The h Bar

8:00 PM

Think about it like this: Since the oscillation of $\sin(x^2)$ grows faster for large $x$, the sine traces out almost no area anymore, so the amplitude of the oscillation of the integral gets smaller and smaller. The plot that the top right of the Wiki page nicely illustrates that

0celou7

@Slereah I'm fairly sure $\delta*\delta=\delta$.

$(\delta*\delta)(\varphi)=\delta*(\delta*\varphi)=\delta*\varphi=\varphi$.

Or maybe it's the identity

Who knows

ACuriousMind

@Slereah Clearly, $\int \delta(x-y) \delta(x)\mathrm{d}x = \delta(y)$, what's the issue? :P

0celou7

christ

ACuriousMind

Don't make me Taylor expand the $\delta$!

0celou7

The definition here is actually quite strange. You don't get a distribution, you get a map $\mathscr D\to\mathscr C$

I wish he had some examples

@ACuriousMind You're not entirely wrong though, the convolution with $\delta$ is just the identity.

DanielSank

8:06 PM

@ACuriousMind o_O

$$\lim_{a \rightarrow \infty} \int_0^a \sin(x^2) \, dx = \sqrt{\pi/8} \, ???$$

That makes no sense at all.

Oh maybe it does.

Wow.

0celou7

I say it's wrong

fight me

DanielSank

Yes, indeed. I forgot that the "lobes" get smaller as you go.

ACuriousMind

I linked the fricking page on Fresnel integrals, why is everyone so disbelieving?!

DanielSank

@0celou7 Star Wars trivia?

@ACuriousMind Because we're dumb.

0celou7

@ACuriousMind what?

@DanielSank computing curvature tensors

@Slereah will select a random spacetime from Stephani et al

DanielSank

8:08 PM

@0celou7 no

0celou7

fine then

the people next door are either having vigorous sex or playing a video game too loudly

one of them keeps crying out

ACuriousMind

@0celou7 ...why not both?

0celou7

Hmm

test $\darrow$

test $\sarrow$

:(

test $\searrow$

??

DanielSank

@0celou7 How is that confusing?

Those things don't sound the same.

0celou7

I wouldn't know, I don't play video games

OBE

8:14 PM

lol

0celou7

$\downarrow$

for some reason $\epsilon\downarrow 0$ looks like $\epsilon$ is choke slamming $0$

OBE

8:59 PM

@0celou7

0celou7

what the hell is $\mathrm{l.i.m.}_{N\to\infty}$

OBE

how does lee smooth manifolds compare to the other lee manifolds and differential geometry?

i think you mentioned once

0celou7

SM is for beginners

the other one isn't

OBE

oh okay

just to make sure i'm referring to this

amazon.com/Manifolds-Differential-Geometry-Graduate-Mathematics/…

it looked to be the same level as SM to me but you're saying it's not for beginners

thx

0celou7

...does he mean limit in measure?

OBE

9:11 PM

@0celou7 do you get this

last 2

0celou7

second to last yes

last one is just silly

loltospoon

How should I think about the vector potential in magnetostatics?

0celou7

looks like nets but stupider

loltospoon

The B-field is already a vector field

(magnitude and direction)

so then what do we need the vector potential for?

Does it physically represent anything?

OBE

it's used to define the field strength tensor.

ACuriousMind

9:15 PM

@OBE what?

OBE

what am I wrong? there's a definition for it in terms of the vector potential.

0celou7

@yuggib Wtf is this l.i.m. thing in Yosida on page 149, 154, and later? It's not "limit in measure," at least I hope not. Control F gives no result for it other than when it's used and he doesn't say what it is

@OBE Can you explain what the field strength tensor is?

ACuriousMind

@loltospoon Actually, the magnetic field is not a vector, it's a pseudovector, or more precisely a 2-form, cf. my answer at physics.stackexchange.com/q/313091/50583

And what you "need" it for...well, what do you need any potential, no matter whether scalar or vector, for?

OBE

@0celou7 depends. my explanation will probably not satisfy you lol

0celou7

@ACuriousMind any clue what l.i.m. is?

ACuriousMind

9:18 PM

@OBE Sure. But the more "physical" reasoning is that you first put $F$ together, then observe $\mathrm{d}F=0$, then define $A$. You can define the field strength tensor without needing the potential.

OBE

@ACuriousMind okay sorry.

:'(

ACuriousMind

@0celou7 nope

0celou7

"$\lim_{h\to\infty}||\hat f_h-f||=0$, that is, $\hat f(x)=\mathrm{l.i.m.}_{h\to\infty}\hat f_h(x)$ a.e."

OBE

@ACuriousMind can't you say another reason is that it combines electric and magnetic potentials?

which is useful

0celou7

very strange...

the use of "that is" implies it's an iff condition

but convergence in measure is weaker than $L^2$ convergence

ACuriousMind

9:21 PM

@OBE Since loltospoon said "vector potential in magnetostatics", they're talking about the magnetic potential, not the combined 4-potential.

OBE

oh sorry I misread that.

ahhhhh there's something wrong with me

i'm so bad at life

ACuriousMind

@0celou7 I would not read that as "iff", it might just be "if", i.e. the part after "that is" is implied by the preceding part, but not the other way around

OBE

everywhere

loltospoon

@ACuriousMind why DO we need any potential?

(for reference, I'm an undergrad physics student)

0celou7

@ACuriousMind The context here is Fourier transforms of $L^2$ functions, i.e. taking the integrals over balls $|x|\le h$, then $h\to\infty$

that limit should be in $L^2$, not in measure

OBE

9:24 PM

@loltospoon well in e&m introducing potentials makes it easier to solve maxwell's equations

potentials just make physics simpler to work with

@ACuriousMind am i wrong please don't destroy me

loltospoon

@OBE Yea the book I'm going through kinda says the same thing. Potentials help find the E and B fields

ACuriousMind

@loltospoon Well, generally a "potential"does two general things: It allows you to write down a Lagrangian formulation, and to encode the information of the physical fields in a more "efficient" manner.

0celou7

@ACuriousMind What's wrong with you, why are you destroying the boy?

loltospoon

(Introduction to Electrodynamics by Griffiths)

ACuriousMind

Consider a scalar potential - a vector field has three independent components, but the scalar only one. Expressing a vector as the gradient foa scalar therefore is more "economic"

loltospoon

9:25 PM

@ACuriousMind wait whaaaa we haven't talked about the Lagrangian yet in my E&M course!

OBE

@ACuriousMind yay so I'm not wrong that's what I was saying too

Bernardo Meurer

Good evening folks

OBE

hi bernardo

0celou7

@ACuriousMind economical

Bernardo Meurer

Hope we are all free of proprietary malware today

OBE

9:26 PM

how's 'loo? admission*

Bernardo Meurer

@OBE All done & submitted, now just have to sit and wait :)

ACuriousMind

For a vector potential, it first seems that you have gained nothing, but then you remember that the curl of a gradient is zero, so actually your vector potential has only two degrees of freedom since you can add arbitrary gradients to it without changing the magnetic field it implies

OBE

@BernardoMeurer good luck!

ACuriousMind

@loltospoon That doesn't mean it's not useful for that! ;)

Bernardo Meurer

@OBE Thanks man, I'll need it :)

@0celou7 Tell Reb to pray for me

ACuriousMind

9:27 PM

If you want a least action formulation of EM, you gotta use the potential, and view it and not the fields as the dynamical variables.

loltospoon

Hmm I see

0celou7

@BernardoMeurer she's not religious?

Bernardo Meurer

@0celou7 She should become and then pray, easy

OBE

@ACuriousMind but can't that work both ways since the fields can be defined by the potential and vice versa.

0celou7

@OBE yes but then you need constraints

that's what two of the Maxwell equations are

OBE

9:30 PM

yeah but potentials aren't unique

so why don't you use the fields instead

ACuriousMind

@OBE What? If you take the usual EM Lagrangian and vary it with respect to $A$, you get Maxwell's equations (two of them, the other two are the Bianchi identity $\mathrm{d}F=0$ for $F=\mathrm{d}A$ and not equations of motion). So if you view the fields as the dynamical variables, you can't get that Bianchi identity, and need to add it in by hand

0celou7

that's what I just said

does no one listen to me?

Bernardo Meurer

@0celou7 I do

ACuriousMind

@0celou7 You did, but terser, and I had started typing that anyway :P

Bernardo Meurer

I just filter out most of it

0celou7

9:31 PM

you're just like my mother

Bernardo Meurer

I don't even know her

OBE

@ACuriousMind okay thanks.

0celou7

god dammit this is why we need a notation index on books

what the hell is l.i.m.

OBE

@ACuriousMind just a tip you shouldn't start almost every reply before you explain something to someone with "what?" because it makes them feel bad.

3

or maybe it's just me

0celou7

what?

OBE

9:34 PM

yeah like that.

what?

what?

blocked

¯_(ツ)_/¯

@ACuriousMind The Fourier coefficients of $f\in L^2(0,2\pi)$ are $\hat\varphi(k)$, then he writes $f(x)=\text{l.i.m.}_{s\uparrow\infty}\sum_{k=-s}^s \hat\varphi(k)e^{ikx}$. Any idea?

...limit in metric?

Limit in metric!

@ACuriousMind Sound legit?

ACuriousMind

9:45 PM

Honestly, no idea

0celou7

where is the yugoslavian when you need him

ok Russian servers, can you deliver me the original book?

@ACuriousMind well I'll be goed to hell

in Reed & Simon they say l.i.m. means $L^2$-limit

and they say that $$\text{l.i.m.}\int$$ is a common shortcut for an $L^2$-limit over increasing balls

Would have been nice for Mr. Yosida to say something about that!

skill patrol

@OBE it's not just you pal, that would be true about anybody in real life too :-)

0celou7

if you're talking to someone and they feel bad, maybe they should

skill patrol

10:01 PM

-_-

uryga

Hi! Asked a related thing a few days ago. I'm still looking for intuition on Clifford/Geometric algebra stuff - bivectors and trivectors.
https://en.wikipedia.org/wiki/Exterior_algebra#/media/File:N_vector_positive_png_version.png

an explanation of bivectors i heard here: the bivector u ^ v can be thought of as "rotating from u to v". If that's the case, how should i think about a trivector, which the wikipedia picture above shows as a spinny ball thing?
Intuitive ways of thinking about the wedge product appreciated too.

skill patrol

Ask on main.

0celou7

I'm done with analysis

uryga

@skillpatrol okay!

0celou7

Maybe I should be a physicist

or a physician

Danu

10:07 PM

hey @0celou7 you spend all your time on Riemannian geometry, right?

0celou7

no?

Been doing nothing but analysis lately

Danu

Why is a ''ball'' $\{x\in M\mid d(x,p)\leq c\}$ for some fixed $p$ compact, in a complete Riemannian manifold?

0celou7

Hopf-Rinow

Completeness is equivalent to the Heine-Borel property, i.e. closed bounded sets are compact.

Danu

I thought Hopf-Rinow is that geodesic completeness $\Leftrightarrow$ metric completeness

0celou7

Hopf Rinow has about 4 $\Leftrightarrow$s :)

Danu

10:10 PM

Not in Milnor's book ^^

So can you explain the implication I'm asking for? You can start from either metric or geodesic completeness (I suspect you should start from metric completeness) if you like

0celou7

Of course

Danu

Thanks! I'm listening

0celou7

You're asking for the hard direction :)

Danu

Really? That surprises me

Milnor just says "since M is complete, the ball is compact"

no explanation (and it's not included in his version of H-R)

0celou7

Let $p\in M$ be fixed. By assumption $T_pM$ is defined everywhere.

If we have a Cauchy sequence $(x_n)$, we can find a minimizing, unit-speed geodesic $\gamma_p^{x_n}$ from $p$ to $x_n$.

Danu

10:15 PM

What do you mean by "$T_pM$ is defined everywhere"?

0celou7

Now we see that $\ell_n=L(\gamma_p^{x_n})$ is Cauchy in $\Bbb R$, hence converges to some $\ell\in\Bbb R$.

@Danu oops, I meant $\exp_p$ is defined everywhere on $T_pM$

Danu

Oh, sure

@0celou7 Why is that?

0celou7

Let me get out Milnor to see what you're looking at

@Danu Theorem 10.9

Take the minimal geodesic connecting $p$ and $x_n$, then reparametrize to get a unit speed one

loltospoon

$V(\vec{r},t)=0$, $\vec{A}(\vec{r},t)=-\frac{qt}{4\pi \epsilon_0 r^2}\hat r$. I found the corresponding fields to be $\vec{E}=\frac{q}{4\pi \epsilon _0 r^2}\hat r$ and $B=0$. Now, to find the charge and current distributions, I used $\nabla \vec{E} = \frac{\rho}{\epsilon _0}$ and $\nabla \times B = \mu _0 \vec{J}$. I found that there is 0 charge and 0 current. But the author finds the charge distribution to be a stationary point charge $q$ at the origin. Why is this??

Danu

@0celou7 Alright

ACuriousMind

10:19 PM

@loltospoon Because you did not treat the singularity of $E(r)$ at $r=0$ properly.

loltospoon

Or, asked in a better way, what reasoning did he follow to stray away from the answer $\rho = 0$?

Danu

Dirac delta stuff

ACuriousMind

@loltospoon See e.g. math.stackexchange.com/q/1339360/143136

loltospoon

@ACuriousMind so then should I have noticed this when I derived $\vec{E}$?

Danu

It's in Griffiths' book

0celou7

10:21 PM

@Danu Oh, no, here's what I want to do. Take $K\subset M$ bounded and compact. Consider the function $f(x)=d(x,p)$, for $x\in K$. By compactness, this has a maximum on $K$, call it $r$, and the maximizer $y$. Now we can connect $y$ and $p$ by a geodesic $\gamma$.

Danu

No, wait, I want to get compactness as a result

not the other way around

loltospoon

@Danu yep, I remember the earlier chapter now. So basically, one notices the need to deal with a singularity from looking at $\vec{E}$? Or could I have looked at $\vec{A}$ and said "oh I'm going to need dirac delta here because there is an $r^2$ in the denominator"?

ACuriousMind

@loltospoon Well, you should have noticed that the function is singular at $r=0$ and therefore you cannot take the ordinary derivative in the usual sense there. I would not expect anyone to figure out it's a $\delta$-function purely mathematically, but as a physicist it might have given you a clue that this field is precisely the field of a point charge the Coulomb law gives, so the answer $\rho=0$ is physically non-sensical

0celou7

@Danu $K$ is closed and bounded. For $x\in K$, let $f(x)=\ell(\gamma_x^p)$, the length of the minimizing geodesic from $p$ to $x$.

We know that this is equal to $d(x,p)$ by Hopf-Rinow

Now I'm on the right track

ACuriousMind

For how you actually figure out that the derivative is a Dirac delta mathematically, see the math.SE post I linked

0celou7

10:23 PM

We also know from properties of the exponential map that this length is equal to the initial speed of $\gamma_x^p$, agreed?

loltospoon

@ACuriousMind which function? - " the function is singular at... "

$\vec{E}$ or $\vec{A}$?

ACuriousMind

@loltospoon The electric field $E(r)$

loltospoon

oh ok

thanks

Danu

@0celou7 Yeah, I think so

ACuriousMind

'course, the potential is also singular, so same difference, I guess

0celou7

10:25 PM

@Danu Now we're going to do a little triangle. Pick $y\in K$. Then $f(y)\le f(x)+c$, where your $c$ is from above.

In other words, $\sup_{x\in K}f(x)=R<\infty$. Thus $K\subset \exp_p(B_R)$, where $B_R$ is the ball of radius $R$ in $T_pM$

Since $\exp_p$ is continuous and $B_R$ is compact, $\exp_p B_R$ is compact. As a closed subspace of a compact set, $K$is compact.

@Danu Ah, but in your case one just has that the ball is the image of a compact ball in $T_pM$ under a continuous function, so you can ignore this :P

Danu

@0celou7 You're just usingthe triangle inequaity? If you assume $x,y\in K$ (that ball of mine) then you probably want $2c$ sine that's the diameter of $K$

0celou7

@Danu Ignore all of it. $\{x\in M:d(x,p)\le c\}=\exp_p\{v\in T_pM: ||v||\le c\}$.

Danu

@0celou7 No, I don't think so

Gotta take into account geodesics that loop around

0celou7

In a complete manifold...

@Danu Oh yeah, $2c$

Danu

On a sphere the great circles can be traversed any number of times

So then the preimage is not compact

0celou7

10:31 PM

I'm saying the image is compact

loltospoon

When we we use a gauge function to transform the electric potential and the magnetic potential, the electric and magnetic fields remain unchanged, correct?

Danu

@0celou7 Oh, right

Riemannian geometry is like 3/10

fucking analysis man; I just don't get excited about it

ACuriousMind

@loltospoon yes

0celou7

¯_(ツ)_/¯

loltospoon

@ACuriousMind ok thanks

0celou7

10:34 PM

@Danu For future reference: for each $p\in M,$ $\exp_p$ is defined everywhere $\Leftrightarrow$ closed and bounded sets are compact $\Leftrightarrow$ $M$ is complete as a metric space $\Leftrightarrow$ $M$ is geodesically complete

any any of those implies for $p,q\in M$ there's a geodesic $\gamma$ from $p$ to $q$ with $\ell(\gamma)=d(p,q)$

Secret

10:54 PM

Car race goes nano on April 28-29

Argon doping makes hydrogen stay molecular at higher pressures

0celou7

@ACuriousMind @Danu that MO question in the feed looks like something for you

loltospoon

11:16 PM

Why do we think that an electromagnetic wave has electric and magnetic components perpendicular to each other? See this image from wikipedia: en.wikipedia.org/wiki/Maxwell's_equations#/media/…

Did we experimentally find that for an electromagnetic wave they are perpendicular?

Danu

@0celou7 I can't see the feed

@loltospoon I guess you just get it straight from Maxwell's equations in a vacuum.

0celou7

@Danu mathoverflow.net/questions/265630/…

Danu

Welp, no idea

0celou7

@Danu what are you working on in Milnor that requires you to know those balls are compact?

Danu

I'll give a seminar talk in 3 months on sections 16-18

(I'm procrastinating; avoiding more important stuff...)

dmckee

11:28 PM

@loltospoon You can test that quite easily in the radio band. You need a couple of dual-dipole antenna, a couple of circular loop antenna, a high-power source and and an osciliscope.

Danu

^down & dirty

0celou7

@Danu 16 is pretty neat

loltospoon

@Danu ok

dmckee

It's one of the things I did with my first oscilliscope.

loltospoon

@dmckee I'm leaning towards theoretical physics hah....

So all of that just went wayyy over my head

dmckee

11:33 PM

Take a two conductor wire and peal the two bits apart until you can tape them to a dowel with the main wire coming off the center. That's a dual dipole. It responds to electric fields in the direction of the dowel. Ptug the other end of the wire into the oscilliscope.

0celou7

@dmckee lol what a nerd

dmckee

Similarly, twist a metal coat hanger into a circle and attach the two conductors of a dual conductor wire into the two ends (which you need to hold apart with an insulator of some kind). That's a circular antenna. It's sensitive to magnetic fields normal to the plane of the circle. Again you plug the wire into the oscilliscope.

@0celou7 It gets better. The oscilliscope in question ran on vacuum tubes.

Now by twisting the two antennas around you can determine the direction of both the electric and magnetic fields due to your transmitter.

Done and dusted.

loltospoon

11:59 PM

$\nabla ^2 V = \rho / \epsilon_0 $ and $ V(\vec{r},t)=\frac{1}{4\pi \epsilon_0}\int \frac{\rho (\vec{r'},t)}{\mathscr r}d\tau '$

The first equation is supposedly the second derivatve of $V$

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