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01:00 - 21:0021:00 - 00:00

1:08 AM
@ACuriousMind what kind of shoes do you wear
 
@0celo7 Answering what you asked me before: New Balance, Oxfords , and these
 
1:46 AM
I wear these shoes.
the look more blue in real life.
 
2:31 AM
Hey there, gnite
Can someone help me with an exercise? I have te answer but I cant get to it by my own
 
@Chinatsu-creepy-chan Don't ask to ask a question. Just ask it
 
Ok then.
I have to find space, given that $F(v) = -Cv^2 = m\frac{dv,dt}$
IDK how to solve the diff eq and playing with the infinitesimals didn't help much. :/
$F(v) = -C v^2 = m \frac{dv}{dt}$
 
3:08 AM
@Chinatsu-creepy-chan that equation is separable...
 
@0celo7 ikr. but i get something like $ln\,t$ in one of the sides
which doesn't look right heh
 
let's see...it's $mv'=-cv^2$?
 
So that's $$\frac{\mathrm{d}v}{v^2}=-\frac{c}{m}\mathrm dt$$
Now $$\int\frac{\mathrm{d}v}{v^2}=-\frac{1}{v}$$
I think
So you get $$-\frac{1}{v}=-\frac{c}{m}t+k$$
 
the exercise actually asked to prove (using the conditions) that $x - x_{0} = (m/c)\,ln\,(v_{0}/v)$
 
3:13 AM
what are the conditions?
 
21 mins ago, by Chinatsu-creepy-chan
$F(v) = -C v^2 = m \frac{dv}{dt}$
 
@Chinatsu-creepy-chan ok, now determine $k$
then solve for $v$ and integrate wrt. $t$.
That gives you $x$, and if you do the definite integral $\int_0^t$ you get $x-x_0$ and your expression should fall out
$k=-\frac{1}{v_0}$ btw
 
@0celo7 how'd you get that?
 
@Chinatsu-creepy-chan set $t=0$ on both sides.
 
@0celo7 I'm so dumb...
 
3:16 AM
You get $$-\frac{1}{v_0}=-\frac{c}{m}\cdot 0+k$$
so you have to do an integral
 
right. then $ln\,v = -(c/m) t^{2} - \frac{1}{v_0}t$
 
umm where did that come from
 
(I hate LaTeX...)
 
no you don't get $\ln v$
 
oh
yeah, I don't lol
 
3:19 AM
You get $$v(t)=\frac{1}{ct/m-k}$$
Then you have $$x-x_0=\int_0^t\frac{\mathrm dt}{ct/m-k}$$
 
hmm...
I also got, using $F = ma$, that
 
what does that integral actually give?
it's a $u$-sub integral
but I'm not good at integrals
 
$x - x_0 = (m/c)\int_0^{t} \sqrt{a}\,dt$
 
oh god where did that come from
 
that's not the result of the previous one lol
 
3:24 AM
well I don't see where that came from
 
the last one i posted is wrong too but because i remembered it
the actual one is from $v^2 = (-m/C)a$ which means that $v = \sqrt{(-m/C)a}$
integrate and you'll get that
$x - x_0 = \sqrt{(-m/C)}\int_0^t \sqrt{a}\, \mathrm{dt}$
 
oh what are the log rules?
$\ln x-\ln y=\ln x/y$?
 
yay I was right
 
3:29 AM
that integral evaluates to $$\frac{m}{c}\ln (\frac{k-ct/m}{k})$$
The numerator is $-1/v$ and the denominator is $-1/v_0$ so their ratio is $v_0/v$
 
And you get $x-x_0=(m/c)\ln(v_0/v)$
 
<3 thank youuu
 
np
 
 
2 hours later…
user116211
5:30 AM
Yes, thanks, now all clear; $\Huge\textbf{THANK YOU SO MUCH!!!!!}$ — user210673 Jan 26 '15 at 17:22
 
user116211
Hmm...
 
9:37 AM
@0celo7 Um...the kind without distinguishing features?
They're simple leather shoes (not the kind you'd wear to a suit), no idea how one calls that.
 
 
2 hours later…
11:47 AM
Problem of being a perfectionist: Won't answer a question unless the answer is a perfect one
 
@Secret You do know that you can post links to questions here, right? Why would you choose to post a screenshot over a clickable link?
 
because I sometimes annotate. Thus it became a habit of me to screencap
Will be mindful of this in the future
http://physics.stackexchange.com/questions/258578/is-it-possible-that-propagation-of-acustic-waves-leads-to-emission-of-radiation
There
 
12:25 PM
hey
 
12:40 PM
hey
 
hey
 
user116211
Hey @DanielSank; I've edited your posts at the meta requesting only to upvote the post if they deem that format apt for the session; that would make it a poll-style post and easier to decide; however, that's your post; so if you are not pleased, you can rollback the edits :))
 
user116211
1:11 PM
Why did this post come in review?
 
Fellows
If I know what a function should be asymptotic to
Can I write a generic form of it
Like
$f(x) \asymp g(x) \rightarrow f(x) = a(x) g(x) + b(x)$
With $\lim a(x) = 1,\ \lim b(x) = 0$
Or something
To use as an ansatz
 
@Slereah What? Sure, for any $a,g$ you can write $f = ag + b$ with $b = f-ag$.
 
Well the thing is, I know $g$
 
And if you pick $a\to 1$, then you have $ag \to g$, so $b = f- ag \to g - g = 0$
 
1:21 PM
But I can only guess at a decent form for $f$
 
That's not an "ansatz", it always work for any choice of $a$.
 
So I'm trying to find a decent form for $f$
Based on physics wizardry
 
"decent form" is not a particularly well-defined term
What do you know about f?
 
Well it's several functions, actually
The metric components for a wormhole
In bipolar coordinates
So at $\xi \rightarrow \infty$, it should reduce to flat space bipolar coordinates
Then there's a bunch of conditions for wormhole stuff, I suppose
No horizon
Minimum throat radius
etc
So I'm thinking of currently just applying $a(x) g(x) + b(x)$ for each component, and find some amenable functions
Like the exact form of the solution doesn't matter
 
@Slereah You can to that for any function, as I said.
 
1:25 PM
Just that it should obey those properties
And ideally maybe not be too difficult, analytically
 
If you can't find an analytical solution for something in the literature, it's rather unlikely you will find one on your own :P
 
Also maybe $u$ should be a Killing vector to simplify things
But that's a big maybe
I'm not sure it makes much sense here
True
Maybe I should go back to the "simpler" version
The $S^2 \# T^2$ wormhole
Since it is just $T^2$
Much simpler to deal with
much more easy to make symmetric, too
$S^2 + T^2$ isn't too hard to deal with for some basic results, but
It's hard to find much interesting to say about it without modifying it in a way that makes it hard to solve
I think my problem is that I have a hard time doing like
Analysis things
Without solving
Just be able to say properties of a PDE without solving it
I'm missing a few theorems on that
what are some good important theorems on like
Bounds of a function obeying some differential equation?
 
@Slereah You mean stuff like Schauder estimates? (I don't know this stuff, I'm just googling here :P )
 
The schnauzer estimate, yes
Let's see if it's good
>elliptic partial differential equations
Elliptic
No thanks!
Could be interesting, though
 
2:31 PM
@ACuriousMind how is Wick rotation rigorously defined
 
@0celo7 $t\mapsto -\mathrm{i}t$ :P
You'll have to be more specific about what you want to know for me to give a better answer
 
Isn't Wick rotation just analytic continuation
Or something
 
@Slereah Well, one part of it is just considering all expressions with Euclidean instead of Minkowski metric, another part is analytically continuing the Euclidean results back to Minkowski space. Both might be called "Wick rotation".
 
2:46 PM
@ACuriousMind why do you put {} when you don't need them
 
what?
 
There is no what
You could do \mathrm it and it works just was well.
@ACuriousMind is there a canonical way of turnin a Lorentz metric into a Riemannian one
 
@0celo7 I also don't write $\sin\theta$ when I mean $\sin(\theta)$. I hate ambiguous brackets
@0celo7 If you're trying to ask whether you can Wick rotate GR, then the answer is no, at least not in a way that would be apparent from the SR procedure.
I also have the feeling we discussed this before
 
I can't remember yesterday, much less anything before my long ass ban
 
Ah, yes:
Feb 22 '15 at 23:36, by 0celo7
@ACuriousMind In learning topics in QFT such as CFT, we Wick rotate. Why can't we Wick rotate in general?
Feb 22 '15 at 23:49, by Chris White
If you try Wick rotating in GR to change the coefficient in front of $dt^2$ from negative to positive, what, pray tell, are you going to do to the coefficient in front of $dt\ d\phi$?
It was only 1 year and three months ago!
 
2:55 PM
Holy fuck that was a long time ago
Also I don't even know what CFT means
I used to be smart
 
@0celo7 Conformal Field Theory?
 
@ACuriousMind hmm, what's that
 
I can't tell whether you're serious :P
You read books that contained it! :P
 
You can't Wick rotate a Lorentz metric and hope to get a euclidian one
You will just get a complex metric
Most of the time
 
@Slereah How are you even defining "Wick rotating" there?
 
3:03 PM
Hm, what's a good example
Well, I can just play it fast and loose and say "pick a timelike coordinate and do the Wick"
 
Do you just pick a random timelike coordinate and do $t\mapsto-\mathrm{i}t$? That doesn't seem a particularly well-defined operation
 
I think any metric with cross timelike terms will make it fuck up basically
$g_{xt}(x) dx dt \rightarrow i g_{xt}(x) dx dt$
If we had to wait to have well defined operation before using them we'd never do any physics :p
 
But what physical motivation do you have to pick one timelike coordainte over the other?
 
Well
 
The whole point of GR is kinda that none is better than the others, isn't it? :P
 
3:07 PM
Hopefully
It would be independant of that choice
But that is probably too hopeful
 
I'm 99.9% sure it won't.
 
The idea behind euclidian gravity was that, once integrated over, there wouldn't be any difference
I don't think it panned out all that well
 
@ACuriousMind meh
 
Then I'd be more on boat with treating GR as a special relativistic field theory ala Pauli-Fierz and Wick rotating that.
 
The whole point of GR is the equivalence principle
@ACuriousMind well if the spacetime has a timelike killing vector you can rotate that coordinate
@ACuriousMind I couldn't tell you anything about it
 
3:10 PM
I think that if you have a static stationary spacetime, you can wick rotate it
 
And then you run into the trouble what if your manifold isn't covered by a single chart...I think there's no way around that, the Wick rotation just doesn't glue
 
Without problem
 
I actually think when one computes gravitational anomalies in SUGRA theories and such we do Wick rotate the SR field theory version.
 
hm, sugar theory
Static stationary is just $$ds^2 = - a(x) dt^2 + h_{ab}(x)dx^a dx^b$$
Pretty sure you can wick rotate it and the result is independant from the foliation
 
@ACuriousMind real manifolds have one chart
 
3:19 PM
@0celo7 Then most manifolds are unreal, I guess :P
 
@ACuriousMind Air Force 1 or Air Jordan 1? (both low)
white, of course
 
@0celo7 what?
 
I'm going shoe shopping
I need white shoes
 
I wear my shoes till they get a hole in them. Then I buy another pair. I don't care for shoes, at all.
 
whew lad
is that some kind of fashion statement?
 
@0celo7 It's more a statement that I'm the last person you should ask for fashion advice on shoes, I guess
 
Get these shoes
You'll be the coolest kid
 
 
my brother used to wear those
@Slereah no
@Slereah no?
 
Trust me
I'm a shoe expert
A shoeologist
 
3:27 PM
@Slereah shoe fetishist?
 
@MAFIA36790 I like it in principle buy unfortunately the UI subtracts the down votes from the up votes to compute the number it displays.
 
Depends, do you think all gynecologists are in it for the vagina
 
Are you suggesting we just ignore the down votes and use our superpowers to inspect the number of up votes?
 
@Slereah Where did you get your degree?
 
Foot Locker
 
3:30 PM
Foot Locker is ok
Champs and Finish Line are better
@ACuriousMind going to be building a PC this summer
putting in a new GTX1080
can it run TW3?
 
My GTX660 handles that just fine, so most probably yes
 
gut
And I'll replay Fallout 4 with good graphics
 
why would you replay Fallout 4 tho
It's boring
 
4:24 PM
@Slereah I thought it was good
A lot better than New Vegas
 
I dunno
It felt very short and samey
There was like three towns, most missions were just to go to a place and kill everything
Only 4 factions to apply to
Might get better with DLCs but I dunno
 
4:39 PM
I wonder how easy it would be to find a CTC solution on $\Bbb R \times T^2$
Just a timeshift between the two mouthes I guess
Bigger than their separation
Maybe pick some promising looking closed curve on it and find a metric that makes it causal
 
 
1 hour later…
5:55 PM
May someone please tell me what does this sentence mean? "I can't side with anybody"
 
@lucas To "side" with someone means to support them. That sentence means that the person saying it is not of the same opinion as anyone else in that discussion.
 
@ACuriousMind Thank you.
 
6:31 PM
@Slereah what does that thing look like
 
A shape that you can't represent in 3D
 
proof?
 
Well it's 3D and it's not $\Bbb R^3$
 
@ACuriousMind I got the J 1s
@Slereah So?
 
I don't think you can immerse a 3D manifold in $R^3$ outside of itself
It's just a foliation with a torus as a leaf
 
6:34 PM
@0celo7 Now you can add them to get all $n\in\mathbb{N}$! (I have no idea what you're talking about)
 
@ACuriousMind Jordans
Goes through $n=23$
 
@Slereah I would get those in the gray but they're expensive
 
I do the same thing every time I need shoes
I don't buy any
Then after a while my mom says that my shoes are unacceptable and she sends me new ones
 
good strategy.
 
6:40 PM
@dmckee: If I delete the solution and maintain the figure, is it OK?
@dmckee: Can I do that?
 
@lucas He'll answer when he sees the ping. No need to keep pinging.
Online chat is an asynchronous communication protocol.
2
 
@DanielSank Dunno, chat is a synchronous protocol in my book.
 
@DanielSank Thanks because of teaching:-)
 
Wow, I think that's the fourth time someone has starred me saying that chat is an asynchronous communication protocol.
 
If you remember to ping one of the mods in a couple of weeks we'll undelete the post for you in its current state. This policy is meant to allow you to get credit for the work you put into that and still make it harder for askers to benefit from asking about solutions to exercises (which we don't want to be what this site is about). I don't really approve and would rather that answering such questions was equally fruitless, but I've been out-voted.
 
6:47 PM
@0celo7 Sometimes, it's also more of a miscommunication protocol ;P
2
 
@0celo7 That's because you don't understand programming ;P
 
Of course, I understand the desire to help someone who is struggling. I get it myself. But this is a tragedy of the commons issue.
 
@DanielSank Wrong.
 
@DanielSank Indeed, it is the fourth
 
Even though every person asking such questions has no ill will and does little enough damage to the site to be ignored, if we answer such questions that will become the bulk of what the site does.
 
6:48 PM
It's basically an ancient tradition by now
 
If anyone cares, the new Pokemon Mystery Dungeon game isn't that good
 
@ACuriousMind ^^
 
They changed the formula and it's not an improvement
 
user54412
surprisingly, @DanielSank has only said "asynchronous" 12 times in this chat out of the 32 times is has been said. Then again, most of the others are us quoting him.
 
@ChrisWhite That's surprising?
 
6:51 PM
@dmckee Isn't it better that Mods or high rep users prevent these questions before they be visible for every one?
 
user54412
@DanielSank I take it back. Nothing about this room surprises me.
 
@lucas Nah, Stack Exchange has shown that with enough dedicated community members, it works to have everyone self-govern.
In fact, in my estimation the self-governance style of Stack Exchange breeds better users.
@ChrisWhite I expelled air through my nose at a higher than usual rate for a short amount of time, and felt mildly amused.
 
@lucas Do you really want to wait until someone reviews your question before any users get a chance to answer you?
 
^ Also that
@lucas what's with the 1984 quote?
 
@dmckee I don't. But when my answer is deleted, I get uncomfortable because of my time and effort.
 
6:55 PM
@lucas Someone deleted your answer?
That usually only happens if it's offensive.
 
@DanielSank or when it is an explicit answer to a homework question.
 
@ACuriousMind Ah. I see now.
I didn't realize that was site policy.
 
Yeah. I think that why the rule that you can have in undeleted eventually. But we would really rather that you learned to not answer questions which would result in your answer being deleted.
 
This is probably good.
 
@ChrisWhite Is that a challenge?
 
6:57 PM
@DanielSank It's buried in the usual homework policy post and is just as open to discussion as the rest of the policy.
 
@DanielSank What do you think about this quote?
 
If you think it should be changed.
 
@lucas What do I think about it? Uh, not much I suppose.
 
user54412
@ACuriousMind Are you threatening more bestiality trivia? :p
 
@ChrisWhite o_O
 
6:59 PM
lol
maybe
 
@DanielSank Have you read that book?
 
::winks suggestively::
 
Yo @dmckee, is there a way to make special chat rooms without the usual rep requirement?
@lucas Yes.
 
@ChrisWhite huh?
Where was I for this
Link pls
 
@0celo7 Use search.
It's easy.
 
7:00 PM
what am I searching for
 
@DanielSank So, why do you ask about that quote? If you have read the book, you certainly know its mean.
 
@lucas I wonder why you put in on your profile.
 
To subvert the latter two I suppose you ask a higher rep user to help you out.
 
@dmckee I'm interested in whether or not to bother inviting my friends if I do an AMA.
 
7:02 PM
Or, I suppose you could create a pointless comment thread so long the system automatically moves them to chat. But I didn't say that.
 
@dmckee Interesting...
 
@DanielSank Ah, so you want to subvert the "talk in chat" limit. I don't think there is a way to do that.
 
@dmckee ok
 
@DanielSank Because I believe it as George Orwell.
 
Yes, the question is how to allow folks (both guests and speakers) to participate in the AMA if they don't have teh repz.
For example, if I were to get some bigshot physicists to show interest but they don't have an account, what to do?
 
7:04 PM
I think that would be something to bring up on the mother meta.
 
@dmckee I see.
 
It's not a use case that was originally envisioned.
 
@dmckee Indeed not.
 
user54412
@DanielSank Yeah -- currently they could listen but not speak.
 
> Even when this room is read-only or their reputation is too low, these users will be able to talk in this room.
 
user54412
7:05 PM
I suspect this was implemented to stop the network from becoming just a chat network.
 
@ChrisWhite How to listen?
42
Q: Allow temporary guest accounts in chat

mbqWhen organizing an event on chat, it is sometimes nice to invite an expert or a guest. Yet such person must go through the process of making OpenID, registering account, going to chat, setting write privilege -- this is complex and may effectively repel such person, especially if he/she is not fr...

 
@ACuriousMind I don't know what I'm searching for!
 
user54412
@Loong Oh yeah. That seems to be a mod privilege -- even room owners can't do it.
 
@ChrisWhite but you can see the option?
 
user54412
7:08 PM
@DanielSank The rooms are completely public. Try visiting in a private browser.
 
user54412
@Loong Yes. It even looks like I can fill out the form, but then it doesn't work.
 
@ChrisWhite What form?
@ChrisWhite Oh, neat.
 
user54412
@DanielSank The access page looks different for room owners:
 
user54412
 
@0celo7 Try "pig orgasm" or "octopus sperm" :P
 
user54412
7:11 PM
But it's a lie.
 
@ACuriousMind I've tried one of those.
 
user54412
@0celo7 Also elephant I think
 
Oh you meant in search
 
@0celo7 ew
 
Neither gives results.
 
7:12 PM
@ChrisWhite Is there someone to whom it is not a lie?
 
God the chat search is terrible
 
user54412
@DanielSank Real mods
 
We have unreal mods? Ghost mods?! :O
 
user54412
They should be able to add people one-by-one, which of course means those people need some sort of SE account.
 
Fadeev Popov would be proud
 
7:13 PM
@ACuriousMind The residual mod theorem allows you to integrate around them.
 
Faddeev-Popov mods are probably needed to make the chat decouple from the main site
 
@ChrisWhite Ok. Well at least that's not much work to add a guest speaker.
Cool.
 
@ChrisWhite I just tried. Seems to work.
 
 
1 hour later…
8:31 PM
Asynchronous
 
@BernardMeurer <3
 
8:46 PM
"There are as many as eight different double covers of O(p, q), for p, q ≠ 0, which correspond to the extensions of the center (which is either C2 × C2 or C4) by C2. Only two of them are pin groups—those that admit the Clifford algebra as a representation. They are called Pin(p, q) and Pin(q, p) respectively."
Whaaaat
 
@DanielSank <3
 
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