The h Bar - 2015-02-27 (page 4 of 5)

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6:03 PM

@ChrisWhite so, sounds like different interpretations? ;)

@Sean Physics is rough ;)

@GlenTheUdderboat The pot calling the kettle black

@vzn I did some serious research into this stuff lol

It's from the spelling of sowa; a Russian would spell it 'sova', apparently

whoever he is, he seems quite brilliant & well conversant with very deep theories. & quite political wrt mathematics culture.

also a cautionary tale for ppl who want to "rank" this or that mathematician etc... hes nearly "obsessed" with math awards & "rankism" etc

He's kind of a d!ck, basically

And thinks very little of many important fields of research, notably combinatorics

it seems possible he might have quite high credibility/ reputation/ accomplishments under his real name. which reminds me of an earlier comment wrt vixra "jekyll and hyde" etc

Anyone else think this question is a little opinion-based?

0

Q: Do most physicists believe that axions exist?

Is the theoretical justification behind proposing axions sufficiently robust, like, say the Higgs boson prior to its detection, that one should consider the existence of axions as generally accepted (though unconfirmed) physics, or are they more speculative than that?

particle-physics dark-matter axion

6:10 PM

Yeah, I think it's kinda cowardly of him not to put his real name on there

lol

the internet/ cyberspace is the ultimate in "diversity" ... caveat emptor :|

RIP Leonard Nimoy

8

:(

> "live long & prosper" —spock

@DavidZ is this acceptable?

@KyleKanos I don't think it's too bad. Like any area of research, Physics has stuff that is generally accepted, stuff that is earnestly being investigated, and stuff that is highly speculative (to be kind). There is some fuzziness at the boundaries, but I don't think that's a bad question. If he asked 5 years ago about the Higgs, that would be reasonable as well.

6:18 PM

But the question is seems to be more about the belief of the majority and not anything physics

I mean, it's a valid and researched theory

So if he asked about that aspect of axions, it'd be a straight-forward "yes"

But he doesn't

@KyleKanos He asked (or is asking now) if the theoretical basis is robust. It sounds like it is. Maybe the title should change.

@KyleKanos I can't make my mind up on this one

@KyleKanos Yes, I do! Unfortunately, I am out of close votes

Again!?!?

@Jiminion But that's just a bad question because the answer is really Yes, it is robust

6:22 PM

@KyleKanos But a modification to make it more interesting would be easy, so that'd be ok

The question is bad because the Higgs and axions are not really comparable

The initial monopole comparison was much better, imo

The axions question actually smells like the original version of the tachyon question to me

Both monopoles and axions are possible. Neither are necessary for other parts of the theory to work, and hence most people don't give a flying f**k about their existence until experimental evidence comes in

lol, the comment thread is growing hilarious

@ACuriousMind "Pulling something out of my ass" is totally normal as far as I know

@StanShunpike I'm always out of close votes, it's nothing special

@Danu Yes, it is

6:25 PM

@Danu What are you proposing?

Those damn Brits

But Brits would a) have said arse and b) been more subtle in the first place ;)

@KyleKanos If you'd want to make the factual question more interesting? Just add something like: If the theoretical justification is deemed to be robust, then could someone present the most important theoretical arguments for their existence?

Sounds could. Why not post that as a comment for OP to add? Otherwise, I think I'm alright with closing it

Especially since hte comment thread seems to be disagreeing, POB seems like a fine choice!

@Danu I'm not sure if there are theoretical arguments beyond "Hey, we can write down this theory and it's interesting and consistent", but at least that'd be an objective and answerable question.

6:29 PM

@ACuriousMind I love how Wilczek named the axion after his laundry detergent

myfoxdc.com/story/28220591/…

How does that happen?

@StanShunpike lol, what? I always thought it had something to do with some "axial" symmetry

But laundry detergent is far better :D

@StanShunpike Sauce?

@ACuriousMind nope! He said in his book The Lightness of Being when he was a kid he had this box of laundry detergent made by some company called axion

And he always thought it would be a good particle name

So when he got the chance, he used it

@KyleKanos NASCAR people are pretty retarded, on average, no? That's the image they have in any case

@StanShunpike epic

6:32 PM

@StanShunpike That's even better than the story of the penguin diagram

@Danu I know the fans of NASCAR are generally regarded as unintelligent, I'm not sure that the drivers themselves are.

@ACuriousMind What's the story behind that? Because that stuff DOES NOT look like a penguin lol

@Danu Read the Wiki article

It consists almost entirely of the story :D

@KyleKanos Fair enough. Probably biased because of watching the first 10 minutes of 'Taladega Nights' before I turned it off because it was awful

And has a picture rather proving your point

6:33 PM

@ACuriousMind lolol I've never seen that before. Cracking up. People think of everything

@Danu That movie had its moments

@ACuriousMind That story is better than the axion one, lol

@Jiminion you get your girlfriends @ my bachelor's degree?

Good choice, my liberal arts and science nonsense degree is a guarantee for many hot girls :D

Back to Chemistry again I see.

My bookshelf is starting to become just math and physics.

Previously I had all sorts of subjects. It's becoming a theme.

Woit's paper has attracted a Milgrom enthusiast....

http://fqxi.org/community/forum/topic/2366

@StanShunpike Good :P

6:43 PM

@Danu No, WAG = Wild Assed Guess.

@Jiminion Beautiful misunderstanding

@Jiminion lolol

@Danu What does WAG mean to you?

I am totally lost what this conversation is about

you should put that one on urban dictionary

6:44 PM

Googled it, got Wives And Girlfriends

luls

WAGs (or Wags) is an acronym used to refer to wives and girlfriends of high-profile sportsmen. The term may also be used in the singular form, "WAG", to refer to a specific female partner / life partner. The term was first used by the British tabloid press to refer to the wives and girlfriends of high-profile footballers, originally the England national football team. It came into common use during the 2006 FIFA World Cup, although the term had been used occasionally before that. The acronym has since been used by the media in other countries to describe the female partners of sportsmen in general...

@Jiminion Figured you would probably qualify as a "high-profile sportsman" ;)

comment thread in...
http://physics.stackexchange.com/questions/167479/do-most-physicists-believe-that-axions-exist

@Jiminion like the ref to number theory in quantum theory :)

I see

lol

Lol, that Milgrom enthusiast: "The law of conservation of gravitational energy is wrong (in the Newtonian approximation)"

@ACuriousMind lol Danu's first comment on that question is exactly what I was thinking

"Hey, I understood some of those words!"

6:49 PM

To me, it is kind of telling that someone would participate in a contest like that - it seems to indicate that they don't have something better to do.

@StanShunpike Heh, well, I'm also not sure if I understood every word correctly, that's part of why I asked the question - if it's egregiously stupid, someone would correct me ;)

LOL

I'm about to make homemade granola. Should be epic.

How does one proceed with that? Link to recipe?

3 cups ( 1 box ) almonds, 5 cups oats, ¾ cup brown sugar, ½ cup honey, ½ cup water, ¼ cup canola oil, 1 tbsp. vanilla, 2 tsp. cinnamon, 1 tsp. salt

Preheat the oven to 250 degrees F.
Combine the oats and almonds into a large bowl. In a separate smaller mixing bowl, combine the brown sugar, honey, canola oil, vanilla, cinnamon, salt and water. Pour the liquid mixture over the oat mix and stir until combined. Pour onto three baking sheets and spread evenly.

Bake 40 minutes, stirring halfway through to allow for even browning. Raise the oven temperature to 300 degrees F and continue to bake and golden brown, for 20 minutes.

Pour the cooked oat mixture into a bowl and stir in the golden raisins, regular raisins, and cranberries. Let cool. The granola will keep in an airtight container at room temperature in a cool dry place for up to 3 weeks.

Add chocolate chips and coconut for serving.

It's soooo good. I got it off Food Network like 2 years ago.

Ah yeah, makes sense

Soooo unhealthy though, so I shouldn't eat it :P

6:56 PM

It's not!

It's quite healthy actually. My family eats it too and they are a bunch of health nuts

Just don't make it 80% chocolate chips :p

lol

Meh, I'd just eat it all in one go, making it unhealthy

@Jiminion a lot of other interesting stuff on that site. did you guys here about this essay contest? Trick or Truth: the Mysterious Connection Between Physics and Mathematics . looks like ~$50K prizes. alas only ~1wk left before deadline. :(

@Danu Amen! lol best way

you guys wondering about the pt of "QM foundations/ interpretations", this guy Chiribella just got a $50K grant to study it... Purifying Physics: The Quest to Explain Why the “Quantum” Exists / fqxi

You're really pushing the quantum interpretations agenda huh, @vzn?

7:01 PM

what is fqxi? that's twice now it's come up in the last hour

its indeed an "agenda". its called "physics" :p

-1

Q: Dress color - explanation?

I'm having a dispute with some friends in school about the color of this dress (many people on the Internet are disputing this as well). We cannot agree on the color of the dress - is it blue/black or white/gold. Could this be due to saturation of the colors? I'm interested in understanding why t...

optics visible-light

It came back up as another question...

@Sean : Where's the first black/blue/gold/white dress question? Ah, found it. Here. And now here for the 3rd time. Fourth time. (Only visible to 10k+ users.)

omg I thought DavidZ was kidding someone would post a question. I didn't realize someone would actually do it....

I asked it this morning. I was closed so I deleted it

7:04 PM

@vzn I disagree :P

whatever

I figured the question was going to get asked sooner or later, so I tried to ask a well written one about color. It wasn't well received, so I deleted

This one is just like "ZOMG dress!!" though

Why is everyone talking about that dress? The first time I saw it was in the xkcd, and I was like...um, 'kay Randall, I have no idea what this comic is supposed to be.

How did I mistype saw as thought?!

@ACuriousMind it was a viral picture from last night

Perhaps I'm really going insane

user54412

7:08 PM

@Danu You're not the only one. Identical to me.

When I went to bed last night it wasn't even a thing, but this morning it had been shared literally 20 million times

@ACuriousMind I don't understand it either. It's a blue and black dress. That's all there is to it.

To a lot of people, myself included, it appears white with gold trim

So it became a big deal

I suspect before the day is over we'll probably see another question or two about it crop up

So far, I've now spoken to people in China, Jamaica, the Netherlands, Germany and Japan who all are talking about it.

Quite amazing.

It depends on which device I look at the image

It looks one way one my phone, another on my laptop

7:11 PM

@tpg2114 Really? I tried to look and I just couldn't myself to see white, gold no matter what I did

It's actually a cool way for people to learn something new about how vision, perception and devices work but as usual it's just people being stupid about things

@StanShunpike I also have known color deficiencies in my vision, so ambient lighting, screen backlighting, and probably a dozen other factors will affect how I see it

We were talking about colorblindness in here the other day, I have similar problems with other colors. Brown with a red background looks green to me, brown with a green background looks red

Which is what the xkcd is referring to -- how you see colors depends on how saturated the color receptors in your eyes are

I don't want to disturb anyone, but I see a blue and gold dress. There's nothing black or white about it

theconcourse.deadspin.com/…

@ACuriousMind You're actually closest to being correct

CW said there are slightly different images going around. seems theres some pranking involved.

Photoshop colors indicate it's blue and brown

user54412

7:15 PM

@ACuriousMind You, me, and @Danu are the only sane people left, it seems.

@ChrisWhite Except the original poster of the photo said it's definitely blue and black in person

user54412

@tpg2114 Well, as an amateur astrophotographer, I know all too well that what colors the camera sees can have no relation to what colors your eyes perceive :)

I'm really more puzzled why everyone is freaking out about it. That our visual perception is easily fooled shouldn't be news, right?

user54412

user image

I mean, there are dozens of these optical illusions where you think two things are a different color but they are the same, and vice versa

user54412

7:18 PM

Look at the gray dots!

@ACuriousMind Have you ever talked to anybody outside of a physics chatroom? You're talking about a collective group of people who spend all day making cat pictures to send to one another.

2

user54412

Look at them!!!

Yeah, that, for example :D

@ACuriousMind People are so uneducated. I was horrified to learn there are kids in my local neighborhood who can't add or subtract.

@ChrisWhite Ummm... I see orange dots. ;)

7:19 PM

LOL

@tpg2114 I'm not really chatting to strangers on the internet outside this room, no, and I have a hard time believing the people I know in person would find this so surprising.

(And I know plenty of non-physicists)

FWIW, the photo everybody is worried about does not do the dress any justice. It looks pretty ugly in that photo, it's actually not bad looking:

user image

@ACuriousMind Consider yourself lucky.

user54412

So... the conclusion here is that the photographer is just really bad

Seems to be

It was posted by two girls or something like that

7:22 PM

Ooo,

user image

Yes! That's the classic one

@vzn Lee Smolin also has an entry. That's too much firepower for me....

on this?!?

@StanShunpike Do you know you can click the little arrow to the left of the @ to jump to the message that is being referred to?

Oh, really? No I didn't. Thanks!

Wow, that's really convenient!

I scrolled through an entire day's worth of chat yesterday to find one comment. Good to know!

7:31 PM

So I clicked on QuantumMechanic's links, and was able to see my dress question (that I deleted myself) and now it appears blue and black. It was gold and white this morning

I just don't know what to believe anymore

user54412

@tpg2114 There's a GR analogy in that one. Local color (vector) comparisons are the only unambiguous ones. If you want to compare distant things, you need to transport one to the other. You could keep the color aligned relative to its neighborhood at each infinitesimal step (parallel transport), or you could hope the coordinates are sensible and just take the components in your coordinate system to be unchanged.

Further confirmation that I should stick to hard physics I guess

@ChrisWhite Colors are just vectors after all, eh? That's a good analogy.

But I don't understand @tpg2114's picture. I mean, there's actually a shadow there, right?

@Sean Because of the shadow, you think B is lighter than A

However, they are the same shade

Your brain makes you think it should be lighter because of the surrounding shadow -- ie. color perception is local

7:37 PM

@ACuriousMind RGB values appear to be a vector space. As do HSV :D

@KyleKanos Yep, that's why the analogy is good

Holy crap everyone at school is talking about that dress.

US Education system at its finest!

Although we sit here making fun of people for talking about it. Yet, we've been talking about it for awhile now.

I have no idea how anybody uses Windows for any kind of actual development of anything outside of using Visual Studio

@ACuriousMind has informed me curves on a manifold are invariant under coordinate transformation. why? I thought nothing would be invariant under coordinate transformation. But he said that, suddenly a whole bunch of things made sense. Yet I still don't understand why curves would be invariant under coordinate transformation.

Finding and installing libraries is a huge PITA

7:46 PM

@StanShunpike I'm not sure if you understood me correctly, but a coordinate transformation is just that...a coordinate transformation. It changes nothing except your personal description of the manifold. In a sense, coordinate transformations are gauge transformations.

Hey guys, why does conformal invariance show up in string theory? Is it because a string is analogous to a 2-d cylinder in 3-dimensions and the energy given off by the string is symmetric around the cylinder?

@bolbteppa The string worldsheet action is conformally invariant.

(If you think of conformal transformations as angle-preserving transformations)

Yeah but why is it conformally invariant physically?

@bolbteppa If you take $n$-dimensional objects an let them trace out $n+1$-worldvolumes, the action that is just the integration over the worldvolume is conformally invariant for $n=1$.

@StanShunpike Why would the curve change just because our arbitrarily drawn coordinate lines change?

user54412

7:51 PM

user image

user54412

@StanShunpike The homeomorphism you use to describe the manifold doesn't affect the curve $\gamma$

The Nambu-Goto action is just integrating the surface area, doing a Legendre transform gives the Hamiltonian and conformal invariance basically says that the energy of the string doesn't depend on the angle you look at on the surface right?

@StanShunpike coordinates parametrizing a curve on a manifold are an artificial construct, they are just a way to talk about an invariant geometric object on a surface called a curve, whatever definition we cook up better respect this intuitive property of curves on surfaces right?

user54412

If you want your curve to be from the interval to U itself, then sure the curve will change when you change coordinates, but the compositions of these alternately-defined curves with their respective maps back to the manifold will be the same

@ChrisWhite Ah, well said. I think that's what I wasn't understanding.

user54412

user image

7:56 PM

@ChrisWhite I will need to think about that and read a bit more. But that's very helpful.

@ChrisWhite What is $\alpha$ supposed to be?

user54412

The confusion I think comes from the fact that we often interchange the open subset of $\mathbb{R}^n$ with its image under the coordinate chart

user54412

@0celo7 The chart.

@bolbteppa Have you tried doing that? The Hamiltonian of diffeomorphism-invariant systems vanishes identically :)

user54412

homeomorphism/diffeomorphism/whatevermorphism it's supposed to be

7:58 PM

So talking about the "energy of the string" is not really a sensible thing to do

@ChrisWhite Uh, don't charts take you from a subset of the manifold to the subset of the real numbers?

Really?

user54412

@0celo7 It's a simple manifold. One chart in the atlas :)

@bolbteppa Yep.

Right, but charts don't take you to the manifold.

7:59 PM

That's why GR don't plays well with the Hamiltonian idea, either

They take you from the manifold to the real numbers.

@ACuriousMind There is a Hamiltonian formulation. Just not pretty.

user54412

@0celo7 Whatever. They're perfectly invertible, and I drew the arrow going both ways :P

The problem is that you have to do a 3+1 split.

@ChrisWhite So is $U$ a subset of the real numbers?

@0celo7 I said "don't plays well". And that Hamiltonian formulation doesn't come from a Legendre transform of the Lagrangian in the action, I hope?

user54412

@0celo7 A subset of $\mathbb{R}^n$ -- see the previous image

8:00 PM

@ACuriousMind It does.

Does it vanish identically?

You have to do a 3+1 split of spacetime. Then the field is the spatial metric and the momentum has something to do with extrinsic curvature.

Looking, don't recall.

The thing I'm not sure about is whether the vanishing constraint is on- or off-shell

user54412

@ACuriousMind I thought it does. $\pi^{ij}$ is canonically conjugate to $g_{ij}$ (the spatial metric)

Yes, okay, I backpedal

8:03 PM

@ChrisWhite It does, Wald confirms.

You can do the Hamiltonian formulation, but you have a constraint that says the Hamiltonian vanishes

I have some lecture notes on it somewhere.

That's where the level matching condition on the string comes from in the first place

So, @bolbteppa, you can get the Hamiltonian, but it has no intuitive meaning because it is constrained to vanish

@ACuriousMind Are you sure it vanishes?

I might want a source/proof of that.

Yes, I'm reasonably certain

Give me a sec

8:06 PM

I'm not sure about any of that tbh I think that is a misunderstanding of Hamiltonians but I've seen it said before in places but also contradicted

Hamiltonian constraint

The Hamiltonian constraint arises from any theory that admits a Hamiltonian formulation and is reparametrisation-invariant, although the Hamiltonian constraint of classical general relativity is included as an important non-trivial example. Something of note in the context of general relativity is that the Hamiltonian constraint technically refers to a linear combination spatial and time diffeomorphism constraints reflecting the reparametrizability of the theory under both spatial as well as time coordinates. However, most of the time the term Hamiltonian constraint is reserved for the constraint...

...is at least a starting point that seems to agree with me

But it seems it is not as simple as the Hamiltonian itself vanishing every time

On general principles an action in Hamiltonian form is just a Legendre transform of an action $S(x^i,u,\partial_i u)$ in Lagrangian form, which is just another way to say that instead of generating the action surface S with coordinates ($x$, & $u$) we can instead generate the surface with coordinates $x^i$ and (tangent) lines $\partial_u S$, like it's literally just a mathematical claim, however convexity issues mean it's not always possible, so that might be the explanation

@ACuriousMind OH. In string theory, the vanishing of the energy momentum tensor implies the vanishing of the Hamiltonian.

I think.

In other words, generate a surface with points and lines instead of points and points

@0celo7 Yes, and that's a consequence of the diffeomorphism invariance

8:11 PM

But that's actually a gauge choice.

The energy-momentum tensor being identically zero is a gauge choice.

Conformal invariance tells us it must be traceless.

Wait, no. This is confusing.

Yeah this has always bugged me

That actually follow from the equation of motion of the string worldsheet.

Conformal invariance implies tracelessness of the EM tensor

But using the EOM as a constraint on the Hamiltonian doesn't mae much sense.

@bolbteppa What ST book are you looking at?

ST? String theory lol

8:15 PM

What string theory book are you looking at?

ST = string theory

Wait, the equations of motion for the world-sheet metric - which was forcibly introduced to be dynamic in the first place - are $T = 0$, or am I really going insane now?

That is, the original Nambu-Goto action has $T=0$ always, because it is Polyakov with metric on-shell

@ACuriousMind BBS says that the worldsheet EM tensor is identically zero because the string-sigma model action has no kinetic term of the worldsheet metric.

The vanishing of the Hamiltonian is given here books.google.com/… but...

@0celo7 Yeah, and you can't add one without breaking a symmetry

I don't see why having no kinetic term for the metric means the EM tensor vanishes.

8:18 PM

Conformal Invariance implies tracelessness, bottom of p. 1 & top of p. 2 here nikhef.nl/~t58/CFT.pdf

@bolbteppa We all agree on that I believe.

@bolbteppa: I think the "energy" of the string would, anyway, more properly be the timelike component of its momentum in the target space

@StanShunpike Heh, yeah... chat is complicated sometimes

@ACuriousMind You can get the 4-momentum using Noether.

@0celo7 You mean the 10-momentum?

Or, the 26-momentum? ;)

8:20 PM

26 lol

Sure.

D-momentum

Sure, it's conserved by the target space-time symmetries.

I didn't say anything against that, did I?

@ACuriousMind Why does no kinetic term for the metric imply a vanishing energy-momentum tensor?

I find it very interesting that in a CFT video lecture the guy said Polyakov derived his path integral (sum over surfaces) as a way to try to talk about string theory in non-critical dimensions

@0celo7 No idea why they phrase it like that.

8:23 PM

@ACuriousMind So why does the energy-momentum tensor vanish?

But, probably, because that'd be a term that would definitely contribute, while the ones we already have don't

@0celo7 Well, just look at the equations of motion for the metric!

That's $T=0$

Shouldn't this be Bohmian Gravity??
https://www.youtube.com/watch?v=2rjbtsX7twc

This is, by the way, different from the tracelessness, because the tracelessness holds off-shell

@ACuriousMind Uh. The EM tensor is $T\sim\frac{\delta S}{\delta\gamma}$ and the EOM is $\frac{\delta S}{\delta\gamma}=0$. Thus $T=0$?

Stupid trivial crap.

@0celo7 Sounds silly, but it's true

This all is confusing because you have essentially three different laws for the tensor, which all hold on different shells - off-shell, you only have tracelessness, on-shell for the bosonic fields, you have conservation, and on-shell for the metric, it just vanishes.

8:29 PM

The reason I was confused is that in GR we only vary the matter action to get the energy-momentum tensor.

Yeah, GR is a bit different.

In ST there is only one term in the action.

Else it wouldn't have much to do if $T=0$ would be true :D

@ACuriousMind I was seriously worried about the Einstein Field Equations: for a second I thought they were identically zero.

woah, so if you take the derivative of the special relativity action w.r.t. the metric it's zero because it's the fixed Minkowski metric?

8:30 PM

@Danu it needs to be edited to transcribe the math. If that's done, I guess it might be an answer, though it doesn't look like a good one.

@bolbteppa What SR action?

@bolbteppa The metric is not a dynamical field in SR, you can't define a derivative w.r.t. to a constant

@ACuriousMind Formally, we can. Just make it a variable and then take derivatives while evaluating at the constant.

That's how we calculate energy-momentum tensors in SR, after all.

I never calculated a EM tensor in SR in my life, so I'll believe you

But special relativity is just general relativity where you choose a particular fixed value for the $g_{\mu \nu}$ in the action $S(x^{\mu},A_{\nu},g_{\mu \nu})$ so taking the derivative of a function where the variable $g_{\mu \nu}$ always stays fixed will obviously be zero, that is strange...

8:38 PM

@ACuriousMind N. Straumann, General Relativity (2013), p. 92.

Note that this gives us the Belinfante-Rosenfeld tensor, not the canonical one from Noether's theorem.

So wait, why does the Hamiltonian vanish for special relativity and string theory, yet you can use it in quantum field theory for KG or Dirac?

@DavidZ Why did you expect to get a question about the dress? Is it common for trendy things like that to end up on SE?

@bolbteppa What is the Hamiltonian for SR?

@StanShunpike My reason for expecting it is that certain users seem to feel that if there is anything at all sciencey about a question then they should bring it to Physics SE.

And they get standoffish when someone suggests that maybe it isn't really the right place for it.

And some even tell me that the [whatever] they are asking about is made out of atoms. So, there.

::sigh::

@ACuriousMind S. Carroll, Spacetime and Geometry (2004), p. 176.

8:47 PM

@dmckee Interesting. I usually don't pay attention to poor questions because I know they aren't supposed to be here. So I guess I didn't realize how off-topic some things could get.

@0celo7 sorry p. 11 here books.google.ie/… has the SR hamiltonian vanishing, p. 14 has the Nambu-Goto one vanishing

@0celo7 What am I supposed to do with the references you send me? I said I believe you :D

@ACuriousMind I actually looked in 5 books. I'm horrified I only came up with two references.

@StanShunpike It's probably not that bad, really. But part of moderating is looking at all the real stinkers...

Heh, then it is perhaps not so unusual that I never saw a EM tensor in SR

8:49 PM

@bolbteppa I can't view that page. You mind writing the relevant equations here?

@ACuriousMind I don't think people care about them at all.

@dmckee: I don't know if you read it further upstream, but I have the suspicion that one of our users has been hacked. I raised a mod flag for that, was that the correct thing to do?

@ACuriousMind R. M. Wald, General Relativity (1984), p. 427. Took me a while to find that one, but it's the most authoritative.

@ACuriousMind I saw your flag, but I am uncertain what to do about it.

Possibly it should be escalated to the team. Plain 'ol moderators don't have sufficient power to to anything about it.

Given $S = \int \mathcal{L} d \tau = -m \int ds = - m \int \sqrt{- \eta_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}} d \tau$ we have $p_{\mu} = \tfrac{\partial \mathcal{L}}{\partial \dot{x}^{\mu}} = \tfrac{m \dot{x}_{\mu}}{ds}$ so that $H_{canonical} = \tfrac{\partial \mathcal{L}}{\partial \dot{x}^{\mu}} \dot{x}^{\mu} - \mathcal{L} = 0$

That's what the book says

@dmckee I kinda suspected that, and I didn't really know what to do, either, but I felt I couldn't simply let it be, so I passed the decision on to...you, it seems

8:56 PM

26

Q: How to investigate Stack Overflow account hacking that resulted in answering some question inappropriately?

About an hour ago, from my account, an inappropriate message was posted as a reply to a Stack Overflow question. I have noticed it, because I started getting downvotes. To my understanding my Gmail account wasn't compromised (I reviewed my account security settings and login history). I am from ...

discussion login hacked

@bolbteppa Huh? Unless I'm doing that thing where I forget math, that's not zero...

I think your momentum is wrong.

$p_{\mu} = \frac{m \dot{x}_{\mu}}{\sqrt{-\eta_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}}}$

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Feb '1527

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