2:28 PM
@JohnRennie any relation to one of en.wikipedia.org/wiki/John_Rennie ?

He's clearly the MI6 director who faked his death and now answers physics questions on the internet.

@ACuriousMind but why would he still use the same name ?

Vanity

he is British after all

No one will believe he's a retired spy anyway, so why hide?

2:37 PM
@ACuriousMind have you seen Secret's search starred above?

@EmilioPisanty Yes
SE is working on a method to remove these spammers in bulk, it's infeasible for individual mods to try and combat them

@ACuriousMind a'ight

Russian bots?

For instance, the poor SU mods would have to remove so many users that it would take them more than a day at 1 deletion every 5 seconds :P

@ACuriousMind oooof
I guess that, so long as those users aren't posting, they're not really that harmful, and their removal can wait until SE has a solid automated solution

2:42 PM
@EmilioPisanty yup

like, faster than @ACuriousMind fast
and purely community action too

Nope, it was my spam flag that deleted it :)

huh
then why doesn't it attribute it to you?

@EmilioPisanty Because I didn't delete by "mod deletion" but by casting a spam flag (which is binding when coming from a mod)

or were there more flags before yours?

2:47 PM
@EmilioPisanty You can usually assume that Smokey will raise flags on obviously spammy posts before even I get to it ;)

@ACuriousMind ah, fair enough
-1 no hand drawn circles on the graphs, I don't know where to look. — Bhargav Rao Feb 20 at 15:29
lolz

Ah, meta.
I'll bet there's a "-1 not enough JQuery" somewhere there, too :P

@ZeroTheHero Not that I know of.

@JohnRennie Your namesake was an MI-6 officer? :o

One of them ...

2:56 PM
JR is a killer

There were lots of engineers called Rennie
It's a Gaelic name

@0celo7 he is a spy.

@JohnRennie That's exactly what the MI6 director would say!
4

@ACuriousMind this is one of those no win situations isn't it :-)

yup :)

2:59 PM
this is how I concluded ACM is Alicia Keys

I wonder if being the top rep user on physics.SE fulfills Wikipedia's criteria of notability...
@0celo7 No, you concluded that because I was on fire ;)

The most closely related famous Rennie is Michael Rennie who's a second cousin several times removed or something like that. But then he's not really very famous ...

Ah, you hail from a "space visitor"

John Rennie is an alien.

3:04 PM
Sure, try to throw us off the spy trail with an alien conspiracy theory

Explains a lot really :-)

@ACuriousMind Well, he's the only soap scientist in the world who is also an expert in GR and Javascript! :P

[citation needed]

I know all about bubbles in spacetime.

I can derive the distance traveled by a point on the circumference of a pure rolling circle in one revolution, it makes a cycloid and the arc length comes out to be 8R but all this is considering constant angular velocity. What if the circle starts from rest being accelerated by a constant tangential force?

3:07 PM
@ACuriousMind lol

Will the distance traveled by the point in one revolution change?

@MasterYushi Yeah, in the case of sliding on surface, that distance traveled should increase. Suppose you apply a very large force it will travel a large distance before it completes one full rotation.

Assuming that I know the force and other info about the circle, how can I find that distance?

Who would have thought car insurers were so interested in physics :-)

Looks like the same person made all those accounts XD
@MasterYushi I'm thinking
Will let you know if I come up with something
Meanwhile try searching the net

3:17 PM
@Blue I imagine it's part of the big spam accounts attack on the SE
What surprises me is that it's worth the time taken to do it.

@Blue I did, couldn't find anything for accelerated motion

@JohnRennie I'd wager they just have a botnet doing that, no? But I don't really know how spammers work

@ACuriousMind Yes I'm sure the attack was automated, but will anyone actually get any extra business as a result of all this?

It's Q.83

@JohnRennie Does anyone get extra business from all the other spam I find in my mailbox daily? If yes, then why wouldn't they get some from this, if no, why do any of this in the first place? :P

3:20 PM
@JohnRennie That's not at all surprising. Spammers and trollers find spamming and trolling to be a refreshment or a "nice" method for procrastination. Also, I don't think that's any real company. Probably a troll who got bored and decided to play pranks on PSE.

@ZeroTheHero Rennie was originally a clan name, so in principle all Rennie's are related though you'd probably have to back a long way. I think the name originated in about the tenth century in the Kingdom of the Gaels.

@MasterYushi Do you know the derivation of the general cycloid equation?

@JohnRennie Oh, that guy

@JohnRennie All Europeans are related to basically everyone who lived in Europe around 1000 CE :P

3:28 PM
@Blue
x = r(t-sint)
y = r(1-cost)
This?

@ACuriousMind well humans will keep shagging each other. Disgraceful really - no better than rabbits.

@JohnRennie Eh, we build nicer abodes than rabbits :P

That's what the rabbits say when they see us
I need to keep up with British slangs...

@MasterYushi Yeah, now try to find the $x$ coordinate when there is a force applied at the top
Any guesses from this picture? @MasterYushi
@BalarkaSen Tell me about the fractal stuff you learnt!
We could shift to the other room if you wish :P

@Blue the x coordinate would be the same in terms of t, it's just the angle(t) that will change according to angular acc.
wait
x = r(t-sint) + (aT^2)/2?

3:44 PM
@Blue Well there's not much on my end to say so I'll just say it here

Wait a bit. The question asks only for the distance travelled by the "sphere". Not a point on it's surface. That is a way simpler problem

Say $f : \Bbb C \to \Bbb C$ is a rational function of $z$

@BalarkaSen Okaies

Fact: There is an open dense subset (not necessarily connected) $U$ of $\Bbb C$ such that $f(U) = U$. So under successive iterations of $f$, $U$ is left invariant

What does iteration mean in this context?

3:49 PM
Oh, like, $\{f, f\circ f, f \circ f \circ f, \cdots\}$
It's a discrete dynamical system

I thought you were gonna talk about the fun part of fractals
@Blue Oh right, that's easy

where each time interval is taken to be 1 and what happens after each interval of time is an extra iteration of $f$

Alright, gotcha

$\Bbb C \setminus U$ is usually the fractal-like subset that comes out of this system
This is called the Julia set, I think

Dense set just means that all the limit points are contained within it, right?

3:51 PM
Oh, no, dense means every point of $\Bbb C$ is a limit point of $U$
Rationals are dense on the reals for example

@BalarkaSen But $5$ is not a limit point of the rationals, no?
It is a rational!

sure it is
it's limit of the sequence $5, 5, 5, \cdots$ :P

@BalarkaSen Oh, lol
Got it

So yeah I have been given an example I don't really understand
And that's $f(z) = z^2 + c$
For large enough $c$, apparently it's Julia set is supposed to be a Cantor set

3:56 PM
Only today did I find out why my college leaves were not getting sanctioned..

"The Cantor ternary set ${\displaystyle {\mathcal {C}}}$ is created by iteratively deleting the open middle third from a set of line segments." and "Julia set consists of values such that an arbitrarily small perturbation can cause drastic changes in the sequence of iterated function values"

I just told you what a Julia set is

@BalarkaSen Uh, where?
Did I miss something?

Yes. :P

12 mins ago, by Balarka Sen
Fact: There is an open dense subset (not necessarily connected) $U$ of $\Bbb C$ such that $f(U) = U$. So under successive iterations of $f$, $U$ is left invariant
This?
If we change values of that open dense subset...
The successive iterations will no longer be invariant

4:01 PM
10 mins ago, by Balarka Sen
This is called the Julia set, I think

Oh, damn. I dunno how I missed that message
Let's see..hmm
I don't know the notation C/U

C \ U. Setminus

C-U. Oh, alright, got it
Interesting. This makes more sense
The complement of an open set is closed

What fractals are you talking about?

@MasterYushi Fractals in mathematics...

4:06 PM
These are not the fractals I've worked with lol

This is indeed related to the fractal geometry you worked with, I guess. Unless you are talking of fractals in art
@BalarkaSen Silly question. C-U would be the complement of U, or no?

Yes it is

Phew. It's making sense. Go on

I don't have anything else to say, really

I am talking about procedurally generated graphics, I've coded a few. L-systems and stuff

4:09 PM
I'd like to understand the picture for $z^2 + c$, like I said

@MasterYushi If you are talking of these fractal images, the yes...it is related to mathematics
In fact it IS the mathematics we are talking about

I know this one, I've written code to generate it

@BalarkaSen I see. Interesting. Will think about it
In fact the Mandelbrot set is the set of complex numbers for which the function $f_c(z)=z^2+c$ does not diverge when iterated from $0$ (as wiki says)
hmm

@blue ping me whenever you are free.

@JohnRennie Got an exam tomorrow. Confused about some concepts. Are you busy?

4:16 PM
@BalarkaSen Did you check the Wiki page on Mandelbrot set? That function looks exactly similar. But I think there is a more generalized version

I did

@Sid Ping!

@Blue What all things do I have to revise for that ML Course that you had convinced me to sign up for?

@Sid I think one of those two courses are already online (as one of them is a re-run of a previous course). Try to go through the course material once. Also complete the Python (Algo and Data Structures) lecture series by them
I think you can do that by December

4:20 PM
One buggy line and you end up with spiderman instead of mandelbrot

I'll start as soon as my exams end
@MasterYushi That's characteristic of Julia sets it seems. I'm going through the Wiki page

@BalarkaSen Is the scalar curvature of $H^n$ like $-n$ or is it just $-1$

@Blue Could you help? Or are you busy?

@Abcd A bit busy. But you could ask...someone might be able to answer...

@0celo7 i have forgotten scalar curvature
can't help

4:23 PM

In $y = y_m \sin (kx-\omega t+\phi)$, could anyone tell me the significance of $k, \omega$ and $\phi$?

phi is the initial phase

@Abcd $k$ is the wave vector. $\omega$ is angular frequency. $\phi$ is the initial phase.
Check those terms on Wiki

tells you about the particle at x=0 at t=0

@Abcd I've just finished lunch so I'm happy to chat now. What's up?

4:26 PM
k is wave number, it is the number of waves in unit length

@MasterYushi "trying to get into a good university to pursue cse without having to burn my nerves on pcm". Story of every kid nowadays :) Good luck :P

@JohnRennie what is wave vector, and phi?

@MasterYushi Nice. Will watch

@Blue Haha, thanks!

@Abcd Let's take $\phi$ first because that's quite easy.

4:27 PM
Ok

Suppose you look at the point $x=0$ and $t=0$ and ask what is $y$ at the time and place.

@JohnRennie Is this the correct use of phi : desmos.com/calculator/ojbbwm5tjw

We know the values of $y$ range from $-y_m$ to $+y_m$, but there's nothing in our equation to say what the value of $y$ is.

@JohnRennie I've seen k as number of waves in unit length so that should be 1/wavelength but somewhere I've seen it written 2pi/wavelength also. What's up with that?

The value of $y$ is what we call an initial condition for our equation.

4:32 PM
@BalarkaSen Got the PDE book by Farlow's, today. It looks quite thorough and readable! So all book-buying done for the next one year! XD Hirsch-Smale, Farlow, Shakarchi-Stein, Gallian/Artin, Nash-Sen and Arfken. I really hope I can complete all of them. Lot of interesting stuff to learn. Just waiting for 14th Dec :p

Nice

@MasterYushi in these sorts of situations you sometimes see angular frequency $\omega$ used, and you sometimes see frequency $f$ used. They're basically the same thing with just a factor of $2\pi$ difference.

Hello, everyone :-)

Well we get the same thing with the wave vector. Sometimes we include the factor of $2\pi$ and sometimes we don't. The difference is that people tend to be sloppy and use the same symbol for both.

@Kaumudi.H Hey. Getting enough food? ;)

4:34 PM
@JohnRennie I can't understand.

@ACuriousMind Hi :-) Yes, today was a good day, in terms of the amount of food I was able to eat :-P How's it going?

@Abcd that was an answer to @MasterYushi ...
@Abcd suppose we take your equation $y = y_m \sin (kx-\omega t+\phi)$ and set $x=t=0$ then we get $$\frac{y}{y_m} = \sin\phi$$

Yes

Also, @ACuriousMind: boom, double rep-cap =D

4:36 PM
@Kaumudi.H Going well, just watching the chat while listening to music and playing Dungeon Crawl
@EmilioPisanty grmbl

@Abcd so if $\phi=0$ that means $y(0,0)=0$, or if if $\phi=\pi/2$ that means $y(0,0)=y_m$

@ACuriousMind ::grin::

You're much more profilic in answering than I am, currently

@ACuriousMind Sounds fun :-)
I am slowly dying.

@Kaumudi.H We all are, it's called living.

4:37 PM
@Abcd The parameter $\phi$ just determines our initial value for $y$.

:-P Right, of course.

it might yet be rolled away by the system's anti-systematic-voting defenses, I suspect
yesterday I got one vote each on my top-scoring answers
I appreciate the gesture from whoever it was, but the system does tend to not like that

I meant to emphasize the fact that I feel like I am dying :-P

@Kaumudi.H Well, that sucks. What's up?

Engineering Drawing exam tomorrow :-/

4:39 PM
@EmilioPisanty The reversal script runs each day at 3 am UTC, so yesterday's serial votes should already be reversed if it wanted to reverse them

@Kaumudi.H if everything else fails, bring some crayons and a crayon sharpener

@JohnRennie So we use omega in the wave equation if we use k = 2pi/wavelength, and if we use k=1/wavelength, it should be Asin(2pi(kx-ft)) ?

::grins::
aaaaand
::ducks::

:-)
@EmilioPisanty It's OK, I lack the energy to come at you :-P

I rate my day as NOT GOOD/10

4:41 PM
which one was it?
this one?

@BalarkaSen :-/ Damn, what happened to you?

@JohnRennie Understood. Wait a minute. Let me note that.

Allergy
General idleness

Ah :-/ Dang, I hope you feel better soon.

I don't

4:43 PM
@Kaumudi.H Well, that's tomorrow-you's problem ;)

I like to live in perpetual misery

Ah, right, I'd forgotten.

@MasterYushi Yes, I suppose so, but no-one will applaud you for using $f$ in a wave equation, or for using $k$ that doesn't include the factor of $2\pi$.

no, that's not it

@EmilioPisanty Whatcha looking for?

4:44 PM
@JohnRennie Ok, what next?

May 15 at 9:41, by Balarka Sen
@Kenshin I don't want to feel good! That's the point!

I think I know most Minchin clips by heart :P

@ACuriousMind Tim Minchin going "aaaaaaand, duck"

I remember this well enough :-P

@Abcd can't remember, were you asking about $k$?

4:44 PM
then says he's just practicing for his American tour

@JohnRennie $k$ and $\omega$

@EmilioPisanty Ahhh
Hm

@ACuriousMind exactly
it's one of those, innit
the question is which one

@Abcd shall we go to the problem solving room? It's getting noisy in here ...

@JohnRennie Well I'm confused now, according to the definition of k, shouldn't the proper way of writing the equation would be using ft?

4:46 PM
@JohnRennie Yes, sure. Even I was getting confused by the parallel discussion on the same topic with Yushi.

@MasterYushi I would define k as $2\pi/\lambda$ ...

@JohnRennie But it is defined to be the number of waves in unit length

In physics, a wave vector (also spelled wavevector) is a vector which helps describe a wave. Like any vector, it has a magnitude and direction, both of which are important: Its magnitude is either the wavenumber or angular wavenumber of the wave (inversely proportional to the wavelength), and its direction is ordinarily the direction of wave propagation (but not always, see below). In the context of special relativity the wave vector can also be defined as a four-vector. == Definitions == There are two common definitions of wave vector, which differ by a factor of 2π in their magnitudes....
> Its magnitude is either the wavenumber or angular wavenumber of the wave

@EmilioPisanty Well, unfortunately there are a lot of spots in his routines where he might want to duck in front of certain audiences ;P

@Kaumudi It's a good ontological exercise to defend my position about misery against the sword of "enjoyment of misery is also enjoyment of a sadistic sort". I leave this as an exercise to the reader.

4:49 PM
So I don't recall either where exactly that was

@ACuriousMind I'm pretty sure it was religion-related
which narrows it down by maybe 10 to 15%

@BalarkaSen -.- I'll pass, thanks.

@JohnRennie So technically it is both, but angular wave number is usually used in physics

@MasterYushi usually, yes

4:51 PM
@ACuriousMind Well, depending of course, upon which end of the action that inflicts the pain you are in. In this case, uh, both?

@JohnRennie thanks

@MasterYushi yes. In signal processing, as I understand it, they do use the FT normalized as $\int f(x) e^{2\pi i k x}dx$, but that would just get you some very puzzled looks if you used it anywhere I've been in physics

@Secret There's an user who well known for using that phrase. Hehe

@ACuriousMind Coordinates are $\{x^i\}$ or $(x^i)$?
Probably the latter because we use parentheses for vectors

4:54 PM
@Blue not that I knew for h bar. h bar recently is relatively free of help vampire infestation

hbar is full of trolls though

@Secret I was talking about Math SE chat
hbar did have some vampires previously
Now it's just trolls

not sure that's any better than help vampires

Trolls are at least funny
Vampires...are well....just annoying

and help vampires at least ask productive questions :P
even though, well, in a vampirish manner

4:56 PM
Well, lol

I have been observing Maneesh Narayanan for more than 2 months already, and he is not showing signs of improving. Mary Star is only a bit better than him. It seems he will become the next koolman
This is one reason I hate the downtimes in maths chat, because a chat inhabited with vampires is so boringand annoying I just want to spam it to nonexistence

@Secret Maybe someone should point that out to Maneesh directly. Apparently he is preparing for some entrance exam and needs help every now and then.

@Blue But, as the pejorative against the vampire culture goes, "Still a better love story than Twilight"
So I prefer trolls anyway

@Blue I know. but for other people (recall the numerous JEE people a year ago), at least they are showing effort