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

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

Truly so

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

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

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.

Look up "derivations"

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

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

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

@BalarkaSen So it chews functions then.

(einstein notation because I'm lazy)

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

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

Yes

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

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

@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 $\Bbb R^n$.

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

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 $\mathbb{R}^3$."

Einstein notation is pretty convenient. So not lazy

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 $\alpha'(t)$.

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

Is that a fair understanding of it?

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

Hi Hippa

I mean a hilbert basis

@Semiclassical That's perfectly fine.

Oh, I guess that makes more sense

I'm not sure my question makes sense though

mmkay.

It probably doesn't

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

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.

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

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

That should work

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

Hmmm...

You should probably be careful with your notation here

That's what I'm wondering.

@Steamy! Do you have a sec?

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 $t$?

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

Yep

Hmm.

@ShaVuklia Just a min

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

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

Yeah, in this setting.

@SteamyRoot Still better than an inf

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.

Well, the acceleration will still give information

hmmmmm

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

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.

That's correct.

mmkay.

(never mind btw @Steamy I got it)

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

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)

@Steamy actually I don't :P

I just realised that I still don't see it

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

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

mmkay.

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

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

I did not say anything

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

@ShaVuklia Meaning you ended up understanding it anyway?

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

Well, whatever works for you :P

indeed

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."

hello everyone

Ohi @SoumyoB

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

@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" :/

Yeah.

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

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

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

could you link it to me?

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

It's in French, but sure

denischoimet.com/Exercices_files/proba_1617.pdf

Solutions not included

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

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

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

Isn't probability pure math ?

hmm it's kinda borderline I'd say

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

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

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

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

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

@Astyx is it exercise 81 you were talking about?

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

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

@SoumyoB No, 57

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

right.

Ew, rot

Curl ftw

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

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

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

Uh, sure there is. Pick Coulomb gauge.

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

@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

@Astyx :(

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

I hear you on that.

Never thought of it that way before o.O

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

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

I am, but it'll get better

But math/physics as part of academia...

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

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

I've no idea what that is @Semiclassical

hi everyone

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

Hi @Adeek

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.

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

that's where I'm going.

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

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

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

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.

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?

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

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.

@Alex TI sells cables to do it

can anyone help me

No idea

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

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

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

And, unlike some classmates, I never owned the cable

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

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

mmkay.

I'm fine with above and below

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

what is a TI-83?

It's a calculator

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

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

I think so, yes

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

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

I understand it as both

Yeah.

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

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

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

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?

Could you $ that please ?

sure

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

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

Yup

$\text{rot }\vec{A}=\vec{B}$ 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

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

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

div rot = 0 right ?

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 $\vec{A}$?

div A = 0

Right.

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

You mean it implies div A=0 right ?

right.

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

I'll blindly trust you on that

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.)

I'm back!

@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

How does one write rot and div in Latex ?

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

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

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

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

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

@astyx Right. But what's div B?

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

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

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

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

Right. Maxwell's equations yet again.

Oh yeah I'm silly

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

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

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

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

Well, let's find out.

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

It'll turn out to be fine, actually.

@Daminark Happy birthday!

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?

Yup

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

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

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

That's where B is "constant" right ?

Right.

Along with another condition.

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

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

Agreed

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

(That's even equivalent)

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

Hmmmmm

I'm forgetting.

I'll check

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

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

The normal component is always continuous

Okay.

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

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

Agreed

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?

Being constant below 0 means something above 0 ?

Does it have something to do with the divergence

No, it has something to do with symmetry.

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

Oh I misunderstood your claim

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.

Isn't that possible ?

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.)

That doesn't seem obvious to me

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.

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

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?

Does a function being harmonic imply it being holomorphic

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

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

Yes, my message was linked to another message of yours

This one :

@Astyx But that's what I just said.

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

Right.

Yeah I agree

mmkay.

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

Or rather, I trust you :p

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.

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

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.

I like nice conclusions :)

yep.

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

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 $z<0$.

Oh you're making an analogy with electrostatics then ?

A bit of one, yes.

Does my last comment make sense?

<

Yes

@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.

Right

So, what's the solution?