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Secret
Secret
12:41 AM
Last night dream, somewhere in a dark coloured theme version of h bar, couple of people are discussing about an esoteric theory called flow string theory. A cutscene showed an animation consists of circular objects being guided towards some kind of ring attractor and following trajectories in the process
I then argued about how I dislike them using the word "quantum" in anything that is not even physics. Another user who is some kind of astroparticle physicists then pointed out that the trajectories are way too precise than those allowed by quantum mechanics, even if we assume each circular object and their vibrations give rise to elementary particles and their interactions
i added onto this point that the closest physical analogue to this esoteric model is bohemian trajectories (which is called quantum bom theory in the dream) but even those are probabilistic and not that precise
One of the spiral trajectories which is said to have a slope of a transcendental number, which is way too precise for the model to be a physical theory
Reality check: In the dream, Flow String Theory postulates that every field and particle is generated by both the vibrational modes of the individual strings and the trajectories they are being guided on as well the ring attractors they are guided to. Whether the real life counterpart of string theory and M theory can exhibit such complicated fluid like turbulence I don't know
 
 
5 hours later…
John Rennie
John Rennie
5:48 AM
@SirCumference when you open Word it briefly displays a banner that tells you what version it is
 
Sir Cumference
Sir Cumference
6:23 AM
@JohnRennie Guess it doesn't on mac :/
thanks for trying tho
 
John Rennie
John Rennie
Ah, the Mac version. I don't have a Mac to hand to test it on I'm afraid.
@SirCumference use the system information?
If you look in the Software section I think it should show the version of the apps.
 
Sir Cumference
Sir Cumference
@JohnRennie Not sure why it seems harder to find the software version on word than most other software
But granted it still looks the same, I probably redownloaded the 2016 version. Weird but kind of expected since I clicked "get the latest Office" only 2 weeks after it was released
 
John Rennie
John Rennie
@SirCumference does this help?
 
Sir Cumference
Sir Cumference
@JohnRennie "the version number is 16.18 and the license is a one-time purchase of Office 2019 for Mac."
Huh, that matches what I have. But I'd assumed "16.18" meant "Office 2016"
not to mention the lack of design changes, weird
 
John Rennie
John Rennie
16.18 is Office 2019/365 ...
 
Sir Cumference
Sir Cumference
6:35 AM
Welp guess I'm updated
thanks for the help btw
 
Nobody recognizeable
Nobody recognizeable
7:00 AM
@Johnrennie are you around.
 
 
4 hours later…
Slereah
Slereah
11:20 AM
hey
3
 
 
4 hours later…
ACuriousMind
ACuriousMind
2:55 PM
@Slereah hey
 
Slereah
Slereah
Gotta write a personal statement for that Cambridge PhD
 
ACuriousMind
ACuriousMind
Just talk about how you want to flee the EU :P
 
Slereah
Slereah
God save the Queen!
I'm trying to convey my enthusiasm for the field without coming off as deranged
 
Slereah
Slereah
3:26 PM
"There are certain things best left out of personal statements. For example, references to experiences or accomplishments in high school or earlier are generally not a good idea. Don't mention potentially controversial subjects (for example, controversial religious or political issues)."
I'd better not say that 9/11 was an inside job
 
ACuriousMind
ACuriousMind
Well...if your prospective supervisor agrees, you'll instantly have the job :P
 
Yashas
Yashas
4:17 PM
How can I show that the following equations are mathematically consistent $\cosh \alpha = \gamma,\space \sinh \alpha = \gamma\frac{v}{c}$
 
Slereah
Slereah
@Yashas Hyperbolic identity
 
Yashas
Yashas
Those were guesses but there I have assigned values to $/coshh$ and $/sinh$ (i.e. did not get one from the other). I need to show that these definitions are mathematically consistent.
 
Slereah
Slereah
$\cosh^2 - \sinh^2 = 1$
Or whatever
 
Yashas
Yashas
@Slereah Is that sufficient to claim that it's mathematically consistent?
 
Slereah
Slereah
You can look up what $\sinh(\arccosh ())$ is, if there's such an identity otherwise
Also better tip : go to exponential forms
Things are usually simpler
 
Yashas
Yashas
4:20 PM
Hmm, wait. I think I could've simply got $\cosh$ using $\sinh$ from that identity. Silly me.
 
Slereah
Slereah
There it is
 
Enzo
Enzo
@ACuriousMind Hi, where can I look up the chat rules?
 
Slereah
Slereah
There are no rules
it's a wild wasteland
 
Yashas
Yashas
"In above question show that the Lorentz transformation corresponds to a rotation through an angle $i\alpha$ in a four-dimensional space."
 
John Rennie
John Rennie
 
Yashas
Yashas
4:25 PM
I am not comfortable with hyperbolic functions. Where do I start?
 
Slereah
Slereah
@Yashas convert to exponentials
it should be easy to see then
 
Enzo
Enzo
4:42 PM
Suppose that a particle with energy E and impulse p is decaying into 2 photons. Is it possible to determine the minimal angle between 2 photons?
 
Slereah
Slereah
@JohnRennie I need your stiff british upper lip
From Fewster regarding funding :
"I suggest you write that you would hope to obtain a Departmental studentship; if you know of other funding opportunities [e.g. scholarships from France] that you could apply for then you could also list them."
"Departmentla studentship" isn't a listed choice on the application, though
 
John Rennie
John Rennie
Departmental studentship presumably means the university provides the funding ...
 
Slereah
Slereah
yeah but I have to pick an option from a big ass list on the application
 
John Rennie
John Rennie
It's hard to comment without seeing the application form and the options available.
 
Slereah
Slereah
not sure which one to pick
Lemme get it
Sorry I meant big arse list
 
John Rennie
John Rennie
4:49 PM
Your English is improving already :-)
 
Yashas
Yashas
I appear to have understood hyperbolic rotations in 2D. How do I extend this to 4D? I have managed to show that with the way I defined $\sinh$ and $\cosh$, it corresponds to a Lorentz transformation in a hyperbola defined by $1 = x^2 - (ct)^2$. How do I extend this to 4D?
 
John Rennie
John Rennie
@Yashas In 4D you can always rotate your axes so that the rotation is in the t,x plane, so the proof for 2D is all you need.
 
Slereah
Slereah
Self Funding
Family
AHRC
Bolashak
CONACyT Estatal
CONACyT Nacional
CONICYT
Canadian Centennial Scholarship
Canadian Student Loan
Chevening
China Scholarship Council - UoY PhD Programme
Commonwealth
DC Hub Studentship
EPSRC
ERASMUS
ESRC
EU Horizon 2020
Egyptian Cultural Bureau
FIDERH
FIIDEM - Mexico
FONCA - Mexico
FUNED
Fulbright
Global Education Program (GEP) - Russia
HEC
HYMS 20% NHS fee discount
HYMS 50% staff fee discount
HYMS Alumni Discount
Kuwait Cultural Office
LPDP (Lembaga Pengelola Dana Pendidikan)
Do I just pick "Other" and write what he said
 
Yashas
Yashas
Ah, I just realized the question asked me to prove by a rotation by $i\alpha$ and not $\alpha$ which I think is for Euclidean space since the minus disappears off the time component?
 
John Rennie
John Rennie
@Slereah I can't see anything that's a match. I think you're going to have to choose "other"
 
Slereah
Slereah
4:52 PM
Alright thanks
 
Yashas
Yashas
Oh my. The hyperbolic ones get converted to regular trig functions and it works for 2D!
I learn more mathematics solving physics assignments than math assignments 0o
 
parvin
parvin
Hi
 
Yashas
Yashas
Are rotations always in a plane?
 
parvin
parvin
I was wondering how diffusion current hapeens in semi conductors?
Does that happen the moment we put two parts together?
I know that happens when the carrier density is uneven and carriers are not uniformly spread, but what causes that form of uneven spreading?
 
Slereah
Slereah
5:29 PM
So who are the crazy people starring my "hey"
 
bolbteppa
bolbteppa
So the canonical example of an adiabatic invariant is allowing $\omega$ in the Harmonic oscillator Lagrangian $L = mv^2/2 + m\omega^2 x^2/2$ vary slowly during a time period $T$ ($d \omega/dt << \omega/T$)
It's just a small step from this to superintegrable systems, gateway math :o
 
vzn
vzn
5:57 PM
@bolbteppa lol rats missed all the tumultuous energy fun yesterday wrt hossenfelder, really wanted to say something at the time :( fyi there is not exactly a cabal or conspiracy but there is something called groupthink among scientists, studied in psychology/ sociology, and its hossenfelders main point! & history tells us scientists are not immune ... hossenfelder is really much like woit/ smolin who criticized string theory over a decade ago, not really a new story....
 
Semiclassical
Semiclassical
@parvin fun fact, one of the profs at my uni is a superintegrability guy
 
vzn
vzn
@Secret lol bohemian mechanics luv it! :P =D (that wasnt a typo was it?)
 
bolbteppa
bolbteppa
@vzn it's just lukewarm iconoclasm
The Lisi comments section is Hilarious, the guy who wrote the paper destroying his paper back then was commenting
 
vzn
vzn
@bolbteppa lol its grade A iconoclasm face it. you just cant stand iconoclasm of any type, youre highly allergic... have you looked at hossenfelders credentials? ok she does sound shrill and burned out lately! "siren", lol! was also thinking femme fatale :P
 
Semiclassical
Semiclassical
@parvin oops, that was supposed to be a reply to bolbteppa
 
bolbteppa
bolbteppa
6:04 PM
@Semiclassical ah, this stuff seems to get pretty nuts fast alright
 
Semiclassical
Semiclassical
Yeah
 
bolbteppa
bolbteppa
Apparently there's an analogue of the LRL vector for the Harmonic oscillator and it's a tensor because of this stuff (Goldstein comment)
 
Semiclassical
Semiclassical
I never got into the subject myself
 
bolbteppa
bolbteppa
@vzn words like shrill are another example of yesterday's issue, you know the dealio
I'm now wondering whether the whole $SO(4)$ hydrogen atom stuff is a false way to look at it, seems like degeneracy and superintegrability are the answer, maybe they are related not sure
 
Semiclassical
Semiclassical
As regards the subject at hand, I’m cynical about string theory, but I’m also cynical about most of the supposed alternatives
i find it hard to take Lisi very seriously for instance
 
bolbteppa
bolbteppa
6:11 PM
 
vzn
vzn
@Semiclassical honestly not a lisi fan either but think his "heart is in the right place™"... fluid dynamics paradigm still needs a few yrs to catch on it seems :)
 
bolbteppa
bolbteppa
Search 'Dist' and just bathe in it
'I am shocked that you have still not understood this after more than a decade.'
 
Semiclassical
Semiclassical
@bolbteppa ?
Oh
 
deadpool
deadpool
Hi @JohnRennie , can you please take a look at this question ?? physics.stackexchange.com/questions/442562/…
 
bolbteppa
bolbteppa
The Distler comments seem to be the end of his latest ideas too, seems like Lisi really has no idea what's going on
 
ACuriousMind
ACuriousMind
6:19 PM
Alright, this has gone on long enough. Everyone stop insulting people. To be clear: This includes, but is not limited to, statements questioning an author's mental health.
5 messages deleted
 
bolbteppa
bolbteppa
It's amazing how people seem to go for decades in physics getting basic things wrong and don't respond to hints at trying to correct things, in this case he seems to be trying to fix his false intuition on $E_8$ by finding a way to 'deal' with mirror matter and seems to not realize how it's all linked to real, pseudoreal and complex reps and even seems to admit in his post he misunderstood the paper he thinks answers everything
9
A: Why do we need complex representations in Grand Unified Theories?

Edward HughesThis can be explained by thinking about the coupling of fermions to the $SU(2)$ weak gauge field. Let's recap what we know Weyl fermions necessarily appear in two complex representations of the Lorentz group $L$ and $R$. Only fermions in the $L$ representation of the Lorentz group couple to the...

Seems to be related to the issues described here
'You have spent more than a decade attempting to start with fermions in a real representation of the gauge group, wave a magic wand, and end up with fermions in a complex representation.'
 
Yashas
Yashas
What does rotation by an imaginary angle mean?
 
Semiclassical
Semiclassical
6:35 PM
@Yashas context?
 
Slereah
Slereah
@Semiclassical Lorentz transform
 
Yashas
Yashas
I had expressed the Lorentz transform as a hyperbolic rotation with real angles. When I tried to express it as a Euclidian rotation, I had to rotate by an imaginary angle.
 
Semiclassical
Semiclassical
Ah
 
Yashas
Yashas
but one of my axis is $-ict$ but I can't relate it to angles being imaginary
 
bolbteppa
bolbteppa
From $ds^2 = - dt^2 + dx^2$ if you set $t = i t$ this becomes $ds^2 = dt^2 + dx^2$ and you can derive Lorentz transformations using rotations instead of hyperbolic rotations
 
Semiclassical
Semiclassical
6:37 PM
$t=it\implies 1=i$ ? :P
(I know what you meant)
 
bolbteppa
bolbteppa
haha
tilde...
 
Yashas
Yashas
@bolbteppa but what does it mean to rotate by an imaginary angle?
 
bolbteppa
bolbteppa
We need the primes separate
I don't think you're rotating by an imaginary angle here
 
Yashas
Yashas
I rotated by $i\alpha$ to derive the Lorentz transform with $\sinh$ and $\cosh$ as parameters
Well, I eventually used $i\sinh \alpha = \sin i\alpha$
I am actually not sure what I just did
I ended up with the expressions derived using hyperbolic rotations by using regular rotations except that I multiplied the angle $\alpha$ (the angle that would be used in hyperbolic rotation) by $i$ and used it in the regular rotation
 
bolbteppa
bolbteppa
If $\cos(x) = \frac{e^{ix} + e^{-ix}}{2}$ and $\cosh(x) = \frac{e^x + e^{-x}}{2}$ then setting $ix = y$ we see $\cosh(y) = \cosh(ix) = \cos(x)$
 
Sir Cumference
Sir Cumference
6:42 PM
\
 
bolbteppa
bolbteppa
So rotating a trigonometric function with an imaginary angle is equivalent to a hyperbolic rotation, $\cos(ix) = \cosh(x)$
I think that's what's going on
 
Yashas
Yashas
 
Anonymous
 
Anonymous
See the last paragraph
 
bolbteppa
bolbteppa
Yeah you just show that the cos and sin functions with an imaginary number in there results in the cosh and sinh functions
 
Yashas
Yashas
6:44 PM
I don't know how to interpret what I just did though
it feels more like I hacked to get the answer
 
bolbteppa
bolbteppa
Replacing a purely real angle $\alpha$ with a purely imaginary angle $i \alpha$ converts the trigonometric cos and sin functions to hyperbolic cosh and sinh functions by doing what I just did above, the exponentials get extra $i$'s (or get eliminated), that's all
@Semiclassical the fascinating thing about superintegrability to me is you can get the LRL vector theoretically and explicitly i.e. predict it should exist and it's form
Laplace–Runge–Lenz vector
In classical mechanics, the Laplace–Runge–Lenz (LRL) vector is a vector used chiefly to describe the shape and orientation of the orbit of one astronomical body around another, such as a planet revolving around a star. For two bodies interacting by Newtonian gravity, the LRL vector is a constant of motion, meaning that it is the same no matter where it is calculated on the orbit; equivalently, the LRL vector is said to be conserved. More generally, the LRL vector is conserved in all problems in which two bodies interact by a central force that varies as the inverse square of the distance between...
 
Semiclassical
Semiclassical
I imagine part of it is simple unfamiliarity
Angular momentum conservation is familiar and therefore not surprising
 
bolbteppa
bolbteppa
Vaguely, if a system allows separation of variables in more than one system of coordinates, then because you end up getting non-trivial expressions being equal to separation constants, you'll have new invariants, so the Kepler problem separating in spherical (as usual) and parabolic (as in that post) results in the parabolic separation constants, the LRL vector comes from playing with the parabolic separation constants as in that post, which is cool
 
Semiclassical
Semiclassical
Whereas LRL is not
@bolbteppa yeah
 
bolbteppa
bolbteppa
I think that's just so cool, took so long to see
So the question is, when does a problem separate in more than one coordinate system, at least for finite motion systems it's related to action angle variables and from this you find concrete conditions when it will happen (end of Landau)
 
Semiclassical
Semiclassical
6:59 PM
iirc, one basic difference between the spherical vs. parabolic solution is whether the motion is a rotation or a libration
which i think in turn is why the quantization condition is different when working in spherical coordinates vs. parabolic coordinates
 
bolbteppa
bolbteppa
I don't see how to apply this to the harmonic oscillator though, let alone end up with a conserved tensor :p
 
Sir Cumference
Sir Cumference
8:00 PM
Making sure I'm not insane: the only proper subfield of $\mathbb{C}$ containing $\mathbb{R}$ is $\mathbb{R}$ itself, right?
 
ACuriousMind
ACuriousMind
@SirCumference Right. $\mathbb{C}$ is a degree 2 extension of the reals.
 
Sir Cumference
Sir Cumference
Yep. Sometimes I feel I'm losing my touch with basic algebra :/
 
 
1 hour later…
enumaris
enumaris
9:25 PM
@danielunderwood cuda10 came out in August I think...
@danielunderwood high order optimizers exist, but are generally too computationally intensive to use for neural nets
 
danielunderwood
danielunderwood
Ahh I see. Just finished up the tuning course that for whatever reason contained stuff about softmax...so now I actually know what softmax is
 
bolbteppa
bolbteppa
10:22 PM
 
dmckee
dmckee
11:14 PM
@JohnRennie While the joke remains funny, I actually asked a few theorists about their funding situations. In addition to the classical requirements they seem to need computing support (lots for some of them), and to attend more conferences than experimenters ( per capita travel budgets ran about the same as the experimenters, but they didn't have to do on-site work so the money all went into conferences and the like).
Still cheep compared to buying 1 kiloton of LOS or 500 bespoke DAQ boards, but more than just wastepaper bins.
 

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