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3:00 PM
Yep
I just did that
neat
 
3 mins ago, by Yashas
$x = x\tan \theta - \frac{gx^2}{2u^2 \cos^2\theta}$
LHS must be $y$ not $x$ :P
in JEE Preparation, 8 mins ago, by Programmer
Which is better : Genius in Chem and absolute noob in Physics or Genius in Physics and absolute noob in Chem.
dono what to reply when people ask questions like these
 
@Yashas Can you compute and see what you get for $t$?
 
@0celouvskyopoulo7 Anyways, I don't think so: the initial topology $\mathfrak{T}_{\lVert \cdot\rVert}$ on a Banach space is finer than the weak topology $\sigma(X,X')$, and therefore $f\in C((X,\mathfrak{T}_{\lVert \cdot\rVert}),T)$ may very well be discontinuous in $(X,\sigma(X,X'))$, and almost surely not even semicontinuous.
 
@BernardoMeurer not possible
You need at least 3 points to fully determine an equation of second degree
 
@Yashas
 
3:08 PM
@AlbertEinstein ?
 
@Yashas, what is the difference between heat and internal energy?
 
@yuggib from X to R
@yuggib Initial topology?
 
@AlbertEinstein Heat is the flow of energy.
Internal energy is the total energy contained in the system.
Sum of kinetic + potential + everything else
 
@0celouvskyopoulo7 the topology of the banach space
 
@yuggib For reference, I mean: $x_n\to x$ weakly, then is $\lim |f(x_n)|\ge |f(x)|$?
 
3:14 PM
@AlbertEinstein Saying the body has a lot of heat is a completely meaningless statement.
 
where $f\in X'$
 
heat = transfer of energy
not energy itself
 
@Yashas, oh. Thanks
 
@0celouvskyopoulo7 wait
that is not what you asked before
 
Is that not what we mean by weak lower semicontinuity?
Lower semicontinuous wrt. weak convergence
 
3:15 PM
@
 
yeah but your $f$ is in the continuous dual
now
before, it was a generic function
 
Yeah I want $f:X\to\Bbb R$ continuous
@yuggib ohh, sorry
 
linear or not??
 
Oh, not linear :P
Sorry I just ran across campus
 
no prob
so, as I said, the weak topology is coarser than the norm topology, do you agree?
 
3:17 PM
@yuggib It's the function $(\int |Du|^p)^{1/p}$ on a certain sob space
 
in general, you can't expect that function to be weakly continuous
 
@yuggib I can never get these things straight
 
(or any strongly continuous function)
 
Coarse means it has less open sets?
 
@0celouvskyopoulo7 yes
 
3:19 PM
I think that's right then
@yuggib Ah well by Rellich my sequence converges strongly in $L^p$ so I get that part of the norm does converge
so from $\lim ||u_n||\ge ||u||$ I get a comparison for the first derivative part
i think that works out
 
I don't know your setting, anyways in general it is not true
 
I believe that
@yuggib I'm trying to prove a Poincare inequality on manifolds with boundary
 
@0celouvskyopoulo7 something I don't really want to understand ;-P
 
me neither, as it turns out
I like analysis on $\Bbb R^n$ faaaaar better
 
Just do analysis on $\Bbb Z$
It is much easier
 
3:28 PM
@Slereah don't say that...abstract harmonic analysis has huge applications in number theory / discrete mathematics
but it is far from easy
 
What about analysis on $\varnothing$
It's the best manifold
 
I wish I knew some harmonic analysis and number theory
Completely disjoint from my main interests but it seems neat
*analytic number theory
 
harmonic analysis is interesting
I may work a bit on that in the near future
I have some ideas and I have to see if they work out
 
This summer I ought to read Chrisdoulou (who knows how to spell that)
 
you forgot a bunch of letters
actually maybe a couple
he works here where I live now
(if he did not retire)
 
3:35 PM
@yuggib I'm writing my thesis on perfect fluid spacetimes
undergrad thesis
The conjecture is that a static, compact star is rotationally symmetric.
 
Quite hilarious
What kind of topics are other undergrads doing, paper maché volcanoes?
It seems fairly complex for an undergrad
 
Why hilarious?
I don't expect to solve it, I want to summarize a 2007 "proof" of it
There's a lot of details left out of the proof
Well, not summarize
The opposite
 
I dunno, I guess that nobody in my undergrad class would have the means to work out such a thing?
 
There's a book on Dirac operators written by a German guy who says it's for second year undergrads
It involves bundles, sobolev spaces, sheaf cohomology
 
I don't think I even saw the Dirac equation until my master years
 
3:44 PM
Oh and algebra that I don't dare understand
The hardest part of the project is that some of the work was done by physicists in the 80s and they made some mistakes.
I'd also like to find the metric of a cubic star and see how it evolves
 
lol
Even the Newtonian field of a cube is terrible
 
For that I have to read Yvonne
@Slereah link?
 
Why are electrical engineers writing a paper about this?
 
Boredom?
It's not a terribly interesting solution
What are you gonna do, calculate the gravitational field of the Borg cube?
I remember a paper calculating the metric of an ellipsoidal object in GR
slightly more interesting
 
3:53 PM
What about a cube in GR?
 
dunno
Sounds fairly hard
I guess it's doable in linearized GR
A cube has no killing vectors except for time
I mean, there's a reflection symmetry, but that's about it
I guess that means that there are no cross terms
It's some diagonal metric $$ds^2 = -f(x,y,z) dt^2+ g(x,y,z) dx^2 + h(x,y,z) dy^2 + k(x,y,z) dz^2$$
 
Hey, can I ask another time about something really basic that I can't get my head around
 
I think that maybe, the fact that it has some discrete rotational symmetry might mean that $f = g = h$?
 
@Slereah wow that really narrows it down
 
Since basically $xyz$ all play the same role
So if I had to guess I'd say $$ds^2 = -f(x,y,z) dt^2+ g(x,y,z) (dx^2 + dy^2 + dz^2)$$
 
3:59 PM
I understand Partitions, just imposing restrictions like "r = some constant" in a space with co-ordinates $(r,\theta)$, and then you're left with different sets of points for different r that, if Union'ed, give you the whole space.
But I still don't understand Equivalence relations
and it's really annoying me
 
where both $f$ and $g$ are symmetric wrt permutation of $x,y,z$
And the stress energy tensor is a cube
I guess solve for that???
Also 'course it's gonna be $\approx $ Schwarzschild for large $r$
And it should be $\approx$ that Newtonian solution
Considering what the Newtonian solution looks like, it's gonna be very ugly
That's not even considering the rotating cube, aka the Kerr cube, or krube
 
I need to start compiling a bibliography for this.
Get organized
 
@0celouvskyopoulo7 In the context of rotations is linear velocity the same as tangential velocity
(as opposed to angular velocity)
 
No, linear velocity should be the velocity of the center of mass
 
you can put Stephani on the pile
He's got a whole bunch of fluid spherical solutions
 
4:12 PM
@0celouvskyopoulo7 And what's the tangential then?
 
@BernardoMeurer $r\omega$, no?
I'm probably forgetting something, it's been over a year since I've done this.
 
Errr
Yes
That's reasonable
 
that is that, yes
the acceleration is the tough one
Wait, no
I forget
It's been a while
 
@Slereah Yeah I'm looking at a formula in my notes for acceleration
But it makes no sense to me
 
Show what formula that is
 
4:16 PM
@BernardoMeurer do you mean a point object rotating around something else, or an extended object like a rotating sphre?
 
@JohnRennie Like a sphere or cylinder, yeah
 
oh thank god someone who knows this has shown up
@Slereah Is Stephani expensive?
 
With an extended object the linear velocity is normally the velocity of the centre of mass i.e. the average velocity of the object as a whole.
I think that's what Ryan meant.
 
I remember Stephani being average price
 
@JohnRennie Yeah, I agreed with him on that much :P
 
4:18 PM
Like 60 bucks or so
 
Now what I can't understand is this:
 
But if you're talking about some point object in circular motion then yes the linear velocity of that point object is the tangetial velocity.
 
@JohnRennie who is Ryan?
 
@0celouvskyopoulo7 you might also want to check "Exact solutions and scalar fields in gravity"
 
your daddy, boy
 
4:18 PM
There's a lot of fluid stuff in it
 
$a = r\frac{d\omega}{dt}û_t + \omega^2 R û_v$
 
I don't remember why I bought this book
75 is alright
 
The hell are those u's
 
@Slereah Do you want to send it to me?
 
4:19 PM
Basis vectors
Nah
 
I mean it's not a great book
 
It say sthe first part of the sum is tangential acceleration
 
But it has neat parts?
 
and the second is centripetal acceleration
 
4:20 PM
yeah rotations have that thing where you need to also differentiate the basis vectors
So you get those weird terms
 
@BernardoMeurer OK. This is now a point object rotating around something.
 
@JohnRennie What does that formula mean?
 
Where $r$ is the distance to the pivot point.
 
I LEAVE FOR MY EXAM IN 10 MINUTES THIS I AM DYING
 
4:21 PM
To be fair I also don't remember why I bought Stephani
but I don't regret it
 
@BernardoMeurer are you happy with the first term?
 
@JohnRennie What are the funny u's
 
I need to buy "Friedrich, Thomas: Dirac operators in Riemannian Geometry" as well
God damn so many books
 
$\hat{u}$ is the unit vector of velocity. It's just there to give the direction.
 
@BernardoMeurer unit vectors...
 
4:22 PM
Ah! Okay!
 
So $\hat{u}_t$ is the tangential component of the unit vector.
 
Okay, yeah, the first term makes sense
 
Without fail
 
Well, what else are you gonna put in
it's like if you're writing a book on Christianity and don't put the bible in the bibliography
 
@Slereah Here's the problem. Let $M^n$ be a compact $\partial$-manifold, $1\le p<\infty$ and $f\in C_c^\infty(M)$ (that is, $f|_{\partial M}=0$). Is there a constant $C>0$, independent of $f$, such that $$\int |f|^p\le C \int |\nabla f|^p?$$
 
4:27 PM
Okay everyone
Exam time!
Thanks for the help these days!
See you all later
@0celouvskyopoulo7 Do you have time to explain me manifold parametrization today?
 
Probably?
 
Cool
Wish me luck
See ya
 
Good luck!
 
gl
quavo be with you
 
@0celouvskyopoulo7 Not a clue
 
4:39 PM
Anyone know a journal where one could send a short paper showing how to process a particular type of data?
No new science, just "If you take this data, here's how to process it".
 
Specifically science or just any old type of data?
 
It's a very particular type.
 
Hm
I dunno
I guess there's some physics data analysis journal
There's a journal for every field
 
If you have a sequence of 0's and 1's coming from a quantum two level system, this processing method pulls out fluctuations in the system parameters.
 
it's true!!!
 
4:42 PM
is this relevant?
 
functional analysis is so useful
truly the best math
 
@Secret Could be.
 
In other news, I think I need to go to sleep. Not eating lunch plus sleep deprivation due to work have caused my brain to be not really functioning
 
Also it is finally the week end
Back to writing GR
 
4:52 PM
I should probably decide on the convention to use for the Riemann tensor
 
I need a lemma for the rigorous proof
It's clear for smooth functions if that if $f=0$ on the boundary and $f=\mathrm{const.}$, then $f\equiv 0$.
But that's not clear for Sobolev functions.
Conjectured lemma: If $f\in H^{1,p}_0(M)$ is constant a.e., then $f=0$ a.e.
Hmm. Why wouldn't I do something with a bump function to avoid that...
Push it out from the inside
Smush up against the boundary
 
and what if we drop $\text{div}\, K=0$?
ie, a conformal Killing vector
 
wot
That's some string theory shit
 
$\nabla_{(\mu}K_{\nu)}-\frac{2}{d}g_{\mu\nu}\text{div}K=0$
I get $\Delta K_\nu+R^\mu{}_\nu K_\mu-\frac{2}{d}\nabla_\nu\nabla^\sigma K_\sigma=0$
 
ask @ACuriousMind
 
5:00 PM
nah I dont like that guy
 
rip
There's nothing worse than doing some Sobolev thing and then remembering everything is wrong because Sobolev functions are god awful
 
Oh boy
 
I got an answer from Сергей Красников
 
Krasinov!?
He's alive!?
 
5:03 PM
He confirms that the star operator is indeed a relation and that $A \star B \neq A$ means indeed $A \star B$ and $B \neq A$
He also adds "Anyway, I would discourage you from wasting you time on reading this part of the paper. All worthy was extracred from it, reformulated, clarified, and published in arxiv.org/abs/1408.6813 "
 
Ahhhhhhhhh
THE POINCARE INEQUALITY IS TRUE ON MANIFOLDS WITH BOUNDARY
 
He signed off his email with "С уважением"
Is that some russian polite phrase
apparently it means "yours faithfully"
I'm on a roll
Quick, who's another famous GR guy I could email!
 
Hitler is not a GR guy
 
Ask Wald about the proof of Lemma 8.1.
 
5:05 PM
I bet he's still alive
 
GR is Jewish science
Was Riemann an Jew?
 
I don't think so
He has a very German name
 
Lutheran
 
David Hilbert also wasn't, despite being called David
I could send an email to Harold White
Asking him what the fuck his experiment is supposed to be
you crazy crazy man
 
Hii @JohnRennie
 
5:08 PM
What are you talking about
 
@Koolman Hi
 
@JohnRennie today I have a physics question
 
Harold White is the guy who claims to be running an experiment on the warp metric
Despite calculations showing that it's impossible to test for with current technology
The fun thing about the Alcubierre metric is that Alcubierre is the guy who seems to care the least about it
He did the original paper on it and I think he never made another after that
He just went back to doing cosmology papers
 
@JohnRennie
 
5:13 PM
@Slereah If Harold White is a crackpot (which is debatable) then at least he's a nice crackpot.
@Koolman that looks straightforward enough. What bit of it is giving you problems?
 
you know who else seems insane?
Eric Davis
He has a Thing with aliens
 
(4x7) - (7x5.5) - (2x1) . @JohnRennie I think this should be the answer
 
that is what drives his wormhole research
 
@Koolman almost ... remember that two helium nuclei are formed.
 
But then energy is absorbed @JohnRennie
 
5:19 PM
wait, no
it's not true
I hate this
 
My favorite sentence in a Davis document :
"Puthoff’s polarizable vacuum general relativity model is the only alternative theory of gravity that has been successfully applied to explain the physical, anti-physical and physiological characteristics & performances of UFOs"
 
^ That's a riot.
 
@Koolman Binding energy can be a bit confusing. Remember that the binding energy is the energy released when the nucleus is formed. That it, if we take a proton and a neutron and bind them together to make a deuteron then it releases 1MeV per nucleon.
 
The binding energy conventions so regularly confuses people that it should could with big, neon, blinking warning-sign that says "You're not going to like this, but stick with it and do it our way."
 
@JohnRennie but mathematically binding energy should be negative for the energy to release
 
5:22 PM
@Koolman Imagine taking the deuteron and lithium nucleus and splitting them up into individual protons and neutrons. To split up the deuteron takes 1 x 2 = 2MeV i.e. we have to put in 2MeV.
 
Can anyone please just help me with one thing? Would the equivalence class that describes the partition $D$ of a space, where $D$ is the set of all circles, be $E = [x,y \subset AxA \vert \text{x and y are the same distance from the origin}]$
whats the \ for element of?
I didn't mean to use subset rip
and A X A is meant to be the cartesian product
 
@JohnRennie ok
 
@Koolman To split up the Li takes 5.5 x 7 = 38.5MeV i.e. we have to put in 38.5MeV
 
k
 
So we've put in a total of 40.5MeV to dismantle the starting nuclei into protons and neutrons.
 
5:24 PM
yes
 
The question is how much energy do we get back if we now let the protons and neutrons assemble into two He nuclei?
 
yes
 
And the answer is ... ?
 
56
 
So what we're saying is that we put in 40.5MeV and got back out 56MeV so we ended up with 15.5MeV more energy than we started with.
Hmm, the first time I did this I got the answer 16MeV, which is option (B). I've lost half a MeV somewhere.
 
5:28 PM
actually the answer is also 16 MeV
I think there would be typo
 
OK, I'll have to figure out where the 0.5 MeV went, but do you see the basic idea?
 
yes
 
by the way my science folder has a subfolder for crazy papers
they're all great
 
@dmckee too right. It confused the hell out of me in the innocent days of my youth :-)
 
Transdimensional unified field theory is very annoying because it's by a crank that used to be on a chatroom I was on
 
5:32 PM
@dmckee I sometimes think the best qualification for being a teacher is having made all the mistakes yourself!
 
@JohnR: How goes it?
 
Hi!
 
Hi :-)
 
@Kaumudi.H Hi. I'm just about to collapse for the evening. I've cycled miles today!
 
Ah, OK :-)
 
5:37 PM
But that's OK. It's a bank holiday weekend in the UK so that's three days off :-)
Monday is a holiday.
 
Wow. Nice!
Any special plans?
 
One day I should read the full Davis crazy document
"Teleportation physics study"
 
@Kaumudi.H no, not really. I'll see how I feel on Monday and what the weather is like.
 
@JohnRennie Oh, wokay...
 
5:39 PM
@Kaumudi.H there is a shop near me that has an offer on Indian (ready made) meals.
I might have an Indian feast on Monday.
 
Ooh. Are u planning on getting something? U seem to love offers :-P
 
@JohnRennie I'm teaching thermal physics this semester and the parts I am most successful at teaching seem to be the bits that made me cry in the misty days of yore. The few bits that came easily have been the ones I've struggled to convey clearly.
 
@JohnRennie :-) Nice! Do tell me what u buy and all.
 
So it goes.
 
@Kaumudi.H tomorrow I plan to have carrot cake with icecream.
For dessert - that isn't the whole meal :-)
 
5:42 PM
Dammit Sir, didn't I ask you not to tell me about delicious food while I'm out here?
 
Oops :-(
@Kaumudi.H oh well, I'll try to keep the really delicious meal for Monday, by which time you'll be home.
 
I will :-)
I should also be heading off to bed. See you tomorrow, bye!
 
Goodnight
 
I need a different sobolev inequality
my current plan of attack is doomed
this is awful
 
Why is this in MTW
 
My face right now
 
@Slereah Can I write part of your book?
 
Well you probably helped a fair bit on it already :p
 

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