00:00 - 14:0014:00 - 17:00

2:06 PM
my eyes get so tired for reading on the computer screen.

Do we know any example of a theory that actually has the Hilbert space $$\mathcal{H} = \mathbb C \oplus \mathcal H_{p^2 = m} \oplus \mathcal H_{p^2 \geq M}$$
Sine-Gordon sounds like it should
But I still don't know if its Hilbert space is known
It seems to always be semiclassical stuff

2:40 PM
@ACuriousMind I too had a confusion that how come instantaneous acceleration be perpendicular to the velocity vector, that's why I asked whether it happens in general. I think it's applicable for only uniform circular motion and projectile motion.

It's not true for projectile motion
And for uniform circular motion tangential acceleration is 0, no?
There is centripetal acceleration, yes
And that is perpendicular to velocity vector

Projectile motion has its acceleration perpendicular to the direction of the velocity, isn't that?
@Blue

Isn't the acceleration always $g$ downwards in projectile motion?

Oh. I think I got u. At maximum height there only remains horizontal component of projectile motion and the vertical component is zero, only at that point gravitational acceleration is perpendicular to the velocity vector @Blue, am I right?

@ffahim Yea

2:49 PM
But I wonder why it's said in our textbook that instantaneous acceleration is perpendicular to the velocity vector. Where most of the Google searches showed that's not true in general.

Can you quote your book exactly word by word?
Or upload a picture of it

Actually it's written in our Bengali language. So u can't understand for this. But that really meant that instantaneous acceleration is perpendicular to the velocity vector.

I know Bengali, go on

Really? Great

Yep

2:52 PM
if the velocity is normalized, this is trivial
if it's not, then it's not true
are you talking about fluid mechanics?

what on earth is $\mathrm{Lt}$?
I can guess but it's horrifying

@0celo7, no only for kinematic
@Blue u wanted to see the picture

@0celo7 limit in Indian

@ffahim Yeah, I'm reading it. Give me a minute

2:56 PM
@BalarkaSen is there a special reason why they can't be like the rest of the world

What means if velocity is normalized

@ffahim arc length parametrization

@0celo7 Old Russian books use $\text{Lt}$

@BalarkaSen we both know I don't consider old Russian math worth considering

2:59 PM
Sorry I am a beginner for learning calculus. I don't have enough idea what you are referring to :) @0celo7

@ffahim The sentence you underlined just says that at some point of the trajectory magnitude of instantaneous acceleration is equal to $|\frac{\Delta v}{\Delta t}|$ as $\Delta t \to 0$.

I can't understand the mathjatex

It doesn't say anything about perpendicularity. It just says "barabar", meaning "equal" or "equivalent".

No, barabar in hindi means equal, in bengali it means towards that direction

You're a PDE theorist, you'll appreciate

3:05 PM
@ffahim Even then it just says that $\vec{a}$ is in direction of $d\vec{v}/dt$. Where did they mention perpendicularity?

@BalarkaSen pls, I'm a geometric analyst
I also do gmt
@BalarkaSen oh my god

Told you
It's golden

The line means that Instantaneous acceleration at a given point is perpendicular to the direction of velocity of that point @Blue

@ffahim Which word in that sentence stands for "perpendicular"?

@0celo7 I mean it's certainly better than this fucking shit: youtube.com/watch?v=NPN4m0d6PvI

3:10 PM
লম্ব stands for perpendicular @Blue

@BalarkaSen can't do analysis hw pls help

The white guy doing the flabby thing at the back cracks me up tho

No, Balarka is from India, I am from Bangladesh, we're neighbors

Honestly Oppa nabla style should just be the rightful winner of Dance your Phd
I haven't seen a single good thing from the top ones

@ffahim I think they're using it in some other sense. But as far as that is concerned, instantaneous acceleration is perpendicular to velocity iff the velocity vector has a constant magnitude at all times (which is true in case of uniform circular motion as you noticed).

3:14 PM
he mentioned complex but didn't really do anything about it
should have put the residue theorem in there

he mentioned Sobolev spaces too
as a passing line

yeah I noticed

I see. So that's no true for an objective moving at straight line in uniform acceleration right? @Blue

@ffahim Right
"constant magnitude of velocity vector" is the required condition

3:17 PM
Ok. Thanks a lot :) @Blue

@BalarkaSen If a function $f\in W^{2,p}_\text{loc}(R)$ has a second weak derivative in $L^p_\text{loc}$, then it has a first weak derivative in $L^p_\text{loc}$
Problem is, I have a proof, but it has to be wrong
clearly one should define $f'=\int f''$, but there's a constant missing
my proof doesn't seem to care about that constant, which is strange

integral of f'' over [something, x] i presume?
(I do not know these spaces)

well, fix a finite interval $(a,b)$ for definiteness
then define $f'(x)=\int_a^x f''(t)dt$
the idea is then that $df/dx=f'$ weakly

gotcha

but clearly this is only true up to a constant, because I didn't write constant in that integration
but moving on, for any test function $\phi$ some real analysis shows $d(\phi f')/dx=\phi'f'+\phi f''$

3:21 PM
uh, constant of integration appears only when you take indefinite integral? i dont see why you get fucked by constants here
might be stupid qn

@BalarkaSen omfg...what's that

@Blue truly a masterpiece

@BalarkaSen Well the correct formula for C^1 functions is $$f(x)=\int_a^x f'(x)dx+f(a)$$

That surely deserves an oscar, yeah.

@0celo7 Oh i see ok

3:23 PM
so I have $d(\phi f')/dx=\phi'f'+\phi f''$
then integrate to get $$-\int f'\phi'=\int \phi f''$$

mmk

and then use the definition of $f''$ to write $$-\int\phi'f'=\int \phi''f$$
so this is almost done, but the problem is that not every test function $\eta$ can be written as $\phi'$
to remedy this, let $\psi:(a,b)\to R$ be smooth with $\psi=0$ near $a$ an $\psi=1$ near $b$, and define $$\overline \eta=\int_a^x\eta-(\int_a^b\eta)\psi$$

Cool trick

this has the advantage of being in $C_c^\infty$, and the derivative is $$\overline \eta'=\eta -(\int\eta)\psi',$$ with both terms being test functions

True

3:27 PM
so you can plug this into above formula, and then use it again with $\phi'=\psi'$
so you end up with $$-\int f'\eta=\int \eta'f$$
but this cannot be true because $f'$ might start at the wrong value

@0celo7 what do you integrate this over?

@BalarkaSen $(a,b)$

my first reaction would be some constant gets ignored in your manipulation during integration
(of course i could be wrong)

well what's interesting is that in $$-\int_a^b \phi'f'=\int_a^b\phi''f,$$ you can redefine $f'$ by a constant and not change anything

Trying to detect it with limited understanding of things
Ah...

3:34 PM
because $$\int_a^b\phi'=\phi(b)-\phi(a)=0$$
so you start with something agnostic of the constant, and end up with something that detects it

Weird... \

4:05 PM
@Slereah you truly have the best countrymen
god bless the french

that we do

I understand this stuff now

do you mean french patriotism

Can anyone tell me how to share an image here ?

@SkyWalker are you on a phone?

4:13 PM
Yes
But desktop mode is on.

use imgur or something equivalent

Ah. Just click the upload button.

I can't see it.

You need 100 rep for that I think

4:15 PM
On my phone screen, it is showing only the send button.

52+23+6+1+1=83
So get another 17 rep points, and then you can upload pictures here

The workaround is to upload it to some other site like imgur and share the link.

@SkyWalker try refreshing the web page now and see if the button appears

Actually, I am not going to post in physics meta. You people always downvote my questions. I am on the verge of getting banned from asking questions.

4:19 PM

Go on and please look at my profile.

3 duplicates!

@JohnRennie No.

Most newcomers get downvotes. That's the only way they learn to follow rules and ask good questions.
Or else it would become something like Reddit or Yahoo

4:23 PM
@Blue I don't take personally. But, I am going to get banned. How can I edit my answers ?
I mean, how to improve them ? Specially the duplicate ones ?

@Blue I am not a spammer or a trash writer. I am a student who is interested in physics.

To be blunt, your questions look pretty straightforward and lack research efforts.

rekt

For example:
1

https://youtu.be/wvJAgrUBF4w In this video, the sand particles form beautiful patterns on the vibrating metal plate on different frequencies. Can anyone explain what is going on there mathematically ?

No mention of how you tried to find the answer
And where exactly you got stuck

4:27 PM
@Blue Yes. I ask questions without much details. I want to get the view of the answerer. I don't want him to think like me.

"Can anyone explain what is going on there mathematically ?"
That's one of traditional examples of lazy questions
@SkyWalker I don't buy that reason, sorry.

this is a beat down

@Blue Yes. I wanted the mathematical analysis as I lack mathematical skills to analyse the patterns.

@SkyWalker No one has the time or energy to explain all the mathematics behind formation of resonance patterns. You are expected to search the net and read relevant books and ask where exactly you're getting stuck. Ask more specific questions and if you don't know the math used in those books, learn then from other books first and ask if you get stuck there.

Presentation is absolutely crucial in SE. No matter how good of a question/answer it is, you have to present it the right way to not be seen as a bad question
Recently there was a good question which should have a deep answer that got closed down in math.SE... nothing to be done about that
(To illustrate that it's deep, I'm still studying stuff to find out the "right" answer)

4:32 PM
One good thing about SE is that they expect you to be very professional on the main site, which imo is very good. Laziness has no place here.
However, on chat, such questions may be fine once in a while.

It's not always very good.

@BalarkaSen You are from MSE. What do you know about Cleo ?

@BalarkaSen Well, depends. But most of the times

There's no way to get around it, but it's certainly not the best system

I've had my share of bad experiences too. But it's the best that exists on the internet

4:34 PM
Also it doesn't stop the tons of homework questions that barrages MSE

Personally I do have some issues with the policies, but I can live with that I guess.
@BalarkaSen True

The question I am speaking of should be MO-level when appropriately generalized
@SkyWalker Nothing, like everyone else