The h Bar - 2016-09-18 (page 1 of 6)

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12:05 AM

Screw this

I missed the Harvest Moon

It's overcast

user228700

12:51 AM

Hello all :) If anybody has the time, please answer yet another thermo. question of mine:

user228700

1

Q: Specific heat capacity of a body?

My textbook states that "If the temperature of a substance changes without the transfer if heat ($Q=0$), then $S=Q/(m$∆$T)=0$ Thus, when liquid in a thermos flask is shaken, its temperature increases without the transfer of heat and hence, the specific heat of liquid in the thermos flask is ...

user218912

3:16 AM

@0celo7 u there?

for the moment.

@0celo7 He probably includes the document as a suitably cropped image.

I would use the adjustbox package so that I only need one command (\adjustimage; albeit with lots of options) to achieve all the necessary manipulations.

INcluding a PDF in the document is easy when you are using pdflatex.

user218912

@0celo7 when you're computing the commutator $[\hat{x}(t), H(t)]$ do you not include the position operator in the definition of the hamiltonian?

That sounds like a homework problem

user218912

no

3:30 AM

Also lol

Shankar explains this.

user218912

what page?

user218912

D:

Beats me. But I know he explains it.

user218912

ah fml

I know the section

user218912

3:31 AM

I need to know this

user218912

to finish a problem

user218912

is it true though?

No. It just doesn't matter whether you include it or not.

user218912

okay so I can remove it?

It's there, but irrelevant.

user218912

3:32 AM

great

user218912

thanks :D

You have to explain why first.

If you can't figure this out...not good.

user218912

hmm

user218912

does it have to do with time dependence?

What do you mean by $H(t)$

I might be lying to you.

user218912

3:35 AM

I mean the hamiltonian is now in the heisenberg picture

Ah, that makes things harder, and I'm inclined to help.

Note that the Hamiltonian is unaffected by the S pic --> H pic transformation.

I'm assuming $\partial_t H^{(\text S)}=0$?

user218912

yes

Right, so $H^{(\text H)}(t)=U^\dagger(t) H^{(\text S)} U(t)$.

But what is $U(t)$?

user218912

why do we need to know that?

user218912

is that the propagator?

3:38 AM

Do you not know what the general form of $U(t)$ is?

How are you going to the Heisenberg picture without knowing that?

@IceLord Yes.

user218912

yes I do.

What is it?

user218912

$e^{-iH(t - t_0)}$

$t_0=0$ for simplicity.

user218912

okay

3:42 AM

So what is $\mathrm e^{\mathrm i Ht}H\mathrm e^{-\mathrm iHt}$?

it's on that thing I sent you hours ago

@IceLord what?

user218912

wait

user218912

I didn't see the $H$

user218912

it's $H$

Yes. So $H(t)=H(0)$

user218912

yes

user218912

3:43 AM

basically you can write

The Heisenberg picture does not affect the Hamiltonian.

user218912

$H(t) = e^{-iH(t)}H(0)e^{iH(t)}$

What are you doing?

user218912

writing it out explicitly

I just said $H(t)=H(0)$

user218912

3:45 AM

oh

user218912

didn't see

You said "yes"!!!

user218912

idk what I'm doing

user218912

i just say stuff

user218912

anyway we don't need this stuff to solve my problem

3:46 AM

Then you know why $[x(t),H(t)]=\frac{1}{2m}[x(t),p(t)^2]$?

user218912

yes

Why?

user228700

@JohnRennie: Morning sir :) Was wondering if you could tell me something about freezing of water. Just a quick doubt.(Also, do let me know if I'm bothering you too much :P )

user218912

i'm thinking

You're right we don't need that, but I figured you should understand it anyway.

user218912

3:50 AM

is it because

user218912

one of the operators commutes with H?

one of the operators?

user218912

like $p(t)$

$p(t)$ does not commute with $H$

user218912

true

user218912

3:52 AM

okay i'm sorry

?

user218912

please let's continue it then

user218912

Idk why

What's your Hamiltonian?

user218912

in my case it's

user218912

3:53 AM

$H = \omega\frac{x^2 + p^2}{2}$

you could have said SHO

user218912

or that

but this works for any Hamiltonian $H=p^2/2m+V(x)$

We have that $H=H(t)=p(t)^2/2m+V(x(t))$

user218912

one thing though

the picture change does not actually change the hamiltonian as a whole

but it sure does change $p$ and $x$

user218912

3:57 AM

my position and momentum operators are $X = \frac{x}{\sqrt{m\omega}}$ and

user218912

$P = \sqrt{m\omega}p$

ok?

user218912

so what I did was solve for $x$ and $p$

user218912

and then use them in the hamiltonian

user218912

for the commutator

user218912

3:58 AM

is that okay?

yes

user218912

and so what I'm stuck on is why I can ignore the position one in the momentum commutator and viceversa.

user218912

right?

@IceLord tell me

Have you written out $[x(t),H(t)]$

actually written it down

with your $H(t)$

and then played with it?

user218912

sure

user218912

4:00 AM

I did

what did you write down

user218912

well I wrote it in the form where I include both the position and momentum operators

user218912

is that wrong?

user218912

I want to remove one of them so I can get something out of the commutator

user218912

because I can use $[x, p] = i\hbar$

user218912

4:02 AM

so I want to write

So you have $[x(t),H(t)]=[x(t),p^2(t)/2m+V(x(t))]=[x(t),V(x(t))]+[x(t),p(t)^2/2m]$?

user218912

$$[x(t), H(t)] = \bigg[x(t), \frac{\frac{P(t)^2}{m\omega} + x(t)^2 m\omega}{2}\bigg]$$

user218912

ok what do I do with that big thing?

user218912

do I just

sigh, did you even look at what I wrote

user218912

4:05 AM

split it

YES

user218912

and then do what you wrote?

user218912

okay

user218912

makes sense tbh

user218912

and now does the first term at the right hand side equal to 0?

user218912

4:06 AM

in your equation?

you tell me

user218912

it should

user218912

for this to work

user218912

unless i'm being dumb like always.

give a proof

user218912

4:09 AM

easy

user218912

if you use my hamiltonian the commutator of that is $0$

user218912

i did it

user218912

and same with yours

user218912

general case works so it's true

user218912

@MAFIA36790 hi

user116211

4:18 AM

hey @IceLord

user218912

what courses are you taking?

user116211

@IceLord Although the first semester focusses on numerical analysis and complex analysis and abstract algebra, I'm myself studying real-analysis and calculus.

user218912

oh okay

user116211

Physics course is bleh... Newtonian Mechanics.

user218912

so you have analysis in sem2?

user116211

4:21 AM

@IceLord 3.

user218912

@MAFIA36790 I know right, I couldn't wait so I'm just destroying myself taking QFT in sem 1.

user218912

so far so good though.

user116211

I'm myself studying Variational Principles of Mechanics by Lanchzos; the book is far better than Goldstein.

user218912

I used tong's notes

user116211

@IceLord ohh.

user218912

4:22 AM

they are really concise and detailed at the same time.

user218912

has all the important things you need from mechanics

user218912

for qft and such

user116211

Also, as yuggib dared me, I'm studying Jach's set theory and Bourbaki 1 ;)) Although at a much slow pace.

Jech Set Theory is a meme text.

user116211

@0celo7 You didn't like it?

user116211

4:23 AM

There is Halmos' Naive Set Theory also.

Jech is not naive set theory.

Jech is set theory for freaks.

user116211

@0celo7 yes, I know that.

user116211

They don't touch logic in any way just like Bourbaki did; he defines set in the third chapter and ordered pair much much later ;P

user116211

9 hours ago, by MAFIA36790

@icelord, Did you guys, start there from QM course? Just saying by the by ;)

user116211

that was what I was saying yesterday.

user218912

4:25 AM

what do you mean by that?

user116211

@IceLord I was bit surprised as why such things are taught at sem 1; but as you said you are self-studying....

user218912

they're not.

user218912

I enrolled in a graduate course xD

user218912

and I am self studying too

user116211

@IceLord O.o

user116211

4:27 AM

@IceLord kudos

user218912

nah

user218912

it's not really an achievement yet

user218912

it will be once I get above 70% in the course

user218912

by december

user218912

I'll try my best.

4:27 AM

I say kudos when the boy solves quantum gravity.

user116211

Anyways, I did a MathJax exercise today:

user218912

@0celo7 which boy?

Until then he's no better than Jamal, who was basically a string theorist at 15

user116211

1

A: Are electric fields produced by static electric charges different from those produced by time-varying magnetic fields?

$$\begin{array}{|c|c|} \hline\textrm{True in Statics} &\textrm{True in General}\\ \hline \mathbf F = \mathbf E~= \dfrac1{4\pi\varepsilon_o}~ \dfrac{q_1q_2}{r^2}~\mathbf{\hat r} & \mathbf F= q(\mathbf E+ \mathbf v\times \mathbf B)\\ \hline \nabla \cdot \mathbf E = \dfrac{\rho}{\varepsilon_0} & \...

So he's actually much less better

@IceLord You!

user116211

4:28 AM

@0celo7 Lumo?

user218912

yes Jamal is my hero.

user218912

xD

user116211

@IceLord sure ;)

Ooooh, Jacobi fields with torsion.

Maybe I should try to analyze that equation.

What does smelly pee signify?

user218912

ask biology se

user116211

4:33 AM

maybe health se

I'm...asking for a friend

user218912

whatever you do

user218912

don't google it

user116211

Ask it

@IceLord why not

user218912

4:36 AM

because you'll be misdiagnosed.

user218912

with bad diseases

"If you catch a whiff of something really strong before you flush, it might also be a sign of a UTI, diabetes, a bladder infection, or metabolic diseases."

FUCK

user218912

xD

user218912

you care so much for your friend?

yes

user218912

4:40 AM

I keep messing up this calculation

user218912

hey @0celo7 can you do me a favour and calculate the commutator using my hamiltonian for both position and momentum operators?

user218912

for position I get $\frac{P(t)}{m\omega}$

user218912

and for momentum I get $-m\omega X(t)$

user218912

is that right?

user228700

Is @JohnRennie usually away on Sundays?

4:42 AM

Ok, let's look at $$D_t^2J+R(\dot c, J)\dot c+D_{T(\dot c,\dot c)}J=0$$

Hmm

Why isn't the last term 0

$T$ is antisymmetric

user218912

wow nice ignore.

@IceLord no

user218912

:|

@KaumudiHarikumar No.

He's probably dead, holy crap.

user116211

He wakes up early generally; hmmm.

user218912

4:43 AM

@0celo7 don't say that...

user228700

@0celo7 -_-

user116211

Okay, got the line-element and now he says it doesn't preserve the Euclidean structure.... damn ;(

Aha.

$$D_t^2J+R(\dot c,J)\dot c+D_{\dot c}T(J,\dot c)=0$$

there we go.

@IceLord Solve that on a sphere please.

user218912

::dies::

@KaumudiHarikumar Morning. What do you want to know about the freezing of water?

4:47 AM

I've done it without the $T$ term @IceLord

it's not that bad

assuming you know the Riemann tensor on a sphere

user218912

I forgot all of GR

Haha, this is not GR.

user218912

then why do you expect me to know?

user218912

I'm a pure physicist.

user218912

well in progress

4:48 AM

pure physicist?

not for long

user218912

why?

user116211

@IceLord Applied mathematician for sure.

@0celo7 I'm a bit late this morning. We had the monthly SF reading group meeting yesterday.

no one can resist the draw of mathematics

@JohnRennie Ah, you're alive. Good.

user116211

@JohnRennie I knew it!

user218912

4:49 AM

@JohnRennie can you help me?

@0celo7 alive, but a bit the worse for wear.

@IceLord my chances of doing maths effectively in my current state are slim, but I'll help if I can.

user228700

@JohnRennie Lakes freeze from the top-bottom, yes?

user218912

@JohnRennie I want to calculate $[x(t), H(t)]$ and $[p(t), H(t)]$ in the heisenberg picture.

user218912

with my hamiltonian as $H = \omega\frac{x^2 + p^2}{2}$

user228700

@JohnRennie Oh, are you unwell? If you're tired and all, I'll ask later...

user218912

4:51 AM

HOLY FUCK

user218912

I forgot the $\omega$ term in the hamiltonian

user218912

that's why

user218912

@JohnRennie it's all good.

@IceLord I should flag -.-

user218912

why?

user218912

4:52 AM

does it offend you?

it's not appropriate

yes.

user218912

@0celo7 chat.stackexchange.com/…

@KaumudiHarikumar Yes

@IceLord Not me.

amazon.com/Spaces-Constant-Curvature-Chelsea-Publishing/dp/…

@IceLord let's read

user218912

boring af.

4:53 AM

@KaumudiHarikumar Though that just means the density of ice is less than the density of water

user228700

@JohnRennie Right, but this is the case only for water...yeah?

I think there are metals where the solid is less dense than the melt. Antimony possibly?

But ice/water are the only common example.

user228700

Oh, okay...

@IceLord boring

AF?

what does AF mean?

user228700

My doubt was this; when other liquids freeze, they freeze from the bottom up?

4:57 AM

@JohnRennie how do I figure out what TeX symbol I need for something

I need a circle with a wedge inside

I wonder if there's a thing for this already

user218912

detexify?

aha

user218912

detexify.kirelabs.org/classify.html

@KaumudiHarikumar In a lake freezing occurs at the surface because that's where the cold air cools the liquid. If the lake were e.g. alcohol then the crystals formed at the surface would sink to the bottom

user228700

@JohnRennie Ah, okay.

4:58 AM

So the solid builds up from the bottom to the top, but you need to be a careful about saying it freezes from the bottom up.

user image

:(

user228700

@JohnRennie Because it actually freezes at the top but since the solid is denser, it sinks to the bottom, giving the effect that freezing starts at the bottom.

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Sep '1618

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