2:04 AM
@EmilioPisanty that's pretty arrogant though, isn't it?

2:31 AM
@EmilioPisanty That is gold

2:50 AM
When a vibration is created, there are usually multiple modes, such as the fundamental, first harmonic, second, third and so forth.

When light is sent through a diffraction grating, the spectral components are dispersed into the zeroth order diffraction, first order diffraction, and so forth...

Are these two concepts (harmonics and order of diffraction) the same thing? Or am I confusing terminology?

2 hours later…
4:26 AM
anyone?

Anonymous
4:37 AM
@user400188 Not really

@vzn clickbait
$-1 \star$

5:26 AM
@user400188 no, they are different concepts

@JohnRennie Good morning John :)

Morning :-) Quiet in here this morning ...

Indeed, not many people around
Your amp is making it's way to me this monday
Or well, starting its trip

Aha, you got your stuff shipped then :-)

Yeah, finished arranging it today
I'm super excited
I can't wait to listen to Whole Lotta Love again

5:39 AM
Yes, it'll be fantastic to have your hi-fi back :-)

Yeah, it really will
and I got a couch now too so everything is coming together
I just need a rack, which I will soon have

So overall, switching to the US has worked out OK?

It's been okay, yeah, I miss Lisbon a bit but it's been fine
It's good to have Dan around, it makes life much better having a friend
Santa Barbara is kind of boring though, which is okay, although I miss the quiz nights I did in Lisbon
Not being able to drink is a major issue for me; I really miss it

@BernardoMeurer it is good for your overall health

@Sid No it's not, I just drink soda now
It's just as bad if not worse and it doesn't make me a better person
I am such a sweetheart after a couple beers, you would love me

5:56 AM
@BernardoMeurer don't drink that too. Drink water.

6:43 AM
everything is bad for you
yolo

1 hour later…
8:01 AM
fock what just happened
drop the @Bass drum
sorry i tried my best
bye

8:18 AM
@DanielSank yes, yes it is

It is currently snowing in Chester

@JohnRennie obviously. It is winter
(And Chester is in the northern part of UK, no?)

It's not too bad down south...
Certainly not anywhere near snowing.

@JohnRennie weather forecast tells me that the temperature in my hometown is 19 degrees. Pleasant weather...
(That is in celsius)

8:36 AM
@Sid Chester isn't that far north.
In fact Chester rarely gets snow because it's on the west coast so it's kept warm by the Gulf Stream. We rarely get snow more than once or twice a year, and some years there is no snow at all.

So.. climate change is real then..

I count Chester as very North...
Maybe that's cause it's like a 6 hour drive

@CooperCape from where? I can get to London in under four hours ...

@JohnRennie About an hour west of southampton and a bit more south
(although you can't get that much more south)

Ah OK, yes, that's a long drive, but mainly because you don't have a motorway conveniently to hand.

8:49 AM
Yeah the A34 does a lot of the work (I think?) but it's such a boring road...

But even from Southhampton Chester is only about two thirds of the way to the Scottish border.

This is why I didn't apply to any Scottish universities...
S'pose it's all relative innit...

If you're significantly west of Southhampton it might be quicker to head for Bristol and pick up the M5 ...

Possibly... can't say I go that far North too often to test it.
Although google maps usually doesn't take that way and it's usually pretty intelligent about it.

9:22 AM
@0celo7 did you ever get a response on strong causality

Anonymous
9:43 AM
@BalarkaSen Are you able to understand the solution to 2.3 (a) here ? It is basically a probability problem, but I'm not sure what the significance of $dx/d\phi$ in the denominator...

Working for Unilever wasn't all bad!

Anonymous

Anonymous
"The details are endlessly fascinating if you're a colloid scientist (or just like ice cream)."

Anonymous
Hehe

Anonymous
"For example sugar poisons the surface of the ice crystals and changes their morphology. In ice cream the crystals tend to be rounded blobs rather than the jagged crystals ice usually forms." Basically some lowering of freezing point stuff going on due to presence of "impurities". Cool stuff

Anonymous
9:51 AM
That effect is called something

Anonymous
I forget the name

Anonymous
It also happens when you supercool water and then suddenly it hits a surface

Anonymous
Ice sort of needs a nucleus to form the crystals (iirc)

@Blue Say $X, Y$ are random variables and $X = \varphi(Y)$. Then $\Bbb P(f(a) \leq Y \leq f(b)) = \int_{f(a)}^{f(b)} \text{p.d.f}_Y(y) dy$, and $\Bbb P(a \leq X \leq b) = \int_a^b \text{p.d.f}_X(x) dx$.

@JohnRennie can I help you in any ice cream experiment

9:55 AM
Can you relate these two? How are the pdf of $X$ and $Y$ related?

@Slereah actually i rarely buy any Unilever ice cream as ... well ... it isn't very good :-)

Anonymous
@BalarkaSen $X$ is the displacement $x$ (say) and $Y$ is the initial phase angle $\phi$, ok?

I don't care about that. These are arbitrary random variables (of certain regularity, but let's not think about that)
This is not a physics question

@JohnRennie obviously you didn't do a very good job then!

Anonymous
$\Bbb P(f(a) \leq \phi^{-1}(X) \leq f(b)) = \int_{f(a)}^{f(b)} \text{p.d.f}_{\phi^{-1}(X)}(y) dy$

Anonymous
10:01 AM
Hmm thinking

Anonymous
Wait...$f=\phi^{-1}$ I guess

Yea that's a typo; I meant $\varphi(a) \leq Y \leq \varphi(b)$

Anonymous
You meant $\varphi^{-1}(a) \leq Y \leq \varphi^{-1}(b)$

Actually I meant $Y = \varphi(X)$ but sure :P
Lots of typos there

@Slereah to make good ice cream requires spending a lot of money on good ingredients, and Walls tend to sell to the mass market where price is an issue.

Anonymous
10:08 AM
Hmm...so some change of variable will do I think

Anonymous
$y=\phi (z)$

@JohnRennie such capitalistic views!
All you need is put love in the ice cream

that or some finest communist flavors
Stalin's icecream
bestselling icecream of the gulag camps

Anonymous
Thus $\text{p.d.f}_Y (Y)=\text{p.d.f}_X(X)/\phi'(X)$

Ironically the cheap ice cream has been the subject of much more research effort than expensive ice cream. If you can afford good ingredients it's relatively easy to make ice cream that tastes great and has a good texture. With cheap ingredients it's far harder to keep the quality acceptable.

10:15 AM
like this beautiful brand of tea
@Blue what is $f$?

Anonymous
$\text{p.d.f}_Y (y)=\text{p.d.f}_X(x)/\phi'(x)$

Where $y = \phi(x)$? That sounds about right
How did you arrive at that

Anonymous
@BalarkaSen Yes...I did the substitution

Actually it should be $|\phi'(x)|$ because you want to keep the pdf positive

Anonymous
Oh, right. Got it!

Anonymous
10:20 AM
$|\phi'(x)|$ just acts as a scaling factor

Anonymous
To move from one p.d.f to another

Anonymous
Thanks!

Just seen on Facebook:

@BalarkaSen Closed timelike curve tea?

10:23 AM

10:48 AM

11:00 AM
John I'm subscribing immediately.

@BalarkaSen is the local trivialization $M \times F \to \pi^{-1}(U)$ or $\pi^{-1}(U) \to M \times F$
I'm seeing both
I guess both work since it's a homeomorphism but what's the most common convention

you mean $U \times F \to \pi^{-1}(U)$ and $\pi^{-1}(U) \to U \times F$
I use the first
I think the second is the most common though

From a non-knowledgeable viewpoint the first is more aesthetically pleasing.

11:35 AM
@Kaumudi.H Hi!

I ended up in the 8th position on the last exam I wrote. Not bad, considering I don't study
I am content

Anonymous
@JohnRennie Do you remember the derivation for $\left(\frac{d\mu}{dT}\right)_{P,N}=-s$ ($s$ is molar entropy and $\mu$ is chemical potential) ? I forgot...and can't seem to find it...

Now I shall do some math

Anonymous
@BalarkaSen 8th in school?

Anonymous
11:39 AM
Noice-noice :)

@Blue I'm afraid not. I could possibly find it if I grub through my old physical chemistry books ...

Anonymous
Ah...well...:/ Trying to search it on the net....

Anonymous
Let's see...I need a potential function of $T,P,N$ for differentiating to get that formula...

@Blue That would be the Gibbs free energy, yeah?

Anonymous
@Samarth Nah, that doesn't work. I need $\mu(T,P,N)$

Anonymous
11:46 AM
I need to find $d\mu/dT$ from there

Anonymous
Keeping other parameters constant

Anonymous
Some Legendre transform should work

Ef would do more work because it is at a higher pressure?
Or would it do the same work because the area under the curve is the same for both
wait nvm

@Blue Perhaps. How about taking $\mu$ as a function of G, P and T? Does that even make sense?

Pretend that instead of "AC" it says AD
Would AD and EF have the same amount of work just with different signs?

Anonymous
11:49 AM
Let's do this. Find a potential function $X(T,V,\mu)$ by Legendre transform on $U(S,V,N)=TdS-PdV+\mu dN$

Anonymous
$X=U-TS-\mu N$

Anonymous
$dX=-SdT-PdV-Nd\mu$

@Blue praying please don't manipulate differentials as fractions, please don't manipulate differentials as fractions

Anonymous
$d\mu=-(1/N)dX-sdT-(P/N)dV$

@Blue yeah, alright then.

Anonymous
11:52 AM
$\frac{\partial \mu}{\partial T}=-s$

Anonymous
Ah, so that does it!

@Blue Huzzah

Anonymous
@Samarth Well, that almost always works in physics! ;)

@10Replies the work done is the area inside the cycle. From a glance they look the same to me.

@JohnRennie Thanks. Since they are cycles they both do work to the system and get work done on them by the system, right?

Anonymous
11:57 AM
@Samarth Anyway, welcome. I haven't seen you around, before

@Samarth Well. It can be done.
There's an extra bit of rigorous machinery that needs to be put on before justifying those manipulations
As the wise men say, r/iamverysmart

12:12 PM
Also I think you can pretty much manipulate 1-forms like fractions?
Like I think there's mostly the proper identities if you do $df = dx \to f' = 1$
At least if $f$ is a $0$-form

$df = f' dx$ by definition
so yeah you can write $df/dx$

so yeah, no worries for that

Anonymous
The only place where someone might make a silly mistake is the implicit function differentiation....but well, that is quite a different case

there's a minus sign hanging out there

the real question is
How do $n$-forms work in thermodynamics
Like how does it relate to $dW = \delta Q + \delta W$
What object is $\delta Q$

12:17 PM
it's a

I know the "physics" version of the object

non-exact form

but what's the rigorous one
Yeah, but then does that mean this is all on some weird manifold?

has to be

it's a contact form on a (1,\infty) topos category jet

12:18 PM
Aren't all forms exact on $\Bbb R^n$
or something
What's the manifold of thermodynamics

K(G,3)

wot wot

@Slereah poincare lemma yea
is there any K(G, 3) that's a manifold?
no, absolute garbage

@BalarkaSen That only works for closed forms.

O
Is $dE$ closed?

12:20 PM
oh i keep missing the closed thing

He's claiming that every form is closed, which is garbagio.

ur garbagio

is it me, or does the ISS really look like a TIE fighter?

@EmilioPisanty looks F A K E to me

12:20 PM
but @0celo7
you can see it in the night sky

@0celo7 of course it would

@Slereah garbagio / i can't speak english but here's some random latino / don't get into an internet beefio ...
sorry for bad rhyming but that's despacito for you

@Slereah I haven't
That image could very well be photoshopped
I mean, there certainly aren't 9 of them irl
so the image is already altered, even if you believe it's real

@0celo7 yeah. OP was all "I hope this hasn't been photoshopped..."
uh...

the moon might be a disk
love it

12:27 PM
Maybe there are 9
and the government is lying to us

@0celo7 Flat Earth isn't a particularly tenable theory
but Flat Moon?

if the ISS is real, it's probably a missile silo

much less evidence against it

@EmilioPisanty it could be a disk
Pointed toward us

why don't we ever see the dark side of the moon? Czechmate, atheists

12:28 PM
also : flat sun???

though I guess lunar phases would require some contortions

the sun is probably round
@EmilioPisanty the moon is made out of pliable mozzarella

@0celo7 why are you so certain all of a sudden

@EmilioPisanty because we go around it and can determine that it's an ellipsoid to a high degree of accuracy
the sun being round has never been a question

@0celo7 why are you so certain that the Earth orbits the Sun, all of a sudden?

12:30 PM
Huh? Heliocentricity is clearly true

Is free RQM equivalent to the quantization of a Polyakov 0-brane action

I only doubt the shape of the Earth
and now the moon

I thought you were firmly in the "this is all received wisdom, there's no way to be truly sure" camp
@0celo7 how so?

@0celo7 that's not diffeomorphism invariant

how is heliocentrism any less received wisdom than round Earth?

12:31 PM
well that's true but I think that it's more believable wisdom than the round Earth
I'm trying to become more mainstream

@0celo7 ¯\ _(ツ)_/¯

@EmilioPisanty also GR predicts the orbit of mercury nicely, so Mercury probably orbits
so we probably do

@Slereah I know neither what "free RQM" actually is, nor what you mean by "quantization of a Polyakov 0-brane action". (For comparison, string theory is not merely "quantization of a Polyakov string/1-brane action".)

@Slereah please let's learn LQG

12:35 PM
@0celo7 ... and you have no problem buying all of that but you still have trouble with results dating back to Erastothenes?
¯\ _(ツ)_/¯
to each their own

Exactly
why should I believe Eratosthenes over modern evidence
the Erathosthenes result needs an understanding of how EM waves travel over large distances
and I'm not sure we have that
also I'm sure the Flat Earth Society has an explanation even without reworking EM

¯\ _(ツ)_/¯

but remember, I don't believe them either
I try to be neutral

¯\ _(ツ)_/¯

¯\ _(ツ)_/¯¯\ _(ツ)_/¯¯\ _(ツ)_/¯¯\ _(ツ)_/¯¯\ _(ツ)_/¯¯\ _(ツ)_/¯...
and I have no idea wtf just happened

12:39 PM
really, my biggest issue with flat Earth is that I can't believe airlines purposefully waste fuel trying to convince people the Earth is round
but I haven't quite worked out that conspiracy fully
Now the most effective part of the whole thing is that if you doubt the shape of the earth, you’re immediately branded a lunatic

@ACuriousMind You know, basic RQM
Klein Gordon equation style
With $\hat x$ and $\hat p$ operators

@Slereah I don't know basic RQM - I went from standard QM straight to QFT :P

For shame

RQM to me is a weirdo ad hoc construct that's not really a consistent physical theory if you look to closely at it :P

It's the usual thing, $$(\hat p_x^2 + \hat p_y^2 + \hat p_z^2) \Psi = \partial^2_t \Psi$$
or something like that
@ACuriousMind It doesn't hold up for the general case but it mostly works for the free case IIRC
It bothers me that RQM is apparently Verbotten but then you get shit like quantization of point particles working okay in string theory intros

12:50 PM
Haven't we been over this?

Probably!
I do have that paper on the BRST quantization of point particles somewhere
But I wonder what the fundamental difference with RQM is
I'm sure Lubos could answer me!

@Slereah I actually don't think there is one - the quantization of the relativistic particle yields free RQM in the sense that you get "particle states" and a time-evolution that's the KG equation.

Well that is good to know

But the point is that free RQM is completely uninteresting, but you cannot easily incorporate interactions in this "quantization of the relativistic particle" framework

Yeah, that I get
but how do they avoid the whole negative probability state thing
Especially for scalar particles, since I don't think you can use the Dirac sea trick

1:00 PM
What "negative probability state thing"?

Oh wait, does the antiparticle business also work for RQM?

When you quantize properly, your physical states always have finite positive norm.

I'm not sure
@ACuriousMind Well it was one of the big issue with RQM

Note that I said "yields free RQM in the sense that you get particle states and the KG equation", not that the quantization of the rel. particle reproduces other features of RQM

When quantized they had negative probability solutions
I forget if there was a solution to that in RQM though
Let's see if there's a good rigorous source on RQM
^the Polyakov one btw

1:09 PM

Doesn't help that I'm still not 100% clear on how BRST quantization work
I think Henneaux is a bit too much for me for now
I need to find a more gentle intro
@ACuriousMind Seems good, thx
What is even the gauge of the point particle, is it like transformations on 1D space?

I don't know what you mean

Isn't BRST generally for gauge symmetry
I think the paper says it's reparametrization of the curve maybe?

Sure

which is a choice of the einbein

1:14 PM
But I don't know what you mean by "the gauge of the point particle"

What group would that be

Do you want to know what the gauge transformations are?

The gauge group acting on the action of the point particle
yes
the paper you link does seem to discuss the gauge

There's just one constraint (the KG equation itself), so the group of gauge transformations is just a group of real functions
And indeed these transformations are just the reparametrizations.

I think I should look back to Weinberg this weekend
I think it's the only book I have with some extent description of RQM
@0celo7 hey do you wanna study LQG

1:34 PM
@Blue I haven't joined in before. Always looked a little too chaotic

Anonymous
chaotic?

@Blue Chat can be rather confusing if you come in at a time where there are multiple conversations going on ;)

Anonymous
True, hehe

Anonymous
I remember the feeling when I visited this chat the first time :p

> Given an even higher level of control over the interactions between spins, as already demonstrated for smaller numbers of trapped-ion qubits, this same system can be upgraded to a universal quantum computer.
cough hype cough