The h Bar

right, that does makes sense.

@DanielSank, but it is multiplied by the 0.

@heather The notation $U'(0)$ doesn't mean that you take $U(0)$ and differentiate it. It means that you take $U(x)$ and differentiate it to get a function $U'(x)$ and then plug in 0 into that function.

^

Very, very important point.

12:16 AM

i'm pretty confused right now.

$U$ means "the potential energy". You can ask for the potential energy at some point $x$ by evaluating $U(x)$.

$U'$ means "slope of the potential energy. You can ask for the value of the slope at a point $x$ by evaluating $U'(x)$.

@heather Oh, I might've misunderstood what your problem is, sorry

$U'$ is a function.

Same story for $U''$.

You want to differentiate $\frac{1}{2}U''(0) x^2$ as a function of $x$ - that means you treat $U''(0)$ like any other constant, and you know the rule $(a f(x))' = a f'(x)$, right?

sure, yeah.

oh...2x would be the result?

or, wait, no.

it'd be x because of the 1/2?

12:20 AM

@heather The result of what?

of the derivative of 1/2U''(0)x^2

Yes, except @ACuriousMind wants you to write the full result...

Don't ignore the constants.

@heather Then it's close, but where did the U''(0) go?

2x/2 +0

or x+0

or x?

Huh

what?

12:22 AM

@ACuriousMind That's not useful.

i don't know what i'm thinking

sorry

=/

@ACuriousMind "Huh" and "what" are not helpful.

@heather don't apologize, please.

@DanielSank @heather Sorry, that's true. Did you mean to write the '+0' for the derivative of the U(0), i.e. the first term we had?

We have $$\frac{dU}{dx} = \frac{d U(0)}{dx} + \left( \frac{d}{dx} \right) \frac{1}{2}U''(0) x^2$$

As @heather said, the first term on the right hand side is zero.

Now the question is, what is the second term.

@heather, you're just taking the derivative of a bunch of constants multiplied by $x^2$.

well the derivative of x^2 is 2x

and then you, um...i can't remember all of a sudden...you multiply the constants by 2x?

12:26 AM

yes

so you would multiply by 1/2, getting x

and then you'd multiply by...0?

getting 0 again?

i'm sorry i have to go

ok, see you later.

seen this one?

I shoulda brought one of those in, just in case.

MY RESULTS ARE A SIGNIFICANT IMPROVEMENT ON THE STATE OF THE AAAAAAAAAAAART

xkcd did it first (?)

@ACuriousMind facebook.com/legogradstudent/photos/…

12:33 AM

Ah, alrighty then

> In tribute to xkcd (https://xkcd.com/1403/): the final slide of my defense tomorrow.

> Into the abyss I go.

=P

@ACuriousMind wait, xkcd oneboxes?

Yes, it always has

@ACuriousMind huh

well, at least SE has its priorities right

funnily enough, xkcd/1404 is this

I don't understand the things you do, and you therefore may represent an interaction with the quantum vacuum virtual plasma.

HDE 226868

12:48 AM

@DanielSank It's a terrible question, but I thought the premise of this question might amuse you. It's also why people need to have your talk.

1:47 AM

I had a mentor at a national lab who introduced all grad students as "The Future Doctor SoAndSo".

Which was comforting, in a way, when you were brand new at the lab and lost in every context and totally uncertain of how you will ever become like those godlike, nearly done grad students. Much less like a postdoc.

Uh, Emilio is not defending tomorrow, that's just a LegoGradStudent post he liked.

Oct 11 at 15:51, by Dr. Emilio Pisanty

@MAFIA36790 Yeah, defended last week.

^he defended some time ago

He hasn't updated his profile then...

2:35 AM

Well, just bring up inconvenient facts, then. See if I care. ::sticks out tounge and goes pssththththttt!::

Ya ::pssththththttt!::

3:20 AM

@dmckee You around? or @ACuriousMind?

@BernardMeurer I'm allegedly writing tests, but I'm here.

@dmckee Could you help me fix a screwed up in this chat I have with Daniel? We're kind of using it as a reference and I wouldn't like the equation to be wrong. A few lines above there's an inner product where I replaced some pluses with commas on the TeX. it should be:

$\langle x, y\rangle = \langle\alpha_1 b_1 + \alpha_2 b_2 + \ldots + \alpha_n b_n, \beta_1 b_1+ \beta_2 b_2+ \ldots+ \beta_n b_n\rangle$

Done. But you already knew that.

Y U post stuff here?

Oops

3:44 AM

sub.uni-goettingen.de/en/news/details/…

5

4:18 AM

@Secret "We regret that this page is not available in English." ?

From their "Ask a librarian" link at the bottom.

It appears to be something about eh scientific publisher Elsevier from the link, but my German isn't up to it.

OK. I just followed the link and found an English text:

>
No full-text access to Elsevier journals to be expected from 1 January 2017 on

12/13/2016 General news Holdings

More than 60 major German research institutions are to be expected to have no access to the full texts of journals by the publisher Elsevier from 1 January 2017 on, among them Göttingen University with 440 Elsevier journals. There will be access to [most archived issues of journals](https://www.sub.uni-goettingen.de/fileadmin/media/texte/informationsversorgung_z/Titelliste-Elsevier_mit_archiv.pdf) (PDF 95 KB), but there may be [no access to individual e-packages for the econo…

Thanks

Harumph. Can't use markdown links in quoted text. Which is probably the right thing, but is no fun at all.

It's the thought that counts :-)

4:35 AM

Hi, everybody.

Important question: "algebras" or "algebrae"?

@ACuriousMind ^

The latter is Latin, no?

I would vote for algebras.

Somehow I get a full english text

formulas or formulae @DanielSank

:-)

5:05 AM

@Pissedofflayman either

Why are you pissed off?

looong story

@Pissedofflayman Long, or you don't feel like telling it?

@dmckee "Give the governor a harumph!"

Name that film.

Mostly long.

Blazing Saddles?

^ Correct.

@DanielSank Let's leave at "That was close. Darn near lost a $400 hand cart."

5:14 AM

@dmckee Hahahaha, YES!

Swing low? Sweet Chariot?

:D

vzn

5:27 AM

South Korea comes a step closer to LIMITLESS energy: Country's fusion reactor sustains plasma for more than a minute in a new world record / dailymail

Q: Does the top of a spherical ball having a forward velocity and reverse spin appear to be at rest?

5:44 AM

@DanielSank Most of the wonks say use English pluralization these days. Which I do for some things, but others I learned long enough ago that it is now habit. So it is always novae and nebulae with me.

And for now spell check is OK with that.

user246160

Can someone here help me with this question...it doesn't feel very intuitive actually....physics.stackexchange.com/questions/299058/…

user246160

0

Suppose we have a spherical ball with radius $R$. We give it a velocity $v$ towards left and a clockwise spin (angular velocity) $w$. The $w$ is given in such a way that $v=Rw$ is satisfied. In such a situation the net velocity at the top of the sphere comes out to be $v-Rw=0$. So, suppose we m...

6:00 AM

@Doraemonドラえもん you remember that mass on a spring problem we were playing with yesterday?

user246160

6:23 AM

@JohnRennie yes obviously ! I don't think i'll ever forget it :-D...but how is that related to this question ?

user246160

good morning btw :)

@Doraemonドラえもん You remember how the velocity of the block, as viewed from the ground, oscillates between $0$ and $2u$?

user246160

@JohnRennie Yes, I do

Well the velocity of a spot painted on your ball oscillates between $0$ and $2v$ in exactly the same way.

This is because circular motion and SHM are very closely related.

user246160

@JohnRennie Okay, so you mean it is like superposition of linear motion of a particle along with circular motion of particle ? Like the SHM problem ...right ?

6:32 AM

Yes

Suppose the ball was standing still, and you were looking from above as it rotates. Then the spot would appear to you to be performing SHM (assuming you can see through the ball to the spot).

I can draw a diagram if that would help ...

user246160

@JohnRennie yes..please :)

OK, a moment while I do the diagram ...

user246160

@JohnRennie Sure, take your time. Meanwhile I am trying to understand this sentence

hey hey

@Doraemonドラえもん: there's your diagram!

user246160

6:41 AM

@JohnRennie Yes, I sort of get it now. I suppose we can write $X(t)=vt-R\cos(wt)$ for X coordinate of the particle....

Spinning Tube Trick Explained

Yes

And that's exactly the same motion as the block and spring.

Though I'd have written $\sin\omega t$ rather than $\cos\omega t$ but that's a detail.

user246160

Okay, now this thing is understood. But something related to this has been bugging me. I saw a video of the spinning tube trick youtube.com/watch?v=7rAiZR_zasg

user246160

►

user246160

Here they used a similar concept

user246160

But

user246160

6:44 AM

They said that when we label one end of the tube with a letter X and

user246160

the other end with O

user246160

and give the end X a velocity v and a reverse spin w

user246160

The end X seems to move slower than the end O

user246160

They say that this is because the reverse spin slows down the velocity of the end X

user246160

But from the equations it doesn't really seem so

Escape The Kugelblitz Challenge | Space Time | PBS Digital Studios

6:46 AM

hey Doraemon and johnrennie, wanna try this challenge after you guys finishes?

17 hours ago, by Secret

►

user246160

@Secret Umm, sure :)

user246160

Suppose the position of the x coordinate of X is taken as $X(t)=vt-R\sin(wt)$, then the velocity of end x as a function of time still is $v-Rw\cos(wt)$

user246160

And the velocity of x should oscillate between v+Rw and v-Rw just like the end O

user246160

But why is it that the end X seems to turn slower than the end O ?

user246160

@JohnRennie

6:56 AM

@Doraemonドラえもん It's because the ends of the cylinder have to be rotating the same way. Obviously, because they're part of the same cylinder.

So as seen from above, at any moment in time the two letters have the same velocity in the rest frame of the cylinder.

But because the whole cylinder is spinning its two ends have opposite velocities.

Hmm, this is hard to explain. I wonder if I need to draw another diagram ...

user246160

@JohnRennie Umm. I do understand that. But, for end the spin is forward spin and for the other end the spin is backward spin. How to mathematically represent the trajectory of the letter "X" and the letter "O" as a function of time? It doesn't seem mathematically though, that one end will move faster...

user246160

(For the time being let us assume that the tube is completely touching the table and not one end moving up)

user246160

@JohnRennie Yeah, exactly...they should have same velocity which oscillates between v-Rw and v+Rw ...isn't it ? ....

user246160

The explanation given in the video doesn't seem very convincing to me

Suppose the cylinder is rotating anticlockwise with angular velocity $\Omega$ then the ends of the cylinder are moving at $u = \pm d\Omega$.

user246160

7:09 AM

@JohnRennie Yes...okay!

And the cylinder is also rotating about it's long axis with angular velocity $\omega$, then the letters written on the cylinder have a velocity $v = +r\omega$.

So the net velocities of the letters at the two ends are $r\omega + d\Omega$ and $r\omega - d\Omega$. So they are different.

You can see immediately from the arrows I've drawn that at one end the velocities add and at the other end they subtract.

user246160

@JohnRennie Ok, so this is at the instant when the letters come to the top ?

Yes

user246160

Aha.. I get it now...we have to only consider the net time when each letter comes at the top!

user246160

I was thinking that the whole velocity throughout the rotation is slower on one side

7:13 AM

Yes. The velocities switch over when the letters are at the bottom, but if course you can't see them when they're at the bottom!

So one letter is fast at the top and slow at the bottom, and the other is slow at the top and fast at the bottom.

user246160

And that explains why when we see from the bottom of the transparent table it gets switched...yay! I got it now :) Thanks a lot...you are great :D!

user246160

BTW @JohnRennie Can you suggest any one good book to learn rotational dynamics ? I feel I am quite weak in that topic :P

user246160

(For undergrad level)

resource-recommendations faq

@Doraemonドラえもん I'm out of touch with books these days as I haven't taught physics for decades (apart from here :-). Have you seen the book recommendations FAQ?

217

Q: Book recommendations

Every once in a while, we get a question asking for a book or other educational reference on a particular topic at a particular level. This is a meta-question that collects all those links together. If you're looking for book recommendations, this is probably the place to start. All the question...

Q: Separation Between Two Objects Falling Radially towards a Planet

0

The Setup: Two objects are placed in an elevator, which is located at some distance radially outward from a planet, as shown here: Over time, as the elevator, and the two contained objects, follow paths which lead radially inward, the separation between the two objects decreases, as witnessed...

general-relativity spacetime

an interesting question with a simple concept

user246160

7:22 AM

@JohnRennie Umm, some of them seem good. I will have to see the customer reviews online I suppose. Anyway thanks for the link.

7:34 AM

Hi, is Sakurai's Modern Quantum Mechanics suitable as a companion to a second course in QM?

I've taken QM1, atomic physics, nuclear and particle physics, but in some places they refer to Sakurai as a graduate level book, and in others they do not.

Previously I have been using Griffith's Intro to QM

"QM1" isn't a standard thing. No two schools are the same.

I used Griffiths in my undergraduate course, and then Sakurai in graduate school.

The distinction between graduate and undergraduate is mostly in the mathematical sophistication.

Fair point. In QM1 we covered lots of the classic examples such as square wells, reflection/transmission, the coulomb potential in 3D etc. Lots of time independent perturbation theory. Spin and symmetry

QM2 introduces Dirac's formulation, higher order perturbation theory, WKB approximation, field quantisation, scattering, intro to relativistic QM

I think I would still be ok with Griffiths when comparing the two

7:50 AM

one day I should find out how to do QFT particle in a box

It's not an easy one

What is the 'simple' example when you start looking at QFT?

8:12 AM

Well, free particles

Also hydrogen atom

In the limit of the proton and EM field being classical

The classical QFT problems are like

Using QFT to describe bound states is simple?

@JohnRennie It is simple if you assume a static classical EM field

It's basically a slightly more complicated version of the non-relativistic one

Classic QFT problems are free field, Casimir effect, hydrogen atom

I think that's about it

Beyond that it's mostly scattering

Mew

who would win in a fight, a person wielding a samurai sword or a person weilding a baseball bat?

Do either of them know what they're doing?

I'm fairly sure a samurai would beat a pro baseball player

Well a baseball bat is kind of an unwieldy weapon

You can't really hurt someone with the tip of a baseball bat

you can stab someone with a sword, though

Mew

8:25 AM

assuming it's an average untrainerd person

for both the sword wielder and the bat wielder

user246160

@Mew ah, but why are you asking this question ? :-P

We start by assuming both combatants are uniform spheres ...

2

@Doraemonドラえもん jee pad launch

I would pick up the samurai sword

It has advantages in being an effective slashing/stabbing weapon, and upon looking up the typical masses of both objects, parrying a plow from a baseball bat shouldn't be problematic

Yes @Mew why are you asking such questions

Mew

8:30 AM

@koolman so i know which weapon i should keep in my car

user246160

Better to assume the fighters as point objects and write their equations of motion XD @Mew

@Mew you probably won't be arrested for keeping a baseball bat in your car.

@Mew keep both

user246160

Solve the equations to find out which is better :D

@Doraemonドラえもん which equations

8:32 AM

Back in the 70s, when life was considerably more violent than it is today, a lorry driving friend used to keep a large monkey wrench in his cab. "It's for repairing the lorry officer."

You hit someone with a large monkey wrench and they're unlikely to want to continue the fight.

Not that I recommend this!!

user246160

This gun is good for self defense amazon.com/Mace-Brand-Pepper-Spray-Gun/dp/B004YWKF8C

user246160

Looks fancy images-na.ssl-images-amazon.com/images/I/…

user246160

:P

@JohnRennie Did you carry a large slide rule

@Slereah I'm not built for fisticuffs. I concentrated on learning to run fast :-)

8:42 AM

If you carry a samurai sword you have the advantage of surprise, though

Since your opponent will start laughing

Beginner's question I know, but can someone remind me what $\partial_\mu f$ means. IIRC it's $(df/dx^0, df/dx^1, df/dx^2, df/dx^3)$. Is that right?

Roughly, yes

The "real" definition is a bit more complicated

With maps and all

OK, thanks. Don't bother with the detail :-)

I just wanted to make sure this made sense:

Also this should be $\partial f / \partial x^0$ etc

A: What is the value of the Riemann tensor at the center of Earth and why is it not 0?

0

The notation: $$ \partial_\mu\Gamma^\rho{}_{\nu\sigma} $$ means: $$ \frac{d}{dx^\mu}\left(\Gamma^\rho{}_{\nu\sigma}\right) $$ so we are taking the derivative of $\Gamma^\rho{}_{\nu\sigma}$ with respect to the coordinates $t$, $r$, $\theta$ and $\phi$. The fact a function has the value zero at...

I had a sudden crisis of confidence

But now I have a crisis of no coffee. Where does it go? I swear someone is siphoning it off when my back is turned.

9:02 AM

The first pie in space!

Well, in the upper atmosphere.

Who says the UK Space Agency is underfunded?!

9:15 AM

@Mew One more suggestion for the site

9:25 AM

@Danu Nope, I just never saw your post, sorry.

@G.Bergeron You found it now, though ;)

@ACuriousMind Agreed, but my point was mainly about the fact that the determinant will be given by the alternating sum of the component is evident, form the structure of the exterior product, when seen as the scaling factor of a volume form under change of coordinates, whereas as seen as a cube without the structure of the exterior algebra, it seems more out of nowhere.

@Danu Yes, when seeing a reply on a comment, I just saw your message...

@Danu So hi!

Hi indeed

@Danu I see your studying geometry/topology. Is it from an algebraic point of view, or more analysis?

@G.Bergeron Neither, I guess.

Mostly from a topological point of view :P

I like pictures.

But I am mostly interested in questions of cohomology right now, so if I'd had to choose I'd say algebraic.

I'm not a big fan of analysis for its own sake

9:35 AM

@Danu Me neither! But I feel topology is lacking in my background. I'm thinking about taking an algebraic topology class next semester. I did some homological algebra at the end of my differential geometry class, but I'm not sure I was following everything.

@G.Bergeron Algebraic topology is really really great.

I think alg. top. 1&2 were the most useful courses I've taken in terms of applicability

Even in physics, people care about cohomology ^^

It's not something I think I would use in my research, but you never know! Still, I feel I should

@Danu True, and I don't like that now, it sometimes feels like hocus pocus

@G.Bergeron Homology is more intuitive and picturesque, I guess.

At least in manifolds you can often think about homology classes as embedded submanifolds

@Danu Hmmm... We defined first the deRham homology as the quotient of closed to exact k-forms

That's cohomology

An easy way to tell them apart is that cohomology comes with a natural graded ring structure, which in de Rham cohomology is described by the wedge product.

Homology is more picturesque---it's about sums of simplices that "close up" (i.e. have no boundary)

9:41 AM

And then, he went on to construct homological algebra with simplex

Yeah ok

Yeah, the singular chain complex? Or the simplicial one?

Did he allow arbitrary continuous maps from simplices into the space?

He called singular homology

Right, so arbitrary continuous maps

Built from singular k-simplex

Yeah

Formal sums of maps from simplices into your space

9:43 AM

Yeah

But see, that was the part of the course I did not all understand

If you have any specific quesitons, I could perhaps help out

I could prove theorems and all, but I felt I did not quite "get it"...

I think the best way to go is usually looking at smooth manifolds

Hmm.. why not: Is there an easy way to explain the "co" in cohomology as a proper categorical dual? It wasn't entirely clear to me.

Yeah

So you have the singular chain complex right

9:47 AM

yes

Now you just apply a contravariatn functor to it

Screw Elsevier

Namely $\operatorname{Hom}(\cdot,G)$

The boundary operator gives you a natural way to go "the other direction" in the resulting (cochain) complex

ok

(by precomposing)

Now just apply the homology functors to this complex, and obtain cohomology

9:53 AM

Ok, I see this, but how will the notion of a differential form on a manifold appear from that? I mean, I can understand a formal "dual" of a singular chain complex, but is this dual precisely the deRahm cohomology, isn't there several possible cohomology I could define?

Right, there are many cohomologies

The game is to prove that they compute isomorphic cohomology rings

Aha!

So this is the assertion of de Rham's theorem

that de Rham cohomology computes singular cohomology

It's not an easy theorem---the proof I know is based on sheaf cohomology.

This requires the machinery of sheaves and their cohomology to be developed first (though, once that language is established, the proof is very short)

I think de Rham proved his theorem "by hand", so without the sheaf cohomology stuff, but I think the proof is complicated.

This short paper by Samelson gives references to different types of proofs: sciencedirect.com/science/article/pii/004093836790002X

We did a not fully formal proof implicating decomposing the variety and then ending up with an object diffeomorphic to $\mathbb{R}^n$ then used a result on the equivalence of the two for $\mathbb{R}^n$.

Okay, cool

9:59 AM

But isn't this only proving that the deRham cohomology computes singular cohomology, not for all possible cohomologies?

Yeah, right. So there are other cohomology theories that compute other things.

But that's a good thing: More invariants!

Thanks already, this was not entirely clear... -_-

So I spent a bunch of time learning about complex geometry

The book by Huybrechts

It's essentially all about developing different cohomology theories (cohomology of sheaves associated to your space)

Then these give a powerful way of computing stuff/distinguishing spaces

Like the rest of the course was all clear, but then it was: "Oh so we want to compute these integrals so let's define this weird structure you probably heard about and know that it encodes some topological information and then let's toy around with it until we prove it's related to this other weird structure..." If you see what I mean.

@Danu He cited it

But as I don't like this state of affair in my understanding at all, I want to take the algebraic topology course...

Yeah, I recommend it

Though you might end up first working on the fundamental group.

Which is also cool!

10:05 AM

@Danu This is the source of my problem! I side-tracked topology all along learning what I need in different classes, but never on it's own and I am starting to feel the consequence now >:(

And as I don't use topology at all in research, I never had to delve into the topic

Only pure algebra

I guess it's mostly used in geometry and more-abstracted versions thereof (homotopy theory n stuff)

@Danu Yeah, but having the basics and a proper understanding of the language would be very appreciated in some talks :)

I also find that many, many people talk about cohomology ^^

@Danu And in any case I am slowly coming back to some form of geometry, although on the algebraic side, through non-commutative geometry

@Danu Are you sure they always know what they're talking about? ;)

@G.Bergeron Oh, nice. Spectrum of the Dirac operator, huh?

10:10 AM

More like quantum groups

Can you say something about how I'm supposed to reconstruct a manifold from that kind of data?

through the study of integrability

@Danu To me the "manifold" you end up with is the dual of the deformed Hopf algebra

But you kind of loose any notion of locality

I mean... I really want a manifold (locally $\Bbb R^n$, Hausdorff, second countable topological space)

@Danu In the normal case, you consider the local rings of functions on the manifold, and these are commutative.

my understanding was that Connes claims that you can reconstruct the manifold from this other algebraic data

(and I'm sure he's right :D)

10:14 AM

They can be seen as the local dual of the manifold and modulo some topological data, you can reconstruct the manifold from these. But then when you deform this algebra and drop commutativity, it is not clear what is to be done

You can take the formal dual and declare this as being the manifold, but there is no, locally $\mathbb{R}^n$, Hausdorff, second countable topological space

So not really a manifold anymore ;)

Actually, I found it quite nice that this is exactly the procedure of quantization when applied to the phase space manifold, in my case, we apply it to Lie groups

So with this extra structure you get Hopf algebras when considering the dual.

@Danu So yeah, you can reconstruct it, but is it really a manifold anymore?

@BernardMeurer Why now?