The h Bar

1:46 AM

@ACuriousMind sigh It actually kinda bothers me when people don't understand metaphysics.

Also: what has been going on the last few days that we've consistently had 30+ items in the review queue (which I've cleared & can't access, meaning it's you other guys)

ACuriousMind

2:07 AM

No idea what's going on there

I guess many posts are left hanging in the queue with 3-4 close votes, but I'm not sure

Kyle Kanos

Hmm, there are quite a load in the 3-4 range: physics.stackexchange.com/tools?tab=close&daterange=

ACuriousMind

I forget I could've checked that myself. In fact, I periodically forget that the 10k tools are there at all

Kyle Kanos

Yeah, they've not seemed entirely useful

DanielSank

2:31 AM

@DavidZ I don't get why people say this about quantum mechanics.

It's not that weird.

It's certainly no weirder than magical invisible electric fields which are spewed by particles and can push on things at a distance.

ACuriousMind

@DanielSank Because the founders of quantum mechanics themselves admitted being confused by it, I guess, and because "understanding" for many means somehow identifying an Aristotelian cause for every effect, something which quantum mechanics generically doesn't have (hence the desire for hidden variable theories)

DanielSank

@ACuriousMind Yes, indeed.

Let me rephrase: I don't get why contemporary physicists who have the benefit of the last 80 years of history say that anyone who says they understand QM is lying or clueless.

Kyle Kanos

@DanielSank It seems to me that David's just using that colloquial phrase about C++ and isn't (necessarily?) applying it to QM

ACuriousMind

One possibility is: They don't understand it, and don't want to say they are clueless :P

DanielSank

@ACuriousMind ;-)

Kyle Kanos

2:41 AM

Also: I think I've learned way more about my research the last ~3 weeks of dissertation writing than the last 3 years of doing the research

Stan Shunpike

Learning is life :D

And writing is a great way to learn

DanielSank

@KyleKanos Hey, congrats on being in the writing stage!

Stan Shunpike

Yeah, I meant to say that as well. That's awesome!

¡felicitaciones!

@ACuriousMind what's the best video game (or PC, etc) you've ever played?

Kyle Kanos

@DanielSank 2.# days left

Not sure how to count the remainder of today and the amount of Wednesday I have

But thank you

@StanShunpike While I'm not ACM, I'd say FF7. You can get it on Steam and hack the crap out of it (e.g., beat the nominally 20+ hour game in 4 hours)

DanielSank

@StanShunpike I have opinions on this.

Have you ever played Super Metroid?

Kyle Kanos

2:56 AM

^Also a good one

(though I wasn't really a fan of Metroid Prime, I think the platformer version is way better than the 3D version)

Stan Shunpike

No, what's super metroid?

DanielSank

@KyleKanos: Did you like Super Metroid a a lot?

ACuriousMind

@StanShunpike That's like asking a parent for their favourite child ;) But I will always love the RPGs of old like Baldur's Gate, Morrowind and Planescape (and the new Pillars of Eternity is looking like a faithful hommage to BG so far). Of the newer ones, The Witcher 2 was great, and I also very much liked both Dark Souls.

DanielSank

@StanShunpike It's the best video game ever made.

It was a Super Nintendo game.

The reason I asked is that if you are a fan (like me) who has played it over and over and over, you would be quite interested in a fan hack of the game called Super Metroid Redesign.

Kyle Kanos

@DanielSank I played the GBA ones (Fusion & Zero Mission) and enjoyed those , but I have not actually played the SNES games (though I believe I've seen some speedruns of those)

DanielSank

3:00 AM

It's bigger, more complex, and much harder than the original. For people who've played the original to death this fan hack was a godsend.

@KyleKanos: Well, from one video game aficionado to another, I recommend Super Metroid with all my heart.

Do not look up walkthroughs if you play it.

Kyle Kanos

@DanielSank That sounds like gjoerulv's hard core mod for FF7

@DanielSank After the defense, FF5 is first up. But I will certainly put it on the list

(and not a typo, I mean FF5)

Stan Shunpike

So people take these old games and format them for modern systems?

DanielSank

Super Metroid Redesign is a completely different map than the original.

@StanShunpike You can just download emulators.

snes9x.com

Stan Shunpike

Ah, that's cool.

DanielSank

zsnes.com

Then you find memory images ("ROMs") of the games.

Kyle Kanos

3:03 AM

I must also admit that I sucked at both Metroid's I played. I died. A lot

DanielSank

@KyleKanos Well... that's the life of a space bounty hunter.

Kyle Kanos

Had to watch youtube tutorials on boss strategies

All that stuff

DanielSank

@KyleKanos What's "the defense"?

By the way, for anyone hanging out here who wants to play a silly and ridiculously fun/nostalgic game, try this:
http://explodingrabbit.com/pages/super-mario-bros-crossover/

You get to play Super Mario as one of several different classic Nintendo characters.

It's ridiculously fun.

Kyle Kanos

@DanielSank Where I defend my work against (to?) the committee

DanielSank

Oh oh right right right.

Kyle Kanos

3:05 AM

PhD defense?

Okay

DanielSank

Dude, you need to change your attitude.

It's a thesis OFFENSE.

Kyle Kanos

I should think of it that way

DanielSank

They're the ones who are lucky if they pass.

ACuriousMind

Obligatory xkcd.

Kyle Kanos

Be proactive about making the one guy who's going to be the pain of the committee convinced and won't need to say a damn thing

DanielSank

3:08 AM

@KyleKanos It's interesting that you didn't like Metroid Prime.

I think like many Super Metroid die hards, I was just so impressed that it wasn't horrible that I was happy.

I've actually replayed it a few times and still like it.

Kyle Kanos

I just think 3D games like that are too much for my brain to handle casually, so I reject any fun had playing

Like, I'm supposed to get up there? How the hell does that work and then YouTube shows like 4 carefully executed bomb jumps later you're there

Too complicated for me. I like to vege out when playing.

Stan Shunpike

I can't stand having to do anything when I'm playing video games besides just twiddle my thumbs

Kyle Kanos

3:30 AM

Noooooo.....I just replaced an image I spent 15 minutes working with another image I spend 15 minutes working on. No CTRL+Z to save me :(

Stan Shunpike

@KyleKanos screw that. I feel your pain though. Been there and it sucks.

Kyle Kanos

Actually, I hadn't refreshed the PDF, so I just took a screenshot of the image off the computer and am running with that

Stan Shunpike

That's convenient.

@KyleKanos that top answer of yours about the rocket fuel is cool. But its also insane how much fuel rockets take.

Kyle Kanos

(btw: the "replaced" I state above mean, I copied image A from folder C into folder B where a needed image A existed)

@StanShunpike Well that's why they're so big! :D

I also need 24 more votes on it

So if you have 24 friends, let them know

Stan Shunpike

I'll see what I can do. Ohhh, right. You need that to get gold right?

:D

Kyle Kanos

3:45 AM

I need that gold

I'm like the Aliens from Cowboys & Aliens, gold is some sort of rare mineral in the universe that I've gotta collect it all

Stan Shunpike

Lmaooo

Omg the other day, there was this question at -3. I read it and decided to edit it just because I liked the question. And after I did that, the question jumped onto the HNQ and rode to +60 for both question (dont remember who) and answerer (John Rennie). I was like.....man that's a lot of points.

Kyle Kanos

Which one?

Stan Shunpike

0

A: Could we send a man safely to the Moon in a rocket without knowledge of general relativity?

Absolutely we could, and in fact, I strongly suspect that General Relativity was never used in the Apollo program. for one thing, the on-board navigation computers were nowhere near powerful enough to perform any useful calculation with GR. on the other hand, it's possible to measure the positi...

Kyle Kanos

Ah, that one

That was a really interesting question and I had no idea why anyone started the close votes on it

Stan Shunpike

Yeah, I thought so too! I don't make a habit of editing (just flagging) but I couldn't resist. And it got some nice answers that was the cool part. I had no idea Newtonian gravity only was off by a few cm.

@KyleKanos do downvotes bias ppl to vote to close?

Kyle Kanos

3:59 AM

@StanShunpike I cannot speak for anyone else, but I do not let it bother me.

There was a recent Q about fluid dynamics that had a -1 or -2 with a close vote (unclear), but I stated that it wasn't at all unclear and it got 2 good answers (one coming from me, of course)

Stan Shunpike

@KyleKanos Are you a fluid dynamics guru? I noticed on your profile you said its one of four SE tags your answers tend to belong to.

Kyle Kanos

I'd not call myself a guru in anything really. I know fluid dynamics because my research depends on it

But my interests are in the evolution of supernova remnants, so I'm more of an astro guy than a fluids guy

Stan Shunpike

I remember when I learned that fluids are useful for studying GR. I thought that was so cool. Superfluids are awesome too. I don't really understand them, but the videos are cool

Kyle Kanos

Well I've gotta get to bed. I'm starting to drift off to sleep whilst typing and that's not really a good idea

Stan Shunpike

Gnite!

Stan Shunpike

4:43 AM

@DanielSank I forgot about Metroid!!!!!! I loved that game.

That was when I was reaaaaaally little.

Wow, totally forgot about that.

Physics Meta

5:58 AM

0

Q: Borrowing images

Can you tell me if it is allowed to borrow a diagram from another post in this site? Is it necessary to cite both the source and the post from which it was borrowed? Thanks

feature-request

DanielSank

6:11 AM

@StanShunpike If you liked Super Metroid, you really should check out Super Metroid Redesign.

Stan Shunpike

9:38 AM

@DanielSank yeah I plan on it now. I'm really curious what SMR will look like

ACuriousMind

2:48 PM

0

Q: Horrible crossword

There's this physics crossword that is giving me a lot of trouble Water waves do this when they pass from deep to shallow water (7) _ e _ r _ _ _ Carries information by total internal reflection of light (7,5) _ _ t _ _ _ _ _ i _ _ _ Angle of incidence when angle of refraction is 90 degrees ...

visible-light waves water reflection

^really?

Kyle Kanos

And those comments are precisely why I think I need to write an answer to John Rennie's recent Meta post about HW

ACuriousMind

@KyleKanos You mean, it's obviously off-topic, but the asker still gets the answer in the comments?

Kyle Kanos

@ACuriousMind Exactly. Rennie made a statement that said something like, "Some people think VTC + answer/hint in comments is okay" and I believe that's entirely wrong

If we do not want HW questions, we should, collectively, stop even giving hints

We should point them to the Meta post about where they can get actual HW help and VTC

Jimnosperm

These homework issues will be the death of us

Kyle Kanos

Probably so

Also, is Valter serious here?

I don't even know what a "partition of unity" is here, but the Wikipedia article makes it sound like VM's got some crazy scheme that's probably far more complicated than the simple Jacobian transformation would require

Jimnosperm

3:00 PM

Okay yeah, I am the opposite of him there

I can see how to do it with anything except a partition of unity

Kyle Kanos

I guess it's just one of those things that happens when one becomes so entrenched in (more) complex mathematics that they forget the trivial things

ACuriousMind

@KyleKanos I think Valter is serious in that a rigorous treatment of this delta probably requires looking at patches of the space where the $\phi = 0 = \2\pi$ or the $z=0$ with $\phi$ arbitrary ambiguities in the cylindrical coordinates are not there, and then defining the delta properly requires a partition of unity

Kyle Kanos

$\int\int \delta(x-x_0)\delta(y-y_0)\,dxdy=\int\delta(r-R)\,rdr$ and you're done. Without invoking anything except the definition of the delta function.

ACuriousMind

I bet you that the subtlety of the definition of the delta as a distribution rather than a fucntion makes this not as simple

You get the correct result, but it's not rigorous

Kyle Kanos

IMO: anything beyond that is over-complicating the situation

@ACuriousMind And how is it that the way I did it not rigorous?

Like not being defensive/offensive here, I just don't understand how/when something is or isn't rigorous

0celo7

3:07 PM

I think that to be completely rigorous, you always need a partition of unity when spheres are involved.

Kyle Kanos

AFAIK, as long as it follows the mathematical logic, it should be rigorous

ACuriousMind

@KyleKanos As a distribution, you cannot integrate over the dirac delta alone - you need to multiply it with a function of compact support, which is where the partition of unity comes in - $f(x) = 1$ doesn't have that compact supports, but the elements of the partition will have.

0celo7

@ACuriousMind Do you have any clue how to calculate the holonomy group of the Poincare half plane? Nakahara kinda dumps this exercise on the reader without much explanation of what to do.

I vaguely know that I have to determine how a vector is parallel transported around closed loops.

The metric is $g=y^{-2}(dx\otimes dx+dy\otimes dy)$

ACuriousMind

@0celo7 You mean the holonomy of the Levi-Civita connection on it?

0celo7

Yeah

He does the calculation for the 2-sphere, but there the loop is just a circle...on this space I have no idea what to do.

Kyle Kanos

3:12 PM

@ACuriousMind So basically Valter is saying the same thing as me, just using a more formal terminology?

0celo7

@KyleKanos Valter's book has "review of measure theory" or something of the sort as the first chapter.

ACuriousMind

@KyleKanos You both will have the same idea of proof, but your integrals do, formally, not exist, I think.

0celo7

He's probably a bit more pedantic than most of us.

@ACuriousMind So should I determine the Christoffel symbols to begin with?

Kyle Kanos

@ACuriousMind IMO: the integrals totally exist because Wikipedia says so (though it does also say that it's heuristic)

0celo7

@ACuriousMind I cheated and found a list of the Christoffel symbols online. What now?

ACuriousMind

3:18 PM

@0celo7 You'll need them, I think - you need to determine the parallel transport map that the connection induces along the closed loops. But it could be you don't need to calcuate anything - you know that the parallel transport maps will be isomorphisms between the fibers (tangent spaces, here), and you know that Levi-Civita preserves the metric. So, what are the linear isomorphisms that preserve the metric?

0celo7

@ACuriousMind Constant $y$ preserves the metric.

ACuriousMind

@0celo7 Huh? What do you mean with that?

0celo7

@ACuriousMind You don't mean isometries, do you?

I misread "linear isomorphisms"

ACuriousMind

I mean isometries, but on the tangent spaces, not on the manifold

0celo7

I'm not sure what you mean.

ACuriousMind

3:22 PM

The tangent space is a vector space with metric - which of the $\mathrm{GL}(n)$ additionally preserve its metric?

0celo7

Ok, if it preserves the metric, then it preserves lengths as well, right?

ACuriousMind

I'm not yet sure if you can say that all of these will indeed be realized as holonomies, but these are the only ones that can occur as holonomies

0celo7

If lengths are preserved...then we must have some $\mathrm{SO}(n)$, right? And in this case $n=2$?

ACuriousMind

@0celo7 Is the metric on the tangent space the one that induces the ordinary Euclidean length here?

0celo7

@ACuriousMind I do not know.

ACuriousMind

3:25 PM

(I'm not saying it isn't, I'm saying you have to check)

0celo7

How do I check?

ACuriousMind

@KyleKanos "heuristic" means "strictly speaking, this doesn't work" :P

@0celo7 Look at it. That is, what is the metric on the tangent space $T_{(x,y)}M$?

0celo7

If it's not the same as the metric on $M$, I don't know. I'm not even sure what that means then.

ACuriousMind

@0celo7 How is a metric on a vector space given, and how is a metric on a manifold given?

0celo7

@ACuriousMind The metric on a vector space is an isomorphism between the vector space and its dual.

Manifold...something with lengths?

ACuriousMind

3:30 PM

@0celo7 That would be the thing that would turn the manifold into a metric space, but usually, one gives a "metric" on the manifold by giving the metric tensor, right?

0celo7

Of course.

ACuriousMind

So, what is the metric tensor in terms of the tangent spaces?

0celo7

$y^{-2}(dx^2+dy^2)$?

ACuriousMind

...well, that's not wrong :D

0celo7

I'm still not sure what that means.

ACuriousMind

3:32 PM

The metric tensor is a thing that assigns to every point a tensor that eats two tangent vectors and spits out a number

A metric on a vector space is a thing that eats two vectors and spits out a number.

Hence, the metric tensor is an assignment of a metric to the tangent spaces at every point

0celo7

Oh of course. I didn't know that's what you were asking.

ACuriousMind

Ordinarily, you would represent a metric on a vector space as a matrix $G$ such that $g(x,y) = x^T G y$, right?

0celo7

Yeah.

ACuriousMind

And the group of isometries or "linear isomoprhism that preserve the metric" is then the group of matrices for which $M^T G M = G$.

By the properties of the Levi-Civita connection, you know that the holonomy group must be at least be a Lie subgroup of this group, and I suspect it is already the whole group.

0celo7

Yes. So I have to find the matrices $M$ that preserve the Poincare metric $G$?

Elementary inspection shows that all orthogonal matrices do this.

ACuriousMind

3:37 PM

Alright, so $\mathrm{O}(2)$ is the maximally possible holonomy group.

Now, one should probably try to find a loop for every generator of this group to show that all of them occur as holonomies (or not)

0celo7

How the heck do you do that?

We should only have one generator, right?

ACuriousMind

Well, I'd try infinitesimal squares first and look at what holonomies they give me, I think

@0celo7 Why that? The Lie group with one generator is $\mathrm{U}(1)$

0celo7

Well, the text actually gives the answer as $\mathrm{SO}(2)$.

He's pretty good about distinguishing between $\mathrm{O}$ and $\mathrm{SO}$.

ACuriousMind

Ah, yes, if the manifold is orientable you won't get a reflection as a holonomy

0celo7

And $\mathrm{SO}(2)\cong\mathrm{U}(1)$, so it has one generator.

ACuriousMind

3:41 PM

@0celo7 Oh. Yes. You're right

So the only task left is to show that there is at least one loop with non-trivial holonomy.

0celo7

So what now? We need the Christoffel symbols, right?

Do we really have to do that though?

Isn't nonvanishing curvature enough?

ACuriousMind

@0celo7 Yes, you need to pick a concrete loop and calculate the parallel transport along it, I think

@0celo7 If you know that "curvature gives infinitesimal holonomy", then, yes :P

0celo7

So we need to pick a vector and four points you say?

ACuriousMind

For example, yes. I guess, I don't think I've ever calculated a holonomy :D

0celo7

So something like $\partial_x$ and $(1,2),(-1,2),(-1,1),(1,1)$

@ACuriousMind Good exercise for both of us :P

ACuriousMind

3:44 PM

Yep, and then we translate it around the square and if it comes back not the same, we should be done

0celo7

Ok, so along this path the parallel transport equation is not obeyed?

Or is it and we just show that $V_i\ne V_f$?

ACuriousMind

The latter

We parallel transport the vector and show that the result is not the same vector, but (hopefully), one rotated by an element of $\mathrm{SO}(2)$

0celo7

Ok, so we make a vector $A=A^x\partial_x+A^y\partial_y$ with the initial value $A_i=\partial_x$ at the top right corner?

ACuriousMind

Yep

@KyleKanos: It seems people can't resist. Someone deleted the answer comments and there's already a new one :D

0celo7

What is the vector $V$ that I am transporting around in the transport equation?

Constant coordinate for each side of the square?

ACuriousMind

3:53 PM

@0celo7 Uh...the $A$ is the vector we want to transport, isn't it?

0celo7

@ACuriousMind Yeah, poor choice of wording. Assume the notation $\nabla_VA=0$.

ACuriousMind

@0celo7 Ah, I see. Then it's the vector parallel to the side of the square, yes, which has constant coordinates.

0celo7

Ok, so I have the two equations now (x is the 1 coordinate and y the 2 coordinate) $$\dot A^1-\frac{2}{y}V^1A^1=0,\quad \dot A^2+\frac{1}{y}V^1A^1-\frac{1}{y}V^2A^2=0$$

I'm really bad at summation, so I might be missing a 2 somewhere...I hate my life.

No I think I'm good.

So now what? I solve these equations?

That seems really hard.

Jimnosperm

Hey guys, I'm trying to answer the conservation law question and I'm at a point where I'm explaining how something can be globally conserved but not locally conserved and I'm trying to think of an example of that. But I can't. That is, I can't think of a case where you could violate local conservation but maintain global conservation by, say, destroying part of something in one place and creating an equal amount in a separate location. I can't see how that's possible

0celo7

@Jimnosperm Is energy in GR globally conserved? (I know it isn't locally.)

Jimnosperm

3:59 PM

@0celo7 Most would say no, and it is conserved locally

0celo7

Ok, bad example then :P

How can it be conserved locally?

I might have something mixed up.

Jimnosperm

energy-momentum is locally conserved. That's where the classical law of energy conservation comes from

0celo7

Yeah I got that backwards.

Jimnosperm

My thinking is that anything conserved globally should be conserved locally too. If something can just disappear from one place and simultaneously reappear in another, then you could find another observer, to whom it would look like it's not simultaneous and all of a sudden the global conservation is temporarily violated. But I may be overthinking this. I thought I'd ask if there's any good examples you guys know of

ACuriousMind

@0celo7 I think we indeed should solve that. Parallel transporting is ugly, I remember almost going crazy translating something along a triangle on a sphere

@Jimnosperm I...don't think there is something conserved globally but not locally, usually.

All the global conservations come from integrating the local conservations in a sense, I think

0celo7

4:07 PM

@ACuriousMind Ok, so I'll try to do this along the first leg of the square, which has $y=\text{const.}$

I meant side, not leg. Whatever.

Jimnosperm

@ACuriousMind What? I thought they came from global symmetries. Perhaps we're thinking the same thing though... bottom-up vs top-down

0celo7

@ACuriousMind Lol I got an exponential for $A^1$.

ACuriousMind

@Jimnosperm Heh. I think we're not disagreeing - the global symmetries induce conserved Noether currents (which is the "local conservation law"), whom we integrate to get the conserved Noether charge (which is the "global conservation law").

@0celo7 Yeah, looked like that to me, too :D

0celo7

@ACuriousMind I don't recall $\mathrm{SO}(2)$ giving random exponentials.

ACuriousMind

@0celo7 Perhaps you can sneak an $\mathrm{i}$ in there to give us sine and cosine? :)

0celo7

4:13 PM

lol

ACuriousMind

Honestly, after finding my messy (and probably wrong) sphere calculation, I'm not sure if I know how to parallel transport stuff :D

0celo7

Ugh, I don't think this will be particularly enlightening.

Ok, a more general question. Is the holonomy group always $\mathrm{SO}(n)$ or a subgroup thereof?

(Assuming an orientable, Riemannian manifold.)

ACuriousMind

For a Riemannian manifold with its Levi-Civita connection, yes, there are theorems that say that.

0celo7

Do you know the proof?

Does it have to do with diagonalizing the metric or something?

ACuriousMind

I think the mere statement "$\mathrm{SO}(n)$ or subgroup" follows from that the parallel transport map is an isometry of the tangent space, and the tangent spaces of a Riemannian manifolds are isometric to Euclidean space, so the group of isometries is isomorphic to $\mathrm{SO}(n)$.

And, yes, that's "diagonalizing the metric", basically, I think

0celo7

4:18 PM

@ACuriousMind Lol, physics.stackexchange.com/questions/173108/….

ACuriousMind

@0celo7 At the question or at my comment?^^

0celo7

Question.

Also "dit product"

OHHHHH

I know what he's asking.

ACuriousMind

Really? Are you a psychic?

0celo7

He wants to know why $\langle u,u\rangle=1$ implies $u^\mu\nabla_\nu u_\mu=0$.

;)

I'm unsure about the second part, however.

ACuriousMind

Ah. So the question is "Why is the derivative of a vector orthogonal to the original vector?", but in a covariant context?

0celo7

4:22 PM

Yeah

You have to feel the words, man.

ACuriousMind

lol

0celo7

He's just some poor chap lost in the world of GR.

ACuriousMind

That's not a bad question, in its essene then. Although it's math rather than physics, probably, and it's expressed horribly unclear.

0celo7

Now for the second part...

A geodesic satisfies $\nabla_{u}u=\alpha\gamma$.

Is that what he's asking?

"the result" is really vague.

ACuriousMind

It really shouldn't be a guessing game. Why do these askers think we're psychic and always know exactly what they're talking about?

0celo7

4:28 PM

@ACuriousMind Because in school they had teachers who had heard the same (possibly dumb) question a billion times and therefore seem psychic.

@ACuriousMind I guarantee I knew this at one point. The group of isometries in $\mathbb{R}^n$ is the rotations ($\mathrm{SO}(n)$) plus the translations (insert brain fart here).

Is there some way to write the Euclidean group as a product or sum of $\mathrm{SO}(n)$ and insert brain fart here?

Jimnosperm

@ACuriousMind Are you not?

Kyle Kanos

4:46 PM

@Jimnosperm If you were psychic, I do not believe you'd need to ask this question.

Jimnosperm

@KyleKanos I ask so that you aren't unnerved by my powers

Plus it's rude to read minds without permission

Kyle Kanos

-4

Q: How Could Remote Viewing (If real) Relate to Physics?

Remote viewing, however controversial in the mind of some has been said and in some cases supposedly proven to work. Is there anything at all in physics that could account for this strange phenomenon? Most would consider remote viewing to be in the realm of "Metaphysical" but that's just a word f...

experimental-physics

ACuriousMind

@0celo7 The translations are usually written $T^n$ or something (but are just $\mathbb{R}^n$ with addition as group operation), and the Euclidean group is the semi-direct product of the orthogonal group and the translations

@KyleKanos Why does that garbage have five upvotes?!

0celo7

@ACuriousMind Is it bad I don't know what a semi-direct product is?

ACuriousMind

Sometimes, I really wonder what people think physics is.

@0celo7 No :D

It's a group theory construct that you don't really need to understand, I think

Kyle Kanos

5:02 PM

@ACuriousMind I did not realize it has any upvotes

But maybe the guy's got some friends?

Gowtham

or people just like upvoting , regardless of crap or not :P

Gowtham

5:32 PM

can some highly insulating material be used as earth instead of the actual 'earth'

Gowtham

5:54 PM

scratch that question , i got my doubt cleared

0celo7

6:11 PM

@ACuriousMind For a good 15 minutes I thought I was going insane. I didn't realize the trafo property of the Weyl tensor depends upon the position of the first index.

0celo7

6:42 PM

@ACuriousMind Why don't we define the Laplacian on forms $(\mathrm{d}+\mathrm{d}^\dagger)^2$ with a minus out front so it gives $\partial^i\partial_i$ on functions instead of the negative?

ACuriousMind

@0celo7 The Wiki article tells me that it is so that the Laplacian on forms is a positive operator.

But it's a choice, after all

0celo7

Oh, makes sense

Phonon

hey :)

Physics Meta

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Q: Deleted question with positive score: when does my rep update

After a (for me) unusual spate of unpleasant back and forth with several users, I decided to delete an answer I wrote although it had a +11-1 vote count. However I am not seeing the negative impact on rep count that I was expecting. I just hit the rep cap for the day - which is why the timing of ...

discussion

ACuriousMind

@Phonon Heyhey - how's it going?

Phonon

6:58 PM

@ACuriousMind not bad mate, studies going well, just got home, relaxing with some beer till next day basically... how about yourself? any new endeavors for you lately? :D

ACuriousMind

@Phonon Nothing new, I'm still just having lazy days ;) (Two weeks of them left till university starts again)

Phonon

@ACuriousMind hehe :) at least you're resting, never a bad thing. Any more concrete ideas for your thesis?

ACuriousMind

@Phonon Not yet, but time's not pressing, after all. I'll see what catches my eye in the next months

Phonon

@ACuriousMind ah that's good

ACuriousMind

@PhysicsMeta Reading a bit, the people bugging Floris are the same as the ones who are taking offense at every utterance of high-rep users on meta lately. Suspicious behaviour.

0celo7

7:13 PM

@ACuriousMind Why are people bugging him?

ACuriousMind

@0celo7 I'm...not sure. It's not clear to me from their comments what the issue really is, but they seem to think he didn't deserve the upvotes his answers got at some questions compared to others.

0celo7

So why did he delete the answer in question? (I can't see it because rep.)

Just to stop the annoying messages from them?

Oh god...Nakahara uses $Ric_{\mu\nu}$ for the Ricci tensor...

ACuriousMind

Yep, mainly to appease them, it seems to me, but of course I cannot speak for him (and I don't think he wants to discuss this further).

@0celo7 Beautiful notation. But I think using $\mathrm{Ric}$ for the tensor is rather usual, isn't it?

0celo7

At least it's better than MTW's $\boldsymbol{Riemann}$ and $\boldsymbol{Faraday}$

@ACuriousMind $\mathrm{Ric}\ne Ric$

ACuriousMind

@0celo7 Ah. Okay, that's ugly

0celo7

7:19 PM

@ACuriousMind It's curious since he uses \mathrm for differentials and groups.

etc.

Also he uses $\mathcal{R}$ for the scalar.

@ACuriousMind Do you like when authors use the \var capital Greek letters?

I think it looks atrocious for a lot of them.

ACuriousMind

Well, I guess there are only so much $R$s out there to denote all the $R$ quantities^^

@0celo7 Yeah, the \var letters are terrible

And $\varsigma$ is actually the sigma you would write at the end of a word as opposed to $\sigma$ at its beginning or somewhere in the middle

0celo7

@ACuriousMind In index notation it is obvious which one is which. And in coordinate free I like $\mathrm{Riem}, \mathrm{Ric}$ and $R$.

Well $\vartheta$ is nice.

$\varrho$ looks like a tadpole.

ACuriousMind

Right, $\vartheta$ is pretty, but the "handwriting" version of the print-type $\theta$. I would not use it in print.

Just like $\varphi$ is the handwriting version of $\phi$

I just don't get why I should deviate from the print versions

But the biggest abomination is when $\varphi$ and $\phi$ actually denote different things

0celo7

@ACuriousMind I was talking more about the capital \var letters like $\varGamma$ and $\varPhi$ versus $\Gamma$ and $\Phi$.

ACuriousMind

Oh. The capitals just look...not right

0celo7

7:28 PM

@ACuriousMind Straumann, for instance, uses $\phi$ for the Newtonian limit of $\varphi$ in the simple cosmological model $\mathbb{R}\times\Sigma$.

I think slanty capital letters are a Springer approved font...several Springer texts use them, but no one else does.

ACuriousMind

@0celo7 As I said, abomination - when handwriting Greek, you always write $\varphi$, and when typesetting it, you always write $\phi$ for the minuscule phi, at least I was taught that.

0celo7

@ACuriousMind I'm not disagreeing with you, but most people don't follow that.

I happen to like the looks of $\varphi$.

@ACuriousMind Lol, every QFT text I have handy uses $\varphi$ except for Peskin and Weinberg.

ACuriousMind

@0celo7 Yeah, I had lecturers using $\varphi$ for the field and $\phi$ for its expectation value. Halfway through the lecture, they usually forgot that they were meant to distinguish between the two, and all phis looked the same :D

Phonon

haha :)

0celo7

Another pet peeve of mine is when authors don't use \text for textual labels. For instance, Straumann doesn't seem to know what \text is. $\pi(D_d\theta_{arc})^2\varSigma_{crit}$ looks horrible.

Phonon

7:39 PM

wth is this physics.stackexchange.com/review/suggested-edits/77898

0celo7

@Phonon "spam and vandalism" :)

Phonon

@0celo7 oh man some people don't have anything better to do with their time...

0celo7

@ACuriousMind Probably dumb question. Does every Lie algebra have structure constants?

ACuriousMind

@Phonon Did you see the "cheese edit" under "um what?" in the starred comments? ;)

@0celo7 Well, every Lie algebra has generators, and the structure constants are simply obtained from the commutators of the generators, aren't they?

(So, yes, I think every Lie algebra has structure constants)

0celo7

(One moment while I get a shot of the section that makes me ask this.)

wat

m.imgur.com/srbxg5j

(7.224)

That doesn't look like a structure constant to me...

@ACuriousMind Also, how do we get (7.223)? I've never really understood where this comes from.

ACuriousMind

7:52 PM

@0celo7 It doesn't because it is written in "matrix form". When you write the $\Sigma_{\mu\nu}$ as generators with a single index as is usual for the Lie algebra, $[\Sigma_i,\Sigma_j] = f_{ijk}\Sigma_k$ for some $f_{ijk}$ (simply because the $\Sigma$ are a basis of the algebra as a vector space).

Phonon

@ACuriousMind haha I just did, crazy...

0celo7

@ACuriousMind I see.

ACuriousMind

@0celo7 Which part of 7.223? The thing behing the arrow or behind the approximate equality sign?

0celo7

@ACuriousMind Arrow. I obviously know Taylor :)

ACuriousMind

@0celo7 Juuuust checking ;)

The thing behind the arrow is the definition of a spinor - a (bi)spinor/Dirac spinor is a thing that transforms like that under the Lorentz Lie algebra

0celo7

7:55 PM

That's thoroughly unsatisfactory.

Why the $1/2$?

ACuriousMind

@0celo7 Because the sum over $\alpha$ and $\beta$ counts the $\Sigma$s twice otherwise

0celo7

Ok. It might be wise to reread the relevant section(s) in Weinberg before I continue.

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