The h Bar

12:47 PM

in Mathematics, 28 secs ago, by s.harp

I am having some trouble with understanding the sentence, because $C_0^\infty$ is not contained in any Hilbert space that I know, and self-adjoint is a qualifier that only makes sense for operators on hilbert spaces

1:04 PM

Not contained in any Hilbert space? Except for like all of them

@Semiclassical Yeah so when I split up the square root then my CAS calculator gives a sensible limit at infinity. There's something very strange going on here!

1:23 PM

@0ßelö7 it does all seem to come down to the difference between $\sqrt{z-1}\sqrt{z+1}$ and $\sqrt{z^2-1}$.

Hello citizens

For $z>1$, Mathematica concludes both are positive. But for $z<-1$, it gives negative values for the first and positive values for the second.

actually, I don't even need the $\pm 1$. Mathematica gives $(\sqrt{z})^2=z,\sqrt{z^2}=|z|$ for all real $z$.

in the case, though, I guess it makes sense. If $z<0$, then $\sqrt{z}=i |z|^{1/2}$, so squaring that gives $-|z|=z$. By contrast, $\sqrt{z^2}=\sqrt{|z|^2}=|z|$.

I dunno. I guess the point is that if $y=\sqrt{z^2-1}$ then $y^2=z^2-1\sim z^2$ for large $z$, so for large $z$ we can take $y\sim z$ in a single-valued way.

or, to put it another way: we usually would think of the $\sqrt{}$ in $\sqrt{z^2-1}>0$ as being the principal root i.e. positive both $z>1$ and $z<1$.

But while it's certainly positive for $z>1$, if we analytically continue $\sqrt{z^2-1}$ around the branch cut then it ends up being negative for $z<-1$.

1:41 PM

@Semiclassical amazing

It does suggest that writing it as $\sqrt{z^2-c^2}$ is misleading, though

@Semiclassical this is making my brain hurt. How am I supposed to evaluate the far-stream velocity?

I'd say you evaluate it like this. For large $z$, you're away from the branch points and $\sqrt{z^2-c^2}$ should be an analytic function of $z$. Hence $\sqrt{z^2-c^2}\sim z$ for large $z$

What are we doing

conformal mappings

1:44 PM

@BalarkaSen g******* branch cuts

but I think it could be boiled down to "how does $\sqrt{z^2-1}$ behave for large complex $z$?"

Naively, you'd have $\sqrt{z^2-1}\sim |z|$ for large $z$. But then it's not analytic at infinity.

@Semiclassical yeah, and the answer depends on how one interprets it...wonderful

Q: "Purpose" of torque in projectile motion

0

Torque, $\tau$ is defined as: $$\tau = \vec{r}\times\vec{F}$$ Intuitively, it's the rotational effect of force. However, using the above definition of torque we can calculate the torque due to a force acting on any particle about a reference point. Consider projectile motion for instance...

My ruling is that it should be $\sqrt{z^2-1}\sim z$ for large $z$ if you insist that it's analytic at infinity.

Thanks so much for the help :) I'll talk to a guy in my elliptic PDE class who's also in fluids and see what he got

1:45 PM

@0ßelö7 How does the far-field velocity come out if you assume that?

@Semiclassical I'm in class right now

ahh

It's a function of a+b and U, however, where U is the velocity for the disk flow

@Semiclassical now physically I'd hope that as a-->b the far-stream velocity goes to U, right?

That seems like a good sanity check.

looks like it comes out as $w(z)\sim Uz+U\left(\dfrac{a+b}{2z}\right)^2$ for large $z$

which is to say, at infinity it behaves like the flow around a disk of radius $\frac{a+b}{2}$?

The dominant term in there is $Uz$, which corresponds to a far-field velocity of $U\hat{x}$.

@Semiclassical oh also there's a complex conjugate in here somewhere ::eye twitch::

1:51 PM

yeah, that's a pain

but I think the above asymptotic for $w(z)$ is a good sanity check.

I want to say there's a further sanity check as well. If one thinks of this $w(z)$ as the complex potential of an electrostatics problem, then the problem is one of an elliptic cylinder with finite charge per unit length in an electric field.

@Semiclassical yeah because it's just a complicated Laplace equation

The $Uz$ term confirms the presence of an electric field. The second term should then correspond to the electric field generated by the cylinder itself at large distances.

But why $\frac{a+b}{2}$? If anything I'd have expected $\sqrt{ab}$, since a cylinder of radius $\sqrt{ab}$ has the same cross-sectional area as an ellipse with semi-major/minor axes $a,b$.

oops. should've been $w(z)\sim Uz+U\dfrac{a+b}{2}\dfrac{b}{z}$

which is still pretty weird.

Hmm, maybe I get it. The surface of the elliptic cyilnder is curved, so the surface charge density would vary across the surface. So I'd need to be more careful about the relation between $U,a,b$ and the charge on the cyilnder.

Hm

I need to put a symbol for adjacency via mathjax

But mathjax doesn't have $\Ydown$

What to use

$Y$?

$\operatorname{Y}$ maybe

2:16 PM

@Slereah what is that?

Two points are adjacent if every of their neighbourhoods contain both points

samuel-lereah.com/articles/Mathematics/…

cf this extremely great site

Subscribe and I'll send you a pizza roll

@Slereah what kind of non-Hausdorff BS is that

@Slereah how do I sub?

just kidding

no subscription yet

2:44 PM

in Mathematics, 57 secs ago, by Secret

ugh, f*** distributivity

lollololol

samuel-lereah.com/bbos

The page where to put thems pdfs

how many chapters have you written

so far

Define "written"

All of them have things in it

Not necessarily a lot of things

latexed

appropriately formatted

All of them are latexed

2:54 PM

@ACuriousMind is the burping in Rick and Morty supposed to be funny or cringey?

If you mean completed then 0 right now

A few are pretty close to being completed

@Slereah well i'm DISAPPOINTED

Rome wasn't built in a day!

@BalarkaSen do you want to write a book?

Differential topology without analysis

@Slereah This is my reaction to you

I give your work a light 0 out of 10

you hear me

2:59 PM

Nooooooo

@0ßelö7 no

or at least i wouldn't publicly accept that offer

Hi @Avantgarde. How do you like my new gravatar?

It's worse than Corey Feldmans album

@BalarkaSen what?

Why do people act so strangely about stuff like that

You should listen to Fantano's review of the new brainless brain album

"I am brain" or some shit like that

it was beautiful

Avantgarde

3:16 PM

@BalarkaSen Ingenious, but disappointing

come on why does everyone keep saying that

i put a lot of work in making this you know

Meh

Avantgarde

and I appreciate that, sure

We expect better from the legendary Balarka

actual 6 minutes of my life were spent

England is my city, MS Paint is my Photoshop

2

3:25 PM

@Semiclassical so when I add the swirly term I get the 1/square root thing which doesn't give a pole...

There's still no lift

Or maybe I'm wrong and just need to do the integral

@BalarkaSen very catchy :P

the huge content created by that one phrase is unimaginable

and awesome

in Mathematics, 1 min ago, by Secret

Actually what happens when we have a set theory where we are unable to have a choice function on finite sets?

in Mathematics, 1 min ago, by Secret

e.g. imagine a universe where we can only count to 3

Set theory with finite sets already exists

It's basically equivalent to Peano's axioms

Of course this has finite choice

Finite choice derives from just a handful of axioms, though I'm not sure which

3:42 PM

@0ßelö7 well, you're putting an elliptic airfoil in a stream and then adding vorticity

If you don't have finite choice and you still have a coherent theory I'm not sure it would be a very rich system

By symmetry, I think you still won't have any net force in that case. (You'll have a net torque but not a net force.)

On the other hand, if you displace the center of vorticity a bit then I think you would get a net force

Q: Which axioms of ZF are used for finite choice?

7

Apologies if this is a silly question, not an expert in set theory but just wondering about it. ZF implies finite choice. But let's suppose one wanted to work without it. The thinking here is being uncomfortable about sets where none of the elements are computable/definable, therefore one couldn...

set-theory

Seems that would need induction to be nuked, something that is not very useful for most of the common maths we do

@Semiclassical no, the vorticity will produce a force (in the circular case) because there's a torque, and then the far-stream velocity causes it to move

You get a 1/z term in the complex velocity of you put in a vortex

But now I have that awful reciprocal square root which is actually impossible to integrate

Hmm, fair

3:50 PM

Just out of curiosity, what is the integral of 1/sqrt z around the unit disk?

I'm assuming zero but I can't into calculus apparently

No, it's not zero

You'd need to go through an argument of 4pi to get a zero integral

Huh! Maybe I should break up the integral into two square roots and use the residue theorem

Yeah I got -4 for the integral

Did it by hand, not sure how right it is

(If you think of your function living on a branched cover of C, then going once around the origin in C actually puts you on the second sheet

So in that sense it's not really a closed contour

Right so to do that integral you have to go through a branch cut

Right

3:55 PM

I'm going to office hours. This is ridiculous for a class that doesn't have the grad complex analysis as a prereq

The guy in my PDE class dropped the fluids class :(

most elementary way to do the square root integral is to parametrize by $z=re^{i \theta}$

Yeah i did that by hand

Check out this baby

But I don't want to use Residue theorem around a branch cut

I'm not sure what I'm doing here is legitimate

@Slereah geez, how much did you pay for a hard cover?

4:00 PM

Nothing

Birthday gift baby

You can't use the residue theorem on a branch cut as such, no

@Semiclassical but to evaluate the integral of 1/sqrt(z^2-c^2) around the ellipse id have to use the residue theorem

I can't do that integral by hand...but residue theorem probably doesn't apply

Use the physicist way @0ßelö7

Use the residues

nor directly. But if the function is analytic in the complex plane away from the cut, then you can view the integral as winding around the point at infinity

Where exactly is the branch cut here

From -c to c?

4:03 PM

And you can use the residue theorem there

Yeah

More formally, try subbing w=1/z

Then do a w integral around the unit circle?

Q: i have a differet answer from what is written (always)

Then f(z)dz = f(1/w)(-dw/w^2), the minus sign reflecting that z and w wind in opposite directions

Yep

Physics Meta

0

in my book of general relativity it says that T=ict ,$ {x^\prime}$ = x $cos\theta$ + T${sin\theta}$ and x=vt ,${x^\prime}$ =0 and he conclude that ${tan\theta}$=iv/c and the problem is i always get ${tan\theta}$=-v/ic so how did he get there

discussion homework chat

It's still a nasty square root...

Key point is that f(1/w)/w^2 should have a pole at w=0

Well, the pole at w=0 should be the leading term for small w

So it may be simpler than it looks

4:12 PM

Is a branch cut just fancy complex analysis talk for a specific type of discontinuity btw

@Slereah we know

It's still an issue when trying to do integrals

That was a question

It's a discontinuity in how you represent a multi-valued function

Ah k

Is it general to all multivalued functions?

The multivslued function itself changes smoothly across the cutt

It's pretty generic, yeah

Though the structure you get as a result of the branch points is weirder for some cases than others

4:23 PM

@Semiclassical he delayed the homework and said he'll do most of it in class tomorrow

@0ßelö7 is the last paper in the book by Geroch the one with the awful GR from gauge theory

Yeah I tried reading it

Very philosophical

@0ßelö7 lol

@Semiclassical he looked at his notes on the problem and didn't understand what he did

😪

What a waste of time

So @Semiclassical

what semiclassical theory are you

Are you semiclassical gravity

Or semiclassical electromagnetism

4:33 PM

Quantum

@JohnRennie so what's the formula.... ?

What does "semiclassical quantum theory" mean

WKB approximation usually

And it must include the velocity of individual particles.

o

4:34 PM

@Abcd Suppose you have a ball of gas. The gas molecules/atoms are all whizzing around at high speed so each molecule/atom has a kinetic energy.

Q: Why is translational kinetic energy defined only for the centre of mass' velocity?

1

$$K_{\mathrm {translational}}= \frac{1}{2} Mv_{\mathrm {com}}^2$$ Why does the term for translational kinetic energy include only the velocity of the centre of mass of a rigid body? How can we ignore the velocity of the different particles constituting the system? Can someone prove this to me...

energy energy-conservation

@JohnRennie Yes,I know.

If we add up the kinetic energies of all the molecules we get the internal energy.

What about the potential energy

@JohnRennie Consider a rolling wheel instead of gases.

The internal energy is just the sum of all the kinetic energies, but we don't call it that.

4:35 PM

I thought that was part of the internal energy too!

@JohnRennie I hope you didn't assume that my question referred to gases.

@Slereah ideal gas! :-)

I just said that google showed results only for gases.

@Abcd With a rotating object the kinetic energy of all the bits is called the rotational kinetic energy and given by $\tfrac{1}{2}I\omega^2$.

I guess I should remove that comment lest everyone should think that I am confused about the translational KE of ideal gases.

@JohnRennie what about their linear velocities? the formula you stated doesn't account for that, does it?

Oh!, $v=r\omega$

4:39 PM

That works if you've got an object rolling without slipping

So @JohnRennie I have inferred this: The T.KE of a rolling body is due to the velocity of centre of mass while the RKE is due to the linear velocity+angular velocity (not in the addition sense)of the constituents. Am I correct?

"+" used above in the "and" sense.

But suppose you toss a frisbee. Then there's no obvious relation between the rotational velocity and the linear velocity

The proper sign for and is $\wedge$

@Slereah Both are acceptable.

@Abcd I wouldn't say it's due to linear velocity and angular velocity, because as you say they are related by $v = r\omega$.

4:40 PM

&& works too

@JohnRennie Accepted. So can I at least conclude that all the particles are contributing to the kinetic energy during rolling?

@Semiclassical yes.

Anonymous

@Abcd Rolling motion is a superposition of translation and rotation. So, your first statement is wrong. For T.K.E all the particles have same translational velocity.

Anonymous

Oho! How confusing!

@Blue Picture isn't clear.

Anonymous

$(1/2)\sum m_i v_i^2$

Anonymous

4:43 PM

$v_i=v$ for all particles $i$

Anonymous

And you know $\sum m_i = M$

Anonymous

So, you get $T.K.E=(1/2)Mv^2$

Anonymous

Where $v$ is translational velocity of COM

@Blue @JohnRennie My book derives: $$K = 0.5 I_{com}\omega^2 + 0.5 Mv_{com}^2$$

Anonymous

@Abcd That's right.

Anonymous

4:45 PM

@Abcd yes, that's just the sum of the rotational and translational energy

@Blue translational velocity of com = translational velocity of each particle?

Please verify this^

Anonymous

@Abcd Yes. See the picture ^

pretty simple. Understood @JohnRennie and @Blue. Easy.

Thanks.

Anonymous

@0ßelö7 My gravatar is dedicated to you. :D

4:49 PM

That's nice I guess?

Anonymous

Depends on how you take it. I was looking for something more weird.

Anonymous

But for now this one looks nice. :P

@blue A question for you here :-)

Anonymous

Woah. Huge question. I think I can answer it. :P Thanks to ACM

Anonymous

Gibbs paradox

In statistical mechanics, a semi-classical derivation of the entropy that does not take into account the indistinguishability of particles, yields an expression for the entropy which is not extensive (is not proportional to the amount of substance in question). This leads to a paradox known as the Gibbs paradox, after Josiah Willard Gibbs. The paradox allows for the entropy of closed systems to decrease, violating the second law of thermodynamics. A related paradox is the "mixing paradox". If one takes the perspective that the definition of entropy must be changed so as to ignore particle permutation...

Anonymous

4:53 PM

Oh, he's talking about Gibbs paradox.

Anonymous

Thanks a lot, I am truly reassured by your statement about the microcanonical ensemble being observer-independent. Let me ponder about the matter a bit. It seems that, as @By Simmetry was maybe hinting at, the fact that Gibbs noted that the definition of entropy works only if we include indistinguishability was an ante litteram evidence of the need for QM. — Smerdjakov 10 mins ago

@ACuriousMind should I skip algebra for an AdS/CFT talk?

@JohnRennie We defined velocity only linearly (this doesn't have relation with the previous conv.) right?

$\vec{v}$ is always linear?

Therefore we try to condense everything into $\vec{v}$'s form?

if anyone knows the answer please let me know.

Though I feel that the answer is obviously "yes".

even displacement is always linear

Anonymous

5:11 PM

@Abcd Yes

Anonymous

It's displacement per unit time

5:25 PM

@Blue you're name no longer matches youre avatar, change it to S.Cat,

pretty classy name i think :P

Anonymous

My shirt is still blue as you can see from the picture. There's always a tinge of blue associated with me. ;)

your shirt????

what shirt?

Anonymous

Zoom into the profile picture

oh what the hell

thats too small :P

Anonymous

It still exists :D

Anonymous

5:29 PM

I selected the picture very carefully

apparently...

hows you're project going?

Anonymous

@PrathyushPoduval I'm learning the theory portion at the moment. First chapter of the book by Naoki Shinohara is over. Also, I've progressed a bit with Arduino. I can write code for LEDs, Switches, etc now

nice, you bought that arduino pack?

Anonymous

I'm getting slow due to academic load

Anonymous

@PrathyushPoduval Yup

5:32 PM

how's it?

@Blue Dump the academic load :P

Anonymous

It's good. Has everything to get started

which book is it you're using?

Anonymous

I'm using Wikipedia and the book by Naoki Shinohara. For Ardunio I'm using the book named "Getting Started with Arduino"

Anonymous

Search on Google

Anonymous

Massimo Banzi

5:35 PM

@BalarkaSen what's a circle group?

@Blue I meant the one by Naoki Shinohara

Mithrandir24601

@Blue somewhat unhelpfully, there's a user called @green who currently has a blue profile pic...

is it semiconductors?

Anonymous

as.wiley.com/WileyCDA/WileyTitle/productCd-184821605X.html

Ah okay :P

dmckee

5:43 PM

@Abcd It is worth your time to work through the completely general factorization of the kinetic energy of a system of particles into the CoM part and the relative part. (Of course it must be this way to satisfy the principle of relativity (which was present in classical physics from Galileo on)).

But to your specific question, the mass element at the point of contact in a rolling-without-slipping situation is momentarily at rest and so contributed nothing individually.

Which doesn't change the fact that you can write the energy in terms of Com and relative parts.