The h Bar

Physics Guy

00:02

knzhou

hey!

Physics Guy

Hey !

knzhou

The atmosphere in this room has really changed in the past few months...

It's gone from mostly technical physics discussion with ACM, to hating on ACM and talking about 'math-free' physical theories.

4

Physics Guy

Yeah. That could be. i dunno it because I'm just here since June.

ACuriousMind

At least I'm not imagining that, what a relief :P

knzhou

00:05

Like, if you make a connectivity graph of who talks with whom, there are clearly two completely separate components.

Except with attacks from one component on ACM in the other. :P

Physics Guy

Tell me about those good ol' days

knzhou

Well, I'd come in, read about the latest problem 0celo7 is working on, and leave. Much more pleasant!

Physics Guy

@knzhou I wont say who is the generator of this theme-changing here in h bar. We all know the answer.

But Ocelo7 still talks about his (mathematical) ideas and problems. Leastways he does.

vzn

00:22

@Mast thx for your attn! & what was your vote? & maybe you prefer the 2 paragraph 69v top answer with no citations. so go vote for that one instead. (its definitely much easier/ quicker to understand! if one can say that its actually an understandable explanation...)

> "What is a photon?" we need to be honest and say that we don't really know.

Mast, Netherlands

107 6

Sir Cumference

Does anyone think it'd be possible to become a doctor while also learning GR and QFT?

vzn

← now taking physics advice from some with less than ½ my own rep :P

ACuriousMind

00:38

@SirCumference If you like neither sleep nor a social life, possibly :P

bolbteppa

@SirCumference Yeah, just go through these lectures theoreticalminimum.com/courses and/also these youtube.com/channel/UCLjdjfV5J-xsb2KP4AO-bZw/playlists as you study and forget about society ;)

Sir Cumference

Sigh... ;-;

bolbteppa

@0celo7 did you get the sine derivatives

vzn

@ACuriousMind lol you say that like its a bad thing :P

Sir Cumference

I know being an astrophysicist makes no money, but I also really want to learn everything

DanielSank

00:39

'sup y'all?

bolbteppa

Those theoreticalminimum videos are for someone like you dude, no reason not to try them out, has books on the first two courses out already and a third coming soon

Sir Cumference

Still tho...

I also don't really wanna be crushed by the stress

There really is no right solution...

bolbteppa

Getting past med school ultimately requires memorizing 3 books, the First aid books, not expected to think don't worry, save that for physics ;)

Sir Cumference

I guess I could keep majoring in physics and apply to med school, right?

Would that give me enough time to become fluent in GR and QFT?

bolbteppa

Are you in your undergrad?

Sir Cumference

00:46

Yep

ACuriousMind

@SirCumference Walk before you can run. Are you perfectly comfortable with SR and standard QM?

Sir Cumference

@ACuriousMind As of now? Not at all.

bolbteppa

I would say do your med school stuff as best you can and don't let physics distract you, use those theoreticalminimum videos if you haven't done the groundwork courses

If you actually sat through those theoreticalminimum videos zero, one or two a day for the next six months and did them all, or most of them, you'd get enough of a feel to know whether you want to re-do every one of those mini-courses in more detail and go through all the problems, all the technicalities etc... or not, those videos are made for people who can't put in the time of wading through the pedantic details but want to get the real jist of what's going on

They wont give you the best way to think about these things, but they will give you the in-depth jist to have a real feel for what's going on, which is most likely what you're looking for

Learning QM well enough to learn statistical physics well enough to learn chemistry properly to learn biochemistry properly would be well worth the effort though

0celo7

@SirCumference fluent in GR doesn't make sense

It's a huge field

@bolbteppa nope

Too advanced

bolbteppa

What's wrong with www-math.mit.edu/~djk/18_01/chapter05/proof02.html

If you can't remember $\sin(x + h) = \sin(x) \cos(h) + \sin(h) \cos(x)$ then use Euler's formula to prove it, there is a proof and visual explanation of the meaning of it here betterexplained.com/articles/…

If those two limits are a problem, proofs here en.wikipedia.org/wiki/…

If you want to cheat, define sin and cos as solutions of a 2nd order ODE, solve it using power series to define sin and cos using power series, prove the above identity using Euler's formula (or your power series expansions), then you can treat the limit proof as an exercise, but proving those two tough limits then becomes easy because you can use the series expansion to verify them

It's not a cheat, but it's not as nice := cheat

0celo7

01:11

@bolbteppa Euler's formula is proved using Taylor series

no cheating

Sir Cumference

@0celo7 True dat

@0celo7 All right, knowledgeable enough in GR that I can hold lengthy, in depth conversations about it with people here

Or regularly answer the questions about GR on the site

0celo7

@bolbteppa tbh I want to see trig explained from ZFC

bolbteppa

I did too, the best thing I've come up with so far is to use the integral definition of arcsin as your starting point

I've found 2 books doing it this way

0celo7

Hmm, $\int\sqrt{1-x^2}\mathrm dx$?

I don't know calculus

bolbteppa

If you wanted to define trigonometric functions like $\sin(\theta)$ you need to know what $\theta$ is first, how it relates to a circle, what $\theta = \pi$ is, and how this number really relates to a circle, and you need to do it in a way that links to abstract mathematics not 2-D Euclidean geometry where you're using pictures and not really sure how to justify things right?

You know that $\ln(x) = \int_1^x \frac{1}{t}dt$, i.e. the stupid Logarithm is defined using integrals in calculus books and you can take that as your starting point with this function

Why not do the same thing with trig functions

0celo7

01:24

so what's the integral

I'm not sure what's wrong with defining $\sin t$ as the unique solution of $x''=-x, x(0)=0,x'(0)=1$.

You get the series by using standard ODE methods

In fact

$x''=-x$ is Sturm-Liouville so you can use all of those things

you get the oscillatory behavior that way.

hmm, no, it's not SL

bolbteppa

The way I remember to derive the integral is to use $\theta = \arcsin(x)$ so that $\sin(\theta) = x$ and you have $\cos(\theta) \frac{d \theta}{d x} = 1$ so that $d \theta = \frac{1}{\cos(\theta)}dx = \frac{1}{\sqrt{1-\sin^2(\theta)}}dx = \frac{1}{\sqrt{1-x^2}}dx$ so that $\theta = \arcsin(x) = \int_0^x \frac{1}{\sqrt{1-t^2}}dt$, you can get it just by drawing the circle and using the arc length formula too I think

0celo7

I think getting periodicity from the power series is hard

bolbteppa

Another way to get this starting definition is to use $s = r \theta$ (can't see a way to avoid this) and the arc length method $ds = \sqrt{dx^2 + dy^2} = \sqrt{1 + (\frac{dy}{dx})^2}dx$ so that for a circle of radius $1$, $s = \theta, ds = d \theta$ and $x^2 + y^2 =1 \rightarrow y = \sqrt{1 - x^2}$ along with $y' = \frac{- x}{\sqrt{1-x^2}}$ implies $ds = \sqrt{1 + \frac{x^2}{(1 - x^2)}} = \sqrt{\frac{1-x^2 + x^2}{1 - x^2}} = \frac{1}{\sqrt{1-x^2}} = d\theta$ so that $\theta = \int d \theta$

There's nothing wrong with defining $\sin$ and $\cos$ that way using differential equations

There is even an amazing visual way to link it to circles using complex differential equations in Tristan's book

0celo7

01:41

how do you get periodicity?

bolbteppa

If you define it using ODE's then the domain of definition is $[0,1]$ and you have no idea what's beyond that, then you simply define the "periodic extension" of $\sin$ to satisfy periodicity by definition

0celo7

why $[0,1]$

bolbteppa

Lets see, you can define most functions a) analytically using integrals, b) iteratively/approximation-eley using series expansions, c) geometrically using graphs/pictures, d) constrainedly using differential equations, e) producteley using infinite products, anything else?

0celo7

c) is bullshit

a) and b) are probably the same?

bolbteppa

Sorry $[-1,1]$

0celo7

01:48

why $1$ tho

bolbteppa

b) is a Taylor/Laurent/Fourier/AHH expansion of a)

Trigonometric functions are also called "circular functions", they are defined with respect to the unit circle $x^2 + y^2 = 1$

The domain of definition of $x$ is $[-1,1]$ via $y = \pm \sqrt{1-x^2}$

0celo7

what

bolbteppa

(Hyperbolic functions are "complex circular functions")

0celo7

I meant using the ODE definition

bolbteppa

Why do you use $[-1,1]$? Because this gives the right answer

If you use $[a,b]$ then you'll get $\sin(\omega x)$ right?

For some $\omega$

(where you define the boundary conditions in the middle of the interval)

0celo7

01:54

dunno

bolbteppa

Alright, well if you visualize the complex number $z = x + iy$ as an arrow, i.e. a vector $z = (x,y)$ in the plane, then you know $z'(t) = \frac{d}{dx}(x(t),y(t))$ is a tangent vector, orthogonal to $z$ right? Then $z''$ is orthogonal to $z'$, i.e. $z''$ is parallel to $z$, that is $z''$ is a multiple of $z$. In general you'll have $z'' = \lambda z$, and you want to find the arrow $z$ such that it's second derivative is a multiple of what it was originally, $z'' = \lambda z$.

What about the special case where $z'' = - z$?

Can you visualize that

What would the motion look like in this case?

What figure would it generate (if there was any motion at all, what is the only way the arrow can move?)?

0celo7

02:34

@dmckee Do you know Sakurai's QM book?

dmckee

@0celo7 No. Keep meaning to look at it. In my copious spare time.

0celo7

@dmckee You're the first prof I've met who has copious spare time

I'm taking a course on QM based out of it

I'm currently trying to decipher physicist mumbo-jumbo regarding the math

also it has really wide margins and sticks out in my bookshelf >:(

what is it with physics books labeling every equation

bolbteppa

02:51

@0celo7 can you prove $\lim_{\theta \rightarrow 0} \frac{\sin(\theta)}{\theta} = 1$?

dmckee

@0celo7 In Jackson's case he's anticipating referring back to them ten chapters later.

bolbteppa

(Obligatory: youtube.com/watch?v=mm-4PltMB2A )

DanielSank

@0celo7 Why?

bolbteppa

@0celo7 page 2 web.auburn.edu/holmerr/1617/Textbook/derivsincos-screen.pdf has the nicest pic of $\sin(\theta) < \theta < \tan(\theta)$ I can find, the proof is immediate from this, that's the hardest part of the proof done

I mean you can even see why the limit is 1 looking at this pic

0celo7

@DanielSank why what?

On mobile

@bolbteppa No, not without using sin(x)<x, which I have no reason to believe.

bolbteppa

03:06

Can you not see the picture in that link?

0celo7

Picture, smicture

bolbteppa

Using $s = r \theta$ with $r = 1$ you have $s = \theta$ and so in the picture you immediately see $\sin(\theta) < \theta < \tan(\theta)$

0celo7

I'm talking analytically

bolbteppa

What does analytically mean?

0celo7

Using the ODE/power series def

No picture

I'm going to sleep, cheerio

bolbteppa

03:09

Getting a bit circular now

You have some random power series which you randomly labelled sin and cos, the derivative of one gives the other instantly, so you're done. Why you called those power series sin and cos, and how those random power series link to circles, has yet to be explained by you.

If you want to use those in the limit proof above, you have to make sure those crazy power series satisfy the trigonometric addition theorem too, then those limits should be obvious immediately, again to be explained by you, don't know what the issue is anymore

I recommend putting down the QM or GR books and picking up the calculus books, no reason why you can't zoom though them if Wald is so easy

Sir Cumference

05:17

You know what I love about the Universe, if it's infinite? There are a finite number of quantum states that matter in a given volume can have.

So if you travelled far enough, you'd start seeing repetitions of our observable universe.

Meaning that there would be observable universes slightly different from ours. We basically have a multiverse inside a universe.

Secret

05:33

That's a multiverse level 1, and what you said is related to poncaire recurrence

peterh

05:45

@ACuriousMind If you found my opinion false, you could correct it, you didn't do. I want to understand you, I explained it not once but many times. You choose to go into silence, and I not "I am not listening".

@ACuriousMind You are over-sensitive, I never had any intent to accuse you, or have any conflict with you! I want(ed) only to understand your vision and... well... voting patterns. If I formulated incorrectly, it is my mistake, and I am sorry for that. But I am sure, the statement that "I am not listening", is simply exactly the opposite of the reality.

peterh

06:03

@knzhou On this ACM thread, please read my previous 2 chat messages. I still honor ACM, the physicist. I simply can't understand why do you consider "hate" that I am just curious to somebodys views. Now I think the best if also I prefer a read-only mode here, although I am a little bit disappointed in the treatment I've got.

John Rennie

06:20

@SirCumference There's a big difference between learning GR and/or QFT just enough to take pleasure in it, and learning it well enough to do research in it.

GR in particular is a beautiful theory and it's possible to learn enough to appreciate that beauty without having to dedicate your life to it. That's pretty much what I've done.

With just basic calculus you can understand what the metric tensor means and how to use it to understand all the weird stuff that Discovery Channel programmes make sound like magic.

I think you could reasonably hope to do this and still pursue some other career.

Sir Cumference

@JohnRennie Ah, thanks

John Rennie

At the risk of putting you off, it took me several years to get to where I am now with GR, and I still know next to nothing about QFT.

But it's been a lot of fun for those years, and after all the point is to enjoy physics not make a religion out of it :-)

John Duffield

07:34

@bolbteppa : money is merely a tokenised agreement of value that doesn't really exist but we think it's the most important thing in the world.

@bolbteppa : I'm not averse to learning the maths. The point I make is that the maths isn't much help if you don't understand the terms. Like E and m and t and c and C.

Slereah

@JohnRennie Nobody knows anything about QFT

We only pretend that we do

John Duffield

@SirCumference : we have no evidence whatsoever to support that assertion.

Slereah

I know enough about QFT to pretend I know about QFT

John Rennie

QFT seems to bring a whole new level of meaning to the term unintuitive.

Relativity is notoriously uninuitive, but actually it's quite easy to get an intuitive grasp of it without getting bogged down in the maths. But QFT seems a whole different game. I wonder if there is any way to get an intuitive grasp of it, even when you understand the maths perfectly.

Slereah

07:49

Well lattice field theory is not too hard to grasp

you have values at points at a time, and you have to compute the transition to values at another time

John Rennie

08:01

Take for example that what is a photon question. The answer is straightforward: it's the quantum of the vector four potential field when you quantise it. But try to link that concept with e.g. the thing that ejects a photoelectron and no-one seems able to do it in simple terms.

Slereah

08:13

Well in general in QFT there is a bit of a confusion between the operators and the states

Some books are pretty bad at disambiguating what they're talking about

That's why I still write my operators with hats

Overall I'd say a photon is a a wavefunctional packet of the EM field

It butts head a little with the rigged Hilbert space definition but those aren't extremely useful intuitively for a particle

Well, a packet peaked around 1 for the number operator, too

That way it is close to the usual definition

David Z

08:28

@peterh no more discussion of this issue in this chat room. (And @ACuriousMind too, not that I really think I need to tell you)

Slereah

I think the big problem of defining a photon is that QFT doesn't really have a native notion of "particle", for a start

although you can approach it

peterh

08:51

@DavidZ O.k.

Noah P

09:08

Hi all

Qmechanic

-1

Q: What if Earth spinning decelerate to 168 hour (one week)?

What if Earths spinning suddenly decelerate to 168hr per rotation? and how it will effect to the things (live non-live) on surface? does it make end of the world? or just some small incidences here and there? please explain me with classical mechanics equations.

Too broad? Off topic?

Noah P

I'm writing an essay, and seem to keep repeating the word manouvere when describing the Oberth effect and similar, any alternatives?

@Qmechanic It seemed too broad when I read it

David Z

@Qmechanic Both, I think.

Noah P

09:33

Anyone here able to help me with a quick question on black hole formation? Not complicated

John Rennie

I can try :-)

Noah P

Great! I'm just finishing off my essay, and I'm doubting what I've written: As a star ages it's supply of hydrogen begins to run out. The star will burn the remaining elements such as
helium into heavier ones, including oxygen, yet these reactions do not release as much energy. It is at this
point that the thermal pressure generated by a star is no longer enough to balance the gravitational forces
and prevent it from contracting. If the mass of the collapsing star is not less than approximately 2 solar masses, the pressure from the thermal reactions will be insufficient to prevent it from …

I know this isn't always the case, but is it a valid explanation?

John Rennie

09:46

This isn't an area I know well, so the usual disclaimers apply ...

Noah P

Any help would be appreciated!

John Rennie

It's true that fusion of heavier nuclei releases less energy per fusion event, but the fusion of heavier nuclei is much faster than hydrogen fusion. That means when the star switches to heavy nuclei fusion it releases more energy not less.

Noah P

Ok

John Rennie

For example even low mass stars are expected to produce an outburst of energy when helium starts. This is the helium flash.

It's when the heavy nuclei fusion ceases that all hell breaks out!

Noah P

Right okay

John Rennie

09:49

For a modest star like ours it settles down to a white dwarf. But for a heavy star fusion ends abruptly and the outer lays fall back into the star and kick off a supernova.

Alex

Hi all

Noah P

I'm following so far

John Rennie

Any star heavy enough to produce a black hole probably had a dramatic end to it's career in fusion.

Noah P

What do you mean by that?

John Rennie

I don't know if all black holes involve a supernova, but you would expect something fairly dramatic. The star doesn't just settle down gently into being a black hole.

Noah P

09:51

No, I can't imagine it would

You say any star heavy enough - could you quantify that?

John Rennie

The point is that while there is a neutron star mass limit, the Tolman–Oppenheimer–Volkoff limit, that is like the Chandresakhar limit you shouldn't think of a star gradually being compressed by its own gravity to form first a neutron star then a black hole.

Noah P

ok

John Rennie

The black hole formation will generally be exceedingly violent.

I don't know the details, but I think we are talking about a star much heavier than the Sun e.g. ten times heavier. During the collapse most of the star will be blown away to leave a much smaller residual mass as the black hole.

So a two solar mass star will not form a black hole because much less than two solar masses will be left after the collapse is complete. It will form a white dwarf or neutron star.

Noah P

Ohh okay

John Rennie

But I'm saying this from memory so some Googling would be in order to make sure what I'm telling you is true.

Noah P

09:57

Okay, thanks

John Rennie

Hopefully what I've said gives you a good starting point to Google from.

Also, black hole formation is a somewhat ticklish subject.

For an outside observer black holes take an infinite time to form.

That is, the formation of an event horizon takes an infinite time.

Noah P

It sounds like im in for some fun then on google

John Rennie

So you need to be cautious about statements like collapsing to a single point of infi nite density, forming a singularity

Noah P

ok

John Rennie

The simplest model for black hole formation is the Oppenheimer-Snyder metric. This assumes the collapsing material is a uniform pressureless dust, which obviously isn't really the case, but we expect it to get the general features right.

Noah P

10:01

Ill have a look at that then

John Rennie

Again some Googling might find you reasonably accessible articles on this, and it's worth searching this site as I'm sure it's been mentioned several times.

Noah P

Brilliant, thanks for your help

John Rennie

But then you might decide this is all a bit much for a few lines in an essay :-)

Noah P

I'll see :)

John Rennie

Anyhow, you're welcome to come back and ask if you want any more info.

just recently the chat has been deperately short of conversations about physics :-)

Noah P

10:05

Thanks! I might do when I've made some progress with it

Alex

10:26

Hi @JohnRennie would you be able to assist with a quick electromagnetism question.

John Rennie

@Alex you can ask, but electrodynamics was the subject I liked least at university so I've forgotten most of the little I ever knew about it.

Alex

@JohnRennie Okay thanks. If you have a positive charge above a grounded conducting plane then the charge above the plane induces a negative charge in the region closest to the plane. In this region then we have $\rho \neq 0$, but $V = 0$ since it is grounded hence $\nabla^2 V = 0$. Can I conclude then that Poisson's equation ($\nabla^2 V = -\frac{\rho}{\epsilon_0})$ does not hold in this region?

John Rennie

@Alex Have a look at section 3.2 of this article. from my quick reading of it the articles discusses exactly the situation you describe.

Alex

@JohnRennie That's funny, I just opened that link a few minutes ago. Thanks.

Secret

@Slereah I think the gist of QFT might be not about the notion of particles, but the whatever concept that derives classical fields and particles. That might be the fundamental entity we need to understand. Just as in quantum mechanics, to really understand, we have to forget classical notions and intuition, and learn how states actually superimpose and interact in the Hilbert space

John Rennie

10:41

Whoever write the article knows a lot more about electrostatics than I do :-)

Secret

I think an axiomatic approach to QFT will flesh out all the weirdness it has to offer, and once understanding them in every level of the mathematical workings on how it lead back to the physics, we can understand QFT

The good thing about the abstract is it often lay bare the nature of something, without any interpretation or contamination by a framework we are familiar with

If you want to know what a crazy person is thinking, you need to understand why and how they think

That's the gist of understanding a completely alien worldview

You need to learn how to think like them, without your identity being consumed

Bernard Meurer

11:10

@JohnRennie Halp

Do you know about audio equipment at all?

Slereah

There is totally a limit for black hole formation

Like a hard limit

Oh wait, I think that only applies to hydrostatic equilibrium, though?

Or something

Something like 9/8th of the Schwarzschild radius isn't stable

ACuriousMind

11:23

@Slereah Well, I tend to think about the "sharp" momentum states as standing for packets narrowly peaked around that momentum, anyway. I would just be more annoying to calculate everything with packets.

Secret

Acuriousmind, I have a weird question for you. For the examples where the questions I ask you that you do understand what and why I am looking for in the question, how do I usually phrase them, do I have some kind of pattern in the way I speak for those questions that makes it more easy to comprehend?

Slereah

@ACuriousMind Well I dunno about any too sharp state being a photon, either

I'm not sure I can call something smeared over the solar system a "particle"

Bernard Meurer

@ACuriousMind Hai

ACuriousMind

@Secret My problem in understanding you is not a matter of phrasing, I think.

@Slereah Well, but localization is notoriously not a good concept in QFT, is it?

Like, you may use the Newton-Wigner operators, but the notion of being localized is not observer independent, iirc

@BernardMeurer heyhey

Slereah

@ACuriousMind Well you know

It's a semantic issue

That's why I tend to go for like

Gaussian states

That's an objective notion

Minimal uncertainty and closest to "classical" particles

Bernard Meurer

11:33

@ACuriousMind How have you been dood?

ACuriousMind

@Slereah maybe you're right

Slereah

But in the end it's all for naught because there isn't much point in defining what a particle is

ACuriousMind

@BernardMeurer fine, enjoying the summer while it lasts

Slereah

Well experimentally, maybe

I do wonder what states are produced in those famous "one photon" experiments

Bernard Meurer

@ACuriousMind Oh dammit

I forgot I'll be over by the time I get there

I need to go to Wiesbaden soon

Secret

11:36

I am not sure what that is, because for 95% of the cases where my questions turns out to be invalid because of a wrong premise, you often can see it and point it straight out (take the group theory questions as example, you can often correct the terminology I misuse, or provide the correct and rigorous way to phrase something that is close to what I thought about the question or issue), so I guess the *correctness* of the topic is also not the major reason of the confusion

Perhaps one way that will help finding out, is maybe you can tell me how does my confusing type question look to you i…

ACuriousMind

@Slereah I think those are better described by the hybrid QFT/QM description of quantum optics than the full-blown QFT, but I don't actually know anything about it

@BernardMeurer why's that?

Bernard Meurer

@ACuriousMind Women

ACuriousMind

To get away from or to get close to one? :P

Bernard Meurer

Both

:D

Does everyone still hate you Jolly Little German?

ACuriousMind

@BernardMeurer The best reason :D

Bernard Meurer

11:45

Lol, I guess that's accurate :p

In any case

If/when I do go I need to also go to Mannheim

So I'll pay you the beer I owe you

b/c it's close to Heidelberg

ACuriousMind

It is!

Bernard Meurer

Then I have to go to Bohm too

Sigh I shoudl've just moved to Germany lol

ACuriousMind

I think Bohm is dead :P

Bernard Meurer

Tsc

Bonn

:D

ACuriousMind

Aha, thought so. Why do you need to visit so many places here?

So many girls? ;)

Bernard Meurer

11:48

Lol I have a book in front of me by Bohm, that's why

I wish! Ha!

I need to go to Wiesbaden for a girl, then Mannheim for an old friend, Heidelberg to see you, Bonn for another friend that recently got back from Kenya, Bremen to visit a school

ACuriousMind

@BernardMeurer Hmmm, a school? Weren't you going to Lisbon last I knew?

Or are you already looking to switch?

Bernard Meurer

@ACuriousMind I am going to Lisbon, but I'm already doing research as to alternatives for a switch if I don't like it

I want to have a plan B ready

ACuriousMind

I see

@JohnRennie What's going on with this answer? Starved for rep? :P

Sir Cumference

12:21

@JohnDuffield Which assertion? That the Universe may be infinite? Or that we might have a type 1 multiverse?

Danu

The phrase "size of a particle" does not have any meaning in quantum field theory. Please give references that use these terms. — ACuriousMind Jun 25 at 18:43

@ACuriousMind Isn't it pretty normal to take "size" to mean "the scattering experiments give results that look like what would happen with a classical particle of size X"?

ACuriousMind

@Danu In some circles, it might. In others, it means that the energy scale at which the substructure of the particle can be resolved translates into a length scale then seen as "size". In still others, it has something to do with the magnetic moments or something like that

That is, there are notions of "size" but there is no universal one, which leads to a lot of confusion

Danu

Right. Seems like overkill to say "does not have any meaning". I assume that, especially in experimental circles, this is simply not true.

@ACuriousMind That's a more nuanced and better response :)

Also, I don't wanna pile on, but the chat room has gotten completely #rekt over the past few weeks.

I'm seriously considering setting up an alternative, with moderated access. Do you have any thoughts on this, @ACuriousMind?

John Rennie

@ACuriousMind that's a little harsh :-)

Danu

@JohnRennie Lol, that is a weird answer though. Also seems like a homework question to me.

John Rennie

12:34

@BernardMeurer I used to be heavily in to audio engineering, though that was several decades ago and I'm out of touch with current hi-fi. What did you want to know?

ACuriousMind

@JohnRennie I was just wondering why you posted the answer without any explanation or guidance whatsoever

Since you usually make a point of trying to explain things rather than just giving the answer

@Danu Yeah, I'll concede that "not have any meaning" might've been too strong

Danu

@ACuriousMind Time for some "obsolete" flags :D

ACuriousMind

BTW, you don't need to convince me about that question, if you look you'll see I initiated the reopening process ;)

Danu

@ACuriousMind Oh, I wasn't worried about that. I just encountered it in the queue.

John Rennie

I think I'm a little grumpy at the moment, mainly because I've managed to hurt my back, so I'm a bit short on patience. The result is that when someone asks what does this look like as a matrix I take them literally.

I probably ought to go back to that answer and make it a bit less terse.

ACuriousMind

12:43

@Danu I'm not sure. I actually like that the free access means that we get just people stumbling in from time to time, and I don't think we should "give up" on the "general chat for Physics Stack Exchange" by just leaving it to the people making it uncomfortable.

John Rennie

Actually, of todays answers I most expected to be flamed for:

3

A: Why can only gravitons or gravity pass through to other membranes?

The idea is that gravitons are represented by excitations of closed strings, while the other particles are represented by excitations of open strings. An open string has its ends pinned to the brane so it cannot migrate between different branes, however a closed string has no ends and is therefor...

On the grounds that it's so vague as to be meaningless.

ACuriousMind

wait, I thought I left a comment on that!

Danu

@ACuriousMind I'm afraid it's spiraling, though.

John Rennie

@ACuriousMind I should have kept my mouth shut :-)

Danu

Wait, what? Closed string excitations include other particles, too!!!

John Rennie

12:46

OMG!

ACuriousMind

@JohnRennie To the effect that you can have electrically and magnetically fluxed branes, i.e. there is non-gravitational interaction between branes, and that you can't distinguish the world sheet of an open string going in a circle from that of a closed string going from one brane to the other

I must have typed it and not hit enter

John Rennie

Well that's what I read in my Disney Guide to String Theory

ACuriousMind

After Star Wars, Disney now also bought String Theory? ;)

Danu

Am I taking crazy pills? How are there not many other excitations in the closed string spectrum?

ACuriousMind

@Danu There are

Danu

12:48

So how is this answer not completely wrong?

John Rennie

@Danu the chat hasn't been a particularly cheerful place for the last few days, but I think this comes down to a select few users who haven't got the hang of being nice.

Bernard Meurer

@JohnRennie Here's my question, Headphones, speakers, and so on are analog devices and, in the digital era, rely on a good source to convert digital to analog and, if need be, amplify it, the usual DAC + Amp combo. Now, the typical L/R connector, as well as the 3.5mm Jack are both of classical usage for analog signals, which makes sense because they're simple and work nicely.

Danu

@JohnRennie Yeah, and they're dominating the chat room.

ACuriousMind

@Danu The closed spectrum has no massless vectors in it

John Rennie

@ACuriousMind see Danu. Oh my God didn't you know that?

ACuriousMind

12:49

So the only "force particle" that is associated to the closed string is indeed the graviton, while gauge bosons belong to open strings

Bernard Meurer

In it's latest iPhone revision it is rumored that Apple will get rid of the 3.5mm jack and support Bluetooth & Lightning headphones only, now, both of those are digital things. A lightning connector is just a Thunderbolt 3 iirc, and Bluetooth is very much digital. So wouldn't this mean that any of those headphones would need a builtin DAC?

Danu

@ACuriousMind Eh, are we doing something uncompactified here or something?

I just assumed we're compactifying and then there are the $\alpha_i\alpha_\mu|0\rangle$ which are certainly massless vectors

ACuriousMind

@Danu Unless something is said otherwise, I always assume we're on $\mathbb{R}^{1,9}$.

John Rennie

@BernardMeurer err, yes, I think USB headphones have a DAC built in.

Danu

@ACuriousMind OK, I guess I can relax then ;)

Bernard Meurer

12:52

@JohnRennie But that's so stupid

ACuriousMind

Compactifications bring a lot of other features with them

John Rennie

@BernardMeurer Why?

Danu

Sure do

Bernard Meurer

My source is guaranteed better than any DAC that is builtin on a headphone

And I don't mean my desktop source, I mean my portable one

John Rennie

@BernardMeurer I'm not sure that makes sense. Isn't your source digital?

Bernard Meurer

12:53

Heck, the DAC on that thing is just a delta-sigma half a dollar IC anyway, it's probably already on your phone

@JohnRennie Yes. Well, some of it. My point is: I have a very good DAC, USB headphones seem to throw that away

Danu

By the way @JohnRennie I suspect that with relatively little effort, you can get a decent low-level understanding of string theory by just reading the first few chapters of e.g. Blumenhagen, Luest, Theisen's book

If you just sorta skim over the harder stuff it's like QFT 101

John Rennie

@Danu thanks, I'll permalink that for when I've finished discussing headphones :-)

Bernard Meurer

@Danu I'm fairly sure you can get a low-level understanding of string theory by just reading XKCD

Danu

(they teach it, here, to students that know neither QFT nor GR)

@BernardMeurer I doubt it

Bernard Meurer

@Danu I didn't say quality understanding did I?

ACuriousMind

12:55

@BernardMeurer Abstruse Goose is the better choice ;)

(click on the image to advanced through the steps)

John Rennie

@BernardMeurer expensive USB headphones will have an expensive DAC, and even for cheap headphones I think you underestimate how good even moderately priced DACs can be.

Yes, if you want ultimate quality you'll buy a top end hi-fi DAC for several grand and use analogue headphones.

Danu

@ACuriousMind Neat

Bernard Meurer

@ACuriousMind Lel

John Rennie

But for most people in most circumstances all but the cheapest and nastiest of DACs will do the job.

Bernard Meurer

@JohnRennie I just don't like where this is going... The source and the headphone should be two totally separate devices

I quite like my current setup, the HD650s are amazing

John Rennie

12:58

@Danu that's kind of down to us. If we aren't discussing cool physics then we leave a vacuum in which some unwelcome fluctuations will occur.

Danu

I think I'll spend some time chasing & policing comments by CuriousOne

Bernard Meurer

They leak like a motherschrebbles though

Danu

@JohnRennie Yeah, but I, for one, don't really find myself spending time here anymore because, besides ACuriousMind, you, DanielSank, ChrisWhite and maybe a few others, there is no fun or interesting content to be found here anymore.

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