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1:03 AM
@Mr.Feynman i think stuff is like in usual QFT when you map to the cylinder via the transformation $w = e^z$. While you work on the two punctured sphere it has some differences
oh but maybe you are already working on the cylinder since you speak of fourier modes
otherwise you would have coefficients of the Laurent series
@DIRAC1930 i guess its because those are the stuff you are most interested in. It is the same for me. also welcome back!
 
1:58 AM
the fake scorseze movie is definitely not related
 
 
3 hours later…
123
5:05 AM
Hello Everyone...
 
 
2 hours later…
6:56 AM
@123 hello
 
7:14 AM
hello
 
7:24 AM
@lucabtz I'm not sure what you're talking about but I solved the problem
So this is the mode expansion
Damn, I forgot the constant before the sum up there
$$X^\mu=x^\mu+\alpha' p^\mu\tau+i\sqrt{\frac{\alpha'}{2}}\sum_{n\neq0}\frac{\tilde{\alpha}^\mu_n \mathrm{e}^{-in\sigma^+}+\alpha^\mu_n\mathrm{e}^{-in\sigma^-}}{n}$$
The analogy with the KG scalar product was not working because of the normalization choice in the sum. That $i$ factor ruined it
Basically, because of the $i$, I had to replace $$f\overset{\leftrightarrow}{\partial_0}g=f\partial_0g-g\partial_0f\mapsto f\partial_0g+g\partial_0f$$
Other than that the analogy works perfectly fine, just noting that the role of negative/positive energy modes of QFT is played by left/right movers here, so we take the scalar product with $\mathrm{e}^{im\sigma^\pm}$, depending on whether we want $\tilde{\alpha}_m$ or $\alpha_m$
 
8:02 AM
I mean yeah technically the X are just massless scalars in 2d, it's more than an analogy it's actually a Klein Gordon field. Just you have way more symmetry due to the fact that you're in two dimensions and the field is massless
 
Sure, the thing was less deep than you'd expect. I was just puzzled by a sign in the product
The conformal gauge Polyakov action is the action of $D$ massless real scalar fields, of course
 
what r some good sad movies
 
does anyone know if fulton and harris has the following theorem in it
for some reason i cannot even find "compact" in the index
 
123
8:17 AM
@RyderRude Hello
What are the evidences for heat is due to the movement of molecules/atoms not calorie like field?
 
also ACM: I think I understand what you were helping with yesterday better now. I guess the two short exact sequences can be summarized as being consequences of (1) the identity component $G_0 \subseteq G$ is normal and (2) if $G$ is compact, then $G_0$ is also compact. Since it is a connected component, $G_0$ is also connected. And, there is a well known classification of compact, connected Lie groups.
i wonder why the notes put it that way :P i guess i will see hopefully
 
@SillyGoose page 439, end of the "Real Groups" section in §26.1
 
oh thank you
 
@SillyGoose for people used to the notation, writing "$1\to G_0 \to G \to \pi_0(G)\to 1$" is shorter and cleaner than spelling it all out, e.g. saying "We have an inclusion $G_0\to G$, a projection $G\to\pi_0(G)$ and $\pi_0(G) \cong G / G_0$"
you shouldn't suspect some deep reason why sequences are "better" here, it's just notation
 
@Mr.Feynman yep
@RyderRude this isn't Tumblr
 
8:33 AM
@ACuriousMind hm but in this particular case, isn't this all summarized by $G_0 \leq G$?
or maybe i should ask is this all summarized by $G_0 \leq G$. (given the first isomorphism theorem)
 
@SillyGoose that's just another notation you have to get used to
 
@lucabtz I don't know that movie, where can I find it?
 
took me a bit to remember that $\leq$ means "is a normal subgroup of" to some people; I never use it
 
i only know it from dummit and foote :P im not sure if it's usual notation
 
@ACuriousMind It's a pretty bad notation indeed :P
 
8:35 AM
@Mr.Feynman it ran in a double feature with Goncharov back in the day
 
You made me discover a not-so-old meme
 
@ACuriousMind in this claim by F&H is it implied that the $G_i$ are connected? I am trying to match the claim stated in F&H to the theorem seemingly applied in the chern-simons notes
 
@SillyGoose if the group you're decomposing is connected, then so are the $G_i$ - the discrete quotient cannot turn a disconnected Lie group into a connected one
 
@Mr.Feynman well since < means subgroup it isn't terrible but I prefer the triangle
 
ooh
wait i think the triangle underline is the notation in dummit and foote
i guess i lost a line in my memor
 
9:12 AM
rich capitalists are killing the world
after hell breaks lose, maybe people will think of a system to keep rich capitalists in check. but probably not
 
Let $\theta$ be a differential $1$-form. What is the differential of it $d\theta$?
Also, how do I tell if $d\theta$ should mean exterior derivative or differential
 
I mean if it's a form it will almost always be the exterior derivative
 
or is the exterior derivative and the differential the same thing 0.0
do they mean to call this a differential then?
 
(also that's why I write the differential as $D\theta$ or $T\theta$ (the latter since it's the result of the tangent functor $T$ acting on $f$))
 
hm wiki says it should be the exterior derivative
 
9:17 AM
@SillyGoose it's the exterior derivative
you will almost never see a form and take its "differential" in the ordinary sense, to the degree that whoever wrote this didn't even consider this a potential confusion :P
 
and it was a mathematician who wrote this!
 
also note that the differential of a 0-form $M\to \mathbb{R}$ is the same as its exterior derivative, so that's why one might also call the exterior derivative "differential"
 
isee ~
ah yes
 
123
10:15 AM
We know that N2L $F = ma$ and Law of gravitation $F = G \frac{m_1m_2}{r^2}$
What is the explanation to make these equal $mg = G\frac{mM}{R^2}$
 
@SillyGoose ass alg
@ACuriousMind what's the differential of a form
 
@lucabtz as forms are sections $\omega : M\to T^\ast M$, you could take their differential in the ordinary sense of smooth maps, viz. $\mathrm{D}\omega : TM \to T(T^\ast M)$. I assumed that was the notion of differential the silly goose meant
 
@ACuriousMind oh okay yeah I thought this is what you meant, but yeah there is no chance of confusion because it seems pretty weird to consider that
 
@123 take $M =$ mass of the earth and $r =$ radius of the earth. this makes a constant we call $g$.
Hence, $mg$ is an approximation of the force of gravity on an object close to the surface of the earth due to the earth
I am a little bit confused here. this states that $\theta$ is $\mathfrak{g}$-valued. But isn't it $\theta_g$ that is $\mathfrak{g}$ valued?
 
10:32 AM
what's the difference?
 
i would think $\theta$ is $\text{Hom}(T_gG, \mathfrak{g})$ valued. where i am slightly abusing notation
 
I recommend to let go of this pedantry with respect to notation/terminology :P
when someone says that a form is $\mathfrak{g}$-valued, what they mean is what you understood as "$\theta_g$ being $\mathfrak{g}$-valued"
there is no potential for confusion here because there is no meaningful interpretation of the phrase "$\theta$ is $\mathfrak{g}$-valued" other than this
 
123
@SillyGoose My what is the argument at which we can make newton's law of gravitation (NLG) and newton's second law N2L are equal
Like we know N2L F = ma where F is the summation of all external forces. These external forces could be friction, drag etc.. how law of gravitation be external force
 
@123 $ma = G\frac{mM}{R^2}$ simply holds for objects under the influence of gravity and nothing else.
there's nothing complicated about this
 
123
@ACuriousMind Hello. are we taking law of gravitation as external force on "m"
 
10:42 AM
where can i find a definition of the exterior derivative $d\theta$ and also of the lie bracket of the wedge product $[\theta \wedge \theta]$?
i have one definition of exterior derivative like
 
and what about that definition doesn't work for $\theta$?
 
i guess i am wondering if that just is the definition of an exterior derivative in genreal
 
@123 yes
@SillyGoose yes
 
123
Thanks ACM.
 
okay so if i naively try to express $d(\theta(g)dg) = \frac{\partial \theta}{\partial g} dg \wedge dg$ but this seems a bit wrong :P.
 
10:52 AM
@SillyGoose what is $\mathrm{d}g$ supposed to be? You have to express this in coordinates!
you have $\theta = \theta^a_i T^a \mathrm{d}x^i$ where the $x^i$ are coordinates on the manifold and the $T^a$ a basis of $\mathfrak{g}$
then $\mathrm{d}\theta = (\partial_i \theta^a_j)T^a \mathrm{d}x^i \wedge \mathrm{d}x^j$
and likewise $\theta\wedge\theta = \theta^a_i \theta^b_j T^a T^b \mathrm{d}x^i \wedge \mathrm{d}x^j = \frac{1}{2}[\theta_i,\theta_j] \mathrm{d}x^i \wedge \mathrm{d}x^j$
 
@SillyGoose do you have a problem with the definition depending on coordinates?
 
(depending on conventions, there can be various factors of 1/2 or $\frac{1}{n!}$ here, but otherwise there is nothing unusual happening here)
 
@Mr.Feynman oh no i am just unfamiliar with the form stuff
@ACuriousMind are repeated indices summed in your expressions?
 
@SillyGoose of course
 
is it a nontrivial result that if an equality holds in local coordinates that it holds in general (i.e. when re-expressed in coordinate free langauge)?
 
11:00 AM
I mean, it's not trivial, but it's a foundational part of differential geometry
it's part of why physicists - always working in coordinates - emphasize that all the things with indices have to "transform properly", since equations where one (or both) sides transform "improperly", i.e. not as tensors, will not hold in all coordinate systems just because they hold in one
3
 
oh in teresting
 
I don't get Bohmian mechanics. Wouldn't the pilot wave of a photon have to move faster than light during a double slit experiment?
 
hm i think im still not understanding $\theta$ conceptually. at a high level it is defined to take in a single element $g \in G$ and spit out a $\mathfrak{g}$-valued 1-form. so then i am trying to parse your written local coordinate form of (what seems to be the reasonable interpretation) $d\theta_g$. okay the $dx_i$ form a basis of the cotangent space (?) of $G$, so $d\theta$ eats a vector in the tangent space of $T_gG$ and spits out a vector in $\mathfrak{g}$?
 
Since the pilot wave would have to move through whatever slit the particle-component of the photon doesn't move through, then return to interact with the photon.
All while the photon moves in a more-or-less straight line at c.
 
@SillyGoose physics only gets better with forms :P
 
11:10 AM
@Mr.Feynman that's one reason i am excited to work through these notes :D it seems to be a pointer towards relevant lie and bundle theory and differential geometry relevant for the physics i am interested in hehe
 
@SillyGoose Yes, that's how forms work - the $\mathrm{d}x^i$ eats a tangent vector and returns a number, and since the $\theta_i^a T^a$ in front of it is an element in $\mathfrak{g}$, the full result is something in $\mathfrak{g}$
 
Also, so if i wanted to prove that $\theta$ satisfies the maurer-cartan equation, would i prove that in local coordinates the equation holds for $\theta_g$ for all $g \in G$?
 
I mean...that depends on your definitions :P
I'm sure some proofs proceed by explicit computation and others by abstract arguments
 
oh i see. i mean by explicit computation
although...an abstract argument would also be nice
but i think iw ould probably benefit from writing up a concrete computation since i don't know much about forms and how to work with them :P
 
given that you seem to have trouble with the coordinate expressions, I would not recommend sticking to abstract arguments :P
 
11:14 AM
right
 
because sooner or later you need to have a firm grasp on how to convert beautiful abstraction into coordinate messes in order to connect any of the formalism to what you will see in examples and applications
while I don't like doing stuff in coordinates, it is very important in a practical sense
 
is $[\theta \wedge \theta]$ is literally just the wedge product and [] is solely for decoration? from your answer yesterday this seems to make sense, but i am wondering why there is a distinct wiki page dedicated to these $\mathfrak{g}$-valued forms: en.wikipedia.org/wiki/… and a section for their wedge product
well i guess there is no the wedge product it has to be specified
 
Maybe it would be wiser to take the time to learn some DG before delving into Lie theory, which is a part of DG
There are many geometric ideas you may not know and are quite relevant to understand Lie groups: vector fields and Lie brackets, pushforwards, flows
 
11:40 AM
@Obliv log is intimately related to the trig functions, via the complex numbers. In fact, log is an inverse tan. For historical reasons, we normally start learning trig before we learn complex numbers, so we don't learn about these important connections until later.
The key equation is De Moivre's identity. $e^{i\theta}=\cos\theta+i\sin\theta$ The exponential function is well-behaved over the entire complex plane, so we can use that identity to extend the definitions of the trig functions to the complex plane. When we do that, we find that sin & cos are also well-behaved over the entire plane.
Let $(t+1)(x+1)=2$, so $t=(1-x)/(1 + x)$.
Then it can be shown that $\ln(x) = 2i \tan^{-1}(it)$
So using complex arithmetic you can calculate logs using an arctan algorithm. Or vice versa.
In practice, if you're working with real values, you can do the computations using real numbers, you just have to tweak the algorithm slightly. Eg, take a look at the Taylor series for arctan(x) and for log((1-x)/(1+x))
 
11:55 AM
@WaveInPlace why is this
 
?? Doesn't the pilot wave has to cover more distance than the particle portion? That seems like straightforward geometry.
 
idk..
the introductory Bohmian mechanics only works with the non relativistic Schrodinger eqn
in qft, u hav a lack of position basis and hence a lack of Bohmian mechanics, I think
 
Almost all of the discussion I've seen has focused on massive particles. But the double-slit works with both light and massive particles, so the same underpinnings should apply.
Also, weak measurements show Bohmian-like trajectories for light: doi:10.1126/science.1202218
 
the trajectory eqn of Bohmian mechanics is for the non rel Schrodinger eqn
@WaveInPlace oh
 
@WaveInPlace standard Bohmian mechanics is not for relativistic particles, and even in standard relativistic QM the question of what constitutes a "wavefunction" in the positional sense for massless particles is a subtle and thorny issue
The double slit can be explained with the classical wave theory of light much easier than by talking about the wavefunctions of individual photons
 
12:14 PM
@ACuriousMind is the second equality obtained by splitting the coefficients into $\frac{1}{2}$ of itself plus $\frac{1}{2}$ of itself and then doing some relabeling and using $dx^i \wedge dx^j = -dx^j \wedge dx^i$?
@Mr.Feynman that would be ideal indeed :P
 
@SillyGoose let's take abstract algebra out of this please
 
@SillyGoose yes
 
yay :D
 
I love spring, best time to do physics
Best time to do everything I guess
 
12:18 PM
@ACuriousMind, it's more a question of validity, I guess. I understand that they are usually understood through different frameworks, but if there's no real difference in outcomes between relativistic and non-relativistic particles it seems like the theories that work for one should work for the other (with tweaks, of course).
 
everyone should plant trees wherever they can
 
@Mr.Feynman, spring is the season of renewal. It almost feels like getting a second wind, sometimes.
 
@ACuriousMind okay last thing before i sleep :P how do i make sense of $(\partial_j \theta^a_i)$? it seems we would like to show that this expression is equal to $-\frac{1}{2}[\theta_i, \theta_j]$ and probably using one of the defining properties of $\theta$
 
1:12 PM
@SillyGoose what do you mean by "make sense of"?
@SillyGoose You are not showing that equality!
$\partial_j\theta_i$ is contracted with an object antisymmetric in $i$ and $j$. What those lines imply is that $1/2[\theta_i,\theta_j]$ is the antisymmetric part
 
1:59 PM
@SillyGoose yes
 
2:10 PM
@SillyGoose technically speaking it's different than the wedge product because you can't really take the product of two elements of a lie algebra
 
@lucabtz that's not really an obstacle - either work with matrix algebras or with the universal enveloping algebra to make the product make sense
 
@ACuriousMind well yeah but the forms are $\mathfrak{g}$-valued not $U(\mathfrak{g})$-valued
and in $\mathfrak{g}$ there is no product
 
@lucabtz but $\mathfrak{g}\subset U(\mathfrak{g})$ canonically and the antisymmetry ensures the resulting expressions still end up in $\mathfrak{g}$
 
@ACuriousMind yeah i mean i agree it works and one can think of it that way
just wanted to point out that there are some technical subtelties hidden behind the fact that one can do as if it was a regular wedge product
@Mr.Feynman im confused the equation in coordinates is $(\partial_i \theta_j +\frac{1}{2}[\theta_i, \theta_j]) dx^i \wedge dx^j = 0$
 
 
2 hours later…
4:08 PM
Hi everyone, does anyone has some good references for starting to study quantum spin liquids? I have found and red mainly review articles which are great for general knowledge but nothing that actually explains results and stuff.
 
4:23 PM
to clarify for @SillyGoose: the spirit of @Mr.Feynman is saying is right, I was just confused with the factor of 1/2. The reason in more mathematical terms is that $dx^i \wedge dx^j$ are not all linearly independent but they are independent only for say $i > j$ so you can't conclude $(\partial_i \theta_j +\frac{1}{2}[\theta_i, \theta_j]) dx^i \wedge dx^j = 0 \implies (\partial_i \theta_j +\frac{1}{2}[\theta_i, \theta_j]) = 0$
you can though express the form in the basis $dx^i \otimes dx^j$ to get $(\partial_i \theta_j - \partial_j \theta_i + [\theta_i, \theta_j])dx^i \otimes dx^j = 0$ and thus $\partial_i \theta_j - \partial_j \theta_i + [\theta_i, \theta_j] = 0$
which is the correct formula for the vanishing of the curvature
however i think there should be a factor of 1/2 difference wrt what Feynman said
 
5:15 PM
What's the bare minimum essentials to derive everything for classical EM
(I don't mean math btw)
Like can you derive everything just by using maxwell's equations or do we need something more? (say, for circuits)
 
do u think big physicists behave like in the Oppenheimer movie
 
I didn't see that movie but almost certainly not.
 
yeah... probably not
@Obliv in principle, all of classical EM should come from Maxwell eqns and Lorentz force law
 
I mean in some ways yes, probably.. but typically movies (especially mainstream ones) aren't very honest in their depictions of people
 
in the movie, Oppenheimer sees complex physics visualisations in his head
ooh and u also need special relativity technically
 
5:23 PM
I don't think SR falls under classical EM but I could be wrong
 
but it comes built into Maxwell's eqns
if u want to know the behavior of EM from a different frame, u must use SR
becuz SR comes built into Maxwell's eqns. SR was discovered because of them
 
I'm not 100% but I thought SR came independently after maxwell's eqns
after the search for the aether was inconclusive
 
Maxwell's eqns played a big part in the derivation of Lorentz transforms
i wud say this was a huge contribution toward SR
see the history section :) en.m.wikipedia.org/wiki/Lorentz_transformation @Obliv
these transforms arrived 10 years before SR becuz of Maxwell's eqns
 
so basically it was just math
lol
 
yeah
 
5:28 PM
then einstein made it sound more physicy
 
probably the earliest example of investigating symmetries
@Obliv yeah
Lorentz had his own theory where space shrunk and clocks slowed down
but it was messy i guess
Lorentz probably read this from the transforms
 
were u the one who recommended griffiths
or was that mr. feynman
 
it must be mr. feynman
 
5:51 PM
@lucabtz mhhh I just had a quick look at the equations above taking them for granted
No wait, I got confused with two different lines xD
Yeah I have no idea what was going on in my mind while writing that lol
@Obliv Electrodynamics or QM?
 
I think it was for ED
for some reason I thought it covered circuital stuff but I was mistaken
 
I think everyone advices that book
@Obliv yeah it doesn't, just a quick mention
 
I think it goes over the basics though, which is sufficient for me
 
Griffiths is more a book about ED than E&M
 
6:15 PM
@SirCumference hi. apart from cosmology, are you also interested in quantum gravity and measurement problem
 
 
2 hours later…
8:01 PM
can a closed system have an infinite amount of microstates
by closed I mean finite energy, volume, #of particles
 
@RyderRude TBH I normally would love to get deep into theory, but as a career I'm probably gonna have to say no to it
just so much harder to get work or a research position compared to observational physics
I am totally up to learning things like that for fun though
But I'm not at that point yet
 
seems like classically it's uncountably infinite but in QM it's countable. That makes sense for energy but i wonder why that's the case for space
griffith is so silly
 
9:07 PM
@Obliv wait until that comparison he makes to describe radiation
Iirc it compares EM radiation to flies on a garbage truck
 

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