The h Bar

yep

i've read the poem =)

Good.

i didn't realize it was the only thing that really referenced Heart of Darkness.

user228700

Ohh, @Balarka: You're back! \o/

user228700

Are you busy?

3:18 PM

The Hollow Men is a strange poem. It's not Eliotesque at all

@Kaumudi Nope

user228700

Great! I was wondering if you could help me with something...

@heather were their negative markings on the questions?

Go ahead

user228700

Given a function such as $F(x,y,z)=x+y+z$, how the heck am I s'posed to visualise what the level curve of this function would look like for, say, $F(x,y,z)=1$?

@Sid hmm?

what do you mean?

user228700

3:20 PM

Not only that, how am I s'posed to figure out what the projection of this curve is, onto one of the coordinate planes?

@Kaumudi.H It's $x + y + z = 1$! It's a hyperplane!

Also level "curve"???

user228700

@BalarkaSen ...first time I'm hearing that term.

It's a level surface

user228700

Oh, crap, yes, sorry.

Anonymous

3:21 PM

@Kaumudi.H What term? $x + y + z = 1$ is equation of a plane in 3D

@heather For instance, if you buzzed and said the wrong answer, would they deduct points?

user228700

@Blue Oh, oh my God, yes, THANKS! <3

@Sid no, but your team is then locked out from answering the question, so the other can wait until the very end to answer.

which means they get a bit of an advantage.

Hmm, these questions are hard to get until the very end unless you are really familiar with some really obscure references

yeah =P

that's the point.

for example, a question about the book "The Giver" the other day started "In this book, the character Asher confuses the term 'smack' with 'snack'..." which if you've read the book is a nice tip-off.

or, i can't remember it, but a question on subatomic particles the other day i was able to get pretty quick.

user228700

3:30 PM

Hmm, I was able to do a question involving a parabaloid fairly easily, but a cone :-/

Can I hear the question?

user228700

I'm unable to figure out what its projection would be, given that it's represented by $z=\sqrt{x^2+y^2}$

@heather I got one question regarding some Governor General pretty early but had to read till the end to get most of others.

user228700

And lies b/w the planes $z=1$ and $z=2$.

Have you drawn the object?

3:34 PM

@Sid yeah, in practice we'll read about various artists, books, etc, to try to learn enough strange stuff to get the questions earlier.

Do you see that the projection to the $xy$-plane should be $1 \leq \sqrt{x^2 + y^2} \leq 2$?

user228700

No :-/

the bonus questions are easier to get, especially as you can work as a team to answer them, but you only can answer them if you get the tossup question associated with it, so .

user228700

Give me a mo...

Anonymous

@Kaumudi.H Did you try using geogebra 3d plotter? It might help a bit in visualizing..

Anonymous

3:37 PM

geogebra.org/3d

Anonymous

Plot $z=\sqrt{x^2+y^2}$ there

Anonymous

And then the planes $z=1$ and $z=2$

user228700

Oh, hey, I get it! But, hmm, my book says that the projection would be the annulus or whatever, which, I Googled it and it's exactly what you say it is, but how is it that? Wouldn't it be all of the circle $x^2+y^2=4$?

Anonymous

The portion between $z=1$ and $z=2$ is projected on xy plane

Anonymous

Which is indeed an annulus

user228700

3:39 PM

I don't get why only this portion is the projection of the whole thing onto the x-y plane :-/

user228700

Where's the notion of a portion coming in?

Anonymous

@Kaumudi.H It's mentioned in the question that you're only projecting the portion between z=1 and z=2

Anonymous

6 mins ago, by Kaumudi. H

And lies b/w the planes $z=1$ and $z=2$.

user228700

Those two planes cut off a frustum, no?

Anonymous

Yes, in 3D space. But what happens when you project all the points of the frustum on the 2D xy plane?

Anonymous

3:41 PM

Think of it as the shadow of the frustum on the xy plane, when you're shining a torch from above

Anonymous

The shadow will be an annulus

user228700

In real life though, the shadow would be a circle of radius of the base...

No, it wouldn't

user228700

Or am I not thinking straight?

user228700

:-/

user228700

3:44 PM

My brain, sigh.

Anonymous

It's a hollow cone....

The little surface above is a slanted annulus.

If you perpendicularly project that you get an annulus below

user228700

:-/

user228700

Never mind, I'll come back to this later, I guess.

Anonymous

3:46 PM

What do you see when you shine a torch from above in this ^ ?

user228700

Yes, I got this, but...

user228700

@Blue Yeah, see, my answer hasn't changed :-/

user228700

My brain's clearly f*ed up.

Anonymous

I repeat the cone frustrum is hollow.

@Kaumudi.H You do see a circle. But, you see two concentric circles. That is an annulus

user228700

3:48 PM

I don't get how the other circle's also visible.

user228700

Oh, crap, crap, crap.

user228700

I am such an idiot, oh God.

user228700

Yes, I get it. I just...I was thinking about the shadow alone, not all the crap I'd be seeing.

user228700

I'd be seeing the above circle+the shadow, which is the bigger circle, so the projection would be the annulus.

user228700

Yeah, I'm an idiot. Thanks for putting up with me, you guys.

user228700

3:53 PM

Ugh, Red Bull tastes exactly like cough syrup, blegh.

Try Beer. That's way better

user228700

Beer to keep me up? Yeah, right :-P

...I meant coffee, of course. :P

Try Syrup of ipecac

Bit of an experiment today ...

3:57 PM

@JohnRennie What's that?

Anonymous

@JohnRennie Are those chicken cubes?

I made a sauce with cream and some scraps of chicken left over from another meal, then stirred it into Maggi noodles.

Anonymous

ah! It does look like maggi

Anonymous

A bit too much gravy though

Quite nice, but a little bland.

3:59 PM

@JohnRennie Mush?

Anonymous

I like a bit dry maggi

Anonymous

you would need some more maggi masala perhaps :P

...that looks like something you excrete out in the morning.. :/

Thanks for that thought :-)

Anonymous

Shush. JR hasn't eaten yet XD

4:01 PM

Fortunately I have finished :-)

Anonymous

lol

Also... why does your keyboard have 2 windows button on it?

Anonymous

it's okay then

Anonymous

@Sid One for right hand and one for left

@Sid if that looks like what comes out of your bottom then you urgently need to see a doctor!

4:03 PM

lmao

lol.

@JohnRennie Nah, I am good. I am at home. Doctor says I am fine

Q: 3 dimensional is length x width x hight"depth". Wouldn't this make the 4th dimension length x width x hight x π?

Anyhow, a failed experiment I think. Not one to repeat.

Oh well ...

Can anyone come up with anything constructive to say about this?

-3

In accordance to the order 1d = length, 2d = length x width, 3d = length x width x hight. The next plane i believe should be total circumference or π. As well as possibly being I know most would say time but i believe time to be a count of cycle/ sequence from ones perspective. Meaning that time ...

I've tried, but ... well ... it's total nonsense.

Anonymous

"Exasperating farrago of distortions, misrepresentations"....:p

4:07 PM

@Blue Man, I was just about to write that

I love that

Shashi Tharoor is a meme god

Anonymous

@Sid Probably your thoughts got transferred to my brain through some higher dimension, ya nauw XD

user228700

@Balarka: You remember how I asked about spherical coordinates and then said "Never mind"?

I do

user228700

Yeah, so it turns out that I shouldn't have said that :-/

4:19 PM

So what do you want to know?

@Slereah we should really learn LQG

it seems quite a bit more analytical than string theory

Slereah

should we

I tried reading Rovelli

But then my brain just goes bluh bluh bluh

Also Lubos wouldn't approve

user228700

At this point, I don't have the time to understand the actual theory, so can you please just tell me how they went from $\phi=$ in rectangular coordinates to spherical coordinates? What substitutions have they made? Here:

user228700

That's better, I hope.

user228700

4:23 PM

(I'll figure out the limits myself)

It should be a variation of $(x, y, z) = (\rho \sin \theta \cos \phi, \rho \sin \theta \sin \phi, \rho \cos \theta)$

You have to be careful because $dV \neq d \rho d\phi d\theta$

(There is the Jacobian too...)

Precisely that

Q: Why does the volume element in spherical polar coordinates contain a sine of the zenith angle?

@Kaumudi.H see:

0

The volume element in Cartesian coordinates is $dx dy dz$, the volume of a rectangular prism with side lengths being the length elements along the three rectangular axes. In spherical polar coordinates, however, the infinitesimal volume element is $r^2 \sin\phi dr d\theta d\phi$. Why is the $\p...

coordinate-systems

In polar coordinates $dV = r^2\sin\theta dr d\theta d\phi$

user228700

Thanks, John, but I'm afraid I don't have the time :-/ I'll learn this by heart, keeping in mind what Balarka just said.

user228700

4:27 PM

('Cause I know it'll take me far too long to understand-at least 15 minutes, I'm sure)

And $x^2 + y^2 + z^2 = r^2$, which is just Pytharoras' theroem.

Unfortunately this cannot be grasped quickly

Anonymous

Understanding this is actually far easier than trying to keep in mind.

It takes a nontrivial amount of time to understand spherical coordinates and what happens to the volume element after change of variables

Anonymous

4:28 PM

Anonymous

This is essentially the element you'll be using to integrate in spherical coordinates

user228700

Gosh, you're all so great <3 Still, @Blue: I've already learned those by heart, man, I'mma move on... :-/

@Kaumudi.H how will you know the limits of the variables?

user228700

Jugaad :-/ Well, not entirely; I've understood how it works for a hemisphere, so analogously...

4:34 PM

@Slereah no we should read the book by Thiemann

@Kaumudi.H Well, okay. That is slightly risky. Think if the hemisphere is cut at the top by a line of some equation. How will you integrate the volume if the upper bound is the line and the lower bound is the XY plane?

Anonymous

Came to know a senior at my uni who's working on application of fractal geometry of music (over Facebook). I should learn about fractals at some point. Looks like cool stuff.

user228700

@Sid I'm gonna have to calculate $\rho$ for this setup and the other two limits remain the same.

Anonymous

I guess it's quite related to complex analysis

Dynamics on the large

Anonymous

4:39 PM

@BalarkaSen Umm?

Anonymous

It's related to dynamical systems you mean?

I mean they come up whenever you have dynamics there's nothing special about complex analysis

Anonymous

Oh

Anonymous

Fractal

In mathematics a fractal is an abstract object used to describe and simulate naturally occurring objects. Artificially created fractals commonly exhibit similar patterns at increasingly small scales. It is also known as expanding symmetry or evolving symmetry. If the replication is exactly the same at every scale, it is called a self-similar pattern. An example of this is the Menger sponge. Fractals can also be nearly the same at different levels. This latter pattern is illustrated in small magnifications of the Mandelbrot set. Fractals also include the idea of a detailed pattern that repeats itself...

Anonymous

"As mathematical equations, fractals are usually nowhere differentiable."

Anonymous

4:45 PM

Ah, noice

Not noice

It's awful

Anonymous

Noice to read about it. Difficult to handle. :P

Anonymous

"Mandelbrot himself summarized it as "beautiful, damn hard, increasingly useful. "

Anonymous

Okay, you are gonna teach me this :P

I don't know much about these

user228700

4:51 PM

Can anybody please help me to understand this question:

user228700

So, I'm s'posed to use Stokes' Theorem to do this.

Right

user228700

How's that s'posed to be obvious, BTW? :-/

What's the formula for work? :)

user228700

4:55 PM

:-/ Aggh, right.

@Kaumudi.H Stokes gives you a line integral around a boundary

Work is a line integral

user228700

Hmm, right, right.

user228700

OK, so I don't get this part:

user228700

I don't get most of it, but firstly, hmm, I don't understand what they mean to say here: "Assigned a downward orientation to make the orientation of C positive"

user228700

5:00 PM

Isn't the downward orientation negative?

IIRC, Stokes said that the Region R to be integrated should be on the Right hand Side of the direction in which you integrate

(Or something along those lines)

user228700

:-|

user228700

Erm...hm.

user228700

On an unrelated note, how are we s'posed to find $n$?

user228700

..? Anybody?

5:07 PM

its normal to the surface

Anonymous

$\frac{\nabla f}{|\nabla f|}$

user228700

That...hmm, that doesn't explain how they've done in in the solution above :-/

I have to work on some other things now. I am sure Blue et al would be able to help you

user228700

OK. Thanks so much! :-)

user228700

@JohnR: Are you around?

5:12 PM

@JohnRennie I see your meal and raise you:

@Kaumudi.H yes ... ?

@ACuriousMind Oooooooh :-)

user228700

@JohnRennie Oh, thank God. Right, do you know how to find $n$ while solving Stokes' Theorem problems?

The unit vector normal to the surface?

user228700

Precisely.

@Kaumudi.H geometry

5:16 PM

Not unless there's a symmetry to help you e.g. for the trivial case of a sphere it's always radial. What's the application? The problem above?

for the plane it's just trig

also there's a formula

user228700

@JohnRennie Yepp.

user228700

@0celo7 Yeah?

@Kaumudi.H in this case you can take the vectors for the two edges of your rectangle and take the cross product. By definition that will be normal to the plane. Does that help at all?

@Kaumudi.H mathworld.wolfram.com/NormalVector.html

equation 3

user228700

5:22 PM

Hmm, this is quite confusing :-/

user228700

OK, I'll get back to this later. Thanks :-)

user228700

@JohnR: For how much longer are you going to be around?

@Kaumudi.H In that question the plane is just $y=z$ i.e. it's at 45º to the $xy$ plane. So the normal is just rotated 90º anticlockwise from that. I can dash out a diagram if that will help ...

user228700

No, no, that's OK :-)

@Kaumudi.H only another half hour or so I'm afraid.

user228700

5:24 PM

Ah, dang, OK...

The normal would be $n = (0, -1, 1)$ with some sqrts in there to normalise it.

user228700

Ah, hmm, OK, thank you :-)

"The Death of one man is tragic. The death of millions is a statistic." - Man, Stalin knew what he was talking about there...

user228700

5:46 PM

@JohnRennie @JohnR: Goodbye, then :-)

@Kaumudi.H I'm here for a little while yet ...

user228700

Cool :-) I've got two chapters left to revise and they're both relatively OK, actually, so I hope I'll skate by...

5:59 PM

@Kaumudi.H Which chapters?

user228700

Some more vector calculus and multiple integrals :-)

user228700

Taking a quick break now...

@Kaumudi.H Lagrange's multipliers?

GDT?

user228700

@Sid Nup.

user228700

@Sid Huh?

6:01 PM

@Kaumudi.H Gauss Divergence Theorem

user228700

@Sid Ah, yep.

Leibnitz Theorem?

user228700

Nup!

Meh. You have got the easy part then...

user228700

:-P I guess.

7:13 PM

The number of valence electrons is the number of electrons in the last energitical level, right?

dmckee

@Curio Mostly. The rare earths are weird.

But if we consider an element which ends in 3d10, do we have to do 2+6+10=18?

No wait, it doesn't make sense

There would be 4s2

8:01 PM

So it is between 1 and 8

hello

And the number of oxidation is the number of electrons which has to get or leave to reach the octect

Hello all...

8:16 PM

Is it right? @CooperCape

Hmmm/

I recently learnt that Feynman was once called a "sexist pig" by some people

Oh yeah @Blue perks of having 4 brothers - A £60 textbook turns into a £15 from each brother textbook for christmas... ahah.

(For reference I mean the Artin but I just want it in paper cause an actual book is kinda hot)

8:39 PM

@Sid I've heard him described as a 'womaniser'. I've heard stories of him "chasing women around the lab", so yeah... I'm not surprised

(*Makes executive decision not to make 'fine man' joke*)

(and yes, the "chasing women round the lab" is a direct quote)

Lab... lab hide n' seek?

9:01 PM

well. i just got my hair cut today.

i haven't cut for ~13 yrs. it's a bit different =P

@heather err.. That's so long. Are you sure you hadn't turned into a something like early humans?

lol no

9:27 PM

@heather :o That's... A while... In other news, the quantum computing proposal is only 2 questions with 10+ upvotes away from the commitment stage :)

@Mithrandir24601 =D i am super excited for the qc proposal

and yeah, it is kinda a while. i was a bit shocked at the end of my hair cut but i am already getting used to it.

ah, so basically just 14 more votes, so 3 more people.

@heather Not quite - 2 questions with 7 votes each, so 7 more people

wait, what?

oh, duh.

::facepalms::

Well, there's more than enough votes left in the people that followed the proposal, so technically, we don't actually need any more people. We just need the people who've already followed the proposal to vote

(not that I'm complaining or anything if we get more people)

@heather :P

Still, I'm hoping for an increase in activity now we're in December and especially once Winter Bash starts :)

vzn

@Semiclassical btw thx again for sharing that, was musing on it some, & find nonreductionistic povs in science esp wrt physics not-so-common theses days. intend to blog on it (further). was reminded of it reading this book, some similar sentiments, you might enjoy it... Overcomplicated by Arbesman amazon.com/Overcomplicated-Technology-at-Limits-Comprehension/…

9:43 PM

@Sid Hair stops growing at a certain point, dependent on the individual. I haven't had a haircut in ages, either, and it doesn't grow longer than the middle of my back.

@ACuriousMind Mine seems to just get very thick instead of growing longer

Huh, that's odd

To be fair, I do get it cut every couple of months, but it (apart from my fringe for some reason) doesn't seem to grow downwards in that time

So maybe I'm just not leaving it long enough or something

10:04 PM

@ACuriousMind mine still grows, though obviously at a bit of a slower rate than when i was younger.

or at least, it grows enough to stay proportionally the same length to my height.

10:17 PM

Yeah, mine grew faster/longer when I was younger, too.

10:55 PM

This is gloriously fantastic :D

hard

Hey, I've been working on a special relativity question for like a week, but I am having trouble solving it. Although I think I've made some progress, I am unable to show the relation between the magnetic field and electric field, and I would appreciate it if someone could please help me.

Hmmm...I just realized a question I've never had before...

What is the space of null vectors a null vector can be orthogonal to? In 4-D space time, a null vector is 1. orthogonal to itself and 2. can not be orthogonal to time-like or space-like vectors right. But are there other null vectors a null vector can be orthogonal to? I've never pondered this...hmm...

wait, can it be orthgonal to space-like vectors...

I'm confusing myself now -.-

11:10 PM

@enumaris It won't be orthogonal to any other null vector that's not collinear to it, go to the frame where the null vector is $(a,a,0,0)$ to see that.

so the space of vectors a null vector is orthognal to is vectors which are colinear to it?

@hard As it says in the room description: Don't ask about asking, just ask. If someone knows the answer and has time to help you, they will

@enumaris No, as you correctly noted it is also orthogonal to spacelike vectors, those of the form $(0,0,b,c)$ in the frame I chose above.

Q: Relation of Magnetic field and electric field

I see

yeah I carried over the proof for time-like vectors wrongly

so it is in fact orthogonal to a 3-dimensional family of vectors

that makes sense

but does that mean I can't construct a orthonormal set of basis vectors where two of the vectors are null

hard

1

There is a wire in which the protons are non-moving with a charge density of $\lambda_+$ and the electrons move at a velocity u to the right with a charge density $\lambda_-$ in a rest frame. A particle of q+ is r away from this wire and moves at a velocity of v to the right. Q1. What is the to...

homework-and-exercises electromagnetism special-relativity

I suppose it would have to mean that

hmmm interesting. I never considered this lol.

11:16 PM

@enumaris Indeed. The null vector that completes $(a,a,0,0),(0,0,b,0),(0,0,0,c)$ to a basis is $(-a,a,0,0)$, which is not orthogonal.

Did someone claim you could do what you're trying to do?

no

It's just some musings by myself lol

I always had the subconscious misconception that null vectors dot null vectors were 0

lol

@enumaris If that was true, then that would mean that the sum of two null vectors is null again

hmm

@ACuriousMind what a physicist proof

I guess I just never did enough problems to encounter this lol

johnnynelson

11:28 PM

Can somebody help me understand this solution? I don't under how he derives the $x^{(1)}$. The equation of motion is that of a the damped and driven oscillator.

Mozibur Ullah

@enumis: maybe that why they're also called lightlike vectors to dispell that unconcious bias.