12:01 AM
It is

3 hours later…
3:10 AM
Hi everyone !
Can someone tell me how to get an notification on my mobile whenever someone ping me on stack exchange chat ?
Is there anyone active on this chat room at now ?

5 hours later…
8:15 AM
i am amazed we live in a time where e.g. in-browser emulators for the DS or N64 exist. Makes me feel less bad about Flash dying
curious if this is the future of browser games

hey
I am thinking what "physically" the exchange operator does?
is exchange operator something that happens during scatterings of particles?

9:11 AM
what function does the firewall of a black hole serve?
by preventing information from falling into the black hole so that the information just leaves on the event horizon?

@CaptainBohemian the firewall idea was invented to deal with conceptual problems caused when one out a pair of entangled particles crosses the horizon leaving the other one outside.
It is on the speculative side of extremely speculative.
4

@JohnRennie does the firewall ensure the two particles, one outside and the other inside the black hole, to keep entangled?

No, it ensures the first particle never crosses the horizon and avoids the problem that way.

@JohnRennie by crosses, do you mean entering or exiting (the horizon)?

9:28 AM
@CaptainBohemian yes
The firewall means anything trying to cross the horizon into the centre of the black hole is destroyed

1 hour later…
10:37 AM
The whole firewall thing kind of depends on your quantum gravity theory
so it is kind of a hassle to deal with
Also presumably depends on the type of black hole

1 hour later…
11:41 AM
Yo @ACuriousMind
I need some help understanding this:
what the hell is this, why is it in google scholar, and why is it in the citing articles of the Ballantine paper?

uhhhhh
looke like very non-mainstream physics frmo a rather prolific author, but no idea why Scholar thinks it cites Ballentine
maybe the author managed to get their site indexed as a "publisher" like arXiv? :P

12:03 PM
@ACuriousMind yeah, who knows
maybe Scholar did it on its own?
let's see if they act on it

I don't know enough about biochemistry to judge, but the fact that $G$ is in there seems suspiscious
"binding of dark protons to formamido-pyrimidine"
Secret magic
Also every article in the bibliography was written by him
I'm guessing he's the foremost authority on dark biology
"DMT, pineal gland, and the new view about sensory perception"
He seems bananas

Hi everyone !

hello

Will someone look at the problem I am good ng to share ...
@Slereah hi
I think that the potential at centre and at B is same because there is no electric field inside the sphere. But the book says that the options A and C both are correct .

@Slereah have a look at his About Me on his website

12:17 PM
@EmilioPisanty I can't get over his webdesign choice
But then again, it's not much worse than actual physicist webdesign
He seems to have actual credentials
But I think he got the physicist disease bad

Did someone look at my problem ?

Sorry, not in the mood to do someone's homework

@Slereah I am not saying to solve it. I am just asking that should not the potential at centre and at point B be sam ?

12:37 PM
@ronakjain found this online, does this help?
actually it seems to depend on whether the charges sit at the surface or are uniformly distributed see here.

@Slereah but did you look what the book is saying. The books is saying that the potential at the centre and B are different ? And that's the problem . Why they are saying so ?

@ronakjain Please be mindful who you ping (address with the @). Slereah has already said he's not interested in the problem.

I @Charlie I know that very well. Problem is why the book says that the potential at centre and B are not equal ?

oh sorry that's a graph of $\vec E$, it seems like the potential is discontinuous at the boundary but the electric field is smooth except at the boundary

12:41 PM
@ACuriousMind that was just my mistake. I did not notice that

If that's what the book is telling you I don't know, I can't see what else is written
also my EM knowledge is patchy at best, maybe ask on the main site physics.stackexchange.com

Ok
Did anyone else try ?

Come on, you've already been told several times now that if people want to answer your question they will. The main site is where you should go primarily if you have questions

^that
There's also the problem solving chat room for such questions about exercises

Maybe we should open a problem solving business
Five bucks to solve

12:55 PM
You'd have to pay me more than five bucks to solve these exercises :P

If you can do 20 a day it's not bad
Although depends on the problem, I guess
If someone submits "What's a target space compactification for the standard model" I might have to bump it to $20 1:23 PM Suppose a wood cube, a plastic cube, and a steel cube all of the same volume are all submerged at the same depth in a container of water all of them feel same buoyant force why? the wood and plastic cubes float while the steel cube sinks Is the force of gravity on them not different? but isn't the mass different and$F_g=\frac{GMm}{r^2}$, no? @SpecterProphet "the wood and plastic cubes float while the steel cube sinks" or "all submerged at the same depth"? you cannot have both @FadedGiant yep 1:27 PM surely if the upwards buoyancy force is less than the force of gravity the net force will be upwards, and vice versa? unless I'm missing something I mean it doesn't depend on mass difference ? assmuning they have same dimensions what doesn't? three of them steel,wood.... so the buoyancy force depends only on the volume of the object? 1:34 PM so if all of the blocks are the same volume, the buoyancy force is the same what does change is the force of gravity they feel, the balance of the two then determines whether something sinks or floats at least that's what I can conclude well looking at formula I can conclude it to be true my book starts with A cube feels forces from the top from the bottom and from the sides. Their difference is the source of the buoyant force ( the total upward force that the cube feels in water) yes, mass doesn't enter into the buoyancy formula If the pressure is greater at the bottom of the object then the force applied to the bottom of the object is greater than the force applied to the top, so there is a net force upwards it just started with letters. may be it wants me to formulate it it's similar to saying that the force you apply to something is independent of its mass. if I push a car and a football with the same force they both feel the same force @Charlie wow that was straightforward why didn't i thought that 2 hours later… user434058 4:10 PM Just noticed that ja72 changed their username to John Alexiou. So there is no real difference between formulating Maxwell's equations as: $$\partial_\mu F^{\mu\nu}=\mu_0 J^\beta,\quad \partial_\mu\left(\frac{1}{2}\epsilon^{\mu\nu\alpha\beta}F_{\alpha\beta}\right)=0$$ and formulating them as: $$\text dF=0, \quad \text d^\star \bar F=4\pi ^\star \bar J$$ in other words exterior calculus is just some convenient notation for working with anti-symmetric tensors? those two sets of equations may be using different units, idk Your first sst of equations is just the second set, but in local coordinates So yes, there's no difference Is local coordinates also known under another name? I can't find a reference to it @Charlie What I mean is that$\mathrm{d}F=0$is a "coordinate-free" expression, but the ones with the$\mu$s and$\nu$s is one in a particular coordinate system. With "local coordinates" I mean a chart oh 4:23 PM In practice, if you have an$F$and want to check whether$\mathrm{d}F = 0$holds, you will compute that by picking "nice" coordinates and computing the expression from your first set of equations or you know that$F=\mathrm{d}A$and then it follows directly from$\mathrm{d}^2=0$, which is harder to see in coordinates [but not impossibly hard]. Sometimes the coordinate view is better, sometimes the coordinate-free view is better oh I see it's just going from the geometric object of the tensor to it's compoenents which requires picking a coordinate system yeah, exactly do we require exterior calculus to formulate maxwell's equations in component free notation? Or is it just considerably more compact than going to the trouble of formulating them in regular tensor form I'm still under the impression that it's just notational convenience, not anything particularly deeper than that What is "regular tensor form"? I think there's two different things to exterior calculus - one is the notation as you say ah I mean as in if we used$\vec u \otimes \vec v - \vec v \otimes \vec u$instead of$\vec u \wedge \vec v$4:30 PM but the other is things like$\mathrm{d}^2 = 0$and the Poincaré lemma for forms, which I think is more than just notational convenience and then you have stuff like deRham cohomology but that's probably not something you should worry about now I guess i'd then ask if things like$\text d^2=0$require new tools that exterior calculus provides or if it's really just a general tensor equation in very condensed notation the thing is that$\mathrm{d}$can only act on forms, not on general tensors ah of course the special thing about forms is that you have this exterior derivative that works without needing a connection contrast this with derivatives on generic tensors, where you need Christoffels to define a properly covariant derivative Forms have many properties that a generic tensor does not, and that's what exterior calculus exploits ok that makes sense does exterior calculus find much use in physics outside of electromagnetism? I mean I'm reading gravitation atm but from what I've learned in GR so far there isn't a whole lot of anti-symmetric tensors or any that I've seen for that matter, except the riemann tensor but even that's not totally anti-symmetric so I guess doesn't count 4:43 PM The generalization of electromagnetism, Yang-Mills gauge theories, are quite popular ah so scary qft stuff But as you already said, most things can be stated without explicitly using the language of "forms", so it depends a lot on the authors' preference how much of it you see in any given context I often like it much more to talk in terms of forms than in terms of components, but not everyone prefers the coordinate-free approach at least it follows relatively painlessly from having learned about tensors, which was a whoole thing 5:05 PM Exterior algebra has its perks Once you have the properties of the various operators down, it's easy enough to compute various identities if I've understood what I've just read, if I take a tensor space$T$, decompose it as$T=T_{a-sym}\oplus T_{sym}$, take the symmetric part and equip it with the exterior product, this is an "exterior algebra"? The antisymmetric part That's what the forms are ah yeah mistype this stuff is actually not that scary well, yet It can get, but the basic stuff isn't too hard Especially on flat space 5:21 PM Earth is flat, prove me wrong i mean, nasa is obsiously lying.. if earth was round, then how come water from seas doesn't fall... Basically you have to look at the wedge product, exterior derivative, the Hodge star and the interior product those are all the big ones @Charlie ur pic looks something from a game that I forgot the name of @JingleBells That's totally a Runescape character 5:49 PM @JMac I see you are a man of culture @JingleBells It's the default skin from the old version of runescape 6:04 PM @FakeMod please do not revert the edits to the spherical shell question again. The author and I have discussed it in detail and we are agreed on how we want the question to be. user434058 @JohnRennie Sure thing. I did that in hindsight, and will be more careful when doing it again (of course, for any other question, this one doesn't need to be reverted), if needed. 6:45 PM Does$\text d^2=0$have a name? @Charlie It's the chain complex Or... cochain complex? One of these I've seen it written twice now that$\text d^2f=0$for$f$, a 0-form, which seems to imply that this doesn't apply to higher rank differential forms, is this true? the wikipedia page for "chain complex" looks scary so I'll leave that for now lol It applies to all$n$-forms ok nice ty This is just due to 1) the antisymmetrization 2) the fact that derivatives commute 6:58 PM @Slereah Actually to this point, is$\text d\tilde\omega$necessarily zero for a$p-$form$p\geq1$? Why would it be? sry i forgot that edits ping, mb doesn't this imply that there are$(p\geq1)-$forms that can't be written as$\text df$? Yes, those are called inexact forms I mean just consider it in one dimension I might need to be clearer, if$\text d(\text d f)=0$, this seems to imply that only special$1-$forms that can be written as$\text df$where$f$is a$0-$form have vanishing second exterior derivative ohh this only applies to closed forms The$1$-form$\alpha = x dx$Errr Not that one @Charlie Well, for a$k$-form, there are two types of forms you can consider First, exact forms, which are such that, for some$\alpha$a$k-1$-form,$\omega = d\alpha$And then you have inexact forms, which cannot be written as such. This means that if a$k$-form is exact, then$d\omega = 0$but otherwise, it may not be 7:04 PM is$\alpha$necessarily exact too? Well no hmm actually that solves the problem I was thinking of if$\alpha$can be non-exact then it doesn't follow that all exact$(p\geq1)-$forms are necessarily zero. Since given an inexact$0-$form$f$we can take the exterior derivative$n$times until we hit an exact$n-$form at which point$\text d^2=0$applies Well, you can only take the exterior derivative twice hmm ok so for an inexact$p-form\text d^2\tilde\omega=0$and for an exact$p-form\text d\tilde\sigma=0$that's the bottom line and also that there are$p+1-$forms that can't be written as the exterior derivative of a$p-$form Yes 7:17 PM ok nice thank you If every$p$-form was exact, then there wouldn't be any$p$-form for$p > 1$Outside of zero, anyway ok good that's the problem I thought existed earlier @Charlie yes, Runescape A simple example : Take a$2$-form$f(x,y) dx \wedge dy$Either that form is inexact, or it is exact, in which case$\omega = d\alpha$But then$\alpha$itself cannot be exact So that would be an example of an inexact form the 2-form you've given, is that$\alpha$or$\omega$? 7:31 PM it is$\omega$bruh and your 3rd line follows from the fact that if$\alpha$were exact then$\text d\alpha=0\neq\omega$8:09 PM If$\alpha$were exact, then$\alpha = d\beta$, which means that$\omega = d^2 \beta$So that it can only be zero. 0 When i joined stackexchange I started answering questions and gained a positive response from the community. But these days, I don't feel like my answers help anyone anymore. If you look into my physics SE profile, you will find that almost 90% of my answers I have provided have neither been acce... Ok that makes sense thank you In mathematics, especially vector calculus and differential topology, a closed form is a differential form α whose exterior derivative is zero (dα = 0), and an exact form is a differential form, α, that is the exterior derivative of another differential form β. Thus, an exact form is in the image of d, and a closed form is in the kernel of d. For an exact form α, α = dβ for some differential form β of degree one less than that of α. The form β is called a "potential form" or "primitive" for α. Since the exterior derivative of a closed form is zero, β is not unique, but can be modified by t... It's pretty nice to have some light shed on what exactly the$\text dx$-like terms in integration actually are, been wondering about that for a while 8:25 PM I sort of think it's just defining away the problem, e.g. in$df = f'(x)dx$it's saying$dx$is a function in the tangent line to$f$at$x$giving the length of a vector in this tangent space, but that's what people are thinking anyway when they manipulate these things 9:14 PM Just so I'm clear, when we integrate say a 2-form over some region$\partial V$: $$\int_{\partial V}f(x,y)\text dx\wedge \text dy,$$ the surface over which we're integrating$\partial V$has to be writable in the form$\vec u\wedge \vec v$, we then contract$\vec u\wedge \vec v$with$f(x,y)\text dx\wedge \text dy$? and the resulting scalar is the value of the integral More or less yes, not sure about the$u \wedge v$bit, it's like saying a surface like$f(x,y,z) = x^2+y^2+z^2=c$has to be writable a$u \wedge v$Would$\langle g(x,y,z) \vec v\wedge \vec u,f(x,y)\text dx\wedge \text dy\rangle\$ be more correct?
I'm not sure how else we get the degree of freedom necessary to define a surface like that
that actually gives me another thought, this integral is just a single tensor contraction, right? Instead of becoming a weighted sum as it usually is introduced it's just a single coordinate independent scalar

9:38 PM
There's a nice easy/explicit example on page 5 here of integrating a 2-form over a surface which may help?

9:53 PM
hmm it's kind of clear but still a little advanced for me, i'll just keep reading, thanks though i'll save it