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00:00
I'm young and foolish, two years seem like a really long time to me
Seriously?
I guess I'm a big exception in this sense... I'm looking very far ahead (let's say ~5 years at least).
Yes, I'm out of school for four years now, and I feel I've been here in Heidelberg for an eternity
@Danu You're wise.
And although I like it here, I'd have no qualms about uprooting myself and going elsewhere
@ACuriousMind Married? Children?
00:03
@ACuriousMind Well I have actually done that, and note that it kind of sucks.
Everyone has to balance their own life according to their priorities.
For me, family would always be #1.
easily.
Academia is demanding. Those who burn for it should go for it, but I think everyone of us should at least be aware of the price and the options.
@Danu In that case, academia could be rather difficult. Lots of people do make it work, however!
I really want to do this though; I don't see any other reasonable-sounding alternative in terms of work-life.
@Danu Just try to remember that there is a huge world out there. It's easy to forget that.
00:05
Irrelevant side note: This a good start for "chat reform". Interesting conversation.
@DanielSank Of course.
I really have to sleep now guys, sorry :| Would love to continue this conversation some other time.
@DanielSank Neither. And I won't get married or have children while being "on the move", but I don't think I want to get married or have children in the next ten years.
Really?
Welp, I guess I'm just crazy ;)
Cya guys.
@Danu Nah, you're alright.
g'night
@Danu ?
00:20
I should go to sleep too, gotta catch a train tomorrow and I've a lot of things to do that I'm not doing tonight in any case...
00:56
Anybody here?
01:15
0
Q: Is over-vigilant monitoring of questions driving people away from the site?

matsciencemanNow, I know the need to create standards for asking questions. However, particularly for first time users, the ability to ask questions in a way that this forum expects can be difficult, often due to a lack of language surrounding the topic they are asking about. As a result, their question get...

 
2 hours later…
03:10
hello,somebody there?
 
1 hour later…
04:16
0
Q: Questions related to geopolitical events

Brian M. HuntI've raised a question that was put on hold, here: Could a falling office tower cause the melting of its iron or steel support structure? I wanted to post about a topic for discussion about why the question was put on hold. I would preface that I don't want to offend anyone, raise defences, br...

 
2 hours later…
06:14
@0537 Yes
Currently in Ramstein
06:26
I have a feeling this will be deleted...
06:48
@skillpatrol what's up
@skillpatrol what did you remove -.-
Why bother arguing with him? @0celo7
it's my only joy in life
sad life pal
:-/
yup
You might as well have been a raiders fan over the last 12 years.
06:51
once the heroine tolerance started...I knew I needed something more powerful
Addiction can get ugly quick.
can it
Ask any recovering addict.
Oh, were you asking me?
:P
 
2 hours later…
09:17
@Danu yap, great thing such a physics chat! very helpful when I'm too dumb to recognize a simple Taylor expansion :D
@0celo7 do you know what triggers it?
 
3 hours later…
Huy
Huy
11:51
@0celo7: I have a book for you (and your friend): "SJWs Always Lie: Taking Down the Thought Police"
you should send it to her
 
2 hours later…
13:33
0
Q: My question was held off by vote for wrong reasons

DShaI was asking for a clarification My question was not Homework type I had worked out a solution myself

 
1 hour later…
15:01
Hey @ACuriousMind
3
Q: Diffeomorphism group vs. $GL(4,\mathbb{R})$ in General Relativity

MarionI am quite confused with the groups Diff$(M)$ and $GL(4,\mathbb{R})$ in the context of general relativity. I understand that the symmetries of GR are the transformations that leave the equations invariant under under continuous smooth coordinate transformations. That is, automorphisms $f:M \to M$...

^that was the thing we discussed and all
And I did not know the answer
 
2 hours later…
16:41
@JohnDuffield Did you delete that answer or was it the site?
17:05
Hey @Huy what's the style over in the math chat?
Seems pretty dead most times I go over there.
Huy
Huy
@DanielSank: You're probably there at wrong times.
@DanielSank: There's very often really bad drama because some people can't stand each other. But also often there are rather vivid discussions. Not sure if they are about anything you're interested in, though.
@DanielSank: What math topics do you like talking about? From my experience, it's a lot less dead than this chat, to be honest.
@Huy That sucks.
@Huy I need to ask some questions about delta functions in greater than 1 dimension.
Think you can help?
Huy
Huy
@DanielSank: Not sure, just ask though. You can also try pinging DanielFischer if you don't think that's rude. He knows almost everything.
@Huy I don't like to ping until I've given to the community a bit more.
My rep there is low.
Huy
Huy
then you should frequent the chat and soon you can ping. :P
just ask the question in the chat. a lot of people are actually around but don't talk if nothing interesting is going on (or nobody interesting is around)
17:11
@Huy Yeah, I need to answer some stuff there...
Huy
Huy
for example, I'm online all day but right now I'd like to talk about some geometric topology but neither Balarka nor Mike are on so I don't really say anything.
It's hard, there's not a lot of math where I'm more expert than all those other users.
Hi everyone, Just reading about Floer homology within symplectic geometry. Anyone give me some pointers or basic idea to help? :)
Huy
Huy
@DanielSank: we don't just talk about maths all the time. just see if there's something interesting going on every now and then and you'll see :)
@Danu In a sense that is strange because there was never any guarantee (or even a good odds) of finding evidence for string theory at LHC energies.
On the otherhand, a lot of students needed thesis projects so there were a lot of "well, if we assume that [parameter] has [the right range of values] then we might see [effect] at LHC energies..." going on.
The important thing here is what they call you after you turn in a dissertation based on a null result ...
Doctor.
Then they shake your hand and go back to calling you by your first name, because that's the culture in particle physics.
17:25
heheh
@dmckee Wishful thinking
It's like how Ronald Mallett thinks he can create CTCs by changing the speed of light and thinking nobody will notice
17:45
@dmckee Eh... I think you're being a bit too cynical, but okay.
I think there were a lot of people who were genuinely hoping (and still are!) that something SUSY-related may show up.
I'm convinced nothing will show just as much as you, but no need to talk down that much :P
@Danu Sure. But there has never been an apriori reason to expect it. To many order of magnitude between the Tevetron and the plank scale.
Especially since I assume that you have never really worked on it yourself.
Well SUSY was possible
String theory was a bit of a reach
@dmckee There is no reason to believe that one can only do things with string theory at the Planck scale (as you probably know).
I don't claim these analyses weren't worth doing. The data needed looking at, but the driving reason for the LHC was a much more general search for new physics.
17:48
Of course.
I agree that SUSY was never a main goal.
@Danu Of course not, but that leave a heck of a lot of phase space for the effect to hide in.
But, if someone had to guess a priori what kind of beyond SM physics to find...
Of course, I wasn't around back then so I don't really know.
@Danu explain multidimensional delta functions to me.
@DanielSank You mean.. Delta's of the type $\delta^{(4)}(x^\mu-y^\mu)$?
I can prove that $\delta(\vec{x} - \vec{\mu}) = \prod_i \delta(x_i - \mu_i)$ using Fourier transform.
17:51
okay, good :P
How do I deal with $\delta(f(x))$ where $f$ is $N$ dimensional?
Ah, yes
That stuff is annoying.
You have to work with the zeros of $f$
In 1-D I can prove $\delta(f(x)) = \sum_{\text{roots }r} \delta(x-r)/|f'(r)|$.
But in $N$ dimensions the equation $f(x)=0$ defines an $N-1$ manifold.
17:53
Right... I think you should try to reduce it to 1-D $\delta$'s first and then work with the function
is that impossible in your case?
@Danu Great.
I'm trying to figure out how to do that.
That's at least how I derived cross-sections in QED
We can try this: $\delta(f(x)) = \sum_{\text{"roots" }r} \delta(\sum_i (\partial_i f)(r)(x_i - r_i))$.
See what I mean?
$r$ is not a single coordinate, but a full position vector (or whatever analogous thing)?
@Danu Yeah.
The thing in the $\delta$ function is the first order term in the Taylor series of $f$ expanded around $f$.
The zeroth order term doesn't appear because it's zero by construction.
17:58
Right, that's how to derive that formula I guess.
you should have some kind of a product in your last formula, no?
Well actually I'm not sure what do do next.
If I use the Fourier transform trick we get:
$\sum_{\text{"roots" }r} \int \frac{dk}{2\pi} \exp (i k \sum_i (\partial_if)(r)(x_i - r_i))$
This is no bueno because I only have one integral.
In the $\delta(\vec{x} - \vec{\mu})$ case I had $N$ integrals which factorized.
I'm thinking that my error here is this:
With $\delta(\vec{x} - \vec{\mu})$, the argument of the $\delta$ function is a vector.
How can that be (recalling that $f=\operatorname{id}$ should work)?
However, I've written the first order expansion of $f$ as a scalar.
Wait... $f$ is a vector too, no?
I think this is wrong. I think I need to respect that the derivative of $f$ is actually vector-like.
18:04
I mean, $f:\Bbb R^n\to \Bbb R^m$ right?
@Danu Not sure what that meant.
Yeah, exactly, if $f$ maps to more than one dimension
@Danu I was assuming $f$ goes from $R^N$ to $R$ for the moment.
Oh, so $f$ is real-valued.
Then you are dealing with a 1-D Dirac delta, for sure.
@Danu Ok let's do the more general case.
I like where this is going.
18:05
Okay, so Taylor gives $f=f_0+(x-r)\cdot \nabla f$ right.
Yes...
for a real-valued function.
So what's the problem with this?
$\delta(f(x))=\delta(f_0+(x-x_0)\cdot \nabla f+\dots)$
I already wrote that.
Remember?
18:07
Yes... But what's the problem with $\delta (f(x))=\frac{1}{|\nabla f|}\delta(x-x_0)$
I agree actually that this is not nice, since how did this suddenly turn into a multi-dim one
@0celo7 entertain me
@Danu I'm convinced that's right... but why?
Oh wait...
18:10
@Huy That would be more helpful if I knew what 6.1.1 were...
Implicit function theorem?
and what the heck is $\psi$?
What's the book? *
Huy
Huy
check the reference I gave you in maths chat
it's available on springerlink
"if you want to read about it yourself, see Theorem 6.1.5 on page 136 in Hörmander's Analysis of Linear Partial Differential Operators I."
Pfft secretly having double-convo's huh, Daniel ;)
I appreciate it, and if we can't figure this out here, I'll read it.
But I am pretty damn sure we don't need a book on analysis to understand this.
Huy
Huy
:P
no comment
18:15
@Danu I tried the math chat, but frankly physicists often think about math better than straight-up mathematicians.
@DanielSank Ehhh :P different
Whatever.
@DanielSank Well, at least more directly and in a more solution oriented way.
In any case, the result is also on wikipedia in pretty accessible language...
But it's not that simple.
To really understand the Wiki thing one needs a bit of measure theory, it seems.
Well...
We had all the ingredients, I think.
I'm 100% sure you don't need measure theory.
You can always write a delta function as a limit of something else or just write it as a Fourier transform.
18:19
Well,
The "derivation" can be done from measure theory
But I'm pretty sure we can come up with it ourselves
let me first state teh result I guess...
$f:\Bbb R^n\to \Bbb R$, then we have:
No need for measure theory for derivatives
@Slereah I've never seen anything where measure theory was needed or even helpful.
What about measuring
$\int_{\Bbb R^n} g(x) \delta(f(x))\mathrm d x= \int_{\{x\in \Bbb R^n | f=0\}} \frac{g(x)}{|\nabla f|}\mathrm d \sigma $
It is mostly useful for weird integrals?
18:22
So basically just the domain is a bit tricky---but you had the right idea already.
I have also seen it used for probability theory
Basically, it generalizes the idea of summing over the roots to integrating over the zero surface
actually, it's just an obvious generalization.
@Danu You pulled that denominator out of your butt though.
@Danu Yes, I agree with that statement.
@DanielSank Hmm?
It's the one we had already.
@Danu No... you just wrote it down before ;)
18:27
@DanielSank Ehh?
@Danu We never derived the $1/|\nabla f|$ part.
You want to write this basically, yes:
20 mins ago, by Danu
Yes... But what's the problem with $\delta (f(x))=\frac{1}{|\nabla f|}\delta(x-x_0)$
@DanielSank Yes we did!
We just didn't know how to deal with the 'remaining delta'
But it's clear in retrospect.
What, no. You "derived" it by symbolic analogy but blithely ignored the dot product etc.
The remaining delta part I understand.
@DanielSank Haha, okay.
You just integrate over the $N-1$ manifold.
18:29
fucks given < 0
I see your point :P
I'm annoyed that when I try to use Fourier transform it doesn't work.
Or maybe it does and I'm too dumb...
But really though, I think factoring should definitely work.
Remember I wrote out that whole Fourier integral?
Factoring what?
18:30
The delta
What you tried to do
@Danu Again, I can prove that $\delta(\vec{x} - \vec{\mu}) = \prod_i \delta(x_i - \mu_i)$ via actual math (i.e. Fourier transform).
So yeah, that factors.
Yeah, but something like that should work here too... somehow.
But it's tricky.
That "somehow" is precisely why I'm here...
heheh
:D
Sorry for not being too helpful.
That's cool.
It's sooooo close though.
18:39
first day of holidays here
@0celo7 I didn't delete it, see the review:
Jon Custer reviewed this 5 hours ago: Recommend Deletion
Kyle Kanos reviewed this 7 hours ago: Recommend Deletion
The Dark Side reviewed this 8 hours ago: Recommend Deletion
Danu reviewed this 11 hours ago: Recommend Deletion
John Rennie reviewed this 11 hours ago: Delete
user36790 reviewed this 11 hours ago: Recommend Deletion
18:56
@Slereah That really didn't add anything (of value).
@Slereah I never was.
19:44
@JohnDuffield I can't see the review.
@FenderLesPaul what?
@JohnDuffield Well...at least I gave you a reason for my vote to delete. If you didn't/can't see them, I have it saved.
@0celo7 : can you see this?
I can't see your vote to delete.
@JohnDuffield Yes, I can.
@JohnDuffield Oh, maybe I misclicked? In any case, did you read my list of reasons?
@0celo7 : yes I read your list of reasons. Picking one at random: curved space and curved spacetime might be used interchangeably in your conversations, but they are not the same. The distinction is crucial. All the votes in the world won't make them the same.
@JohnDuffield It's completely irrelevant.
In particular, it's irrelevant for any answer to my question.
@JohnDuffield Suppose I replace "curved space" with "curved spacetime" and "follows a null geodesic" with "the integral curves of the wave vector are null geodesics"
@JohnDuffield What is the remaining content of the deleted post?
20:17
what is the correct way to take the exterior derivative of the product of two functions? i.e $d(fg)=?$ Here is my attempt, but I am left with a result that doesn't look too good imo... $d(fg)=d(f\wedge g)=dg\wedge f+g\wedge df $
@0celo7 are you free rn?
i don't understand some pde derivations.
@0celo7 : just carry on as you were. Ignore my answer. Don't worry about it. Now if you'll excuse me I've got work to do.
20:34
@AngusTheMan Hint: $\mathrm{d}f\wedge g=g\cdot\mathrm{d}f$
@0celo7 There we go! :D Cheers!
@AngusTheMan Or just use $\mathrm{d}(f)=\operatorname{grad}(f)$
then use the standard product rule ;)
@0celo7 it is all so obvious now! Using your identity I now have $(fdg-gdf)\wedge dt$ which all appears to work out :D
20:51
@AngusTheMan Note: $g\wedge \mathrm{d}f\ne -\mathrm{d}f\wedge g$
IIRC $d(fg) = f \wedge dg + (-1)^p g \wedge df$ for pee forms
@Slereah Yes and $p=0$ for functions.
I remember it because it is so grossly offensive
A derivative that doesn't obey the Leibniz rule
@Slereah yes
it's an anti-derivation
v. shameful
Even the boolean derivative obey the Leibniz rule
21:04
IIRC $\mathrm{d}$ is the unique degree 1 antiderivation on diff forms that is also the gradient on functions
wtf is a boolean derivative
@Slereah sounds made up
It's a concept used in like
CPU design
The boolean derivative tells you if changing the value from false to true will change the expression's truth value
21:35
@Slereah is there a boolean covariant derivative
only if you have curved logic
Interesting...
@Slereah is there a logical manifold or something
Hm
Obviously not with classical logic
But I wonder what fuzzy logic looks like
Obviously?
Classical logic is discrete
21:46
Let $M$ be some diff. manifold. Then take, locally, for $U$ an open subset of $M$, $M\cup\{0,1\}$
piece together these bits
what is $M$ supposed to be tho
and you get a fiber bundle $\mathscr{L}$, the logic bundle
@Slereah some space
go from there
the boolean derivative is in the space of truth values, though
@Slereah idgaf
 
1 hour later…
22:59
Hai
Meta-physics! Very appropriate :D
Hmm, is there a dramatic mask take-off in the new SW, @Huy
and it's a spoiler if you see it
@Huy pretty sure I know what happens in the new SW
I don't see the thrust of that thread @Danu
typically people don't take QFT until after having taken QM sooo
I guess I don't understand why Anna has that concern
@FenderLesPaul I can attest to this claim.
=(
23:21
@0celo7 I hope you don't mind me asking you one last question :p if I want to take $d(fg)$ where $f$ is a function and $g=dx/dy$ how do I do this ?
@AngusTheMan Not sure what you mean...what's the manifold?
So in classical mechanics I want to take the exterior derivative of $p\dot q$
I have the dt on the bottom of the dot q which I don't know how to deal with
@Danu : anna asked for opinions, I gave mine. And it looks to me as if you're touting for serial downvotes. If you've got an issue with my answer point it out and let's talk about it. If you dare.
ok that $\mathrm{d}t$ is not the same as the 1-form $\mathrm{d}t$
23:24
@JohnDuffield I don't think he dares.
@AngusTheMan Do you know what the result is
I think (emphasis think) it is $\dot qdp-\dot pdq$
but I'm not sure
It would make sense if it was
Like a hell of a lot of sense, but I can't make it equal that
@0celo7 : no I don't either. Because I back up what I say with good references. I don't just make stuff up.
@0celo7 when are you coming back to Murrica
to freedom kingdom
well we see that $\mathrm{d}(p\dot q)=\dot q\mathrm{d}p+p\mathrm{d}\dot q$
hmm, that does not match your result in the slightest
I know
:(
23:27
@AngusTheMan are you trying to calculate $\mathrm{d}H$ or something
sort of... its just random thinking at the minute
@AngusTheMan well...
is $\dot q$ even a scalar
it might be a tangent vector
@FenderLesPaul why?
ah thats true ...
well then $H=p\dot q-L$ doesn't make sense
actually since $p$ is a cotangent vector...that might make sense
@AngusTheMan sorry, I don't know right now.
have you checked in Arnold? that's where I would look
ah no problem :) Its just me working through stuff, cheers for looking though :)
I have a copy upstairs I will go check !
I checked foundations of mechanics but noti=hing I could see
23:32
maybe chap 8 or 9...
@AngusTheMan I think someone, somewhere, has calculated $\mathrm{d}H$...so you just have to find that calculation :)
23:47
@FenderLesPaul are you knowledgable in fluid mechanics? D:
23:57
@0celo7 can't a guy ask?
@0537 not beyond the basics, sorry bud
@0537 what do you need

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