« first day (2528 days earlier)      last day (2408 days later) » 

3:00 PM
Isn't the derivative just $\alpha '(t)$? Unless I'm missing something blindingly obvious
 
No.
This is just a calculation. I will not do it for you.
 
$x'(t)^2+y'(t)^2+z'(t)^2$?
 
No, you have forgotten how to differentiate.
 
kek
 
(I hope by $x, y, z$ you mean the component functions of $\alpha$ as a map to $\Bbb R^3$)
 
3:03 PM
Yeah I did
 
Ok. What you wrote is garbage
Fix it
 
Sir Sen is so aggressive today
 
I'm so lost today
this? Yeah I gave up...
 
@CooperCape what is $$\frac{d}{dx}(f(x))^2$$
 
$2f'(x)$
 
3:05 PM
You need to learn single variable calculus before doing differential geometry of curves/surfaces, seriously.
 
That's a fair shout
Whoopsie
maaa bad
Oh wait chain rule
 
This is probably in Federer...
 
@SirCumference The SMBO only makes a detectable difference very near the centre. For more on this see Kinematics of the Nuclear Ionized Gas in the Radio Galaxy M84, which I think was one of the first SMBO detections.
As soon as you're a few kPa away the black hole mass disappears into the general mass of the galaxy.
Ah, that paper is on the Arxiv. Here's a better link
 
It's 2f(x)f'(x) right? Cause $nf(x)^{n-1}f'(x)$
 
yes it is
 
3:09 PM
and n=2
 
So what's the derivative of $\|\alpha(t)\|^2$ if you write it down in terms of it's components?
 
use Einstein notation
 
yeah fair
was about to type it all oot
$2(x_i(t)e_i)(x'_i(t)e_i)$
That seems wrong
but it would make a dot product i think?
 
I have no problem with whatever coordinates you use until you get the right answer, but if you're struggling with a simple computation stick to normal people methods. $\|\alpha(t)\|^2 = x(t)^2 + y(t)^2 + z(t)^2$.
Differentiate that, you get $2(x(t)x'(t) + y(t)y'(t) + z(t)z'(t))$.
That's $2\alpha(t) \cdot \alpha'(t)$.
But $\|\alpha(t)\|^2$ is minimized at $t = t_0$, so the derivative at that point is zero. That in turn implies $\alpha(t_0) \cdot \alpha'(t_0) = 0$.
 
that was what I was looking for...
isnt $2(x(t)x'(t) + y(t)y'(t) + z(t)z'(t))$ equivalent to $2(x_i(t)e_i)(x'_i(t)e_i)$ though? when $(i=1,2,3)$.
 
3:18 PM
You missed a dot product between the two vectors.
@0ßelö7 Ricey says disstrack comes out in saturday
 
Oopsie yeah...
 
@BalarkaSen how much calc do you need to know?
Because I'd be interested in learning but idk when I could start
 
for what?
 
diff geo
 
Hm, well, depends on what kind of diff geo you'd like to do
You definitely need to know multivariable calculus to various degree
 
3:25 PM
What like multiple integrals and partial derivatives?
or more nuanced stuff
 
inverse function theorem
 
What's that?
Just that f-1 f x = x?
 
google
 
idek what a jacobian is
rip
 
The high points in multivariable calculus are the Jacobian formalism, second derivative test, inverse/implicit function theorem and change of variables theorem in my opinion
 
3:29 PM
Does anyone actually prove the change of variables theorem for Riemann integrals?
It's legitimately easier to invent measure theory
 
That is true.
I never worked out the proof myself. It's in Ted's book
I know the basic idea
 
So I need those for Diff Geo?
 
Yeah, approximate with piecewise affine maps
 
Honestly the theorem is more important than the proof, and I'd argue that with inverse function theorem too
 
But the estimates you need in the Lebesgue case are easier
 
3:30 PM
Is there like a
 
@0ßelö7 Write, the |det Df| factor comes from the stretch factor on the affine pieces
 
Turbocourse in Diff Geo online that has bare prereqs?
 
@Phase I would say yes
 
@BalarkaSen Agreed. IVT is based on Banach's theorem but beyond that I've got no clue
 
I know the proof by heart because I spent a nontrivial amount of time understanding it
 
3:32 PM
If you go back in a time machine 150 years that might be useful
 
Basically if you have full rank $Df(x_0)$, then to prove surjectivity of $f$ in a neighborhood of $x_0$ is given by solving for $f(x) = y$ for some given $y$ in the codomain neighborhood of $f(x_0)$.
That's a fixed point of $\phi(x) = y - f(x) + x$
 
@BalarkaSen OTOH, most people prove the area formula for Hausdorff measure and then notice it's the change of variables formula in a special case. The proof of that is nontrivial, although the basic idea is the same
You need these things called Federer approximations
@BalarkaSen I don't know how people can get invested in GTM as a career. The high level ideas are nice, granted, but the details...
 
Hey @0ßelö7, check it out
6 hours ago, by Slereah
user image
B)
 
is that the one with the Visser article?
 
among other things, yes
Also Alcubierre wrote in it
quite nice
 
3:46 PM
@Slereah I'm thinking about putting Wald on the second shelf. I hardly use it anymore
 
is that the shelf of shame
 
Nah, it has lots of good books I just don't use as often
 
My shelves are sorted by topic
And within the shelf itself, by book height
 
On the main shelf (right above the desk) I have geometry/analysis on one side, GR on the other.
The lone algebra book in the middle
 
There's so many publishers and book style that a science shelf will basically never look consistent, as far as style goes
unless you get nothing but books from the same publisher but snore
 
3:49 PM
 
It's not like fiction books where you can basically get all the famous authors from the same publisher
 
all the AMS books are the same size
it's pleasing
 
it is nice
The worst books are the ones with weird proportions
 
Wald
 
Especially if it's length
They stick out awkwardly
Wald isn't the worst as far as proportions go
As far as GR goes the worst is probably Carroll
 
3:51 PM
I don't have it
oh, I do have a book where the spine is printed upside down
 
Carroll's a nice introduction but yeah, it's not really indispensible if you know GR
Though it has a nice section on QFT
But then Birrell is much better for this
 
I should learn QFT one day
 
@0ßelö7 I'm honestly surprised you haven't already learnt at least some QFT for some reason :o
 
I hope you like distributions
 
@Mithrandir24601 I know a fair amount
But it was long ago.
Renormalization group flow is what made me decide to not do physics
 
3:56 PM
@0ßelö7 Ah. Yeah, I know the feeling, although it was only 2 years ago :P
 
I figured any field that put up with that crap was not worth my time
 
There's rigorous use of renormalization if you want
I think Nguyen has paper on the topic
also this I guess
 
One of my QFT lecturers does renormalisation theory
 
if I decide to relearn QFT it would be through Folland
 
(or rather, once-QFT lecturer)
 
3:58 PM
but that's a big if because QFT is far removed from my interests
Straumann has a good book on it but it's just QED
 
QED has most of the weird shit of QFT
 
QFT would be easier to dismiss if it didn't work so damn well :P
 
Except for non-commutative gauge group stuff
 
@0ßelö7 I don't really know how I feel about it - it was so long ago, my brain's very, very dusty on it, but at the same time, it was a fair amount of time and effort put into it, so it now feels like a bit of a waste :/
 
@Slereah eh, it's also in general harder to compute stuff in QCD than in QED
 
3:59 PM
that too
QFT is quite frustrating because the scope of most QFT book is so small
nothing but scattering
 
ugh, yeah
I think condensed matter field theory books aren't quite as dominated by that
 
Well you can check out specialized books yeah
Quantum optic books also cover a bit more
It's pretty hard to find books on like bound states
I'd like a good book on QFT bound states
 
i think the condensed matter books are somewhat closer to that
 
yeah, that too
I'm interested in QFT as an idea, maybe, but everyone likes scattering
it's meh
 
bloody cross-sections
if i never have to compute another scattering cross-section again it'll be too soon
 
4:03 PM
I got the feeling in QM that everyone pretends to know what's going on when it comes to scattering
 
One of my internship was nothing but cross section calculations
 
well, i think we do know a lot about scattering if only because that's the main vehicle for studying the standard model
 
Yeah, but you know
Know one problem that's in almost no QFT book?
Fucking particle in a box
 
With the Klein paradox and whatnot
Also most QFT books don't do the hydrogen atom with a classical potential
They don't do coherent free states
Or squeezed states
They don't do the whatchamacallit method for bound states
Bethe-Salpeter or something
 
4:06 PM
you might appreciate Altland and Simons' Condensed Matter Field Theory book
 
I'll check it out
 
they do coherent state quantization
and even spin coherent states
 
I mean those aren't terribly complex things
But no, more scattering
Only two days of class left
About fucking time
Next time I think about going to a class to find a job again just kill me
 
hah, the word scattering doesn't even show up in the index of Altland and Simons
nor cross-section
 
Then how do they do light reflection on solids
that is v. much scattering
 
4:09 PM
they probably don't?
 
Book is gotten
Let's do one simple test to see if it's good
"The metric $\mathrm{diag}(1,-1,-1,-1)$"
TO THE TRASH
 
@Slereah omg why
 
That is fairly common of QFT books
I guess because $(+---)$ means that timelike quantities are positive
Which are the main interest
 
yeah actually I was thinking that
you look at particle interactions as time evolves
so better keep it the positives
 
4:20 PM
hey everyone
 
ayy
 
i'll say i do prefer the mostly-positive convention in principle, since then you only need to analytically continue time to get a Euclidean theory
 
seems like some interesting discussion happened a while ago
 
and yet when I think about spacetime intervals, my knee jerk reaction is to have timelike separations be positive and spacelike separations be negative.
 
@Semiclassical is that the complex time bullshit
how does it work
 
4:22 PM
Today in a solution I got angular velocity as magnitude+$\hat{j}$
I can't understand how angular velocity can be $ \hat{j}$
 
in special relativity, neither distances nor times are invariant quantities. however, $ds^2=dx^2+dy^2+dz^2-c^2 dt^2$ is invariant under Lorentz transformations
 
@Semiclassical you're ill
 
@Abcd angular velocity is an axial vector
 
@BalarkaSen are you taling bout proper time?
 
I don't know what I'm talking about lol
 
4:24 PM
@BalarkaSen Okay thanks. Let me search about it.
 
okay, I never heard 'bout complex time :P
 
Anonymous
@Abcd Read
 
I have seen some shit about "making the time variable complex"
 
so that you have a Euclidean metric
 
4:24 PM
ah okay
 
yeah. you analytically continue your time variable from the real axis to the imaginary axis
 
what does it mean though
 
Please tell link formula again @Blue (link with title in a message)
 
in which case you can write $t=i\tau$ with $\tau$ be real and therefore the metric becomes $ds^2=dx^2+dy^2+dz^2+c^2 d\tau^2$
 
Anonymous
4:25 PM
@Abcd What?
 
@BalarkaSen Proper time has a meaning (i.e, your time, literally), if youre interested
 
that's not what he's after
he's after imaginary time
 
Anonymous
@Abcd Use [] instead of {}
 
yeah I understood that
 
The way I understand the Minkowski metric is as a Lorentzian metric on a real 4-manifold $M$
 
4:26 PM
I was just telling it, since I'm bored and have nothing else to do now :P
 
what does it mean, mathematically, to make time a complex variable?
 
best way to get at it mathematically, I think, is to note that in QFT you'd write observables by integrating over both space and time
 
@BalarkaSen as one should
physicists do some other shit
best to ignore it
 
so you end up with an integral like $\int_{-\infty}^\infty f(t)\,dt$
 
lol
 
4:27 PM
as Hilbert said, physics is too complicated to leave to physicists
6
 
@PrathyushPoduval Tell me about it
@0sselo7 +1
 
well, if $f(t)$ is well-behaved in the complex plane, then you can deform your contour to point along the imaginary axis
 
(did you get that ping, btw?)
 
I did
 
and there are reasons why that's a good idea, even just on the grounds of making the integral easier to handle
once you've deformed it as such, you can express it as an integral over real values if you substitute $t=i\tau$
 
4:28 PM
@0ss excellent
 
so you start with an integral over real time, deform it to an integral over imaginary time, and then parametrize it appropriately
what's more, you're actually more-or-less forced to do this by the math: the initial integrals would have poles along the real axis
so you need to deform the contour to avoid them. and by analyticity, this turns out to be equivalent to integrating along the imaginary axis.
 
@Semiclassical Aha, I finally found what I needed.
 
so to compute an integral that doesn't make sense you change the integral completely
got it
physicists are crazy
 
It follows from the proof of a previous result, not the statement
wtf
 
4:31 PM
...lool
 
@BalarkaSen now you understand what I've been saying for years
 
@BalarkaSen there's other ways to come at it. for instance, you can also chalk this up to the fact that the photon has zero mass
 
in string theory you get $1+2+3+\cdots$ and that doesn't make sense, but $-1/12$ does, so let's go with that
 
so one way to make it work is to pretend that the photon does have a finite mass, and then take the limit at the end of the calculation to see what you get
and this is equivalent to changing $f(t)$ a little bit so that the poles move off the real axis.
 
...has anyone checked that the limits can be interchanged like that?
 
4:33 PM
shrug
I know such things by concept, not by rigour
 
yeah i suspect all of this would make sense if a bunch of analysts went down and rewrote string theory completely
or qft or whatever
 
there's some dreams of making sense of QFT purely by understanding analytic continuation better
Witten has done stuff in that vein, for instance
 
ugh no the thing I need still doesn't happen because I can't control the L2 norm
 
i am convinced that everything in this world is a system of PDEs and we try to extend solutions beyond their domain
 
there is something conceptual I don't get
@BalarkaSen There was a post on PSE where someone argued existence/uniqueness for a bunch of PDEs based on the fact that they represent physical systems
So just set up the physical system and there's your solution
 
4:36 PM
@BalarkaSen I'ma sorry I can't, since I don't understand it properly yet. Me explaining it may only confuse both of us :P
 
lol
well it's a cute fact though
 
@0ßelö7 Whats wrong with that??
 
not a mathematical proof but an illuminating point
 
it's a pseudovector @Blue en.wikipedia.org/wiki/Pseudovector
the name makes it sound like a "false vector"
 
@BalarkaSen Except for the fact that everything in nature is nonlinear and these PDEs are all approximations
 
4:37 PM
@PrathyushPoduval That's OK
 
maybe some other time
 
So why do we call it vector in the first place?
 
@0ßelö7 Oh speaking of I heard some sketchy thing about that if you have a nonlinear ODE $x' = f(x)$ w/ initial condition $x(0) = 0$ say and you pull a first order Taylor to get the new system $x' = Ax$ where $A$ is the derivative, if $A$ is hyperbolic then these two systems are semiconjugate or something
That sounded fun
 
I have no idea what that means
 
I'll have to pick it up from Hirsch-Smale
 
4:39 PM
Does lagrangian have a capital L?
 
@0ßelö7 semiconjugate means there's a map between the phase spaces that makes the flows of the two systems commute
$f \circ \varphi_t = \psi_t \circ f$
 
Oh, I see it gives the direction of the axis @Blue. Am I right?
 
Anonymous
@Abcd It's a vector which is invariant under inversion.
 
I Like To Capitalise Every Word So That Everyone Knows What I'm Saying Is Important Especially With lagrangian So They Pay Attention
 
Anonymous
@Abcd Right
 
Anonymous
4:40 PM
It's not the typical physics vector
 
@JohnRennie Isn't it a name? So I guess it wshould be
 
the tricky thing with vectors vs. pseudovectors is that, from the perspective of vector calculus, there's not much difference
 
@PrathyushPoduval abelian is lower case
typically
 
who decides it in the first place?
 
1 min ago, by Abcd
Oh, I see it gives the direction of the axis @Blue. Am I right?
 
4:42 PM
Yes, and I said that way back above
It's an axial vector
 
it's only when ask how they behave under transformations, specifically spatial inversions, that the distinction between vector and pseudo-vector matters
 
@JohnRennie Unless you're of the French school, yes.
 
@PrathyushPoduval I think it's just habit
 
@PrathyushPoduval If you were referring to the person Lagrange then yes it is capitalised. But L/lagrangian isn't a proper noun.
 
4:43 PM
@JohnRennie See my authoritative message.
 
I think Lagrangian looks nicer
 
Thanks balarka and blue. I didn't know anything about axial vectors till now.
 
@0ßelö7 you don't cite your sources!
Chrome agrees with you though ...
 
@JohnRennie Life experience. If you want to name something after someone, you capitalize it unless you like the way the French do it.
 
Anonymous
4:44 PM
@Abcd Unfortunately they don't teach you these things properly in class
 
Sid
@JohnRennie there is your source
 
@0ßelö7 You mean like the way french do it?
 
Yes
 
@Sid :-) Good luck citing Chrome in a scientific paper!
 
@JohnRennie Lagrangian, Riemannian, Lorentzian, Hamiltonian...
 
4:46 PM
wait
Hamiltonian isn't capitalised in English usage?
 
are you even listening
 
Sid
English is a messed up language. Nothing ever makes sense in it
 
I thought you said the French do it with capitalisation
 
I said unless you're French.
 
Yeah but I'm not
 
4:47 PM
So you capitalize it
 
Anonymous
@Sid Name a non-messed up human language
 
Oh
RIP i cant read
Maybe that's a prereq for diff geo too
 
One day conversations in this room will make sense ...
 
Sid
@Blue Latin.
 
Anonymous
@JohnRennie But on all other days it will not :P
 
Sid
4:48 PM
No one knows what it is but everyone respects you if you know it
 
@JohnRennie Yes, if we analytically continue to imaginary time.
2
 
@Blue :-)
 
I found myself absent mindedly singing Darude Sandstorm again
what is wrong with me
am i going insane
 
@BalarkaSen good track!
 
@BalarkaSen going or gone? :P
 
4:50 PM
I am become insane
 
destroyer of sleep
 
@PrathyushPoduval I was quoting Oppenheimer
 
Ah okay
@Semiclassical Now that makes sense :{
 
@Semiclassical From the infamous discord where I and Daminark live: "I am become lol, destroyer of kek"
 
4:53 PM
Inspired from past comment:
I am become physics, destroyer of maths
 
I am become Big Shaq, destroyer of quick mafs
skrrrra
boom boom
 
Mans not hot
just ketchup no sooce
ting goes boom boom
 
he protecc, he attac, but above all.... you dont know big shacc
 
@JohnRennie I agree. Maybe when these kids start paying taxes
 
4:56 PM
@BalarkaSen big shaq need to go against Suppa hot fire one day
 
::stares blankly::
what are you people talking about
 
Forget about it, @0ßelö7. It's meme-town.
 
some memes
 
La meme chose
 
@Semiclassical Ok here's another approach.
Same setup, remember that $N>0$
Suppose I have some functions such that $\|f_i\|_N=1$ for all $i$, BUT the supports are shrinking
I suppose that means $\max |f_i|$ has to blow up
Otherwise the shrinking support would kill the integral
It's then feasible that the $L^2$-norms have to blow up or go to zero, right?
It can't be a good situation for them
 
5:00 PM
that seems reasonable enough
 
@Semiclassical Yeah, so when you take the $N$-th power you get $f_i$ just large enough so that the integral is 1
But when it's taken to the second power, it's not good enough
So the shrinking support kills the integral
I'm a genius
 
so pick a point and take $f_i$ to be supported only near that point
 
work on $\Bbb R$ for simplicity
Take the supports to be $I_j$, uh, shit
what's a good way to write shrinking supports
Let $I_j$ be a symmetric interval with length $1/j$, say
So we suppose that $$\int_{I_j}f_j^N=1$$
I need to think more about the details
 
i think it should be enough to have $f_i=0$ unless $|x|<\epsilon_i$
or dist(x,x_0) if you want to have an arbitrary point
 
@Semiclassical Oh I've got some other constraints I have to work with :/
 
5:05 PM
ahh
yeah
 
I want this to be general
 
what sorts of constraints?
 
The idea is that $|f_j|^N$ "beats out" the shrinking supports, but $f_j^2$ doesn't do it
$N$-th power beats out the square in large $|f_j|$ regions
@Semiclassical These should be smooth functions that minimize a certain localized optimal Sobolev inequality.
 
5:18 PM
@Semiclassical Holder's inequality ;3
If the integrations are over the region $\Omega$, we have for $p\le q$ $$|\Omega|^{-1/p}||f||_p\le |\Omega|^{-1/q}||f||_q$$
Set $p=2, q=N$, $f=f_j$, and $\Omega=$ geodesic ball of radius $1/j$ based at some point $x$, call it $B_{1/j}(x)$
Then $$||f_j||_2\le |B_{1/j}(x)|^{1/2-1/N}$$
$1/2-1/N$ is positive, and that volume goes to zero
so $||f_j||_2\to 0$
 
 
2 hours later…
6:59 PM
@BalarkaSen Hmm. Turns out Yamabe flow needs the positive mass theorem too.
 

« first day (2528 days earlier)      last day (2408 days later) »