The h Bar - 2016-07-28 (page 3 of 6)

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4:01 PM

@heather Yeah, there's something to that. Have a look in the mother meta to see if it's been proposed and what the team said then. It's bound to have come up at some point.

@EmilioPisanty I'd rather suggest that your assessment of the audience itself is not completely correct :/

@EmilioPisanty, okay, I'll look into it and come back when I have something.

@lucas Your votes are for you to use.

(In that particular case, I really don't see the issue. OP asks "what's the term for X?", answer says "the term for X is $S_z$". But again, your votes are ultimately for you to use.)

@heather your comment is not clear to me. What are you responding to, exactly?

@ACuriousMind: it was an attempt at a Discovery Channel level explanation, but since you've seen fit to post an answer I'll delete mine.

4:04 PM

Also, to make sure we're on the same page: the solution that you say completely solves the problem is to have a queue for flags of pseudo-answers, which would be handled by regular users posting those pseudo-answers as answers and flagging the comments, right?

@ACuriousMind which is a very common way for authors to get their text wrong. If you misjudge who will read your stuff, and that makes your text not suited to its readers, then that's what you have.

@EmilioPisanty In my case Op asks "What is entropy really?" And answer says "Entropy is ..." I see no problem here also.

@JohnRennie ???

@DavidZ God, please, no. I certainly don't have the energy to go through another depressing review queue.

Hello friends

4:05 PM

@ACuriousMind well I had a feeling you and others would think so, but I'm discussing this with heather and I think we're currently ignoring the issue of people's reluctance to go through queues for simplicity

Who's Heather?

@EmilioPisanty I think one person here down-voted my answer!

https://en.wikipedia.org/wiki/Mass_in_special_relativity

Ouch... I guess I will leave $mU^{0}$ uninterpreted as otherwise I fear I will be confusing myself with the notion of relativistic mass

I do remember $mU^{\alpha}U_{\alpha}$ is the energy of the object, but I might need to revise my GR books to see what type of energy it is

@lucas Yeah, that's one way to see it. I disagree with your assessment of what entropy is, and I disagree that that's all that the OP is asking (i.e. they're not soliciting any perspective on entropy, they're asking for a physical explanation of a concept). If I thought that your post was so off-base that it did not actually attempt to answer the question, I would have flagged it as NAA. As it is, it's just an answer I disagree with.

@JohnRennie And here I was thinking you had stumbled upon a deep insight into string theory ;/

4:07 PM

@lucas three people, as of right now. +2 -3.

@EmilioPisanty No, I mean right now.

Seriously. If you're upset about downvotes on that answer, consider the possibility that people simply disagree with you.

@EmilioPisanty: meta.stackexchange.com/questions/180325/…

@EmilioPisanty ACM is stalking me - he keeps on trumping my answers with a better answer of his own :-)

@EmilioPisanty I know and I respect all of them. But I will never change my answer.

4:09 PM

@EmilioPisanty: meta.stackexchange.com/questions/180082/…

@lucas That's your prerogative for sure. Just stop caring about downvotes so much. Que se te resbale. Laissez-les glisser hors de vous.

@EmilioPisanty I don't care about down-votes. But I will be happy if they say their reason. Just this. And I read English hardly and you post an Italian sentence!!!

@heather That's a good find. But then the thread is three years old and SO-specific.

@lucas You're French, if I remember correctly, right?

@heather And also with no replies from the team. Look for questions with the [feature-request] tag.

@EmilioPisanty: meta.stackexchange.com/questions/219612/…

@EmilioPisanty No, you cannot remember! Because I never have said where do I come from :-)

4:14 PM

@EmilioPisanty: meta.stackexchange.com/questions/213691/…

@lucas Then you'll have to forgive my faulty memory.

OK, I have to leave chat for now, I'll come back and check in later

@EmilioPisanty I am worst than you. I sometimes forget $F=ma$! ;-)

@heather Consider proposing it as a feature here meta.stackexchange.com/questions/252690/…

Though I'd put it at 15k

Bigger than 10k, but a good extra to motivate you past that milestone.

@lucas OK, if the discussion is only smalltalk then I do have other things to do.

@DavidZ, okay, just one final thing to address above comments: yes, the solution would be an all-purpose comment flag review queue including specific flags for pseudo-answers. The posted links are precedent for this as EmilioPisanty thought such a queue would be useful to take weight off moderators.

@EmilioPisanty, sure, I think that'd be perfect.

4:19 PM

@EmilioPisanty Sorry, I couldn't get this comment's meaning.

@secret does light have mass in its own reference frame?

the only mass it gains is from relativistic effects, correct?

@EmilioPisanty OK. Thanks again!

Ah, that's heather

nope. It has no invariant mass, but its energy can differ with reference frames

@EmilioPisanty, I'll update my physics meta post with this conversation info and I'll post the feature request. Otherwise, I have to leave chat. Thanks for your help!

4:21 PM

@DavidZ : are there many answers in comments anyway? I thought it was only CuriousOne doing it.

@heather No worries. Remember: if you're talking about any changes to the core Q&A engine, check the mother meta first. SE's been around for eight years, so most of what you're thinking has already been brought up elsewhere.

@JohnDuffield I don't know about here, but it's very common on SOverflow

@ACuriousMind Thank you very much! But I will try to get an up-vote from you in the future. :-)

http://physics.stackexchange.com/questions/253372/zeroth-component-of-4-momentum-and-relativistic-energy-momentum-relation

Ok, so $p^0$ is the energy

@Secret I was reading this wiki article I was wondering, if the variant mass changes in an observers reference frame, does its speed also change?

4:25 PM

@EmilioPisanty : I'm sure he did. By the way, if you've got something to say that concerns me, please do mention my name.

...oh god. Is there a way to ignore comments from a specific user on the main site? I'm being spammed right now.

link?

nope, the variant mass (relativistic mass) is an artifact of the energy of the object (even for massless objects like light), and the electromagnetic wave's energy is determined by its momentum

@Obliv what purpose would that serve?

I want to see what you're dealing with

4:27 PM

It's not the funny kind of troll :P

@JohnDuffield No thanks. I've seen to what depths of unconstructive discussions can go when you're involved, and I'd rather not have one of those. I've generally been wary of blocking any comments on chat, but if you really feel like starting a fight, then I'm happy to flick some switches there.

(I refuse to say frequency, in light of earlier discussion not to say a light wave is simply made of photons (honestly, I still trying to understand the specifics, light is not a straightforward thing))

@ACuriousMind link?

PS I suck at electromagnetism

@secret "being determined by its momentum" doesn't that include its velocity?

4:29 PM

@ACuriousMind I think all trolls are funny

@0celo7 Then your humor is not the funny kind of humor :P

Says the German

@Obliv a photon's momentum has nothing to do with its velocity (that's the bit I am confident with). @Acuriousmind may be better than me on explaining the other half of this sentence I am trying to explain without making the photon=light wave misconception

@ACuriousMind Has great humor

@BernardMeurer <3

4:31 PM

suck up

@ACuriousMind be careful around that one

remember Brutus

@ACuriousMind :)

@0celo7 Around that ö?

@secret light's velocity is constant in a vacuum, right? From a non-stationary reference frame, light's velocity either increases or decreases. Does anything else change?

@0celo7 : here's the link. I think.

@Obliv See this Q&A

4:33 PM

@ACuriousMind on that subject, is Schroedinger an acceptable way to spell Schrödinger for those of us using keyboards without accented characters?

@JohnRennie If and only if you lack the umlaut, but yes.

I see that spelling used, but I've never been clear if was an acceptable alternative.

@Bernard re that hackerspace what are ppl working on? went to a maker club at museum recently, liked it... lots of 3d printing stuff... someone had a 3d metal driller that had good lateral force, new technology, very impressive

ö,ä and ü may be written as 'oe', 'ae' and 'ue' in general

@ Acuriousmind, obliv

*If you consider an accelerating reference frame with respect to Rindler coordinates (where time is measured by idealized point-particle accelerating clocks, and objects at different locations accelerate at different rates in order to preserve proper lengths in the momentarily comoving reference frames),, then light may not move at c, and can in fact even stop*

Ok that is totally NEW to me!

4:35 PM

@ACuriousMind Lucky Germans. áéíóúñ are way harder to substitute.

@JohnDuffield Thanks!

à

pretty easy in Chrome

@secret I almost understood that. What do you mean by preserving proper lengths

@vzn There's a group working on PaperX, a website with high school physics/maths/chem/bio questions to help students practice and study in general. Another group has a startup called Botfy works on what they call "triggers", they just did a campaign with Coca Cola where when you open their fridges or pass near one it'll play different sounds ( I generally help them out the most). And then you have other people not in a group working on all kinds of personal projects really.

@JohnRennie Have you thought about the adoption thing?

aren't proper lengths by definition preserved in the proper reference frame

@BernardMeurer cool is it mostly CS or do any do electronics/ hardware?

4:38 PM

@obliv That's not my answer, but proper lengths are lengths as seen in the rest frame of the object, which is not accelerating

@vzn PaperX is just web dev and CS, Botfy has a lot of harware, the rest of the people mostly do CS apart from one guy who messes with FPGAs which I guess are kind of electronics

@BernardMeurer I have been busy doing other more important things - combing my hair, cutting my toenails, etc

@BernardMeurer oh yeah fpgas are awesome have read a few articles on new applications of those (maybe have em in my bookmarks or blog somewhere...) so do you think youll spend less time in chat after you start school?

@Secret see physics.stackexchange.com/questions/242043/…

In that answer I explain exactly what it means for the coordinate velocity of light to change for an accelerating observer

Actually understanding the nature of the Rindler horizon is an excellent way to get an intuitive grasp of other coordinate singularities like a black hole event horizon.

@JohnRennie I gotta inherit all that money SE is giving you :p

all the millions

@vzn Can't tell, but probably yes

4:42 PM

Ever since I discovered SE points could be swapped for bitcoin I've been a happy man

2

specially because I don't have a laptop as of yet

so I may be incomunicable for a while

@BernardMeurer what about your potatos?

@BernardMeurer ok re fpgas, just found/ ran across this a few days ago on reddit re robotics control, its way cool, think its likely to catch on/ spread eventually hackaday.com/2016/07/23/…

@JohnRennie Really? how?

@JohnRennie exchange rate?

4:43 PM

Can Intel laptops run POS (Potato Operating System) ?

@vzn They where used to break the DVD encryption I think

and/or Blue Ray

@BernardMeurer I would tell you but I fear it would affect the exchange rate and decrease my wealth

@ACuriousMind Leaving them behind :/

@JohnRennie :)

@EmilioPisanty : it isn't me starting a fight. I'm not going round downvoting users then putting up a list of downvoted users and nagging the moderators for "bitier downvotes" or "kill-it-with-fire".

@BernardMeurer no kidding hadnt heard that. also think they were used for quite awhile in bitcoin mining before asics took over.

4:44 PM

@JohnRennie It must be really cool to be able to convert this answer into a computer generated animation to show the formation of the apparent (?) event horizon

@JohnRennie I'm actually beginning to write my own OS, Daedalos, which I hope will be able to run on potatoes & intel laptops

Years (decades) ago I decided I would name my operating system XOU.

@JohnRennie But the potatoes currently run ALARM, which is ARM-only, your laptop is most likely x86_64 so no :(

You can still run Arch Linux though!

It's basically the same thing

XOU?

When NT was first released the main architect was Dave Cutler who used to work for Vax, and there was a rumour that Windows NT was named so that WNT would be one better than VMS.

@vzn They were, back when GPUs where still very popular the FPGAs came around and took the hill

4:46 PM

Go one better again and you get XOU.

Hahahaha

WOW

@JohnDuffield ...aaaaand there we go. This is a polite notice that I won't respond further on chat. At some point later I may discuss some of my position on meta but for now if you think my behaviour falls outside the standard 'be nice' rules, I'll just refer you to the mods or the SE team. (As a final note, do notice that kill-it-with-fire was actually against such a suggestion. But whatever.)

@JohnRennie How do I sell my points? I need a laptop

@BernardMeurer Dell Latitude E6330s sell for around $200 on ebay and they are stonking good laptops.

@JohnRennie You mean you've been happy because you used your immense wealth to buy enough points to be the top user here? ;)

4:50 PM

In fact I'm typing this on an E6330 equipped with a 3GHz i7.

Find a vendor willing to ship to Brazil and you're away.

200 is like a thousand bucks here :/

Still gotta earn that

I'll get it when I'm in portugal, nothing makes it to brazil

For you guys interest, Background: Actually it is this completely fictional stuff I read on the other day that caused me to become very interested in coaccelerating frames and noninertial frames in general

http://powerlisting.wikia.com/wiki/Motion_Paradox

In short, this fictional stuff mentioned in this wiki imagined some weird object such that if you don't match the velocity the object 'has' you will not be able to catch the object (even if it stays put in your perspective)

This fictional stuff then create a 'jolt' of memory recall as I recall how back in classical mechanics we are trai…

@ACuriousMind :-) Paying all those graduate students to write answers for me costs a lot as well.

@JohnRennie I knew you were a slave master! That beard and mustache don't lie!

@JohnRennie I see. I can only afford two scrawny undergrads and a disillusionized postdoc

4:57 PM

@ACuriousMind: while you seem to be in a reasonably good mood: if I start with a (non-interacting) QFT vacuum and apply the creation operator once, do I end up with a plane wave?

Basically yes

Although it will be difficult to make precise how exactly that's a plane wave

Since QFT states are not wavefunctions.

Ahhhh, coffee

@ACuriousMind do you like coffee

@secret What about having the momentum in his frame remain 0, after using the ability, and having his momentum in everyone else's pov increase. This breaks the conservation law though

@ACuriousMind why not

OK, so if we regard particles as the energy exchanges associated with creation operators then the particles are basically just waves?

4:58 PM

@JohnRennie What the hell, I can't find on the webz how to sell points :(

@0celo7 Yes, black. We've had this conservation at least twice already

@JohnRennie Can only badasses like you do it?

@ACuriousMind How do you know?

maybe his momentum would be in different directions to retain the symmetry/conservation law @secret for the other frames of reference

I lied about the bitcoin.

They're actually exchangable for Air Miles.

5:00 PM

@JohnRennie -.-

You're lying to me

I don't think we have, @ACuriousMind

try not to break Noether's Theorem

Also according to the article, his momentum is constant in any orientation and directions, it's like a vector always pointing to some direction in space (which also calls into question of the issue of absolute space)

perusing the chat log now

@BernardMeurer Me? !! :: John does his best to look innocent ::

@JohnRennie Well, like the plane waves in QM, these things aren't actually states (they're non-normalizable). The actual states are superpositions of them

5:01 PM

@JohnRennie I'm storming Chester with bloody automated vegetables

Why is she smiling like that

@ACuriousMind agreed. So particles in QFT are basically just as elusive as particles in non-relativistic QM?

My mind work somewhat like this: https://www.youtube.com/watch?v=C5LrS4faC3g
There's some knowledge or experience I learnt, and then it result in recalling of some random memory (often look completely unrelated to the knowledge or experience). This memory then recall further memories in some really unpredictable fashion and this is then repeated as it spreads. Eventually all these memories resurface and mix at the same time in some random way, and then suddenly the result intrigues me and then I ask question

@JohnRennie More elusive because you can't actually localize them well (there's no straightforward position operator).-In non-rel QM one can at least simply write down a Gaussian wavepacket that's sharply peaked in position space that will reasonably act like a particle localized at the peak.

Nov 5 '15 at 22:25, by ACuriousMind

@0celo7 I do...still no idea what that means, my coffee order is always "Coffee, black" :P

@ACuriousMind so if I were to persist in my quest of explaining the difference between particles and light rays (an explanation that no-one seems to want) I have the problem that particles basically don't exist - except in the sense that they're what the creation operator creates.

5:04 PM

@JohnRennie Yes, indeed

I really, really want to write a canonical Q/A about all this ...

cononical?

Ah

colonical

as in irrigation

SLAVE MASTER

Everytime I use the word canonical people give me crap

@JohnRennie I still think that we have a suitable canonical question. The answers might be lacking, but I think posting a new question is unnecessary

5:05 PM

@ACuriousMind Can we localise a QFT particle in spacetime by using a spacetime version of a gaussian wavepacket, e.g. $Ae^{-(x^2-t^2)}$?

"Cano who?"

@Secret I said that it's not possible because there's no straightforward position operator, so it is a bit difficult to tell what "localized" means.

@ACuriousMind hmm, yes, maybe I'll post a second separate answer to that question.

In particular, if one uses the Newton-Wigner operators to define localized particles, you have an issue with this notion not being invariant under boosts - a particle that is localized in one frame won't be in another

@ACuriousMind I have no recollection of this.

@ACuriousMind What does Newton have to do with QM o.O

5:12 PM

Theodore Duddell Newton, not Issac (forgot) Newton

Isaac

@secret what is your country of origin? or is that a secret too ;)

I am Australian

but born Hong Kong

oh I thought you were Taiwanese since you had that art book in Taiwanese @secret

Hong Kong has a bookstore called eslite, which sell taiwanese books

Do people give speeches when they receive a Nobel?

5:23 PM

@BernardMeurer yes

@ACuriousMind What are you going to say in yours?

I know what I'll say when I get the Nobel Peace award

@BernardMeurer "All hail our new potato overlords"?

2

@ACuriousMind Yes, that's when I start the attack

@BernardMeurer He has like 2 years to make a discovery before he passes the peak age

@ACuriousMind get to work

Don't worry about me, I'll get a rubber duck

@0celo7 And then it'll take decades till I actually get it, cf. e.g. Higgs

5:26 PM

@ACuriousMind Honestly you'll probably die before then :/

@0celo7 i think I read a study where 20th century and onwards, people began receiving nobel prizes/did their most significant work later in their life (40s+)

I cried when I saw Higgs crying when the LHC announced the confirmation of the Boson

I cried when I saw how much the thing costs

@ACuriousMind: since I've managed to ask a question about QFT that wasn't complete gibberish let me try another ...

I should prove that the tangent bundle is always an orientable manifold

@ACuriousMind Have you done that calculation?

5:28 PM

The usual introduction to a scalar non-interacting field starts by Fourier transforming the classical field. So is this just describing the classical field as a sum (well integral) of plane waves i.e. a Fourier synthesis?

Can we rename a mole to a shitton?

I think it'd make chemistry a lot more fun

@ACuriousMind If particles aren't states, why does Weinberg use states when talking about scattering ing QFT

how many shittons in 5g of carbon?

or wavefunctions

or doesn't he

@BernardMeurer 5 grams.

@secret have you determined the time-travel model you're going to be implementing in your wiki?

5:30 PM

@0celo7 because in a scattering calculation the particles are plane waves?

Why are you doing basic chemistry

@JohnRennie no, they're wave packets.

By scattering calc do you mean calculating the S matrix?

@JohnRennie Yes. Weinberg actually does it the other way around - he first defines the creation/annihilation operator and then constructs the field as that sum.

@0celo7 Because the calculation for wavepackets is much more complicated and the result may be computed from knowing the results for all hte sharp momentum states.

@ACuriousMind OK so after quantisation is the operator field expressed as a sum of all the separate creation/annihilation operators?

And since in collicion experiments the momentum is usually sharply peaked, the approximation as sharp momenta is already good in many cases

@JohnRennie yes

5:36 PM

Aha! So the value of the quantum field at any point in spacetime is a weighted sum of creation/annihilation operators at that point. Its value is an operator that gives the probabilities for creating/annihilating particles of all possible wavelengths. Yes?

Well if you can define creation and annihilation operators, sure

Also the field is gonna be like

The expectation value of those operators

wrt some state

@JohnRennie No, the operator does not represent a probability.

that was the whole point of QFT, really

Although you can do operators as wavefunctions in RQM

In the limit of low energy

If I start with a plane wave that passes through some point in spacetime then at that point in spacetime the quantum field has a value that is expressable as a sum of creation/annihilation operators. Is that right so far?

What do you mean by "the value of the quantum field"

5:40 PM

@JohnRennie What do you mean "if I start with a plane wave"?

heh

@ACuriousMind the thing that gets created by a creation operator.

I mean what are you measuring here

Field mode?

Be careful not to confuse operators and wavefunctions, that can be tricky in QFT

5:40 PM

@JohnRennie Ah, but that's a state - the value of the operator has nothing to do with specific states.

Yes, the states are independant of the operators

The "plane wave" that the creation operator creates is not something that depends on spacetime.

If you want to express a state, for a free field, you can just use the Fock space thing

The field is an operator that does depend on spacetime.

The Fock space $\bigoplus_n \mathcal H^{\otimes n}_{Sym}$

5:42 PM

@ACuriousMind OK. So if we take QED (and ignore interactions for now) a photon creation operator doesn't create a propagating light wave?

@JohnRennie not really, no

@ACuriousMind Is Čech sensitive to the twist of bundles?

Is that why you want to use it for that proof?

What do you mean by "creating"

I'm not sure what exactly happens in quantum optics, where $E$ and $B$ are quantized directly as far as I've heard.

Bott & Tu show that $H^*_{dR}(E)=H^*_{dR}(M)$ for any vector bundle $E\to M$.

5:43 PM

Ah, OK, the house of cards that was my understanding of QFT has just collapsed. That canonical Q/A has just receded into the distance.

A creation operator will take the vacuum state and turn it into a one particle state

@0celo7 The isomorphism classes of $k$-vector bundles are in bijection to the first Cech cohomology of the manifold with values in $\mathrm{O}(k)$.

...ok then

@Slereah with a well defined momentum?

$\hat a^\dagger_{k,s}|0\rangle = |k, s\rangle$

Yes

5:44 PM

You probably don't know the proof of that?

The momentum distribution is a delta at $k$

Stuff is currently on hold because of the bottleneck identified by Slereah.

If our observed reality is actually based on entangled systems, then the way decoherence and von neumman measurements destroys superpositions might provide a way out of the bottleneck by simply considering every reality that has an alternate outcome corresponds to an eigenstate of the system is located in a different set of einselected states that is formed by the instrument's states entangled with the quantum system.

OK I guess I don't understand why a one particle state with a well defined momentum doesn't correpsond to a light ray of the same momentum.

Well it's not at all localized, for a start

@0celo7 It first associates the corresponding $\mathrm{O}(k)$-principal bundle to the vector bundle and then proves that principal bundles are classified by that Cech cohomology. Once one unravels the cocycle definition of a principal bundle and the definition of Cech cohomology there actually is not that much left to prove

5:45 PM

It's a monochromatic wave that exists everywhere in space for all times

It's also not a proper state, since it's not normalizable

A light ray is better modeled by a Gaussian state

Doesn't a plane wave exist everywhere in space at all times?

Yes

@ACuriousMind Yeah, I think the cocycle approach is what one has to do for the BT exercise.

I wouldn't call a plane wave a "light ray"

Not very ray-like

@Slereah OK, yes, a real light ray is constructed from a sum of plane waves.

5:46 PM

Yes

That's why plane waves are well studied

But for most purposes a real light ray is approximately a plane wave isn't it?

Maybe one can show explicitly that any $k$-bundle $E_k$ over $S^1$ is either $S^1\times\Bbb R^k$ or $(S^1\times\Bbb R^{k-1})\oplus M$, where $M$ is the Mobius bundle.

Quantum states can always be constructed as a sum of plane waves

Seems like a tough proof.

Sure, you can say that

5:47 PM

So does the operator create an infinite plane wave?

Yes

i.e. approximately create a light ray?

@JohnRennie stop right there

ACuriousMind, John Rennie, 0celo7, garyp, knzhou 2

ACM IS MAD

5:48 PM

@ACuriousMind oh no :-(

You're constantly switching between two notions of "plane wave"

Plane wave is $\mathrm e^{\mathrm ikx}$

The thing the creation operator creates is in no precise sense a plane wave because it is not a function of spacetime

@Isomorphic Hi. Come to berate ACM? :-)

@ACuriousMind ...what does the creation operator create

5:49 PM

^

No why would I berate anyone?

The classical plane light wave is a functon of spacetime

You all are rational people

@Isomorphic I'm not.

@Isomorphic that's a bold assumption :-)

5:49 PM

I've just put some questions there in the comments section

But @ACuriousMind, in the Fock description of EM fields, isn't a one particle state precisely a plane wave?

I have acute schizophrenia

of my question

They are all math related

@0celo7 A state in the Fock space. Which can be seen as a function of field configurations, but not as a function of spacetime.

Please vote to close them

5:50 PM

Function of field configurations?

@Isomorphic I'm not rational

@0celo7 "wavefunctional". We've been over this.

@ACuriousMind The fuck space?

@Isomorphic link?

@0celo7 I could have guessed

5:50 PM

@ACuriousMind I don't remember that

But the Fock space is a sum of $L^2(\Bbb R^n)$ states

2

Q: What is "a vector of $SO(n)$"?

I'm watching (or trying to watch) this lecture from NPTEL on classical field theory. I've understood everything in the series up till this point, including the first half of the lecture on elementary group theory. However, at a certain point he begins talking about a "vector of $SO(d)$". He essen...

terminology vectors group-theory group-representations tensors

@JohnRennie That turns out to be impossible. I see I have that user on ignore from a previous interaction.

38

A: What is a tensor?

A (rank 2 contravariant) tensor is a vector of vectors. If you have a vector, it's 3 numbers which point in a certain direction. What that means is that they rotate into each other when you do a rotation of coordinates. So that the 3 vector components $V^i$ transform into $$V'^i = A^i_j V^j$$ u...

Especially this last one

about tensors

@Slereah Ehhhhh

5:51 PM

@BernardMeurer A room owner reaches for Cancel stars button.

someone asked what is tensor as a whole

Not in the standard physicist way, it isn't

BUT YOU TOLD ME SO A FEW DAYS AGO

@JohnRennie I know you laughed :p

WHEN WILL YOUR LIES CEASE

5:51 PM

In short, until I can figure how to solve that bottleneck (and reproduce all known physics (or at least GR) in the non time travelling limit), I am going to stick with CTCs for now else I will be kicked out from the physics community due to being accused a crank

alternately, I can simply forget abotu making it consistent and just use it as it is, like anyone else who have done back to the future time travel plots in the past, but I don't like that, because it is *not real enough*

Or well, it is isomorphic to such a thing, but that doesn't actually tell you anything

What is happening

Too many conversations

@Slereah I also told you that abstract isomorphism of Hilbert spaces don't actually mean much

Hey @Slereah What's your favorite potato?

I like the big potatoes

5:52 PM

OK, so all this means that my dreams of explaining the link between a light ray and photons are all dashed to pieces.

That I can stick in the oven

Baked potatoes are great

@JohnRennie ?

@ACuriousMind You're good at explaining math, but not good at explaining QFT.

@Isomorphic What's your favorite potato?

@Slereah Maybe let me think a bit more about that $L^2(\mathbb{R}^3)$ business and don't believe me yet either way :P

5:54 PM

@Slereah Those don't make good computers sadly

@ACuriousMind : Would you disagree that you can express a free field wavefunction as a member of $\bigoplus_n L^2_{Sym}(\Bbb R^3)^{\otimes n}$

@Isomorphic the world needs a really good explanation of the relationship between photons and a light ray i.e. (approximately) a plane wave. I thought I understood enough about QFT to write one, but it appears not.

Although there appears to be some disagreement between our resident experts on this subject.

Are you calling Colombeau a liar

@Slereah no, because as an abstract separable Hilbert space, the one-particle space is of course isomorphic to $L^2(\mathbb{R}^3)$. But what tells you that the coordinates of that space have anything to do with the spatial coordinates?

@JohnRennie I don't what you're talking about

5:55 PM

My gut, that's who

@BernardMeurer I am not interested in talking to you

3

@secret Can't you just use BST as the model? I think every interpretation on this wiki page is consistent with our results so using one shouldn't be inconsistent for a sci-fi wiki I hope.

@Isomorphic Oh wow

I'm pretty sure it should do this to some extent, because otherwise the RQM limit wouldn't make sense

@Isomorphic wow

5:57 PM

@JohnRennie Can you please dumb down your last comment here

@BernardMeurer I still love you

@Isomorphico $S A V A G E$

@Isomorphic the one about the photon and light ray?

@JohnRennie You deleted the fuck space!

@0celo7 Thanks <3

@JohnRennie Why did you delete that >:(

5:58 PM

Yes, you tagged me in that comment

@BernardMeurer :: John cackles manaically ::

How many times?

@JohnRennie : Although one thing

BEWARE

Beware of the eigenstates

@0celo7 11 I think

@Isomorphic I thought you were asking what I was talking about in:

6 mins ago, by Isomorphic

@JohnRennie ?

5:59 PM

Because for a start, $\langle k | \hat A^\mu | k \rangle = 0$

The EM potential of a "plane wave" state is 0

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