The h Bar - 2018-02-05 (page 1 of 2)

« first day (2650 days earlier) ← previous day next day → last day (2269 days later) »

12:19 AM

@navnav Also you may find this interesting.

Bernardo Meurer

@BalarkaSen Objections?

Bernardo Meurer

1:10 AM

@ACuriousMind Do you want to help your boy out

Your boy being me

2 hours later…

3:00 AM

user image

Let $\beta_1>\beta_2$

Then $\lambda_1>\lambda_2$ it means first fringe should be red and second fringe blue

i don’t understand how fringe closest to central white fringe is red 😩

3:25 AM

Lol typo again. I mean If $\lambda_1>\lambda_2$ then it means fringe closer to bright central fringe should be blue and away from centre should be red

4:09 AM

If you take a metal into a vacuum chamber and blast it with high energy photons, you'll get a bunch of electrons to leave the metal via the photoelectric effect right? What happens to the metal... it just builds up a static charge or what

and then once it makes contact with something it'll recover some missing electrons?

3 hours later…

6:58 AM

@pentane as the metal becomes positively charged it attracts the electrons back towards it, so there is a maximum charge at which the photoelectrons can no longer escape from the metal.

And yes, if you keep the metal completely isolated it will just sit there holding that positive charge.

7:18 AM

@BernardoMeurer what's up?

user image

A person starts from C and reaches A at speed 1000m/s. An object starts from B simultaneously as C at same speed. Will he see object at 1 sec or not?

7:47 AM

@0celo7 ? Embed it in some R^N. Compose with a homeomorphism of R^N to a ball of finite radius inside R^N.

It's an absolutely trivial question unless your embeddings are special

@Bernardo I haven't had the time to write a sensible counterargument. I'll postpone it until I have read your argument more carefully. Maybe this weekend.

hey hey

Ah, apparently the Fourier transform of a measure is just some bounded continuous function

It's not like Schwartz where it's an automorphism

@0celo7 In particular if you are thinking about isometric embeddings, curvature gives restrictions, which you already should know

8:05 AM

Slereah: All measures or just lesbegue? Bounded sounds pretty strong a condition if the measure is arbitrary

(Well, curvature gives restrictions on the size of the ball if you have a compact manifold beforehand. Completeness gives restriction on bounded embeddings)

O wait I need to check how measures on distributions are defined

Positive measure on some Banach space

(bounded measures are the topological dual of Banach function spaces)

Ah I see

Sci-hub seems to not work too well

I hope it didn't get blocked

ah, apparently Cloudflare got a court order for sci hub's activity

so it is not working too well because of this

8:48 AM

Yeah it can get wonky at times, links are down more frequently than before

There is an onion link, but I think you won't want to touch that unless you want to increase your chance being tracked by the CIA (if I recall correctly)

Ah, I managed to sneak my way onto the paper

thank you poorly secured websites

Wiley library books are almost impossible to get onto since not many Aust uni have subscriptions for them

Captain Bohemian

9:50 AM

I found conference papers seem to indeed have more typos than journal papers, as I have heard from the web.

Bernardo Meurer

@BalarkaSen Cool, I'm almost done writing the real essay

Captain Bohemian

10:17 AM

the weather is uncomfortably cold currently.

too cold to think physics well.

11:06 AM

Guys, I have a very important question.

They say when a rocket is launched from Earth, that time moves more "slowly" for the observer in the rocket, because it's moving faster.

Now in a relativistic sense, that rocket is moving faster... than us. Which means, that to the rocket, we are the ones that are moving away at the exact same rate.

So does that mean that time is, from the rocket's perspective, slowed down for us too?

No

Acceleration from a rocket and acceleration due to coordinates are two different things

I mean, why doe? Is it something to do with like potential energy or something?

2 hours later…

1:21 PM

physics.stackexchange.com/questions/383248/… holy shit this guy had the exact same question

it's a common question

where is johnrennie's essay on symmetric time dilation again? It might be good to refer him to that

67

Q: How can time dilation be symmetric?

Suppose we have two twins travelling away from each other, each twin moving at some speed $v$: Twin $A$ observes twin $B$’s time to be dilated so his clock runs faster than twin $B$’s clock. But twin $B$ observes twin $A$’s time to be dilated so his clock runs faster than twin $A$’s clock. Eac...

special-relativity inertial-frames coordinate-systems time-dilation observers

especially it is important to recognise that it is different from gravitational time dilation, which is asymmetric

Super interesting

1:27 PM

4

Q: Gravitational time dilation, does time of the observer at a lower gravitational potential looked slowed down in the frame of the higher one

This question is mainly inspired after watching the movie known as Interstellar We knew that for time dilation caused by relativistic motion between A and B. A will measure B's clocks slowing down, and B will measure A's clock slowing down by the same rate, while they both measure their own cloc...

general-relativity

For the gravitational bit

aw yeah gimme dat good good

Hi. Large aperture cameras are used for more light to reach sensors or to capture high resoulution images?

or both means same?

1:43 PM

Guys I have an idea. We make a photon calculate values for us. Then, we let it travel at 'c'. Then, we use it as a turing-complete machine somehow. Then we ask it to simulate the universe.

5

wot

2:06 PM

@BernardoMeurer project for you above

2:29 PM

Hi all... another funnnnnnnnnnnn GR question

So I have a 2+1 d spacetime here. obviously the Weyl tensor vanishes

This feels a bit peculiar to me

seems pretty normal to me

$$ds^2 = - \frac{r^2 - r_+^2}{\ell^2} dt^2 + \frac{\ell^2}{r^2 - r_+^2} dr^2 + r^2 d\phi^2 $$

Is that the BTZ black hole

So there's no singularity at $r=0$?

Idk I'm just a bit confused xD

Yes it is

BTZ black hole does have a singularity, IIRC

Compute the determinant and check $r = 0$?

2:32 PM

The BTZ BH is only defined for 2+1 d?

Honestly I'm supposed to use the fact that Weyl vanishes

Well there's an equivalent for every dimension, but they have different properties

The $3+1$ equivalent is the Kottler metric

aka Schwarzschild-de Sitter

I'm just a bit insecure about the physical meaning of the Weyl tensor

The Weyl tensor basically describes the propagation of gravity in a vacuum

If the Riemann is non-zero it's kinda obvious that $r=0$ is not a singularity

what?

2:34 PM

If I can show that Riemann is e.g. M log (r + 1000) then at r=0 there's obviously not a singularity right

(not a real example obviously)

Showing singularities is a bit complicated but in the case of black holes it's just a divergence in $r$, yes

Ok, so suppose I show that Weyl is not divergent at $r=0$ but instead it vanishes at $r=0$. Is that sufficient to show that $r=0$ is not a singularity

The Riemann tensor is roughly composed of three parts

The Weyl tensor, the Ricci tensor and the Ricci scalar

I don't think the Weyl tensor is the divergent one?

I mean the Ricci scalar for Schwarzschild is $R\approx \delta(r)$

It's not terribly well-behaved, even without looking at the Weyl tensor

I'm supposed to show that $r=0$ is not a singularity by using that the Weyl tensor vanishes for any metric in three dimensions

I think Schwarzschild is conformally flat, so no Weyl

2:41 PM

Hm

Weird if true

how do they define "singularity"?

The $2+1$ D version of the Schwarzschild metric (the non-BTZ one) has a singularity

It's a conical singularity but still

I'm as surprised as you are

It's supposed to hold for the BTZ one only

But for that specifically the Weyl tensor apparently plays a role

I guess maybe the Ricci tensor and scalar are all well-defined for it?

As to your question: there are multiple definitions but I guess that it's a good start if the metric determinant is regular at that point

Well

One of the common argument to show that the Schwarzschild metric is singular is the use of the whatchamacallit scalar

Kretschmann scalar?

Yes

2:45 PM

Which I think is built of the Weyl tensor

Since the metric now reads
$$
\begin{pmatrix} -\frac{\ell^2}{r^2 - r_+^2} & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & r^2 & 0 \\ 0 & 0 & 0 & r^2\sin^2\theta \end{pmatrix} $$

It’s the full Riemann

So if it's identically $0$, it would show that the Kretschmann scalar is well behaved

After the E-F coord transformation

That's not proof it's not singular but it's a hint

2:45 PM

I'm guessing all is well?

@user55789 that’s a very bad thing to look at.

Polar coordinates are singular in that sense

ah, there's a relation

$$R_{abcd}R^{abcd}=C_{abcd}C^{abcd}+{\frac {4}{d-2}}R_{ab}\,R^{ab}-{\frac {2}{(d-1)(d-2)}}R^{2}$$

Honestly I'm not sure, and since this was an exam question and my professor isn't being very nice about explaining it

@Slereah his name is Weyl not Ceyl

I don't think we're supposed to use insanely complicated formula's and Kretschmann scalars which generally aren't in an introductory GR course

2:47 PM

Cherman Weyl

If you're doing the BTZ black hole this ain't an intro GR course :p

Idk, we've been doing 6 weeks of GR, and then that was on the exam

Kek

Can you show us the exact question?

Maybe he’s doing German GR. First class is bundle functors

user image

The first sentence is "A spacetime is a paracompact Hausdorff $n$-dimensional manifold"

2:50 PM

That’s the Straumann approach

@Slereah connected!!

Hold on while I copy problem 1.4

@0celo7 You can have a disconnected spacetime

It doesn't change much

You just have to treat every connected piece in its own right

Lmao that’s introductory?

user image

Surface gravity of black holes?

Good lord. It would take me 6 week to teach enough geometry to even talk about GR

2:52 PM

So showing that the Weyl tensor vanishes wasn't a biggie

Surface gravity is like $k^a\nabla _{a}k^{b}=\kappa k^{b}$

I just don't see how I should relate that to singularities, I can't remember talking about Weyl in the context of a singularity

For a Killing vector $k$

Perhaps I can relate the fact that the Ricci tensor and the Riemann have the same amount of independent components?

In D=3?

Can't think of anything else

yeah you can express the Riemann tensor just using Ricci tensors

I guess you could argue that since $R_{ab} = \Lambda g_{ab}$ is well-defined everywhere, then the Riemann tensor never diverges

but that's pretty handwavy

You just have to show that $g$ doesn't diverge

2:56 PM

With $g \equiv \lvert g\rvert$ the determinant of the metric?

$g$ the metric tensor

But of course you can have well-defined tensors that have singular spacetimes

and divergent metric tensors for non-singular spacetimes

If you choose your coordinates poorly enough

I have already removed the singularity at the event horizon with the E-F coordinates

So I think some conclusions can be made at this point already :p

oh singularity analysis is very tricky, if you allow really stupid spacetimes

@Slereah I know what it is

@user55789 so he' not telling what he means by singularity?

Geroch built the worst spacetime of all time to make a really tricky singular spacetime

(also Beem)

2:58 PM

which one is that

@0celo7 Geodesically complete but singular

ah right

Random question for the really pure mathematician here:

I think any "reasonable" spacetime will be non-singular if all its scalars are well-behaved

What is the gist of "completeness" that unify all notions of completeness from functional analysis, to geometry and vector spaces?

3:01 PM

@0celo7 Not really, I'm not sure. One could argue that a divergent curvature would be considered "singular"

at a point

@Secret the notions of completeness in Lorentzian and Riemannian geometry are completely different

@Secret Sequences converge in the space

@user55789 Have you computed the Riemann tensor?

@Slereah Cauchy sequences.

yeah

@user55789 there are mathematica programs to compute the curvature

also, are we helping you with a current exam question?

3:02 PM

@0celo7 I have not, but again, I don't think that this is a useful thing to do (although I would have done it if it were homework). This question was one of four exam questions, a question where somebody would have 45 minutes for

It's not a current exam question, it's from fall last year, an exam I failed

It's quite rare for me so I want to understand it inside out now

@0celo7 By intuition, I will imagine the null vectors will be the ones causing trouble, so uh, can we still denote completeness in lorentizian geometry in terms of sequence convergence?

Yes

@Secret no, because there's no natural distance function on a Lorentzian manifold

@0celo7 If you're uncertain I can send you the full pdf with date included ;-)

You don't have to take my word for it

But you have to do it in the frame bundle

3:04 PM

@user55789 no that's fine

A singularity is if the frame bundle isn't complete

The completion of the frame bundle will have points that project to the singularity

All right

You can do that because you have a Riemannian metric on the frame bundle

@Slereah this isn't a canonical choice

Sure, but it works

3:05 PM

Sooo although it is fun that I sparked discussion, I think you're assuming I have more knowledge than what Carroll offers me :-)

Carroll as in the book :-)

Lewis Carroll?

Sean Carroll

Shocking!

Baffling.

Carroll's pretty good but it doesn't really have any discussion on singularities I think

3:07 PM

christmas carol?

Ok, so bear with me here. Suppose Weyl vanishes. Wouldn't that automatically imply that either both Riemann, Ricci tensor and/or scalar pair-wise diverge, or they are all regular at a point

Why would computing Weyl be easier than computing the full Riemann

It's a much more complicated formula

Well you don't need to compute it

It vanishes in 3D

Since the number of independent components of Riemann and the Ricci are the same

I think the trick is just that $R_{ab} \approx g_{ab}$ for the BTZ black hole

3:09 PM

We really only have 6 independent components, such that pairwise divergences are impossible

But that assumes prior knowledge of the BTZ black hole

Hence, they are all regular

Idk

I don't like this question :-)

Although that's pretty bogus since in the limit $\ell \to 0$, it converges to the cosmic string metric

Which has a quasiregular singularity

but quasiregular singularities can't be shown from the components

Sounds more like String theory, also not a part of an intro GR course :p

Cosmic string metric is all classical

It's just the metric associated to an infinitely dense string

Which is equivalent to the Schwarzschild metric in $2 + 1D$

It's some weird identification of Minkowski space with a conical singularity in the middle

3:16 PM

Almost sounds like condensed matter with delta function potentials...

But I'm a bit of a CM geek...

stm.sciencemag.org/content/10/426/eaan4488

Wow

@user55789 In $2+1$ D it's just $T_{ab} = \delta(r) u_a u_b$

or something like that

the metric is just $ds^2 = -dt^2 + dr^2 + r^2 d\varphi^2$

But $\varphi$ has an angular deficit

$\varphi \in [0, 2\pi - A]$

so that the spacetime has the topology of a cone

Ok but suppose we would leave out these extra constraints, such as this $T_{ab}$ one

Does Schw. have a singularity in (2+1) D?

wonder if layer cake rep is in Federer

By the way

On the question of singularities

3:26 PM

@user55789 Depends what you call a singularity :p

Quasiregular singularity are locally extendible

I think that the working definition in this course is:

Check that the curvature does not become infinite.

As simple as that..

Yeah there's no singularity if you go by that

It's just $R_{abcd} = 0$

The canonical way to check it in Sean Carroll's method is to investigate the contraction $R^{\mu\nu\rho\sigma}R_{\mu\nu\rho\sigma}$.

@user55789 well then you have your answer

$$R_{abcd}R^{abcd}=C_{abcd}C^{abcd}+{\frac {4}{d-2}}R_{ab}\,R^{ab}-{\frac {2}{(d-1)(d-2)}}R^{2}$$

The first term vanishes trivially

3:29 PM

Since for the $BTZ$ metric, $R^2$ and $R^{ab}R_{ab}$ are both constants

And $C$ is $0$

It's all regular

How am I supposed to show that $R_{ab}R^{ab}$ is a constant, and R^2?

Is it obvious? Without prior knowledge?

Kinda

Your exercize says that it stems from the cosmological constant

if you know that BTZ is a soln of the EFE with $\Lambda<0$

then contract the field equations

So $G_{ab} = \Lambda g_{ab}$

You can show that both $R_{ab}$ and $R$ are constants from this

you know that BTZ is an Einstein metric

3:31 PM

I think that's supposed to be the solution

Remember that there's another version of the EFE

$R_{ab} = T_{ab} - \frac 12 Tg_{ab}$

or something similar

which can be shown through some contractions

Hmm ok but why is $T$ regular ?

The metric in my coordinate system is obviously regular

$$ R_{\mu\nu} = 8\pi G \left(T_{\mu\nu} - \frac12 T g_{\mu\nu}\right)$$

@user55789 Well $\Lambda$ is a constant. Showing that $g$ has no divergences is up to you!

Yes, the metric is non-divergent by a suitable redefinition of coordinates

then you're golden

I chose the obvious
$$
v = t - \frac{\ell^2}{r_+} \operatorname{arctanh}\frac r{r_+}
$$

3:37 PM

plus the metric doesn't matter anyway

Since you're only computing scalars

$R_{ab} R^{ab} \approx \Lambda^2 g_{ab} g^{ab} = \Lambda^2 n^2$

Ok so let's go back to the contraction you gave

@Slereah This one

Is this obvious or just a grind?

Weyl isn't a terribly pretty expression to contract with itself

In your exam you seem to already have the decomposition of the Riemann tensor

So just compute its dual by raising all indices

then the product is obvious

It was an open-book exam yeah, we only have
$$
C_{\rho\sigma\mu\nu} = R_{\rho\sigma\mu\nu} - \frac{2}{n-2} \left( g_{\rho[\mu} R_{\nu]\sigma} - g_{\sigma[\mu} R_{\nu]\rho}\right) + \frac 2{(n-1)(n-2)} g_{\rho[\mu} g_{\nu]\sigma} R
$$

Shouldn't be too hard to compute from this, yeah

just a lot of contracting with the metric tensor

All right. Hold my beer -

Ah yes I can see it now

yeah and with $R_{\mu\nu} - \frac12 R g_{\mu\nu} = \Lambda g_{\mu\nu}$, contracting with $g^{\mu\nu}$ makes the whole thing trivial

Snap

3:47 PM

spherical rearrangements are confusing

@Slereah for a borel set $A$ in $R^n$, define its spherical rearrangement by $A^\star=\{x\in R^n:\omega_n|x|^n<\mathcal L^n(A)\}$

and given a function $f$ tending to zero at infinity, define $$f^\star(x)=\int_0^\infty\chi_{[f>t]^\star}(x)\, dt$$

like, wtf this this even supposed to be

What is $\mathcal L^n$

Lebesgue measure

Oh so it's the set with same measure but spherical?

yeah

just a ball with the same measure

I think imma gonna try to write a paper on thin-shell wormholes

It seems very vaguely defined

I want to do it in all proper manifold gluing

I suspect that you can have accelerating wormholes which do not have a time shift, but then the stress energy tensor will be weird

3:52 PM

I need to reunderstand the layer cake representation

user image

is this it

Oh it's the Riemann sum-ish rep of functions

with measures

it's the "Lebesgue integration uses horizontal slices" meme

it's not very intuitive

as seen in the "Riemann integral sucks" paper

level sets are crazy shit

Hm

If I have a manifold glued along some boundary $S$

Such that there's a coordinate patch $S \times (-1,1)$

What should I call this patch

I'm thinking $J$ for junction maybe

Also I probably shouldn't call the boundary $S$ since I will also have spheres involved

It's pretty hard to use standard names for everything since everyone calls their functions $f,g,\varphi$

So if you have more than 3 functions it starts to get confusing

user image

plz rate my collar

Is there like a paint program that converts automatically to Tikz

it would be nice

tikzedt.org/doc.html

Yesss

4:08 PM

@Slereah whoa what

what does that do

Anonymous

I'm confused about something. In general, is direct product of infinite groups of infinite order, a group too? Is $\Bbb R^{\infty}$ considered to be a group? Or do the properties differ for the infinite case?

Dunno yet

I'm installing it

The proper name is $\mathbb{R}^\omega$

It is a group too, yes

Anonymous

@Slereah Nice! And is that true for the general case: "direct product of infinite groups of infinite order" ?

Errr I would worry about generalities, but I'd say yes?

Infinities have a tendency to screw things up but I can't think of something that would prevent it

user image

@0celo7 seems nice so far

Anonymous

When I asked the prof he said that such direct products can be considered to be infinite sequences. That is a map from $\Bbb N \to X$

Anonymous

4:12 PM

And sum of two sequences is also a sequence

Anonymous

etc

yes

Anonymous

But then I was worried what would happen in the case when the infinity is not countable

Anonymous

How does one map from natural numbers in that case

Anonymous

Or is that an invalid case?

4:14 PM

Well you just do the product $G \times \Bbb R$

Then you have continuous sequences of group elements

Or is it $G^{\Bbb R}$

Anonymous

@Slereah I'm not exactly sure what you mean by that

I think $\mathbb R^{\mathbb R}$ is just the set of real functions defined on $\Bbb R$

Which indeed form a group

Sum of two functions is a function, zero function exists and there's always a negative

So good group structure

Anonymous

Interesting. Makes sense :)

you just associate a value of $\Bbb R$ for every point on the real line

Anonymous

Yup

Anonymous

4:21 PM

This is good way to look at groups

it's a pretty shit set of functions, though

Since most of them are very discontinuous

Anonymous

"most"? Numbers of discontinuous real functions on $\Bbb R$ is greater than number of real continuous functions on $\Bbb R$ ?

I think so

-4

Q: Why does this not solve all Millennium Prize Problems problems?

Since the Millennium Prize Problems were created a few years ago, their description of the current universe possesses uncertainity. It's possible to calculate that uncertainity, but we would need to unwind the events of a few years ago. If an intelligent being unwound those events, it would pro...

quantum-mechanics

wow

that's something

Cardinality of $C^0$ is just $\mathfrak{c}$ I think

Anonymous

4:25 PM

This keeps getting more interesting. Heh :P

Anonymous

I'll read more about this. Thanks

41

Q: Cardinality of set of real continuous functions

The set of all $\mathbb{R\to R}$ continuous functions is $\mathfrak c$. How to show that? Is there any bijection between $\mathbb R^n$ and the set of continuous functions?

real-analysis elementary-set-theory cardinals

4:41 PM

What's the name of a function $f$ such that $f =0$ is the submanifold

Is it just the level set

4:52 PM

@Blue I have a doubt in chemistry. Where do I ask it?

@JohnRennie Hi

« first day (2650 days earlier) ← previous day next day → last day (2269 days later) »

Transcript for

Feb '185

The h Bar

General chat for Physics SE (physics.stackexchange.com). For M...

join this room
about this room

00:00

06:00

12:00

18:00

all times are UTC

site design / logo © 2024 Stack Exchange Inc; legal

mobile