$\vec B_1e^{\lambda z} + \vec B_2 e^{-\lambda z}$ where the $\vec B_i$ are integration constant we could eventually simplify by expecting continuity at z=0

right.

Now, you were earlier worried about this solution diverging.

But keep in mind that this solution is only supposed to be for z>0.

So $\vec B_1 = 0$

Right.

Which greatly simplifies the expression of $\vec B_2$ too

And if you demand continuity of the normal component at $z=0$, you get $B_{2z}=0$.

Yeah, sniped

You can't, however, demand continuity of the tangential components. Can't rule out surface currents.

...I think.

hmm, that doesn't sound right.

Anyways, the main lesson is that $\vec{B}=\vec{B}_2e^{-\lambda z}$ for $z>0$.

So whatever the magnetic field is, it'll be exponentially decaying for z>0.

And now I can connect back to the London equations: Those are intended to describe how magnetic fields work inside of a superconductor.

The last question concerning the exercise was about finding $\vec j$ everywhere

I guess we just apply $\vec {\text{rot }}\vec B = \mu_o \vec j$ ?

And what we've just obtained is the Meissner effect: If we put a magnetic field above the surface of a superconductor, then that field will only penetrate a distance $1/\lambda$ into the superconductor.

So a superconductor resists the introduction of a magnetic field.

@Astyx Yeah.

For volume current, at any rate.

For surface currents it's something else.

What's the difference ?

It's something to do with the discontinuity in the tangential components of B, if memory serves.

But something seems wrong there.

One thing you can expect, though: If $B$ decays exponentially with z, then so will $\vec{j}$.

So the current will be confined to being only near the surface of the 'superconductor.'

Here's a relevant quotation from the "London equations" page I linked earlier, btw

"For an example, consider a superconductor within free space where the magnetic field outside the superconductor is a constant value pointed parallel to the superconducting boundary plane in the $z$ direction. If $x$ leads perpendicular to the boundary then the solution inside the superconductor may be shown to be $B_z(x)=e^{-x/\lambda} B_0$".

(they've got a different orientation of things than we've been doing and they also have $1/\lambda$ where you have $\lambda$.)

That's basically the setup you've been working through.

Indeed, it's very interesting

Thanks a lot for your time

np

We can't get more information about $\vec B$ from the information we have though right ?

Doesn't seem like it.

I feel like there's a boundary condition being missed.

My feeling is that the field should actually be continuous in all components across z=0.

There are boundary conditions regarding surfacic currents

So that $\vec{B}=\vec{B}_0$ for $z<0$ and $\vec{B}=\vec{B}_0 e^{-\lambda z}$ for $z>0$.

If memory serves $\Delta_{12} \vec B_T = \mu_0 \vec j_S \land \vec n_{12}$

I think that, in the absence of any surface currents, this is what the field will be everywhere.

That looks right.

So yeah, if $\vec j_s = \vec 0$, we indeed have continuity

I think that's what's going on, yeah.

I mean, if there's a surface current, then that'll produce a magnetic field independent of whatever is going on below.

But I'd need to see a full and clear statement of the problem to be sure.

Yeah the exercise seems very weirdly put

I'm pretty sure what I wrote is the desired solution.

So am I, and I don't really care about the actual answer to that specific exercise anyway

Your insight is much more valuable

With the lesson being: The London equation has the implication that a superconductor will exponentially expel any internal magnetic fields and volume currents, and only 'penetrate' some depth into the superconductor.

A bit like Skin effect for metals and electromagnetic waves ?

Right.

Another small question that has nothing to do with the rest : do you know about laser power ?

not enough to say anything useful.

More specifically : Is 10W electromagnetic output a lot for a laser ?

The internet seems to be quite contradictory on this topic

Not a clue.

"Meissner effect: When a magnetic field is applied to a superconductor, current flows in the outer skin of the material leading to an induced magnetic field that exactly opposes the applied field."

Right, thanks

I guess I have to go and sleep now

I mean, googling "10-watt laser" brings up a bunch of laser pointers.

Thanks again for your patience and time

But it may be that a 10-watt laser pointer is not quite what is intended.

Yeah but if you read wikipedia it seems lasers above 100mW put things on fire and cause blindness and burn people alive

So I'm not too sure about what they mean by "10W"

Yeah.

I mean, there are regulations that prevent selling lasers above 5mW if I read correctly so I dunno

Anyway, good day to you, I'm off to sleep

night

Hm. Suppose $X$ is a CW complex with nontrivial homotopy groups $\pi_1$ and $\pi_2$ and zero all above.

$\tilde{X}$ is then simply connected, and the rest of the homotopy groups are the same as $X$, so only nontrivial homotopy $\pi_2$, implying $\tilde{X} = K(\pi_2, 2)$.

On the other hand, $K(\pi_1, 1)$ can be constructed off from $X$ by adding enough $3$-cells so that $\pi_2$ gets baleeted ($\pi_1$ remains, because 3 cells do not affect 1-homotopy type) and add higher cells to deleted all the extra homotopy groups higher up.

So we get a series of maps $K(\pi_2, 2) \to X \to K(\pi_1, 1)$, the first being covering projection and the second being subcomplex inclusion. I would think this is a "fibration" in some sense.

So that $X$ is a "twisted product" of $K(\pi_1, 1)$ and $K(\pi_2, 2)$, upto homotopy equivalence.

Maybe the word I am looking for is "homotopy fiber"?

Hi @EricSilva

@BalarkaSen Can we start with something higher than $\pi_1$? Fundamental groups are always scary.

Hm, fair suggestion. Let's see.

hey, is there a lemma or theorem states that $f(x)>0$ for $a<x<b \Rightarrow \displaystyle \int_{a}^{b} f(x) \mathbf{d}x>0$

@AbdullahUYU yes if $f$ is continuous

i forgot to say that, can you give a webpage for it?

Hi @BalarkaSen

@AbdullahUYU I don't know if it has a special name so I don't really have a page in mind, but I can sketch a proof if you want

pleasure, i also look around

The nonstrict case comes from the fact that $f(x)\ge g(x)\forall x\in(a,b)$ implies $\int_a^b f(x)\ge\int_a^b$ (take $g(x)=0$)

@BalarkaSen I got my conditions today and apparently there are none. The basis for admission was my high school diploma (64% avg) lmfao.

I guess I just got lucky.

Let $f((a+b)/2)>\epsilon>0$. Since $f$ is continuous, we have $\exists\alpha,\forall x\in ](a+b)/2-\alpha,(a+b)/2+\alpha[, |f(x)-f((a+b)/2)|<\epsilon$. Therefore $\int_a^b f\ge\int_{(a+b)/2-\alpha}^{(a+b)/2+\alpha} f\ge 2\alpha[f(a+b)/2-\epsilon]>0$ @AbdullahUYU

@Dodsy Which conditions?

(I should mention here that $]a,b[$ is French notation for $(a,b)$)

Well usually there are conditions

(in case you're not familiar with it)

since I'm taking courses it should really be to maintain an 82.5 percent

and they basically said no conditions since the basis for admission was my hs diploma

which I did 4 years ago

Huh!

yeah so I didn't even need to do these courses.

@Hippalectryon You should write \in { ] stuff [ } (spaces optional), so that the spacing works out better

They basically just give me the prerequisites at this point.

Or \in {} ], I think

I don't think I'm going to finish chemistry.

Yeah **** chemistry

Otherwise, LaTeX thinks that the $]$ is glued to the $\in$

lmfao seriously tho m8

@AkivaWeinberger Thanks for the tip :-) unfortunately I can't edit anymore

Oh :(

But the point is, near any point where a continuous function is (strictly) positive, you can find an interval around it and a positive constant on which the function is greater-than-or-equal-to that constant.

So, on that interval, it's $\ge$ the length of the interval times that constant, which is ${}>0$.

@Dodsy You're in UWO now and that's priority. don't let the trifles bother you, patron

(And for the rest of the interval, use the nonstrict case from above)

true :}

mention is appropriate, i don't know the notation so indeed

Instead of (a,b), (a,b], [a,b), and [a,b], the French notation uses ]a,b[, ]a,b], [a,b[, and [a,b]

It doesn't conflict with the notation of the ordered pair (also $(a,b)$), so that's good

i haven't take math101 so i don't know the $\epsilon$ definition of continuity, but i'll take a look at it. thank you so much @Hippalectryon

@AbdullahUYU Don't worry, all what I wrote says is basically what Akiva said. Since the function is continuous, if at some point it's higher than 0, then its area under the curve will be positive locally around that point.

yes, it comes me so intuitive.

@Hippalectryon That leads to a proof of the intermediate value theorem, by the way: If it starts at $1$ and ends at $0$, take the infimum of all the times it's negative. The value at the infimum can't be positive, because of that property, and it can't be negative, because of that property applied to the negative of the function

@Hippalectryon Is there a discontinuous counterexample? I can't think of any

Of $f(x)>0$ on $a<x<b$ but $\int_a^bf(x)\operatorname d\!x=0$

I'm pretty sure there are, if we're using Lebesgue integration. But I've got no example :( let me think

Well, I'm wrong

As long as $[a,b]$ has a positive measure the integral is positive

Oh, wow

Cool

$$\int_a^b1\operatorname d\!x\le\sum\int_{\{x:f(x)>1/n\}}1\operatorname d\!x\le\\\sum\int_{\{x:f(x)>1/n\}}nf(x)\operatorname d\!x=\sum0=0$$

^That's essentially the contradiction, I guess

The essential idea is that $\int_{\bigcup A_n}\le\sum\int_{A_n}$, even for infinite sums

Or, rather, $\displaystyle\lim_{n\to\infty}\int_{\bigcup_{k=0}^NA_k} =\int_{\bigcup_{k=0}^\infty A_k}$

So, actually, it's more like, if we had such an $f$, we would have:$$b-a=\int_{\{x:f(x)>0\}}1\operatorname d\!x=\lim_{n\to\infty}\int _{\{x:f(x)>1/n\}}1\operatorname d\!x\\\le\lim_{n\to\infty}\int _{\{x:f(x)>1/n\}}n f(x)\operatorname d\!x\le\lim_{n\to\infty}\int_a^bnf(x)\operatorname d\!x\\=\lim_{n\to\infty}0=0$$

@Hippalectryon

Contradiction

(That was just me rewriting the answer given in your link to make sure I understand it)

@AkivaWeinberger yep :-) that's a nice way to show it

I think this falls into the category of the "Approximate $0$ by $1/n$ and then use the fact that there's only countably many of those" trick

which I see every so often in these kinds of things

Hi, folks. Anyone know if there is an easy to learn/use online facility for quickly creating weighted graphs?

@Jeff Do you know python ? Otherwise, graphonline.ru/en

i don't know python. but i used to program. is it easy to learn?

It's rather easy, but if you don't already know it it's gonna be a bit overkill just to do graphs

this seems nice so far. but i can't drag the nodes around separately to make the graph 'regular'. i see a video tutorial. i'll try that when i get home.

@Hippalectryon Nice! TY

:-)

Okay so I'm having a bit of trouble with this

Let's say $G$ is a finite abelian group

Define $\hat{G} = Hom(G,S^1)$

I'm trying to show that $\hat{G}$ forms a basis for $\mathbb{C}\langle G\rangle = \{f:G\to\mathbb{C}\}$

I get spanning, but now I'm trying to do linear independence and I can't for the life of me conclude yet

@Daminark How does spanning work?

So the idea is that you define $\delta_g(h) = 1$ if $h=g$, and 0 otherwise

Then you can express any function from $G$ to $\mathbb{C}$ as $f(h) = \sum_{i=1}^n f(g)\delta_g(h)$

Where $|G| = n$

Now, you can express $\delta_1(h) = \frac{1}{|\hat{G}|}\sum_{\chi \in \hat{G}} \chi(h)$

For $h = 1$ this is clear

For $h\ne 1$, the idea is that you know there exists some $\chi_*$ which doesn't annihilate $h$

Since our group is assumed to be abelian, consider a list of generators including $h$ and mod out by the subgroup generated by all the other stuff

So that'll get you a homomorphism $G\to \langle h \rangle$