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01:00 - 14:0014:00 - 17:0017:00 - 00:00

vzn
vzn
1:52 AM
 
DanielSank
DanielSank
2:04 AM
@EmilioPisanty that's pretty arrogant though, isn't it?
 
Bernardo Meurer
Bernardo Meurer
2:31 AM
@EmilioPisanty That is gold
 
user400188
user400188
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…
user400188
user400188
4:26 AM
anyone?
 
Anonymous
4:37 AM
@user400188 Not really
 
ooolb
ooolb
@vzn clickbait
$-1 \star$
 
John Rennie
John Rennie
5:26 AM
@user400188 no, they are different concepts
 
Bernardo Meurer
Bernardo Meurer
@JohnRennie Good morning John :)
 
John Rennie
John Rennie
Morning :-) Quiet in here this morning ...
 
Bernardo Meurer
Bernardo Meurer
Indeed, not many people around
Your amp is making it's way to me this monday
Or well, starting its trip
 
John Rennie
John Rennie
Aha, you got your stuff shipped then :-)
 
Bernardo Meurer
Bernardo Meurer
Yeah, finished arranging it today
I'm super excited
I can't wait to listen to Whole Lotta Love again
 
John Rennie
John Rennie
5:39 AM
Yes, it'll be fantastic to have your hi-fi back :-)
 
Bernardo Meurer
Bernardo Meurer
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
 
John Rennie
John Rennie
So overall, switching to the US has worked out OK?
 
Bernardo Meurer
Bernardo Meurer
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
 
Sid
Sid
@BernardoMeurer it is good for your overall health
 
Bernardo Meurer
Bernardo Meurer
@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
 
Sid
Sid
5:56 AM
@BernardoMeurer don't drink that too. Drink water.
 
ooolb
ooolb
6:43 AM
everything is bad for you
yolo
 
 
1 hour later…
ooolb
ooolb
8:01 AM
fock what just happened
drop the @Bass drum
sorry i tried my best
bye
 
Emilio Pisanty
Emilio Pisanty
8:18 AM
@DanielSank yes, yes it is
 
John Rennie
John Rennie
It is currently snowing in Chester
 
Sid
Sid
@JohnRennie obviously. It is winter
(And Chester is in the northern part of UK, no?)
 
CooperCape
CooperCape
It's not too bad down south...
Certainly not anywhere near snowing.
 
Sid
Sid
@JohnRennie weather forecast tells me that the temperature in my hometown is 19 degrees. Pleasant weather...
(That is in celsius)
 
John Rennie
John Rennie
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.
 
Sid
Sid
So.. climate change is real then..
 
CooperCape
CooperCape
I count Chester as very North...
Maybe that's cause it's like a 6 hour drive
 
John Rennie
John Rennie
@CooperCape from where? I can get to London in under four hours ...
 
CooperCape
CooperCape
@JohnRennie About an hour west of southampton and a bit more south
(although you can't get that much more south)
 
John Rennie
John Rennie
Ah OK, yes, that's a long drive, but mainly because you don't have a motorway conveniently to hand.
 
CooperCape
CooperCape
8:49 AM
Yeah the A34 does a lot of the work (I think?) but it's such a boring road...
 
John Rennie
John Rennie
But even from Southhampton Chester is only about two thirds of the way to the Scottish border.
 
CooperCape
CooperCape
This is why I didn't apply to any Scottish universities...
S'pose it's all relative innit...
 
John Rennie
John Rennie
If you're significantly west of Southhampton it might be quicker to head for Bristol and pick up the M5 ...
 
CooperCape
CooperCape
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.
 
Slereah
Slereah
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...
 
John Rennie
John Rennie
Working for Unilever wasn't all bad!
 
Anonymous
Interesting answer :)
 
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)
 
Balarka Sen
Balarka Sen
@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 $\Bbb P(a \leq X \leq b) = \int_a^b \text{p.d.f}_X(x) dx$.
 
Slereah
Slereah
@JohnRennie can I help you in any ice cream experiment
 
Balarka Sen
Balarka Sen
9:55 AM
Can you relate these two? How are the pdf of $X$ and $Y$ related?
 
John Rennie
John Rennie
@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?
 
Balarka Sen
Balarka Sen
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
 
Slereah
Slereah
@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
 
Balarka Sen
Balarka Sen
Yea that's a typo; I meant $\varphi(a) \leq Y \leq \varphi(b)$
 
Anonymous
You meant $\varphi^{-1}(a) \leq Y \leq \varphi^{-1}(b)$
 
Balarka Sen
Balarka Sen
Actually I meant $Y = \varphi(X)$ but sure :P
Lots of typos there
 
John Rennie
John Rennie
@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
$y=\phi (z)$
 
Slereah
Slereah
@JohnRennie such capitalistic views!
All you need is put love in the ice cream
 
Balarka Sen
Balarka Sen
that or some finest communist flavors
Stalin's icecream
bestselling icecream of the gulag camps
 
Anonymous
Thus $\text{p.d.f}_Y (Y)=\text{p.d.f}_X(X)/\phi'(X)$
 
John Rennie
John Rennie
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.
 
Balarka Sen
Balarka Sen
10:15 AM
like this beautiful brand of tea
@Blue what is $f$?
 
Anonymous
$\text{p.d.f}_Y (y)=\text{p.d.f}_X(x)/\phi'(x)$
 
Balarka Sen
Balarka Sen
Where $y = \phi(x)$? That sounds about right
How did you arrive at that
 
Anonymous
@BalarkaSen Yes...I did the substitution
 
Balarka Sen
Balarka Sen
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
 
Balarka Sen
Balarka Sen
ye
 
Anonymous
Thanks!
 
John Rennie
John Rennie
Just seen on Facebook:
 
Slereah
Slereah
@BalarkaSen Closed timelike curve tea?
 
John Rennie
John Rennie
10:23 AM
 
John Rennie
John Rennie
10:48 AM
 
CooperCape
CooperCape
11:00 AM
John I'm subscribing immediately.
 
Slereah
Slereah
@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
 
Balarka Sen
Balarka Sen
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
 
CooperCape
CooperCape
From a non-knowledgeable viewpoint the first is more aesthetically pleasing.
 
Balarka Sen
Balarka Sen
 
John Rennie
John Rennie
11:35 AM
@Kaumudi.H Hi!
 
Balarka Sen
Balarka Sen
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...
 
Balarka Sen
Balarka Sen
Now I shall do some math
 
Anonymous
@BalarkaSen 8th in school?
 
Balarka Sen
Balarka Sen
Ya
 
Anonymous
11:39 AM
Noice-noice :)
 
John Rennie
John Rennie
@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...
 
Samarth
Samarth
@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
 
10 Replies
10 Replies
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
 
Samarth
Samarth
@Blue Perhaps. How about taking $\mu$ as a function of G, P and T? Does that even make sense?
 
10 Replies
10 Replies
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
$dX=-SdT-PdV-Nd\mu$
 
Samarth
Samarth
@Blue praying please don't manipulate differentials as fractions, please don't manipulate differentials as fractions
 
Anonymous
$d\mu=-(1/N)dX-sdT-(P/N)dV$
 
Samarth
Samarth
@Blue yeah, alright then.
 
Anonymous
11:52 AM
$\frac{\partial \mu}{\partial T}=-s$
 
Anonymous
Ah, so that does it!
 
Samarth
Samarth
@Blue Huzzah
 
Anonymous
@Samarth Well, that almost always works in physics! ;)
 
John Rennie
John Rennie
@10Replies the work done is the area inside the cycle. From a glance they look the same to me.
 
10 Replies
10 Replies
@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
 
Balarka Sen
Balarka Sen
@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
 
Slereah
Slereah
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
 
Balarka Sen
Balarka Sen
$df = f' dx$ by definition
so yeah you can write $df/dx$
 
Slereah
Slereah
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
 
Balarka Sen
Balarka Sen
there's a minus sign hanging out there
 
Slereah
Slereah
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 $\delta Q$
 
Balarka Sen
Balarka Sen
12:17 PM
it's a
 
Slereah
Slereah
I know the "physics" version of the object
 
Balarka Sen
Balarka Sen
non-exact form
 
Slereah
Slereah
but what's the rigorous one
Yeah, but then does that mean this is all on some weird manifold?
 
Balarka Sen
Balarka Sen
has to be
 
0celo7
0celo7
it's a contact form on a (1,\infty) topos category jet
 
Slereah
Slereah
12:18 PM
Aren't all forms exact on $\Bbb R^n$
or something
What's the manifold of thermodynamics
 
0celo7
0celo7
K(G,3)
 
Slereah
Slereah
wot wot
 
Balarka Sen
Balarka Sen
@Slereah poincare lemma yea
is there any K(G, 3) that's a manifold?
no, absolute garbage
 
0celo7
0celo7
@BalarkaSen That only works for closed forms.
 
Slereah
Slereah
O
Is $dE$ closed?
 
Balarka Sen
Balarka Sen
12:20 PM
oh i keep missing the closed thing
 
0celo7
0celo7
He's claiming that every form is closed, which is garbagio.
 
Emilio Pisanty
Emilio Pisanty
 
Slereah
Slereah
ur garbagio
 
Emilio Pisanty
Emilio Pisanty
is it me, or does the ISS really look like a TIE fighter?
 
0celo7
0celo7
@EmilioPisanty looks F A K E to me
 
Slereah
Slereah
12:20 PM
but @0celo7
you can see it in the night sky
 
Emilio Pisanty
Emilio Pisanty
@0celo7 of course it would
 
Balarka Sen
Balarka Sen
@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
 
0celo7
0celo7
@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
 
Emilio Pisanty
Emilio Pisanty
@0celo7 yeah. OP was all "I hope this hasn't been photoshopped..."
uh...
 
0celo7
0celo7
the moon might be a disk
love it
 
Slereah
Slereah
12:27 PM
Maybe there are 9
and the government is lying to us
 
Emilio Pisanty
Emilio Pisanty
@0celo7 Flat Earth isn't a particularly tenable theory
but Flat Moon?
 
0celo7
0celo7
if the ISS is real, it's probably a missile silo
 
Emilio Pisanty
Emilio Pisanty
much less evidence against it
 
Slereah
Slereah
@EmilioPisanty it could be a disk
Pointed toward us
 
0celo7
0celo7
why don't we ever see the dark side of the moon? Czechmate, atheists
 
Slereah
Slereah
12:28 PM
also : flat sun???
 
Emilio Pisanty
Emilio Pisanty
though I guess lunar phases would require some contortions
 
0celo7
0celo7
the sun is probably round
@EmilioPisanty the moon is made out of pliable mozzarella
 
Emilio Pisanty
Emilio Pisanty
@0celo7 why are you so certain all of a sudden
 
0celo7
0celo7
@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
 
Emilio Pisanty
Emilio Pisanty
@0celo7 why are you so certain that the Earth orbits the Sun, all of a sudden?
 
0celo7
0celo7
12:30 PM
Huh? Heliocentricity is clearly true
 
Slereah
Slereah
Is free RQM equivalent to the quantization of a Polyakov 0-brane action
 
0celo7
0celo7
I only doubt the shape of the Earth
and now the moon
 
Emilio Pisanty
Emilio Pisanty
I thought you were firmly in the "this is all received wisdom, there's no way to be truly sure" camp
@0celo7 how so?
 
Slereah
Slereah
@0celo7 that's not diffeomorphism invariant
 
Emilio Pisanty
Emilio Pisanty
how is heliocentrism any less received wisdom than round Earth?
 
0celo7
0celo7
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
 
Emilio Pisanty
Emilio Pisanty
@0celo7 ¯\ _(ツ)_/¯
 
Slereah
Slereah
 
0celo7
0celo7
@EmilioPisanty also GR predicts the orbit of mercury nicely, so Mercury probably orbits
so we probably do
 
ACuriousMind
ACuriousMind
@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".)
 
0celo7
0celo7
@Slereah please let's learn LQG
 
Emilio Pisanty
Emilio Pisanty
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
 
0celo7
0celo7
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
 
Emilio Pisanty
Emilio Pisanty
¯\ _(ツ)_/¯
 
0celo7
0celo7
but remember, I don't believe them either
I try to be neutral
 
Emilio Pisanty
Emilio Pisanty
¯\ _(ツ)_/¯
 
Secret
Secret
¯\ _(ツ)_/¯¯\ _(ツ)_/¯¯\ _(ツ)_/¯¯\ _(ツ)_/¯¯\ _(ツ)_/¯¯\ _(ツ)_/¯...
and I have no idea wtf just happened
 
0celo7
0celo7
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
 
Slereah
Slereah
@ACuriousMind You know, basic RQM
Klein Gordon equation style
With $\hat x$ and $\hat p$ operators
 
ACuriousMind
ACuriousMind
@Slereah I don't know basic RQM - I went from standard QM straight to QFT :P
 
Slereah
Slereah
For shame
 
ACuriousMind
ACuriousMind
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
 
Slereah
Slereah
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
 
ACuriousMind
ACuriousMind
12:50 PM
Haven't we been over this?
 
Slereah
Slereah
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!
 
ACuriousMind
ACuriousMind
@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.
 
Slereah
Slereah
Well that is good to know
 
ACuriousMind
ACuriousMind
But the point is that free RQM is completely uninteresting, but you cannot easily incorporate interactions in this "quantization of the relativistic particle" framework
 
Slereah
Slereah
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
 
ACuriousMind
ACuriousMind
1:00 PM
What "negative probability state thing"?
 
Slereah
Slereah
Oh wait, does the antiparticle business also work for RQM?
 
ACuriousMind
ACuriousMind
When you quantize properly, your physical states always have finite positive norm.
 
Slereah
Slereah
I'm not sure
@ACuriousMind Well it was one of the big issue with RQM
 
ACuriousMind
ACuriousMind
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
 
Slereah
Slereah
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
 
ACuriousMind
ACuriousMind
1:09 PM
 
Slereah
Slereah
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?
 
ACuriousMind
ACuriousMind
I don't know what you mean
 
Slereah
Slereah
Isn't BRST generally for gauge symmetry
I think the paper says it's reparametrization of the curve maybe?
 
ACuriousMind
ACuriousMind
Sure
 
Slereah
Slereah
which is a choice of the einbein
 
ACuriousMind
ACuriousMind
1:14 PM
But I don't know what you mean by "the gauge of the point particle"
 
Slereah
Slereah
What group would that be
 
ACuriousMind
ACuriousMind
Do you want to know what the gauge transformations are?
 
Slereah
Slereah
The gauge group acting on the action of the point particle
yes
the paper you link does seem to discuss the gauge
 
ACuriousMind
ACuriousMind
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.
 
Slereah
Slereah
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
 
Samarth
Samarth
1:34 PM
@Blue I haven't joined in before. Always looked a little too chaotic
 
Anonymous
chaotic?
 
ACuriousMind
ACuriousMind
@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
 
Emilio Pisanty
Emilio Pisanty
> 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
 
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