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John Rennie
John Rennie
10:02 AM
@Kaumudi.H right, that's my filial duties for the day taken care of. Back to dielectrics!
 
user228700
> filial
 
user228700
New word!
 
John Rennie
John Rennie
Latin - filius = son
 
user228700
Right.
 
John Rennie
John Rennie
In my day we were taught to speak posh! :-)
 
user228700
10:04 AM
You don't say! :-) Does my statement look OK?
 
John Rennie
John Rennie
I've been going downhill since!
Can you link to a derivation?
So I can see what argument is being used.
 
user228700
I will type.
 
user228700
The electric field inside the capacitor is changed as $E=E_o/K$ where $K$ is the dielectric constant of the material.
 
John Rennie
John Rennie
Presumably you could treat the capacitor as isolated, in which case the charge is constant and the voltage changes, or connected to a voltage source in which case the voltage is constant and the charge changes.
 
user228700
($E_o$ is the electric field when the dielectric is not present)
 
user228700
10:07 AM
Now, this is the argument that has been presented by the book:
 
user228700
 
user228700
(Excuse my rampant highlighting)
 
user228700
It starts on the 6th line from the top of the page.
 
user228700
Right, so these charges interact with those on the plates (opposite in sign) to partially nullify some of them and hence the charge on the capacitor decreases.
 
user228700
...is this incorrect?
 
John Rennie
John Rennie
10:12 AM
Yes, that makes sense.
Although the charge on the capacitor plates isn't reduced by the charge on the dielectric because the charges on the dielectric are bound i.e. it's an insulator.
So charge doesn't flow off the dielectric onto the plates.
 
user228700
What about the nullification by virtue of being very close to the plates?
 
John Rennie
John Rennie
Assuming the capacitor isn't connected to an external circuit, i.e. charge can't flow on or off the plates, the number of electrons on the -ve plate doesn't change.
However when you draw a Gauss' law type surface inside the plates the net charge on the plate side of the surface is reduced.
So charge is neutralised in that sense.
 
user228700
Right...
 
John Rennie
John Rennie
If you shorted the capacitor then the amount of charge that flows off it is unchanged by the presence of the dielectric.
 
user228700
Gimme a second...
 
user228700
10:19 AM
Alright, never mind this line of reasoning. Can u please explain this to me:
 
user228700
(Sorry about that; I have about 5 books and a laptop on this small table of sorts)
 
John Rennie
John Rennie
:-) Incidentally what prompted:
2 hours ago, by Kaumudi. H
Crap. @JohnR: Halp!
I was worried you were about to tell me the laptop had died.
 
user228700
Nah, from now, you should be more worried about me telling that I'm about to die :-P
 
John Rennie
John Rennie
No-one dies from exams - they just wish they could!
 
user228700
 
John Rennie
John Rennie
10:24 AM
OK
 
user228700
Oh, never mind. For a moment there, I thought maybe you'd read the whole thing in 3 seconds :-P
 
John Rennie
John Rennie
I did ... I was waiting for the question ...
 
user228700
:-| Did u really read and understand all of that in 3 seconds?
 
John Rennie
John Rennie
Yes. But then I already knew it so I didn't have to work at understanding it. That's a pretty straightforward derivation.
 
user228700
...OK. I suppose the confusion starts at 2.49
 
user228700
10:28 AM
(I don't understand how they arrived there)
 
John Rennie
John Rennie
Do you mean you're not sure how 2.49 follows from 2.48?
 
user228700
Yep.
 
John Rennie
John Rennie
OK $\sigma$ is the charge when no dielectric is present, so $\sigma = E\epsilon_0$. Yes?
 
user228700
Yep.
 
John Rennie
John Rennie
The electric field $E$ causes an induced surface charge $\sigma_p$ on the dielectric. We assume the induced charge is proportional to the field so we get: $$\sigma_p = kE $$ for some constant $k$ that we won't worry about for now.
 
user228700
10:37 AM
Yes, OK...
 
John Rennie
John Rennie
So: $$\sigma - \sigma_p = (\epsilon_0 - k)E$$ OK so far?
 
user228700
Yes...
 
John Rennie
John Rennie
And $E = \sigma/\epsilon_0$ so we substitute for this to get: $$\sigma - \sigma_p = (\epsilon_0 - k)\frac{\sigma}{\epsilon_0} $$
 
user228700
Uhh, OK...
 
John Rennie
John Rennie
And we end up with: $$ \sigma - \sigma_p = \frac{\epsilon_0 - k}{\epsilon_0}\sigma $$
Which we can write as $$ \sigma - \sigma_p = \frac{\sigma}{K} $$
where: $$ K = \frac{\epsilon_0}{\epsilon_0 - k} $$
 
user228700
10:43 AM
Ohhh :-| OK...
 
John Rennie
John Rennie
I suspect you were looking for something deep and meaningful rather than some rather trivial substitutions ...
 
user228700
I was :-/ I'm sorry, I should've been able to figure out that on my own; my brain is slowly dying. Thanks very much.
 
John Rennie
John Rennie
The notation is a bit unhelpful.
 
user228700
Huh?
 
John Rennie
John Rennie
$\sigma$ should really be $\sigma_0$ to make clear it's the charge density when no dielectric is present.
 
user228700
10:47 AM
Right. Yeah, no, that's OK. It's just that I didn't know that this was so simple. Thanks so much and I'm sorry I wasted ur time on this :-|
 
John Rennie
John Rennie
S'OK. I'm kind of chilling at the moment anyway. My servers are all OK so I'm just killing time until I can reasonably make lunch :-)
 
user228700
OK :-)
 
John Rennie
John Rennie
Listening to an album by Air - which is pretty chill ...
 
user228700
Wait, so the charge on the plate doesn't ostensibly change when a dielectric is introduced?
 
John Rennie
John Rennie
In this working they are using $\sigma$ to represent the initial charge. They then (in 2.50) eliminate that initial charge from the equation to get the final equation they want.
 
user228700
10:57 AM
No no, not here.
 
John Rennie
John Rennie
You mean in general?
 
user228700
Yeah.
 
John Rennie
John Rennie
That depends. If the capacitor plates are isolated, i.e. not connected in a circuit, then charge can't flow on or off the plates. That means the charge can't change. What happens instead is that the voltage between the plates decreases as the dielectric is inserted.
 
user228700
Right. What if it is connected to a battery?
 
John Rennie
John Rennie
Then the voltage between the plates is held constant, and charge will flow from the battery onto the plates. So in this case the charge on the capacitor will change.
 
user228700
11:00 AM
How so..?
 
user228700
(I bet you can tell that I'm really struggling to understand this)
 
John Rennie
John Rennie
Imagine a two step process: (1) keep the capacitor disconnected and insert the dialectric, then (2) connect to capacitor.
Let $V_0$ be the battery voltage, so we start with the capacitor connected and with a voltage $V_0$. Then we disconnect it from the battery.
Now we insert the dielectric. The charge on the capacitor doesn't change, because the capacitor is disconnected, so the voltage between the plates falls to some voltage $V \lt V_0$. OK so far?
 
user228700
Uh, yes...
 
John Rennie
John Rennie
You sound uncertain?
 
user228700
No no, please go on...
 
John Rennie
John Rennie
11:06 AM
OK. Now the dielectric has been inserted the voltage across the capacitor, $V$, is less than the battery voltage $V_0$. So if we reconnect the capacitor to the battery a transient current flows.
Charge from the battery flows onto the capacitor to increase its voltage back to $V_0$.
 
user228700
A transient current?
 
user228700
@JohnRennie Right, right.
 
John Rennie
John Rennie
That's why asking whether the charge is constant can't be answered unless you specify whether the capacitor is connected or not.
 
user228700
OK, never mind, then. I now understand how it works...
 
user228700
Thanks very much :-)
 
John Rennie
John Rennie
11:08 AM
Cool. Now time to go shopping for lunch I think ...
 
user228700
OK bye :-)
 
0celou7
0celou7
1:16 PM
What's up
 
 
1 hour later…
Slereah
Slereah
2:42 PM
@0celou7 how would you prove that a contractible curve can be entirely within one coordinate patch
Krasnikov wrote a paper called "Even the Minkowski space is holed"
How scandalous
"a postulate is definitely too strong if it excludes the Minkowski space."
"Recently, Manchak has constructed a spacetime which is inextendible and globally hyperbolic but fails to be hole-free"
Dammit
"This spacetime possesses a nasty singularity"
"However, it was suggested by Ellis and Schmidt (1977) that every inextendible, globally hyperbolic spacetime is hole-free (a proof of this claim was later given in Clarke (1993))."
WHO TO TRUST
 
0celou7
0celou7
3:12 PM
@Slereah hmm
 
Slereah
Slereah
I'm pretty sure it's true
since a contractible curve is basically in some open set of $R^n$
 
0celou7
0celou7
that's a good question, I have to think about it
 
Slereah
Slereah
Also do you know if quasiregular singularities are a math thing
I seem to only find them in GR
 
0celou7
0celou7
might be a mathy GR thing, I don't know
 
Slereah
Slereah
they are not that great at explaining them rigorously, though
I think it's supposed to be a singularity you can't extend away because the tangent bundle wouldn't mesh together properly
 
0celou7
0celou7
3:17 PM
That doesn't seem very precise
By a chart you just mean a set diffeomorphic to an open set of Rn, right?
 
Slereah
Slereah
the precise definition is just "a non-removable singularity where the curvature is everywhere well defined"
yes
For instance a conical singularity
but I'd like some mathy reasons that apply generally for it
also they are locally extendible, in the sense that you can form a submanifold from open sets such that the singularity is removable from those submanifolds
I think it might be something like if you have two such open sets $U$ and $U'$ the transition on $U \cap U'$ might fuck up somewhere
Considering what the paper goes on about I'm guessing it fucks up on the transition function of the tangent bundle or somesuch
 
ArmaGeddON
ArmaGeddON
3:44 PM
anyone online?
A student has to determine the focal length of a convex lens by obtaining the image of a distant object on a screen.
Which of the following measures should he take to obtain better results?

a. Select a lens of smaller diameter( say 3cm)
b. Select a lens of larger diameter (say 5cm)
c. Select an object very far from the lens
d. Select an object far,but not very far,from the lens
e. Keep all the lights of the room on.
f. Keep minimum lights of the room on.

(A) a,c and e
(B) b,d and f
(C) a,d and f
 
Leonhard Euler
Leonhard Euler
4:17 PM
Hello
 
Fawad
Fawad
Hi
 
dmckee
dmckee
@ArmaGeddON Questions like that often involve understanding trade-offs.
Consider just (a) and (b) for instance.
The image needs to be bright enough that you can see it clearly in order to determine when it is focused. That might argue in favor of (b) because a larger diameter collects more light.
However, it is only an approximation that spherical lenses focus to a point and it only works properly when the diameters is small compared to the radii of curvature of the faces. This argues in favor of (a).
But you can't have both.
Which is more important in a particular case? Depends on other factors.
If you control those other factors you can probably rig it to allow you to prioritize reducing the spherical aberation, but the question doesn't let you resolve that ambiguity.
 
ACuriousMind
ACuriousMind
> Political science how to use it in bengali language
...why would anyone post that as an answer to a physics.SE question?
 
dmckee
dmckee
What?
 
ACuriousMind
ACuriousMind
It doesn't even make sense as a comment, and it doesn't relate to the question at all, and it's not spam
@dmckee Posted here, new deleted
One of the odder NAAs I've seen
 
dmckee
dmckee
4:27 PM
And the user doesn't even have an account in a site where either political science or Bengali might be on topic.
 
0celou7
0celou7
4:39 PM
@ACuriousMind See Sam's question above about contractible curves
I am willing to believe it in 2 dimensions
 
Slereah
Slereah
plz
Call me Dr. Lereah
 
0celou7
0celou7
In 2 dimensions you can probably construct the chart using the homotopy
it will probably be star shaped even
maybe
 
Slereah
Slereah
I don't think so?
Like $R^2 - \{p\}$ I think has curves that can't be held within a star shaped patch
For instance a loop shaped like a bunny head with the singularity between the ears
 
0celou7
0celou7
but those curves are not nullhomotopic, no?
 
Slereah
Slereah
that one would be, anyway
I can't confirm that the patch isn't starshaped but I suspect it's not
 
0celou7
0celou7
4:47 PM
 
Slereah
Slereah
Mathematica is pretty annoying for that
You have to specifically say N[19/4]
Companies complaining about the shortage of engineers always makes me laugh
Hire them, maybe???
 
0celou7
0celou7
@Slereah Aha
 
Slereah
Slereah
Plenty of out of work people looking to get hired
 
0celou7
0celou7
Take a tubular neighborhood, then use the contractible base theorem
that may or may not work
 
Slereah
Slereah
I'll have to look it up
"No results found for "contractible base theorem"."
 
0celou7
0celou7
4:51 PM
a vector bundle is trivial if the base is contractible
 
Slereah
Slereah
could b
 
Secret
Secret
[Insert random caption]
Alien scifi like
 
Slereah
Slereah
that is a giant todger
 
Secret
Secret
 
Secret
Secret
Weird idea/question: Is it possible to construct a spacetime such that as it evolves, it behaves as if the CTC is translating forward in one of the spatial directions. (Naively did not see any topological issues here as the overall topology is not changed as you still have a single closed curve in spacetime)

but anyway, I am still suck at discussing about CTCs in any meaningful way, bleh...
 
Slereah
Slereah
5:01 PM
Take any CTC, add a boost???
I dunno what you mean
 
0celou7
0celou7
@Slereah So if you have a loop $\gamma:S^1\to M$, then you actually get an embedded $S^1\times D^3$, where $D^3$ is the open solid $3$-disk
and the curve will be contained in the core
 
Ben Niehoff
Ben Niehoff
in fact, constantly complaining about a shortage of engineers is doing two things: allowing them to hire more H1B's, and making more people go to engineering school, creating a labor surplus and depressing wages. You might call it a brilliant plan
 
0celou7
0celou7
this is assuming there are no crossings
and that does give a chart around it
this might all be a lie, I'm not sure
 
Slereah
Slereah
Wouldn't $S^1 \times D^3$ be a torus?
 
0celou7
0celou7
solid torus is covered by a single chart
 
Secret
Secret
5:07 PM
Basically, a CTC that trace out a worldsheet in spacetime
So when taking any xyz slices, it looks like as if the CTC is translating along some direction in space
 
Ben Niehoff
Ben Niehoff
@Secret but those coordinates don't mean anything...what you need to look at are the possible paths a timelike observer can take
and having CTCs means exactly what it says: closed timelike curves
 
0celou7
0celou7
@Slereah Your curve question is really good
 
Slereah
Slereah
Thank u
I need it to show that CTCs aren't timelike contractible to a point
 
Ben Niehoff
Ben Niehoff
what does "timelike contractible" mean?
 
Slereah
Slereah
Well at least it would be easier to show this way
Contractible in a way that all intermediate curves are also timelike
 
0celou7
0celou7
5:11 PM
@BenNiehoff each intermediate curve in the homotopy is timelike
 
Ben Niehoff
Ben Niehoff
ah, ok
it seems that that would require two time dimensions
 
Slereah
Slereah
it's basically intermediate value theorem + normal neighbourhoods are globally hyperbolic
and it would help if the curve was in a single coordinate patch
 
0celou7
0celou7
Umm, is any smooth curve contained in a coordinate chart?
 
ACuriousMind
ACuriousMind
@Slereah Well, once it's close enough to the point you're contracting to it will be, since it will lie inside a patch around that point
 
0celou7
0celou7
I shouldn't think so
 
ACuriousMind
ACuriousMind
5:13 PM
I'm not sure what the problem is supposed to be
 
0celou7
0celou7
Wind a curve around the torus enough times and it should be imporrible
 
ACuriousMind
ACuriousMind
@0celou7 Of course not, otherwise all loops would be contractible!
 
Slereah
Slereah
@ACuriousMind Could b
Oh well
 
0celou7
0celou7
@ACuriousMind in more than 2 dimensions that's probably true
 
ACuriousMind
ACuriousMind
@0celou7 what?
 
0celou7
0celou7
5:14 PM
but in 2 dimensions you can have $\pi_1$-crazy charts
 
ACuriousMind
ACuriousMind
I don't know how that's supposed to work - a chart is a homomorphism $\mathbb{R}^2\to U$, no?
 
Slereah
Slereah
Spacetime topology is really a neverending topic
There's like 10 authors and I still find papers I've never heard of
 
0celou7
0celou7
@ACuriousMind it's a homeomorphism $U\subset M\to \hat U\subset\Bbb R^2$
 
Slereah
Slereah
It's always Hawking, Ellis, Geroch, Krasnikov or something
 
0celou7
0celou7
It's easy to find a noncontractible curve contained in a single chart in 2 dimensions
Take $M$ to be the open annulus
 
Slereah
Slereah
5:16 PM
true
 
0celou7
0celou7
In 3 dimensions I think every open set is simply connected?
 
Slereah
Slereah
I think so yeah
 
0celou7
0celou7
Should be: factor the induced maps on homotopy $S^2\hookrightarrow U\hookrightarrow \Bbb R^3$
 
ACuriousMind
ACuriousMind
Ah, you don't define your charts to be contractible
 
0celou7
0celou7
@ACuriousMind Um, no? Who does?
 
ACuriousMind
ACuriousMind
5:17 PM
Is there a manifold you can't cover if I demand the charts to be contractible?
 
0celou7
0celou7
@ACuriousMind I have a math SE post on this
It's a nontrivial theorem but can be proved
 
Slereah
Slereah
stop making up conditions @ACuriousMind
 
0celou7
0celou7
It's implictly assumed if you do Cech cohomology but good luck finding the proof anywhere
 
ACuriousMind
ACuriousMind
@0celou7 You're right it's not usually demanded in the definition, but I've seen people assuming them to be contractible without loss of generality so often that I tend to view the other case as pathological :P
 
Ben Niehoff
Ben Niehoff
it still blows my mind that you can have manifolds that are not triangulable
 
0celou7
0celou7
5:19 PM
5
Q: Why are geodesically convex sets diffeomorphic to $\Bbb R^n$?

0celou7In the construction of a good cover for a manifold $M^n$, Bott & Tu use the fact that each point in $M$ is contained in a geodesically convex set (after picking a Riemannian metric). They then claim that the intersection of geodesically convex sets is a geodesically convex set, and that makes sen...

differential-geometry algebraic-topology differential-topology riemannian-geometry
@ACuriousMind You're looking for "good covers"
As far as I know, that post is the only complete proof in existence
 
ACuriousMind
ACuriousMind
Yeah, I remember that now
 
Ben Niehoff
Ben Niehoff
@0celou7 do you have any intuition about how triangulability can fail?
 
0celou7
0celou7
@BenNiehoff No, I haven't had to deal with triagulability outside of 2-manifolds, where it can't fail
 
Ben Niehoff
Ben Niehoff
I think I read that the lowest dimension in which it can fail is 4, but not entirely sure
 
0celou7
0celou7
Yeah that's right
In 3 dimensions you can have topological manifolds that can't be smoothed
 
Ben Niehoff
Ben Niehoff
5:22 PM
as in no smooth structure? I thought that required 4 as well!
 
0celou7
0celou7
@ACuriousMind I think I was subconsciously plugging that proof. Of course you can take the charts to be contractible, last month we decided we could take a locally finite chart consisting of balls and half-balls
My proof above controls the intersections as well
 
Slereah
Slereah
is "unwrapping" a thing outside of some weird GR papers
Removing a point and going to covering space to make closed curves not closed
 
0celou7
0celou7
@BenNiehoff maybe, this is not my area
I'm trying to goolgle, but I'm failing!
 
ACuriousMind
ACuriousMind
@0celou7 Is it really non-trivial? The "good cover" is non-trivial, but that you can cover all manifolds with contractibles is immediate - if one of your charts is not contractible, replace it by its restrictions to open balls around the points in $\mathbb{R}^n$. Sure, the atlas becomes uncountable, but that's not forbidden, is it?
 
0celou7
0celou7
@BenNiehoff you're right, smooth and top agree in 3 dimensions?
@ACuriousMind Did you not see my message just now?
4 mins ago, by 0celou7
@ACuriousMind I think I was subconsciously plugging that proof. Of course you can take the charts to be contractible, last month we decided we could take a locally finite chart consisting of balls and half-balls
 
ACuriousMind
ACuriousMind
5:28 PM
Yes, but that thing about the manifolds with boundaries was also non-trivial
I'm just saying there's a really dumb way to get a cover by contractibles :D
 
0celou7
0celou7
I hate chat, my fingers don't work today
What you just said was step 1 in proving the locally finite covering
And you can take a countable subcover by second countability
Any open cover can be restricted to a countable one
@Slereah O'Neill has such a terrible looking cover
they could have at least put a torus on it or something to spruce it up
 
Slereah
Slereah
It's quite minimalist
not even a border around the title
It's a cover that tells you we're not here to have fun
so
Is the deck transformation named after a mister Deck
or is it a noun here
 
0celou7
0celou7
it's from German
covering is deckung
@Slereah Spivak lets you know that LSD is a prerequisite for the book
 
Slereah
Slereah
Straumann tells you "I use the standard cover of the editor"
 
0celou7
0celou7
wat
"if $\lim_{h\to 0}x^h=x$"
 
Slereah
Slereah
5:38 PM
quite an odd limit
 
0celou7
0celou7
$x\in\Bbb R^n$, what is this nonsense?
 
Slereah
Slereah
Maybe it's a typo
$h \to 1$
 
0celou7
0celou7
@Slereah what is $x^h$ supposed to be if $x$ is a vector
 
Slereah
Slereah
A valid question
 
0celou7
0celou7
I think he means $h\mapsto x(h)$
 
Slereah
Slereah
5:41 PM
what is $x(h)$
 
0celou7
0celou7
a function of $h$ with values in $\Bbb R^n$
@Slereah all of the GTP and GTM books use the standard covers
there are some like "Classical Fourier Analysis" that have little designs
wonder whom they have to pay off to get that privilege
the designs do change throughout the years
the new design is very nice actually
oh noooo
I have to prove that Riemann sums work in $n$ dimensions
Oh wait, these functions are continuous. Crisis averted!
@Slereah how do you like this cover
 
Slereah
Slereah
I think my Jost has the same one
Or maybe my Hilbert space book
 
0celou7
0celou7
I don't think so, Jost is not GTM
@Slereah link?
 
Slereah
Slereah
Weidmann's linear operations in hilbert spaces
Same cover
oh wait
Not quite
 
Ben Niehoff
Ben Niehoff
deck the halls with non-contractible charts!
 
0celou7
0celou7
5:54 PM
Yeah that's the terrible late 90s-early 00s GTM cover
I refuse to buy those books on principle
 
Ben Niehoff
Ben Niehoff
oh my, I hope "Hilbertriiumen" is due to some horrible OCR and not anybody actually typing that
 
0celou7
0celou7
@Slereah do you understand Appendix A?
 
Slereah
Slereah
I did not read it yet
you know what else has the same cover, except for colors?
 
0celou7
0celou7
Hmm?
 
Slereah
Slereah
Topology of fiber bundles and Non-standard analysis
 
ACuriousMind
ACuriousMind
5:56 PM
@BenNiehoff Pretty sure it's OCR, mistaking an ä for a ii is pretty weird otherwise but youc an see how a computer might have trouble distinguishing them
 
Slereah
Slereah
both Princeton books
 
0celou7
0celou7
I can't put another picture on imgur
why is imgur limiting me?
same covers, different printers
 
OBE
OBE
nice books
 
0celou7
0celou7
@Slereah all AMS books have the exact same covers too
 
0celou7
0celou7
6:35 PM
@Slereah what illustration would you even put on a Hilbert space book
not exactly something you can see in the night sky
 
OBE
OBE
@0celou7 that proves they don't exist hehe
 
Slereah
Slereah
the classic cover for an abstract math book is to put like either a grid or random formulas
Plus $\Bbb R^n$ is totally a Hilbert space
 
0celou7
0celou7
7:27 PM
@Slereah what's a good formula for hilbert spaces?
 
DanielSank
DanielSank
Anyone know this one:
$$\int \exp(i x^2) \, dx$$
It's oscillatory, yeah?
 
0celou7
0celou7
@DanielSank ...oscillatory in what variable?
If you're doing a physicist, use the integral for $\int e^{-ax^2}\,dx$ and put $a=-i$
 
ACuriousMind
ACuriousMind
@DanielSank Have a look at Fresnel integrals
 
DanielSank
DanielSank
@0celou7 Oh that's cute.
Is it right?
Doesn't look right.
 
0celou7
0celou7
7:42 PM
probably not right
in any case, the integral should be badly divergent
you get $e^{ix^2}=\cos x^2+i\sin x^2$
 
ACuriousMind
ACuriousMind
...I just linked to the Wiki page that computes that explicitly
There's no divergence
 
0celou7
0celou7
I doubt that
 
ACuriousMind
ACuriousMind
@DanielSank The solution of that integral is $\int_{-\infty}^\infty \exp(\mathrm{i}x^2)\mathrm{d}x = 2\left(\int_0^\infty \sin(x^2)\mathrm{d}x + \mathrm{i}\int_0^\infty \cos(x^2)\mathrm{d}x\right) = \sqrt{\pi/2}(1+\mathrm{i})$, where the integrals over $\sin(x^2)$ the cosine are computed as limits of the Fresnal integrals as in the section I linked to
 
0celou7
0celou7
doubtful
 
Slereah
Slereah
isn't that a thing for the stationary phase method
 
ACuriousMind
ACuriousMind
7:49 PM
@Slereah That would give some estimate, not an exact result
 
0celou7
0celou7
convolution of distributions is awful
who thinks of these things
 
Slereah
Slereah
you can't convolute distributions!
 
0celou7
0celou7
yeah you can
if one has compact support you can
 
Slereah
Slereah
What is $\delta \star \delta$
 
DanielSank
DanielSank
I see, so if you treat the integral as a limit as the bounds go symmetrically to infinity, then you get a sensible answer, @ACuriousMind?
 
0celou7
0celou7
7:52 PM
@Slereah first, what is $\delta\otimes\delta$?
 
Slereah
Slereah
I do not know
 
ACuriousMind
ACuriousMind
@DanielSank I don't think you need to take it symmetrically. You could split $\int_{-\infty}^\infty = \int_0^a + \int_{-b}^0$ and take $a,b\to\infty$ separately.
 
0celou7
0celou7
the defintinition of distribution convolution is fairly horrible, I'm not sure how to actually compute it
@Slereah So if $T_1\in\mathscr D'$, and $T_2\in\mathscr C'$, then you define $T_1*T_2$ by $\mathscr D\ni\varphi\mapsto T_1*(T_2*\varphi)$.
$\mathscr C=C^\infty(\Bbb R^n)$ with the seminorm topology
 
DanielSank
DanielSank
@ACuriousMind That seems odd because then both of those integrals oscillate.
 
ACuriousMind
ACuriousMind
@DanielSank So what? The integrand oscillating does not mean that the integral cannot exist. We know that $\lim_{a\to\infty} \int_0^a\sin(x^2)\mathrm{d}x = \sqrt{\pi/8}$, and that's all I used - I used the symmetry of the integral to reduce it to this asymmetrical case!
So instead of $\int^\infty_{-\infty} = 2\int_0^\infty$ in what I did, you might as well write $\int_0^\infty + \int^0_{-\infty}$, it is of no import.
 
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