« first day (3735 days earlier)      last day (39 days later) » 

12:01 AM
Well, what's the main advantage of the Lorenz gauge over the Coulomb gauge? The former seems to be used more frequently.
 
the equations of motions are more symmetric - both are just wave equations with the charge density and current as the sources respectively
so in Lorenz gauge, you can combine them into the 4-potential and have the equation for it just be the wave equation with the 4-current as the source, i.e. the Lorenz gauge is what you want to use if a relativistic/covariant formulation is important to you
 
Interesting.
 
 
3 hours later…
3:14 AM
Has anyone read Loop Spaces, Characteristic Classes and Geometric Quantization by Jean-Luc Brylinski
 
3:45 AM
@NiharKarve Nope but from the introduction it sounds ambitious
 
4:17 AM
@NiharKarve /ˈteɪb(ə)l/ t before stress mark (') sounds ट
more like टै
lol
 
 
1 hour later…
5:28 AM
@JonCuster @JohnRennie For finding the possible dislocation reaction (breaking into partials)is it at all necessary to check x y and z.Or are the square of Burgers vector sufficient?
@JohnRennie Did I explain the problem properly?
 
I don't know I'm afraid. Sorry :-(
 
ok
Can you explain the methodology in zone refining
?
 
Zone refining? There must be a Wikipedia article on that ...
Zone melting (or zone refining or floating-zone process or travelling melting zone) is a group of similar methods of purifying crystals, in which a narrow region of a crystal is melted, and this molten zone is moved along the crystal. The molten region melts impure solid at its forward edge and leaves a wake of purer material solidified behind it as it moves through the ingot. The impurities concentrate in the melt, and are moved to one end of the ingot. Zone refining was invented by John Desmond Bernal and further developed by William Gardner Pfann in Bell Labs as a method to prepare high purity...
 
ok
thanks
 
5:50 AM
Which direction is the impurity supposed to flow?
So where is diffusion easier to take place?Is it the solid phase or the liquid phase?
I need some mentorship about this.
@JohnRennie Sir can you help me with this...
It seems more of phase transformation to me
 
The impurities are more soluble in the liquid than in the solid, so as the liquid region travels along the rod the liquid becomes increasingly enriched in impurities.
The impurities don't flow anywhere. They just concentrate in the liquid.
Then when the liquid zone reaches the end of the rod it is allowed to solidify, and the end of the rod with all the impurities is cut off.
 
6:16 AM
morning
 
6:35 AM
@schn Choosing a gauge is like choosing a coordinate system. Always remember that V and A are not physical. The Coulomb gauge is designed to make electrostatics easy.
Hi @JohnRennie, can I ask about the conversion of the quadrupole moment tensor into TT gauge? From what exactly are we converting, and why does it involve the projection tensor?
 
@VincentThacker no idea. Sorry :-(
 
Oh well
 
7:06 AM
I am having a tough time to understand as to why $g_{00}$ is taken to be 1 in the FLRW metric. The book says something like “ we can always choose our time coordinate to be the proper time”. Why then we didn’t do that in Schwarzschild metric? Would, anybody here like to explain that?
 
You can change the time coordinates in Schwarzschild, if you so choose
but this will affect the other coordinates
Remember that the coordinate transformation is the old coordinates derived by the new coordinates
if you have $t \to t'$ and $r \to r'$, you will get $$g_{r't'} = g_{\mu\nu} \frac{\partial x^\mu}{\partial r'} \frac{\partial x^\nu}{\partial t'}$$
Both $r'$ and $t'$ depend on $r$, so it will be non-zero
 
@JohnRennie This guy is a professor at TIFR!! (Tata institute of Funadmental Research which is quite reputable in India) I'm surprised by the behavior of someone so senior D:
 
I was surprised. If he was referring to a peer reviewed publication it would be different, but it's just a self publication on ResearchGate.
We don't know for sure that it's him though.
 
psa
7:21 AM
that'd be odd if someone posed as him and then linked his very recently self-published paper
 
True ...
 
@JohnRennie Thanks, I understand it better now.
 
@VincentThacker I'm glad it helped :-)
 
One more question though, by projecting it, why can we simply ignore the non-transverse part of the tensor? Basically I have the same question as physics.stackexchange.com/questions/320889/…
Maybe @Slereah can help?
 
heck I dunno
 
7:32 AM
:56854201@Slereah so does that mean we could have a “type” of Schwarzschild metric written in coordinates where $t$ is the proper time and $r$ is the proper distance, of course by having some other terms in the metric like, $g_{tr}$ and $g_{r\theta}? If that is correct why don’t we do so? The interpretation of $t$ and $r$ coordinates would be so nice. Also, can we then choose $t$ to be the proper time in any metric in GR ?
 
You misplaced a $ there
Also yes, you could
It's not wise but why not
 
@Shashaank the proper time is not a coordinate
 
there's always a ton of coordinates you can put on a metric
 
You can choose a time coordinate that corresponds to the the proper time for some selected observer. The Gullstrand-Painleve metric does this.
 
@JohnRennie oops, sorry I meant to say- “ choose the time like coordinate to be such that $g_{00}=1$
 
7:37 AM
If you want an example of Schwarzschild with $g_{tt} = 1$, look up the Lemaître coordinates
 
@MoreAnonymous I dunno what the status on non-peer-reviewed papers is, but the main thing is in physics.meta.stackexchange.com/q/4538
 
His Twitter account looks fake.
 
@Slereah @JohnRennie So is there a special reason to choose the $t$ and $r$ coordinates such that $g_{00} \neq 1$ $\g_{rr} \neq 1$ in the Schwarzschild case, when in fact we could and then also have a nice interpretation of the $t$ and $r$ coordinates? Is that more beneficial in some way?
 
I don't know if it is "nice"
I mean, specific coordinates have their uses, but I've never seen Lemaître coordinates used for calculations
 
Ahhh maybe the circumference of a sphere won’t be $2\pi r$ in that case. But I don’t see that to be a problem
 
7:44 AM
@Shashaank it's nice to have the metric in a diagonal form, and setting $g_{00} = 1$ generally produces non-zero off-diagonal values.
 
@JohnRennie @Slereah Ahh I see. Thanks ! I was thinking that the only possible choice of coordinates in the Schwarzschild case was the way they are chosen. Now, at least theoretically it is clear we could have in fact chosen any other coordinates too which would respect spherical symmetry.
 
In general relativity you always have the choice of picking whatever coordinates you want
That's general covariance
although some will be more practical than others
 
@bolbteppa really?
"general covariance"
 
Hi pal @RyanUnger skullpatrol here
 
But is there any specific reason to choose $g_{00}=1$ in FLRW? There we would be having a diagonal metric nevertheless (once we have the metric for the space like hypersurface). So we could very well choose not to have $g_{00}=1$. And then plug the metric in the Einstein equations to see what all possible values could $g_{00}$ take. Is that right?
 
7:53 AM
The FLRW metric is often written using conformal time so $g_{00} \ne 0$. It's just whatever is convenient for doing calculations.
 
@RyanUnger that's its name
 
Ahh I see. Yes I got it now. Thanks @Slereah and @JohnRennie for helping me out.
 
FLRW in conformal time is nice because then you're basically doing the product of the space with just $\mathbb{R}$
 
Do you know any reference ( like the Lemaitre coordinates one that you provided) where Cosmology is done using some other coordinates
 
Other than what
 
8:04 AM
Sorry, but would anyone like to tell me the issue with Tejinder P Singh. I happened to met him a few weeks gaon in a QM foundations conference. He works with Bassi on the GRW theories and such. Was just curious to see the same person surface here
@Slereah where the time coordinate is not the conformal time, where $g_{00}\neq 1$
 
I think most cosmology books use both versions?
 
Oops I have just referred Caroll and Hobson and a bit of Narlikar. There wasn’t any
 
8:29 AM
@Shashaank I've seen it written in non-comoving coordinates. Give me a moment and I'll have a look ...
 
@JohnRennie sure, please. Thanks
 
This is the de Sitter metric not the FLRW metric, though we expect the universe to asymptotically tend to a de Sitter geometry so it is related.
Anyhow it discusses various coordinate systems that can be used.
 
@JohnRennie Can I ask you some help in GR? I've seen you help other people on the chat before?
 
8:45 AM
@MoreAnonymous I doubt I can help. I'm only an amateur in the field and my knowledge of it is pretty sketchy.
 
@JohnRennie Sarcasm?
 
@JohnRennie what toothpaste do you recommend, then
 
@Slereah Go for the one with the small hole!!
Big holes are scam!!
:P
 
@MoreAnonymous No, I'm serious. If you look at my answers on GR you'll see that they are mostly explaining GR in a non-technical way, or at least with the bare minimum of maths.
 
9:01 AM
@JohnRennie the same could be said about Feynman. But noone would say he didn't know physics. In fact it takes more skill to do that
 
I wanted to learn enough GR to understand the principles by which it worked, and I managed to do that. But my grasp of the intricacies of differential geometry is very limited.
 
For some reason I find this a consolation (I'm not the only one)
 
@Slereah Buy the supermarket own brand! 99.999% of the effectiveness of toothpaste comes from brushing well, and the stuff in the toothpaste makes little difference. Most of it is just abrasive (silica), surfactant and flavouring anyway.
 
Understood, I'll just use sand
 
If your teeth are still forming then the fluoride in toothpaste can help make teeth stronger, but for those of us long paste those years even this doesn't matter.
@Slereah The silica in toothpaste is a silica precipitate with a low hardness. Sand is much harder than teeth so if you use sand you will quickly have no teeth left to clean.
 
9:05 AM
@MoreAnonymous that doesn't mean Feynman didn't also have a deep technical understanding of what he was doing - beware of the fallacy that because some people are good at explaining things non-technically in a way that sounds sensible, a non-technical understanding suffices to argue/reason about the topic
 
Also thinking Feynman didn't use a lot of math is a bit silly
Feynman's books are horribly mathy
Less than some perhaps, but still
 
9:20 AM
@Slereah lol, Where you get such ideas?
haha :-)
 
I am Richard Feynman. I was bitten by a radioactive colloid while working at Los Alamos and it turned me into a middle aged colloid scientist.
 
Legit
 
Now the secret is out I suppose people will be asking what I had been smoking when I invented the path integral.
 
what?
 
My master thesis was on the path integral and for some reason some people in the jury didn't seem to know about it and were surprised?
It was a bit weird
They were like particle physics people
you'd think they would know it
 
9:22 AM
That's a good thing. It's easy to escape that way...
I used Euler's Identity to solve question of superposition of infinite SHM waves and the teachers don't seem to know that way of doing it.
That makes things good. I'm still worthy.JPG
 
@Slereah I've found that phenomenologists and other physicists focused on applying QFT to specific scenarios mostly focus on clever methods to compute how the QFT amplitudes translate to stuff in colliders and as such usually care very little about anything in "theoretical" QFT beyond how to use Feynman diagrams - maybe they were like that?
 
yeah but still
It's in the basic QFT books
Although to be fair, I didn't see it in class, either
 
@Slereah sure, but usually at the end and usually it's not part of the syllabus of the intro to QFT courses
so they might have read it once decades ago :P
 
phenomenologists... Okay, that's a new thing for me...
Hey! Does anyone here believe in Vitalism?
Or any proper argument to dismiss it?
 
9:27 AM
@RewCie I'm not 150 years old, so no
 
@Slereah Argument?
 
You can clearly see my birth year on my ID
it is not in the 19th century
 
But that's not a good argument to dismiss it either.
 
@RewCie I have no need of that hypothesis
 
@JohnRennie Okay...
I was looking at Russel's and Vithoulkas' arguments on it.
Seems something very useful if formulated, can help in construction of AGI/ASI
 
9:32 AM
@Slereah Wow! Kind of shows how overspecialized academia can make you :/ (in a bad way)
"So many people today—and even professional scientists—seem to me like somebody who has seen thousands of trees but has never seen a forest"
 
@RewCie What about the Wiki article is unclear to you? This has been considered pseudoscience ever since we understand enough biochemistry to see that the same reactions can happen in a test tube that happen in a cell.
 
Where are arguments against it?
I see criticism section that's isn't a disprove either....
 
I think u can prepare an organic molecule from an inorganic one? (might be wrong it was ages ago)
 
Indeed you can
 
Organic Molecule is totally a lifeless thing
I'd define vitalism something like this : A vital force/field that animates the carbon based life to intelligence
 
9:36 AM
@MoreAnonymous This isn't "overspecialization" - not everyone has to be interested in theoretical considerations.
 
@ACuriousMind Then what were doing judging Sleareh's thesis?
If they weren't interested in theoretical considerations
 
@MoreAnonymous that's a failure of the system, not of the people
 
@MoreAnonymous Can't have everyone be a specialist on every topic at the university
Especially for a thesis, which is potentially on an entirely new idea
 
@Slereah True but relevant experts for relevant thesis?
 
Well, if you're willing to pay to invite those people in
 
9:38 AM
@RewCie Oh dear! D:
 
It would be appreciated
 
Hey uni! is super expensive
Having paid for 2 masters as an international student
I can vouch for that much
@RewCie You will never find this field
 
@MoreAnonymous Okay, then we can create a rough model for it without digging either (in case it doesn't exists)
Mathematical model of the field.
 
@MoreAnonymous many unis have small research groups under a single professor, and that professor is the only specialist in their field at that place - if you need to get a committee together, you have no choice but to pick non-specialists
 
like chemistry to QM
 
9:42 AM
@RewCie physicists are ultimately pragmatists. If a theory works it's a good theory. If vitalism improved on our current theories we'd adopt it. But it doesn't. And that's all there is to it really.
 
if the vital fluid doesn't exist, how do you explain this :
 
@RewCie I think you can model fairies existing when I close my eyes. But what's the point?
 
well, physicists mostly except for biophysics don't care about living systems to begin with :P
 
@JohnRennie Sounds good
 
@ACuriousMind Wigner did! (and so did his friend :P )
@Slereah Fake news!! I smell a russian conspiracy!
 
9:45 AM
Can unit vectors in curvilinear coordinates also be defined as $ \vec{e_i}=h_i\nabla x_i$? I'm only familiar with
$ \vec{e_i}=\frac{1}{h_i}\frac{\partial\vec r}{\partial x_i}$, and I'm unable to prove the former using the latter.
 
I sometimes get the impression if you arent going to Cambridge or Oxford u have a significantly smaller chance of success in quantum gravity academia
(from undergrad)
 
@Slereah the lack of a hyphen between "man" and "eating" there bothers me - I clicked on it expecting to see a man eating a toaster
2
 
LMAO!
Now I can't unsee it
 
@MoreAnonymous uh, "quantum gravity academia" is a very small field to begin with
 
@ACuriousMind $\sim$ "Let's eat, grandma!" vs "Let's eat grandma!"
 
9:47 AM
the popularity of quantum gravity in popular perception is extremely disproportionate compared to the fraction of physicists working on it
 
(secret)
 
@ACuriousMind Yea its kind of sad though because you miss out on Oxford and Cambridge by a hairs width
And then that has such bigggg implications
 
what about Imperial
it's not like Oxbridge are the only big players
 
It's totally a different feeling when you miss class lectures and don't listen to teachers, solve questions on board and get appreciated in front of everyone.
different feeling! Gangsta!
 
I don't agree that Oxbridge have any monopoly of quantum gravity or indeed any area of theoretical physics.
 
9:50 AM
@MoreAnonymous if you're international anyway there's no need to focus on the UK, e.g. there's several unis with good string theory research in Germany, too (e.g Munich, Heidelberg, Hamburg (DESY))
I don't know the situation in the UK but I suspect you're blinded by PR there, too, and there's more places where you might find the kind of research you're looking for
 
More physicists work on like rheology or antenna theory
 
@NiharKarve I was part of Durham (considered as good as imperial) and man I hated that place
 
But it is less spooky
 
Interestingly I loved Nottingham a lot more
 
@MoreAnonymous really? Why?
@MoreAnonymous my mother went to Nottingham.
 
9:53 AM
@JohnRennie There's a lot of stupid politics going on. For example, if a prof sets an exam which students are finding difficult to pass without memorising the students remove him via the means of negative feedback
 
These days Nottingham (the city not the university) has a major crime problem. I used to go there on business, and the local radio was full of stories of how many people had been stabbed the previous day.
 
@MoreAnonymous academic cancel culture?
 
@JohnRennie I have a friend who did some drugs there and choked another student ... Lost his Phd Offer and sent back to India
@NiharKarve Well thats not the only problem with Durham. Durhams way handling students with disability sucks
They basically want to know everything about ur illness
while the NHS is trying to give you privacy
So thats another thing
In Durham scoring good marks does not mean understanding the subject. You are incentivized to memorise alot of past papers, etc.
Lots of BS
 
That surprises me. Durham has a very good reputation in the UK.
 
@JohnRennie Indeed. It surprised me too ://
I mean the students are smart
But they;ve gamed the system
too much against my liking
 
10:04 AM
To be fair, if I was paying £9k/year to go to a university I'd be pretty demanding of it.
 
Suckers
I paid like 400 bucks a year
 
@Slereah Where??
 
France!
 
@Slereah I got paid to go to university.
 
nice
 
10:08 AM
Not only were there no fees to pay, but I got a grant to cover living expenses.
 
cries in USA
 
@JohnRennie Also maybe I'm wrong but compared to my classmates they had far inferior math abilities to mine. Which suggested to me something went terribly wrong in UK schooling eduction system
 
Of course there was less to learn back in those days.
@MoreAnonymous are you from India?
 
@JohnRennie Yes
 
I don't know why so many people go to uni in the US
 
10:09 AM
For all it's failings the JEE system turns out students with awesome maths skills.
 
It would probably be cheaper to move to Europe for a few years to get a degree
 
@Slereah The american dream
@JohnRennie Ah yes
I did prepare for that exam
 
The JEE students have stopped asking me to help with maths questions because I can't answer them :-)
 
@JohnRennie What???? Your a Phd!!!
 
He's not a PhD in "bag-of-tricks-needed-to-solve-JEE-maths-problems"
 
10:11 AM
To be fair a lot of the JEE maths depends on knowing tricks to solve problems, and unless you do maths all the time you very quickly forget them.
Plus the availability of online integrators has devastated my ability to do integrals manually :-)
 
@JohnRennie Not true... I've seen some kids to whom all this comes very naturally
Also did I ever tell u i invented my own math formula?
 
@JohnRennie I still mostly use Gradshteyn
 
13
Q: The Definite Integral Problem (with a twist)?

More AnonymousThe Definite Integral Problem (with a twist) In the Riemann integral one essentially calculates the area by splitting the area into $N$ rectangular strips and then taking $N \to \infty$. Here's something I asked myself related to the Riemann integral. Let's say I split the area into say $3$ st...

@NiharKarve I reckon I could still do well in the JEE math section if I prepared a week for it
 
Now or back then?
 
@NiharKarve Now. But with a job and all I doubt I'll have the time
 
10:26 AM
@JohnRennie @JohnRennie Thanks for the article. The time coordinate in that link is not, the proper time, right? And also is $H$ there a constant or a variable? In one of the previous comments above you mention that in the Gullstrand Painleve metric the time coordinate for some observer is the proper time. But in that metric $g_{00}$ is not equal to 1. So why does the time coordinate there correspond to the proper time.
 
H is the Hubble parameter so it is time dependent not a constant.
In the GP metric the time coordinate is the elapsed time for an observer falling freely into the black hole i.e. it is the proper time for that observer.
 
@JohnRennie And in the PDF you linked, eqn 1.38, the time coordinate is not the proper time ( $g_{00}\ neq 0), but it’s just the Schwarzschild metric with M=0 and a cosmological constant. Does that eqn suggest an expansion of the universe ?
@JohnRennie and will the t coordinate in that linked answer correspond to the proper time for any observer (i.e one for whom H =0)?
 
123
10:59 AM
Hello guys..
 
 
2 hours later…
12:41 PM
3
Q: Variant on divergence theorem

custodiaIf I want to prove that for any scalar field $f:\;\mathbb{R}^3\to\mathbb{R}:$ $$\int_V \boldsymbol{\nabla} f\;\mathrm{d}V=\int_{\partial V} f\;\mathrm{d}\mathbf{S}$$ Can I apply the divergence theorem to $\mathbf{a}_1=(f,0,0),\;\mathbf{a}_2=(0,f,0),\;\mathbf{a}_3=(0,0,f)$ and then stack the equal...

?Does this form of the divergence theorem have a name
 
 
1 hour later…
2:08 PM
ok, BPST instantons:
principal $\rm G$-bundles over $X$ are classified by homotopy classes of maps from $X$ to $B\rm G$
We love $\mathbb{R}^{3, 1}$ which we can shuffle around to $\mathbb{R}^4$ and then $S^4$ for kicks, and $\pi_4(B\text{SU}(2)) = \mathbb{Z}$, call this number A, which classifies isomorphism classes of principal $\text{SU}(2)$ bundles over $X$
Then there's also the second Chern class which sends $c_2 : B\text{SU}(2) \rightarrow K(\mathbb{Z}, 4)$. Since we're only working up to homotopy equivalence, principal $\text{SU}(2)$ bundles over $X$ can be classified by elements of $H^4(X, \mathbb{Z})$
Two questions here, is the Chern class map also preserve homotopy classes? I would assume it does, since the composition has to respect that.
Second, taking $S^4$ again, the "instanton number" is an element of $H^4(S^4, \mathbb{Z})$. Is this the same thing as "A" in the previous (less general) derivation?
This might be really obvious but I'm blanking out here
 
2:26 PM
@NiharKarve This is the universal 2nd Chern class. Ordinary Chern classes are the corresponding elements in $H^4(X,\mathbb{Z})$.
I'm not sure what you mean by asking whether the universal Chern class "preserves homotopy classes". It's a continuous map, if that's the question
 
2:37 PM
For the second question, in the case of $X=S^4$, you have $H^4(S^4,\mathbb{Z}) = \mathbb{Z}$ and so the Chern class is just a number we call the instanton number
 
@ACuriousMind ah yeah, whoops
@ACuriousMind does that coincide with the integer that's an element of $\pi_4(B\text{SU}(2))$
Side question: if I were to ask something like this on the main site, would that be on maths SE?
or physics SE with [mathematical-physics]
 
Depends if you can tie it to a physics problem
 
2:52 PM
I mean, it's about BPST instantons which I guess are physical
but the actual question is more about the mathematics
 
3:05 PM
@NiharKarve Not obviously, no
 
Today, I realized that Wikipedia isn't more reliable than some random tweet
with a webpage mention on it
 
@ACuriousMind all right, thanks
 
@NiharKarve In this case I'd probably rather try it on math.SE since it's much more likely to be answered by a "mathematical" gauge theorist than a "physical" one
 
makes sense
 
you can also try physics overflow
I don't know if people still go there though
 
3:22 PM
@NiharKarve Here's an MO answer that spells the maps in question out in more detail - the latter number can be the negative of the first number
 
@ACuriousMind Any idea of what theorem he uses to go from 32 to 33? arxiv.org/pdf/1201.2714.pdf
He expands around multiple saddle points, but I'm not 100% sure where that comes from
Steepest descent I guess, but it's hard to find examples online about the case with multiple saddle points
 
@ACuriousMind very nice, that's a good find
 
@Slereah I don't think it's a theorem :P
 
then what is it
Did it fall from the heavens
 
Laplace something?
 
3:29 PM
In mathematics, the method of steepest descent or stationary-phase method or saddle-point method is an extension of Laplace's method for approximating an integral, where one deforms a contour integral in the complex plane to pass near a stationary point (saddle point), in roughly the direction of steepest descent or stationary phase. The saddle-point approximation is used with integrals in the complex plane, whereas Laplace’s method is used with real integrals. The integral to be estimated is often of the form ∫ C...
I think it may be this
 
Looks kinda like Laplace's method
oh yeah
 
@Slereah I think it's very obviously the method of steepest descent just applied "around each saddle point"
I'm not sure what sort of theorem you're looking for for that
 
I guess so, but I'm having troubles finding how to apply it in such a case
I mean there's that wiki section, but there's no bibliographical reference
 
Again, what do you need a reference for here? What statement needs to be proven?
 
That one I suppose
I'll check De Bruijn
 
3:34 PM
I'm not sure what about the section of the Wiki article you think is incomplete
sure, it should probably have a reference to comply with Wiki's standards, but the partition of unity argument is sound
 
Not incomplete, just would like it explained in more details
 
what additional details would you like? The partition of unity trick reduces the integral into a sum of integrals of functions that each just have a single saddle point, uses the original asymptotic for each of these, then sums them together again
what's missing?
 
Nvm
 
This might just be a matter of preference, but what is the instanton? Is it the principal bundle or the curvature or something else?
 
@NiharKarve It's the classical solution/vacuum field configuration for a particular bundle
 
3:40 PM
What is a vacuum
I'm a mathematician
 
it's a classical solution to the equations of motion :P
 
Let's say all I know is that $F_A = F_A^\pm$
 
Believe me when I say that it was that answer that got me interested in instantons in the first place
I'll have a closer read, cheers
 
4:34 PM
when is Urs Schreiber publishing a book anyway
He keeps rewriting fancier version of his introduction to prequantum field theory
 
4:49 PM
can't wait for sheafified ∞-quantum field theory
 
Or its sequel, the $\infty+1$-theory
 
5:03 PM
seriously though, what on God's green earth is a categorical sieve
 
I assume it's like Eratosthenes' sieve, but for categories
 
 
3 hours later…
7:48 PM
@NiharKarve the objects exist not on green Earth, but only in platonic Heaven
 
8:29 PM
Is the path integral formulation in QM a way to avoid invoking an interpretation like the Copenhagen interpretation?
As in, Copenhagen and path integral are mutually exclusive interpretations of qm
 
No, the path integral is a formalism, not an ontology
or an interpretation
 
hmm
oh I think I see
 
I would say the 'axioms' in a Copenhagen approach can be replaced by equivalent path integral 'axioms' i.e. it's just Copenhagen
 
the path integral is just a way to calculate the amplitudes, the copenhagen interpretation is still possible, since it just states that that particle collapses to such a state with the given probability
 
@Slereah any idea if we can get back schrodinger’s eqn from dirac’s eqn. Dirac proceeded the other way round, I know. But I was trying to see if under proper non relativistic assumptions, the opposite is possible. I got into trouble because Dirac eqn is 1st order is the time as well as space coordinates
 
8:36 PM
The Schrödinger equation would be the non-relativistic limit of the Dirac equation
Also removing the spin dependence
 
In non-QFT, the path integral is just a different way to compute amplitudes and expectation values. In QFT, the waters get a bit murkier since it's technically unclear whether or not it is equivalent to the operator formalism in general, but in neither case does it have anything to do with interpretation
 
It can suggest an interpreration
but that is a different thing
 
oh, I wasn't aware that canonical quantisation and path integral were anything but equivalent
 
although of course you might argue it lends itself a bit more naturally to interpretations like MWI or consistent histories than it does to Copenhagen or other collapse-y interpretations
 
that's just what I'd been told though
hmm yeah
 
8:39 PM
@Slereah but how do I formally get back Schrodinger’s eqn from Dirac’s eqn
 
@Charlie the path integral over arbitrary fields is not mathematically well-defined so you cannot ask whether it is equivalent to the operator formalism before you've figured out its definition :P
 
@Shashaank Some kind of Wigner contraction, I would guess?
 
Oops I have no clue about what a winner contraction is :)
 
Take the limit $c \to \infty$
 
@ACuriousMind :0
 
8:42 PM
and it's a bit difficult to tell the difference anyway since in many cases you end up computing the same Feynman diagrams perturbatively either way
 
But in Schrodinger the Order of t derivative is one but of “x” derivative 2. Whereas in Dirac, the order of both t and x derivatives is 1.....
 
but the usual physics assumption is that they are indeed equivalent and you can switch between the formalisms at will
I don't know of a case where it goes obviously wrong, so it's not a bad assumption :P
 
Remember that solutions of the Dirac equation are also solutions of the Klein Gordon equation
 
Hmm yeah
But then how does it help, Klein Gordon becomes 2nd order in t. If I take c to infinity the t derivative becomes 0.
 
Heck I dunno
the limit isn't gonna be so trivial I'm afraid
 
8:48 PM
Hmm yeah I was wondering that..
 
5
Q: Non-Relativistic Limit of Klein-Gordon Probability Density

SimonIn the lecture notes accompanying an introductory course in relativistic quantum mechanics, the Klein-Gordon probability density and current are defined as: $$ \begin{eqnarray} P & = & \dfrac{i\hbar}{2mc^2}\left(\Phi^*\dfrac{\partial\Phi}{\partial t}-\Phi\dfrac{\partial\Phi^*}{\partial t}\right) ...

 
@Shashaank can you derive the non-relativistic $L = T - V$ for a point particle from some relativistic action?
 
@Slereah looks interesting. Let me see. Looks interesting.
@bolbteppa but that doesn’t mean that the correct non relativistic should always be T-V. Further I can always (should always) be able to derive the correct non relativistic eqns from the relativistic ones (which I get by the relativistic action)
 
9:10 PM
Have you ever tried to consider an interacting relativistic point particle apart from the EM case
1
A: Lagrangian of a Relativistic Harmonic Oscillator

AndreasThis is a problem that requires the extension to general relativity. It is not possible to solve this problem remaining in special relativity. Make the relativistic ansatz (in 2d spacetime) $L = - m c \sqrt{-g_{\mu\nu}(x)\dot{x}^{\mu}\dot{x}^{\nu}}$ where the dot is with respect to the geodes...

If you can get a free particle Schrodinger equation from a free particle relativistic equation such as Klein-Gordon or Dirac that's enough before things go absolutely haywire
 
Sorry to bother u all, Could u please check the statement mentioned below and make the possible correction, {statement: MM experiment: Results and conclusion: (I) No deflection detected in the fringes pattern, So there is no 'constant relative speed of light' with respect to any medium hence, No Aether Medium exist. (II)The speed of light will be the same in every direction in a particular inertial frame of reference.}
 
9:33 PM
@Shashaank see (2.103) to (2.105) here for the KG case
Slide 57 on for the Dirac equation (set the EM potential to zero)
 

« first day (3735 days earlier)      last day (39 days later) »