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5:08 PM
@BalarkaSen same with Star Wars
It’s all a reflection of the mental illness of Hollywood
 
Fun fact: It turns out I am mostly talking to myself a few minutes ago in math chat
 
You do that all the time
 
Don’t you have him blocked?
 
he isn't, he use selective ignoring, thus bypassing the limitation of the ignore button
You can show that by a continuity analysis of the transcript
 
Using the amazing power of my imagination
 
5:17 PM
I wonder if it's possible to ignore yourself ...
2
 
If only.
 
:-)
 
r/2meirl4meirl
Maybe there’s underscores there
 
5:46 PM
@ACuriousMind seriously, though ─ when people used to shoot with bows and arrows, there were people whose job it was to put arrows on bows, no?
Fletching is the fin-shaped aerodynamic stabilization device attached on arrows, crossbow bolts or darts, typically made from light, semi-flexible materials such as feathers. Each piece of such fin is a fletch, also known as a flight or feather. A fletcher is a person who attaches fletchings to the shaft of arrows. The word is related to the French word flèche, meaning "arrow", via the ultimate root of Old Frankish fliukka. == Description == As a noun, "fletching" refers collectively to the fins or vanes, each of which individually is known as a fletch. Traditionally, the fletching consists of...
 
@ACuriousMind why does Green define Poisson brackets for classical string modes
 
Wikipedia informs me that the name in English is "fletcher"
 
Arn't they just constants
 
which is also a surname, no?
so Federer is, in the end, much the same name as Fletcher?
 
@EmilioPisanty Again, Federer is not a German word I know. The proper translation of "fletcher" is Pfeilmacher. However, some googling for German sources reveals that Federer might have been an old term for people who trade in feathers, or who manufactured pillows and the like stuffed with feathers.
However, results in German not already in relation to the surname are exceedingly rare, and I am not fully confident in their veracity
 
5:57 PM
@ACuriousMind ah, gotcha
@ACuriousMind =P
so what you're saying is "folk etymologies are only as good as folk etymologies"?
 
@EmilioPisanty Heh, I guess
 
boring pde question
Suppose I take the 1+1D heat equation with initial condition being a box function (i.e. 1 on a certain interval and 0 outside). Then evolving that forward in time (numerically) is easy enough. Am I right in thinking that evolving it backwards will by contrast be baaaad?
It's not smooth function, so my understanding would be that it's not going to be stable.
okay, yeah, Mathematica is Not Happy (tm)
 
6:13 PM
@Semiclassical yes
In mathematics, the Gibbs phenomenon, discovered by Henry Wilbraham (1848) and rediscovered by J. Willard Gibbs (1899), is the peculiar manner in which the Fourier series of a piecewise continuously differentiable periodic function behaves at a jump discontinuity. The nth partial sum of the Fourier series has large oscillations near the jump, which might increase the maximum of the partial sum above that of the function itself. The overshoot does not die out as n increases, but approaches a finite limit. This sort of behavior was also observed by experimental physicists, but was believed to be...
basically
but worse
 
yeah, that's about what I concluded
I mean, it makes sense that trying to do the heat equation backwards is not going to be a fun time
 
there are conjectures about backwards heat flows
if you know that your function was obtained by heat flow for time $T$, you can hope to evolve backwards by $T$ in a unique way
Kotschwar answered this question for Ricci flow
 
Yeah, that makes sense.
But you still shouldn't expect to evolve forward in time via the heat equation and get discontinuous data
 
yeah, that's not good
probably doesn't even work with distributional data
 
this is pertinent: "The smoothing effect also shows why the backward heat equation is ill-posed. Indeed, there can be no solution to the backward heat equation with an initial condition that is not C^∞, since an initial condition for the backward heat equation is a final condition for the forward heat equation. It is not even clear that all C^∞ functions can be reached by the evolution of the heat equation. Therefore, time is irreversible in the heat equation."
 
6:25 PM
My colleague just wrote to me: "No rush. I use [online chat] to schedule asynchronous call backs."
IT'S NOT JUST ME
 
so much for CPT symmetry
 
lol
I was handed this problem and asked to Mathematica it up, so I'll have to bug them
I should note as well that this was intended to be the special case of something more involved, namely:
 
I think this is the mathematical analogue of finding the elbow grease or whatever
the breastplate stretcher in GoT terms
 
...oh, deeeeerp
not $\partial_x^2 \phi$, $(\partial_x \phi)^2$
I misread what I was given. no idea how
 
6:31 PM
@Semiclassical so this is a first order thing...might want to learn about first order hyperbolic equations now
 
yeah
mathematica doesn't seem to like either forward or backwards time evolution for this one...yay :/
 
I dunno
 
yeah, I do not trust what mathematica is giving me here
 
@Semiclassical Have you looked in evans for equations of this type
 
nah. I just dove in
I think this should be something like an NLS?
hmm, no
 
6:39 PM
@Semiclassical Ch 3 of evans
it's probably in there
can you solve it with characteristics?
 
not when you have the inhomogeneity of course
 
tbh I was assuming I'd have to do it all numerically so I wasn't thinking about that
 
gotta bust out the undergrad PDE notes
 
looks like this should really be thought of as a Hamilton-Jacobi equation
at least if i ignore that extra term
oh, and if I include that second term then that's what corresponds to a potential here
nice
@0celo7 looks like this was on the money
Example 3 + Example 6
 
6:53 PM
@Semiclassical Hi ! Could you tell me what this means "Electric field at the surface of a hollow spherical conductor is discontinuous" ?
Oh , nvm , I got it . @Semiclassical
 
I think that counts as a discontinuity in conceptual understanding :)
(i.e. not understanding -> understanding)
 
@Semiclassical haha
@JohnRennie Are you free right now ? Or is it too late in the day and you just wanna chillax ?
 
Just passing through. Back tomorrow.
Actually I have a few minutes.
What did you want to discuss?
 
@JohnRennie uhm , for a few minutes ,
I don't get why A is not true.
 
If you're rolling without slipping, then what kind of friction do you have?
Static or kinetic?
 
7:03 PM
Didn't we do something very similar to this? Or maybe it was with someone else.
 
not sure.
should it be kinetic ?
 
if it's rolling without slipping, then is the bottom of the cylinder ever sliding across the ground?
 
@Semiclassical oh yes , it is at rest always
 
not sure I'd word it quite like that. but the upshot is that it's static fraction, not kinetic
 
I can't remember the details. I would have to put pen to paper and work it out, and it's too late in the day for me to want to start that.
 
7:06 PM
@Semiclassical yes got it .
 
and the thing with static friction is that it's always expressed as an inequality, i.e. F_f <= mu*N
 
@JohnRennie it's okay , don't worry.
@Semiclassical yes
 
Hence the friction force needn't be $\mu mg \cos\theta$ here. That's the most it can be, but that's not what it will typically be
 
but the values of $\mu$ and $N$ certainly remain constant don't they ?
 
@Tanuj ah, of course. Consider a cylinder rolling horizontally i.e. on a horizontal plane.
The frictional force is zero.
 
7:08 PM
Sure? But you don't have $F_f=\mu N$ here. You have $F_f\leq \mu N$
 
@Semiclassical hmm , but why isn't it always the equality case ?
 
1 min ago, by John Rennie
@Tanuj ah, of course. Consider a cylinder rolling horizontally i.e. on a horizontal plane.
 
@JohnRennie okay , but why is friction 0 there ?
 
Not sure I'm right about what I said there.
 
@Tanuj if the cylinder is rolling at constant speed on a horizontal plane it is neither accelerating nor decelerating.
That means the torque on the cylinder must be zero. But the torque is provided by the frictional force, so the frictional force must be zero.
 
7:11 PM
@JohnRennie hmm , but why are we assuming it must be under constant speed and no acceleration ?
 
@Tanuj it's on a horizontal plane ...
 
@JohnRennie no external force , right ?
 
And it's not sliding so the frictional force cannot be doing any work.
So the energy of the cylinder cannot change
 
@JohnRennie okay . So the frictional force is decreasing on decreasing the angle of the incline , how do I see it mathematically ?
Also , why isn't friction always $\mu mg \cos\theta$ ?
 
Semiclassical put his finger on it. $\mu mg\cos\theta$ is the maximum value the frictional force can reach, not the actual value of the frictional force.
 
7:14 PM
I mean, suppose you replaced the rolling cylinder with a block.
Then if the angle is small enough, you expect the block to remain stationary on the ramp and not slip.
 
@Tanuj you'd have to do the calculation. The cylinder rolling down an incline is a standard problem.
 
It's only once you increase the angle sufficiently that the friction force required to maintain that position exceeds $\mu N$ and starts sliding instead
 
But like , while doing problems in which some block is going up or down the incline with some acceleration , we indeed take the maximum value of friction and solve , why is that ?
 
If you've got a block going up and down an incline, it's sliding
 
@Tanuj If the block is sliding the frictional force must be at its maximum value.
 
7:16 PM
Hence you have kinetic friction, and there indeed $F_f=\mu N$ is an equality
 
i.e. the weight of the block has overcome the frictional force and made the block slide.
 
@Semiclassical okay , big confusion cleared out . Thank you guys !
 
The case of a cylinder is not intuitive in that you have motion but not sliding
The difference, I'd say, is that static friction is always there to model some condition/constraint---in this case, the condition being that the cylinder isn't turning faster than rolling without slipping would allow
Hence it's not saying "the force is X" but "the force can't exceed X"
 
Guys , another theoretical question
 
hence why it's easy to get confusing
 
7:18 PM
in option (c) , what does "visible region" have to with in the question ?
 
@0celo7 the hopf-lax formula is new to me, but i'm not entirely surprised to see something like that here
@Tanuj Not sure. My guess would be that it's sort of a trick question, e.g. that the energy radiated per unit time over any frequency range would be going up
 
@Semiclassical Don't even know what that means.
 
well, suppose you weren't talking about the visible spectrum. Suppose you picked some wavelength range like 1000-2000 nm (I pick wavelength here b/c I don't remember typical visible frequency)
 
yeah
 
My guess would be that the energy radiated per unit time over this wavelength range would also be going up.
 
7:24 PM
okay
 
i.e. it doesn't matter what part of the E&M spectrum you look at; the energy is going up regardless
But that is not a certainty on my part.
It just seems strange to talk about the visible spectrum in particular?
 
I know right !
 
@Semiclassical you found it in Evans? Where?
 
A few places. Main thing was for me to realize what I have is just the Hamilton-Jacobi equation with $H(p)=p^2$
 
Yeah, so definitely characteristics, right
 
7:30 PM
starting from page 124 he's got a whole discussion on HJE
including the Hopf-Lax formula which is pretty great
(pretty weird, but pretty great)
the only discrepancy is that really I've got $H=p^2$. But I can get rid of that by doing $t\mapsto -t$ in my original equation, in which case I also just have forward time evolution
There's still something hinky, though, at least re: what solution I think the Hopf-Lax formula is giving me
 
@0celo7 There's a good question flying around in the math chat if you want to catch it
 
7:59 PM
@0celo7 hmm, the hopf-lax formula assumes initial data which is C^2
mine isn't even C^0 everywhere, so I shouldn't be shocked that I'm not getting a unique solution
 
8:16 PM
@0celo7 well
I think I can verify that my implementation of NDSolve isn't good enough to do this hyperbolic PDE
namely, I can see that it works okay-ish for small t (in the sense of agreeing with the Hopf-Lax formula) but fails badly as t increases
 
8:39 PM
Dependency hell is the bane of existence!!!
I just had to say it...
 
 
2 hours later…
10:09 PM
@Semiclassical can you program MMA to give you Hopf-Lax?
 
Insofar as I can ask it to do NMinimize, yes
 
@Semiclassical ew
 
e.g. NMinimize[t*L[(x-y)/t]+g(y)]
where L is the Lagrangian and u(x,0)=g(x)
gotta pick specific values of x,t in order to run that, of course
main problem I have right now is that it's just so slow
which probably means I'm not doing things as cleverly as I could, but still
 
@Semiclassical if MMA ever did anything quickly I’d doubt the correctness of the result.
 
10:44 PM
@0celo7 I think I'm also making things too complicated: It might be cute to do a smooth bump function, but a piecewise linear one is probably fine for now
 
11:18 PM
@EmilioPisanty Wow, has he gotten back to you?
Sigh, I'm blind, missed the comment
 
Hit the 200 rep cap.
 
11:44 PM
@BalarkaSen what don't you like about that terminology?
@Slereah LOL
@0celo7 What?
 

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