The h Bar

0celo7

12:00 AM

Yes, that should work

Bernard Meurer

12:21 AM

@DanielSank Yes master

0celo7

@Slereah what is that

Slereah

Hopefully ladder operators expressed in the Schrodinger wavefunctional representation

0celo7

why are you doing that

Slereah

I am curious of what it looks like on the vacuum

My guess is that in the end the ladder operators look something like $$a_k(x) = \frac{\delta}{\delta \varphi_k(x)}$$ $$a_k^\dagger(x) = \varphi_k(x)$$

0celo7

12:37 AM

Does this raise the GDP?

Slereah

where $\varphi_k(x)$ is the one particle state wavefunction

Well it's basic QFT

It might possibly raise the GDP

It's not too fanciful physics

0celo7

Probably more GDP than the algebra shit I'm doing

Slereah

most certainly

You could probably do some condensed matter physics with this

0celo7

uhhh

this book is a troll

Secret

I see unions, intersections, direct sums, direct products, cartesian products, maps and vector spaces

0celo7

12:43 AM

tensor products, not direct products

Secret

ok

5

Q: Schrödinger's cat and the difficulty of macroscopic superposition state

The Schrödinger's cat was regarded as peculiar since we seldom encounter a superposition state in macroscopic scale: $$ \mid \mathrm{dead \,\,cat} \rangle + \mid \mathrm{alive \,\, cat}\rangle $$ We more often describe an unknown cat as $$ \mid \mathrm{dead \,\,cat} \rangle \langle \mathrm{dead ...

Now to do some exercise and monitor what happens to the full density matrix and the reduced density matrix as I do a rotation in order to see how exactly what is called "spin polarisation" by susskind dissapears

that might help me to have a better grasp on how entangled states change expectation values as a function on how entangled it is

Secret

1:18 AM

1

Q: Could a strongly enough charged black hole allow for a strongly charged particle to escape the event horizon?

If there is a strongly charged black hole, and a particle with a strong charge of the same sign goes inside of the black hole's event horizon while moving quickly, is there a reason why it would not be able to come out the other side of the black hole's event horizon?

electromagnetism black-holes

Fun with light cones

0celo7

@ChrisWhite What's the best algebra book

user54412

1:33 AM

@0celo7 Dummit & Foote

0celo7

@ChrisWhite is that an undergrad book

I need to learn multilinear and homological algebra

user54412

You said "best"

0celo7

Does it cover those topics?

Aha, it does

jesus christ 900 pages

user54412

is multilinear algebra something other than linear algebra?

0celo7

maybe not

I want to do it via functors

user54412

1:36 AM

reminds me of when people define tensors as multilinear maps, as though extra syllables make everything better

0celo7

That's a bad definition

The universal property is hidden

Maybe I should just read MacLane

learn some category theory

user54412

::looks up MacLane::

user54412

categories right in between rings and modules??

0celo7

what

right in between?

user54412

III. Rings
IV. Universal Constructions
V. Modules

0celo7

1:44 AM

what

user54412

poor group theory doesn't actually happen until XII, but I guess pairing it with Galois... sort of makes sense

0celo7

amazon.com/…

user54412

oh that maclane

user54412

apparently he wrote multiple algebra books

0celo7

o

amazon.com/…

Wonder if this is accessible

user54412

1:48 AM

almost without exception, all and only the most advanced, inaccessible courses in my undergrad were titled "an introduction to..."

0celo7

Lee's books are all "introduction to" and they're good

Bernard Meurer

1-800-CHRIS-WHITE

Have you ever messed with OpenMP?

0celo7

oh god a Brazilian

@ChrisWhite Spivak's "comprehensive introduction to diff geo" is pretty good

maybe geometry is an exception to your rule

BBS's intro to string theory is complete ass

user54412

@BernardMeurer all the time

Bernard Meurer

@ChrisWhite It's magic isn't it? It's so nice!

0celo7

1:55 AM

@BernardMeurer did you read my Brazilian proof yet

Bernard Meurer

@0celo7 nope

0celo7

do it, and I will help you understand it

user54412

yeah not often you can just put a comment in your code saying "make this loop parallel" and it just works perfectly

0celo7

I have to explain it to my Brazilian prof in a few days, I want to run it by you

Bernard Meurer

@ChrisWhite Exactly! I was totally amazed!

It made my multithreaded ackermann much easier to make

@0celo7 Link

0celo7

1:57 AM

1

A: Why are geodesically convex sets diffeomorphic to $\Bbb R^n$?

We provide a careful proof of Theorem 5.1 from Bott & Tu. The main body of the proof is taken from this MO post. Let $M^n$ be a smooth manifold. An open cover $\{U_\alpha\}$ is good if each nonempty finite intersection $U_{\alpha_1}\cap\cdots\cap U_{\alpha_k}$ is diffeomorphic to $\Bbb R^n$. The...

Bernard Meurer

why is the manifold denoted $M^n$ and what does it mean for it to be smooth?

0celo7

n dimensional

smooth is technical and not used in the proof explicitly

Bernard Meurer

What's a cover and what does it mean for it to be open?

And what does it mean for it to be good?

0celo7

cover means the union of things in the cover contains whatever you're trying to cover

@BernardMeurer I define that right there

open means the cover is comprised of open sets

Bernard Meurer

What's diffeomorphic?

0celo7

2:02 AM

isomorphism in the smooth category

Bernard Meurer

What's isomorphism?

0celo7

Ask bob

Seriously, ask him

Tell him I told you to

Bernard Meurer

Is he even up at this time?

0celo7

...maybe

Bernard Meurer

Put us in an iMessage groupchat

0celo7

2:09 AM

No

Bernard Meurer

Okay

What's a Riemannian metric?

0celo7

Oh geez

It's a smooth assignment of an inner product $\langle-,-\rangle_p$ to each tangent space $T_pM$

Bernard Meurer

No clue what you mean

0celo7

oh, that wasn't wrong

@BernardMeurer did you not read Shankar?

Bernard Meurer

No.

user218912

3:06 AM

how dumb can religious people be? Lol

0celo7

Dawkins is an ass

user218912

how?

Jordan

5:56 AM

@JohnRennie after looking over my previous work, I'm fairly sure that my equation $$v_c^2 = \frac{GM_E(2h+R_E\lambda^2)}{(R_E+h)R_E}$$ is correct. However, do you get the same result using the vis-viva equation?

lucas

6:56 AM

What does this sentence mean? "You made my day"

Sanya

7:31 AM

@lucas you probably either entertained or helped someone a lot

lucas

@Sanya Thanks a lot! And what should I say in response?

"You'r welcome!" is suitable?

lucas

8:29 AM

Hi @JohnRennie

Sanya

9:15 AM

@lucas should be; in general

lucas

@Sanya Thank you! :-)

Secret

sciencedaily.com/releases/2016/07/160722104135.htm

Three articles, thus posting one link to show them all

Main topic: Multilevel quantum systems (qudits) instead of $2^n$ qubits

David Z

12:38 PM

@lucas I would suggest "thanks" or perhaps "my pleasure" or "glad to help" or "good that you understand" (depending on why it was said to you). "You're welcome" sounds a little weird as a response, to me.

@DanielSank sweet :-)

@DanielSank Y'know, I could twist that to imply that you're a sloppy coder ;-) but seriously, it's just the nature of Python that it draws the boundary between compile-time errors and runtime errors a little differently than some other languages, so that it transfers some work from the compiler to the test suite. But the issue where you can run a program for hours and not find an error until the end is bigger than this. It happens in all languages.

Jim

@lucas "don't mention it" or "happy to help". Definitely my top two choices. Unless it was a lot of effort on your part and you feel they said it in a way that implied a "I owe you one". In that case, you could always try the "Buy me a cold beer and we'll call it quits". That can work nicely too

David Z

If you have problems with syntax errors, maybe try pylint, if you're not using it already? I haven't used it much but I would imagine it catches a lot of syntax errors for you.

innisfree

12:56 PM

Pylint is handy, though quite pedantic on matters of style. E.g. use of map or anonymous lambda functions

lucas

@DavidZ @Jim Thanks a lot for teaching me! Both of you made my day! :-)

innisfree

What's new here? How did the q&a idea go? Feels like ages since i logged in

Jim

@lucas happy to help

0celo7

1:30 PM

What

Slereah

1:44 PM

@yuggib halp

Jim

@0celo7 I must be missing something because, to me, that seems like fairly standard internet comments. Why are you confused and/or showing this to us?

0celo7

@Jim It's exceptionally cringey

Jim

I'm sorry, I'm not especially familiar with the different gradations of cringiness; I've been meaning to look that up. What about it specifically stands out or makes you cringe more than usual?

0celo7

1:59 PM

all of it

ACuriousMind

You must not read internet comments often :P

@Slereah QFT trouble?

Emilio Pisanty

@ACuriousMind Man, dupe-hammer is too strong

cf this one

physics.stackexchange.com/posts/269563/revisions

This is the sort of stuff that it's meant to prevent, right?

Slereah

@ACuriousMind indeed

Emilio Pisanty

Also cc. @JohnRennie, I don't know if you'd get pinged by mi comment there

ACuriousMind

@EmilioPisanty Hm, no, I don't think it's there to prevent that. I'm actually not sure what its purpose is except to give gold badge holders a feeling of power :P

Emilio Pisanty

2:08 PM

@ACuriousMind Hah, yeah

I guess what I meant was - the reason it's only after a gold tag badge is that presumably there's no way for people with "so much experience with the tag" to disagree about a dupe close

Which is demonstrably not the case

0celo7

@ACuriousMind Let $N\hookrightarrow M$, where both are manifolds and $N$ has codimension $2$. Is it possible for $M-N$ to be disconnected?

Emilio Pisanty

In this one it just came down to "is not", "is too", with the question flapping open and closed at the whim of the wind

But now I can't even vote to reopen without the vote being chunkier than I want it to

0celo7

It does not seem to be true in 3 dimensions

ACuriousMind

176

Q: Increase close vote weight for gold tag badge holders

In order to increase the efficiency with which poor quality questions are closed, it could make sense to have weighted close votes for a small subset of qualified users. A very good way to measure the qualified users would be to leverage the tag badges. However, it needs to be rationally limite...

feature-request status-completed vote-to-close tag-badges dupehammer

I actually can't find any other argument than "hey, gold badge holders know what they're doing so let's give them votes that count more"

And yes, I also find it rather annoying that my vote instantly closes and reopens precisely because I may not be sure about it, but I now have no way to get a question into the queue.

Emilio Pisanty

@ACuriousMind Maybe it's time to ask for a way to turn it off

Or maybe it's already been asked?

ACuriousMind

2:16 PM

24

A: Let moderators disable the dupehammer for certain tags

We originally thought this would be necessary too. Then we looked at the distribution of gold badges. I think you can probably stand to moderate one person if he decides to go rogue on his own area of expertise. As soon as there are two people, you have a balance - one can revert the other's ac...

lol

I really don't follow the thinking here

"Hey, let's do something for the gold badge holders" - "Okay" - "Hey, there's a bit of a problem, can we change it a bit" - "No, there are too few gold badge holders, so we don't care" ???

More funny things in this MO question

Emilio Pisanty

@ACuriousMind But do you actually need a basis? All you really need is an infinite linearly independent set

And that's sort of the definition of infinite-dimensional, right?

ACuriousMind

Okay, given you have an infinite linearly-independent set - how do you produce a subspace that is not of finite dimension?

Emilio Pisanty

@ACuriousMind just pick one vector and skip it?

ACuriousMind

So, if it is not a basis, how do you know the span of that is not still the whole space?

Emilio Pisanty

2:55 PM

It doesn't require AoC because it's a single set rather than an infinite family of sets

@ACuriousMind If $\{v_1,v_2,\ldots\}$ is linearly independent, then the span of $\{v_2,v_3,\ldots\}$ can't include $v_1$, because that would make the original set linearly dependent.

ACuriousMind

Ah

0celo7

@ACuriousMind Can you at least give a yes or no to my exact sequence comment...

ACuriousMind

@EmilioPisanty I'm...not sure you have the existence of an infinite linearly independent set without choice

All you know is that the space is not finite-dimensional

Emilio Pisanty

@ACuriousMind OK, but we agree that if you do then you have a strict subspace?

ACuriousMind

Yes

Emilio Pisanty

2:59 PM

OK, so here's the argument I'm thinking of. You know $V$ is infinite dimensional, so it can't be empty, hence there is at least one LI vector

Now, assume (inductively), that there are $n$ LI vectors in $V$. Their span can't be $V$ or $V$ would be finite dimensional, hence there must be an $(n+1)$th LI vector in $V$, hence there are $n+1$ LI vectors in $V$.

Does that not give you the sequence you want?

Qmechanic

@EmilioPisanty : I think so. Apparently only close-voters who have the same close-reason as the mod have 'signed" the close reason explanation. If there is no mod-close vote, there probably can appear more than one close reason, each "signed" by the respective close-voter.

ACuriousMind

@EmilioPisanty That gives you a sequence of finite-dimensional vector spaces

Emilio Pisanty

@ACuriousMind Sure.

ACuriousMind

How does it give us an infinite-dimensional subspace?

Emilio Pisanty

Does it not also give you a growing sequence of finite-dimensional vector spaces?

$V_1 < V_2 < V_3 < \cdots$?

ACuriousMind

3:03 PM

yes

Emilio Pisanty

Then $\bigcup\limits_{n=1}^{\infty} V_n$ works, right?

@Qmechanic No, I've never seen that. If there are five close votes then all the names are shown with only the majority reason.

It's caused some friction on MSE in the past.

Qmechanic

@EmilioPisanty : Ah, OK.

Emilio Pisanty

@ACuriousMind (But then, on the other hand, I do distinctly remember Zorn's lemma being vital to the existence of a basis. Looking at Friedberg right now to tell what it's used for. Maybe just to prove you can get to the entire space?)

ACuriousMind

@EmilioPisanty Infinite unions behave strangely in the absence of choice, too.

Emilio Pisanty

@ACuriousMind How do you mean?

ACuriousMind

3:09 PM

I.e. I think that although you were able to pick a LI set for each of the $V_i$, that does not mean you can for the union.

Emilio Pisanty

It's just $\{v: \exists n \:\mathrm{s.t.}\: v\in V_n\}$, right?

ACuriousMind

Because in the absence of choice, a countable union of two-element sets can actually be uncountable

0celo7

@ACuriousMind Oh god

Now THAT I cannot handle

Ok, back on the Choice bandwagon

Emilio Pisanty

@ACuriousMind Sure, but it still exists

@ACuriousMind (but yeah, bejeesus, that's a monster)

ACuriousMind

@EmilioPisanty Yeah, but your goal was to construct an infinite LI set

Emilio Pisanty

3:10 PM

@ACuriousMind No, my goal is a proper subspace

David Z

@EmilioPisanty actually I think only the names of those who voted for the final, displayed close reason are shown after the close reasons. In other words, the reason actually used to close the question is displayed, along with the names of whoever chose that reason. Any other reason chosen by someone else is not displayed. And the reason used to close the question is that of the mod close vote, if applicable, otherwise it's the most popular choice among the five close voters.

cc @Qmechanic

0celo7

@ACuriousMind How does choice come in there to save the day?

ACuriousMind

@EmilioPisanty But how do you know that union there is not the whole space?

0celo7

In the 2-element-set example

Emilio Pisanty

@DavidZ No, that's not true. Case in point the linked question, the first three votes were UWYA

Plenty of other examples, but they're hard to document because the actual list of close reasons gets lost

David Z

3:12 PM

How does that make my statement not true?

Emilio Pisanty

@DavidZ "only the names of those who voted for the final, displayed close reason are shown" is not true

The names of all close voters are shown, regardless of what reason they gave

David Z

Oh, I was talking about the names listed after the close reason itself. Not the full list of five close voters.

(now edited to clarify)

ACuriousMind

@0celo7 The axiom of choice pretty much directly says that there are choice function on unions, and you thend just assign odd/even arbitrarily to the two-element sets and this gives a counting of the union, when you send the images under the choice functions to the n-th odd/even number

David Z

I thought that's what you were talking about.

Emilio Pisanty

@DavidZ Huh, oh, yeah, that's pretty new

@DavidZ Naw, so what bugged me was that Qmechanic flipped the reason (which is fine, just noteworthy)

David Z

3:15 PM

@EmilioPisanty oh, I thought it had been there for quite a while (years), but I'm not really sure. Anyway it doesn't matter.

Emilio Pisanty

@ACuriousMind OK, so back from the start.

@DavidZ Agree on that last bit ;)

$V$ is not empty, so it has a nonzero vector $v_0$.

David Z

If I'm voting to close a question which already has three votes for the same reason, I pick that reason, even if I don't agree with it, so that it winds up being the one displayed.

Emilio Pisanty

$\{v_0\}$ is not a basis, so there's an LI $v_1$.

Assume inductively that there exists an LI set $\{v_1,\ldots,v_n\}$ such that $v_0$ is not in its span. $V$ is infinite dimensional, hence there must exist $\{v_1,\ldots,v_n,v_{n+1}\}$ LI and with $v_0$ also not in its span.

John Rennie

The OP couldn't even be arsed to format their homework question properly. What is it with the homework cheats of today? :-)

0

Q: waves motion and sound and organ pipes

in an open organ pipe whose one end is at x=0, the pressure is expressed by P=P0cos(3pi/2)sin(300(pi)t) where x is in meter and t in sec.then the organ pipe can be- a)closed at one end ,open at another with length=0.5m b)open at both ends ,length=1m c)closed at both ends ,length=2m d)closed at on...

homework-and-exercises waves

Emilio Pisanty

Hence by the axiom of induction there exists a set $\{v_1,v_2,\ldots\}$ that's linearly independent and such that $v_0$ is not in the span of any finite set, hence $v_0$ is not in its span.

Take $W=\operatorname{span}\{v_1,v_2,\ldots\}$. That's a proper subspace, right?

Obliv

3:19 PM

@John could you help me out with a gauss' law problem

Emilio Pisanty

And if I haven't messed up, it exists independently of AoC.

Obliv

it seems pretty straight forward but I'm not entirely sure what the question is asking

ACuriousMind

@EmilioPisanty Ah, no, induction gives you only that your statement holds for finitely many $v_i$.

You'd have to take their union and choose the $v_i$ in it again to get the infinite statement.

John Rennie

@Obliv I have long hated and feared electrodynamics (well, electrostatics in this case, but that's just as bad) so probably not. However if you ask the question I'll have a look at it.

Wish me luck, I'm about to bid on a Nexus 9 on eBay ...

Emilio Pisanty

@ACuriousMind That sounds pretty weird to me.

0celo7

3:24 PM

@ACuriousMind oh, right

ACuriousMind

@EmilioPisanty Well, induction say that if $P(n)$ holds for $n=1$ and $P(n)\implies P(n+1)$, then $P(n)$ holds for all $n\in\mathbb{N}$. It doesn't say anything about infinity

Obliv

@johnR I think I understand it actually. I had an issue with the e-field due to a gaussian surface because I thought gauss' law was concerned only with the electric flux with the surface but those equations can be used to determine the e-fields

for those surfaces, which this question was asking about.

John Rennie

@Obliv Cool, that was easy. I can solve problems by my mere presence :-)

Emilio Pisanty

OK, I found it. You do need Zorn's lemma to prove every vector space has a basis, but it's only to get the spanning property. You work in the set of LI subsets of $V$, partially ordered using inclusion. Then every chain has an upper bound, given by the union. (This is because if you pick $n$ elements $v_i$ out of the union, they must belong to $n$ sets in the chain, hence they must belong to a single set of the chain, which is LI.)

You then use Zorn's lemma to get a maximal LI subset of $V$, i.e. such that no superset is LI, and that one is necessarily a basis.

0celo7

@ACuriousMind The codimension 2 thing is false, btw. You can't disconnect an $n$-manifold using an $(n-2)$-submanifold.

Emilio Pisanty

3:34 PM

@ACuriousMind I'm still not sold on that one, though.

0celo7

3:45 PM

@EmilioPisanty You're not "sold" on it?

Emilio Pisanty

@0celo7 I'm not sure that you're not guaranteed an infinite sequence

Or what it actually means

But looking at that MO thread

36

Q: Is the non-triviality of the algebraic dual of an infinite-dimensional vector space equivalent to the axiom of choice?

If $V$ is given to be a vector space that is not finite-dimensional, it doesn't seem to be possible to exhibit an explicit non-zero linear functional on $V$ without further information about $V$. The existence of a non-zero linear functional can be shown by taking a basis of $V$ and specifying th...

axiom-of-choice set-theory linear-algebra dual-pairs lo.logic

definitely makes it look like the result is true (no AoC implies $\exists$ V with no ID proper subspaces)

so there's obviously stuff I'm not getting w.r.t. the induction argument

Little Alien

I do not get it. If QM like says that coordinate distribution uniquely defines the speed distribution in every moment of time. Suppose that coordinate is certainly collapsed in every moment of time. It is certain. Then, speed/momentum is absolutely undefined, which means that speed momentum is arbitrary, which means that coordinate is uncertain in every next moment of time. It looks like a contradiction. Where is the flaw in this logic?

Emilio Pisanty

@0celo7 @ACuriousMind But then I get big ?s instead of pupils when set theorists say "there exists a model where..."

Not sure at all what a model is in that context.

Obliv

@johnr Can you help me again by gracing me with your presence

instead of doing practice problems for the exam i have in ~3ish hours i'm just gonna think about the theory and hope i'm competent enough to apply it. So I'm gonna see if my notions of gauss' law, electric potentials, and electric fields are correct. You just stand there and look pretty, ok? @johnR

John Rennie

OK, though if you have three hours to go to an exam I would sit back, have a light meal and try to relax rather than do any last minute cramming ...

ACuriousMind

3:53 PM

^listen to the ~~old~~ wise man

3

0celo7

@JohnRennie Will freezing a metal with LN2 alter the crystal structure significantly?

Emilio Pisanty

(And yes, that was me starring that, @John. And yes, >50% of people seeing the stars list know who "wise" refers to =P.)

John Rennie

@EmilioPisanty :-)

0celo7

We all know who the old man is

@EmilioPisanty I think you have to define $P(\infty)$ first

John Rennie

@0celo7 if you dunk solid metal at room temp in LN2 it's unlikely to have any effect on the crystal structure

0celo7

3:55 PM

@JohnRennie But it should make it brittle, right?

Maybe brittle enough to give it a good whack and pulverize it?

or at least make smaller bits

@EmilioPisanty In general, $P(\infty)$ has no meaning

I don't think

Like summing the integers from $1$ to $n$

John Rennie

@0celo7 yes, steel will become brittle at LN2 tempertures, but I think that's mainly due to the low temp stopping dislocations from moving. It isn't a change in the crystal structure.

0celo7

@JohnRennie Ok, what I'm trying to crush is not as hard as steel, I think

Emilio Pisanty

@0celo7 yeah, I guess

Nimantha

Help me to solve this question physics.stackexchange.com/q/269860/124403

John Rennie

I think plastic flow in most metals is dur to movement of dislocations, so any metal will become brittle if you cool it enough to stop the dislocations moving.

Nimantha

4:01 PM

Help me to solve this physics.stackexchange.com/q/269860/124403

John Rennie

@Nimantha that question contravenes our rule on homework questions and is likely to be closed at any moment. So drawing our attention to it here (twice) wasn't the smartest move.

0celo7

I like how he didn't even say "please"

Danu

@JohnRennie Lol clearly too long ago you did any exams ;)

Obliv

@JohnR Gauss' law defines the net flux of an electric field going through a closed (gaussian) surface (perpendicular to the surface) being related to the net charge enclosed. Electric potential energy is just an analogue of potential energy in newtonian mech to electrostatics so defining as follows $U = \int F\cdot ds$ $U_e = \int qE \cdot ds = qV$ so these are the core concepts that are required to derive more complicated equations for different arrangements of charges.

The e-field at a point from the dipole axis of an electric dipole can be derived using algebra, the e-field at a point from a solid object like a disc or cube can be derived using calculus, etc. The flux going through a surface by a net enclosed charge can be derived by integrating the E-field due to the charge through an area differential,

the easy way of doing this would be by choosing a surface that is completely perpendicular to the enclosed charge at every point so the integration can be done in one step(?)

John Rennie

@Danu I did some professional IT exams (Microsoft MCSE) about 15 years ago, and I followed my own advice for them. Work hard during the days before the exam then on the day of the exam try to relax. And it worked :-)

Danu

4:14 PM

@JohnRennie That's not the same as an exam week where 5 exams are crammed into 14 days ;)

John Rennie

@Danu I think my finals were more or less just one week, with two exams a day. For those I worked for the preceding weeks then stopped at the weekend and did no further cramming during the week - just light reading.

I agree that if your exams are very spread out then it might be different.

Danu

@JohnRennie Ah, you did multiple exams that time? Fair enough ;)

John Rennie

@Obliv perpendicular to the field lines?

Obliv

I'm not really sure. I know it's defined as $\epsilon_0 \int \vec{E} \cdot \vec{dA} = \phi$ so it looks to be that way

John Rennie

I think in practice you always choose your Guassian surfaces to exploit symmetry ...

Obliv

4:21 PM

wait but

What changes the E-field you derive due to the flux? Can't you consider the charge enclosed to be concentrated at a point?

then can't you always derive the equation for the e-field with a spherical surface?

oh I guess not. There are different e-fields for different symmetric arrangements of charges. For cylindrical,planar, and spherical symmetries there are different equations.

no wait that's only for infinite sheets and lines of charges

and the spherical symmetry only gets a diff. eq if the gaussian surface is enclosed by the spherical charge distribution instead of the other way around

John Rennie

Gauss' law refers only to the total flux through the surface, and for the total flux it doesn't matter what size and shape the charge is as long as it's all inside the surface. However if you look at the flux through a subset of the surface then that will depend on the size/shape of the charge.

Obliv

yeah that explains the infinite cases for sheets and line charges

John Rennie

So changing the charge distribution will increase the flux through some parts of the surface and decrease it in others, but it will keep the total the same.

Obliv

since there cannot be a surface that encloses an infinite dimension

Slereah

4:37 PM

Gauss law more like dross law

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