The h Bar

user228700

16:00

(Oh God, I didn't even question this when we were dealing with that ring or whatever, before)

John Rennie

$rd\theta$ is the length of an arc of angle $d\theta$

0celo7

I honestly don't know what a center of mass is.

user228700

@0celo7 It's very interesting. Google "Fosbury Flop, TedEd", if u want.

John Rennie

So if the area element is approximately a rectangle then $rd\theta$ is the length of the tangential side and $dr$ is the length of the radial side.

user228700

Wait, wait!

John Rennie

16:01

Like any rectangle the area is just the product of these two $dA = rd\theta dr$

user228700

Still digesting everything u wrote before!

user116211

@0celo7 yes.

user116211

$$\varphi_X (t) = \mathrm E\left(e^{\mathrm itX}\right) = \begin{cases}\displaystyle\int e^{\mathrm itx}~f(x)~\mathrm dx &(\textrm{continuous probability distribution})\\ \displaystyle\sum_x e^{\mathrm itx}~p(x) &(\textrm{discrete probabilitydistribution })\end{cases}$$

John Rennie

@Kaumudi OK I'll sit here digesting my lunch :-)

user116211

It's different from mgf by $\mathrm i\,.$

user228700

16:03

@JohnRennie But I'd be more inclined to consider a triangle than a rectangle...

user228700

And hang on, how does looking at it like a rectangle help with multiplying the coordinate..?

John Rennie

@Kaumudi it doesn't. Looking at it like a triangle is the wrong approach.

user228700

*Rectangle

user116211

0

Q: Why can't we make Carnot heat engine in real life?

Question is obvious: Why can't we make Carnot heat engine in real life?

user116211

Insufficient research efforts?

user116211

16:07

I'm seeing no effort at all.

user116211

It's not PhD level thermodynamics.

John Rennie

A crude diagram, but it shows the area element with sides $rd\theta$ and $dr$

user228700

Oh! U mean that! OK, I wasn't able to picture it properly before.

John Rennie

So the mass is $dm = \sigma rd\theta dr$

user116211

yeh, 0celo was right; you all are mad with this PhD CoM; I'm going back to my undergrad studies.

John Rennie

16:10

And the $y$ value is ... ?

user228700

Wait wait! :'-(

0celo7

@JohnRennie Proof?

John Rennie

@0celo7 mass is generally equal to areal density times area ...

0celo7

Why should that be the area though

user228700

OK, got it. What dyou mean "and the y value is..?"

user228700

16:12

Do u want me to integrate now?

0celo7

doesn't seem very Rigorous

John Rennie

@Kaumudi You're going to be integrating $y dm$

user228700

@0celo7 What else would it be?

user228700

@JohnRennie Ah, yes.

user228700

Dammit.

John Rennie

16:13

And you now have the equation for $dm$ in terms or $r$ and $\theta$

user228700

OK, the y value...

John Rennie

So you want $y$ in terms or $r$ and $\theta$

0celo7

maybe if $dm$ is a Radon-Nikodym derivative or something this makes sense.

John Rennie

@0celo7 :: wooshing noise as that sails over John's head ::

user228700

Wait, we're gonna integrate that little rectangle?

John Rennie

16:14

Yes

Well, we're going to integrate the mass of that little rectangle.

0celo7

well maybe not

not sure how to interpret $\int y\,dm$ without physics BS anyway.

user228700

The y coordinate is $rsin\theta$?

John Rennie

Yes :-)

user228700

Phew.

John Rennie

So $y dm = r\sin\theta \sigma rd\theta dr $

user228700

16:15

Alright, now for the integration...

0celo7

maybe this can be done properly with a double integral.

user116211

0

A: What is the difference between Maxwells demon and a refrigerator?

No disrespect intended at all. Maxwell's demon is equivalent to me as "Wow, Man! Ya mean we on this planet are like the atom in our little finger? Whoa that's heavy, Dude!" Use a surfer accent when reading that. I see something like it every day when I open a door slowly. The two rooms temper...

John Rennie

@Kaumudi You can do the integral in either order, but I'd probably integrate wrt $d\theta$ first.

user228700

@JohnRennie Hm...

0celo7

ew, polar

user228700

16:17

So this is the most complicated way to do it, yeah? (:-P)

John Rennie

No, no, it's basically simple. It just looks hard because it's the first time you've done it.

All we're doing is $\int y dm$

user228700

We aren't s'posed to have double integrals in syllabus, which is why my book slyly avoided it by doing it differently.

John Rennie

But we're rewriting $y$ and $dm$ in terms of $r$ and $\theta$

user228700

I learned double integrals awhile back from Khan. Thought it might help and what do I know!

user228700

@JohnRennie Hm, yes...

John Rennie

16:18

Honestly a double integral just means you integrate twice. It's that straightforward.

user228700

Now to integrate.

John Rennie

Your first integral is wrt $d\theta$ and you treat $r$ as a constant.

user228700

@JohnRennie Yes, Ik. Let's see if I can do it right...

user228700

Oh God oh God oh God.

0celo7

this is the test @Kaumudi

John Rennie

16:22

Calm down. Do you want to go through it step by step?

0celo7

if you can do this integral you are PhD level

user228700

No :-P I was just kidding.

user228700

Making sure. I'm integrating $\sigma r^2 sin\theta d\theta dr$, yeah?

John Rennie

Yes. remember that for the first integral $r$ is constant and can be taken outside the integral.

user228700

Yeah, yeah.

user228700

16:24

Must've made a silly mistake as usual, hang on...

John Rennie

So you're calculating: $$ \sigma r^2 \int_0^\pi \sin\theta d\theta $$

user228700

Ah, stupid me. One sec, I think I got this...

0celo7

oh god

I bet the answer is $3$.

John Rennie

@0celo7 now then, be nice :-)

0celo7

Ok, 2

1?

John, can we clean up the star board a bit?

user228700

16:27

Yaaas!

0celo7

Chill dude

user228700

0celo7

that appears to be the right answer

if I remember correctly

John Rennie

@Kaumudi Tada!

0celo7

I would have guessed 3/4 but I guess 4/3 works too?

user228700

16:29

@0celo7 -__- $4R/3\pi$≠1

0celo7

what?

I meant the sine integral.

user228700

Nvm.

0celo7

which I think is 2

or -2

something like that.

user228700

@JohnRennie :-D

John Rennie

@0celo7 Emilio said it was OK to unstar his posts about the closed question, so I'll do those now

user228700

16:30

Okay! So. What did I learn, hm...

0celo7

Who the hell starred all of that?

John Rennie

@0celo7 shall i unstar sorry I'm a troll

user116211

@JohnRennie NOOO!!

0celo7

Yes. That post is a vicious lie.

user116211

._.

user228700

16:32

@JohnRennie: So, can we back up for second?

0celo7

ok @MAFIA36790 let's do some topology

user116211

Go on!

John Rennie

There, the star board is all squeaky clean again

@Kaumudi OK

0celo7

I have $f:\Bbb R\to\Bbb R$ a general function, not necessarily continuous

user116211

okay.

0celo7

16:33

We use the notation $I_\delta(x)=(x-\delta,x+\delta)$ for $\delta>0$.

user116211

okyy

0celo7

We define $\omega_f(x,\delta)=\sup_{x\in I_\delta}f(x)-\inf_{x\in I_\delta(x)}f(x)$.

user228700

To multiply with the coordinate, couldn't we have done the same thing I did with the sector, but then integrate that like I did above?

user116211

gotcha.

0celo7

Claim 1.: $f$ is cont. at $x$ iff $\lim_{\delta\searrow 0}\omega_f(x,\delta)=0$.

user116211

16:35

okay.

0celo7

Define $D_f^\epsilon=\{x\in\Bbb R\mid \omega_f(x)\ge\epsilon\}$.

user228700

I mean, multiply that with $Rsin\theta$ and do a double integral..?

0celo7

Claim 2.: $D_f^\epsilon$ is closed in $\Bbb R$ for all $\epsilon>0$.

$\epsilon>0$.

user116211

yeh; got that.

John Rennie

@Kaumudi In the example above we integrated $d\theta$ first, so we get an integrated element that looks like a half ring. Then integrating $dr$ adds up all those half rings to build up the half disk. Is that OK so far?

0celo7

16:36

Oops.

Claim 0.: $\omega_f(x,\delta)$ is monotone increasing in $\delta$.

user228700

@JohnRennie Yes, OK.

0celo7

But that part is clear.

So 1 and 2 need some proving.

user116211

start!

user228700

(This is exactly what my book did, BTW. Except, they just directly took a half ring)

0celo7

@MAFIA36790 I do not know how to do this yet :P

I am figuring it out

John Rennie

16:37

@Kaumudi Right. But we could have integrated $dr$ first, and that would give us a sector of angle $d\theta$. Then integrating $d\theta$ adds up all those sectors.

0celo7

So, 1 $\Rightarrow$.

user228700

@JohnRennie Oh, yeah...

John Rennie

@Kaumudi Aha, as you say the book was trying to avoid the double integral.

user228700

Then what the heck would I have gotten if I did what I was proposing? Hm...

John Rennie

@Kaumudi have a go at doing the integral in the other order and see how it works out ...

0celo7

16:38

Ok maybe this is nontrivial.

user228700

No, not the order.

0celo7

I have to work it out on paper.

user116211

Ping me when you get a view of the shore @0celo7; I'll be doing some Group Theory now.

user116211

Oh Mike is here! But don't ask for help @0celo7.

user228700

I was thinking I'd take the sector directly. Even if I directly took the sector, I'd need to integrate for $d\theta$ and $dr$ huh.

user228700

16:41

If I didn't integrate with $dr$, I wouldn't cover all the points not on the edge.

user228700

And obviously, I gotta integrate with $d\theta$. Huh. What is happening?

0celo7

@MAFIA36790 One direction is clear for 1.

user228700

Something wrong with basic logic behind integration @JohnRennie?

John Rennie

@Kaumudi multiply by the coordinate remember ...

user228700

Yeah, even after multiplying...

John Rennie

16:45

I'm not sure I understand you. The sector isn't an area element. It's what you get after integrating wrt $dr$

user228700

Ah, yes...so what counts as an area element?

user228700

(These are some of the more fundamental questions I wanna make sure Ik the answers to)

John Rennie

Something that looks like $dx dy$ i.e. an infinitesimal rectangle

user228700

@JohnRennie What about a circle? Or a triangle? Why wouldn't that work?

0celo7

Oh, of course

A 5 epsilon estimate should work

John Rennie

16:47

@Kaumudi An area element has a well defined position i.e. value of $x$ and $y$

user228700

Ah, OK...

John Rennie

Or in polar coordinates a well defined value of $r$ and $\theta$

user228700

And for a circle, it's definitely not wekk defined. For a triangle, well, again, same problem, I guess..?

John Rennie

Yes, a circle has a radius but not a unique value of $\theta$

user228700

Can u help me to see why it's not well defined for a triangle? I mean, just to make sure I'm thinking right...

John Rennie

16:49

As for the triangle, I'm not sure what you mean by that. Do you mean something like a sector?

0celo7

@MAFIA36790 Claim 1 is done.

user228700

Oh, but we could assume, right, that $\theta$ is same?

user228700

Isn't that what we do with the rectangle?

0celo7

@MAFIA36790 Oh, I defined $\omega_f(x)=\lim_{\delta\searrow 0}\omega_f(x,\delta)$.

So $\omega_f(x)=0$ iff $f$ is cont. at $x$.

John Rennie

@Kaumudi you've lost me. Maybe you could draw a diagram showing what your triangular element looks like.

user228700

16:51

That the area is so small that $\theta$ remains constant?

user228700

@JohnRennie No, forget the triangle.

user228700

With the rectangle, we assumed that $\theta$ would be constant, yeah?

John Rennie

@Kaumudi Yes, the area element is so small that both $r$ and $\theta$ are constant across it.

user228700

That's how I got $Rsin\theta$ to be the y coordinate.

0celo7

@JohnRennie eye twitch

user228700

16:53

So why can't we do the same thing with a circle or something? What dyou mean it's not well defined?

John Rennie

In fact the area element has an infinitesimal area

0celo7

@JohnRennie What does that mean?

John Rennie

0celo7 will be happy to give a rigorous definition of what infinitesimal means :-)

0celo7

No, that's @Slereah 's crazyness.

John Rennie

@Kaumudi For a circle $\theta$ isn't constant.

user228700

16:54

@0celo7 You go back ur topology I have to finish some Math too, before bed and it's 10:24 PM gah!!!

user228700

@JohnRennie How so?

John Rennie

@Kaumudi I'll be here tomorrow. Maybe you should sleep on it :-)

user228700

No, no, let's pls get this over with. This has been hanging over my head for like, two weeks now.

user228700

Why is it not constant for a circle?

John Rennie

If you take a circle it has a well defined radius. But what would the $\theta$ coordinate be? Zero? $\pi$? $1.85563554846$?

If it helps, let me make an analogy ...

Take the $xy$ plane. A point on that plane has a value of $x$ and a value of $y$. Yes?

user228700

16:57

Yes.

John Rennie

OK, what about a horizontal straight line? It has a value for $y$. But what is its value for $x$?

0celo7

$\epsilon,\delta,\eta$, what next?

$\mu$ I guess

user228700

@JohnRennie It changes...

John Rennie

@Kaumudi so the horizontal straight line has no unique value for $x$.

user228700

Yeah.

John Rennie

17:00

But an area element must be located at well defined values of $x$ and $y$

In effect the area element is like a point i.e. it has a position $(x,y)$ or $(r,\theta)$

user228700

OK, nvm the thing w/ the circle. For all intents and purposes at this stage, I'll just take it that the area element is a rectangle and nothing else. Is that an OK thing to do?

John Rennie

In 2D it's a rectangle and in 3D it's a cube

In 1D of course it is just a line segment.

user228700

OK, yes.

John Rennie

And in 4D it's a tessaract!

0celo7

@MAFIA36790 Ok. It's just wrestling with sup and inf

user228700

17:02

:-P Enough, enough!

John Rennie

Think of those poor string theorists. They have to work in 11D!!

user228700

OK, I think I got it. At least I hope I do.

John Rennie

One last comment ...

user228700

Yes..?

John Rennie

we've been chosing example where it was obvious what the $x$ value for the COM was. The symmetry makes that easy.

user228700

17:03

Yeah...

John Rennie

So we just had to work out $$y_{com} = \frac{\int y dm}{\int dm}$$

user228700

For $x$, I'd have to do another integral if not symmetrical about y-axis...

John Rennie

But in general we get the x coordinate using $$ x_{com} = \frac{\int x dm}{\int dm} $$

0celo7

what does $dm$ mean?

John Rennie

And in 3d we get the $z$ coordinate the same way.

@0celo7 infintesimal mass element

user228700

17:05

^

John Rennie

@0celo7 Total mass is $M = \int dm $

0celo7

Well the question is, is $x=x(m)$

user228700

Okay! Sir, when's ur birthday? (Just so I can wish u 'cause I want to wish someone who has helped me so much already "Happy birthday" on their birthday :-P) :-) @JohnRennie

John Rennie

All this stuff is well into a first year undergraduate course. Don't be alarmed if it seems hard. It is hard! :-)

0celo7

If so, what does that mean?

John Rennie

17:07

@0celo7 $dm = \rho(\mathbf r) dV $

0celo7

well that's much better

why not just write that instead of the cryptic $dm$

John Rennie

@Kaumudi 9th February. I shall be 56. But my mental age remains approximately 13.

0celo7

That explains your fascination with balls

John Rennie

And fart jokes, and indeed everything else scatological.

I have the rest of my life to grow up in, so why start now?

user228700

@JohnRennie February 9th. Noted. Not that I want u to wish me or anything but I think it would interest u to know that I was born on April fools' day :-P

John Rennie

17:09

Of course this attitude probably explains why I'm still a bachelor :-)

0celo7

@Kaumudi Hmm. Your gender may have been an April Fool's joke. Better check.

Or

John Rennie

@Kaumudi That's a great claim to fame :-) My mother was born on Halloween. She's a witch!!!

user228700

@0celo7: What the heck do u mean?! Like, seriously, what?

0celo7

You were switched at birth by a mischievous nurse

@JohnRennie My dad too.

user228700

@JohnRennie :-D Nice!

user218912

17:10

@0celo7 did you get your book?

user228700

I'm...foolish? :-P

0celo7

No, did it say it has arrived?

user218912

yes.

0celo7

Oh I just got the email

No I'm about to take a topology test and then I have class until 4:30.

user218912

"left with receptionist"

John Rennie

17:11

@0celo7 wow, it's a small world.

user218912

@0celo7 k

John Rennie

There is nothing special about 9th Feb. Well, not until I become famous and they make it John Rennie day.

user218912

lol

user228700

@JohnRennie :-D

user228700

You're pretty famous here!

John Rennie

17:13

It's a small pond!

0celo7

In India?

user228700

Maybe we all could come up with that and have 9th Feb. be John Rennie day?

Slereah

@0celo7 Them's fighting words

0celo7

@JohnRennie say that after swimming it.

user228700

@0celo7 What? No xD I meant here, as in SE.

0celo7

17:13

I demand 0celo7 virgin sacrifice day

user228700

OK I'm gonna go now. THANK YOU!!! @JohnRennie Good night :-)

user218912

@JohnRennie

user218912

lolwut?

John Rennie

Good night. See you tomorrow.

@obe oh yes, that's come up in the chat a few times :-) I think it's quite funny.

To be fair I was a bit rude to the guy so I can see why he got annoyed.

user218912

17:15

lol

0celo7

Why were you rude?

user218912

why is he shirtless lmao

0celo7

Because he's a boss

He wants to fight JR, and quite frankly I think JR would lose

John Rennie

@0celo7 Because I thought he was a tw*t. Nothing I saw in his video changed my mind.

0celo7

Unless he has a knitting needle of course

John Rennie

17:17

But you shouldn't be rude, even to tw*ts. It's bad karma.

user218912

right...

user218912

doesn't that guy kind of look like JD with long hair and no shirt?

John Rennie

Fortunately he and I are on opposite sides of the Atlantic and unikely ever to meet.

0celo7

I'll buy him a plane ticket.

John Rennie

Now where did I put that knitting needle ...

user218912

17:20

qft is getting boring now

user218912

we're doing feynman diagrams and perturbation stuff

Anonymous

Hi

Anonymous

For semi infinite wires

Anonymous

What would be the direction of magnetic field at P?

Slereah

17:21

I hope you like doing muon cross sections

Anonymous

Would it still be into the plane of the paper?

Anonymous

Can someone help me with this Electromagnetism stuff ^ ?

Anonymous

(The wire carries a current)

Slereah

Use the Biot Savart law?

Anonymous

@Slereah By BSL I'm getting into the plane of paper. Am I right?

Slereah

17:26

Yes

Sounds reasonable

Anonymous

Thank you. I just used dl * r and I'm happy that it works :)

Anonymous

18:06

@Slereah Are you stiil there? I have one more question. Consider a long, straight wire of cross-section area A carrying a current i. Let there be n free electrons per unit volume. An observer sitting in the car moving in the same direction to the current with a speed v = (i/nAe) and separated from the wire by a distance r. Then the magnetic field as seen by the observer should be ? Should it be zero ?

Anonymous

I feel the relative velocity is 0

Anonymous

So the magnetic field might be zero

Anonymous

I'm confused though.

0celo7

18:19

Slereah is not actually a physicist

He's a confused mathematician

Anonymous

Ocelo...so could you answer?

Anonymous

@0celo7 @Oce

0celo7

I dunno anything about magnets.

Swapnil Das

18:43

@S007 You could directly use the famous right hand rule, the rest is obvious :-)

Bernard Meurer

@0celo7 U busy?

I still need to vent

0celo7

18:54

In class.

Call Michelle

Bernard Meurer

@0celo7 She doesn't deserve having to listen to my crap

I listen to yours so you gotta listen to mine :P

0celo7

I teach you PhD analysis, which is enough.

DanielSank

You two are cute.

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