The h Bar

user54412

00:16

@nerdy when you suddenly change the left boundary condition, the adjacent bead tries to adjust, which causes the next bead to adjust, and so on. This is a classic example of a shock. It is propagating at the sound speed of the string.

user54412

Now when you slowly change the boundary condition, information about the first little change you do starts propagating at the sound speed right away -- the string doesn't wait for you to finish adjusting the left end before responding.

nerdy

Im still really in doubt about that parameter of the wave .. i cant find a formal name for it or precise definitions . Here is that parameter whenever i do a fast lifting i.sstatic.net/qmjSU.jpg and here its when i do a slow lifting i.sstatic.net/ELi16.jpg

user54412

Well, you could call the shape a wavepacket and talk about its width, though some people like to define wavepackets as returning to the baseline. Perhaps a better term would be pulse and the width you indicate is the pulse width.

user54412

In any event it is given by (sound speed)/(time over which slider is adjusted)

user54412

Though now I think about it some people would say a pulse also should return to baseline, and they would think a pulse width is the width of the top of it:

user54412

00:21

___/---\___

user54412

i guess "leading edge width" is the least ambiguous term -- the pulse consists of a leading edge, the pulse proper, and (if you brought the slider down eventually) a falling edge

dmckee

@nerdy The usual sinusoidal waves get the adjective "harmonic". Non-sinusoidal wave that repeat are "periodic".

user54412

@ChrisWhite correction: (sound speed) * (time over which slider is adjusted)

nerdy

04:48

Sorry being so newbie, but what does sound have to do with my wave propagating through the string ?

I was thinking that the wave-length would be the velocty of the wave times multiplied by the time over the lifting

David Z

05:31

@nerdy sound is a wave too. A longitudinal wave (where the air particles move backward and forward) instead of a transverse wave (where the string particles move sideways), but it generally works the same way.

nerdy

Yeah, but in my example i guess we were considering the variation of heigth of the string propagating as the wave whereas in sound we consider the variation in pressure propagated as the wave

David Z

Yeah, exactly

Still they both work as waves. Both have the same mathematical description.

David Z

06:58

oy... 31 comment flags popped up in the last hour or so

AmagicalFishy

07:23

Anyone here? I have a question that I've been trying to figure out for like two hours

and can't get anywhere—I think I'm just misunderstanding the definition of something

David Z

@AmagicalFishy what is it?

AmagicalFishy

I'm working on some GR homework, am told that $\kappa ^2 = \frac{1}{2}(\nabla^a K_b)(\nabla^b K_a) |_{rH}$

and am asked to show, for some given metric (spherically symmetric static black hole), that $\kappa = \frac{1}{2} \partial_r f(r) |_{rH}$

The thing that is confusing me (and honestly has been for the whole semester) is the statement: Where $K$ is the time-translation killing vector, $K = \frac{\partial}{\partial t}$

This means that $K = (\frac{\partial}{\partial t}, 0, 0, 0)$, right?

Which would mean that $K_x = K_y = K_z = 0$?

(I messed up some of the indexing above, but that doesn't really matter. :D)

David Z

hmm... well, yes, $K_x = K_y = K_z = 0$ is accurate

But I wouldn't write $K = (\partial_t, 0, 0, 0)$

(this is if I remember correctly, which I might not)

AmagicalFishy

Ah. So, how would I interpret something like $\nabla_t K_t$? I end up with a 2nd derivative there acting on... nothing

David Z

I think the partials are meant to be basis vectors of the tangent space. So the notation would be $K = (1,0,0,0)$

which is equivalently $K = 1\partial_t + 0\partial_x + 0\partial_y + 0\partial_z$

AmagicalFishy

07:30

Ah. So $\frac{\partial}{\partial t} K_t = 0$?

(That is the term in the covariant derivative before Christoffel thing)

David Z

That's where my memory kind of fails me

AmagicalFishy

No worries, you've actually helped a lot already. :D

David Z

In flat spacetime, yes, but when you bring in curvature I forget if that changes

@AmagicalFishy ah, well good :-)

AmagicalFishy

I have to remember that $\frac{\partial}{\partial t}$ is a basis vector

and not treated like a derivative

(sometimes)

David Z

Well, it is a derivative, but it's also a basis vector. It's part of the basis of the space of all possible directional derivatives.

i.e. directional derivatives themselves form a vector space.

AmagicalFishy

07:34

Yeah. I guess I meant moreso like, "$K$ is the killing vector $\partial_t$" doesn't mean "Compute the derivative with respect to $t$ of something." It means $K$ should be represented as $(1,0,0,0)$

David Z

@AmagicalFishy BTW that would be a really good question to ask on the main site.

AmagicalFishy

It's been a whole semester and I'm still confused by this stuff sometimes, haha.

Good idea. :v

David Z

@AmagicalFishy yeah, that sounds accurate. I mean, you do eventually wind up calculating the derivative of something, but the point of treating the derivatives themselves as tangent vectors is that it doesn't matter what the derivative acts on in the end

AmagicalFishy

Huh. The Physics SE is pretty harsh when it comes to questions

David Z

Harsh on low-effort questions, yes. We expect a lot of our askers, and it shows because so many people come here asking for homework help, not knowing that we're really not a homework help site.

3

But in your chat messages here, you've done everything we expect from a good homework question: you gave the original problem you were working on, you reduced it to the underlying physical concept that you were confused about and asked about that specifically (as opposed to "how do I solve this")

and you showed that you'd made an effort to work through it yourself

AmagicalFishy

07:53

Ah, excellent.

Thanks for your help. :D :D I posted the question here:physics.stackexchange.com/questions/150579/…

David Z

Excellent :-) I would suggest putting $\kappa ^2 = \frac{1}{2}(\nabla^a K_b)(\nabla^b K_a) |_{rH}$ in display style (use $$ before and after) and the $\displaystyle\frac{\partial}{\partial t}$ in text style, just for readability.

AmagicalFishy

Done!

David Z

Oh, and same goes for the $\displaystyle\frac{\partial^2}{\partial t^2}$ in the last line (i.e. put it in text mode). Not a big thing, but I think it helps maintain the flow of reading.

(I mean, it's up to you of course)

At least in my opinion, there's rarely if ever any good reason to use \displaystyle in MathJax

AmagicalFishy

I think I just have a habit of formatting like I write

... which is unecessarily large

David Z

08:09

Yeah, that works well when writing, not so much when typing. LaTeX and MathJax are already designed to produce pretty good typesetting results when you use their default settings.

Of course it's not like we have a rule against it, so you can format the way you write if you want.

AmagicalFishy

Yeah. I'm too pragmatic to know there's what people consider a more readable format and not follow it. :P

David Z

not a bad thing, as far as I'm concerned

user54412

08:57

Ok -- philosophy time.

user54412

I've seen it said over and over again that the singularities predicted by GR are a sign the theory is breaking down. Fine.

user54412

But in the same breath everyone who mentions this will go on to say "and one day we'll have a quantum theory of gravity that will make the singularity go away" or some such thing.

user54412

Does everyone really believe there is no self-consistent theory of general relativity in a universe without quantum mechanics?

user54412

That's a pretty harsh indictment of a theory, after all.

user54412

To be clear, I'm not talking about how much of real-world physics can be done with or without various theories. I'm just asking how internally self-consistent we believe various proper subsets of physics to be, as mathematical models that may or may not correspond well to reality.

user54412

09:05

Everyone is so certain that quantum shenanigans can solve "problems" in GR, and to me it seems this certainty implies they must believe there is some deep feature of GR that demands it incorporate quantum effects for self-consistency. That claim seems remarkable to say the least.

NikolajK

12:23

@ChrisWhite: What does self-consistent mean here, or rather what does the opposite mean?

The mathematical definition would be that there is a statement P for which you can derive 'P and not P'. I don't quite understand the "the theory breaks down" thing in GR, I've never seen that.

ACuriousMind

13:04

@ChrisWhite I think singularities are really not that "problematic". Singularities are like the tip of a cone - the cone is only a manifold if you exclude the tip, which is a "singularity".

The search for a "quantum theory of gravity" is, in my view, simply necessary because we want to have a theory that includes all known interactions between matter

Yet, it is known that simply turning GR into a QFT by the usual quantization process can only be an effective theory, since it is non-renormalizable, but we want a theory that does not break down at high energy scales, i.e. that has no degress of freedom integrated out and thus hidden from our view

Danu

19:35

This is interesting... Apparently IncnisMrsi thinks I'm 'power-hungry'... lol. I've suddently got a much better idea who is the anonymous downvoter on some of my recent answers... ;)

ACuriousMind

19:49

@Danu Though they seem to hold some grudge, the rant on their profile page seems to indicate that the downvotes are not theirs. At least, if they were, that would be a textbook example of hypocrisy.

Danu

@ACuriousMind What a rant.

ACuriousMind

Vengeful enthusiast has a nice ring to it, though :D

Kyle Kanos

20:10

the ignorant outnumber the knowledgeable and enjoy the same rights. I should hope so!

I'd rather not (a) everyone be the knowledgeable and (b) have only the knowledgeable have rights!

Is it wrong for me to want to eliminate his Ī for I?

ACuriousMind

Not particularly wrong, no

Mostafa

20:58

Hi everyone,
I am looking for a quote that basically says the same thing as the following by J.J Sakurai:

we see a number of sophisticated, yet uneducated, theoreticians who are conversant in the LSZ formalism of the Heisenberg field operators, but do not know why an excited atom radiates, or are ignorant of the quantum theoretic derivation of Rayleigh's law that accounts for the blueness of the sky.

Any suggestions?

Danu

21:36

@Mostafa I have the sad feeling that I belong to that category...

I'm excited: I got invited to be a pro-temp. moderator on History of Science and Math! I guess Inci's worst fears are becoming truth!

sets evil plans in motion

3

ACuriousMind

@Danu Congrats!

You just didn't want to wait to reach 10k to see all the deleted stuff on HSM, did you? ;)

Kyle Kanos

You don't need 10k on beta sites, only 3k

Danu

@KyleKanos Only got 1k though lol

...still I'm #8 as far as rep goes, I think

So, how are you guys

I'm desperately trying to avoid studying for MQM, there's an exam tomorrow at 10 AM

damn saturday exams

ACuriousMind

21:56

@Danu Exams?! At this time of the year in Germany? That's unusual...

Danu

@ACuriousMind Midterm... only this course has one

Herr Doktor Siedentop is relentless :P

Kyle Kanos

I'm trying to figure out how to not get the following ripples in my simulation:

The red lines indicate boundaries between processors

Danu

they look kinda... dynamic

Kyle Kanos

At the vertical line, something unphysical is being passed

ACuriousMind

Ah, always simulations with you :P The only productive thing I did today was cooking a nice steak :D

Kyle Kanos

22:01

And it's annoying me because it shouldn't be there

Danu

I'm afraid my knowledge of computers is entirely insufficient to help almost anyone out with anything programming-based

Kyle Kanos

I mean no offense, but what's the point of accepting an answer you don't really understand:

2

A: Coulomb's Law in the presence of a strong gravitational field

Of course Coulomb's law has to be adapted! And it is therefore fortunate that there exist manifestly covariant formulations of electromagnetism that do not care how spacetime is curved. However, we should first briefly observe that Coulomb's law is not one of the fundamental laws of electromagnet...

Man, I hate overloaded notation. The only use I know for upside-down V is as a logical "and", and super scripts are exponents. But I'm sure that's not what they mean here. — Shufflepants 6 mins ago

ACuriousMind

I was also puzzled by that

Danu

...I'm gonna ask

Kyle Kanos

Anyways, I'm going to head out now....have a good weekend everyone

ACuriousMind

22:05

Probably the non-technical bits saying that Coulomb's law indeed is modified in curved space was enough for them

@KyleKanos Thanks, you too

Danu

This question brought up, again, an old question of mine: Is there a symmetry for entropy conservation?

It seems quite related to the question I linked, but not exactly the same

ACuriousMind

@Danu Is entropy an observable?

That is, can you write down an operator that could be the Noether charge of said symmetry?

Lubos says it isn't

That would mean there is no symmetry of the Lagrangian associated to it

Danu

@ACuriousMind I'm not sure if that's a settled question

But I guess I better take Lubos' word for it ;)

Of course, the observer dependency will mess things up, but maybe there is still something useful that can be defined

ACuriousMind

I think the "problem" is that entropy is really not a property of a state

So there can't be an operator that gives you entropy

Danu

22:24

makes perfect sense

thanks for killing my dreams ;)

ACuriousMind

@Danu Heh. My pleasure

Iplodman

23:02

Heya all! :)

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