The h Bar

Bernard Meurer

02:49

@ChrisWhite It's almost midnight and GMP decided everything is a pointer

Halp

DanielSank

03:05

<3

Bernard Meurer

@DanielSank Hey!

The probability line worked btw, she didn't double check my calculations :)

DanielSank

Nice.

Actually, I'm disappointed. This woman clearly isn't taking the calculation seriously enough.

Bernard Meurer

Exactly what I was about to say! I'm still deliberating whether or not she not double checking is a good or a bad thing

Did you hear the news about Brazil?

DanielSank

No, what happened?

Bernard Meurer

Well apparently our congress thinks impeachment is a recursive function

because earlier today we impeached the impeachment

DanielSank

03:10

heheh

Bernard Meurer

and a few hours ago we seem to have impeached the impeachment's impeachment

So I don't even know what's happening anymore :p

DanielSank

I see.

Well, you could all just eat peaches...

Peaches are delicious.

Bernard Meurer

Oh god

impeachment ~> peaches, what an amazing mind :p

DanielSank

What? The word is literally written in the middle. Impeachment.

Bernard Meurer

Now that you mentioned impeachment kind of sounds like what you'd call stuffing someone with peaches

DanielSank

03:13

@BernardMeurer Yep.

I was going to say you could purchase some peaches and impeach yourself.

Bernard Meurer

"Yeah he got totally impeached at the eating contest"

DanielSank

Do they grow peaches down there?

Bernard Meurer

Good question, I think a few down south

but mostly it's imported for sure

DanielSank

@BernardMeurer Oh then don't bother.

TanMath

@DanielSank hello!

DanielSank

03:15

@TanMath Hello.

Bernard Meurer

@DanielSank Recently I've been stuffing myself with lemons

TanMath

Did you get my email from like a month ago?

Bernard Meurer

and oranges

Oh god

DanielSank

@TanMath Uhhh, not sure.

TanMath

nvm, 2 months ago..

@DanielSank which email do you use? the gmail one or the google one?

DanielSank

03:17

@TanMath I got a note from you but it didn't actually ask any particular question so I figured I'd just wait until you asked something specific.

TanMath

@DanielSank ok... I was awaiting your reply...

@DanielSank I did ask about Prof. Mohseni's email...

Bernard Meurer

@DanielSank I want to buy a plant as my new pet. It has to be small to fit in my window, any ideas?

TanMath

Anyway, I recently got an answer to my question on stack overflow (maybe 2 weeks old...)

DanielSank

@TanMath I'm going to ignore anything you email me or say to me here unless you ask a specific, well posed question. I am not interested in further general discussions.

We've been through that, over and over. I'm surprised you're still approaching things in the same way.

@BernardMeurer Get a cactus or a wandering jew.

@TanMath I will not give you anyone's email. Stop asking for it.

Bernard Meurer

@DanielSank Cactus is a good one! Low maintenance, I like that

DanielSank

03:21

@BernardMeurer Yep. They're very easy.

Bernard Meurer

If I could I'd grow edelweiss

but I don't think that's doable

DanielSank

Never heard of it.

Bernard Meurer

It's my favorite flower

TanMath

@DanielSank I was going to say that it seems that somebody else agrees with me that the programming is not the problem so I kinda wanted to go over whether I am using the right values since the natural units are confusing me...

I will send it as an email...

Bernard Meurer

@TanMath Did you adhere to PEP-8?

DanielSank

03:28

@TanMath Listen to me please. Post your question here on the site.

Bernard Meurer

@DanielSank Think I got my cactus of choice de.wikipedia.org/wiki/Echinopsis_chamaecereus

DanielSank

@BernardMeurer Nice.

Bernard Meurer

@DanielSank Well I'm off to bed now Dan. Goodnight (and I might pass by SB again this year :))

PS. (When's your birthday?)

manshu

04:47

I know the formulas but I am really stuck in the calculus part.

\o

user507974

do you know what a derivative is

user116211

@manshu okay, this question!

manshu

yup

@user507974 I know the high school calculus

user507974

so when its asking as some function varies as a function of another what that is code for in calculus

user116211

This came in AIPMT? I came before it numerous times.... show me your efforts, first ;P

manshu

04:51

@user507974 differentiation

user507974

do you know what v is?

manshu

@user507974 velocity

user507974

its functional form?

manshu

maybe (mv^2)/2

due to work done

user507974

v as a function of x

manshu

04:54

ohh...in that way? I don't really know.

user507974

hint: what type of force is this

manshu

electric force.

user507974

so what is the force between two charged spheres

manshu

kq^2/r^2

user507974

replace r with x in this case

now that gives you force

how can you get velocity from that

manshu

04:56

differentiating it.

-2kq^2/x

user507974

you sure thats differentiating it (you did the right thing though)

manshu

-(2kq^2/x)v

maybe

user507974

well almost the right thing, you are proportionally off

thats not differentiating though

differentiating decreases the order your polynomials right

what increases it

manshu

integration

user507974

yea, what did you integrate to get what

manshu

05:00

integrate dx to get get velocity i guess

no no no

let me think

user507974

take your time (getting you to think is my goal)

manshu

oh damn...I get it now...it was a silly mistake.

-2kq^2/x^3

user507974

colder

integrating is the right call

thats differentiating

but think physically, WHAT are you integrating

and WHAT is the result

and WHAT are you integrating with respect to

type out the equation

manshu

We are integrating force...so we should get change in momentum...So we should get the change in velocity

user507974

well force is the change in momentum (the derivative of it)

manshu

05:03

we are integrating w.r.t the distance between the balls.

force is the change in momentum per unit time.

not just change in momentum

(i guess)

user507974

yea, so you integrate with respect to tine first

time*

manshu

-(kq^2/x)v

user507974

ok, maybe lets take a bit of a different approach

manshu

ohk..

user507974

what is potential energy for coloumb interaction

manshu

05:07

-kq^2/x

user507974

yea

now from conservation of energy what can you say about the kinetic energy

manshu

(mv^2)/2 = kq^2/x

user507974

and potential

well more accurately it equals a constant E;
or $E = KE + PE $

manshu

that's it.

we got it.

user507974

(hint, you have a sign error)

manshu

05:09

v is proportional to x^{0.5}

user507974

yea

manshu

(mv^2)/2 = -kq^2/x

user507974

but yea

thats v dependence

what about as v varies

what do you need to do

manshu

if v increases then x decreases.

user507974

well, with calculus

describe it to me in calculus

OH WAIT

youre good

you have the answer

I was looking at the wording but not the multiple choice

your teacher kinda sucks a bit

manshu

05:12

It was in a national level entrance exam...:p

user507974

well their wording could be better

when you ask how something varies as a function of something else that implies a derivative, but they were asking for the functional form of v

which you have found yourself

now i want to ask you something

manshu

yeah

user507974

calculate what happense when you take the derivative of v with respect to x

manshu

dv/dx=-1/(2(x^{1.5}))

so when x changes then v decreases even more.

user507974

ok, now multiply it by velocity

what is the x dependence

manshu

05:17

I get a = 1/(2x^2)

where a is accelaration

user507974

yea, the factor of two would actually disappear if you carried out all the terms instead of just looking at proportionality

now in general this relationship between velocity, its derivative, and acceleration is true

manshu

So we should get kq^2/x^2 again

right

user507974

yea

your hw assignment is to abstractily show why thats the case

you definitely know how to do it

manshu

one weird question i want to ask.

what does abstractily mean?

user507974

you have $a =$$ v dv\over dx$, show why that works using some simple calculus

so like show without using this specific equation

or show in a way that is more general than just the coloumb problem, a lot of times its easier in a way as long as the ydont demand an explicit answer

so like abstractly speaking $a = dv/dt$

manshu

05:22

a=dv/dt

user507974

so now show me why abstractly this equation works

manshu

multiplying and dividing by dx

we get a= (dv/dx)(dx/dt)

so a=v(dv/dt)

user507974

and a little extra credit, what is this property called?

user116211

@manshu ??

manshu

idk...I am not even sure if we can multiply and divide in this way.

user507974

05:25

@manshu Oh, how did you get that

was that a typo?

manshu

by using dx/dt=v

it's change in position w.r.t. time.

user507974

@manshu but this does doesnt equal

manshu

eh?

ohhh.

yeahhh

user507974

this $a=v$$dv \over dx$

manshu

a=v(dv/dx)

another silly mistake..

user507974

05:27

and its not really multiplication, theres a name for this property you probably learned in your class

you use it all the time when you take the derivative of something like sin(x^2)

user116211

It's a simple trick of operator... and nothing else.

manshu

I know it but I forgot the name a minute ago.

user507974

@MAFIA36790 well kind of, but its not exactly multiplication and getting that distinction down is important

user116211

@manshu: You can derive it by multiplication.... then you would say you are working with infinitesimals so that in the limit it remains the same.

user116211

@user507974 yeh... know that.

user507974

05:29

@manshu chain rule

manshu

@user507974 yeahhh..

substitution was stuck in my mind.

But how are we using chain rule in it?

user507974

care to take over mafia, gotta get back to research

manshu

Thanks a lot.

user507974

yw, good luck on your physics journey friend

user116211

$$a= \frac{\mathrm dv}{\mathrm dt}$$

user116211

05:31

Using chain rule:

user116211

$$a= \frac{\mathrm dv}{\mathrm dx}\cdot \frac{\mathrm dx}{\mathrm dt}\;.$$

manshu

@MAFIA36790 that's clearly multiply and divide by dx

ohhh

I get it..

user116211

@manshu it's not strictly multiplication... you are working with infinitsimals.

manshu

we first find the derivative of v w.r.t x and then using the chain rule we do that dx/dt

right?

user116211

yup.

user116211

05:35

Okay, you can check HC Verma's derivation of SHM..... here he used that. @manshu

manshu

checking

I see it.

But it's nowhere given anything like "I, H.C. Verma, is now gonna use chain rule."

user116211

@manshu hahaha...

user116211

It's implied that the physics author would take it for granted that the readers are well versed in calculus.

manshu

@MAFIA36790 Yeah...it's given in the preface of the book. "The book presents a calculus-based physics course..."

and later it's given that "Almost no knoledge of physics is prerequisite."

user116211

@manshu hahahaha....

user116211

05:42

Sounds funny but well-written.

manshu

True...

ttyl...gotta take bath :p

Slereah

07:38

Hello

yuggib

\o

Slereah

I wonder

what are some quantities that you could calculate for any quantum theory

No matter the formalism and interpretation

I guess cross sections

yuggib

expectation of observables

and then indirectly probability distributions

Slereah

Yeah

yuggib

(at least if you could measure all observables and obtain their expectation value)

Slereah

07:49

Well, they would be rather poor observables if you couldn't :p

yuggib

the fact that you could in principle it does not mean you can in practice

Slereah

But in practice practice

Can you think of an observable you can't measure

yuggib

$(x^2+(-i\nabla)^{18})^{1/5}$

Slereah

Wouldn't it, though

yuggib

or any of its self-adjoint extensions

Slereah

07:52

18th derivative would just be the 18th moment

Tricky to compute I suppose but not terribly hard in principle

yuggib

it is not a problem of computing, but of measuring it

Slereah

Well yes, but then again

yuggib

with an experimental procedure

Slereah

You could say the same thing of like

Charge

You don't directly measure charge

yuggib

maybe; but even if you avoid "dynamical" constants

or quantities

it is already unlikely to measure all "static" observables

Slereah

07:54

But like

What is "measurement", for you

What would you say a direct measurement of a quantity is

yuggib

it is a quantum measuring process

of a given observable

Slereah

yes but in practice

yuggib

in practice it is the realization of a quantum measuring process

Slereah

what's a variable you can measure directly and with what procedure

yuggib

in principle, you can measure every observable with a quantum measuring process

what has been practically done is something you should ask a bloody physicist

Slereah

07:56

Well my point is

Aren't most methods used for measuring variables somewhat indirect

In the end you mostly measure positions and times

and even then smeared positions

yuggib

if "indirect" means that you need an interaction between an instrument and the system, and then you read the outcome on the instrument scale, then I agree measurements are indirect

for that is how measurements are performed

in both practice and theory

for all the gory practical details on the apparati I'm not the person to ask

Slereah

Or Stack Exchange in general, I think :p

Not a lot of experimentalists around

yuggib

yeah, that's definitely true

Slereah

think about those poor dudes that have to look at theoretical papers and then they have to imagine an experiment for it

yuggib

in all likelihood, seen all the theoretical bullshit around, they can pretty much imagine a random experiment

and then find an observable in some theory the experiment is perfect for

I remind you some "brilliant mind" formulated adelic and $p$-adic quantum mechanics

and they were not mathematicians

Slereah

08:05

I am not surprised

It sounds like some shit I'd try

Just look for a random math framework and try to milk it for some physics

I try that a lot but then I find out that some dude always beat me to it

A lot of physicists are on the prowl for math stuff

yuggib

physicists should learn some math before trying to use it randomly

Slereah

Man if I had to learn something before trying to use it

I wouldn't get a lot done!

I wouldn't have a job, certainly

I have no idea what I'm doing at my job

yuggib

and historically, it has almost always worked the other way around: physicists devise some pseudomathematical tool that mathematicians then turn into something rigorous

Slereah

Also knowing math that's like 3 years of studies I'd say

Oh yeah it works both ways

Also sometimes in parallel

Like gauge theory

I am getting better at math but it ain't easy

yuggib

it's never easy

and once you go math you can never go bath

Slereah

08:15

Never go bath?

Are all mathematicians dirty

yuggib

maybe not, but rhymed only that way...

and rhyme is thyme

(supposing that it rhymes...)

bolbteppa

Any thoughts on why we care about Nilpotent lie algebras and prove Engel's theorem for them?

user507974

08:38

@yuggib wth is that thing

like 18th order momentum op or something? what does the 5th root have to do with anything?

Slereah

It's just an example of a weird observable

yuggib

^

I think it has no physical meaning whatsoever

Slereah

I don't think I ever saw any moment over 6th order

and even then that was just the name, not being used

Usually at best you get 4th moment for the jerk

Or the kurtosis, for a statistical distribution

David Z

09:56

home.cern/about/updates/2016/05/…

2

user116211

10:55

Okay, isn't Hamiltonian self-adjoint operator?

user116211

At-least I know so... it has to follow unitarity... isn't it?

user116211

But while googling, I stumbled across this paper cited by over 250:

user116211

Must a Hamiltonian be Hermitian?

Slereah

It is.

user116211

Okay, did I learn wrong?

Slereah

10:58

Otherwise you will have non-real energies

yuggib

11:10

@MAFIA36790 self-adjoint and symmetric(hermitian) are not the same

for an operator to have real numerical range it is sufficient that it is symmetric

for it to be the generator of a unitary group of evolution (dynamics) it has to be self-adjoint

user116211

sorry, got offline... @yuggib o/

yuggib

\o

user116211

@yuggib ooh.

yuggib

@MAFIA36790 and you should take things said by people that do not know the difference between symmetric and self-adjoint with a grain of salt

and very carefully

user116211

I have @yuggib ;D

yuggib

11:17

quantum mechanics is not a theory of matrices

with accidentally an infinite number of entries

user116211

@yuggib o.O

yuggib

it is a wonderful and rich theory with plenty of counterintuitive stuff

that is nevertheless very important

user116211

Thanks @yuggib for the insight o/

user116211

Ah!

user116211

15

Q: Distinguishing between symmetric, Hermitian and self-adjoint operators

I am permanently confused about the distinction between Hermitian and self-adjoint operators in an infinite-dimensional space. The preceding statement may even be ill-defined. My confusion is due to consulting Wikipedia, upon which action I have the following notion. Let $H$ be a pre-Hilbert spa...

functional-analysis

yuggib

11:20

it is just that I find the sloppiness of physicists really stupid in this context

user116211

is hermitian different from symmetruc ;(

user116211

investigating

yuggib

because they often end up overlooking some very relevant features of the theory

user116211

@yuggib ohh.

yuggib

that distinction between hermitian and symmetric is not used in mathematical physics/analysis

at least as far as I know

simply the word "hermitian" does not appear

ever

Slereah

11:23

"T is Hermitian if it is symmetric and bounded."

HINT THOUGH

Most operators are bounded in QM

Although I guess momentum wouldn't be hermitian then

yuggib

@Slereah never read such a definition

Slereah

that's what the math dude said

yuggib

apart from the urreferenced math.se answer

Slereah

What are the definitions you heard

yuggib

I know the definitions of books like Kato, Yosida, Reed Simon, etc.

Slereah

11:25

I do recall working with operators that were self-adjoints but not hermitians, though

yuggib

and they never use the word hermitian

Slereah

Like the momentum operator in a curved space

$p + \Gamma$ or something

that's the one

$-i\hbar(\partial_a- \frac{1}{2} \Gamma_a)$

Slereah

12:08

Hm

When we say that $SO(3) \approx RP^3$

Can we decompose $RP^3$ as like

The set of axis in $R^3$ + an angle

Set of axis would be $RP^2$

I dunno

ACuriousMind

12:55

@Slereah Sort of. You have the relation $\mathbb{R}P^n = \mathbb{R}P^{n-1}\cup_f D^n$ where the gluing map is the projection $f : S^{n-1}\to\mathbb{R}P^{n-1}$.

Slereah

$D$ a disk?

ACuriousMind

Yes

Slereah

what does the radius on the disk corresponds to?

ACuriousMind

Uh, this is a topological statement, it doesn't have a radius.

Slereah

Well what's the difference between a rotation that corresponds to the center or the edge of the disk?

ACuriousMind

12:58

Oh, I have no idea how that relation translates to interpreting $\mathbb{R}P^3$ as $\mathrm{SO}(3)$

Slereah

Hm

I thought it might be more like $RP^2 \times S$

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