The h Bar

12:00 AM

@ACuriousMind Hmm...so even though phantom energy can cause a feedback loop and keep decreasing its pressure forever, it won't necessarily cause a Big Crunch?

Sigh

Bernardo Meurer

Does the Big Rip have something to do with the Big Banks?

ACuriousMind

@SirCumference Uh, you seem kinda unclear on whether you believe phantom energy leads to a rip or a crunch :P

Sir Cumference

@ACuriousMind Er, what part seemed to imply I thought it leads to a crunch?

ACuriousMind

In any case, I believe that if it's only present in "trace" amounts, it just might not affect the evolution noticably at all for at least some initial conditions

@SirCumference "so even though phantom energy can cause a feedback loop and keep decreasing its pressure forever, it won't necessarily cause a Big Crunch?"

Sir Cumference

@ACuriousMind I know, but what part of that?

ACuriousMind

12:03 AM

Asking if it won't cause X usually implies that you think it should cause X.

I have no idea what the latter part of the sentence is doing there if you don't think it leads to a crunch :P

Sir Cumference

Ugh, didn't mean to say "cause"...

Wait, no

I meant to say "Rip"

Ugh, I don't work well when I'm tired

ACuriousMind

@SirCumference Then go to bed, or do something else that's relaxing

Sir Cumference

@ACuriousMind I can't, I got work

user228700

@SirCumference U're always tired .__.

ACuriousMind

@SirCumference Well, your choice, but you should learn how far you can push yourself without losing much of your efficiency and then refuse to be pushed beyond that except for exceptional circumstances.

Sir Cumference

12:06 AM

@Kaumudi.H I am not always tir....zzzzzz

ACuriousMind

I feel you're saying you're tired everytime you're here, which is not a good sign

(I know my sleep cycle is not a shining example, but I'm rarely actually tired :P)

Sir Cumference

Eh, when I'm under a lot of pressure, it's hard to sleep :/

Howdy @DanielSank

user228700

@DanielS: Hello :-)

heather

hello again

@DanielSank hello!

Sir Cumference

@Slereah Say, I've got to ask...what's your gravatar supposed to be?

Slereah

12:17 AM

It is Leonard J. Crabs

Attorney at law

Sir Cumference

Why though?

Is this some meme?

user228700

@Slereah I thought that was you! ._.

I'm no old man

@Kaumudi.H ^

It is an old meme

12:20 AM

There is a picture of you somewhere in this chat though

user228700

@Slereah Thought u were retired .__. What the hell...

ACuriousMind

Hm, no, you removed it from the dropbox link :P

@Kaumudi.H lol

Sir Cumference

@Kaumudi.H haha

@Slereah So wait, how old are you?

If you don't mind me asking

DanielSank

@SirCumference @Kaumudi.H @heather hi.

Slereah

I am as old as time

Sir Cumference

12:25 AM

@Slereah Deep

user228700

Wtf, I just got +25 all at once.

DanielSank

@Kaumudi.H Upvote+accept.

user228700

@DanielSank No, 5 questions of mine were upvoted...

user228700

BTW, @DanielS:

user228700

16 hours ago, by Kaumudi. H

@JohnRennie An hour is too much because I am now contemplating sleeping for 5 hours between 9PM and 2AM because I am running short on time.

DanielSank

12:30 AM

@Kaumudi.H yes?

user228700

Hm? I posted a message, that is why the "BTW".

heather

so, a video is talking about gradients, and it says that if you have for example a 1:10 gradient, where for every 10m you walk you lose 1m in altitude, gradients can be represented by $d\phi=\frac{d\phi}{doc}\,dx$ where $\phi$ is height. Now, I understand that $dx$ is distance walked (change in x position) and the $d\phi$ is change in height, and the guy said the remaining term is the gradient, but I don't understand how it represents the variant, or what the $o$ and $c$ are.

DanielSank

What is $doc$?

heather

@DanielSank i have no idea.

DanielSank

Gradient roughly means "derivative".

There's a slightly richer meaning in the context of functions of multiple variables.

...but it really just means gradient.

heather

12:38 AM

okay

let me listen to the sentence where he talks about that term again

DanielSank

When you do multivariable functions, it's a bit more complex and interesting.

Sir Cumference

@Kaumudi.H Wait hold on

heather

"which we express as $d\phi$ by $dx$, the rate of change of $\phi$ with $x$"

Sir Cumference

You've never installed a userscript before, right?

heather

that's what he said to describe that term.

oh, wait...

that could be a $dx$ on the bottom there...

it looks like $oc$ but it makes more sense for it to be $dx$ so i'm going to assume it's that =)

DanielSank

12:40 AM

It's surely dx.

heather

so then, nvm, it makes sense =)

wow, that was stupid =P

user228700

Quick question! How to write lim from a to b in LaTeX?

Slereah

\lim_{a\to b}

user228700

Thanks!

DanielSank

Interesting.

I usually use \limit_{a\rightarrow b}, which is considerably harder to type.

user228700

12:53 AM

@Sir: Oh, didn't see ur message. If "install" is different from screwing around on Tampermonkey, then no, I have no experience with it.

Sir Cumference

@Kaumudi.H So wait, have you installed stuff like this?

105

Q: "View Vote totals" without 1000 rep

Screenshot About The vote counts are a great tool to determine whether an answer is disputed or not. Unfortunately, not many of us have enough time to join all Stack Exchange websites and get 1000 reputation. This script unlocks the "View Vote counts" feature for those who are not logge...

Those are like the essentials

user228700

@SirCumference Hang on, I'm trying to prove something.

user228700

Nobody's active at the MSE chat so can anybody please help me to prove this:

user228700

in Mathematics, 8 mins ago, by Kaumudi. H

Can anybody please help me to prove that if $f$ and $g$ are two functions which are inverse of each other, then $f^{'}[g(x)]=1/g^{'}(x)$).

user228700

(I have written some steps after this)

Sir Cumference

12:56 AM

Sorry, got hw

user228700

No problemo :-) Good luck with the hw :-P

Sir Cumference

Out of curiosity, does $g'(x)$ refer to the derivative of g(x)?

user228700

@SirCumference Yep.

heather

that's beyond me unfortunately, and i also have "homework" (i.e., i really should work on it but i don't have too, and it's fun homework =P)

user228700

Again, issokay :-P If anybody can, ping me please...

heather

1:10 AM

okay, so imagine that there is a field, and we are looking at a point $p$ in that field. we can use a coordinate system with a $x_1$ axis and a $x_2$ axis. there is also an altogether different coordinate system, with a $y_1$ axis and a $y_2$ axis. and then to find the change in height, $d\phi$, over that field, I have the equation $d\phi=\sum\limits_n\frac{\partial\phi}{\partial x^n}\,dx^n$. if i am trying to go from $\frac{\partial\phi}{\partial x^n}$ to $\frac{\partial\phi}{\partial y^n}$

how would i get the conversion equation $\frac{\partial\phi}{\partial y^1}=\frac{\partial\phi}{\partial x^1}\frac{\partial x^1}{\partial y^1}+\frac{\partial\phi}{\partial x^2}\frac{\partial x^2}{\partial y^1}$?

^the above simplifies to $\frac{\partial\phi}{\partial y^n}=\sum\limits_m\frac{\partial\phi}{\partial x^m}{\partial x^m}{\partial y^n}$

the simplification i understand, i just don't understand how you get to that conversion equation.

Irregular User

1:40 AM

I'm looking at the mathematics Hamiltonian dynamics but I'm not that good or comfortable with physics. What exactly is the Lagrangian? I have it defined as L(q, \dot{q}, t), where q is an n-tuple of variables q_1, ..., q_n, and similarly for \dot{q}. Is this a system of ODEs?

ZeroTheHero

2:56 AM

@IrregularUser: The Lagrangian is a scalar function from with the equations of motion (the Euler-Lagrange equations) for coordinates q_k are obtained. The EOMs are PDE's. The Lagrangian is usually defined in classical mechanics as the kinetic energy minus the potential energy (for conservative systems). In some more sophisticated cases it is gerry-rigged so as to give the known EOMs.

David Z

3:44 AM

@heather wait, height? Where does height come into this?

user228700

5:15 AM

Hello, again. The MSE chat is empty so:

user228700

in Mathematics, 4 mins ago, by Kaumudi. H

I have a quick-ish question about inverse trigonometric functions. The $\sin$ function isn't invertible as it is, since it's not one-one so we restrict the domain to $[-\pi/2, \pi/2]$, correct?

user228700

(My question continues for a few lines after that ^)

user228700

5:35 AM

Me: I see. So even though we need to restrict the domain of the sinsin function to obtain a proper inverse, we go back to defining $R$ to be the domain just as soon as we're done defining the inverse function?

user228700

Kaj Hansen: Yeah.

Secret

Magnetic writing using electric fields

anonymous

6:07 AM

@Kaumudi.H I guess that you will need to start thinking of functions as graphs rather than some algebraic expressions. All these confusions disappear when you draw a graph.

user228700

@anonymous ...I'm not too sure if you fully understand my confusion in the first place because I am quite used to thinking about functions in terms of their graphs.

anonymous

Maybe. I don't know what you mean when you say : "we go back to defining R to be the domain just as soon as we're done defining the inverse function"

For the inverse function of sin(x)

the domain is not taken as R

@Kaumudi.H

user228700

@anonymous I mean that it seems a little bit too convenient to have $\sin(x)$ be a one-one function to define $\arcsin (x)$ and then switch back to $R$, that's all. It's actually not very important, never mind.

anonymous

What do you mean by "switch back to $R$" ? We don't switch back to $R$. The domain of arcsin(x) is [-1,1].

@Kaumudi.H

user228700

I mean $\sin$ not $\arcsin$.

user228700

6:18 AM

@Anonym: Dyou understand what I mean?

anonymous

@Kaumudi.H Just think of arcsin(x) as an angle whose sine is $x$. There could be infinite such angles whose sine is $x$ but for convenience we find only one "basic" angle whose sine is $x$ and we can find the rest easily. The only reason we cannot define a function which gives all angles whose sine is $x$ is because a function cannot have two values at one point (by definition).

user228700

Thanks for ur help, but I don't think you've understood my problem yet and it's OK, really, it's not important (hopefully).

anonymous

Okay. Lets leave it here then :)

user228700

Where is @JohnR?! :-o

Secret

6:31 AM

Squeezed light to cool mechanical objects to levels lower than commonly accepted lower limits

So... we will be reaching absolute zero soon?

Transcript for