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00:00 - 18:0018:00 - 00:00

12:00 AM
Mhhh, I don't think so
oh god, why can $Hv=\lambda v$ be solved
@ACuriousMind I'm hyperventilating, help
what do I do
Because $H-\lambda\mathrm{Id}$ has non-trivial kernel if $\lambda$ is an eigenvalue.
Well, pretty much because the definition of an eigenvalue is that that equation has a non-zero solution
well no shit
so why did you say that earlier
Ok, I've got it @ACuriousMind
We have $H|a\rangle=a|a\rangle,H|b\rangle=b|b\rangle$
since the eigenvalues are real, we have $\langle b|H|a\rangle=a\langle b|a\rangle=b\langle b|a\rangle$
So that means $\langle b|a\rangle=0$ if $a\ne b$.
If $a=b$, then we have a degenerate situation
so just apply Gram-schmidt to those eigenvectors
@ACuriousMind Done.
No spectral theorem.
12:09 AM
That is the spectral theorem, you just proved it.
That's why I was so confused about what you were doing - the spectral theorem is the statement that every self-adjoint operator has an orthonormal eigenbasis with real eigenvalues
Oh, why didn't you say that
I could have looked up a proof
I just cobbled together my own...
@ACuriousMind So is my proof correct
The "hard" part of the proof is that the eigenvalues and eigenvectors exist, i.e. know that a) the roots of the characteristic polynomial are the eigenvalues and b) every non-constant complex polynomial has a root
(a) is trivial?
(b) I proved in GP using oriented intersection theory
12:12 AM
@0celo7 Yes, although it is more customary to proceed by induction.
I also proved it in Milnor using Sard
@ACuriousMind We both know I do not know what induction is.
@0celo7 While "elementary", I do not consider "matrix is invertible iff determinant is zero" trivial.
@0celo7 ::faints in horror::
@ACuriousMind What?
@ACuriousMind Eh, we proved that in linear algebra. Beats me how it worked though.
Something about columns being linearly independent?
@0celo7 Yeah, so a) is not trivial, because you need that afaics :P
Can you give a hint?
12:15 AM
@0celo7 I'm just dismayed at that way of proving the fundamental theorem :P
I have 6 books on by table now
None are linear algebra
I own no linear algebra books
Oh, I am in the library
There might be a linear algebra book here
@ACuriousMind How else can you prove it?
Oh no, do you need cofactor stuff
@ACuriousMind Wow, I did not know the zero matrix is invertible
Good to know
@ACuriousMind I looked up a proof, it's long
I'll just assume it.
@ACuriousMind what's your favorite way of proving the fundamental theorem
@0celo7 I prefer the proof through Liouville's theorem from complex analysis: If $f$ has no root, $1/f$ is a holomorphic function on the complex plane, and $f$ has a non-zero minimum. So $1/f$ is a bounded holomorphic function on the complex plane, and thus constant by Liouville's theorem.
@ACuriousMind Oh
@0celo7 You know what I meant :P
I know that proof too
@ACuriousMind Is there an algebraic proof?
12:25 AM
No pure one, but Wikipedia has some that come close. I don't find them appealing, though
@ACuriousMind why isn't it pure
@0celo7 Because it needs the intermediate value theorem
oh, I see
@ACuriousMind What's "haar measure"
An invariant measure on a topological group
I can read wiki
What does that actually mean
12:30 AM
Well, on the reals with addition, it's the usual Lebesgue measure
On the reals without zero with multiplication, it's $\frac{\mathrm{d}x}{x}$, for $\mathrm{d}x$ the usual Lebesgue measure
I'm not sure what you mean by "what does that actually mean"
If you can read it's the invariant measure, you can also read what invariant means :P
I'm confused by the construction on page 112 of Milnor
I know of a different way of obtaining a bi-invariant metric.
The beauty of the Haar measure is that the method of constructing it doesn't matter, since it is unique up to scaling (which is a somewhat involved proof)
I wonder if the proof I'm thinking about is modern
It's used in do Carmo and Cheeger & Ebin, which are later than Milnor
it avoids Haar measure
@ACuriousMind How do you define the measure
What about "which is a somewhat involved proof" gave you the impression that I would carry the proof out for you?
It's not even in Helgason
12:39 AM
It doesn't matter how the measure is constructed, what matters is that it exists and is unique
@ACuriousMind All of it, honestly
Well I can't use results I haven't proved
We've been over this
Wonder where I could read about it...
you can also use the Killing form
when the group is semisimple
@ACuriousMind I want to ask you what $T_e\iota$ is but you'll tell me to google it
1 hour later…
1:59 AM
@ACuriousMind Trying to prove "a simply connected abelian Lie group is isomorphic to $\Bbb R^k$"
I'm thinking that $\exp:\mathfrak g\to G$ is surjective
oh...maybe it's a homomorphism, too
Now, we know that $\Bbb R^k$ is its own Lie algebra
And certainly $\Bbb R^k\cong \mathfrak g$ for some $k$.
so $G=\exp \mathfrak g=\exp \Bbb R^k=\Bbb R^k$
I never used simple connectedness
actually, maybe you get $\exp \Bbb R^k=\Bbb R^n\times T^l$
I know you can get a torus somehow
yes, apparently $\exp:\mathfrak g\to G$ is a surjective Lie group homomorphism
we also know that $\mathrm{dim}\,\mathfrak g=\mathrm{dim}\, G=:n$.
we can write $\mathfrak g=\mathfrak g_1\oplus\cdots\oplus \mathfrak g_n$ for any 1-dim subspaces.
So it's clear that we have $$\exp\mathfrak g=\exp\mathfrak g_1\times\cdots\times\exp\mathfrak g_n$$
but the only abelian Lie groups obtained in this manner are $\Bbb R$ and $S^1$
simply connected then eliminates the $S^1$ factors
we invoke the classification of 1-manifolds to get the $\Bbb R$ and $S^1$ thing
@ACuriousMind That proof is probably wrong
Seems too easy.
I guess I should prove that $\exp \mathfrak g=G$.
Ok, $\mathfrak g$ is abelian iff $G$ is, so $\exp$ is a homomorphism by...BCH.
BCH is probably too strong there but whatever
surjective though...
clearly $\exp$ is a Lie group homomorphism
so we can use some theorems about that stuff
Given connected Lie groups $G,H$ and a homomorphism $F:G\to H$, if the induced map $F_*:\mathfrak g\to\mathfrak h$ is an isomorphism, $F$ is surjective
Clearly $\mathrm{Lie}\,\mathfrak g=\mathfrak g$ and by definition $\mathrm{Lie}\,G=\mathfrak g$
and we know that $\exp_*=\mathrm{id}$
the identity is an isomorphism, so $\exp$ is surjective
2 hours later…
4:17 AM
What! Ed Witten had a BA in history!!
6:00 AM
@JohnRennie hello
@JohnRennie Good morning!
Morning all
@0celo7: you got the furnaces going then? :-)
I heated some copper to 1100C for lulz and it didn't melt :/
I'll do 1200 on Monday
@JohnRennie,Have you ever visited the defunct Darebury plant or Diamond House, at Oxford? The latter is open next October. If I asked a question, could you find some data from any of them? The usual formula relates power to acceleration, I'd need a curve matching light frequencies and velocities/energies of the electrons (and the angle of curvature of each deviation: 15° (2pi/24) or 7.5*?)
@user104372 Hi.
It's several decades since I last did any work at the Daresbury lab, and I'm afraid I've never been to Diamond House.
6:10 AM
Hi, John, thanks for the kind invitation
I wrote to Diamond House, I hope they'll reply, can you get that info for me?
I'm two hundred miles from Oxford I'm afraid, and have no plans to go there. So I can't help. Sorry :-(
I would contact Diamond House directly, though you say you've already done that.
Are you after technical data on the performance of the Daresbury synchotron? That must be floating around the internet somewhere.
I didn't mean to go there, but to retrieve info on the web, you are far more clever and competent to do that
As I wrote, I just need general info , graph relatinc speed to frequency
The surely have first-hand info at Dicton, but the might have published that somewhere on scientific journals to which I have no access
6:16 AM
@JohnRennie, Let me know if/when I can post a question you might answer. Thanks in advance
@JohnRennie should my next project be Cheeger & Ebin or Bott & Tu?
@0celo7 I really envy you for the amount of energy you have :-) These days I pick up a 500 page maths boks and think maybe tomorrow :-)
Neither are 500 pages
I read about 90 pages of Morse Theory today
I'm on a roll
Even so, I still haven't made a start on any of the QFT books I've bought. One day ...
Bott & Tu is about cohomology and homotopy
It's scary.
not at 2AM please
@ACuriousMind I understand the Mayer-Vietoris sequence
Time to sleep
@ACuriousMind How do we know that the de Rham cohomology groups are finite-dimensional?
6:35 AM
@0celo7 o/
Is this you who stars greetings?@MAFIA36790 ;-)
@ACuriousMind Might be a stupid question
no; why should I?
Just was a guess
@ACuriousMind I'm asking because I want to be careful about using bases and whatnot when proving things about exact sequences
7:29 AM
@knzhou Hi :-)
2 hours later…
9:18 AM
@vzn Yeah that's fine. I'm not around the site much until a few days from now.
No need to multi-ping me :)
2 hours later…
11:10 AM
@0celo7 You don't
It's true for compact manifolds, but false for general manifolds
11:29 AM
Here's yet another case study for why downvotes should carry more weight, at least on answers. If one looks at the reputation timeline, it is pretty clear that the majority of reputation of that user comes from a single other user upvoting each of their posts once (it doesn't really matter if it's more than one user, the important thing it that there is never more than one upvote). Yet no post has positive score.
One should not gain reputation by having no post with positive score.
@lucas hello
11:43 AM
@ACuriousMind May I ask why did you not vote this answer of mine down while you voted both question and the other answer down? Did my answer not deserve to vote down? Or you forgot to vote my answer down? I ask this, because I will feel honor if my answer isn't so bad in your opinion. ;-)
Q: Otto and Carnot efficiency comparison

MotherLandLets say both engines operate between same temperature limits. Carnot eff: $$ e_\text{Carnot} = 1-\frac{T_L}{T_H} $$ Otto eff: $$ e_\text{Otto} = 1-\frac{T_4-T_1}{T_3-T_2} $$ assuming that for Otto cycle points are located as below on a P-V diagram: $$ \array{ 3 & 4 \\ 2 &...

You have no way to know whether I voted on a particular question or answer
Votes are anonymous by design and I do not comment on my decisions to upvote, downvote or not vote at all on specific posts.
I guess, because I was visiting that page when you left a comment and both of those posts were voted down.
Yeah, don't go around speculating on users' voting habits unless they themselves tell you how they voted
Sorry, I didn't want to make you uncomfortable!
@dmckee I stumbled across this question where especially the top answer is pretty non-sensical, shielding itself by saying "the following is simplified" from having to actually back up its claims.
oh god the second one talks about "generating virtual particles"
Really "science based" appears to mean "use your personal misunderstanding of pop science to generate technobabble instead of just making it up completely" especially combined with the "magic" tag...
12:17 PM
Why should I come back to the university while I dislike it :-( I will die there, I know
1 hour later…
1:42 PM
@dmckee I stumbled across this question where especially the top answer is pretty non-sensical, shielding itself by saying "takes every possible path is a mangled statement" from actually answering the question.
oh god the second one talks about "paths which go 'back in time' and 'come forward' again".
@ACuriousMind Can you please check my wall of text above :/
@ACuriousMind I can summarize.
@ACuriousMind Since $\exp_*=\mathrm{id}$, the map $\exp:\mathfrak g\to G$ induces an isomorphism on the Lie algebras. Furthermore, $\exp$ is a homomorphism between these groups (with $\mathfrak g$ viewed as a Lie group wrt. addition) due to BCH. Thus $\exp$ is surjective by a theorem in Lee.
Write $\mathfrak g=\mathfrak g_1\oplus\cdots\oplus\mathfrak g_n$, then since $\exp$ is a homomorphism we have $G=\exp\mathfrak g=\exp\mathfrak g_1\times\cdots\times\exp\mathfrak g_n=G_1\times\cdots\times G_n$.
Each $G_i$ is a Lie group in its own right, and is one-dimensional.
By the classification theorem of 1-manifolds, we have $G_i=\Bbb R$ or $S^1$.
Simple connectedness eliminates the $S^1$ factors, so we have $G\cong \Bbb R^{\mathrm{dim}\,G}$.
@0celo7 No, I'm not your proof checker, especially not in areas where my own knowledge is rather patchwork
Could you please just read it..
2:10 PM
@ACuriousMind What's a 1-dim nonabelian Lie al-jabr?
@0celo7 There is none
mind = blown
I guess by the classification theorem you could have group manifolds $\Bbb R,S^1$
How could a one-dimensional algebra ever have two elements $v,w$ such that $[v,w]\neq 0$?
@ACuriousMind Oh, very good point :P
@ACuriousMind Reading Milnor has made me want to learn alg.top
2:37 PM
@JohnDuffield think youre sniffing on the right track there & have long mused the same. have you ever thought of blogging esp wrt physics/ EM/ QM theory etc?
@0celo7 That's nice :)
Hi all
@vzn : I did a bit of blogging a couple of years back. I was just one of several writers on something set up by a journalist called James Delingpole. But then things got screwed up when somebody changed the hosting, and I packed it in to have more time for other things. One thing I wasn't keen on was that it was like talking at people rather than talking to people.
LOL, and now I'm writing a book, oh the irony!
Is a redshift value of 3000 large?
I mean, excessively large?
::looks around::
I don't think any astro people are currently here, @NoahP
2:47 PM
Ah bummer @ACuriousMind
@NoahP : I think so. The cosmic microwave background radiation has a redshift of z = 1089.
@JohnDuffield huh no kidding. whats it on? blogging can help with that, eg posting work in progress/ TOCs/ outlines/ summaries etc, highly encourage you to restart it to some degree. not lying, its hard at times esp in age of facebook, the 800lb gorilla.
@JohnDuffield So that would make my result reasonable?
@vzn : What's it on? Everything.
@NoahP : no. I don't think so. Can you give some more context?
Given a value of $\Omega_{Rad}=10^{-4}$ and $\Omega_{mat}=0.3$, I need to find the redshift when $\rho_{mat}=\rho_{rad}$
2:52 PM
@DanielSank "multipings"? if you write stuff in chat without acknowledging earlier pings then think its a fair assumption you missed them, have no other way to tell/ know. am trying to limit pinging you to the key session admin stuff. you agreed in principle to slot but not on a date & we have to nail down the date, thx for that :) would still like to do some mgt re prior session documentation but will let it slide for now since you seem very busy, will try to match your pace
@JohnDuffield I then calculated that to be 2999
Essentially 3000
@JohnDuffield any idea on how to publish it? would there be any author bkg in the book? yours seems quite mysterious so far (aka "air of mystery")
@NoahP : sorry, I don't understand what you're trying to do here. Ordinary matter makes up less than 1% of the mass/energy of the universe.
Its a question I was given, part of a project
Ive got $\rho_{mat}=\Omega_{mat}(z+1)^3$
And $\rho_{Rad}=\Omega_{Rad} (z+1)^4$
The value just seemed high when I calculated $z$
Im pretty sure I did it right though
@vzn : yes. But it would be improper of me to talk about it. Besides, if it didn't work out I could end up looking like a chump. I'm not really mysterious. I'm an IT guy who got interested in where physics was going ten years ago when our teenage children gave up all their science subjects.
3:03 PM
@ACuriousMind Do you have access to Bott-Tu?
I know you can't say you don't know anything about cohomology ;)
@JohnDuffield IT huh small world, what area? are you retired? hey lots of ppl have unpublished book manuscripts no shame in that. so your kids dislike science or did something different?
@0celo7 Let's suppose I do
@ACuriousMind On page 24, what exactly do they mean by "choices in this construction"
@vzn : sadly I'm not retired. As for what area, I'm in insurance.
Our children liked science, but science didn't like them.
3:06 PM
The partition of unity?
@0celo7 The $\xi$ as the preimage of $\omega$, for instance
What do you mean
Well, isn't that exactly what I mean by "partition of unity"
@NoahP : I don't know what to say Noah. I think about things like pair production and the wave nature of matter, I don't see a lot of difference between matter and radiation.
@JohnDuffield good movie, more mystery. ok stranger
@JohnDuffield Okay, thanks. No worries
3:09 PM
@0celo7 The operator $\mathrm{d}^\ast$ is constructed by mapping a preimage of $\omega$ somewhere. That's a choice, and you have to show that it doesn't matter which preimage of $\omega$ you choose, i.e. that the all give the same $\mathrm{d}^\ast[\omega]$.
And there's another choice when the preimage of $\mathrm{d}\xi$ is chosen
I have no idea what you mean by partition of unity
@ACuriousMind so $d\xi\in\Omega^{q+1}(U)\oplus \Omega^{q+1}(V)$
Well, after reading a bit backward I do, but that's not what they mean by "choices"
Oh, we have an extension to $\Omega^{q+1}(M)$
and the restriction of that form is $d\xi$?
ah right, and $d\xi$ goes to zero in $\Omega^{q+1}(U\cap V)$
so $d\xi$ is in the kernel of the map $\Omega^{q+1}(U)\oplus\Omega^{q+1}(V)\to \Omega^{q+1}(U\cap V)$?
Yes, $\mathrm{d}\xi$ goes to zero so by exactness it comes from an element in $\Omega(M)$, but there might be more than one element that maps to it, so one has to check that the final cohomology class $\mathrm{d}^\ast[\omega]$ does not depend on which of those elements was chosen
By exactness, $d\xi$ is in the image of $\Omega^{q+1}(M)\to \Omega^{q+1}(U)\oplus\Omega^{q+1}(V)$?
3:14 PM
@NoahP : sorry I couldn't help. Why don't you ask a question about it?
@ACuriousMind Right, that sounds like a worthwhile exercise.
Yes, it is
OK I have to go. By all.
BTW, this coboundary operator is a special incarnation of the "snake morphism" of the snake lemma
@ACuriousMind Yes, they mention that lemma
3:15 PM
I.e. its existence and independence of choices is purely algebraic
or maybe Lee does
I'll probably work on the proof of the snake lemma, zigzag lemma, etc today
Yay, diagram chasing! :D
@ACuriousMind If I have questions, expect pictures :P
Not gonna mess with ticz-cd
@ACuriousMind Is this what's called "homological algebra"?
@0celo7 yes
Who invented it?
3:20 PM
Eilenberg and MacLane, possibly
But I'm no historian
3:43 PM
@ACuriousMind what's your thesis on
@0celo7 I'd love to be able to answer that question :P
At this rate I'll write my BS thesis before you
Time to chase a diagram.
@ACuriousMind Any tips for thus stuff?
follow the arrows?
Makes sense.
so what I want to prove is
actually that's homology
Oh, how about I read the homological proof and do the cohomological one on my own
@ACuriousMind What does one do in "pure" homological algebra
4:02 PM
I'm not sure what you mean by that
Lee says homological algebra started off as algebraic topology but since became a field in its own right
I guess one can do pure homological algebra in the sense of just doing stuff with complexes and their (co)homologies
Proving the lemmata in abstract abelian categories and such
Because it leads to a very powerful abstract toolset of e.g. derived functors
4:10 PM
Oh, topology (Verdier duality/sheaf cohomology), algebraic geometry, abstract algebra (ext/tor functors),...
sounds too abstract for my tastes
and I'm procrastinating on proving this lemma
@ACuriousMind I just used $G(d_p)$ in my proof :D
The hover-over text on today's xkcd is the story of my life.
@dmckee ouch, that's pretty nerdy
4:26 PM
@ACuriousMind Do you know why $$\forall q \in \Bbb N, \int f(x) \frac{1}{\varepsilon} \rho(\frac{x-a}{\varepsilon}) dx = f(a) + \mathcal O(\varepsilon^q) $$
Is different from just $$\int f(x) \frac{1}{\varepsilon} \rho(\frac{x-a}{\varepsilon}) dx = f(a) $$
If it's for all $q$, shouldn't the remainder just be 0
@Slereah I think that there doesn't exclude terms like $\exp(\epsilon)$.
Ah, could be
Would be nice if I could find an explicit calculations with a real mollifier
"Suitable mollifiers" are defined but they never really give an example
Those damn mathematicians and their fear of examples
What kind of mollifier are you talking about?
fuck this proof is long
A "suitable" one
It's like
1) Schwartz function
2) Fourier transform is $1$ in a neighbourhood of $0$
4:33 PM
The ones I know are usually smooth with compact support, which are difficult to give examples of, I think
You usually choose them as normalized bump functions or something, but you won't find an explicit functional form for them
I s'ppose yeah
It helps to reassure yourself that the definition makes sense though
There's the "standard" example built from $\exp(-x^2)$ or something
Also part of the theorem is that all the moments of those mollifiers are $0$
$$\forall a > 0, \int x^a \rho(x) dx= 0$$
and $\int \rho(x) dx = 1$
Does that make sense
I know that functions of the same moments can be different functions, but I'm having trouble visualizing such a function that isn't $\delta$
Having all moments zero looks to be a rather strong restriction
Quite so
4:37 PM
MO has an example, though:
A: Let a function f have all moments zero. What conditions force f to be identically zero?

coudyThe answer to the first question is no. The following example is standard in probability theory, see e.g. Billingsley "probability and measure", example 30.2. $$f(x) = {1\over \sqrt{2\pi}\ x}\ e^{-{(\ln x)^2\over 2}}\ \sin(2\pi \ln x) \ {\bf 1}_{[0,\infty[}(x)$$ You can check that all the momen...

@ACuriousMind $[c_p+c_p']=[c_p]+[c_p']$, right?
Although it stems directly from the Fourier transform being 1 in a neighbourhood of 0
@0celo7 If the brackets denote "taking equivalence class", then yes
Since that means that $(F\rho(x))'|_0 = 0$
@ACuriousMind Yes, (co)homology classes
4:39 PM
@ACuriousMind Ah yes, thanks
That is quite a non-trivial example
I guess I can continue reading on Colombeau algebras with more confidence
@ACuriousMind That proof was annoying.
@Slereah On second thought, it does exclude terms like $\exp(\epsilon)$, since that has polynomial terms. The difference between the two things (let's call the integral $I(\epsilon)$) is that the first one says that $$\forall q\in\mathbb{N}:\lim_{\epsilon\to 0}\frac{1}{\epsilon^q} (I(\epsilon)-f(a)) < \infty$$, which does not imply the second.
So there could be like
At least if that $\mathbb{N}$ does not include 0
Fastly decaying non-analytic functions?
4:59 PM
Looking at it, yes. $\exp(-\epsilon)$ probably works
It has polynomial terms but of all orders
So that's probably fine
No, $\exp(-\epsilon)$ has zero Taylor series at 0
Colombeau algebras aren't easy to do rigorously because there's a lot of shit to carry around
Well yes, but if $\varepsilon = 0$, then it reduces to the delta distribution
So it's all good
What I wanted to say is that it doesn't have polynomial terms - it is an allowed contribution precisely because you don't see it when you go "up to order $\epsilon^q$", since it is zero upon expansion. It's a non-perturbative contribution that you cannot see order-by-order
I wonder if you can define generalized functions as like
Actual functions
Something like $\Bbb R^n \to \Bbb R^*$
Where $\Bbb R^*$ is some non-archimedean extension of $\Bbb R$
I know that you can define one with Colombeau algebras, with generalized numbers
5:17 PM
I was just kicked out of the math chat
@0celo7 No you weren't. You just killed it.
I must be retarded, because I do not see what I did wrong
Please no cross-contamination. Confine drama to one chat :P
I don't talk to Ted any differently than any of my professors at school
Yet somehow I manage to piss him off with 100% chance
I feel sick
@0celo7 Just stay away from him then?
5:20 PM
Why can't someone just tell me what I did wrong
@Danu He's very helpful!
And I want to know why I piss him off so much
And I certainly hope I'm not perpetually pissing off all of my professors
Pretty sure you are
@Slereah I think I saw something like that once but I can't remember the name
I think you're trolling but maybe I am
How do I tell?
But be prepared for that not being a novel idea of yours in any case :D
Maybe Ted is weird, I don't know. But he always flips out, no matter what I say
And he never tells me why
5:25 PM
@ACuriousMind Oh I'm pretty sure
The closest I've ever come to having a novel idea was 8 years too late
Well, I've had ideas that weren't done before, but it's all stuff too hard to solve
science is p. hard
5:49 PM
Thinking aboot it
Any Schwartz function has a Fourier transform, right
Since the big condition is that the Fourier transform is $1$ around $0$, building a suitable mollifier should be as easy as defining some function that is $1$ around $0$, take the inverse FT, and then that should be a suitable mollifier
I'm guessing that the inverse FT of a Schwartz function is a Schwartz function, too
The zero moments derive from that fact
Yes, that sounds reasonable
Are you reading the papers of the Vietnamese guy
I'm guessing that the rectangle function with some smoothed edges would do fine
00:00 - 18:0018:00 - 00:00

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