function $1+nx+...$(negligble terms for small $x$) ?

user228700

> "All functions (whose domain is symmetrical about the origin) can be expressed as the sum of an even and odd function as follows: $f(x)= \{f(x)+f(-x)\}/2 + \{f(x)-f(-x)\}/2$"

user228700

Oh, hang on.

John Rennie

@Kaumudi.H Yes. So in this case you'd get: $$(1+K_2C)^{-1} \approx 1 - K_2C $$

So: $$ \text{rate}=K_1C(1 - K_2C) \approx K_1C $$

user228700

07:24

Oh, yes of course! Thanks! :-D

user228700

And never mind that about the even/odd function thing. I figured it out.

John Rennie

I can't emphasise enough how important a tool the binomial expansion is. As you study physics you'll use it time and time again :-)

user228700

Boy, that chapter Gravitation uses a lot of this for all of its derivations.

Balarka Sen

@Kaumudi.H Be a little careful about that, because it only works for $|x| < 1$. But yeah.

user228700

OK, what my textbook says. I have a really quick question.

user228700

07:31

@BalarkaSen Yes, of course. Thanks for the reminder :-)

DHMO

@BalarkaSen tum yahaan kyon ho?

Tumi ēkhānē kēna?

Balarka Sen

I like physics

DHMO

wow, they are quite different lol

the "tum" is the same though

Balarka Sen

yep

anonymous

@BalarkaSen Binomial theorem works for all $x$ I think. Why should $|x|<1$ ?

user228700

07:33

The 2nd one is in Bengali? 'Cause I don't understand that.

DHMO

yes

user228700

@anonymous U can only neglect all of the other terms if $|x|<1$ in the Taylor series.

Balarka Sen

@anonymous Try nonintegral $n$, or negative.

Secret

Darude - Sandstorm on Eight Floppy Drives

►

DHMO

@BalarkaSen Does the approximation nature of physics disgust you as a mathematician?

e.g. for 0<theta<pi/3, we take sin theta = theta

Balarka Sen

07:35

Not when it can be justified with some work.

When they claim 1 + 2 + 3 + 4 + ... = -1/12, yes

anonymous

@BalarkaSen Yes, true. But in that case we can use the generalized binomial theorem after some manipulation. gradestack.com/JEE-Main-2015-Complete/Binomial-Theorem/… I understood what you meant though.

DHMO

@BalarkaSen I forgot where this result is used in physics

Balarka Sen

Me too.

Shashaank

Could someone have a look at this question math.stackexchange.com/q/1963640/333392 . It's maths but it's about the Lagrangian

@JohnRennie could you please have look a t this question. It has been confusing me since weeks

anonymous

@DHMO String theory :D

DHMO

07:38

I see

I came across this xkcd:

I first saw this problem on the Google Labs Aptitude Test. A professor and I filled a blackboard without getting anywhere. Have fun.

How would I solve the question on the top-right hand corner?

anonymous

@DHMO Easy. Put two current inputs and analyze separately.

By superposition theorem

Of electric circuits

Secret

It seems there's something interesting going on, the same kind of transposition phenomenon I have seen throughout my life: The Maths chat is becoming more like h bar with the addition of the non maths rule. The lack of activity in h bar recently seemed to shift to maths, and the activity in maths seemed to shift to here

I wonder when ocelo7 returns everything can go back to normal. This overall "lack of activity" is like being in a sound proof room

DHMO

@anonymous what is that?

anonymous

Superposition theorem

The superposition theorem for electrical circuits states that for a linear system the response (voltage or current) in any branch of a bilateral linear circuit having more than one independent source equals the algebraic sum of the responses caused by each independent source acting alone, where all the other independent sources are replaced by their internal impedances. To ascertain the contribution of each individual source, all of the other sources first must be "turned off" (set to zero) by: Replacing all other independent voltage sources with a short circuit (thereby eliminating difference...

DHMO

never mind

anonymous

07:41

It is from IE Irodov

that problem

Balarka Sen

@Secret Our definition of "normal" seems to be very different - you're thinking of "0celo-normal".

anonymous

You just need two grounded cells each of potential V.

Balarka Sen

But actually I'd like him to be unbanned. There was more math when he's around.

Secret

Well truth is, a lot of users like the 0celo normal in h bar as evidenced by the many stars about it. And ever since his absense, something is going out of balance in terms of user base, and the activity trend of this room. This room rarely get quiet in the afternoon period corresponding to sydney

I also like his GR discussion. Slereah sticks around more with his QFT stuff when ocelo is around ,so is MAFIA

Currently, the average activity in this room is mainly contributed by kaumudi and johnrennie. WIthout them two, this room is pretty much dead

John Rennie

@Secret that's not really true. Kaumudi and I tend to be very active for a few hours in the morning (UTC) when our time zones overlap. But the other 20 hours of the day there is lots of other activity.

user228700

07:52

@JohnRennie :-| I'm on here for around 8 hours everyday. Oftentimes, I ask random questions during that whole time.

Secret

hmm ok, I guess my bias might be due to my wake up periods are often when the room is having low activity

user228700

(Actually, I'm on here for more than just 8 hours but @JohnR: Our time zones overlap neatly only during those 8 hours)

anonymous

08:07

@DHMO You can solve the problem using recursive methods too mathpages.com/home/kmath668/kmath668.htm

and mathpages.com/home/kmath669/kmath669.htm

It is more of a maths problem :)

John Rennie

@Kaumudi.H Incidentally (and this is not an excuse to procrastinate :-) this is what I ended up eating for lunch yesterday:

It's bread cooked with cheese and chutney. Still warm from the oven!

And it was absolutely delicious :-)

anonymous

@JohnRennie What is chutney ?

John Rennie

Chutney

Chutney (Devanagari- "चटनी" also transliterated chatney or chatni, Sindhi: چٽڻي‎) is a sauce in the cuisines of the Indian subcontinent that can vary from a tomato relish to a ground peanut garnish or a yoghurt, cucumber and mint dip. An offshoot that took root in Anglo-Indian cuisine is usually a tart fruit such as sharp apples, rhubarb or damson pickle made milder by an equal weight of sugar (usually demerara or brown sugar to replace jaggery in some Indian sweet chutneys) Vinegar was added to the recipe for English-style chutney that traditionally aims to give a long shelf life so that fall...

We Brits imported it from India during the days of the Raj.

anonymous

Looks tasty (and spicy)! You know how to cook Indian food ? :) @JohnRennie (I guess you are from UK)

John Rennie

In this case I think it was basically pickle

anonymous

08:16

Oh I see. That was readymade chutney

:)

John Rennie

@anonymous I like my food, but I'm not a very good cook. Indian food is very popular in the UK, so lots of us Brits are familiar with it.

Secret

in Mathematics, 1 hour ago, by user400188

Secret you probaly would find that it was one of them tht put all those ideas in your head

lol

user228700

08:43

@JohnRennie Awesome! :-)

user228700

@Shashaank: Do I know you (IRL)?

user228700

@anonymous :S Aren't u Indian?

Secret

[Is bored]

Ramanujan

@Kaumudi.H after all he wants to be "Anonymous"

Secret

user228700

08:50

@Ramanujan I read the transcript at The JEE Launchpad and he/she said that he/she is preparing for JEE, which is very good reason to assume that he/she is Indian, in my opinion.

Secret

Ramanujan

@Secret why 0x=x?

Secret

oops, that'sa typo

there, fixed

0x=x0=0

(next slide skipped cause not going to to bore people with more meadow stuff)

Ramanujan

@Secret 1/0 ≈infinity I suppose :(

anonymous

@Kaumudi.H My birthplace was India but I live in the Middle East. I visit India sometimes during vacations. I am preparing for the JEE though as I wish to join a college there as there are not many good tech institutes here.

Secret

08:57

@Ramanujan That's how we treat it in calculus, and indeed it will be valid in that context. But we are interested in algebraic structures where you can say there is an elemnt q such that q0=1 and not cause contradictions

Ramanujan

And you want to say q=1?

user228700

@anonymous Oh, interesting!

anonymous

@Kaumudi.H And where are you from ? Which state ?

Secret

09:15

@Ramanujan nope, q can be anything, I only want q0=1

Shashaank

@Kaumudi.H I don't know you in IRL so I guess you might not be knowing too.

@anonymous If you were referring to me then does Nationality holds such a significance for you

@anonymous Sorry! I read that wrong

user228700

09:44

@anonymous I'm from South India.

user228700

@Shashaank Oh, I'm sorry, then. I have a friend whose profile would match yours exactly.

user228700

I have a quick-ish question about the product of functions. This is what my textbook says:

user228700

> "If $f$ and $g$ are both even or odd, then the function $(f.g)(x)$ will be even but if any one of them is odd and the other even, then $(f.g)(x)$ will be odd"

user228700

Is anybody familiar with a proof of this, perhaps?

Ramanujan

+×+=+ and -×-=+

user223506

09:50

@Kaumudi.H how was the tiramisu?

Ramanujan

+×-=-×+=-

Idk just guessing by previous knowledge :D

DHMO

@Kaumudi.H apply the definition of odd and even functions

user228700

@Ramanujan -__- Thanks, but hang on, while I'm trying to work on a more rigorous proof.

user228700

@Doc Oh, it was excellent! :-)

user228700

@DHMO ...doing.

user223506

09:53

@Kaumudi.H thi is good - it is one of my favourite desserts

user228700

@DHMO Because it's the product, I dunno how to prove it for all functions.

DHMO

@Kaumudi.H I'll only prove one case as an example.

user228700

@DHMO Thanks but no, I don't need you to prove any single case--that I am able to do by myself.

user228700

I'm struggling to prove it for all cases.

DHMO

@Kaumudi.H Ugh, prove all cases independently.

user223506

09:55

I got all chatty yesterday - 2 answers!

DHMO

There is just 4 cases.

user228700

Oh, hang on.

DHMO

Your statement basically includes 4 theorems, all of which are independent.

Ramanujan

@DHMO just curious was my proof right?

user223506

I am conjuring up a question - but will take my time to make sure it is a good and well researched one

DHMO

09:56

@Ramanujan yes, if you had elaborated it.

Ramanujan

How I could elaborate it?

DHMO

Do you think "+×+=+" is a valid proof of anything at all?

user228700

@DHMO: I'm sorry, I misunderstood what u meant by "case". One case is absolutely fine, thanks.

DHMO

@Kaumudi.H Alright, so we'll prove this statement:

> If $f$ and $g$ are even, then $(f.g)(x)$ is even.

user228700

Okay.

DHMO

09:57

Consider $h(x) := (f.g)(x)$.

$h(-x) = f(-x) g(-x)$.

user228700

Right.

DHMO

By definition, $f(-x) = f(x)$ and $g(-x) = g(x)$.

Therefore, $h(-x) = f(x) g(x) = h(x)$.

Therefore, $h$ is even.

user228700

:-| That was ridiculously easy. Gosh, I'm an idiot.

user228700

Thanks.

DHMO

You are welcome.

user223506

09:59

@Kaumudi.H you are not an idiot

Ramanujan

chat.stackexchange.com/…

DHMO

Really, all of them are said by Kaumudi

Ramanujan

Yeah,its a habit lol

user223506

dang!

user223506

anyone who likes tiramasu is a legend as far as I am concerned

Secret

10:08

I like mocha cakes and icecream

user223506

@Secret that is legendary also

user228700

@DHMO Well, he searched for "When said by Kaumudi" so of course, all of them were said by me.

DHMO

@Kaumudi.H oh, lol

I'm an idiot.

Secret

Interestingly, I don't like the actual drink itself because coffee is too bitter with a watery taste

Shashaank

@Kaumudi.H Yes I thought that

user228700

10:10

@DHMO x'D It's a bit of a habit, but like I have said before, it doesn't affect my self-esteem or anything. Gosh, I hope Daniel Sank doesn't read all this. He once gave me a lecture about this very habit :-P

anonymous

@Shashaank No, not at all. I was talking about how to choose inertial reference frames if you read my previous message aimed at you.

user228700

@Doc :-) U're kind.

user223506

@Kaumudi.H you have some great contributions

anonymous

@Doc I see you are atmospheric radiation physicist. Do you still work in that field (like academia) or something else ? :) I never heard of atmospheric physics as a different subject. Just curious.

user223506

@anonymous I do research in that field when i can - ofteen making my own tools. For a wage, I work as an editor, programmer, teacher

anonymous

10:24

@Doc Ah, so how do you research ? Are you a member of some university or institute ? Or you have your own labs ?

user223506

I mainly do research with solar UV

user223506

I am an Adjunct for one University, a researcher for a solar UV team and a Skin Cancer research fellow (part time)

user223506

My lab is often where I set down my 'lab-bag'

anonymous

@Doc Fascinating :)

user223506

I have a backpack that I am developing - solar powered, with my equipment in it

anonymous

10:27

@Doc What is the use of the "solar powered" bagpack ? :O

What is the power used for ?

user223506

my electronic devices - some IoT devices for measurements

anonymous

@Doc I see. Wow!

user223506

it's partly a hobby and it's my somewhat vain attempt to get into fulltime research

anonymous

@Doc I guess full-time research gets a bit boring after a certain length of time..(I might be wrong)

user223506

@anonymous perhaps, but it would be a dream for me

anonymous

10:35

@Doc Then why don't you go for full time research ? I think the pay is quite low for full time researchers ( which is very sad )

user223506

@anonymous oh I am trying

user223506

but the compromise I have now is fine

Shashaank

@anonymous Yes I got that ! I read it wrong. That's why I said sorry ! I read it wrong.

anonymous

@Shashaank Okay !

Secret

Example of a vague statement:
"I know nothing about <topic>except anything that I happened to seemed to understand when other people said it"

Secret

11:06

1

Q: Can an atom own a non zero total angular momentum $\bf{L}$ and a zero total magnetic moment $\bf{m}$ at the same time?

In the explanation of diamagnetism, the atoms do not have a total magnetic moment $\bf{m}$. This magnetic moment comes from the sum of all magnetic moments of electrons, that have an orbital $\bf{m_{o}}$ and a spin component $\bf{m_{s}}$. My question is: can the total angular momentum $\bf{L}$ ...

electromagnetism magnetic-fields angular-momentum magnetic-moment

nuclear magnetic moment?

b_jonas

11:42

Am I the only one bothered by the starting assumption of physics.stackexchange.com/q/2328/8851 ? Just what kind of super high resolution CRT with building-sized vacuum tube screen does the OP have at home?

I'd like to have such a television.

Or the textbook author, really.

Is that just a case of bad textbooks?

Slereah

12:11

Oi

Secret

12:33

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