6:00 PM
@JohnRennie The thing I learned doing this is that simultaneity is v*x ... which can be large when v is large OR x is large.

What, can't I recommend a fine science book to a friend

@ACuriousMind so the mathematical framework of decoherance does not fully account of everything that happens in a measurement, and we have to accept that the measurement just magically change the state to eigenstates of the measurement operator with its associated eigenvalue? . That is, we cannot expect to do a matrix multiplication between the state and measyrement) and see how the eigenstate pop out from the computation?

@BernardMeurer Will that settle our debt?

@0celo7 Absolutely
Actually I think I'll be in debt with you by then

Only $63 A bargain 6:01 PM I take it this isn't our @JohnDuffield ? 100% one star review It is our John Duffield He wrote a book? The whole notion of simultaneity exposes the difference between what physicists mean by time and what Joe Public means by time. @BernardMeurer I'll get you a MyCopy version of it It's like$29 then

The hell's a MyCopy?

6:02 PM
If you show me your plane tix to TN, I'll buy it

To me, it means that there's possibly a better way to do relativity transforms.

The question what is the time at x?, where x is some remote position doesn't make sense. We would ask what are the spacetime cordinates at the point P in my coordinate system?

@Secret No, you don't have to accept that, but if you want to describe the measurement within quantum mechanics you have to be a lot more intricate than what you've tried so far. Read the things @yuggib linked, or, for a start, just read about the von Neumann measurement scheme.
The key point is that you have to include the states of the apparatus into your system, you can't give a "measurement evolution" just on the isolated system itself.

I can send you my shankar.

6:03 PM
@JohnRennie Yeah, but this whole time tweaking seems unnecessary.

@Slereah Hmm

@barrycarter It's a coordinate transformation - that's the beginning and end of it

Oh, there's Cheeger and Ebin
And O'Neil

Understand that and you understand relativity

Good old O'neil
GR math and helping the Ninja Turtles

6:03 PM
@JohnRennie I've been doing it that way for ages. Matrix multiplication.

All in a day's work

O'Neil is hella expansive for some reason right now

But for most purposes, how the measurement happens is completely irrelevant - we don't care how our detectors turn quantum states into eigenstates, what is relevant for experimental prediction is that they do.

It used to be like $50 Now it's$75

@JohnRennie BUT... now that I'm actually looking at the understanding behind it, I'm not liking it.

6:04 PM
@barrycarter OK

There... must... be... a better... way.

There's also QoGS
Might be an interesting book

Which one is QoGS

The one ACM reads before bed every night

The Quest of Great Snog?

6:06 PM
Quantization of Gauss Slices or something

@barrycarter I suspect this is what you're used to. To me coordinate transformations (affine coordinate transformations :-) feel entirely natural.

I think I have all of the GR books at this point that are worth a damn

Ok, I should read up yuggib's and the von neumann scheme in more detail to improve my understanding before returning to the general problem of entangled photons

@JohnRennie I love coordinate transforms as long as they stay in bloody space and not time.

Time to buy quantum books now

6:08 PM
Petersen is not released yet and FLP is ignoring my texts

You Euclidean you

@Slereah Meh.
Rovelli?

It's silly for two people seeing the same light beams to assign the universe different times.

Maybe I should buy another copy of Hawking-Ellis

Rovelli is maybe not a good idea if you don't know your shit well

6:09 PM
@JohnRennie Spherical transforms can be non-Euclidean, if you're talking about just the surface.

@Slereah What the hell does that mean

@0celo7 and why would you do that?

Rovelli is not a gentle book

On a different note, not that any of you lot are likely to care, I've finally, finally got my new IP PBX working properly.

Oooh, new question.

6:10 PM
If you don't know GR, QM, QFT and hamiltonian mechanics well

"The longitude is absolute". wowsers.

@Slereah are you saying I don't know it well

I am.

you're the one who can't calculate $a^\dagger a$

You wuss.

6:11 PM
@0celo7 Shit dude, it's depending on how my interview goes today
and tomorrow

At least I'm reading my QM
Unlike you

because if I can assure me a scholarship I'm going over to celebrate with some barbecue

@Slereah Oh fuck, Kobayashi-Nomizu

@Slereah what book?

Well Kobayashi is the standard for diff geo

6:11 PM
@Slereah I'm taking a class next semester

Peskin Schroeder
Redoing all things by hand
No fucking around

hardcore.

@0celo7 that was in Star Trek wasn't it?

@Slereah ttl, PDE class now

Might take a bit

6:12 PM
@JohnRennie it's in every GR book
all the good ones, anyway

Still on chapter 2

@0celo7 missed point error - oh well, it was a silly joke anyway

@JohnRennie 0celo7 is a young fellow
He hasn't seen any old movies
Then again it was in the Star Trek remake

@Slereah Well, it was in the second full length film. They're almost contempory with TOS from today's vantage point.

He's a filthy millenial, though
He doesn't even remember 9/11

6:16 PM
@Slereah My mother was in the Pentagon

Hmm, it looks like you can read a lot of @JohnDuffield's book for free on Amazon's "look inside" thing

Why would you

xD

@barrycarter dangerous waters. I would let the matter drop.

OK... just don't see why people are saying "buy it".

6:18 PM
@barrycarter they are being sarcastic - the lowest form of what passes for wit in physicists.

It's one of the starred messages.
OK, I'm skimming it, but will not comment on it.

@barrycarter want to take odds on how long before a mod unstars and deletes it?

@JohnRennie Aren't you a mod?

@barrycarter Me, no! God forbid!

@JohnRennie We're all aware of the mod's tyranny

6:21 PM
@JohnRennie My idol has feet of clay....

I'm far too opinionated to make a good moderator. I'd be closing 90% of all the posts.

Which would be an improvement on the 95% that are closed now.

@JohnRennie You'd also ban a lot of people
Hopefully not me!

@0celo7 he who shall not be named has been specifically instructed not to engage in discussions of physics in the chat. Given this, it is unfair to start commenting on his book when he isn't allowed to defend it.

Voldemort uses PSE!?
That lag though

6:24 PM

@DavidZ it was me flagged it, so I have an unfair advantage :-)

flagged? I didn't realize there was a flag
I was just randomly checking the room

Checklist on quantum concepts:
Nonlocality=ok and can wrap my mind around (i.e. accept)
Nonrealism=ok and can wrap my mind around (i.e. accept)
*States are not real objects = ok but still reluctant on accepting
Fields are nonlocal = ok and can wrap my mind around (i.e. accept)
Measurement as a black box = Confused on anything that pops out like magic, huge resistance in accepting despite use it in practice
->Measurement in von neumman scheme = working on it
Entanglement = basic concept ok, non communication theorem in brief ok, basic computing experimental methods ok

I want to flag that one about mathematicians...

6:26 PM
@barrycarter what, the I've died and gone to Hell one?
Flag away.

@JohnRennie No, I like that one. The one about pedantry.
Slumming is a mathematician talking to a physicist in a chat
And how does one "behave" like a mathematician. One either carries the mark or does not.
OK, I'm over it.

There's a lot of good natured banter flies to and fro in the chat, and no-one really minds. It's just that every now and then someone manages to make themselves really unpopular and it can get unpleasant very quickly.

The side effect of my above bias is that I like to use the interaction picture (rather than schrodinger or heisenberg) more, because in this framework a lot of things are interacting, and thus doing the maths does not felt like as much a black box where stuff seemed to magically happen

And that spoils the chat for everyone.

@barrycarter It was a mathematician who said that one, though :P

6:29 PM
Maybe I'll just randomly star stuff so it goes down in the list :)

@JohnRennie Flagging isn't for when someone says something that might turn the conversation unpleasant. It's for after the conversation has already turned unpleasant in spite of attempts to remediate.

Someone say something star worthy.

@JohnRennie can you name names

@Iszi noted, but in this case I am unrepentant.

Today I found out I'm not universally loved here, it's shaken me pretty badly (the number of stars on this is the number of people who hate me)
13

6:30 PM
Thanks, Ocelot

@0celo7 You told me I was wrong!
(of course it was true ...)

@JohnRennie Gamma is greater than 1.

Oh, that crap about the geodesic equation I posted. Every now and then I discover I know less about GR than I thought. That's life.

I was wrong??

6:32 PM
Therefore, my notion on what is meant by exist is as follows: That it can be described in terms of interactions and these interaction obeys rules that brought the starting condition continuously or discretely into the final condition. I.e. For every step that is defined in the process, I can compute what happened, even if the stuff in question are just probabilities (because probabilities conserve, thus they act like stuff

Ocelot, are you saying that people who star your message hate you, or merely that there's an isomorphism between the people who star the message and the people who hate you?

@0celo7 No, I meant it was true when you said I was wrong ... oh God, here we go again, another flame war.

Is a 2x2 matrix that have all elements as 0's the null set?

@Obliv A matrix is not really a set. And if it were, it wouldn't be the empty set.

oh right

6:34 PM
@barrycarter The first one.

Ocelot, I only mildly dislike you but starred it anyway as part of my plan to get the message below it pushed down.
Consequently, someone who genuinely dislikes Ocelot should refrain from starring.

Wow what did I do to deserve your dislike

I just find you generally annoying.
And you blocked me once.

In response to you being an ass

And you're an engineer.

6:35 PM
Like right now

Come to think of it, maybe I do hate you.

As long I can understand how a goes from b, I can accept it is real. But if there is something c where it goes to z and there is nothing physical in between (seemed to be suggested by the understanding that the only thing that is real in quantum are the observables) that describe how it got there, I felt like someone is doing magic

@barrycarter then don't talk to me
Wouldn't want you to be annoyed

Well, I never use the @ thing to address you.
So, technically, I'm just talking and you happen to hear it :P
Technically, I am creating electromagnetic waves.
I enjoy being slightly annoyed, Ocelot.

@Iszi no, I wouldn't say that. (Just so that nobody gets the wrong impression.) Our goal here is to prevent the conversation from turning unpleasant, as much as possible. If flagging can be used to support that goal, then it should be.

6:39 PM
@barrycarter what's the matter with the message below? ;-P

Poll on who finds me annoying and wishes I were gone
Star to vote.

@DavidZ The problem with using flags for that in chat is that it brings in a lot of outside third-party involvement for issues that should very much be resolvable "in-house" when the local users are reasonable and at least somewhat non-hostile.

My brain automatically filter away anything I am not interested in
That includes most emotions in chatrooms
which is one reason I am not known to find anyone annoying

@yuggib It sounds like an insult to mathematicians.
Ocelot, gone as in gone from this room, or gone as in dead, or gone as in never existed?

Let's say set $B = \{\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}, \begin{bmatrix} 0 & 0\\0 & 0\end{bmatrix}, \begin {bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}\}$ how can i prove that if $P,Q \in B$, then $P+Q \in B$? I get that the second element is obvious but for the addition of the first and third element I don't think that is in $B$. I summon you @barrycarter

6:42 PM

you are fast.

Damn simultaneity.

@barrycarter From the room.
I'd have to poll the world for the other questions.

@Obliv OK, so if you add any two matrices in B, do you get a matrix in B?

@barrycarter Nah, that would be an auto-insult

6:43 PM
and the reason why I never need to press ignore, because my brain does that subconsciously for me most of the time

@barrycarter I don't think so. the first element + the third element gets $\begin{bmatrix}2 & 1\\0 & 2\end{bmatrix}$ right?

@yuggib Yes, someone mentioned you were one of Us. Are you saying that talking to a physicist requires a different level of speech?

@Secret: This has a nice description of how to mathematically set up measurement processes, and has lots of references. If you really want to see how it's done, read that and follow the leads. It constructs a notion of measurement that has no projection postulate.

@Iszi That is a good point. I certainly don't mean to encourage people just wantonly flagging anything they don't like. But on the other hand, I don't want anyone to get the impression that everything is fine and dandy until after people get in an argument.

6:44 PM
@0celo7 nobody wants you gone. @barrycarter bear in mind chat forums lack the visual queues that tell the other person you're not really being rude.

Ocelot, I'm joking. I love you, man.
@Obliv Unless you're working in mod 2.

@barrycarter nah, I am just saying that mathematicians (myself included) enjoy pedantically pointing out all the mathematical mistakes that physicists do...

@barrycarter that doesn't help me since I have to prove that it is.. :( Maybe my elements in $B$ aren't actually in $B$

@Iszi Honestly, if the local users are always reasonable and non-hostile, we shouldn't ever have arguments breaking out and there would be no need for moderators. But human nature being what it is, that's a little too much to hope for ;-)

and there are a lot ;-P

6:45 PM
And I've finally managed to work out the answer to:
0

I am working on an experiment of rheology and I need to calculate shear stress in order to calculate the viscosity. After some research I found that for the type of viscometer I will be using ( cone-plate), the stress is calculated by dividing the torque given by the apparatus by a form factor eq...

I'm getting slow in my old age.

@ACuriousMind ah, nice! a free ozawa reference...

@yuggib True, but I would say that's more educating physicists than being pedantic.

@yuggib It refers back to the one behind the paywall you linked though for some crucial derivations :/

@Obliv "Maybe my elements in B aren't actually in B" . Wow.

@DavidZ IMHO: When the flagging system is being used properly, nobody from the outside would ever have to look at a flag alert and wonder why the heck anyone would ever flag such a thing. And a lot less flags would be getting raised because people actually try to fix problems before they flag.

6:46 PM
@barrycarter can we agree that $\begin{bmatrix}1&1\\0&1\end{bmatrix},\begin{bmatrix}0&0 \\ 0 & 0\end{bmatrix} , \begin{bmatrix}1&0\\0&1\end{bmatrix}$ when multiplied with the first element are commutable?

@Obliv MathJax didn't render that... ok thanks

@Iszi yeah, sounds reasonable to me.

Still not rendering right.

@ACuriousMind :-/
damn

6:47 PM
The flagging system in chat has issues.

@barrycarter often the line is very thin...

@barrycarter the elements of B are decided if they are commutable when multiplied with that first element.
@barrycarter then I must prove that the addition in $B$ is $\in B$

@yuggib Yes, but pedantic implies we intentionally go overboard. We merely seek to serve, and know not the frailties of those whom we serve.

@barrycarter I was responding to this message but forgot why

@Obliv The identity matrix and the 0 matrix are commutable with everything, so yes.

6:49 PM
@DavidZ Many of those are simply according to its design, and I know SEI is aware of that. But experienced users should be doing what they can to not unnecessarily exacerbate the issues.

My memory gaps are getting worse

@Iszi oh, of course, I agree

I won't be much longer for this world

Stop sniffing the glue in the lab, then!

@barrycarter :-o

6:50 PM
Ocelot, you make me so giddy, I can't even type your @ name.

@barrycarter the first one is the same matrix so obviously $x^2 = x^2$

@yuggib Wouldn't you agree?

@ACuriousMind I think it's the pot I smoke with the experimental biologists

@barrycarter not completely...

@Obliv I think you nailed it. So, yes, for those three matrices, you have commutative multiplication.

6:51 PM
@barrycarter great. so now I have to prove that addition for any two elements in $B$ is in $B$ :D

@yuggib I never intentionally am pedantic towards physicists.
@Obliv You're missing a verb there, but your statement is a tautology the way it stands.

i'm tired give me a break >:(

@Obliv You also said your elements in B might not be in B.

@barrycarter It depends what you talk about

@yuggib I mean, I try to talk down to their level, but it's hard to estimate when it's that low.
@Obliv Which is a pure falsity. Sort of cool, actually.

6:53 PM
You know what else is a pure falsity

"The sphere admits a Lorentzian metric"

@barrycarter the elements in B are definitely in B now because I just confirmed them with you

@Iszi we had the flame war already several times, so I flagged as soon as I saw another fire starting. I note the point you're making, but in this case I consider the flag justified, and so did David otherwise he wouldn't have deleted the post.

@Obliv Are you working mod 2?

6:54 PM
Kinda crazy when you think about it.

@barrycarter what does that mean

@barrycarter it's not a good idea to "talk down to their level"

@Obliv OK, I'm glad to hear that the elements in B are definitely in B.

the devil is in the details

@yuggib But if we talk math, we ARE being pedantic.
@Obliv Are you adding these matrices modulo 2, ie, in binary?

6:55 PM
yes/
(the usual sum of matrices) is what they say

@Obliv No, mod 2 means 1+1 = 0

o
then no

@barrycarter Example - physical statement: all hermitian operators are diagonalizable
what would you respond without being pedantic?

@yuggib that's correct.

@yuggib's all-time favourite: "Every wavefunction has to be continuous."

6:56 PM
@ACuriousMind true for physical reasons.

@ACuriousMind ^

@barrycarter is my textbook just fucking with me because i'm 99% sure this isn't possible to prove..

@0celo7 no

@0celo7 No, not at all. Change a wavefunction by defining different, discontinuous values at countably many points. It still defines the same $L^2$-class, hence the same vector in the Hilbert space.

@ACuriousMind Wrong

6:58 PM
There can't be any "physical" reason for disallowing this because it makes no phyiscal difference at all

You're lost in maths

Oh, for god's sake, stop trolling.

What
How can a physical quantity even be discontinuous

@barrycarter is 2) possible ._.