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5:45 AM
Now I am trying to stop myself from finally reading the BPZ paper mentioned in this pop-sci bootstrap article...
1 hour later…
6:56 AM
When we renormalise the field strength, why do we 'exit' the interaction picture?
7:22 AM
Not sure what you mean
1 hour later…
8:26 AM
Hold on, I'll grab the source
This question: physics.stackexchange.com/questions/3983/…, specifically the answer by user1504. From the looks of it, the applicability of Haag's Theorem seems to be a somewhat controversial point, so I'm not sure if this is mainstream
8:46 AM
@NiharKarve It is definitely "mainstream" that Haag's theorem says the QFT interaction picture doesn't work rigorously
What's controversial is to what extent physicists should care :P
right, so user1504's answer suggests that renormalising the field strength means that you are abandoning the interaction picture
how does that happen?
There's simply no trace of the interaction picture left in the things you compute in the renormalized theory
@ACuriousMind thanks for pointing that out. I think the "actual" world is the real-world we live in; while, "all possible" worlds are the abstractions math makes.
the interaction picture manifests only in the intermediate steps we do to arrive at something like LSZ
The interaction picture is basically just
In mathematics, variation of parameters, also known as variation of constants, is a general method to solve inhomogeneous linear ordinary differential equations. For first-order inhomogeneous linear differential equations it is usually possible to find solutions via integrating factors or undetermined coefficients with considerably less effort, although those methods leverage heuristics that involve guessing and do not work for all inhomogeneous linear differential equations. Variation of parameters extends to linear partial differential equations as well, specifically to inhomogeneous problems...
9:00 AM
Both conform by absolute necessity. @ACuriousMind
If what the answer said makes sense, I guess it's just saying this math trick better not leave any lasting physical impact on the final answer (i.e. it only matters in the intermediate steps)
I think the point about asymptotics that Lubos makes is why all of this stuff is basically irrelevant
Sorry, I don't see the correlation between variation of parameters and the interaction picture
9:21 AM
The thing is the region will never cover "everything imaginable."
as Gödel proved.
Thank you @bolbteppa and @ACuriousMind
10:20 AM
Why do people post on Twitter first, before here that they are quitting?
Must be a SO thing, I guess.
11:10 AM
Hey, I thought I'd ask here first instead of posting as a question, but does anyone have any good resources for understanding multipartite entanglement? I'm well versed with bipartite measures but read this paper (arxiv.org/abs/1106.4774) and i couldn't even understand the first result about all tripartite states being of the form mentioned, so I was hoping for a more pedagogical intro to tis subject.
2 hours later…
12:59 PM
Developing OS is sooo painful
because computers are dumb!
1:39 PM
I often see statements that say Einstein's paper on photoelectric effect proves that light consists of particles. How does it prove or disprove that light is a wave?
My understanding is that the paper proposed a hypothesis to explain Hertz's observations but how does it actually "prove" that light is made up of particles even if the mathematics works out.
2:08 PM
@Yashas It doesn't "prove" anything. It proposes a model of light as particles that explains the specific observations of the photoelectric effect that the classical electromagnetic wave model cannot explain.
in a physics context people often use "prove" when they really mean "provide a convincing explanation", they don't mean that it adheres to any mathematical standard of proof
2:37 PM
hi all
is it correct to say that for a particle like $e^-$, time evolution can bring a $e^-_L$ in $e^-_R$ thanks to the fact that the particle is massive ? (i.e., the weyl equations for the weyl spinors $e^-_L $ and $ e^-_R $ are coupled). And I'd even say that if I imagine absurdly that a $e^-$ is massless than both the left / right chiral states exist but they don't get mixed up like before. (depending on if this is right or not maybe a question on the neutrino will follow)
3:00 PM
@Ratman Yes.
this has nothing/very little to do with neutrino oscillations if that's what you're going to ask
3:16 PM
thanks @ACuriousMind . I actually don't know much about netruino oscillations. My problem with the nuetrino is this, if I assume that it is massless, then there isn't any known process that can create $\nu_R$ ( I think), but conceptually it would be wrong to state that $\nu_R$ doesn't exist, simply for now i can't "see" it. It is kind of a philosofical question maybe, but I got confused about this by a sentence in a paper
@Ratman afaik, no one thinks the neutrino is part of a Dirac field. Even in models like seesaw mechanisms where you have right-handed Neutrinos, they carry Majorana masses, not Dirac masses
i.e. the $\nu_L$ and $\nu_R$ aren't coupled via the Weyl equations even when they're massive - the masses are Majorana masses that don't appear in the Weyl equations
ok, that's really interesting. And at the same time I see I need to study more about neutrinos to understand this. Thanks a lot @ACuriousMind, your help is being very usefull
3:40 PM
@BioPhysicist @VincentThacker Regarding this migration, it's worth keeping in mind that migrations get blocked if the question's tags don't exist in the destination site
ctrl+F for "tag" here for more details
now, as it happens, the maths site does turn out to have an tag, but one can't really count on that
@EmilioPisanty this migration
anyways, I thought it was worth pointing out this quirk of the rules.
huh, I didn't know that either
I can see where this rule is coming from but I can also see why it's silly :P
@ACuriousMind yeah, well, of course you wouldn't, Mr Exception Man
> A migration can be automatically blocked before it even gets migrated if any of the following conditions apply:
> The question does not use any tags that exist on the destination site, with three exceptions:
> - a moderator performs the migration
@ACuriousMind why is it silly?
in larger sites there's a substantial problem of over-migration, i.e., folks migrating crap questions that don't fit the target site at all, just because they don't want them in the origin site
asking that the tags match is a good bare-minimum requirement to ensure this
I don't really envy the sites on the business end of the migration funnels coming out of SO
there's probably a ton of crap flowing there
@EmilioPisanty because it requires the tags on both sites to be aligned but there's no system to align them - what if we have and it's called on math?
3:50 PM
not necessarily as a large fraction of the migration flow, but just out of the sheer volume of the total
(of course if the tag is frequent there will probably be lots of synonyms so this rarely causes issues, but still)
@ACuriousMind that.
and it also means that we have to have a lot of tags like that I don't think we'd necessarily have if we didn't need them to migrate
@EmilioPisanty hmmmm, it appears OP deleted the migrated version.
but since tags don't cost anything that's not really an issue either
is "integration" intrinsic to math?
is intrinsic to math
ah you mean if we just wanted to be able to migrate stuff we could just tag all migration candidates with that
fair enough, then
3:54 PM
if that's the only tag then the migration succeeds (barring any other issue), and it appears in MSE as
@ACuriousMind out of curiosity, what are the intrinsic tags here?
A: Is it possible to find out which are intrinsic tags for a given site?

CaiThere is, as far as I'm aware, no way for regular users to see intrinsic tags*. Moderators can see intrinsic tags in the same place as blacklisted tags though (labelled as "Intrinsic Tag" on the Blacklisted User Input page, visible at /admin/blacklist). For example, on Graphic Design I see (reda...

apparently only mods can see it
it's at /admin/blocklist now and our intrinsic tags are and
eh, fair enough
anyways, it doesn't look like tag mismatch is a significant problem causing failed migrations
@EmilioPisanty Ah I see; I didn't know this rule existed.
@VincentThacker it's an edge case
not many people know it
@ACuriousMind (ahem)
but the way to learn about edge cases it to have them pointed out when one runs into them ;-)
If anything, [limit] should be [limits] and [group-ring] should be [group-rings] on MSE as well. — Asaf Karagila ♦ Jan 2 '14 at 7:50
@AsafKaragila Then there are [definite-integral], [indefinite-integral], [extension-field]... where is Atwood's giant $\Huge{S}$? — Post No Bulls Jan 2 '14 at 8:03
Integrals seem better in singular form; but [extension-fields] does sound better. I only know where is Riemann's giant $\huge S$: $$\Huge\int$$ — Asaf Karagila ♦ Jan 2 '14 at 8:06
4:11 PM
@EmilioPisanty Thanks for the information. I was unaware of this
mathematician humour mixed fifty-fifty with old-timer SE humour, winner combo =P
What came first angular momentum or torque?
the egg
wait wrong question for that answer
I gotta admit it took me a google to recognize "Atwood's giant S" as SE in-joke as opposed to some obscure mathematician
4:14 PM
on physics.SE it might also have been a reference to a weirdly shaped pulley system :P
Hi Guys Yo....!...
@ACuriousMind indeed
Please tell i am asking seriously
@PrateekMourya in historical terms, you mean?
4:15 PM
that's not a particularly easy question
it's more History of Science and Mathematics material tbh
@PrateekMourya while I was making a joke I was trying to point out that this seems like a chicken-and-egg question to me where the relevance of the answer is unclear
Ahh.. i thought it was a common question
I was confirming that angular momentum was first defined
Like Newton's said that any large planets will retain their angular momentum like the law of inertia
@PrateekMourya the history of how the concepts of physics were discovered isn't particularly relevant to how we use them nowadays
So analogous to that will be torque
Uh could I double check I've got something right, a "something"-valued p-form (eg. a vector valued p-form) just means we are defining a tensor $\vec v\otimes \tilde\alpha^p$ where $\tilde\alpha^p$ is a p-form, whose output is a vector if we "partly fill" its arguments with enough vectors to "contract" the p-form, leaving a vector (or something else)?
4:17 PM
there are some exceptions, particularly with quantum mechanics, but this type of history is rarely taught in physics degrees
@Charlie jup
anyways, there's useful links if you look on HSM hsm.stackexchange.com/search?q=torque
and there's a History section in the Wikipedia page for angular momentum en.wikipedia.org/wiki/Angular_momentum#History
note in particular that the term "angular momentum" itself only seems to have come into common usage around ~1850, whereas there's substantial useful understanding towards that term in Newton's Principia form 150 years earlier
the concept of angular momentum did not emerge fully-formed in one single fell swoop
it was built piece by piece over several centuries
but anyways:
Q: Who gave mathematical expression for torque?

Immortal PlayerI am trying to understand the conceptual development of the "torque". I saw that Archimedes was the one from whom this idea took birth. I wanted to know from whom the torque got its mathematical form. I will also be happy to know further reading materials and sources.

there's enough there to make a strong argument that the concept of torque dates back to Archimedes
@PrateekMourya and that's sufficient to answer your question.
I have one more trouble
I am not able to understand areal law
Q: Why is there no named unit for momentum but there is one for energy?

dmckee --- ex-moderator kittenMomentum and energy play very similar roles in mechanics, each being changed by the application of force over a interval. For energy the interval is in space and for momentum it is in time. Both have associated conservation laws. Yet, energy units are named in many systems and momentum units gen...

good question, @dmckee
(wait, dmckee is no longer taggable? oh well....)
4:28 PM
he stopped SE activity pretty much completely after he stepped down
ah well
In first line Newton's says object in first interval move in a straight line
> Momentum was out of luck.
how appropriate would it be to edit the end of that answer to SOL?
I think I'm not the only one that calls the unit "p's"
@NiharKarve I just call the momentum units a.u.
4:31 PM
But the question will be if force is continuously central then why it would move in a straight path
(but then again I do the same with every quantity)
I was under the impression you only spoke GeV
or maybe thats ACM
@NiharKarve I've never (professionally) touched a GeV in my life
I don't think I've ever touched a keV, even
I'm down in the eV regime where all the actual real-world physics happens ;-)
(a.u. standing, of course, for atomic units, i.e. $\hbar=m_e=\frac{e^2}{4\pi\epsilon_0}=1$)
4:35 PM
About the "p's" thing, I only meant classical mechanics - like body 1 transfers 4 of its p's to body 2
@EmilioPisanty I was thinking "arbitrary units" :P
@ACuriousMind we also use those, but only when we can't be bothered to provide an absolute scale for the power spectra of high-harmonic generation
just slap on "(arb.u.)" on the axis and bam, you're done
no need to fiddle with constants or whatever
what is physical in QM? is the wavefunction physical? is there some underlying physical thing which the wavefunction represents?
define "physical"
in classical mechanics, I can say that a ball is physical and x(t) represents its motion
is something really there is the question
4:39 PM
Yes, something is there. What that is is up for debate :P
the standard QM formalism does not commit to any particular interpretation or ontology
that's the realm of quantum interpretations
> is there some underlying physical thing which the wavefunction represents?
there is no accepted answer to that question
@ACuriousMind is there a word for the study of theories? Regarding formulating them, recognising their sphere of application (on a meta-level), etc.
@NiharKarve the term you're looking for is probably the study of quantum foundations
@NiharKarve In general? Philosophy of science, philosophy of logic, epistemology. For QM specifically what Emilio said.
Well, not QM in general, although I was inspired by QFT - yes, epistemology looks roughly correct, thanks
4:48 PM
anyways, it is possible to think about QM strictly in operationalist terms, as described here, where wavefunctions are just descriptors of preparation procedures and "observable" operators are just descriptors of measurement procedures
I don't know how common this view is in practice out in the wild
$\psi$-ontic and $\psi$-epistemic?
or whether it's even held at all by anyone, either by working physicists or by folks in active quantum-foundations work
thanks, but that's a little too meta for my taste
the classic is this survey, and I don't recognize any of the interpretations listed there as purely operationalist (but then again I don't fully understand all the options in the survey)
@NiharKarve yeah, it sounds bonkers at first sight
if you want to get in on the ground floor, I'd recommend this paper
The PBR theorem is a no-go theorem in quantum foundations due to Matthew Pusey, Jonathan Barrett, and Terry Rudolph (for whom the theorem is named). It has particular significance for how one may interpret the nature of the quantum state. With respect to certain realist hidden variable theories that attempt to explain the predictions of quantum mechanics, the theorem rules that pure quantum states must be "ontic" in the sense that they correspond directly to states of reality, rather than "epistemic" in the sense that they represent probabilistic or incomplete states of knowledge about reality...
@NiharKarve (but if I wanted to pull your chain a bit, I'd remark that those terms are only relevant within realist theories, i.e., when you've already committed to there being something real, and not everybody agrees with that =P)
can I approximate any photon travelling in space far away from mass as a free particle?
5:01 PM
@EmilioPisanty Thank you, I'll explore a little quantum foundations
5:15 PM
I saw my question was closed as it was considered off-topic, so I edited it hoping it will be reopened physics.stackexchange.com/q/593321/262619
you don't need to post it here for that, the first edit after closure automatically enqueues it for reopen review
Growing up I used to enjoy reading Halliday and Resnick questions in Principles of Physics. But now I realise that Physics currently cannot represent reality accurately. This seems to me because the subject is hitting a wall which is physical. This wall is the inability of computers to operate on real numbers.
there's lots of problems with the implementation details of numerical algorithms, but rounding errors due to floating point representations are usually rather low on that list, so I'm not sure why you would think that
also, an approximate representation of reality is still better than no representation of reality, so what's the issue?
5:30 PM
Hi, an isometry of a semi-riemannian $f:\Bbb R^{1,1}\to \Bbb R^{1,1}$ is different than an isometry for a Riemannian manifold right? In the Riemannian case distances must be exactly preserved, but for an isometry of a semi-riemannian manifold, the spacetime interval just needs to be preserved?
(btw, $\Bbb R^{1,1} from the textbook is a 2-dim. semi-riemannian manifold)
there's no difference between the notions of isometry for Riemannian and pseudo-Riemannian manifolds, "distance" and "spacetime interval" are just two different names for the metric on such a manifold
we don't call it a "distance" in the pseudo-Riemannian case because it can be zero and negative, which doesn't play well with our intuitive notions of "distance"
ah okay, so that means if you have a mapping that's not linear, it's not going to be an isometry neither in the Riemannian nor semi-riemannian case
depends on the metric you've chosen, but if you're working with the standard Euclidean and Minkowski spaces there, then yes, all isometries are affine
5:38 PM
okay yeah that makes much more sense now. So, I'm trying to work through a proof and show that a nonlinear mapping (a sort of dilation I think) from Minkowski to Minkowski, preserves a "strongly equivalent metric"
because it cannot be a isometry, I now know
Looking at the definition I'm pretty sure it's not even "strongly equivalent" lol
@ShashankVM your edited version now appears to be asking which box falls fast when both boxes are falling freely in air. Is that a fair summary of it?
5:54 PM
What is pain?
Design an OS and a compiler for that OS for x8086 16 bit OS!
6:37 PM
anyone know how to write f(x,y)=(2x,3x) as an explicit function
So $f:\Bbb R^2 \to \Bbb R^2$ I should add
oops. should be 3y
what do you mean by "explicit function" and how is that not one?
1 hour later…
7:52 PM
@geocalc33 That looks like an explicit function to me. At least compared to an implicit one
8:21 PM
There's a user who has placed a 50 point bounty on this question, with not much success, and has written a comment asking for suggestions of other SE sites to ask the question, if the bounty is unsuccessful in getting them an answer. Please take a look if you might know how to help them!
Q: How to convert lab frame quaternions and plane normal to misorientation quaternion and crystal plane normal for grain boundaries?

SterlingBackground Let qA and qB represent unit, orientation quaternions of grain A and grain B of a grain boundary in the lab reference frame, respectively. Let qm be the misorientation quaternion of qA and qB. Quaternion inversion (qinv) and quaternion multiplication (qmult) are discussed in Quaternion...

8:37 PM
I have a quick q about the "choice of connection 1-form". I've seen in the construction of a connection on a principal bundle $P$, the choice of connection comes down to the choice of a 1-form on $P$, this is roughly where the lecture I followed finished. On the other hand a connection on a rank-k vector bundle $E$ involves defining an $E$-valued 1-form such that a connection is a map: $$\nabla:\Gamma(E)\rightarrow \Gamma(E\otimes \Omega^1(M)),$$ so we more or less take a section...
... act with the connection on it, then contract the result with another section (this process amounts to "taking the covariant derivative" of one section wrt. the other) and get a third section. I'm wondering if this "choice of connection on $P$" in the principal bundle is essentially the equivalent of the choice of 1-form in $\Omega^1(M)$ for the vector bundle.
Actually I take it back, I'm not sure ;_;
@Charlie the connection on $P$ induces a connection on $E$ iff $E$ is an associated bundle, see en.wikipedia.org/wiki/…
of course
Does the choice of 1-form field on $P$ induce "the same" 1-form field on $E$ which gets used in the connection on $E$?
I suppose the answer is probably yes, other wise it isn't much of an "induced" connection
well, there is an "induced connection" on $E$ that uses that 1-form field. You could in principle pick a different connection, but why would you?
I see

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