Mathematics - 2017-06-01 (page 4 of 4)

s.harp

17:00

yeah, phase volume was in integer multiples of h or hbar

Semiclassical

One thing to note is that if you don't have the 1/2, you still have $S_{n+1}-S_n = h$.

So the volume enclosed does need to go up in steps of $h$.

But if you want to get the right z-p energy, one needs $S_0=\frac12 h$.

s.harp

let me do a quick calculation

Semiclassical

kk

s.harp

suppose $\omega=1$ then $H=\frac{p^2+x^2}2$, so the volume of phase space so that $H≤r^2$ is $\pi r^2/2$, if you want this to be $h$ then $r=\sqrt{2h/pi}$. Now calculate the expectation
$$\frac{2\pi\int_{0}^{\sqrt{2h/\pi} } dr \, r \frac{r^2}2}{2\pi \int_0^{\sqrt{2h/\pi}}dr\,r \cdot 1}$$

Semiclassical

That's a semicircle in phase space.

It should be a full circle since the orbit goes forwards and backwards.

Though if this is what Planck did I'll wait.

s.harp

17:05

I dont know what planck did, only what the lecturer told us in the correction was he did

why is this a semicircle?

Semiclassical

$\pi r^2/2$.

s.harp

but $H= (p^2+x^2)/2\equiv r^2 /2$

Semiclassical

Oh. So $p_{max}=x_{max}=r\sqrt{2}$.

s.harp

yes

Semiclassical

...which would give the volume of the ellipse as $2\pi r^2$ :/

I think I'm a bit confused, though. Is $H\leq r^2/2$ or $r^2$?

s.harp

17:07

oh

no

wait

Semiclassical

If it's $H\leq r^2/2$ then $p_{max}=x_{max}=r$ and $\pi p_{max}x_{max}=\pi r^2$.

s.harp

the volume of a circle of radius $r$ is $\pi r^2$

if $H≤a$, then this corresponds to a circle of radius $\sqrt2 a$

so volume $2\pi a^2$

kind of embarrassing that I cant do that lol

Semiclassical

If $H\leq a$, that's a circle of radius $\sqrt{2a}$ with volume $2\pi a$.

s.harp

jeeze

Semiclassical

I meant something wrong :P

s.harp

17:12

well

if you do the correct integral correctly you get $hbar/4$

right?

Semiclassical

Yeah.

s.harp

over 4

Semiclassical

Wait, you changed it since I looked.

s.harp

hmmm

Semiclassical

Writing $R=\sqrt{2h/\pi}$, I have $\int_0^R dr r^3/2 = R^4/8$ and $\int_0^R dr r=R^2/2$.

s.harp

17:14

when did I become this stupid

Semiclassical

So the ratio is $R^2/4=2h/4\pi=\hbar$.

s.harp

R should be $h/(2\pi)$

Semiclassical

So you want $H\leq h$?

Oh, wait. I'm being silly.

I feel like we're getting confused, though.

Let's start from the top. We've got $H=\frac12 (p^2+q^2)$, and we want a certain volume in phase space to be $h$.

Which volume are we specifically after?

s.harp

let me write out what we even want to do:
Let $a$ be so that $\Lambda_0:=\mathrm{Vol}\{(x,p)\mid H(x,p)≤a\})=h$. Now $\frac{\int_{\Lambda_0} H }{\int_{\Lambda_0}1}$ is what we want to calculate for the zero point energy

Semiclassical

Okay. That'll be a circle $p^2+q^2=2a$ of radius $\sqrt{2a}$ so volume $2\pi a=h$.

s.harp

17:18

sorry $\Lambda_0$ is the set, not the volume, oops

Semiclassical

Ah. I was wondering about that.

s.harp

$2\pi\int_0^{\sqrt{h/2\pi}} r^3/2 dr =h^2 /(16\pi) = \hbar\, h/8 $

Semiclassical

\hbar

s.harp

divide by $h$ to get $\hbar/8$ instead of the $\hbar/2$ we want

Semiclassical

Hrm.

s.harp

17:24

interesting if you take $H=p^2+q^2$ you have $a=h/\pi$, then the upper integral becomes $h^2/\pi^2 /4$, if we divide this by $h$ we get $\hbar /2$

Semiclassical

Let's turn it around a bit. If the radius is $R$, then the integral becomes $\frac{2\pi\cdot R^4/8}{2\pi \cdot R^2/2}=\frac{R^2}{4}.$

s.harp

yep

so you want $R^2= h/\pi$

Semiclassical

That was too quick for me.

The zero-point energy should be $E_0=\frac{1}{2}\hbar$, yes?

s.harp

wrote $2$ instead of $\pi$

Semiclassical

Ah, okay. So $\pi R^2=h$.

Note, though, that if you want the first excited state $E_1=\frac32 \hbar$ you'll need $\pi R^2 = 4\pi\cdot \frac32 \hbar = 3h$.

So while this may end up reproducing the z-p energy it won't get the first excited state (or any past that) correct.

s.harp

17:28

no, you misunderstood how the higher levles work here

Semiclassical

Is the desire that the increase in area be $h$?

s.harp

yes, but the energy $E_n$ is the expectation over the difference of area, not the entire area

do you know what I mean?

its the expectation of an annulus of area $1$

Semiclassical

Huh.

Hadn't expected that.

s.harp

im really sort of surprised that it doesnt work right now

Semiclassical

But, anyways. If $\pi R^2=h$ then it's a circle of radius $\sqrt{h/\pi}$ corresponding to energy $H=\frac{h}{2\pi}$.

s.harp

17:31

because I dont see any difference inthe formulation to what we did

Semiclassical

Something seems goofy, yes.

s.harp

I'll write it out on paper later today and update you with whats going on

Semiclassical

mmkay.

s.harp

right now im gonna go to the store to get some food

see you, sorry for being incapable atm^^

Semiclassical

kk. I think I still prefer Bohr-Sommerfeld for now :P

Mostly because $S=(n+\frac12)h$ comes out so simply.

s.harp

17:32

(conceptually I think the planck method is very nice, but apparently its not as easy as i thought!)

Monolite

hi fellow members of the chat

is it possible to link a question one has posed on this chat or is it considered incorrect use of the chat?

defalt

17:52

how do i find the multiplicative inverse of a mod b without taking all values from 0 to b-1 and waiting when the result 1 comes?

Astyx

@Monolite As long as you're not interrupting an ongoing conversation and you don't spam the chat with links, it should be fine

@defalt Is $b$ prime ?

Semiclassical

@s.harp just finished writing out the calculation for generic energy levels, and I'm surprised to find that it indeed works :/

defalt

@Astyx it may or may not be a prime number

lets consider b as prime

Semiclassical

If b isn't prime you're not guaranteed to have an inverse mod b for a

But it's enough that they be coprime

Astyx

By little fermat it would be $a^{b-2}\pmod b$

Alessandro Codenotti

17:56

If $a$ and $b$ are coprime you can use the extended euclidean algorithm, otherwise there is no inverse

defalt

@Astyx is it a^(b-2) mod b?

Astyx

Yes

defalt

i think those symbols don't work in chat

BOLD

some symbols work

Astyx

You have to activate mathjax in chat

See the link in the top right

Semiclassical

See the room desc

@s.harp my hypothesis right now is that the two calculations are doing distinct but related things which happen to coincide for the harmonic oscillator

With mine, I'm choosing the orbit such that the volume enclosed is (n+1/2)h and taking the energy of such orbits as the level.

With yours, the relevant volume is that of an annulus with inner/outer areas of nh,nh+h respectively. (Hence my circles always are within your annuli)

And then you take an average over that range of energies

defalt

18:08

should i start chatJax or render MathJax?

Astyx

Start chatJax I'd say

Read the description to know what the differences are

And chose whichever suits you best

user193319

18:25

I recently proved the following: If $A$ and $B$ are compact subspaces of the Hausdorff space $X$, then there exists disjoint open sets $U$ and $V$ containing $A$ and $B$, respectively. So, I am trying to find an example illustrating the necessity of the Hausdorff stipulation on $X$. Would the following work? Consider $\Bbb{R}$ with the following topology: $U$ is open if and only if $\Bbb{R}-U$ is countable or all of $\Bbb{R}$. This is not Hausdorff, since every pair of open sets intersect; and

because of this, it is clearly impossible to find open disjoint sets covering disjoint compact subspaces.

Does this sound right?

iKlsR

Hi guys, could someone explain why this polynomial? expands like this?

Liad

@user193319 im not sure it is correct, but i think $\Bbb R$ and $\phi$ compact in your space.

iKlsR

Basically I have to square the terms/constants and group them, trying to find out why actually. EL15.

user193319

$\emptyset$ is defintely compact; indeed, it is compact in every space.

Liad

yea ,im not sure about $\Bbb R$

arctic tern

18:31

use the distributive property,
(A+B)(C+D)
=A(C+D)+B(C+D)
=AC+AD+BC+BD.
You may be familiar with this as "FOILing."

user193319

@Liad Suppose $\Bbb{R}$ were compact. It isn't immediately obvious why it would be a counterexample.

iKlsR

@arctictern gotcha, the square was what was tripping me up. Also are polynomials like this perfect squares?

arctic tern

@iKlsR polynomials like what? If it's of the form A^2+2AB+B^2, where A and B are algebraic expressions, then yes it factors as (A+B)^2.

user193319

@Liad And anyways, the theorem requires that this is true for all pairs of disjoint compact sets.

arctic tern

For instance 4x^2+12xy+9y^2 = (2x)^2+2(2x)(3y)+(3y)^2 = (2x+3y)^2.

user193319

18:34

But I am still curious to know whether $\Bbb{R}$ in this topology is compact.

iKlsR

@arctictern was referring to the link I posted but I got it now, thanks!

Very helpful. :)

Liad

@user193319 you are right, i thought you tried to prove something else.

Daminark

Hey everyone!

Liad

@Daminark hi!

Astyx

Hi @Daminark

Daminark

18:38

How's it going?

Astyx

Good and you ?

PVAL-inactive

Hi fellow mathers

Daminark

Doing alright, thanks!

Alessandro Codenotti

hi @dami

mathvc_

Hey guys! I have trouble in getting intuition for homology :/ how do u think about homology?

TheGreatDuck

18:50

@mathvc_ homology = hom-ology = home-ology. It's obviously the study of homes. Or is it Sherlock Holmes. I'm not real sure...

Abdullah UYU

$x,y\in \mathbb{R}$. For inequality $|x-y|>|x+y|$, is it valid to take square of both sides to conclude something about $x,y$?

ArtEze

@iKlsR Good avatar.

iKlsR

@ArtEze fight me irl bro.

My S3 main. Thanks! Tho the day I quit that cancer led to the most productive year of my career.

ArtEze

¡ooh!

@iKlsR Keep it up.

Semiclassical

19:07

@AbdullahUYU if $a>b>0$, then $a^2>b^2>0$

So squaring both sides is legitimate.

Abdullah UYU

yeah, i realized just after wrote it

Semiclassical

It's a good thing to be paranoid about

Abdullah UYU

i am sure my last sentence is gramatically wrong

Semiclassical

It works because $f(x)=x^2$ Is strictly increasing for positive $x$

Fargle

Howdy chat.

Abdullah UYU

19:13

and we can consider we have positive $x$ in both sides

i know varibles coincide

Semiclassical

So what conclusion do you draw?

Abdullah UYU

$xy<0$

Semiclassical

Right. And that can be split further.

Pythonista

Dumb question: In differential equations (and other places) say we have e (euler's number) raised to some arbitrary constant. Why is it acceptable to replace this by C denoting an arbitrary constant in the real numbers? I don't think I ever explicitly learned this, just used it as a tool, but feel like it's connected to e being transcendental. Since the constant is arbitrary, and in R, we can make e^(constant) span R? So, replacing e^(constant) by an arbitrary constant (C) in R is equivalent?

Fargle

@Pythonista Well, e to a constant is just another constant, so, yeah, for the most part.

Abdullah UYU

19:22

We can write $(x>0 $ and $y<0)$ or ($x<0$ and $y>0$) if we want.

Semiclassical

Right. So the inequality holds in the second and fourth quadrants, and nowhere else.

SteamyRoot

If you write $C = e^A$, then you do assume that $C$ is positive. But apart from that, you can easily "convert" back from $C$ to $e^A$ by $A = \log(C)$.

@Pythonista What "gaps" ?

Pythonista

Nevermind, did not think that through

Fargle

Right, the caveat is that the new constant is understood to be positive.

Semiclassical

There is a bit of slipperiness in that to get C<0 you'd need a complex a

Fargle

19:25

Because $e^K$ can never be negative for real $K$.

Semiclassical

And C<0 is certainly allowed as part of a solution

Fargle

@Semiclassical If we're talking solutions to $y' = ky$, then I'd say any $Ce^{kt}$ for $C \in \Bbb R$ is a solution, but that falls out of absolute values and the fact that the $0$ function solves it.

Semiclassical

my take is that complex C is allowed, it's just usually not consistent with the initial condition

Fargle

It would be more precise (pedantic?) for me to say that there are 3 families of solutions: $y(t) = 0$, and each of $y(t) = \pm Ce^{kt}$ for $C > 0$.

@Semiclassical Ah. I was just self-reconciling the fact that there are solutions with negative function values with the fact that $C$ must be positive.

Semiclassical

But I think this can't be settled unless one indicates the codomain of y.

I mean, there are contexts where the codomain would be vectors rather than scalar so.

FryanBury

19:31

Hello there

testing $\frac{1}{2}$

:(

Semiclassical

In which case you'd even more freedom in solutions

You need to enable Chatjax, @FryanBury. See the link in the room desc

FryanBury

oh thanks

Simply Beautiful Art

@FryanBury tinyurl.com/cfqcvpc

FryanBury

$f(x) = \frac{1}{x}$

PVAL-inactive

I think often times when you see e^a be replaced by c in a solution to a diff. eq.

the value with negative c is still a solution

Semiclassical

19:33

(Fun fact: if your numerator and denominator are single characters, you don't need the brackets. E.g. \frac12)

PVAL-inactive

you just assumed something or didn't think about all the possible complex solutions

when you solved your de.

Simply Beautiful Art

@Semiclassical Also, things of the form \... do not need brackets. E.g. \frac\pi\phi

$\frac\pi\phi$

PVAL-inactive

Don't you need brackets in TeX

Pretty sure you still need brackets in Tex

even if not in MathJax

Simply Beautiful Art

Only if its multiple characters and not some \ command

Hm, probably

Semiclassical

One way to avoid such issues, if memory serves, is to always do the integration as definite integrals

FryanBury

19:36

\frac1x

PVAL-inactive

TeX generally requires more specific syntax then MathJax

FryanBury

$\frac1x$

Semiclassical

That builds the initial condition right in.

Daminark

Back, and hey @Alessandro!

Also @PVAL, @Semi, @Fargle, @Steamy, and @Simply!

SteamyRoot

ohi

Simply Beautiful Art

19:39

\o/

SteamyRoot

also, RIP climate

Semiclassical

So for instance you'd directly get $x=\ln(y(x))-\ln(y(0))$

No sign ambiguity there since the ratio is definitely not negative

Daminark

Did Trump leave Paris?

SteamyRoot

yup

PVAL-inactive

I mean if America stops using its military to keep the lines of oil to Western Europe clear it wouldn't really be bad for the climate.

It would probably completely destroy

Europe's economy and that hurts the NYSE so I doubt that will happen.

SteamyRoot

19:44

which oil lines to Western Europe?

PVAL-inactive

The ones from the Persian gulf

Fargle

Howdy @Daminark

PVAL-inactive

There are huge deposits of crude in North and South America, but much of Europe is importing its energy (and a whole host of natural resources).

SteamyRoot

The majority of oil in Europe comes from Russia.

If the US somehow cut the lines, that would only allow for Russia to expand its influence over Europe.

Simply Beautiful Art

@AkivaWeinberger Have you seen the answer with the half-iteration functions?

FryanBury

19:52

I just wanted to see if I got something. Let $B_N = {a_N, a_{N+1}, \dots}$ and $\limsup a_n := \lim_{n\to\infty} (\sup B_n)$ and $\liminf$ defined the same way. Then for $a_{2n-1} = \frac1{2^n}$ and $a_{2n} = \frac1{3^n}$ and $b_n = \frac{a_{n+1}}{a_n}$ then $\liminf b_n = \lim_{n\to\infty} (\frac23)^n$ because for all $n$ we have $(\frac23)^n < \frac12(\frac32)^n$ then the inf of every $B_n$ will be of the form (\frac23)^n. The same holds for sup B_n and terms of the form \frac12(\frac32)^n. Is all this correct?

Daminark

Merp

FryanBury

20:05

what?

Semiclassical

@AlessandroCodenotti I meant to ask: When you it's not clear, do you mean on the formal side or to how everything is justified?

I might be able to say something about the formal derivation but on the rigor side of it I probably can't say a thing

Lozansky

20:58

Hi chat

Semiclassical

hi @Lozansky

did you have your exam today?

Lozansky

I sure did

Semiclassical

How'd it go?

Lozansky

I think I got 5 out of 6 questions right

Semiclassical

Nice!

Lozansky

21:01

There was one with a Euler equation that I didn't know how to solve

Semiclassical

Don't suppose you remember what it was?

Lozansky

I think I do

I'll write it down

$\rho (\dfrac{\partial \mathbf{u}}{\partial t}+(\mathbf{u} \cdot \nabla)\mathbf{u}) = - \nabla P$, where $\rho$ is the constant mass density, P the pressure (scalar field) and $\mathbf{u}$ is the velocity field. Show that the density moment vector $\rho \mathbf{u}$ satisfies the differential equation $\dfrac{ \rho \partial u_i}{\partial t}+\partial_j T_{ij} = \mathbf{0}$

Where $T_{ij}$ is a tensor of rank 2 (the tension tensor)

I think that's how it was formulated

Semiclassical

Ahh.

Yeah, that seems pretty nasty.

Lozansky

Yeah, I just wrote some nonsense that I hope render me some points...

Semiclassical

Though, that'd amount to showing that $\rho(\mathbf{u}\cdot\nabla)\mathbf{u}+\nabla P$ is the divergence of a rank-2 tensor?

Lozansky

21:07

Yeah that's what I was aiming for

Semiclassical

Hmm.

Lozansky

But my expression for it got quite messy so I just stopped

Didn't have too much time to mess with it

Semiclassical

Well, term-wise that's $\partial_j T_{ij}=\rho u_i\partial_i u_j +\partial_i P$

Lozansky

Right

Semiclassical

For the second term it's enough to write $\partial_i P=\partial_j P\delta_{ij}$.

Lozansky

21:08

Ohh

I was going for something like that but couldn't remember if that was right

It's the quotient rule or something?

Semiclassical

Nah. It's just the fact that the derivative of P*identity matrix is basically just the derivative of P.

Also, I wrote it wrong above: Should be $\rho u_j \partial_j u_i$ for the first term.

Lozansky

BTW, you might wanna use $(\mathbf{u} \cdot \nabla)\mathbf{u} = \nabla(|\dfrac{\mathbf{u}^2}{2}) - \mathbf{u} \times (\nabla \times \mathbf{u})$

Semiclassical

Nice.

Lozansky

I forgot to say that $\nabla \cdot \mathbf{u} =0 $

Semiclassical

First part of that is fine. It's the second term I'm not sure what to do with.

Lozansky

21:12

Which is important

I tried to somehow get an expression of the kind $\partial_i (...)$ but didn't prove succesful

Semiclassical

In indices, we've got $\rho u_j \partial_j u_i = \partial_j(\rho u_j u_i)-u_i \partial_j (\rho u_j)$

But that second term is again a problem.

You can use $\partial_j u_j=0$ to write that last one as $u_i u_j \partial_j \rho$.

But that seems...eh.

Lozansky

Yeah...

Semiclassical

Oh, wait. Isn't there a condition on $\rho$ as far as imcompressability goes?

Or is that for $\mathbf{u}$ itself?

Lozansky

It's constant

That's about ti

it

Semiclassical

Ahah. If $\rho$ is constant, then we're in business.

That means I can ignore $\rho$ in all of the above.

So then it's just $u_j \partial_j u_i = \partial_j (u_j u_i)-u_i \partial_j u_j =\partial_j (u_j u_i)$.

Lozansky

21:16

Yuuup

Oh wow

Yeah

Semiclassical

So $\rho u_j \partial_j u_i +\partial_i P=\partial_j(\rho u_i u_j+\delta_{ij}P)$.

Lozansky

Brilliant

Semiclassical

And hence $T_{ij}=\rho u_i u_j+P\delta_{ij}$.

Whew.

Lozansky

How could I not see that

Semiclassical

Eh, exams are exams.

Lozansky

21:18

First time I saw such a problem but still... seems kinda obvious now

Semiclassical

Doing it in terms of indices helps.

Daminark

runs away from indices

Semiclassical

And having the presence of mind to do integration by parts.

Lozansky

Yeah, that's what I did, but I mostly messed around with vector identities trying to get a nice expression

Semiclassical

Plus, I'm pretty sure I've had to use that $\partial_i = \delta_{ij}\partial_j$ trick before

Lozansky

21:19

Definitely remember that

Semiclassical

The main reason I'd avoid vector identities here is because it's a divergence of a tensor field, not a vector field.

And I don't know the identities for those myself.

But I do know how to do index gymnastics.

Lozansky

Right. I don't think we covered tensors a whole lot so I wasn't very comfortable messing with that

Semiclassical

Yeah.

It's been a while since I did them, but doing grad-level physics pretty much forces you to get at least some range of comfort with them.

I think I remember doing these kinds of manipulations for a second semester stat-mech / transport theory course, for instance.

Which isn't surprising, given how hydrodynamics fits in there.

Lozansky

Not to mention E&M

The second part question was a bit more difficult I think

Given $\mathbf{\omega} = \nabla \times \mathbf{u}$ show that $$\dfrac{\partial \mathbf{\omega}}{\partial t} - \nabla \times (\mathbf{u} \times \mathbf{\omega} )= \mathbf{0}$$

Semiclassical

ahh, vorticity?

Lozansky

21:26

Precisely

Semiclassical

Yeah.

Lozansky

I wrote two pages trying to solve it

Semiclassical

I guess my first inclination is to put on my E&M hat.

Lozansky

I'm sure you can solve it in two lines

Nooo

Take it off :P

Semiclassical

And to think of $\omega,\mathbf{u}$ as analogous to $\mathbf{B},\mathbf{A}$.

The equation shown then becomes $\frac{\partial \mathbf{B}}{\partial t}-\nabla\times (\mathbf{A}\times \mathbf{B})=0$.

I'm not saying this as "do this" so much as I'm wondering what it's equivalent to in E&M.

To make the analogy work, I need to be in Coulomb gauge so that $\nabla\cdot \mathbf{A}=04.

Lozansky

21:29

Change to B in first term

Semiclassical

Ah, yes.

Lozansky

What the hell is Coulomb gauge? Should I know?

Semiclassical

Well, do you know what a gauge is?

Lozansky

No

Semiclassical

Basically it's a condition you put on your vector potential in order to get a unique result.

Lozansky

21:31

What course would cover that?

Semiclassical

E&M, junior-level probably

Lozansky

Undergrad E&M?

Oh

Then I should know

Semiclassical

I think. But it wouldn't do much with it, tbh.

It's not until grad school that you become really seriosu about gauge stuff.

Matters a hell of a lot for QED.

Also, should've been $\nabla\cdot \mathbf{A}=0$ above.

Lozansky

Yeah that makes sense

Semiclassical

It's the vector-potential equivalent of "if I add a constant to my electric potential, I'll get the same electric field"

Except worse, b/c y'know. vectors

All of this is, of course, rather silly. I'm just curious what it amounts to.

Lozansky

21:34

Sure :)

Semiclassical

Could use Faraday's law at this point: $\frac{\partial \mathbf{B}}{\partial t}=-\nabla\times \mathbf{E}$.

And you'd get $\nabla\times (\mathbf{E}-\mathbf{A}\times \mathbf{B})=0$, huh.

I have no idea why that's true at the moment, but it's a neat statement :)

I'm going to be heading out, but I just realized I'm being potential being silly here.

Lozansky

Loool, that's not their approach

:D

But if it works, it works

Semiclassical

Well, no. Of course not :P

Another way to get one vector field as the curl of another in E&M is Ampere's law

happyEddie

Any easy way to prove that affine varieties over complex numbers containing infinitely many points are always unbounded?

Semiclassical

in which case $\nabla\times \mathbf{B}=\mu_0 \mathbf{J}$.

(in the absence of a time-varying electric field, mind)

So I could instead have taken $\mathbf{u},\omega\sim \mathbf{B},\mathbf{J}$

Lozansky

21:41

Hrm. That reminds me of the flow of temperature, where $\mathbf{J} = - \lambda \nabla T$

Semiclassical

Then $\nabla\cdot \mathbf{B}=0$ is just Gauss's law for magnetic fields, and $\nabla\cdot \mathbf{J}=0$ is presumably just a statement about the continuity equation.

So we want a scenario where there's volume current density but no accumulation of charge.

That'll be directly analogous to the conditions given.

Okay, enough rambling I need to head out

later

Lozansky

Cya!

Captain Bohemian

22:02

positionlessness of graduate school graduates brings disappointment to these graduates

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$Mathematics$ Mathematics