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1:19 AM
Is there any way to create a chatroom for a particular answer to a question? There's more than I can type in the comments for an answer of mine
 
 
3 hours later…
4:13 AM
@doublefelix go ahead :)
 
 
3 hours later…
7:17 AM
@doublefelix If you and the other user post enough comments, the system at some point automatically offers you to move the comments to chat. You can also raise a custom mod flag and one of us will do that.
 
7:53 AM
@Make_Stackexchange_Great_Again lmao!
 
8:32 AM
Now that's what I call music! :-)
More generally it's what boomers with a penchant for weird cacophonous rackets call music.
 
user480696
was LSD that popular?
 
8:52 AM
@BobCoolio no. There were a lot of sensationalist articles in the papers about "the threat to our youth" but the reality was that a tiny proportion of the UK population took LSD.
 
 
2 hours later…
10:25 AM
It was in vogue, certainly
and a lot cheaper than today, from what I hear
 
 
1 hour later…
11:40 AM
What is the formal generalisation of a metric that allows non-positive-definite values, the Minkowski "metric" for instance doesn't seem to actually satisfy the axioms of a metric?
 
Can someone tell me what does the positive and negative lobes of a p orbital indicate ? Is it the spin of the electrons that they indicate ?
 
smoking helps me learn physics
does LSD help with physics?
 
@Ankit No they indicate the sign of the wavefunction
 
Can anyone deduce an expression for work done by friction condition given here chat.stackexchange.com/rooms/115288/…
 
@Charlie can you please elaborate ?
Also why don't each lobe keep on oscillating between positive and negative since each represents a wave ?
Wavefunction ?*
 
11:49 AM
They do oscillate btw + and -. The time-dependent state is the two-lobe state times e^{-itH / hbar}
woops, I mean times e^{ -itE /hbar} where E is the energy of the state
 
The signs indicate whether the wavefunction is positive or negative in that region, since $|\psi(x,y,z)|^2=|-\psi(x,y,z)|^2$ the Born rule is the same for both lobes.
 
@Charlie so isn't the wavefunction time dependent ?
Are they only space dependent ?
 
Well, electron exist in "stationary states", but they do satisfy the full schrodinger equation
Yes I shouldn't have left off the $t$ dependence technically
 
@Charlie So if it depend on t then how can we assign a particular sign to a particular lobe of p orbital ?
 
@Charlie You're right, the "metric" on a pseudo-Riemannian manifold is not necessarily a "metric" in the sense of "metric space". We still call it metric :P
 
11:54 AM
@Ankit Is this helpful?
@ACuriousMind Ah I see, ty
Also note that $e^{-i\hat Ht}$ is a complex number of unit modulus, if $\hat H$ does not depend on time this factor is just some value on the unit circle in the complex plane, which oscillates between $\mathfrak{Re}(e^{-i\hat Ht})>0$ and $\mathfrak{Re}(e^{-i\hat Ht})<0$.
And also that $|e^{-i\hat Ht}|^2=1,\quad \forall t$, hence the state is stationary.
 
@Charlie so both the electrons have opposite stationary state ?
 
I'm not sure what "opposite" means for a stationary state
 
@Charlie I meant to say opposite signs.
 
@Stupidquestioninc Famously amphetamines do
 
I would imagine the answer is no, the exclusion principle prevents fermions having identical quantum numbers, since the spatial wavefunction here is dictating the "sign" I see no reason why two electrons can't have the same spatial wavefunction as long as their spin states are different.
 
12:10 PM
@Charlie Don't confuse the notion of a metric tensor and a metric space
 
@Slereah It should be a universal law in science that "same word = same thing" :(
 
A generic spacetime isn't a metric space of any kind, because it's not even guaranteed that you can define a distance function between any two points
If it's well-behaved enough, you can define a distance function from the metric tensor, which makes it uuuuh
 
is there a simple example where you can't? Or are we talking about singular points?
 
a quasi-metric space?
@Charlie Singular points are an example, yes
 
ok nice
 
12:12 PM
The spacetime distance function is kind of weird, IIRC
it has weird convergence properties in general
 
@Charlie Btw By stationary states do you mean that the spatial coordinates don't oscillate in time ?
 
No "stationary" has a different meaning here
A stationary state is a quantum state with all observables independent of time. It is an eigenvector of the Hamiltonian. This corresponds to a state with a single definite energy (instead of a quantum superposition of different energies). It is also called energy eigenvector, energy eigenstate, energy eigenfunction, or energy eigenket. It is very similar to the concept of atomic orbital and molecular orbital in chemistry, with some slight differences explained below. == Introduction == A stationary state is called stationary because the system remains in the same state as time elapses, in every...
 
although, I guess a stationary state does need to be stationary
 
If you want to look at the distance function in spacetime
 
12:15 PM
otherwise $\langle \hat x\rangle$ wouldn't be constant
oh that looks interesting ty I'll save that to read later
 
also another notion of distance of spacetime is the geodetic interval/world function
which is developped a lot in Synge
 
wait what's synge?
 
An unjustly obscure book
it's a pretty cool ressource
 
@Charlie what does a negative wave function indicate ?
 
Oh nice, I'll add that to my ahem collection of gr books
@Ankit Nothing physical
28 mins ago, by Charlie
The signs indicate whether the wavefunction is positive or negative in that region, since $|\psi(x,y,z)|^2=|-\psi(x,y,z)|^2$ the Born rule is the same for both lobes.
 
12:20 PM
@Charlie are you saying that it had nothing to do with physical appearance ?
 
Remember that the wavefunctions are only defined up to a (complex) constant since they live in a projective Hilbert space, all vectors along the same ray in the Hilbert space correspond to an identical physical state
No, the sign of the wavefunction makes no measurable difference
as a sidenote though I recall the signs being important if we are talking about molecules but for a single hydrogen atom by itself the sign has no physically measureable differences.
Then again we may have just been doing "chemistry physics" which is by no means a careful art :P
 
Well it's important if you consider a superposition of states
 
yeah ^
 
ie $\psi_1 + \psi_2$ and $\psi_1 - \psi_2$ are two different states
 
but if we're just working with the hydrogen atom there's no problem right
 
12:26 PM
@Charlie do the electrons in the two lobes interchange their lobes ?
 
There is just one electron, the two lobes don't contain different electrons
 
What !!?
Each orbital should have two electrons when filled .
 
Two electrons are allowed to sit in, for instance, the $p_x$ orbital, as long as their spins aren't the same
 
So do the two electrons sit in the opposite lobes or can be anywhere ?
 
Yes, sorry if that wasn't clear, what I was saying is that the "lobes" (the regions with positive and negative signs) don't represent two distinct electrons
 
12:29 PM
@Charlie Don't they represent two distinct electrons. ? Distinct in the sense of spin ?
 
yes
 
If you have a $p_x$ orbital with two electrons in it, the two lobes of the $p_x$ orbital don't represent distinct electrons
 
@Slereah was that for me ?
 
Both electrons exist in the entire orbital, this is allowed if and only if their spins are different.
 
@Ankit yes
 
12:33 PM
@Slereah so are you saying that the two lobes represent different electrons ?
 
No
 
The lobes of orbitals are the shapes of probability distributions
for a given electron
 
If you just have one electron in a $p$ orbital, it still has two lobes
 
for a start, for the ground state, there are no lobes
The ground state is just a sphere
but it can still host two electrons
 
I'm a bit confused about mixture distributions. Why isn't a mixture distribution just $p_1(x) + p_2(x)$? Just summing up the two probability distributions and you get a third one that's a mixture, no?
 
12:41 PM
@JohnRennie ahem
 
I was looking at the possible gradient by which we can bend the light like changing the temperature of media by how changing this quantity makes the line bend?
 
@JingleBells Are you asking why they are weighted sums rather than just sums? As in here?
 
@Charlie Hmm, I didn't know they were weighted. My book defines mixture distributions (of a discrete random variable) as $P(x) = \sum\limits_{i=1} P(c = i)P(x | c = i)$ or (I suppose) $P(x) = \sum\limits_{i=1} P(x, c = i)$
 
I'm by no means an expert, that was just what I could gather from the wikipedia page :P
 
@JingleBells The sum of two probability distributions is not a probability distribution.
 
12:55 PM
@Slereah thanks I will try it
 
@ACuriousMind Hmm, why not? If you sum two function f(x) + g(x) you get a third function, no?
 
@JingleBells Sure, but a probability distribution is not just a function, it must have its integral (or sum of values, if its discrete) equal to 1.
 
@ACuriousMind $\frac{p_1(x) + p_2(x)}{2}$ :P
 
Yes, that's a possible choice for combining two distributions - you're saying that it's equally probable that $p_1$ or $p_2$ apply to the case at hand.
 
Ahh, I get it now. The weights make sense
So why does my book define mixture distributions as $P(x) = \sum\limits_{i=1} P(c = i)P(x | c = i)$?
 
1:05 PM
I wouldn't call that a definition of a mixture distribution, that's just a true statement
but you can see the r.h.s. as a mixture of the distributions $P(x|c=i)$ with weights $P(c=i)$
 
Btw, one thing, why is $P(c = i)$ with a $c$ and an $i$ there? $P$ is a probability distribution that we're iterating over so why not just $P(i)$? What's the $c$ doing there?
 
I just copied your notation :P
 
Mkay, nvm, I'mma consult the book
 
but it's probably to distinguish the $P(x)$ on the l.h.s. from the $P(c=i)$ on the r.h.s.
 
Yup, exactly
 
1:10 PM
if $x$ can take a value $i$, writing $P(i)$ for both would be ambiguous (but you should really use a different function name than $P$ rather than this awkward notation :P)
 
blame my book :D
 
I do
 
I think I understand mixture distributions now as the weighted sum of the probability distributions being mixed. But in the statement above $A(x) = \sum\limits_{i=1} W(i)P(x | i)$ what's this "weight distribution" $W$ that we source from and why does the conditional probability of $P(x)$ conditioned on some $i$ give us a new probability distribution that we're weighing with $W(i)$. Sorry if I'm confusing but I'm doing my best
 
I'm just guessing from the initial notation here, but the $W(i)$ are just the probabilities for $i$ happening. It's a fact that if you multiply the probability of $i$ with the probability of $x$ happening given $i$, you get the probability that first $i$ and then $x$ happens. Now if you sum these probabilities over all $i$, then you get the total probability of $x$ happening (since you covered all possibilities for $i$).
 
1:29 PM
Pff, I kinda get it. I'mma stick with $f(x) = \sum\limits_{i=1}^n w_ip_i(x)$ for now since it makes the most sense. I guess I have to refer back to conditional probabilities since they are a bit unintuitive right now
 
2:04 PM
Bruh!
I'm sick of living in this world.
I need my own world with my own rules.
 
@Azmuth I think it's pretty crazy that we're here in the first place, so I'll take what I can get.
 
2:21 PM
Entropy, ammiright?
 
2:34 PM
@EmilioPisanty hmm, that's weird. It's a blatant homework question. Put it down to old age and senility.
 
@JohnRennie Even worse, it is a homework question posted as an answer
 
Does anybody know what version of stokes theorem he’s using here?
 
that's just the regular version of Stokes theorem?
 
Is it?
 
I mean it's expressed in components
 
2:42 PM
I’ve only ever seen it as a line integral form
quite a step up this year
 
Ah, line integral
 
is that not the divergence rule in $\Bbb R^{1,3}$?
 
You mean Ostrogradsky's theorem
Stoke's theorem is the more general version
 
In differential geometry, the four-gradient (or 4-gradient) ∂ {\displaystyle \mathbf {\partial } } is the four-vector analogue of the gradient ∇ → {\displaystyle {\vec {\mathbf {\nabla } }}} from vector calculus. In special relativity and in quantum mechanics, the four-gradient is used to define the properties and relations between the various physical four-vectors and tensors...
 
I thought stokes theorem related to the curl too?
 
2:45 PM
en.wikipedia.org/wiki/Kelvin%E2%80%93Stokes_theorem Is this what you're thinking maybe?
 
@JakeRose There is a general theorem, and then you can apply it to a variety of situations
 
Ah yes it is
 
ie Gauss theorem or Ostrogradsky theorem
Those are all fundamentally the same formula
 
oh so both the divergence theorem and Kelvin-stokes theorem come from the same thing.
 
In this case, it's closer to the Gauss theorem
 
2:46 PM
Do they both not relate to the fundamental theorem of exterior calculus?
 
@Charlie Do you mean the fundamental theorem of calculus
If so, yes
 
Yeah I was only really familiar with the "Kelvin-Stokes Theorem" and I'm pretty sure my profs often just called it "Stokes Theorem"; which now seems like it might be confusing since "Stokes Theorem" is apparently more general. But IDK I was always pretty bad at that math.
 
The fundamental theorem of calculus is just Stoke's theorem in one dimension
 
Yeah, in our lectures we had it that way too. Pretty awkward considering I’m now in the maths department for courses
 
oh fundamental theorem of exterior calculus redirects to stoke's theorem
I thought they were synonymous
 
2:48 PM
the notation or $\partial R$ For the boundary is just convention? Or is it somehow related to the derivative?
 
it's notational convention
 
It is just convention, yes
 
I only ever really saw it in the context of Fluid Dynamics, and I'd be lying if I said I really understood what I was doing in Fluid Dynamics very well.
 
well...the boundary does act like an algebraic derivative on cycles, soo... :P
 
@ACuriousMind let’s take this one step at a time shall we
 
2:49 PM
oh no we've summoned him
 
yeah, I'm fine with "it's just convention" as a white lie here
 
Damn I never realised it was so powerful
I’ll have to rEad into this more when I get some spare time
 
Maybe just a small comment in this direction: $\partial(X\times Y) = \partial X \times Y \cup X\times \partial Y$ does look an awful lot like the Leibniz rule, doesn't it? ;)
 
@ACuriousMind Isn't there also some cohomology for boundaries?
Or is that just for simplexes
 
that's the cycles I was talking about, and it's homology, not cohomology (for cohomology the operator would have to increase dimension)
 
2:54 PM
Your dimensions, give them to me
 
lol
 
you can say that the integral is a pairing ("inner product") between $p$-cycles $M$ and (deRham) $p$-co-cycles $\omega$, and Stokes' theorem just says that $\partial$ and $\mathrm{d}$ are adjoint w.r.t. that inner product.
 
@ACuriousMind Is that some theory of current thing
 
you don't need currents to say this
you just need the notion of (smooth) p-cycles and differential p-forms
 
Can someone explain how to resolve this abuse of notation
$\frac{\partial L}{\partial(\partial_\mu \phi)}$
 
3:07 PM
should there be a field in the denominator there?
 
If you want some intuition behind it try here :
5
A: Notation of derivatives in field theory

SlereahThere are two notions which tend to be a bit confused in Lagrangian mechanics for fields. First, there is the functional derivative. As the name implies, this is a derivative for functionals, such as the action functional. Functionals are maps from function spaces to $\mathbb{R}$ (or some other...

 
it's not an abuse of notation, it's just notation :P
 
All notation is abuse
Leave mathematical objects in their platonic realm
 
but if $L = \partial_\mu \partial^\mu \phi$ then we break the Einstein summation convention
 
See above
 
3:08 PM
you have to choose the free index in the denominator differently from any summed indices inside $L$
 
sorry for slow typing, rsi and ipad keyboard ain’t great
So this is just bad notation on behalf of my lecturer?
 
If you really want the more intuitive way of writing it, it's something like

\begin{eqnarray}
\frac{\partial L}{\partial (\partial_\mu \phi)} &=& \frac{\partial L(f, v)}{\partial v}\Bigr|_{f = \phi(x), v = \partial_\mu \phi(x)}
\end{eqnarray}
No, it's standard notation
 
Still, would you not say it’s hard to keep up with Einstein with it?
 
You have to be careful with your indices, yes
Especially if you do it for like
The electromagnetic field
It's a bit of a mess
 
Your way of writing it seems to just sweep it under the rug the issues with the convention
 
3:12 PM
But overall it's fine
 
@ACuriousMind but then wouldn’t it work out differently?
 
@JakeRose Well the convention may not be ideal, but it's best to get used to it :p
You're gonna see a lot of it
 
I love the convention
just trying to figure out how to mentally place this
it makes maths into more of a game as you can just understand the rules
 
Isn't this just a case of the summation convention not being able to handle triple summations?
 
@JakeRose What is meant by $\frac{\partial L}{\partial(\partial_\mu \phi)}$ for something like $L = \partial^\mu \partial_\mu \phi$ is that you write it as $\frac{\partial(\partial^\nu \partial_\nu \phi)}{\partial(\partial^\mu \phi)}$
i.e. you have to relabel all the summed indices to not conflict with the free index before plugging in
 
3:14 PM
And remember that $$\frac{\partial \partial_\nu \phi}{\partial \partial_\mu \phi} = \delta_\nu^\mu$$
 
this is not unusual. If I have a scalar $v = a^\mu a_\mu$ and I write $b^\mu b_\mu v$ and then want to plug in $v$, I also have to relabel the indices on the $a$ inside $v$
 
Ah I see
this fits with what I was thinking
 
It is slightly tricky to get used to it, but once it's done, basically every Lagrangian is written the same way
 
how do you work out the derivative of a vector wrt a cove for?
*covector
 
It's always some tensor polynomial
 
3:16 PM
i.e the second term in that via product rule
 
@JakeRose $a^\mu = g^{\mu\nu} a_\nu$
 
oh I see
 
If you want the index-free version feel free to check out ncatlab.org/nlab/show/variational+bicomplex
 
My lecturer just wrote it down quite instantly
but took me 3 lines to really be clear on
is this just him being a legend or just not bothering to write the working out?
As in he just wrote this with no working
 
That's basically the variational version of $(x^2)' = 2x$
 
3:21 PM
that’s the way I saw it
and it works out like other Euler Lagrange stuff
but it feels cheating to just quote it when it’s not obvious it would work out the same
 
Yes, if you did Lagrangian mechanics before, this is similar to the case of a free particle
with $$L = \frac{m}2 \dot{\vec{x}} \cdot \dot{\vec{x}}$$
 
Yes but still feels less obvious when playing with tensors
 
@JakeRose it's normal to have to write it out when you see it for the first time, don't worry
 
@JakeRose the classical version is also tensors!
You can rewrite it as $$L = \frac{m}{2} \delta_{ab} \dot{x}^a \dot{x}^b$$
 
Ah I see
Interesting
 
3:32 PM
Just a very short question is my equations correct?
Particle sliding along a quadrant circle
 
If you ever feel a bit too confused by indices, try just writing it out without indices
ie $$L = \partial_x^2 \phi + \partial_y^2 \phi + \partial_z^2 \phi$$
 
@Slereah wdym? Occasionally I do have to go back to linear algebra 101 and think of things as matrices
 
It should at least convince you that the result is correct
Well the Einstein notation is just shorthand for a sum
Write the sum explicitely
 
@sheltonBenjamin What about your work gives you hesitation?
 
3:36 PM
Last equation dtheta/dv one
 
we don't know what v is in your case
 
I am a beginner so i don't easily believe i am correct
V is tangential velocity
 
by the chain rule if you multiply the second and third equations you have $$\frac{dv}{d\theta}\frac{d\theta}{dt}=\frac{dv}{dt}.$$
 
But i divided them
 
Which gives you what you have in the top line
 
3:37 PM
So that means it is correct
Thanks!
 
I mean I don't know what problem you're working on but if you multiply the second and third lines you've written you get the top line
which works out if your derivatives are correct because the chain rule is correct
 
@sheltonBenjamin What are you doing with friction there?
oh I think I was just misreading your writing
Well actually... what are you doing? haha
If you are assuming $N=mg\sin\theta$ then there is a mistake
oh well lol
@sheltonBenjamin Charlie was only commenting on the mathematical validity of your calculus, not your physical correctness
 
rip
 
This was my case as well, but it takes a while to break the habit of "Normal force is just equal to the component of weight in the opposite direction"
And they left. Oh well haha
 
3:54 PM
Sorry my mother called me for some work
I was trying to calculate the work done by friction on a curved surface starting from circle
And so I utilise all the known equations I know and also which are valid
But still wanted to confirm is my work correct or not
 
@sheltonBenjamin So are you assuming that $N=mg\sin\theta$?
 
Are shit here we go again
 
How do I deal with this
Is there any existing answer or any way to find work done by friction on a circular path
 
Notebook is upside down
 
4:01 PM
Is this the same problem as before? Did you change where the angle is 0?
 
Sorry for the handwriting
No I change normal force to MV square by R plus MG cos theta
Angle theta not zero sorry for handwriting
 
Well it should be sine if the angle is 0 at the horizontal
 
But still there are terms such as highlighted term how can I integrate that
Yeah I know that should be sin
 
I don't think this has a nice solution you can write out
 
@JakeRose If $\mathcal{L} = \partial^{\mu} \partial_{\mu} \phi$ then $\frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \phi)} = 0$, surely you meant $\mathcal{L} = \partial^{\mu} \phi \partial_{\mu} \phi$
The safest way to do it is to work out $\frac{\partial}{\partial (\partial_{\rho} \phi)} \mathcal{L} = \frac{\partial}{\partial (\partial_{\rho} \phi)} ( \frac{1}{2} \partial_{\mu} \phi \partial^{\mu} \phi) = \frac{\partial}{\partial (\partial_{\rho} \phi)} \eta^{\mu \nu} ( \frac{1}{2} \partial_{\mu} \phi \partial_{\nu} \phi) = ...$ now it's obvious you need to use the product rule
Set $c_{\rho} = \partial_{\rho} \phi$ and it's just $\frac{\partial}{\partial c_{\rho}} \frac{1}{2} \eta^{\mu \nu}c_{\mu} c_{\nu}$
 
4:43 PM
@Slereah have you gone through this
 
@bolbteppa I have tried
hopefully one day I will
Everyday I learn a little more~
 
"Here, first we consider classical field theory (or rather pre-quantum field theory), complete with BV-BRST formalism; then its deformation quantization via causal perturbation theory to perturbative quantum field theory"
 
It's a fun article to link when someone asks for a tutorial for QFT
 
"Given any context of objects and morphisms between them, such as the Cartesian spaces and smooth functions from def. 1.1 it is of interest to fix one object and consider other objects parameterized over it. These are called bundles" now that's a good sentence
The one thing you have to give them is they sometimes really try to motivate this nonsense
 
The thing I mostly reproach to Schreiber is to not keep a concrete object at hand as an example of the concept he explains
But that's a common issue with math people
cf this joke
 
4:50 PM
 
"Definition 2.3. (octonions)"
 
great joke, very classic
 
"Remark 2.7. (Cayley-Dickson construction and sedenions)" this is in a 'qft' text fyi
 
'Groupoid: Group with a partial function replacing the binary operation'
 
4:57 PM
another classic joke on the topic :
A Mathematician (M) and an Engineer (E) attend a lecture by a Physicist. The topic concerns Kulza-Klein theories involving physical processes that occur in spaces with dimensions of 9, 12 and even higher. The M is sitting, clearly enjoying the lecture, while the E is frowning and looking generally confused and puzzled. By the end the E has a terrible headache. At the end, the M comments about the wonderful lecture.
E: "How do you understand this stuff?"
M: "I just visualize the process"
E: "How can you POSSIBLY visualize something that occurs in 9-dimensional space?"
I'm not sure what "Kulza Klein theory" is but it seems complicated
 
It seems to be the version that can be visualized in $N$ space
 
True of many theories
You can visualize classical mechanics in $N$ dimensions
 
5:11 PM
@bolbteppa Oh no
He doesn't put mathbf for text in equations
everything is askew
turns out I didn't read that article
 
"$Hom_{CartSp}(\mathbb{R}^n, \mathbb{R})$"
 
But Schreiber has been rewriting the same article over and over, kinda
At various levels of rigor and detail
 
This one is supposed to be "a first idea"
What is the psychological justification for saying the BV formalism is the 'ultimate' way to view this stuff
 
Well that's his dream I think
Make his whole BV-BRST pre-quantum geometry deformation quantization renormalization of wavefront sets etc the basis of physics
 
Because ghosts seem to appear out of thin air in Yang-Mills, I guess the idea is you simply have to try to re-do everything so that they are there from the get-go
 
5:16 PM
I'm assuming the people reading this aren't trying to learn Newtonian mechanic
What are the ghost terms of newtonian mechanics anyway
Probably none, I don't think basic Newtonian mechanics has gauge symmetry
 
Ghosts are essentially the conjugate momenta to constraints
 
If you take seriously the notion that ghosts are there from the get-go they should be there in classical mechanics too
 
Although I guess the gravitational field does
What's the gauge symmetry of classical gravity, just $\mathbb{R}$?
Up to a constant potential
Although I think that's assuming the field decaying at infinity
Or maybe not
 
@bolbteppa Ghosts are not "there" or "not there" any more than the gauge symmetry is "there". They're just another equivalent description of the system, and in the case of the quantization of gauge theories the one that allows a proper quantization procedure without ad-hoc methods like Gupta-Bleuler
 
Yeah I feel like there is a very natural easy way to see constrained mechanics with ghosts but everything looks insanely complicated and abstract (to me so far anyway)
 
5:21 PM
Also, don't conflate the nPOV with the (BV-)BRST procedure as such. You can learn the entirety of "practical" BRST from QoGS without ever learning what a category is :P
 
Most physicists do without certainly
 
I mean, there will be a section on homological perturbation theory in the middle but it is very focused on just the stuff you need to understand why BRST works
 
It's really going to be hard to find a better way to introduce ghosts than the way you do in Yang-Mills by simply re-writing the determinant as a fermionic path integral which is actually pretty easy once you get it
 
@bolbteppa that's post-quantization, though!
 
As I said, they are essentially the conjugate momenta when you promote the constraints to phase space variables themselves
and you can do that in purely classical Hamiltonian mechanics, no fancy categories or quantizations involved
 
5:23 PM
Why are they anti-commuting
 
read QoGS :P
 
and BRST I mean I really can't express how good that example in the original paper is to motivate where it comes from
FP and BRST are really just tricks worthy of their iconic status
 
5:54 PM
@JohnRennie is there a reason why we $d_{x²-y²}$ orbital and not $d_{y²-z²}$ ?
This question is for anyone who can help.
 
 
1 hour later…
7:20 PM
@Ankit it's because we use $d_{z^2}$ as the fifth orbital. That forces the other orbital to be $d_{x^2-y^2}$. You could use $d_{y^2-z^2}$ but then the fifth orbital would have to be $d_{x^2}$ instead of $d_{z^2}$.
The axes are arbitrary, it's just convention to use $d_{z^2}$ as the fifth orbital.
 
"The concept of L-infinity algebras was discovered in string field theory ncatlab.org/nlab/show/string+field+theory after the finite case was discovered in Supergravity ncatlab.org/nlab/show/D%27Auria-Fre+formulation+of+supergravity Then string theorists forgot about them, but now they are seeing a revival"
Somehow I am guessing it wasn't first framed in this way!
 
Thales first invented the $L_\infty$ algebra but foolishly called it a triangle
 
"The identification of the concept of (super-)$L_{\infty}$-algebras has a non-linear history: L-∞ algebras in the incarnation of higher brackets satisfying a higher Jacobi identity (def. 3.2) were introduced in Stasheff 92, Lada-Stasheff 92, based on the example of such a structure on the BRST complex of the bosonic string that was reported in the construction of closed string field theory in Zwiebach 92."
It really is insane how this stuff is motivated by crazy things like string field theory, sugra in some advanced formulation etc
"Some people say gravity is gauge theory of the Poincaré group. That's close but not quite true: Gravity is a constrained gauge connection, called a Cartan connection ncatlab.org/nlab/show/Cartan+connection Hence gravity is Cartan geometry." is probably the answer to all these debates on here about gravity as a gauge theory
 
7:41 PM
Until someone finds a flaw in the argument!
 
8:17 PM
0
Q: How can I find a [cosmology] question where I cannot remember any text or title for a search?

BuzzI answered a question and I have saved a copy of the text from my answer. Is there a way I can search for this?

 
9:13 PM
Evening
 
 
1 hour later…
10:33 PM
measuring the hubble constant via the distance ladder method (measuring redshift of a nearby star that we know the absolute luminosity of) gives a figure of 74 km/s/Mpc, and measuring it via the Lamda-CDM method (using the cosmic microwave background, which is ostensibly an image of the early universe) gives a figure of 67 km/s/Mpc. The 2015 method of using the delay between multiply gravitationally lensed supernova gives a similar figure of 64 km/s/Mpc.
Is it possible that instead of some potential crisis in astronomy, that the difference in speed is because of the extreme difference in ma
Is this a question to just ask the site?
 

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