Mathematics - 2017-06-27 (page 5 of 6)

SteamyRoot

7:00 PM

But, it's pretty standard in Riemannian Geometry to introduce tangent vectors as derivatives of curves on the manifold

Astyx

@Daminark Yeah, it's scary isn't it ?

Daminark

Truly so

Balarka Sen

@SteamyRoot $\alpha'(t)$ does act on functions.

Semiclassical

Eh, just because it's standard doesn't mean I understand it.

SteamyRoot

and then consider the derivatives $\alpha'(t_0)$ as a map from the set of differentiable functions in a neighbourhood of $\alpha(t_°)$ to the reals.

Balarka Sen

7:00 PM

Look up "derivations"

Astyx

I still can pretend I know the first chapter of G&P though

Balarka Sen

Tangent vectors can be equivalently defined as an operator (a "derivation") which eats a function and spits out that function, differentiated in the direction of the tangent vector.

Linear + satisfies Leibniz rule

Semiclassical

Sure. $\vec{v}=v^i\mathbf{e}_i\mapsto v=v^i\partial_i$.

Astyx

@BalarkaSen So it chews functions then.

Semiclassical

(einstein notation because I'm lazy)

SteamyRoot

7:02 PM

and said map is given by $\alpha'(t_0)f = \left.\frac{d (f \circ \alpha) }{dt}\right|_{t = t_0}$

Semiclassical

I guess what I'm asking is: Are $\alpha(t)$ and $\alpha'(t)$ different kinds of objects?

SteamyRoot

Yes

Semiclassical

That...bothers me. Probably because the standard vector calc way of thinking is so much in my brain.

SteamyRoot

The derivative in a point is the linear approximation of that thing in that point

But, for like, a curve on a surface, there's no reason why the linear approximation would also lie in said surface

Balarka Sen

@Semiclassical Yes. Because the thing you wrote down as a reply to my message is where you are directly identifying $T_0 \Bbb R^n$ with $\Bbb R^n$.

That is very specific to $\Bbb R^n$.

SteamyRoot

7:05 PM

We interpret the derivative as something pointing in the direction the curve will move in, and the length of the vector tells you how "fast" it'll move in that direction

Semiclassical

Hmm, I think I see how they avoid that issue in the notes.

"Occasionally, we will revert to the position vector notation $\mathbf{x}(t)=(x^1(t),x^2(t),x^3(t))$. Of course, what this notation really means is $x^i(t)=(x^i\circ \alpha)(t)$ where $x^i$ are the coordinate slot functions in an open set of $\mathbb{R}^3$."

Avantgarde

Einstein notation is pretty convenient. So not lazy

Semiclassical

And then when they initially present the velocity vector they write it not as $\alpha'(t)=...$ but as $\mathbf{V}(t)=\dfrac{d\mathbf{x}}{dt}=\left(\dfrac{dx^1}{dt},\dfrac{dx^2}{dt}, \dfrac{dx^3}{dt} \right)$.

which I guess is to emphasize that this really isn't the same as $\alpha'(t)$.

Daminark

Oh also @Balarka get on the NT chat, I'm explaining Emerton's Fourier duality for finite abelian groups on there

Semiclassical

Is that a fair understanding of it?

Astyx

7:10 PM

Is $(1)$ a basis of the infinite dimensional $\Bbb Q$-vector space $\Bbb R$ ?

Hi Hippa

Daminark

@Astyx
>Infinite dimensional
>(1)
Huh?

Astyx

I mean a hilbert basis

SteamyRoot

@Semiclassical That's perfectly fine.

Daminark

Oh, I guess that makes more sense

Astyx

I'm not sure my question makes sense though

Semiclassical

7:13 PM

mmkay.

Astyx

It probably doesn't

SteamyRoot

@Semiclassical I think it's common to first introduce the $\alpha'(t)$ notation, but later on you always end up at your notation anyway

Semiclassical

I'm so used to thinking of position and velocity vectors both being basically just 'vectors' that it's hard to get my brain out of that.

Though another part of my brain says "but you already think of them as behaving differently under coordinate transformations..."

Which is true.

Astyx

I mean that the linear span of 1 is dense in $\Bbb R$

Semiclassical

Here's one thing that does confuse me, though: If $\alpha(t):\mathbb{R}\to M$ (generalizing to a generic configuration space) and $\alpha'(t)\in TM$

Daminark

7:17 PM

That should work

Semiclassical

then what kind of object would the acceleration vector $\alpha''(t)$ be?

SteamyRoot

Hmmm...

You should probably be careful with your notation here

Semiclassical

That's what I'm wondering.

Sha Vuklia

@Steamy! Do you have a sec?

SteamyRoot

The tangent bundle $TM$ is the disjoint union of the tangent spaces $T_pM$

So when you say $\alpha'(t) \in TM$, do you mean the entire derivative, or its evaluation at a specific $t$?

Semiclassical

7:21 PM

Yeah. So I've really got a curve in TM, with each point on the curve lying in a different tangent space?

SteamyRoot

Yep

Semiclassical

Hmm.

SteamyRoot

@ShaVuklia Just a min

Sha Vuklia

okay give me green light when you're ready :P @Steamy

SteamyRoot

But each tangent space can be identified with $\mathbb{R}^n$

Semiclassical

7:21 PM

Yeah, in this setting.

Astyx

@SteamyRoot Still better than an inf

Semiclassical

Which I guess maybe is the point. If I'm a point moving on a sphere, then the acceleration vector really doesn't seem to make much sense.

Hrm.

kinda yes kinda no.

I'm trying to think through the case of M=S^1 as a corrective.

SteamyRoot

Well, the acceleration will still give information

Semiclassical

hmmmmm

SteamyRoot

It tells you how $\alpha'(t)$ changes if $t$ changes.

In a sense, I guess you could say the acceleration is pointless when you're only considering the vectors at one point

Semiclassical

7:26 PM

I'm tempted to say that alpha''(t) should be in T(TM) with appropriate subscripts.

which isn't a huge stretch, given that TR^3 is isomorphic to R^3 already.

SteamyRoot

That's correct.

Semiclassical

mmkay.

Sha Vuklia

(never mind btw @Steamy I got it)

SteamyRoot

Oh, okay

At this rate, I should start ignoring your questions. Seems like everytime I don't see or have time to answer them immediately, you end up finding them yourself :P

Semiclassical

Where I really want to head with this is to actually understand Hamiltonian/Lagrangian mechanics from a differential geometry POV

I'm not sure which of those is easier, though. (Hamiltonian vs. Lagrangian)

Sha Vuklia

7:32 PM

@Steamy actually I don't :P

I just realised that I still don't see it

Semiclassical

With the Hamiltonian approach, everything boils down to first-order equations.

SteamyRoot

Hmmm... I saw Hamiltionian and Lagrangian mechanics in my classical mechanics course, but I never really studied them from a geometric point of view, so I'll probably be of little help there

Semiclassical

mmkay.

SteamyRoot

I do know a crapload about Lagrangian submanifolds of complex space forms, but that's probably not the Lagrangian stuff you're after ^^

Semiclassical

@SteamyRoot I keep wondering what the connection between that terminology and classical mechanics is.

Sha Vuklia

7:34 PM

I did not say anything

Semiclassical

I think this answer here has some of it, though: mathoverflow.net/a/60519/55904 @steamy

SteamyRoot

@ShaVuklia Meaning you ended up understanding it anyway?

Sha Vuklia

lol yea I really confused something @Steamy. apparently, I have to ask you $\leq 2$ times @Steamy

SteamyRoot

Well, whatever works for you :P

Sha Vuklia

indeed

Semiclassical

7:37 PM

ugh, I should probably know what this means: "Occurences of lagrangian submanifolds are indeed manifold: they arise as semiclassical support for certain FIO's and can also be thought of as semiclassical version of states in quantum mechanics via the WKB expansion. This point of view is exemplified a lot in the nice booklet of Bates and Weinstein."

SoumyoB

hello everyone

Astyx

Ohi @SoumyoB

SoumyoB

@Astyx that reminds me, I have not completed your problem yet

SteamyRoot

@Semiclassical Yeah... I know they pop up in Hamiltionian stuff, but I've never been able to find an explanation why exactly they are called "Lagrangian" :/

Semiclassical

Yeah.

SoumyoB

7:38 PM

@Astyx could you tell me which book you were reading from?

Astyx

Not a book, it's an exercise sheet on the internet

But I've seen similar approaches in lots of other exercises

SoumyoB

could you link it to me?

SteamyRoot

In hindsight, it's kind of surprising nobody asked me that at my defence

Astyx

It's in French, but sure

denischoimet.com/Exercices_files/proba_1617.pdf

Solutions not included

SoumyoB

and just curious, are you thinking of switching from pure math to probability? In that case, welcome to the other side :P

Astyx

7:40 PM

The exercise you want to care about are the ones at the end of the sheet

Semiclassical

@SteamyRoot I'm tempted to ask the following question: "What's Lagrangian about a Lagrangian submanifold?"

Astyx

Isn't probability pure math ?

SoumyoB

hmm it's kinda borderline I'd say

Semiclassical

I tend to think of probability as math but not necessarily stats.

Astyx

@Semi What can one say about a vector field $A$ such that $rot rot A = -\lambda^2 A$ ?

Semiclassical

7:41 PM

But I'm not sure that's actually coherent.

That's the vector Helmholtz equation, isn't it?

SteamyRoot

@Semiclassical Yeah, that's what I'm wondering too :P

SoumyoB

@Astyx is it exercise 81 you were talking about?

SteamyRoot

Apparently, they may have been introduced or named in Maslov's "Perturbation Theory and Asymptotic Methods"

Semiclassical

@astyx I remember there's an identity relating rot^2 to the vector Laplacian, but I dont' remember the substance of it.

Astyx

@SoumyoB No, 57

@Semi rot rot = grad div - $\Delta$

Semiclassical

7:43 PM

right.

SteamyRoot

Ew, rot

Curl ftw

Semiclassical

so if A is divergence-free then that's just the vector Laplace equation.

SoumyoB

oh well now I'm going to have to use google translate... cracks knuckles time to learn some French

Astyx

Yeah but there's no reason for A to be divergent free, it's the magnetic potential of a magnetic field

Semiclassical

Uh, sure there is. Pick Coulomb gauge.

SoumyoB

7:44 PM

@Astyx you still have not told me if you're switching to probability

Astyx

@SoumyoB Good luck :p

I don't really know, especially now. I'm not even sure I want to do math at all any more

Semiclassical

@Astyx :(

SteamyRoot

@Astyx Or, curl squared is the commutator of the gradient and the divergence :O

Semiclassical

I hear you on that.

SteamyRoot

Never thought of it that way before o.O

SoumyoB

7:45 PM

you okay man? You sound like you had a rough time @Astyx

Semiclassical

I mean, I can't imagine not having math be a part of my life in a basic way.

Astyx

I am, but it'll get better

Semiclassical

But math/physics as part of academia...

Astyx

Obviously I won't totally abandon doing math, I'm just not sure I want to do it professionnally

Semiclassical

Anyways. Point being, if you're working in Coulomb gauge then the vector potential is divergence-free.

Astyx

7:46 PM

I've no idea what that is @Semiclassical

Adeek

hi everyone

SoumyoB

I faced a similar situation some time ago @Astyx, I know how frustrating that feels

Astyx

Hi @Adeek

Semiclassical

Well, keep in mind that potential are not the physically meaningful object. The electric and magnetic fields are what actually act on particles.

What you care about with the vector potential in magnetostatics is that rot(A)=B.

Astyx

@Semiclassical could you possibly explain it to me in essence (not with too much detail) ?

Semiclassical

7:48 PM

that's where I'm going.

Astyx

I have a specific example to rely on (an exercise one of my classmate had)

Semiclassical

suppose, though, that I were to replace A with A+grad phi

Astyx

I'll type it here, it might be easier if we have something concrete to discuss ?

Semiclassical

well, the rot of any gradient is zero

start typing while I finish my spiel.

since rot(A+grad phi) = rot A +rot(grad phi) = rot(A), both A and A+grad phi are 'vector potentials' in that they generate the same magnetic field.

Alex

This is kind of off topic: but how do I connect my TI-83+ to my computer to back up my programs when all the calculator will output is a headphone size thing?

Semiclassical

7:51 PM

So physically there can't be a difference. It's the vector potential equivalent of shifting your (scalar) electric potential by a constant everywhere in space.

But there's a lot more freedom in this case because I can pick a lot of different phi's.

to distinguish A and A+grad phi, I can compute the divergence of the latter

Astyx

So we divide the space with the plane $z=0$. On the right, we have a constant magnetic field $B_0$ orthogonal to $u_z$. One the left we have a volumic distribution $j = {-A\over \mu_0\lambda^2}$ where $A$ is the potential of some magnetic field $B$. First question is to find the magnetic field at any point. Second is to find wether there is a surfacic current on $z=0$. Last is to find $j$ everywhere.

Zophikel

0

Q: Geometrical Intution behind $\nabla(|f|^{p}) = p(p-1)|f|^{p-2}|\nabla f|^{2}$

In the text "Theory of Functions of One Complex Variable", I'm having trouble verifying the way I expressed the Laplacian Operator, and generalizing $(1.)$ to functions of more then one complex variable following Proposition in $(1.)$ Also I'm having trouble establishing the geometrical intuition...

SteamyRoot

@Alex TI sells cables to do it

Zophikel

can anyone help me

Alex

How much are they?

SteamyRoot

7:53 PM

No idea

Semiclassical

@Astyx A bit confused by your phrasing here: on the right of the plane z=0 ?

SteamyRoot

I still have my TI-83, but I haven't touched it in years.

Semiclassical

I usually think of z=0 being a horizontal plane, so it'd divide R^3 into above and below.

SteamyRoot

And, unlike some classmates, I never owned the cable

Astyx

@Semi it's not mine, I guess it means $z\gt 0$ and $z\lt 0$.

Alex

7:54 PM

I have the cable, but it's 3.5 mm to 3.5mm

Semiclassical

mmkay.

Astyx

I'm fine with above and below

Semiclassical

So in one half-plane you've got a constant magnetic field B_0 which is planar.

SoumyoB

what is a TI-83?

Alex

It's a calculator

Semiclassical

7:55 PM

Am I right in reading u_z as just the unit z vector?

SteamyRoot

Pretty sure the TI-83 doesn't use a 3.5mm jack, but a 2.5

Astyx

I think so, yes

Semiclassical

I guess I'll also read 'constant' as 'spatially uniform' rather than 'time-independent.'

That leads what's happening in the other half-plane.

Astyx

I understand it as both

Semiclassical

Yeah.

Astyx

7:57 PM

The wording is of my classmate, I don't have much more information

Semiclassical

I presume j, A are supposed to be vectors as well. (Wouldn't make much sense otherwise)

Astyx

Yes (I didn't type the arrows on top because I'm lazy, sorry)

Semiclassical

np

So we've got $\vec{J}\propto - \vec{A}$ in one half-plane.

This setup seems familiar, hmm.

ah, yes, it's the setup for the London equation: en.wikipedia.org/wiki/London_equations

And now I can explain the bit about Coulomb gauge.

First, note that we have two equations from magnetostatics: rot A = B, rot B = mu0*j

And then we have this third condition which i'll write as mu0*j = -A/lambda^2.

Agreed?

Astyx

Could you $ that please ?

Semiclassical

sure

Astyx

8:03 PM

It's not even clear we're dealing with magnetostatics here, but I agree with that

Semiclassical

Well, magnetostatics just means the magnetic fields and currents aren't changing in time.

Astyx

Yup

Semiclassical

$\text{rot }\vec{A}=\vec{B}$ is the potential condition, $\text{rot }\vec{B}=\mu_0 \vec{j}$ is Ampere's law, and $\mu_0 \vec{j}=-\lambda^2 \vec{A}$ is the last condition

since I know enough, I'll refer to this last condition as the 'London equation' for convenience

heather

hello, would anyone be willing to take a look at this question of mine?

Semiclassical

Now, here's something to observe: $\text{rot }\vec{B}=\mu_0 \vec{j}=-\lambda^2 \vec{A}$.

What happens if I take the divergence of each of these? @astyx

Astyx

8:07 PM

div rot = 0 right ?

Semiclassical

Right. so div j=0, which makes sense---it's just saying that the current doesn't cause charge to accumulate anywhere.

so it's a condition for magnetostatics to make any sense here.

But what does it tell you about $\vec{A}$?

Astyx

div A = 0

Semiclassical

Right.

So the London equation only makes sense if we require div A=0.

Astyx

You mean it implies div A=0 right ?

Semiclassical

right.

In physics terms, this means that the setup requires that we work in 'Coulomb gauge.'

Astyx

8:10 PM

I'll blindly trust you on that

Semiclassical

Mmkay. It's like a more complicated version of 'set the potential to be zero here'

Except you're working with a vector potential rather than a scalar potential, so there's more freedom in how the vector potential can be chosen.

In order for your equations to be consistent, though, you need to forget about that freedom.

That said, this is all a bit of unnecessary digression, and here's why.

If we go back to the rot B=... condition and now take rot of all three terms, what do we get? @astyx

(well, first and last terms.)

Akiva Weinberger

I'm back!

9 hours ago, by Leaky Nun

@AkivaWeinberger how does it fail if I replace $n$ with $n-1$?

@LeakyNun Take the matrix with $1$s right above the diagonal and $0$ everywhere else

It has the effect of squashing it to a hyperplane, and then rotating the space so a new axis can get squashed

Astyx

How does one write rot and div in Latex ?

Semiclassical

well, above I was cheating by doing \text{rot } and \text{div }

Akiva Weinberger

\operatorname{rot}, \text{rot}, {\rm rot}

Semiclassical

8:16 PM

The former is what I'd use if I wanted to define rot as a new command.

Astyx

$\vec{\text{grad}}\text{ div }\vec B - \Delta \vec B = \text{rot rot} \vec B = \mu_o \text{rot}\vec j = -\lambda^2 \vec B$

Akiva Weinberger

Well, if you want the space, do what Semi did or put a ~ somewhere

Semiclassical

@astyx Right. But what's div B?

Akiva Weinberger

@Semiclassical It doesn't do it for the whole expression; you'd need to retype it each time. You'd want "definenewcommand", I think

Semiclassical

@akiva sure, that's what I'm getting at.

Should've included "...at the start of a latex file"

Astyx

8:18 PM

$\text{div }\vec B = 0$

Semiclassical

Right. Maxwell's equations yet again.

Astyx

Oh yeah I'm silly

Semiclassical

So in particular you've got $-\Delta \vec{B}=-\lambda^2 \vec{B}$.

Astyx

So $\Delta \vec B = \lambda ^2 \vec B$

doesn't this sign mean it'll diverge very badly at infinity ?

Semiclassical

Right. And the vector Laplacian is just "take the laplacian term-wise in Cartesian coordinates"

Well, let's find out.

Astyx

8:20 PM

You'd expect $\Delta\vec B = -\lambda^2 \vec B$ wouldn't you ?

Semiclassical

It'll turn out to be fine, actually.

Akiva Weinberger

@Daminark Happy birthday!

Semiclassical

To sum up what we've found: Each component of $\vec{B}$ satisfies a Helmholtz equation $\Delta B_i =\lambda^2 B_i$ for $z>0$.

Agreed?

Astyx

Yup

(Laplace is =0 right ? or is that poisson ?)

Semiclassical

(=0, yeah. Poisson would be 'inhomogeneous Laplace')

Now, what do we know about B for z<0?

Astyx

8:23 PM

That's where B is "constant" right ?

Semiclassical

Right.

Along with another condition.

Astyx

it's parallel to the plane $z=0$

Semiclassical

Right. Which in particular means that $B_z=0$ for $z<0$.

Astyx

Agreed

Semiclassical

And, if we assume that the field is continuous, it also means $B_z=0$ on $z=0$.

Astyx

8:25 PM

(That's even equivalent)

If the field is continuous that means there is no surfacic current ?

Semiclassical

Hmmmmm

I'm forgetting.

Astyx

I'll check

Semiclassical

I suspect the point is that, whatever the surface current is, it will run parallel to the surface and not perpendicular.

Astyx

Oh no, that's only true for the tangent components

The normal component is always continuous

Semiclassical

Okay.

So we've got $B_z=0$ for $z\leq 0$.

Astyx

8:28 PM

So $B_z$ is continuous at $z=0$

Agreed

Semiclassical

Now, here's where I'm losing track of the details a bit.

If I take the field being 'constant' means uniform as well as time-independent, then that tells me something rather important about B above z=0.

Namely: Can B depend on x or y?

Astyx

Being constant below 0 means something above 0 ?

Does it have something to do with the divergence

Semiclassical

No, it has something to do with symmetry.

Suppose that the field above z=0 depended on x in some way.

Astyx

Oh I misunderstood your claim

Semiclassical

Then I could shift the entire system by some distance in the x-direction and I'd find that the field above z=0 would shift along with it.

But the field below wouldn't have changed, since it's constant and doesn't depend on x.

Similarly with y.

Astyx

8:32 PM

Isn't that possible ?

Semiclassical

I'd argue that whatever B is for z>0, it can only depend on z.

(There's definitely some more formal way to establish this, but I don't want to.)

Astyx

That doesn't seem obvious to me

Semiclassical

Lemme try once more. Suppose I find that the field above has a solution $\vec{B}(x,y,z).$

So I'd have some solution which for $z>0$ is a function of $x$ but not for $z<0$.

Now suppose I consider the field $\vec{B}(x+a,y,z)$.

Since the original solution didn't depend on $x$ for $z<0$, I conclude that this new function of x,y,z will agree with the original one for z<0.

I can similarly show that if the original function satisfies $\Delta \vec{B}=\lambda^2 \vec{B}$, then so does this one.

So $\vec{B}(x+a,y,z)$ and $\vec{B}(x,y,z)$ have the same boundary conditions and satisfy the same PDE.

And if I'm remembering right, this this isn't possible unless both functions are actually the same.

heather

if a bounty isn't really bringing attention to a post, is there anything else that can be done?

Semiclassical

In which case $\vec{B}(x+a,y,z)=\vec{B}(x,y,z)$ regardless of $a$. But that means that $\vec{B}$ is independent of $x$.

@Astyx Does that make more sense?

Zophikel

8:39 PM

Does a function being harmonic imply it being holomorphic

Astyx

@Semiclassical This is what I don't find obvious, but I can accept it

Semiclassical

Well, if $B(x,y,z)$ doesn't depend on $x$ for $z<0$, then that means $B(x+a,y,z)=B(x,y,z)$ for all $a$ when $z<0$. These are equivalent statements.

now i'm being too lazy to include \vec's

Astyx

Yes, my message was linked to another message of yours

This one :

7 mins ago, by Semiclassical

Since the original solution didn't depend on $x$ for $z<0$, I conclude that this new function of x,y,z will agree with the original one for z<0.

Semiclassical

@Astyx But that's what I just said.

Astyx

Oh, I can't read sorry, you mean for $z\lt 0$ only

Semiclassical

8:44 PM

Right.

Astyx

Yeah I agree

Semiclassical

mmkay.

It's like sliding an infinite line of charge up/down.

Astyx

Or rather, I trust you :p

Semiclassical

It's still the same infinite line of charge before/after

So shifting up/down can't change the field.

That immediately tells you in that scenario that the electric field can't depend on the z-coordinate.

Astyx

On another note, I'm not too sure that $\vec A$ is the potential of the actual $\vec B$. My classmate said "$\vec A$ is the potential of some $\vec B$". Don't know if that has any value to it

Semiclassical

8:46 PM

nah.

If it's just the potential of 'some field' then there's nothing to say.

Whereas under the assumptions we're making we'll get a rather nice conclusion at the end of the day.

Astyx

I like nice conclusions :)

Semiclassical

yep.

Astyx

So you're telling me the field for $z\ge 0$ is independant of $z$ ?

Semiclassical

No. Independent of $x,y$.

If I make a shift in $z$, then I'll move my plane $z=0$. The argument won't work.

It only works if I consider shifts like $x\to x+a$ or $y\to y+a$.

One thing to keep in mind: If $B$ is constant for $z<0$, then that means that $B$ points in some particular (horizontal) direction for all $z<0$.

Astyx

Oh you're making an analogy with electrostatics then ?

Semiclassical

8:50 PM

A bit of one, yes.

Does my last comment make sense?

s.harp

<

Astyx

Yes

Semiclassical

@heather My main reaction to the question upon seeing it is: "This is a research question, not a math question." By that I mean that, to make any useful comment, I'd have to read/digest that paper enough to understand what the actual mathematical question is.

And when that's the case, my interest in the problem goes way down. I spend enough time doing my own research questions.

If you want better feedback, then you'll probably need to explain the problem enough that the reader need not consult the paper to understand the problem. (And if doing so would be prohibitively difficult, then probably this is the wrong place to ask the question.)

@Astyx Mmkay.

If that's the case, then the Helmholtz equations become a lot simpler. Namely, the only derivative that'll matter is the z-derivative.

So it'd simplify down to $\frac{\partial^2 }{\partial z^2}\vec{B} = \lambda^2 \vec{B}$

And what we're left with is just three independent ODE's.

And that's got a very simple general solution.

Astyx

Right

Semiclassical

So, what's the solution?

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