The h Bar

vzn

3:14 AM

Thanks To Cosmic Rays, We Finally Know What The Inside Of A Pyramid Looks Like / IFLScience

Bernard Meurer

3:37 AM

@DanielSank!

I saw your recipes, that California Bread Stuffing sounds deliciamazing

amazilicious

dmckee

4:10 AM

@vzn They've been imaging the calderas under volcanic cones that ways for some years. They can reliably measure meter scale changes in the one under Vesuvius.

user116211

5:44 AM

@BernardMeurer He is a nice chef :D

user116211

He does even cook Indian dishes....

user116211

He once shared with me some of his recipes... ;))

user116211

@BernardMeurer: Here is another classic from @Daniel:

user116211

chili.

DanielSank

6:23 AM

Hi, everybody

@BernardMeurer That's my fiance's mom's recipe.

It's ridiculous. No food should taste that good. Except for my chili.

@MAFIA36790 oh hey speaking of which, I have a photo to show you.

One moment.

@MAFIA36790 drive.google.com/file/d/0B3lqYBoO3lLDNUNhc1RtSG9haDg/…

Made that last night. Can I be an honorary Indian?

user507974

7:04 AM

@DanielSank i could've sworn you'd said girlfriend before, congratulations

user116211

7:34 AM

@DanielSank :D

DanielSank

7:57 AM

@user507974 Might have. We're getting married in July.

@MAFIA36790 Does that mean I qualify?

ACuriousMind

9:32 AM

@AccidentalFourierTransform Not sure if that's better, but I'll live with it :D

Ugh, 28 things in the close queue

ACuriousMind

9:50 AM

Artist's comment:

> "Yes, these are hands. No, i'm not good at drawing"

Oddly enough, what bothers me the most about that image is that there are only four fingers on each hand

Is OP a Simpson?

David Z

@AccidentalFourierTransform Understood, but I really do think that writing the question in a way that makes it look like four separate questions is not good.

user116211

@ACuriousMind hahaha..

user116211

Never noticed it changed... damn ;(

ramsay

10:36 AM

1

A: Find the roots of cubic equation.

HINT First you can establish that the graph must have stationary points at $x=2$ and $x=4$ For there to be three real distinct roots, you require the values of $f(2)$ and $f(4)$ to be of opposite parity. This gives rise to the condition that $-20<c<-16$ For each of the extreme values of $c...

can anyone explain this statement "For there to be three real distinct roots, you require the values of f(2)f(2) and f(4)f(4) to be of opposite parity" in above answer

user116211

$Mathematics$ Mathematics

Associated with Math.SE; for both general discussion & math qu...

6

25

user116211

@ramsay: ^^^^^^^

ramsay

@MAFIA36790 i know but there the guys who help me are not present this time :(

ACuriousMind

@ramsay Just try to draw the corresponding function. You know what a cubic looks like, right?

ramsay

@ACuriousMind but '$c$' won't let me do that

user116211

10:40 AM

@ramsay aha!

ACuriousMind

@ramsay I disagree ;P

c can't stop you from doing whatever you want!

(except travel faster than c of course)

ramsay

@ACuriousMind haha

ACuriousMind

Let's try it this way: How does c change the graph of the function?

ramsay

'c' can bring both local maximum or minimum below x-axis

ACuriousMind

Yeah, but what does it do to the graph as a whole?

ramsay

10:43 AM

nothing

ACuriousMind

Well, not exactly nothing, but it just shifts the graph upward or downward.

ramsay

wait! how?

see what stupid thing wolfram is plotting : wolframalpha.com/input/…

ACuriousMind

Um...that's what adding a constant to a function does. For any $f(x)$, if you just add 2 to the function, each point of $f+2$ just lies 2 units higher than before.

@ramsay Well, it's plotting what you tell it to plot :P

ramsay

@ACuriousMind but i needed 2D not 3D

ACuriousMind

Let's try this another way: Do you realize that a cubic has only a few basic shapes it can take? It always goes either from $-\infty$ to $\infty$ or vice versa, and it either has one maximum and one minimum or a saddle point.

ramsay

10:50 AM

@ACuriousMind agreed

ACuriousMind

So, which of these shapes can have three real roots, and why?

Qmechanic

Question: If a user wants to repeat a bounty on a question, the minimum bounty goes up. Does that also apply if another user posts the second bounty?

ramsay

maybe this shape:

but i don't know 'why' it has to make this shape

ACuriousMind

@ramsay Well, let's look at it: It comes from $-\infty$, and it so the first stationary point has to be a local maximum, and the second one has to be a local minimum. That is, the function has three stages. Rising, falling, then rising again. Each stage can only cross the x-axis once, and for the falling and the second rising stage to do so, they must begin above and below the x-axis, respectively.

@Qmechanic "Note that if you offer several bounties on the same question, you will have to double the amount each time (or more). (...)This doubling applies only to bounties by the same user on the same question."

Qmechanic

@ACuriousMind : Ah, Thanks. Everybody: I just noticed that after 18 months nothing has happens on this meta question. If you want to promote this in my view very important post, you are more than welcome to place a small bounty of your hard-earned meta internet points on it.

ramsay

11:02 AM

@ACuriousMind aha! makes sense

thanks once again ACM

ACuriousMind

11:23 AM

@Qmechanic Done. Although I don't think of my meta points I earned by complaining about stuff as "hard-earned" ;)

Qmechanic

11:35 AM

@ACuriousMind : Thanks!

FrancescoS

12:04 PM

Hi everyone! Is there anyone is able to well handle with Fierz identites?

FrancescoS

12:16 PM

Let us consider the left-handend component $\chi$ of a massless fermion field $\psi$ and the operator defined as $\mathcal{O} = \chi^\dagger \bar{\sigma}^\mu\chi (\partial_\mu\partial_\nu\chi^\dagger)\bar{\sigma}^\nu\chi$. If I have use the Chiral Fierz identity I get $\mathcal{O} = 2\mathcal{O}$ where I used $\partial_\mu\partial^\mu \chi = 0$. So, I get $\mathcal{O}=0$. I would like to know if this result is right or not... because it is not long time I started to use Fierz Identities.

The matrices $\bar{\sigma}^\mu$ are $\{ Id, \sigma^i \}$ where $\sigma^i$ are the pauli matrices and Id is the identity matrix 2x2

The Fierz identity I use is $(\bar{\sigma}^\mu)[\bar{\sigma}^\nu] = (\bar{\sigma}^\mu][\bar{\sigma}^\nu) + (\bar{\sigma}^\nu][\bar{\sigma}^\mu) - \eta^{\mu\nu}(\bar{\sigma}^\lambda][\bar{\sigma}_\lambda) + i\epsilon^{\mu\nu\rho\lambda}(\bar{\sigma}_\lambda][\bar{\sigma}_\rho)$ (i used the Takashi notation)

Slereah

1:39 PM

So anyway

The Haag theorem tells us that perturbation theory is not actually Proper, right?

At what point in perturbation theory are we assuming that the operators of the interacting theory and the free theory are the same, though

I also recall that Peskin assumes some overlap of the Hilbert space of the free theory and interacting theory, and I seem to recall that in fact they are not

Is that it

innisfree

2:38 PM

+1'd the arxiv trackbacks idea. I think it would be great

user147690

Does @ACuriousMind teach us things in here? Is he our lecturer?

user116211

2:57 PM

@AlexClark Why are you saying so?

FrancescoS

Is this question really off-topic? physics.stackexchange.com/questions/254329/… Why?

Obliv

3:12 PM

well if your question is of the consequences of some solution I don't think it's off topic. But it seems like you're asking for confirmation that the solution is correct which is what David Z would say is off-topic. @FrancescoS

user147690

@MAFIA36790 Just joking about what I am reading on the star board and around these messages :P

user147690

Although, it would be cool to pay a mathematician to teach me very specific topics

user147690

(of my choosing)

user116211

@AlexClark ohh... phew! Yeh... he helps others :) But so does everyone :))

Obliv

3:56 PM

@barryc you there ?

user116211

3

A: How to derive Schrödinger equation?

You might be interested in this "elementary" derivation of the free particle Schroedinger equation from Maxwell's equations. It seems to be in the same spirit as Schroedinger's original reasoning. The niceness of this approach is that if you also include special relativity, it nets you both the f...

user116211

What to do with this answer? Any idea?

user116211

@DavidZ seemed to tell him, but presently it is a linked-only answer, IMO.

user116211

4:24 PM

@AccidentalFourierTransform: I'm done with these edits ;(

user116211

why would one just tag everything with mathematics ?

Mathematics

Transcript for

$Mathematics$ Mathematics