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01:00 - 22:0022:00 - 00:00

1:57 AM
@JaimeGallego The people is with him. Although I can't see a really dangerous external enemy which could make their this decision rational.
@JaimeGallego Turkey is only around 1/4 in size of the US and they aren't very strong in English. I don't think too many plagiarizer would be from there.
 
 
3 hours later…
5:03 AM
John Rennie! Good morning sir! :-)
@JohnRennie I had an idea. Maybe if BEC could be constructed in the sizes which is enough to measure gravitational effects, it could be used to experimentally test QG.
 
@peterh Such a system might be useful for testing gravitational decoherence, but I doubt it would tell us much about QG. Mind you, just being able to test gravitational decoherence would be interesting.
 
5:21 AM
morning.
 
Morning :-)
 
@JohnRennie I was supposed to go to bed an hour ago, but well, here I am
Stupid sobolev spaces
Trying to prove
$$\inf_{f\in C^\infty_0(M)-\{0\}}\frac{(\int |\nabla f|)^n}{(\int |f|^{n/(n-1)})^{n-1}}>0$$
I think I can prove it
Of course, the point is to assume it's zero and prove a contradiction.
With some Sobolev space magic I think I can do it
 
5:48 AM
Hi, everybody.
 
@heather @BernardoMeurer I designed a working quantum computer
2
 
I was about to say "isn't that your job"
 
6:21 AM
@peterh Nah it's not a pseudoscience. It's just a really confusing philosophy.
Marx never talked about dialectical materialism himself; it's a Soviet interpretation.
 
6:57 AM
Your marble experiment does not violate Bell's Theorem (or any other theorem of classical probability) and is therefore quite entirely irrelevant to the issue here. The wave function does not count as a hidden variable because it cannot *determinstically predict the outcomes of all measurements. In your marble experiment, we can easily explain the result by invoking some determinstic (though mysterious) force. — WillO 15 hours ago
Ok that's annoying, seems there is no easy way to illustrate bell inequlity except that magic square example
2
Q: How to explain in simple terms why Entanglement is more than just complicated hidden variables

Doug CoburnI haven't taken a graduate level physics course on quantum mechanics so I get lost in the strange looking equasions. It's hard for me too see in any of the explanations of how quantum computers and entanglement work without thinking that it looks like a big parlor trick. It's like a magician ta...

I wonder if this will help, or I should just check the maths again and try to make sense of it
At this stage, it is clear to me that entanglement is a correlation, but whether it is possible to say anything more detailed than that I am not sure yet
WillO's explanation suggest entanglement cannot be thought of as a sample space of all possible correlations between the outcomes of A and B assigned to some probability distribution. Somehow I must have accidentally invoked hidden variables there...
 
Happy birthday @BernardoMeurer :-)
6
 
 
2 hours later…
9:04 AM
hey
 
9:38 AM
Proper way to do thermodynamics
 
What is the stress energy tensor analogue in thermodynamics that bent the energy landscape?
and what is the metric signature. Surely it cannot be $\pm (-1, 1, 1, 1) $?
 
There is none.
And the metric is Riemannian.
 
Ok, so the first term in G is basically the change in entropy (although it might took me some reading to understand where those minus signs for the partial derivatives came from), and the second term seemed to be some kind of cross term
So... since it is riemannian, I suppose there is only one type of geodesic?
What is the link to this pdf you are reading?
 
@Slereah Didn't know thermodynamic metric was a thing
 
it's called geometrothermodynamics
And it is absolutely awful
It uses contact manifolds
 
9:51 AM
ack
 
10:11 AM
Can anyone provide me slightest of a hint to progress in this question physics.stackexchange.com/questions/329958/…
 
Thanks for posting this. I have flagged it for closing for being a homework question.
 
But I have shown my attempt.BTW Homework questions are not banned on PhysicsSE
@Slereah thanks for not helping me.
 
No problem ( ͡° ͜ʖ ͡°)
 
$$\Huge{\int_{\sum_{k\in S}{}^kN} ( ͡° ͜ʖ ͡°)^{ ( ͡° ͜ʖ ͡°)^{( ͡° ͜ʖ ͡°)^{( ͡° ͜ʖ ͡°)}}}}d[?]$$
ugh, the eyebrows are still off...
 
user228700
@JohnR: Yello! :-)
 
10:26 AM
Morning :-) Safely home?
 
user228700
Yep!
 
Conclusion: Mathjax wil screw up Lenny Faces
 
user228700
Hadn't turned it on. My reaction when I did: Oh God...
 
@navinstudent see this answer of mine for why your question is likely to be closed, whether you have shown effort or not.
@Kaumudi.H Oh yes! :-)
 
user228700
:-) What's for lunch?
 
10:37 AM
Posh lunch today. Chicken parcels.
 
user228700
Ooh, nice! :-) And ah, carrot cake with icecream for dessert?
 
user228700
Or was that yesterday hmm...
 
I wonder how delicate the balance is, cause if there is even one leg that is subjected to uneven weight, the whole configuration will fall apart (as shown in the end of the video)
 
@JohnRennie Filling the parcels yourself or did you buy them pre-made?
 
Steamed buns not parcels - I had misremembered what they are called.
I have my doubts about how nice they will be, but they were on offer so I thought I'd try them.
Sadly I have eaten all the carrot cake and ice cream, so I'll have to find something else for dessert today.
 
user228700
10:43 AM
@JohnRennie Ah, OK :-/
 
user228700
What ice cream was it, BTW?
 
Buns has a rather different meaning in American slang. Steamed buns sounds like something you'd have to go to hospital for :-)
@Kaumudi.H Mint choc chip!!!
 
user228700
Oh, wow, nice! :-)
 
I love mint choc chip ice cream.
 
user228700
@JohnRennie Haha, yes x'D
 
user228700
10:46 AM
@JohnRennie Sadly, I've never tasted it :-(
 
What flavours of ice cream do you get in Chennai?
 
I will never understand the Brits' love for mint :P
 
@ACuriousMind really? Doesn't everyone love mint?
 
user228700
All of them, I bet! I'd have to visit Baskin & Robbins or iBaco or something to find out if they have mint chocolate chip...
 
@JohnRennie I strongly associate it with "toothpaste" and I can't think of any flavour except bubblegum flavour where I find it acceptable
 
10:48 AM
@Kaumudi.H then buying mint choc chip ice cream seems like an excellent project for a Sunday afternoon :-)
 
user228700
@JohnRennie :-) Ah, but I have so much to do! :-( I'll go after my exam!
 
@Kaumudi.H I guess it's nose back to the grindstone until the JEE Advanced exam is over.
 
when is the JEE anyway
 
user228700
Yes! :-) Eek, nose to grindstone sounds terrible.
 
10:49 AM
Seems people have been preparing it for months
how hard can it be
 
Do you want to go to an IIT? We tend to assume that all Indian students want to go there but I guess that's not necessarily true.
 
i don't
 
user228700
@JohnRennie I...will talk to you about this after my exam :-P
 
@Kaumudi.H I think it means you're bent over with your face close to the grindstone while you're working. I don't think it means you're actually grinding your nose off :-)
 
@Slereah We get a lifetime to prepare for death, so why not a month for JEE.
 
user228700
10:51 AM
@JohnRennie Ah :-P
 
@BalarkaSen I never prepare at all for death
I will just wing it
 
subconsciously you are #deepstuff
#Freudoncoke
 
One of these cases where I'm not sure I want to know why OP wants to know that
 
Strangely enough I have a paper perfectly adapted to that problem
"Mechanical Properties of Biological Materials"
 
Actually I shouldn't say extreme stuff like that on this chat
 
@Slereah Slightly interesting but once again I worry why someone wanted to find these values...
@JohnRennie I'm aware my distaste for mint is not a universal German trait ;)
 
@ACuriousMind I just had a lot of bones in my basement to destroy
For innocent reasons
 
@Slereah that's an extreme way of dealing with people who use the +--- signature
 
how do you destroy bones innocently
 
With a hammer
I'm sorting my science folder
Where do I put "THE SOLAR MOTION THEORIES OF BABYLON AND ANCIENT CHINA"
Physics or humanities
 
11:04 AM
History of science?
 
history of physics probably
Ok, I gotta go.
 
I have a whole book on the sine function in my folder
It's called "The remarkable sine function"
how do you even decide to write that book
 
user228700
Geez.
 
Acuriousmind: The mention of orthopedic instrument might suggest the asker might be a surgeon or something similar. Of course, the info on that question is so limited thus your worries are still justified
 
@Slereah By finding the sine function remarkable and deciding to write about it? :P
 
11:12 AM
user image
2
Suppose the analogy holds, I wonder what is an entire city of people computing...?
 
@ACuriousMind it seems like a fairly mundane thing to write an entire book about
 
If stupid people are so ubiquitous, then surely we could just model them using some average 'potential term', analogous to how in electrodynamics, we model a bunch of charges as some averaged electric potential for some calculations...
Put it in another way, perhaps it might be possible to model the collection of stupid people (stupid as defined in this article) as some background
 
There are no stupid people. That's a over simplification made by ... err ... stupid people. Humans invariably act in ways for which they have a motive that seems good to them. You may not agree with their motives but that doesn't make them stupid.
 
Yeah, and there's also the issue of context, an action can be good for A but bad for B while seemed nothing for C and so on
One thing that seemed clear though, people in groups tend to act in unison. For example:
 
@Slereah It's "The remarkable sine functions"
Not physics. Not even mathematics. Not verified; not even wrong.
 
11:30 AM
"We now want to show that vector fields on a smooth Hausdorff manifolds are the same as derivations on its algebra of functions."
Great sadness
And that's from a paper that talks a lot about non-Hausdorff manifolds
 
11:41 AM
Let's look through the proof to see where the Hausdorffness is important
 
@ACuriousMind Where's my birthday gift?
@JohnRennie Thanks John!
Celebrating listening to my AMAZING STEREO
 
@Secret Did you buy me anything for my birthday?
@AccidentalFourierTransform You?
 
Well, I cannot I possibly deliver you things right on this day given I am living in the opposite hemisphere and delivery will take nonzero days
 
hey Happy birthday!
Hitler died today
he died for our sins
 
But anyway, happy birthday Bernardo.
 
@Secret You would find away if you loved me like you said you do :P
Thanks though :D
@AccidentalFourierTransform Cheers mate
He died from his sins
 
12:11 PM
@BernardoMeurer I'm thinking of buying this to use on my computer. It's just small enough to go on the window ledge and the reviews suggest it's real hi-fi quality sound.
 
@JohnRennie Hmm, I've never owned a sound bar, but monitor audio does some really quality stuff. I considered their speakers when buying
 
12:31 PM
@JohnRennie I have linux running on an FPGA right next to me :P
Are you proud?
 
Happy birthday!
 
1:37 PM
@Slereah I never noticed this on spivak
 
The monkey on the tractor represents the connection on the bundle
 
2:04 PM
It's largely irrelevant to the main question, but your claim in the first point 2 is incorrect. Applying a magnetic field will not entangle the particles because the hamiltonian is a sum of single-particle operators. — Emilio Pisanty 21 mins ago
Ok, that means I need to reread how a CNOT gate is implemented
It is possible I am missing some important detail when computing a transformation on state in $\mathcal{H}^2\otimes \mathcal{H}^2$
 
2:36 PM
@JohnRennie hello
@Yashas hi
 
2:53 PM
 
Hey guys .... I just want to clarify one of my doubts
the hydrostatic pressure just depends on the depth and not on the amount of liquid right?
i am pretty sure that this is the case but a question on this site claims that this is not true
 
Define "amount of liquid"
I mean if you have more liquid above, the depth increases, certainly
 
i.e. the pressure 200m in the ocean and at the bottom of a 200m pipe
they will be same right
 
yes, it should be the same
 
Yes
 
3:06 PM
just look here
Consider this: with gases the force exerted is a function of internal pressure. Why? Property of compressibilty. With spheres against the wall the force is a fubction of the force exerted by the entire mass. Why? Noncompressibility. Now consider liquids. With respect to exerted force they have characteristics of each, differing from the spheres only in that liquids are frictionless. — john 2 hours ago
 
0
Q: Given a wall on planet earth with a constant height is there a point at which a physicist needs to worry about the volume of water behind the wall?

deletingI have a quick question, on the media outlet "Sploid" a post recently came out about this interesting flood wall in Austria that keeps water from destroying a community. The wall seems to contain a massive amount of water but the wall itself looks pretty flimsy. I was wondering if there is a poi...

in this question .. i feel like one of the answers is completely wrong
irrelevent
 
3:33 PM
Morning :)
 
3:45 PM
"First they added in Art to make STEM into STEAM. Now they're adding in Reading to make STEAM into STREAM."
noooo
Bloody Huffington Post
It became a real rag
 
4:06 PM
I'm a STEM person and I have no clue how to read.
Reading is important
 
it probably was a poor idea to call it "reading"
Reading is only an achievement if you are special needs
Literature would be better but I guess STLEAM sounds racist
 
user228700
4:48 PM
Did I tell you, @JohnR, that the headphone jack on this thing is kaput?
 
@Kaumudi.H no. Has it always been bust or did it originally work then stop?
 
user228700
No, I think I may have done some damage yesterday :-/ It used to work just fine before.
 
user228700
Can it be repaired? (I'm guessing no...)
 
I'm not sure whether it can be repaired, but it certainly can't be repaired easily.
 
user228700
Oof. Oh, well...
 
4:51 PM
I suspect the jack is mounted on the motherboard, so replacing it would mean replacing the motherboard and that's a major operation.
 
user228700
Ah, OK...
 
You could use USB headphones with the laptop, though obviously you'd have to buy some USB headphones.
 
user228700
USB headphones?! Didn't know that such a thing existed. Nice. I'll take a pair to college just in case...
 
I'd have to do a bit of research to suggest some you might get. I know they exist because some phones use them. Or you could use Bluetooth hreadphones.
BTW this was lunch:
 
user228700
^ Yes, I was just about to say that I might consider investing in a pair of Bluetooth headphones.
 
user228700
4:55 PM
Ooooooooh, pretty! :-)
 
The weird things that look like some kind of white sea urchin are the steamed buns.
 
user228700
Yes yes, I remember them from the picture of the box.
 
They were rather odd. Not unpleasant, but an odd texture and I don't think I'll be buying them again.
 
user228700
And the other things are toasted cheese rolls? :-)
 
Yes
 
4:56 PM
@JohnRennie I'm writing that app in Rust
I think it will be a nice learning project
 
user228700
@JohnRennie Ah, hmm, OK...
 
@BernardoMeurer which app?
 
@JohnRennie Musync
 
user228700
Ooh, neat name!
 
Ah. OK. Yes, it's a fairly straightforward app but with a lot of features, so it's a good candidate to use for learning a new language.
 
4:58 PM
@JohnRennie Exactly my thoughts
@Kaumudi.H Thanks, I take pride in my naming :P
 
@Kaumudi.H it's an app the synchronises music, so the name is arguably not one of the great innovations of our time :-)
 
@JohnRennie Shut up john, the only other cool name is taken for the chat app I want to eventually make
also in Rust
 
user228700
:-P It's a neat name nevertheless.
 
@BernardoMeurer call it Appchat :-)
 
@JohnRennie Hahahaha
 
user228700
5:00 PM
x'D
 
It's called Oxide
FYI
 
Not obviously related to chat... Anyway isn't that a DJ's name?
 
I don't know
 
Oxide & Neutrino are a British DJ and MC garage duo from London, consisting of Alex Rivers (born 1982 in Isleworth, London) and Mark Oseitutu (born 1982 in Brixton, London). == Musical career == Their first single "Bound 4 Da Reload (Casualty)" entered at the top of the UK Singles Chart in April 2000. It is known for sampling the theme music to the BBC medical drama series Casualty, and also contains samples of dialogue from the 1998 film, Lock, Stock and Two Smoking Barrels. Their music appears in the film Ali G Indahouse in the form of the track "Shoot to Kill", which is played in the movie from...
 
I just think it's neat since it's written in Rust
And that's the only chemistry I know
 
5:02 PM
Ah :-) oxide - Rust. Yes, how did I fail to see that one coming :-)
Do you know Manishearth? He's a mod here and I think on Chemistry. He is very keen on Rust and has blogged lots about it.
 
@JohnRennie I know who he is but I have never talked to him
 
5:14 PM
Was your birthday on February 9, @John?
 
why does my science folder have a book called "Sex and the Intelligent Women"
I am intrigue
"Sex and the Intelligent Woman, for several important reasons, can truly be termed unique and unparalleled. For one thing, this is the only volume presently available that, in a scientific and objective manner, deals specifically with the sexual attitudes, desires, experiences, and practices of women of high intelligence, measured by I. Q."
Well, that's that I guess
 
@John If so, we've got a birthday paradox here :D. I was also born on Feb 9.
 
When it is time for one to review their folders full of their 20 something years of life, one is bound to found something only their past selves will understood
 
5:37 PM
the science folder has about 28000 files
wew
6000 in math and physics
Oh boy it's that paper from Bozhidar Z. Iliev
 
apparently there's flying snakes
 
"Lattice QCD in curved spacetime"
What madman did this
 
@Slereah For $M^n$ a compact manifold with boundary, define the $\alpha$-isoperimetic constant by $$ID_\alpha(M)=\inf_{\Omega}\frac{A(\partial \Omega)}{V(\Omega)^{1/\alpha}},$$ where the inf is taken over all subdomains $\Omega\subset M$ with the property that $\partial\Omega$ is a hypersurface not insecting $\partial M$.
Claim: $ID_\alpha(M)=0$ for $\alpha>n/(n-1)$.
I think the proof should use normal neighborhoods
 
5:53 PM
Don't they always
 
It's true for $\Bbb R^n$ by looking at balls
 
is $ID_\alpha$ supposed to be some measure of the "radius" of the set
 
So it should be true in arbitrary manifolds by looking at geodesic balls as $r\to 0$
Because those are $\approx $ regular balls
@Slereah It measures the ratio between the volume and the area of the boundary, so a measure of the "roundness" of the set
the $\alpha$ is just a technical thing, $\alpha=1$ is the important one
and it's never zero
(I hope, I think I have a proof of that.)
 
one of the file is just titled "spooky"
very concerning
I assupe it's some EPR thing
 
or a ghost thing
-1
Q: Finding a function whose double covariant derievative is delta function

Anupam AhI have a function $f$ defined as $\frac{\overline{z}-\overline{w}}{(z-w)(1 + z\overline{z})(1+w\overline{w})}$. I want to prove that $\partial_{\overline{z}}^2 f + \frac{2z}{1+z\overline{z}}\partial_{\overline{z}}f = \pi \frac{2}{(1+ z\overline{z})^2}\delta(z-w)$. This is basically a double cov...

 
6:00 PM
"QFT as pilot-wave theory of particle creation and destruction"
 
@Slereah for you
 
Lubos repellent
 
6:16 PM
"How to (Path-) Integrate by Differentiating"
And mathematician repellent
"Spacetime is not hermitian"
what
oh, it's about complex structures on spacetimes
 
6:40 PM
23
A: Is 0.24681012141618202224... transcendental?

José Hdz. Stgo.In point of fact, K. Mahler proved in this paper that, if $p(x)$ in a non-constant polynomial such that $p(n) \in \mathbb{N}$ for every $n\in \mathbb{N}$, then the number $$0.p(1)p(2)p(3)p(4)\ldots,$$ which is formed concatenating after the decimal point the values of $p(1), p(2), p(3), \ldots$...

whoa
> ... (in that order), is a transcendental and non-Liouville number.
 
The more I see questions closed due to being Homework related the more confused I get over what conditions make a question closable
 
@Phase unfortunately, due to problems with formulating close reasons, homework is overused for anything felt to be unsuitable
so ... it often is not homework but just a close-worthy question
 
It's definitely a homework question but I feel physics.stackexchange.com/questions/270414/… asks a specific concept, this concept being rotational mechanics, and there's clearly effort there
and that seems to be the main statements that could qualify a post as decent in the bit of text about it being closed
 
Writing tip : never put something like "the myth of X" for the title of your paper
I am instantly suspiscious of such titles
 
The Myth of Pair Production
By John Duffield
 
6:55 PM
The John Duffield part is also a big red flag
 
IIRC he claimed Riemannian manifolds didn't exist
I'm not really sure if he was being ironic or not.
 
well, have you seen one in the sky
 
@Phase there was discussion about allowing reasonably well formulated, well written questions showing effort, but that discussion has, I fear, been forgotten. In general, raising a specific example with numerical values is a red flag for many reviewers. But I agree with you that this question is a bit borderline (if not nice to read)
@0celouvskyopoulo7 how can a mathematical object "not exist"? :o
 
@Sanya ...you're questioning JD's logic?
He also claimed it was futile to prove Stokes theorem on a manifold because I wouldn't be able to measure the boundary or something like that
 
6:59 PM
@Sanya Yeah [it was definitely ugly to sift through though, the diagram was quite a blight on the eyes]
 
I should understand probability theory one of these days
 
@Phase the diagram is fine even if it's not very beautiful; the structure of the question, meaning the structuring of the work for the reader, is a bit lacking I think
in any case, feel free (actually, please!) take this to meta :)
@0celouvskyopoulo7 if you know integration theory, shouldn't that be enough? :D
 
@Sanya make a meta post? I haven't really done that before, besides, aren't there duplicate rules there? I feel like making a post about something probably already posted about will just be flagged
 
@Sanya Ah, well in that context it was the divergence theorem, and there's a subtlety regarding the orientation of the normal vector on the boundary.
 
@0celouvskyopoulo7 that's really a nice statement. A bit like "no reason to work with real numbers, cause we can't measure infinitely many points between two points" :D
 
7:02 PM
@Sanya Ooops.
 
@0celouvskyopoulo7 my statement was about probability theory ;)
 
Wrong message.
 
@0celouvskyopoulo7 :D
 
@Sanya I mean the martingale theory.
 
@0celouvskyopoulo7 ok. That's a different story I fear
@Phase the homework question is an open one, and asking for an explanation for how the guidelines are at the moment employed and thus raising a general issue ... that should be fine
the worst thing is that someone shoots it down, but I actually doubt it
 
7:04 PM
Ok thanks, I'll make a post I suppose
 
"Eigenvalues and the Kinematic Measure"
What on Earth am I reading
 
@0celouvskyopoulo7 link? :D
 
How do I embed like [link]
 
@Phase [description](url)
 
@ACuriousMind are you here?
 
@0celouvskyopoulo7 that's an interesting use of "kinematic"
 
Did you look at the definition?
 
@0celouvskyopoulo7 yeah
I discovered my current university has the ebook access
 
Idk what it has to do with kinematics
But it's useful for analysis so whatever
 
7:13 PM
@0celouvskyopoulo7 me neither ...
 
@Sanya Ah
It's the measure that's invariant under geodesic flow
 
@Phase I like it, I think it is short and understandable and I'm curious to see the replies
 
So it's related to the Lagrangian version of Liouville's theorem in mechanics
(the theorem that the symplectic form is invariant under Hamiltonian flow)
 
@Sanya thanks for answering my questions and helping me with the post : p
 
you've lost me with the geodesic flow - never did GR; but well, I guess that sounds useful
@Phase you're welcome
 
7:16 PM
@Sanya geodesic flow: push things along the geodesic lines
in Hamiltonian mechanics you push things along the solution curves of Hamilton's equations
 
my problem was the "geodesic" part :D but ok, seeing that (if I remember correctly) in GR, the geodesic is the path taken by a particle, then there is a clear parallel of this to Liouville ... that's ... nice, I guess
sometimes I think having done theory would've been nice too :D
 
@Sanya yes
although this has nothing to do with GR
 
1
Q: Homework Policy and what justifies closure

PhaseReading through questions on the Q&A Stackexchange, I came across this Question regarding rotational mechanics. I'd been confused a few other times in the past by homework closures and had come to the conclusion that to merit closure under the homework flag the question would have to show a clear...

 
A free particle in Lagrangian mechanics also follows a geodesic.
 
but with a different metric?
 
7:20 PM
Do you own Arnold?
 
Why do I have a chinese version of Euclid's elements
 
@Sanya A Riemannian metric, yes. One that describes the constraint surface.
$\mathfrak SM$ is terrible
who uses fraktur S unironically?
$\mathscr {SM}$
$\mathscr SM$
 
It is somewhat odd
 
@0celouvskyopoulo7 Arnold ... I think I've got access, yes. Why?
 
What is $\mathscr SM$
Sphere bundle?
 
7:22 PM
@Sanya If you're interested I can tell you which section to look in.
@Slereah ya
Where's my nice cursive S, Latex?
 
@0celouvskyopoulo7 found it; tell me
 
page 247
 
thanks
 
7:29 PM
I'm a little sad that chinese people don't use chinese symbols for mathematics
I want to see something like $$\frac{\partial 意(形)}{\partial 形}$$
 
You can read chinese fluently?
Meanwhile, the "partial" derivative you wrote will probably mean something more complicated than tensors, cause you are asking the change in some function called "meaning" as the shape changes
And shapes can be something as complicated as manifolds or possibly more complicated
Computing that is probably much harder than a variational derivative
 
7:48 PM
I cannot
I just have random papers in various languages, sometimes
Mostly French, German and Portugese
a few chinese papers
some Italian and Russian
oh and some ancient greek and latin
but those tend to be fairly old
 
8:52 PM
The topological derivative is, conceptually, a derivative of a shape functional with respect to infinitesimal changes in its topology, such as adding an infinitesimal hole or crack. When used in higher dimensions than one, the term topological gradient is also used to name the first-order term of the topological asymptotic expansion, dealing only with infinitesimal singular domain perturbations. It has applications in shape optimization, topology optimization, image processing and mechanical modeling. == Definition == Let Ω {\displaystyle \Omega } be an...
 
9:49 PM
@Slereah So if a static spacetime is geodesically complete, are the hypersurfaces Riemannian complete?
 
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