8:00 PM
@Slereah You have to fix that Lorentzian means one negative eigenvalue. That there is an equivalence to metrics with one positive eigenvalue is irrelevant.

Is it because they are NOT EQUIVALENT
(DUN DUN DUUUN)

@0celo7 Are you sure the metric's determinant is negative?

@Bass ACM agrees with me, so yes.
The metric always has one negative and three positive eigenvalues, so the determinant is negative.

@0celo7 Aren't Riemannian manifolds special cases of pseudo-Riemannian manifolds?

For a Lorentzian manifold the determinant is negative.

8:04 PM
@0celo7 Who the heck says the metric has one negative and three positive eigenvalues?

And spacetime is a Lorentzian manifold.
@Bass Sylvester.

@0celo7 Who the heck says my question deals with Lorentzian manifolds?
@0celo7 Who the heck says we're dealing with spacetimes?

fml
it says pseudo-Riemannian
@Bass Ok it's still easy
lucky for you I was proving it for det g < 0, not in 4 dimensions

@0celo7 pseudo-Riemannian could mean Riemannian

@Bass yeah
My proof works for everything
just you wait
all I have to do is turn a negative into an ||
Hmm, $$\frac{\mathrm{d}}{\mathrm{d}x}\lvert f(x)\rvert = \operatorname{sgn}[f(x)]f'(x)$$
That look right to everyone?

8:10 PM
Sounds aboot right
Although it is defined at $0$
at $f(x) = 0$

$0\notin\operatorname{im}f$
how about $sign[f(x)]|f(x)|=f(x)$

@0celo7 I think I basically get his answer, but I'm wondering about some things. First, the indices seem wrong. On the LHS he has the indices $\mu\nu$, on the RHS they're just dummy summation indices. Second, I don't understand what he means by You can admit the penultimate equality.

@Bass working on it
he is not rigorous with his signs!
I am
Or...not
Wow these signs are confusing

Take your time. Didn't want to hurry you.

I'm not hurring
But you might never get an answer

8:19 PM
np

Alright, I have to work pointwise and diagonalize so all the signs cancel
OH the trace is basis indpendent

8:48 PM
@Bass Please alert me of any errors or points that need clarification

@0celo7 Thanks! Gonna take a look at it

@0celo7 : That has everything to do with your question. You asked if light waves follow null geodesic paths in General Relativity, and I said No. The null geodesic maps the path of the light wave through space, but light doesn't follow this path. Instead light curves because "the speed of light is spatially variable". And Einstein said "Eine Krümmung der Lichtstrahlen kann nämlich nur dann eintreten, wenn die Ausbreitungsgeschwindigkeit des Lichtes mit dem Orte variiert".

@JohnDuffield Can you prove that the null geodesic maps the path through space?
If you can, I'd be very impressed/grateful
@Bass you might want me to define my $\delta$ more carefully...it's not immediately obvious that it's the same as yours
or maybe it is

@0celo7 : I don't know if I could "prove" it. But that's essentially the 1919 test.

@JohnDuffield that's an approximation

9:02 PM
Why do you even bother

I know that within that approximation it's a null geodesic
My question is specifically the case when one cannot employ the approximations used by Einstein
@Slereah I don't know

Did you try seeing the path of the center of a wavepacket in a curved spacetime

I could not find one in the night sky unfortunately

It's not too general but gaussian wavepackets do form a basis
Might be interesting to compute

do it

9:08 PM
@0celo7 : I told you the answer is no and explained why. But you didn't like my answer, you were happy it got deleted, and you didn't follow up on my references. Since you got no other answers, might I suggest you move on?

No, you didn't explain why.
You attacked my wording.
If I had asked the question "does a null geodesic map the path through space/spacetime" what would you have answered?

@0celo : I didn't attack your wording, I attempted to clarify.
@0celo7 : I would have made some distinction between space and spacetime.
But like I said, move on.

@JohnDuffield ignore that part

Anyway, I've got to go. Bye.

9:25 PM
1

While checking out some of the user profiles on Physics I have noticed some of the high ranking users with over 20k rep have asked no questions. Why?

Accidentally went into the photon diffraction room
took me longer than I would like to figure out why there was an incredibly long conversation between the same two people on photon diffraction

@ACuriousMind what are you playing these days

@ACuriousMind Take what back :p

15 hours ago, by FenderLesPaul
eh it's not that great anyways

oh
:p
I take it back
it's pretty great I was just in a dismal mood
because it started snowing

9:35 PM
:26763066 nothing to alert yourselves over :)

@0celo7 Various things. Currently either AoE2 with my roommate or Sunless Sea

it was all @ACuriousMind, ban HIM

screams in panic

@FenderLesPaul B-b-b-b-but snow is great!

faints

9:36 PM
@ACuriousMind damn Germans
ACM is a confirmed viking

haha

We haven't had snow this winter yet :(

@ACuriousMind it kept raining when I was there
rain/mist
it was sunny in Bavaria
mist as in English mist, not Mist

It's been raining here constantly during the last week

Sierra mist

9:37 PM
The sound of raindrops on the roof... <3

Man... the new design is shit :\

ACM psych patient confirmed
@Danu new desgin?

on PSE

I thought we're currently in a transition period
Strange...

@Danu This isn't the final one - we've got neither the new profile nor the serif font
(removed)

9:41 PM
ok there is something fishy with my avatar bar

people keep coming in

I'm already here for a while ;)

user54412
The large amounts of whitespace and Arial for everything -- it's currently like a middle school essay that's struggling to meet the minimum length while trying to not be taken too seriously.

then when I refresh they're gone

9:42 PM
@ChrisWhite We've had it on HSM for a while now, and it sucks!
@ACuriousMind What makes you say that?

@0celo7 I have a script on that loads deleted messages - all that happened there was your message changed colour from black to red

@ACuriousMind HSM has had this for a while; what guarantees that we get something custom-made?
@ArtOfCode Neat.

@Danu you're not a mod

@0celo7 lol

what's so funny

9:43 PM
@Danu 1. We don't have the new profile. That's a fact, and that's why I say it ;) 2. meta.physics.stackexchange.com/a/6792/50583

@0celo7 Does that do? :P

does what do

...nothing

@0celo7 He deleted your "prove it", silly :P

@ACuriousMind I recall point 2

9:44 PM
if it's nothign why did you say it
@ACuriousMind uh, proof?

@0celo7 Trying to get "funny"? :P

it could have been @ArtOfCode
@Danu no

@0celo7 read the message history, it'll tell you who deleted it

btw @ACuriousMind is "generalized amalgamated product" a thing, or just something Leeb likes to say? :P

@ArtOfCode darn

9:45 PM
#rekt ocelot

:D
no love

@Danu We called it free amalgamated product, so at least the amalgamated part is a thing ;)

don't worry, be happy

ffs

9:46 PM
@ACuriousMind The thing that is not free but just has the relations that preserve the "homomorphism of directed system" property (to be used to solve the SVK mapping problem)?

still blocked in America

1 min ago, by 0celo7
no love
Got me listening to this again :3
@ACuriousMind ...or is this all Leeb-speak? :P

@Danu Uhhhhh...the thing that appears in the Seifert-van Kampen theorem. I have no idea what a "homomorphism of directed system" property is, but I guess it's a weird way of saying the amalgamated product is the pushout in the category of groups.

@ACuriousMind I wish we could get all three at once.

@ACuriousMind Okay, let me explain the terminology.

9:50 PM
Hey, do stealth pings still work?
damn

maybe

clearly not

well I did get pinged

A directed system in this sense is $(G_\alpha)_{\alpha\in A}$ where $(A,\prec)$ is partially ordered and $\alpha\prec \alpha'\implies \exists \varphi_{\alpha'\alpha}:G_\alpha\to G_{\alpha'}$ (a homomorphism)

aye, but you're not meant to be able to see that not-a-link

user54412
9:51 PM
@ArtOfCode Try random non-printing unicode?

@Danu I know what a directed system is :P

@ACuriousMind oh

@ChrisWhite such as?

A homomorphism of the directed system is a family $\psi_\alpha:G_\alpha\to H$ that agrees with the $\varphi_{\alpha'\alpha}$'s
where $H$ is an arbitrary group

@Danu every time I watch that video I think "that's what you get for wearing that hat"

9:52 PM
(I guess this idea of directed system makes sense in different categories too)
@0celo7 The "what is love"?

@Danu And with "agrees" you mean $\phi_{\beta\alpha}\circ \psi_\alpha = \psi_\beta$?

@Danu "no love"

@Danu Yes, the concept of directed system is quite general

@ArtOfCode still no

@0celo7 hey, that one works

9:54 PM
@0celo7 Accepted your answer. The crucial part is just one line in it, the $\delta \,tr\log\tilde M=tr\,\delta\log\tilde M=tr\,\tilde M^{-1}\delta\tilde M$ part, and the one thing I didn't understand was that the trace commutes with the derivative. Now everything's clear, thanks.

Now it's at least stealthy, question is - is it a ping?

@ArtOfCode nope
I didn't get a ping

@ACuriousMind $\psi_{\alpha'}\circ\varphi_{\alpha'\alpha}=\psi_\alpha$, yes

@0celo7 sigh - you get one thing right and the other one goes wrong. Maybe I should test this on myself.

The free product does not do this (unless either the ordering is trivial or the things that are smaller in the order are all trivial)

9:55 PM
@Danu Arrrgh, I always read the $\circ$ the wrong way :D

@ArtOfCode try it again

@Danu Do what

@ACuriousMind Sorry

@0celo7 did you downvote the other answer? :)

Forgot some step

9:56 PM
@0celo7 Doesn't work for me either

@Bass no

The family $\iota_\alpha:G_\alpha\to K$ where $K$ is the free product
is not a homom of the directed system because it does not agree in general
unless [see above conditions]

@Bass well now you also know that the signature of the matrix does not matter

But the thing we called the "generalized amalgamated product" does the job, by construction, because you just add the relations $\varphi_{\alpha'\alpha}(g_\alpha)g_\alpha^{-1}$
So... is this the thing you called the free am. prod?

user54412
9:58 PM
@ArtOfCode I dunno, U+200B? ->​<-

@Danu Yes, and you're just stating it is the pushout in the category of groups ;)

@0celo7 I never said anything else.

@ACuriousMind So what's the pushout?
...ah, the thing that solves the mapping problem, by def.?

@ChrisWhite would also appear to work

@Bass huh?

10:00 PM
@Danu Well, admittedly, one uses "pushout" usually for the case where you only have 2 $G_i$, but it's just another name (and the categorically/algebraically more common one) for this.

Okay.
So the pushout in a more general sense (can vary the category) is: The object so that the every homom. of the dir. sys. (to be defined in every category in a (slightly) different way) factors through the inclusions into it?

@0celo7 ? quote?

@Bass I have no clue what you're talking about

Your "generalized amalgamated product" is just a weird name for a direct limit of groups

product, not sum
(the amalgamated product is a special case, as is the free product)

10:02 PM
@0celo7 I never said anything about the signature of the matrix. Never mind, thanks for the answer.

@Danu Yes

@ACuriousMind Fuck yeah, I think I'm getting some intuition for this
Looking at the diagram on the wiki page, I see it dun' have to be inclusions per se; how can that be?
Or are they just not saying that it should be inclusions

@Danu (Rigorously you should say "factors uniquely")

@ACuriousMind Yeah duh
Errything unique in category theory, no? :P

@Danu They are "inclusions", but they don't need to be inclusions in the sense of being monos

10:06 PM
@ACuriousMind monos=monomorphisms?
So in what sense are they "inclusions"?

@Danu Yes. They can fail to be injective if the subgroup you have to divide out is large enough.
@Danu Because, at least for most algebraic objects like groups, they descend from the inclusions into the free product (which is categorially the coproduct in the category of groups), since you compute the amalgamated product as a quotient of the free product.

@ACuriousMind nice, thanks.
@ACuriousMind This is a nice statement:
in Mathematics, 4 mins ago, by Balarka Sen
Yep. SVK says $\pi_1$ preserves pushouts.
(you of course already know it)

@Danu Yep, that's the elegant way of stating it ;)

Help!
What do I do if the name of a book on the catalogues doesn't match the cover?
Case in point

@Bass well unless you do exactly what I did you only prove it for the Riemannian case or semi-Riemannian for an even number of minuses
does Nakahara bother with the absolute value?

10:17 PM
@EmilioPisanty What do you do about what?

@Danu How do I reference the thing?

I guess
Just go with the flow, I guess, most people who cite it use the catalogue name
bizarre, though

Quite strange...

Even worse, the chapter I want to cite has a footer with the cover title

10:22 PM
@0celo7 nope

@Bass you asked for general pseudo-Riemannian and that's what mine is

Anyone know of a reference that talks about what the position space analogue of the Regge limit of scattering cross sections is?

@MarkMitchison is here!

@0celo7 Yes I know. I don't understand why you're asking about absolute value, even number of minuses and signature of the matrix. But never mind, I got the answer I was looking for.

Breaking news!
we have converted to the new site format!
@MarkMitchison if you are ignoring me you should at least tell me...
@MarkMitchison it does not normalize automatically...

10:48 PM
@Bass that other dude's trying to pick a fight lol
@Bass ok, I'm being super pedantic: your (1) is not true for all pseudo-Riemannian metrics, only those with an even number of negatives
I proved a generalized version of it

11:16 PM
@DanielSank here is the PEP8 version:
options = Options(nsteps=1000000, atol=1e-5)
controls the accuracy of the result along with:
num_intermediate_state = 100
^ this is probably self-explanatory about what it does
of course this is the end time and therefore controls how long it takes:
endtime = 100
here is the result for time of 2000 fs or 2 ps:
with the following specs:
#OK - lets try evolving the system over time and getting the number you want out ...
starttime = 0 # this is effectively just a label - mesolve alwasys starts from rho0 - it's just saying what we're going to call the time at t0
endtime = 2000 # Arbitrary - this solves with the options above (max 1 million iterations to converge - tolerance 1e-10)
num_intermediate_state = 1000
and same options...
I have noticed this looks like the site 2 population without trapping and loss.
i wonder why, especially since the trapping is accounted for in the hamiltonian...
please ping me if you have questions...
Correction - "almost converted":
1 hour ago, by TanMath
we have converted to the new site format!

11:36 PM
@TanMath Good job. Is the code on github such that I can fork it and try it out myself?
Oh, I see that you still haven't written a simple test case.
Definitely do that next.