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ACuriousMind
ACuriousMind
12:01 AM
It is
 
 
3 hours later…
ronak jain
ronak jain
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…
Sir Cumference
Sir Cumference
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
 
Shing
Shing
hey
I am thinking what "physically" the exchange operator does?
is exchange operator something that happens during scatterings of particles?
 
Captain Bohemian
Captain Bohemian
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?
 
John Rennie
John Rennie
@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
 
Captain Bohemian
Captain Bohemian
@JohnRennie does the firewall ensure the two particles, one outside and the other inside the black hole, to keep entangled?
 
John Rennie
John Rennie
No, it ensures the first particle never crosses the horizon and avoids the problem that way.
 
Captain Bohemian
Captain Bohemian
@JohnRennie by crosses, do you mean entering or exiting (the horizon)?
 
John Rennie
John Rennie
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…
Slereah
Slereah
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…
Emilio Pisanty
Emilio Pisanty
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?
 
ACuriousMind
ACuriousMind
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
 
Emilio Pisanty
Emilio Pisanty
12:03 PM
@ACuriousMind yeah, who knows
maybe Scholar did it on its own?
I just posted a complaint on the Contact Us form
let's see if they act on it
 
Slereah
Slereah
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
 
ronak jain
ronak jain
Hi everyone !
 
Slereah
Slereah
hello
 
ronak jain
ronak jain
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 .
 
Emilio Pisanty
Emilio Pisanty
@Slereah have a look at his About Me on his website
heady stuff
 
Slereah
Slereah
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
 
ronak jain
ronak jain
Did someone look at my problem ?
 
Slereah
Slereah
Sorry, not in the mood to do someone's homework
 
ronak jain
ronak jain
@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 ?
 
Charlie
Charlie
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.
 
ronak jain
ronak jain
@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 ?
 
Charlie
Charlie
see the link I posted
 
ACuriousMind
ACuriousMind
@ronakjain Please be mindful who you ping (address with the @). Slereah has already said he's not interested in the problem.
 
ronak jain
ronak jain
I @Charlie I know that very well. Problem is why the book says that the potential at centre and B are not equal ?
 
Charlie
Charlie
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
 
ronak jain
ronak jain
12:41 PM
@ACuriousMind that was just my mistake. I did not notice that
 
Charlie
Charlie
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
 
ronak jain
ronak jain
Ok
A curious mind . Can you tell me about this one ?
Did anyone else try ?
 
Charlie
Charlie
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
 
ACuriousMind
ACuriousMind
^that
There's also the problem solving chat room for such questions about exercises
 
Slereah
Slereah
Maybe we should open a problem solving business
Five bucks to solve
 
ACuriousMind
ACuriousMind
12:55 PM
You'd have to pay me more than five bucks to solve these exercises :P
 
Slereah
Slereah
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
 
Specter Prophet
Specter Prophet
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
 
Charlie
Charlie
Is the force of gravity on them not different?
 
Specter Prophet
Specter Prophet
but isn't the mass different
 
Charlie
Charlie
and $F_g=\frac{GMm}{r^2}$, no?
 
Faded Giant
Faded Giant
@SpecterProphet "the wood and plastic cubes float while the steel cube sinks" or "all submerged at the same depth"?
you cannot have both
 
Specter Prophet
Specter Prophet
@FadedGiant yep
 
Charlie
Charlie
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
 
Specter Prophet
Specter Prophet
I mean it doesn't depend on mass difference ?
assmuning they have same dimensions
 
Charlie
Charlie
what doesn't?
 
Specter Prophet
Specter Prophet
three of them steel,wood....
 
Charlie
Charlie
so the buoyancy force depends only on the volume of the object?
 
Specter Prophet
Specter Prophet
?
 
Charlie
Charlie
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
 
Specter Prophet
Specter Prophet
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)
 
Charlie
Charlie
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
 
Specter Prophet
Specter Prophet
it just started with letters. may be it wants me to formulate it
 
Charlie
Charlie
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
 
Specter Prophet
Specter Prophet
@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.
 
Charlie
Charlie
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
 
ACuriousMind
ACuriousMind
Your first sst of equations is just the second set, but in local coordinates
So yes, there's no difference
 
Charlie
Charlie
Is local coordinates also known under another name? I can't find a reference to it
 
ACuriousMind
ACuriousMind
@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
 
Charlie
Charlie
oh
 
ACuriousMind
ACuriousMind
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
 
Charlie
Charlie
oh I see it's just going from the geometric object of the tensor to it's compoenents which requires picking a coordinate system
 
ACuriousMind
ACuriousMind
yeah, exactly
 
Charlie
Charlie
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
 
ACuriousMind
ACuriousMind
What is "regular tensor form"?
I think there's two different things to exterior calculus - one is the notation as you say
 
Charlie
Charlie
ah I mean as in if we used $\vec u \otimes \vec v - \vec v \otimes \vec u$ instead of $\vec u \wedge \vec v$
 
ACuriousMind
ACuriousMind
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
 
Charlie
Charlie
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
 
ACuriousMind
ACuriousMind
the thing is that $\mathrm{d}$ can only act on forms, not on general tensors
 
Charlie
Charlie
ah of course
 
ACuriousMind
ACuriousMind
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
 
Charlie
Charlie
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
 
ACuriousMind
ACuriousMind
4:43 PM
The generalization of electromagnetism, Yang-Mills gauge theories, are quite popular
 
Charlie
Charlie
ah so scary qft stuff
 
ACuriousMind
ACuriousMind
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
 
Charlie
Charlie
at least it follows relatively painlessly from having learned about tensors, which was a whoole thing
 
Slereah
Slereah
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
 
Charlie
Charlie
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"?
 
Slereah
Slereah
The antisymmetric part
That's what the forms are
 
Charlie
Charlie
ah yeah mistype
this stuff is actually not that scary
well, yet
 
Slereah
Slereah
It can get, but the basic stuff isn't too hard
Especially on flat space
 
JingleBells
JingleBells
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...
 
Slereah
Slereah
Basically you have to look at the wedge product, exterior derivative, the Hodge star and the interior product
those are all the big ones
 
JingleBells
JingleBells
@Charlie ur pic looks something from a game that I forgot the name of
 
JMac
JMac
@JingleBells That's totally a Runescape character
 
Charlie
Charlie
5:49 PM
@JMac I see you are a man of culture
@JingleBells It's the default skin from the old version of runescape
 
John Rennie
John Rennie
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.
 
Charlie
Charlie
6:45 PM
Does $\text d^2=0$ have a name?
 
Slereah
Slereah
@Charlie It's the chain complex
Or... cochain complex?
One of these
 
Charlie
Charlie
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
 
Slereah
Slereah
It applies to all $n$-forms
 
Charlie
Charlie
ok nice ty
 
Slereah
Slereah
This is just due to 1) the antisymmetrization 2) the fact that derivatives commute
 
Charlie
Charlie
6:58 PM
@Slereah Actually to this point, is $\text d\tilde\omega$ necessarily zero for a $p-$form $p\geq1$?
 
Slereah
Slereah
Why would it be?
 
Charlie
Charlie
sry i forgot that edits ping, mb
doesn't this imply that there are $(p\geq1)-$forms that can't be written as $\text df$?
 
Slereah
Slereah
Yes, those are called inexact forms
I mean just consider it in one dimension
 
Charlie
Charlie
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
 
Slereah
Slereah
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
 
Charlie
Charlie
7:04 PM
is $\alpha$ necessarily exact too?
 
Slereah
Slereah
Well no
 
Charlie
Charlie
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
 
Slereah
Slereah
Well, you can only take the exterior derivative twice
 
Charlie
Charlie
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
 
Slereah
Slereah
Yes
 
Charlie
Charlie
7:17 PM
ok nice thank you
 
Slereah
Slereah
If every $p$-form was exact, then there wouldn't be any $p$-form for $p > 1$
Outside of zero, anyway
 
Charlie
Charlie
ok good that's the problem I thought existed earlier
 
JingleBells
JingleBells
@Charlie yes, Runescape
 
Slereah
Slereah
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
 
Charlie
Charlie
the 2-form you've given, is that $\alpha$ or $\omega$?
 
Slereah
Slereah
7:31 PM
it is $\omega$
 
JingleBells
JingleBells
bruh
 
Charlie
Charlie
and your 3rd line follows from the fact that if $\alpha$ were exact then $\text d\alpha=0\neq\omega$
 
bolbteppa
bolbteppa
 
Slereah
Slereah
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.
 
Physics Meta
Physics Meta
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Vamsi KrishnaWhen 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...

support down-votes accepted-answer
 
Charlie
Charlie
Ok that makes sense thank you
 
bolbteppa
bolbteppa
Closed and exact differential forms
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...
 
Charlie
Charlie
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
 
bolbteppa
bolbteppa
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
 
Charlie
Charlie
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
 
bolbteppa
bolbteppa
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 $u \wedge v$
 
Charlie
Charlie
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
 
bolbteppa
bolbteppa
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?
 
Charlie
Charlie
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
 

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