The h Bar - 2016-09-14 (page 1 of 2)

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12:48 AM

user image

6

1 hour later…

1:51 AM

Trying to get into wireless router.

(not mine)

Don't know password.

Used factory reset button.

Password stated online doesn't work.

Please advise.

@DanielSank Brute-force attack.

explain

@DanielSank Crack the code!

My dad hacked into a neighbor's wifi once.

Useless advice.

That's like me saying "Solve the physics".

1 hour later…

3:12 AM

0

Q: What do answers with zero votes mean?

What do answers with zero votes mean? Can I trust them? Does it mean that there is something wrong but not so noticable in the answer? Does it mean that their answer should not be considered to be a totally complete answer to the question? Or perhaps no one is knowledgeable enough to judge if th...

3:40 AM

@Slereah You are a wicked person. I may have a new obsession. And I don't need that during the semester.

@DanielSank Check for the possibility of variant model numbers with different passwords, or needing to hold the factor reset extra-long to include the password.

Beyond that I fear you may have to visit some 1337 h@ck0r 5!73 and sprinkle line noise in your typing for a while.

@dmckee is it Lagrangian or lagrangian

Hamiltonian / hamiltonian

I capitalize.

I don't know if there is a standard expectation.

@dmckee Are there some authors that like to distance "names" from theorems?

Names as in people names.

1 hour later…

4:54 AM

Wormrot - Eternal Sunshine of the Spotless Grind

It's worth searching for their other tracks. This is an extreme one even by their standards.

user228700

5:27 AM

I found this interesting to read:

user228700

newscientist.com/article/…

This taught us to never overlook the simple things, cause there might be just happened that you have overlooked just one interesting perspective to look an an object which holds most of its mysteries and power

user228700

6:19 AM

@Secret What are u talking about?

The crumbled paper article you referenced

@KaumudiHarikumar: re your question, you write $$\frac Qt = \frac{KA(T_2-T_1)}{d}$$

But really this should be $$\frac {dQ}{dt} = \frac{KA(T_2-T_1)}{d}$$

This second form always applies regardless of how the temperature difference changes

user228700

6:39 AM

Oh, crap. MathJax isn't on :/

user228700

@Secret Yep! :-)

user228700

@JohnRennie Oh, okay...

user228700

But then I can write $dQ=mCdT$ and then cancel right..?

@KaumudiHarikumar you can certainly write $dQ = mCdT$. But what are you going to cancel?

user116211

Just, by the by, the general case $$\mathbf h = -\kappa~\nabla T\;.$$

user228700

6:46 AM

So we've got $mCdT/t=KAdT/L$.

user228700

Can't I cancel the dT's?

No, because $T_2-T_1 \ne dT$

user228700

Oh...

user228700

Um, can you please explain a little bit more?

I'm in the middle of checking some servers, so I'll have to get back to you in 15 minutes or so ...

user228700

6:51 AM

Sure :-)

user116211

7:13 AM

Hey @FrancescoS; after long time.

7:30 AM

@MAFIA36790 Hey

8:16 AM

-1

Q: Turn off Community bumping for meta

Just today, the Community bot bumped a three-year old question. It's also recently bumped a half-year old question. Right now about 1/10 of the posts on the 'active questions' meta tab are bumps from Community. I don't think this helps at all, and it worsens the signal-to-noise ratio. Can this ...

To follow the discovery on seemly mundane things engineers have managed to stop the colourful swirls on a bubble

user116211

An edit is proposed to a question which was surely to be closed. IMO, I would reject the edit since accepting it would make it to the re-open queue; but the post is doomed; no one can save it. So, should I reject or accept the proposed edit? (Better reject it).

user116211

-5

Q: I think I may have found a renewable energy source

Hello guys so I have this idea of mine which I can't get my head around and I need someone to tell me why it won't work, so let's say you have a fully pressurised tank of air and you blow that pressurised air at a turbine connected to a generator, ovisously the generator will then start to create...

electromagnetism energy experimental-physics perpetual-motion renewable-energy

user116211

8:33 AM

I could reject the edit on the ground no improvements whatsoever. For merely rectifying the spelling mistakes doesn't make the question eligible for the site at all..... hmmm.

8:51 AM

same concept similar sound, regardless of language

@MAFIA36790 I really don't think it matters. I would personally accept an edit which honestly does make the post better, on the assumption that it will remain closed as it goes through the reopen queue, but there is something to be said for not using up reviewers' time on a lost cause.

user116211

@DavidZ That's the point.... the post is a lost cause at the very first place. Just by merely correcting the spellings doesn't improve the post at all; of course, it's my opinion. My single vote doesn't reject the edit; it depends on the community.

@KaumudiHarikumar: that was a long 15 minutes :-)

I'm back now if you're still interested.

user228700

9:06 AM

@JohnRennie :P Yes, it was. And yes, I'm definitely still interested.

OK, start with $$\frac {dQ}{dt} = \frac{KA(T_2-T_1)}{d}$$

Potassium isotope ratios being identical in earth and moon suggests the impact is so violent that it vaporise earth into a silicate atmosphere and the moon condensed from it

user228700

@JohnRennie Since I dunno MathJax, my situation is very pathetic. I copy what you've typed into the draft for a new question and obtain the preview to see what you've written.

Go to:

dl.dropboxusercontent.com/u/78279253/mathjax.html

user228700

Okay..?

user228700

9:11 AM

Thing is, sir, I'm chatting using my mobile.

I use the "render MathJax" link on that page. What browser are you using? Chrome, Interet Explorer?

@KaumudiHarikumar that is... not recommended, not if you want to follow math content. Or perhaps it's better to say this: the mobile chat interface is better than nothing, but it still isn't a perfect substitute for the full chat site. We can only do so much to accommodate mobile chatters; the "official" recommendation if you want to participate fully in chat is to use the full site.

I think the Mathjax script should still work in the mobile version of Chrome

@JohnRennie BTW there are better options available, linked at the top right.

user228700

@JohnRennie Nope, not working. Never mind. I'll do the thing with the preview. Please go on...

9:14 AM

:32304513 OK. The equation isn't very useful as it is at the moment because it has $Q$ and one side and $T$ on the other.

user228700

Okay..?

But we can use the definition of specific heat $$\frac{dT}{dQ} = \frac{1}{C}$$

Now $\frac{dT}{dQ}$ is not a fraction. It's a derivative. But physcists tend to be a bit careless about maths and we can treat it as a fraction if we're careful.

That's why we write $dT = \frac{dQ}{C}$

But $dT$ and $dQ$ are infinitesimal quantities. What we are saying is that if we transfer an infinitesimally small quantity of heat $dQ$ the infinitesimally small increase in temperature will be $dT = dQ/C$.

Anyhow, if we go back to our original equation we can replace the $dQ$ on the left with $dT$ to get:

user228700

@JohnRennie What do you mean..? How?

@KaumudiHarikumar Can you clarify what you are asking?

user228700

Sorry, what I meant was, what do you mean by "We can treat it as a fraction if we're careful"?

9:21 AM

It's a limit of fractions. If you're careful with the limits (unlike physicists), you can manipulate it like fractions.

@KaumudiHarikumar Suppose we take a finite heat transfer $\delta Q$. Then if the specific heat is constant we get a finite increase in temperature $\delta T$.

user228700

Okay...

You'll often see $\delta x$ used in place of $dx$ where we are talking about small but finite quantities not infinitesimals.

user228700

Yup...

Anyhow, rearranging gives $$\frac{\delta T}{\delta Q} = \frac{1}{C}$$

user228700

9:23 AM

Okay...

This is just a regular equation so we can multiply both sides by $\delta Q$ to get: $$\delta T = \frac{\delta Q}{C}$$

But, but, but, but, but ... $\frac{\delta T}{\delta Q}$ is not equal to $\frac{dT}{dQ}$

user116211

1525

Q: MathJax basic tutorial and quick reference

To see how any formula was written in any question or answer, including this one, right-click on the expression it and choose "Show Math As > TeX Commands". (When you do this, the '$' will not display. Make sure you add these. See the next point.) For inline formulas, enclose the formula in $......

discussion faq mathjax reference

user228700

@JohnRennie Okay..?

user228700

@BalarkaSen Didn't fully understand, but thanks.

So you can't take $\frac{dT}{dQ}=1/C$ and multiply both sides by $dQ$ because it isn't a fraction like $\frac{\delta T}{\delta Q}$ is.

user228700

9:28 AM

@JohnRennie Okay, I guess that's fair...

But being physicists not mathematicians we say well the difference between $\frac{\delta T}{\delta Q}$ and $\frac{dT}{dQ}$ is vanishingly small if we make $\delta T$ and $\delta Q$ small enough.

So we go ahead and write $dT = dQ/C$

user228700

Okay, sure...

In most cases this works fine, but you have to remember that this is a mathematically suspect procedure and if abused will get you into trouble. In this case however it's fine.

@KaumudiHarikumar Don't really know what is there not to understand, but John Rennie has explained what I was saying.

user228700

@JohnRennie Yeah, OK, will keep in mind...

9:31 AM

Anyway ...

We can use this to replace $dQ$ in our original equation and we get: $$\frac {dT}{dt} = \frac{KA(T_2-T_1)}{Cd}$$

That's because $T-T_0$ is a finite quantity. At the beginning of our experiment it's 100ºC. It can be a large number.

While the $dT$ on the left side is an infinitesimal quantity.

user228700

Ohh, right, OK...

It is not the case that $dT = T_0 - T$

So you can't cancel it from both sides.

user228700

Guess this is what happens when I don't think properly :/

Getting used to this sort of thing is a standard part of learning physics

The point is if you have a thing like $dy/dx$ ($y$ is a function of $x$), then that's the limit of $\Delta y/\Delta x$ as $\Delta x \to 0$, so $\Delta y/\Delta x = dy/dx + \epsilon$, aka $\Delta y = \frac{dy}{dx} \Delta x + \epsilon \Delta x$. Here $\epsilon \to 0$ as $\Delta x \to 0$. It doesn't in principle make sense to write things like "$dx$" and manipulate: what most high school books do is to use that as an alternative for "$\Delta x$", with which you can manipulate.

user228700

9:35 AM

So, OK, that was my basic doubt. Thanks :-) I'll go over the question from the top and check it again.

Schools don't really deal with this aspect of calculus, so you can't be blamed for not being familiar with it.

But trust me, after a few years of doing physics this sort of thing will be engraved upon your soul :-)

user228700

@JohnRennie OK sir, thanks :-) Also, about everything that @sammygerbil was saying...

user228700

Was it not OK to post the question in the first place?

Sammy was being a bit fussy. Your question was a borderline homework question, but it's a judgment call whether it's acceptable or not. I answered because I know from these chats that you're trying to learn physics and not just trying to get other people to do your work for you.

You do need to be a bit careful about posting very specific questions that look as if you've been set a problem as homework and you're too lazy to do it yourself :-)

user228700

Okay, thank you for understanding. Thank you :-) Now I'm off to do some problems!

user228700

9:44 AM

@JohnRennie Yes, I knew that I was running the risk of it being closed because it was a homework question but I couldn't think of any other way to clear my doubt...

user116211

@BalarkaSen There is an chapter dedicated to this named Infinitesimals, Errors in high school maths.

user116211

and it's an important concept in Numerical Analysis.

Chapter in which book?

user228700

Guess I'll ask the specific one line doubts here, from now on. Wasted about 15 mins this morning, commenting back.

user116211

@BalarkaSen Any book certified by CBSE.

user228700

9:45 AM

@MAFIA36790 Actually...not anymore.

OK. I admittedly haven't looked.

user228700

Exactly how long ago were u in high school? :P

user116211

@KaumudiHarikumar O.o

user116211

@KaumudiHarikumar hmm.... I passed my boards exam this year; I'm not that old ;)

user228700

Huh? That's not the name of the chapter! And we definitely don't have all that in math o.O

user116211

9:47 AM

Anyway, that's why I use \frac{\mathrm d}{\mathrm dx_i} to denote an operator and \frac{dx}{dt} to denote a fraction of infinitesimals.

user116211

@KaumudiHarikumar Well, did you follow R.D. Sharma?

I don't really think of $dx$ as an infinitesimal. Those are hard to actually define.

user116211

@BalarkaSen Hmm; the notations vary from books to books; at-least my numerical analysis book follows that.

That's one of the things which nonstandard analysis do, and it requires redefining a lot of set theory, as far as I know.

@MAFIA36790 No, I mean, $dx/dt$ is a fine notation.

user116211

@BalarkaSen ahh!

user116211

9:50 AM

@BalarkaSen ohh.

I just object to "$dx$" used loosely. That's not really very well-defined.

user116211

@BalarkaSen Mathematics literatures generally don't do that; you can get such things in physics book but they know what they are doing; so that's fine till you misinterpret....

user228700

@MAFIA36790 Nope. May I ask which college you attend?

Most of the uses of that in physics are perfectly fine, for sure.

user116211

@KaumudiHarikumar Well, it's ambiguous ;)

user228700

9:53 AM

@MAFIA36790 What is?

user228700

'Cause if you're only as old as I am, you must be a freaking genius! From everything that you were posting here, I figured that you'd be much much older :o

user228700

Are you one of them geniuses who made it into IIT? That's why I asked :P

user116211

Anyways, @balarka, I'm now reading Borel Sets of Real numbers and Countable and Uncountable sets of Royden; will ping you if I face any problem.

user116211

@KaumudiHarikumar No ;)

user228700

Okay. Wow. Anyway.

9:58 AM

Sure.

user228700

Allow me to guess once more please; IISER?

user228700

/IISC

user116211

@KaumudiHarikumar No; c'mon you don't have to dig more ;))

user228700

Sure, okay. Sorry :P I'll drop it.

user116211

@KaumudiHarikumar Just because I write some correct stuffs doesn't mean I'm genius; everything I write can be found in standard books; yes, I can only say, I've not studied conventionally like others in 12th standard do; I got some good books and studied them rigorously to the extent I could and didn't think about syllabus or coaching stuffs. That's it.

user116211

10:05 AM

0

Q: why lean engines are more efficient?

Why lean engines / diesel engines are more efficient? Is it that the higher pressure increases uniformly temperature resulting in more rapid combustion? Just saying. What are the reasons really? thanks!

user116211

Engineering?

user116211

Not sure; so waiting for the community's action.

A really stupid question. I am talking about General Relativity and the metric isometries. Is the Killing vector an element (specially, a generator) of the algebra of the Lie group which describes this symmetry? And what about the covariant derivative of the Killing vector? It is an element of the algebra too?

I like mathematical art because they remind of the surreal worlds in geometric game environments, with its alien logic

1 hour later…

11:45 AM

@bolbteppa : the people who are into TQFT do.

@DanielSank : turn it upside down. The code is on the underside.

Ok guys I would like to repeat my question withe xtension

WHen gtrying to learn more about Complex number---I came across Hypercomplex Number/Quaternion so is any defination probability extended to such models

@ACuriousMind:Any suggestion/advice on the above matter ^

user228700

12:02 PM

@MAFIA36790 I see :-)

user228700

12:42 PM

Wait, so the internal energy of a system is proportional to its temperature only if said system is an ideal gas?

1:26 PM

Symmetry breaking in quantum statistical physics is basically there existing "canonical" (like KMS condition) states that are not invariant under the symmetry right?

@vzn:Just to confirm, I am sadly neither available on the 4th nor the 18th of October. So it seems you'll have to find somebody else...

1 hour later…

2:30 PM

There lurk in the dead yard

To quench the thirst of knowledge

2:42 PM

@Xasel 1. It's "definition", not "defination". 2. I already told you that probabilities are by definition real.

@FrancescoS Yes, the Killing vectors form the Lie algebra of the Lie group of isometries. I'm not exactly sure what you mean by its covariant derivative, but if you mean $\nabla_\mu X_\nu$ for some Killing vector then that's not a vector and not in the algebra.

2:57 PM

@Xasel Of the division algebras only real or complex numbers have found any significant use in physics. Every now and then someone will point out some cool connection to physics but nothing ever comes of it.

Anything observable has to be a real number, and since probabilities are by definition observable they must be real numbers.

@ACuriousMind In Zee's QFT book (2nd edition) in footnote 3 on page 228, he moves the $P^i$ operator across the $e^{-i\vec{k}\vec{x}}$ term without "catching" a commutator term. Why is that correct? After all, $P^i$ is a differential operator in $x$, so it does not commute with $e^{-ikx}$ (AFAIK you have access to the book, tell me if not).

@dmckee Turned out I was pressing the wireless reset button instead of the hard reset.

@JohnDuffield Believe it or not, it's not.

@JohnRennie :Interesting..:"Anything observable has to be a real number"..Is there any reason/proof why itis so?

@JohnRennie [citation needed]

@Bass I would have access were I at home, but I'm currently travelling. However, if $P^i$ is the QFT momentum operator, then it is not a differential in $x$!

3:03 PM

@ACuriousMind Yes, it's the momentum operator. In a prior exercise, it is derived as $P^i=\int d^3x T^{0i}$, and its commutator with a field is $[P^i, \phi(x)]=-i\partial^i \phi(x)$.

Well, I see that it's the commutator that gets us a differential in $x$.

@DanielSank @Xasel proofwiki.org/wiki/Hermitian_Operators_have_Real_Eigenvalues

@JohnRennie Ok, but why are observables Hermitian ops?

Often wondered that myself.

It's an axiom of QM isn't it?

@Martin understood! thx for your response

@JohnRennie : SO why should Reality confirm to our 'axioms'

3:08 PM

@JohnRennie but why

@JohnRennie Is probability a QM "observable" in that sense?

@DanielSank So their eigenvalues are real

Oh my God they're ganging up on me!

whole of QM is ganging up on you :)

You're the celeb here!

3:10 PM

@JohnRennie I've seen quaternionic GR.

Twistors are via quaternions

Ok lets have a hypothetical situation

I don't think probability is an observable because it isn't given by the eigenvalue of a hermitian operator.

Here lets say we build another matematical model of physics which give a limiting case of our current models

Then

@ACuriousMind Proof?

lets say define our model with property

3:11 PM

@JohnRennie The what was this? Hmm, I guess I shouldn't participate in the conversation without knowing what it's about...

that there is probability P(X) that certain axiom will confirm to our realtiy

@DanielSank I've always thought that the "natural" QM axiom is to demand that the possible results of a measurement correspond to a projection-valued measure. It then turns out that "most" such measures are obtained as the projections onto eigenspaces of self-adjoint operators.

@JiK nonsense

with a stability associated S(X)

3 hours ago, by Xasel

WHen gtrying to learn more about Complex number---I came across Hypercomplex Number/Quaternion so is any defination probability extended to such models

3:12 PM

@0celo7 Proof of what?

@ACuriousMind isometries form a Lie group

so any guess why the stability and probability of our mathematical models/axioms is so high?

@0celo7 I'll leave that one to the mathematicians :P

@ACuriousMind Good.

SO what's the opinion of you guys on this?

3:13 PM

@Xasel You lost me along the way I'm afraid. Physicists don't really use axioms - we just pretend to so as not to dismay the mathematicians.

@ACuriousMind do you know why holonomy is a Lie group yet :P

@JohnRennie Good point.

@Xasel We cobble together a theory then test it against experiment and if it works it's a good theory.

@0celo7 I usually assume that any continuous group I encounter is Lie until I'm proven wrong. ;) Topological but not Lie groups are exceedingly rare to stumble across in my experience.

@JohnRennie:Well I am saying that lets say ther'es a certain theory which gives a probability P(X) and stability S(X) of matheatical models that holds true to reality then

3:15 PM

@ACuriousMind Hmm, it is obvious that it's a continuous group?

why a certain models such as ours have such a high P(X) and S(X)

that's it almost(or it certainly true to relaity/truely maps to relaitys..sematics)

@Xasel I don't understand what that means. Presumably you mean some kind of meta theory i.e. a theory that describes theories. To be honest it doesn't sound the sort of thing hard working physicists consider worth their time.

@0celo7 Well, yes - you can write down a Killing vector and generate a continuous one-parameter group of isometries from it

(by scaling the vector, essentially)

@ACuriousMind That'll give you the identity connected component if you're looking at the Lie algebra

Maybe there's some weird pathological disconnected component

@Bass Yeah. Don't fall into the old habit of thinking $p = \mathrm{i}\hbar\partial_x$. That fails in QFT, and, in any case, the analogon to the QM $p$ are the conjugate momenta $\pi$, not the total momentum operator $P$.

3:19 PM

@0celo7 do/did you have any go-to sources in writing a lab-report?

for physics

@Obliv What?

@Xasel I suspect you think of our theories as more special than they are. 99.99% of theories didn't agree with experiment and failed. The ones we have left are the very few survivors that haven't been contradicted by experiment YET.

@0celo7 i'll take that as a no.

@0celo7 Yes, maybe. That's why I would now go looking for the reverse direction, i.e. get a Killing vector from every isometry were I actually trying to prove it

@ACuriousMind I doubt that works.

Is the isometry group connected?

Even then, the one parameter subgroups don't fill out the whole identity connected component.

3:23 PM

So?

So you won't get a killing vector for every isometry

@ACuriousMind But the conjugate momenta $\pi$ are conjugate to the fields $\phi$. How far off is it to say that $P^\mu$ are conjugate to the spacetime variables $x^\mu$? (Zee says so...)

I don't think

Note, btw, that speaking of "connected components" you have already topologized the isometry group. How did you do that? :P

@Bass It's completely off because the $P^\mu$ are operators while the $x^\mu$ aren't. QFT has no position operators, just rather useless replacements called the Newton-Wigner operators, iirc.

@ACuriousMind Well, I know that it's a Lie group

3:26 PM

@JohnRennie: Well not a theory that discuuses theory..but kind of probing why certain mathematical models confirm to reality(or vice versa) and under what condiitons(or what leads to such)

@Xasel Our models conform to reality because they were designed to conform to reality, else we wouldn't call them "models".

3

@ACuriousMind I want to upvote this post.

@Xasel We invented thousands of models and discarded the ones that didn't conform to reality.

No, not star.

Just +1

@ACuriousMind:That's true but I am thinking why a certain 'design' confirms to reality.....and w hy other don't(or some only in a limiting case)

3:31 PM

@JohnRennie thousands?

@0celo7:More that millions..in the form of mythologies :P

@0celo7 millions!

Maybe 100

At best

@Xasel I don't understand the question. Are you puzzled by the fact that some models are more accurate than others?

yeah you can think of it from that perspective too

@0celo7:myths can also be counted as models ain't it?

3:34 PM

But what is there to explain? Some models have different accuracies either because their creator just couldn't think of a better one or that accuracy sufficed for whatever purpose it was needed.

Can a dielectric material insulate a metal sphere so that there is 0 electric field inside of the sphere?

well I would certainly like to discuss more...but lets say I have to answer the call of sirens of real life no..anyway thank you guys for your suggestion/comments

3:47 PM

@ACuriousMind Did you cover "locally compact Banach ==> finite dim" in your analysis class

@Obliv : you want a Faraday cage mate.

@0celo7 No

At least, not that I recall

@ACuriousMind in FA?

Oh, I thought you meant functional analysis anyway. We didn't do anything outside of $\mathbb{R}^n$ in analysis

user218912

@0celo7 wait so if your spacetime has a cauchy surface you can define time on it?

4:03 PM

@IceLord Yes, you can get a continuous time function. You can even make it smooth but I don't know the proof.

@0celo7 @ACuriousMind
In classical mechanics, a space with points (momentum,position) is called phase space. A configuration space is a space where the points are an ordered sequence of positions (q1,q2,q3 ...). Is there a special name for a space where the points are (position, velocities), or more generally (position, velocity, acceleration , ...)?

user218912

@0celo7 how does it look like though?

@IceLord You foliate spacetime and assign a real number to each leaf, roughly.

user218912

okay

Is it even possible for a sphere to have uniform charge density?

4:08 PM

@IceLord Wald chapter 8 has a sketch of the proof.

You need to carefully define a Borel measure on spacetime, and the continuous proof is pretty bad already.

Beats me how the smooth proof works.

Probably uses techniques from geometric analysis.

@Obliv that's a strange question. Why wouldn't it?

@johnR don't they move to occupy the surface of the sphere?

I suppose I should have mentioned it was a metal sphere, or one that allows movement of the charge.

@Obliv ah OK, I assumed you meant a uniform surface charge density. You're quite right that a conducting sphere cannot have a uniform charge density throughout it's volume.

Also, if the amplitude of the vibrations of atoms increase in a solid, does it expand in volume? I believe it is this and not the frequency of vibration, correct?

@JohnRennie why not

4:15 PM

Or must there exist an asymmetric potential about the equilibrium point? does this mean neither amplitude nor frequency is necessarily affected?

@0celo7 stop bugging JR it's my turn

@0celo7 in a conductor electrons are free to move, so the electric field inside a conductor is everywhere zero. If the field wasn't zero the electrons would move until it was zero. One quick application of Gauss' law later the charge density everywhere inside the sphere is zero.

Citation needed on the first claim.

@0celo7 that's how we define a conductor dude.

@Obliv there must be an asymmetric potential

@JohnRennie that the electric field is zero?

@Obliv Suppose the potential was symmetric, then the average position of the atoms wouldn't change. If the average position of the atoms doesn't change the solid doesn't expand.

4:19 PM

guess I have to correct my professor now. HEH

wait but how would heating something increase its potential asymmetrically?

@0celo7 the force on an electron is $eE$ where $E$ is the electric field. In a conductor the electrins are effectively free, so the only force on them is due to the electric field. That means an electron will accelerate if $E \ne 0$. Since at equilibrium the electrons aren't accelerating $E = 0$.

what if you gave every atom an equal amount of momentum, you'd be increasing the avg. KE but it wouldn't have asymmetric potential

so it'd just be hotter?

@Oblive:Never 'correct' you professor..only 'suggest' and ask for 'advice'

Unless he is a very cool fellow

@Xasel until after they've marked your paper :-)

yeah I'm not going to be an ass about it. just gonna mention it in my report

4:22 PM

hahahah..but man take correction to their ego except very cool one(which are very less in quantity)

@Obliv temperature is random thermal motion. Giving every atom an equal momentum wouldn't affect the temperature it would just give your lump of whatever it is a net momentum.

@JohnRennie I'm not convinced.

@0celo7 let me see if I can work balls into it somewhere ...

@JohnRennie Rule 1 of geometry: work in a small ball

then use compactness to cover with small balls

@johnR I have so many questions but I'll save them for now

4:24 PM

@0celo7 ah, now I see why you're ideally suited to geometry

if you don't have compactness, give up; the problem is too hard

@JohnRennie:You are very funny guys(You qualify for cool professor group)\

@JohnRennie Are you saying one of my testicles is undersized?

um ok I have a basic question about coupled waveguides, is it fine to ask it here?

it's ok to discuss here but much betterr if post on forum

4:25 PM

@Eddy yes, though I don't think we have any waveguide experts around at the moment

@Xasel What forum?

@JohnRennie geometric analysis of balls

@JohnRennie I'm getting another book that treats PDEs on balls for like 300 pages :D

well I am just confused on how we could go about characterizing GVD for a coupled waveguide system

come up with a "coupled" propagation constant?

There was a young physicist from Devizes
Whose balls were of different sizes
One was so small
It was no ball at all
But the other one won several prizes

@JohnRennie I'm not a physicist.

@ACuriousMind Elven lord hath bequeathed more homework!

s/physicist/engineer/

s/ball/geodesic ball/

4:28 PM

What?

Hmm, now it doesn't scan :-)

Why would anyone here think I'm a physicist

Engineer, sure. Mathematician, perhaps.

But certainly not a physicist.

I was going to look up some balls

I've forgotten what I wanted to do with them though

Oh!

Topological balls

topology homework

there we go

@Eddy I don't know enough about waveguides to comment. It might just be worth asking on the electrical engineering stack exchange.

Or ask a question on the main site where more people will see it.

Ok yea I guess that's better

thanks anyways

@JohnR web.mit.edu/mbuehler/www/SIMS/Thermal%20Expansion.html down where it says thermal expansion

is this right?

It doesn't sound specific enough. I'm going to find a different source.

4:39 PM

@Obliv yes, that looks a nice summary.

Suppose you have two atoms next to each other in a solid and they are vibrating.

The vibration cannot be symmetric because as the amplitude increases the atoms would bash into each other.

The amplitude of the vibration towards each other is limited but the amplitude in the direction away from each other is unlimited.

So the potential is obviously not symmetric for large amplitudes.

Ohhhhh

you just placed a great image in my head @johnR

My miniature balls?

@0celo7 sigh

You'd think we'd eventually get tired of jokes about balls but no, it is an endless seam of childish amusement :-)

4:57 PM

heheh

balls are funny

5:13 PM

@Secret Position + velocity is the setting of Lagrangian mechanics, and is formally the tangent bundle of the configuration space. You almost never need acceleration or higher derivatives because the equations of motion are typically second-order.

any way I can use latex in word?

does word read xml/mathml/html?

@Obliv LaTeX is a way of writing documents, just like Word, so the question makes no sense ;) If you're asking whether you can use the way of writing math from LaTeX in Word, then no. Word allegedly has its own formula editor, but I've never used it

Once one gets used to TeXed documents the Word versions almost universally look worse

I can confirm the allegation, that Word has its own formula editor, but I rarely use it

@Obliv sourceforge.net/projects/latexinword

I used it in my first physics class because there wasn't heavy derivations or anything. but, now it's picking up and I hate its formula editor

5:17 PM

@Obliv sourceforge.net/projects/texsword

@JohnRennie The horror

They brought together what was never meant to mix :P

@acuriousmind I feel like i've already asked this but what editor do you use?

For TeX? TeXstudio

@ACuriousMind which is?

@0celo7 what?

I am horrified by merging Word and TeX, that is all :P

5:20 PM

@Obliv LaTeX does seem like a much more efficient way to enter math - but only once you learn it well enough that it becomes natural

@ACuriousMind what two things should never mix

Also my keyboard controller died so I can't use my keyboard at home. These computers at my school don't have other programs. I'm pretty much forced into using word.

Oh.

for now at least

@ACuriousMind I have found what I despise most.

People who get on elevators before everyone who wants to leave it on that floor as left.

5:24 PM

I see.

@0celo7 who does that? what kind of barbarians live in Tennessee

At least this time it's a group of people you despise for their actions, not for their height :P

@Obliv some Chinese grad student.

@0celo7 For some inexplicable reason, I have doubts that you know he/she is chinese.

@Obliv I'm sure you do. But you're wrong.

5:30 PM

@0celo7 then I don't understand why you mentioned nationality <|8O)

What?

What kind of stupid face is that

it's a clown

i'm clownin you dawg

You're a clown.

<|8'O(

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