The h Bar

2:03 PM

@ACuriousMind is smoothness crucial to the inverse function theorem?

Does one need many derivatives

Or continuous derivatives?

2:18 PM

no idea

@ACuriousMind one needs $C^1$ crucially

it fails miserably without it

Q: Plausibility of a Dyson sphere

Why do you ask me a question to which you know the answer?

Jim

1

So Dyson spheres are generally considered implausible for two main reasons. One, it would an incredible amount (and type) of resources to build one. And two, a Dyson sphere would be unstable because if it was moved even slightly off center it would collapse into the star. These reasons are spelle...

@ACuriousMind I asked you the question

I did not know the answer

Jim

Anyone with time on their hands and willingness to do math want to tackle this?

2:20 PM

then I though about it, wrote down some things

and figured out the answer

I see

A: Is there a quantitative relation between the correlations at spacelike intervals possible in quantum field theory vs classical field theory?

@ACuriousMind For instance, $x/2+x^2\sin 1/x$ for $x\ne 0$ and $0$ for $x=0$

The derivative at $0$ is $1/2$

But it's not injective on any neighborhood of $0$

It's because the derivative is discontinuous

@ACuriousMind I take it you don't know how to prove especially horrible identities involving Hermite polynomials?

Emilio Pisanty

2:47 PM

facebook.com/thespaceacademy/videos/232402397159285

↑ gotta love it when people take an april fools' and just run with it

user116211

Pseudo answer:

user116211

0

In reply to @Girish's comment on the answer by @flippiefanus: Concurrence is a good measure even for multi-qudit pure states. This has been extended to pure states of arbitrary dimensions (aka multi-qudit pure states) in a very intuitive way by looking at the geometry of the tensor products with ...

user116211

?

@ACuriousMind Can one compute the derivative of $k(x)=\sqrt{f(x)\cdot f(x)}$ (Euclidean norm of diff function $f:U\to\Bbb R^n,U\subset\Bbb R^m$ open) directly from the definition of Frechet derivative?

I think using the chain rule is unavoidable

user116211

Wow! Faster than Barry Allen!!

2:59 PM

@MAFIA36790 Already took care of it (without looking at chat), just flag as "not an answer" next time.

user116211

19 secs ago, by MAFIA36790

Wow! Faster than Barry Allen!!

user116211

@ACuriousMind noted.

ACM is Alicia Keys, not the flash

user116211

:(

user116211

How can Alicia be a speedster ;/

3:01 PM

THAT WOULD EXPLAIN HOW SHE CAN BE ON TOUR AND IN GERMANY AT ONCE

HOLY CRAP ACM IS THE FLASH

user116211

@0celo7 Time-Mirage can only be formed by Eobard Thawne aka Reverse Flash.

Why was this question not closed? physics.stackexchange.com/questions/285528/…

user116211

Barry is still not an expert.

Yes but

She's not in two places at once

She just travels very quickly

user116211

ohh.

3:03 PM

So like when Alicia "goes to the bathroom"

user116211

@0celo7 Then she might be Jesse Quick.

ACM shows up in Germany and chats

user116211

Good for us.

@AnubhavGoel Because the review gathered three leave open votes.

@ACuriousMind just out of idle curiousity, since you had already voted to close that in your pre-moderator incarnation does that mean you can't close it now you're a mod?

3:07 PM

@MAFIA36790 Why did you support it for open? Isn't it against site policy?

@JohnRennie The close vote menu looks as if I could cast another close vote.

@ACuriousMind I suppose there's no way to know except to try it ...

Not that I will lie awake tonight worrying about it :-)

@ACuriousMind Is it possible to prove the Path integral satisfies the Schroedinger equation by just...plugging it in?

The way I know involves some fenageling

What do you mean by "the path integral satisfield the Schrödinger equation"?

@AnubhavGoel: I didn't want to start a discussion about this in the comments:

Why your matrix is Rindler's? You may say he is at rest and me is accelerating and hence has Rindler's matrix.? — Anubhav Goel Oct 10 at 12:45

But I'm happy to discuss it here if you want.

3:12 PM

@ACuriousMind You can compute $\langle x,t|x_1,t_1\rangle$ using a path integral right

Yes!!! But can you wait for 9:3/

9:30

it's $\int_{x_1}^x \mathscr D x(t)\,\exp -\mathrm i\int_{t_1}^t L$ or something

user116211

How should one use his written paper in an answer?

A: If perfect maximal entanglement is never true, does a remainder invalidate the monogamy of entanglement?

@ACuriousMind How do you show that thing satisfies the SE

user116211

I'm talking about this post specifically:

user116211

3:13 PM

0

Maximal Entanglement theoretically cannot be achieved on particular $n$-qu$d$it systems for a specific $(n,d)$ pair. For example, the paper demonstrates this taking the case of a four-qubit system and extends Concurrence to pure states of arbitrary dimensions (aka multi-qudit pure states) in a ve...

@AnubhavGoel What time zone? I'm on UK time and it's currently 16:13 here.

I have to meet some seniors. I was thrown in MBBW

MBBS

45 mins

Please

17:00

@0celo7 Why should that thing satisfy the SE? It's a transition amplitude, not a time-evolving state.

user116211

Shouldn't he need to mention explicitly this paper was written by him?

user116211

I'm not sure though.

3:16 PM

@AnubhavGoel I should still be around at 17:00 UK time i.e. in 45 minutes

@ACuriousMind What?

$|x_1,t_1\rangle$ is a state

Yes

so $\langle x,t|x_1,t_1\rangle$ is a wave function

No

@ACuriousMind Well my homework is instructing me to prove that it does satisfy the SE, and I know the path integral can be used to derive the SE

are you saying it is incorrect?

@ACuriousMind

@ACuriousMind Ok I'm confused

It apparently does satisfy the SE, so why did you say it doesn't?

3:29 PM

@Jiminion: do you want to discuss your question about intrinsic curvature?

user129412

Does using @username still work when the user is not currently in the chat? As in, will they see it once they join? I suppose so

@user129412 it works if they've been in the chat room in the last few days.

user129412

@JohnRennie Can this be gauged by if their name pops up when you start typing @..?

When I say it works I mean they will get a notification that someone has pinged them.

user116211

I think it's 10 days at most.

3:38 PM

@user129412 yes, exactly.

user129412

Hmm, then I guess he hasn't been here for a while. I'll try at a different point. Thanks!

Q: Eventual Formation of Space Junk Disc

Duplicate?

1

I've heard that eventually all the space junk in orbit around Earth will form a disc much like Saturn's rings. If this is true, why? I would expect that that space junk would simply form a spherical shell in orbit over Earth due to the mutual attraction of gravity between the individual pieces o...

gravity orbital-motion

I don't want to dupehammer it without a second (or third) opinion ...

4:06 PM

@JohnRennie I could.

@JohnRennie Sorry I didnt see it earlier.

A: Universe being flat and why we can't see or access the space "behind" our universe plane?

@Jiminion Hi, are you still around?

If so have a look at these questions where I've attempted an illustration of how intrinsic curvature works:

5

When you're talking about curvature it's important to distinguish between intrinsic and extrinsic curvature. I struggled to find a good summary of the difference: page 6 of this PDF discusses it, or Google for similar articles. A very common analogy for spacetime curvature is the rubber sheet, a...

9

A: What is the universe 'expanding' into?

The balloon analogy is useful in some respects, but it is misleading in one important respect. In the balloon analogy the curavture of the balloon surface is extrinsic while in GR the curvature of the universe is intrinsic. Extrinsic curvature is easy to understand. The surface of a balloon, or ...

@JohnRennie OK, will do.

@JohnRennie I am sorry, I will take a little more time . I am in seniors room.

No hurry, I'm here for a while yet ...

Sorry!

4:18 PM

@JohnRennie Is it proven that space in intrinsic? Or assumed so because the extra dimension (extrinsic) has not been sensed?

@Jiminion You can always reproduce the effect of intrinsic curvature by embedding the manifold in some higher dimensional space.

@JohnRennie How does intrinsic curving work? What is the mechanism?

However if you use that explanation you have to explain why we are constrained to our 4D manifold and can't move out of it into the extra dimensions. So while we can't rule out extrinsic curvature into higher dimensions it is a more complicated explanantion.

@Jiminion asking about the mechanism is not a meaningful question. Einstein's equation tells us how the intrinsic curvature is related to the matter/energy distribution.

When you ask about the mechanism I guess you're asking how exactly spacetime curves, and the answer is that we don't know - it just does.

@JohnRennie OK. Perfectly acceptable. Worth a try.... :)

Actually I'm not sure how embedding works with Lorentzian manifolds. I'm not sure if it has been proved that you can always embed a Lorentzian manifold into a higher dimension space. @0celo7 or @ACuriousMind might know ...

4:41 PM

@JohnRennie Yes , Sir!!!!

@AnubhavGoel Hi

user116211

@0celo7, you read Axiom of Choice and infinite sets in Munkres?

user116211

I'll read that now.

Why your matrix is Rindler's? You may say he is at rest and me is accelerating and hence has Rindler's matrix

@AnubhavGoel In our twin paradox we have one twin in an inetrial frame and one twin in an accelerating frame.

The twins can always tell which frame they are in by measuring their proper acceleration.

Suppose you're one of the twins ...

Suppose you are holding an object and you let it go.

If the object continues to float motionless beside you then you are in an inertial frame.

But if the object accelerates away from you then you are in an accelerating frame.

In the inertial frame the metric is the Minkowski metric, while in the accelerating frame the metric is the Rindler metric.

So it is always clear what metric to use for which frame.

dmckee

4:46 PM

To do @John's version of the experiment we should put the stay at home twin in a spacecraft off the planet. Or we have to make a correction to that twins proper time for gravitational time dilation.

But using the kind figured usually assigned for class-room version of the non-paradox that can be neglected.

I use a 1 light-year out and back at half the speed of light for my class. Adds up to about half-a-year difference.

Although we were accelerating, can't we consider ourselves at rest?

@AnubhavGoel every observer is at rest in their own rest frame. That's how the rest frame is defined. What matters is the proper acceleration of the observer in their rest frame.

dmckee

Depends on the precision you need. The analysis of the Hafale-Keeting experiment requires you to take into account not only gravitational time dilation but the rotation of the Earth as well.

And this proper acceleration is different for the two observers. It is zero for the stay at home twin and non-zero for the accelerating twin.

If you're interested see the Wikipedia artice on proper acceleration.

Thanks for link. Please let me read it for a while.

4:51 PM

Okay, at this point I'm pissed

Can someone define what a constant is, in mathematics?

That sounds ridiculously easy, but something's aggravating me

user116211

@SirCumference What?

user116211

Hmm.

Just work with me

user116211

Constants are objects which don't change when you create new statements from older ones.

@MAFIA36790 Yeah, that works.

Now can someone explain why in hell the constant of proportionality in polytropes isn't a nonchanging, easy number?

It seems to be a variable in and of itself!

user116211

4:57 PM

I'm out.

user116211

Don't know about polytropes or what sort of that.

user116211

But I think it's a variable constant.

user116211

@SirCumference A parameter.

@MAFIA36790 Great, so now I'm dealing with what should be called a parameter of proportionality

user116211

Kinda think so.

5:10 PM

How to do two binary operator abstract algebra with diagrams: rules and example proof:
Disclaimer: The diagrams do not provide actual computational advantage, it only make things more organised

@JohnRennie you can

It's a terrible, terrible bound though

Like 259 or something?

and $\Huge{THIS}$, is our enemy:

Don't remember exactly

@0celo7 wow, we could be embedded in 259D space :-)

5:15 PM

[Speech mode] In order to divide by zero, one must break the minimum number of axioms so that nobody can reach to the left or right edge on the top block diagram

However, there are constraints:

One cannot break both distributive laws, otherwise * is no longer mutiplication and + is no longer addition

One cannot break both additive identities AND having no absorbers, else there are no zero elements

One also cannot break both multiplicative identity, else whatever 1 is , it is not defining an inverse element

There are a few options left:

so far this (and the n element generalisation) is the only example of a system having a legimate mutiplicative inverse of zero

Here zero elements do exists because all elements are left additive absorbers

at the same time, * forms a group

+ associativity and distributivity trivally holds because all elements are left absorbers

therefore division by zero is defined here

(not very useful though cause the + is effectively mutilated to unusability)

There might be more division by zero systems out there, but they are going to be not very nice. E.g. nonassociativity is the norm

@JohnRennie not that exact number

But it's big

Nash embedding for Lorentz metrics

@0celo7 arxiv.org/abs/0812.4439

5:33 PM

@JohnRennie I am asking, although Spacecraft twin accelerates , can't we consider a frame in which he is at rest and stay at home twin accelerates?

@JohnRennie are you saying it's not generally possible?

I might be remembering incorrectly...

@JohnRennie Okay , I get it. Thank you so much.

Spacecraft twin himself won't see himself at rest. I get it.

Thank you so very much.......

@dmckee Thank you for showing interest.

vzn

6:07 PM

@rob ok glad to hear you might like or even enjoy it :) whens a better time for you? eve? wknds? are you in US?

6:24 PM

@DanielSank @dmckee "anyone else here". if I recall, only 4 users were actively chatting and I didn't think anyone was a physicist. It's irrelevant though since it was a math question.

user218912

I have my first midterm lol

user218912

I didn't go to any of the classes or do the homework

user218912

because it's linear algebra xD

6:41 PM

@bl00 I just did a linear algebra problem.

DanielSank

6:57 PM

@Obliv oh

@bl00 lol why are you taking a linear algebra class

user218912

@Obliv because I'm in first year.

user218912

I need it for my degree.

@Obliv linear algebra is highly nontrivial

One of the most challenging areas of mathematics

taking it next semester. wish I could take abstract algebra instead though.

7:09 PM

It will be the hardest class you take in undergrad, probably.

user218912

@0celo7 my midterm is literally multiplying matrices and solving linear equations.

user218912

that's like high school.

Solving linear equations is the hardest thing in math

Nothing is harder

Try solving a 7x7 matrix

user218912

not when it's like 3 equations lol

You need a PhD in math

@bl00 that's MS level

user218912

7:10 PM

oh right

@BernardMeurer agrees with me

user218912

I don't agree.

user218912

literally easiest thing ever.

user218912

alright gtg midterm bye!

good luck @bl00

user218912

7:12 PM

thanks

@bl00 Then I'm sure QFT, etc is trivial

@Obliv don't take linear algebra

7:36 PM

@0celo7 in cylindrical coordinates: $ds^2 = dp^2 + p^2d\phi^2 + dz^2$ what is $\phi$? Is it the angle between some chosen reference direction on the plane and the projected line of the point onto the plane?

@0celo7 can you solve a linear equation in any other way besides setting up a matrix & finding inverse

i figured out the cylindrical thing btw. it was what I proposed.

dmckee

@Obliv Ah. I usually take "around" here to mean people who pop in at any time because I treat chat as asynchronous. I think that you're right that everyone then active is still at the pre-proffesional state of their career.

8:17 PM

@Obliv not in general, no

It's the only method guaranteed to work

@0celo7 why would $\nabla f = \frac{df}{ds}$ where $ds = ... $ that thing I put up there

I mean, what would the net change of the function w.r.t the [arc length?] be useful for I guess

for cylindrical coordinates

reading from tong's notes damtp.cam.ac.uk/user/tong/em/prelim.pdf

@Obliv The equation does not make sense - the l.h.s. is a vector, the r.h.s. looks like a scalar.

What page of Tong's did you take that from?

in the link. it's the prelim.

2nd case

I don't see the equation $\nabla f =\frac{\mathrm{d}f}{\mathrm{d}s}$ anywhere there.

well $ds^2 = d\rho^2 + \rho^2 d\phi^2 + dz^2 \to ds = d\rho + \rho d\phi + dz$ then $\frac{df}{ds}$ makes sense

8:23 PM

@Obliv None of that makes sense because $\sqrt{a^2+b^2}\neq a+b$.

cuz then it's $\frac{\partial f}{\partial \rho} + \frac{\partial f}{\partial \phi}\frac{1}{\rho} + \frac{\partial f}{\partial z}$ makes sense

oh shit

@Obliv That's not what is written there for $\nabla f$. Each of the three terms has a different basis vector behind it, they are not simply added.

how is the divergence found then?

By the formula that's written there

Are you asking why that is the formula for the divergence?

but where is that derived from? I don't see where the $\frac{1}{\rho}$ comes from

so it's not $\frac{df}{ds}$?

8:28 PM

@Obliv You derive it by starting from $\nabla f = (\partial_x f)\hat{x} + (\partial_y f)\hat{y} + (\partial_z f)\hat{z}$ and then expressing each derivative in terms of the derivatives $\partial_\rho,\partial_\phi,\partial_z$ and each basis vector in terms of $\hat{\rho},\hat{\phi},\hat{z}$. A more-or-less tedious calculation yields the formula written there.

@Obliv No, $\frac{\mathrm{d}f}{\mathrm{d}s}$ doesn't make any sense - $f$ is a function of $\rho,\phi,z$. What is that $s$ you're wanting to differentiate it with respect to?

$f$ is a function of the position?

Physicists sometimes do some tricks with the metric and functions to get a derivative $\frac{df}{ds}$, but in that case, they're choosing some sort of path and $s$ is the parameter of that path.

@Obliv what did you think it was a function of?

let's not get into that.

Okay :D

so $\nabla f$ yields the change in position w.r.t. $(\rho, \phi, z)$?

oh then $\frac{1}{\rho}$ is involved in $\frac{\partial f}{\partial \phi}$ because the direction changes the position of the point according to that constant factor. makes sense.

8:36 PM

@Obliv Yes. However, you must keep in mind that the gradient $\nabla f$ is a vector. At any point, it points into the direction in which $f$ changes most quicky.

so $\nabla$ is called a gradient? thought it was the divergence

oh $\nabla \cdot f$ would be divergence

@Obliv $\nabla$ is called nabla. Applied to a scalar function $f$, $\nabla f$ is called its gradient. Applied to a vector function $ A$, $\nabla\cdot A$ is called its divergence and $\nabla\times A$ it scurl.

@acuriousmind thanks. I think I can finally go into the EM notes from tong then. After I pick apart those definitions a bit more.

ok, @ACuriousMind

calculus homework or FNV?

FNV, clearly

I don't even know what calculus is

5

8:44 PM

@ACuriousMind analysis on Banach spaces

as usual

I need to estimate some difference quotients on convex sets

@ACuriousMind This might not work on all Banach spaces

but $$\sup_{x\ne y}\frac{||f(x)-f(y)||}{||x-y||}=\sup_{z\in U}||df(z)||$$

operator norm on the right

I said you should play FNV!

Why do you bother me/yourself with the quotients?

I'm not bothering you

if you saw an immediate solution then I would appreciate it

I'm just thinking out loud

rubber duck

@ACuriousMind I want to solve at least half of the problems on the problem set before I play FNV

Aha I think I need the MVT along a line

::opens steam::

@ACuriousMind I can't control myself

rob

9:04 PM

@vzn In all honesty, after the end of the term (first or second week in December) is better for me for a number of reasons.

FNV crashed

@ACuriousMind horrible game

10:20 PM

@DanielSank, I just found out that we might be heading out to California on vacation sometime soon, were you serious when you said

Oct 2 at 17:09, by DanielSank

@heather if you're even in southern California stop by for a lab tour.

10:57 PM

@ACuriousMind OMG why am I in the Legion's camp

Should I join them???

DanielSank

11:15 PM

@heather Yes, I was completely serious.

I have to keep my claim to fame of meeting the most PSE users in person!

So far I have three, I think.

@ACuriousMind HELP

What do I do

@ACuriousMind I destoyed the bunker

I am Caesar's now...

@ACuriousMind Did I screw up?

@ACuriousMind I like the Legion's whole spiel, but I everyone in the game disagrees :(

I don't want them to hate me

11:36 PM

@DanielSank, oh, wow...thank you!

I'm literally jumping up and down right now.

=)

Do you guys force yourself to read a textbook that's boring?

Well, because I'm not required to read textbooks (I just read them because I want too) if I found one boring I'd go and find a better one. But then, our definitions of boring might differ.

I have read this book: amazon.com/Algorithm-Design-Jon-Kleinberg/dp/0321295358

It's hard for me to read because I'm not experienced in algorithms.

In other words, it's too wordy to me.

insert "to" between read and this.

Well, did you ever take a class where you didn't know the material before hand?

That is, you didn't perpare in advance for the class?

11:51 PM

@JingWeng, first, just an fyi, you can edit a comment by scrolling your mouse over the comment, clicking on the down arrow, and then clicking edit. Second, sure, I'm in middle school, so a number of the classes I've taken I don't know the material; the point of the class is to learn it.

But yeah, it's too bad you didn't prepare for it...out of curiosity, what's the class on?

@heather it's on algorithms.

It's called algorithms analysis.

Ah, cool.

Well, maybe you can look around online for resources/courses/such things

Coursera and Udacity have some good stuff.

And if you have specific questions (or more general ones) you could always ask them here in chat or on the main site.

K.