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2:01 PM
More fishy Feynman stuff
8
Q: Feynman's derivation of Maxwell's Equations

Aranya KumarThere are so many times that something leaves you stumped. I was recently reading this paper and the derivation of the Maxwell's Equations from just Newton's Second Law and The Uncertainty Principle really intrigued me. Although they just derived the Bianchi set, yet with slight tweakings with re...

 
he does a lot of it
 
weird paper from an even weirder question
 
Feynman didn't know math
Unfeynman
 
did you read his biography
He basically brags about his bad math
 
Not Fine Man
 
2:07 PM
higher physics without maths is unimaginable
 
Oh he was good at math, just not rigorous
 
yeah
 
I see
 
That's a pretty interesting derivation though, for whatever reason the commutation relations force the force to be a tensor formed from a 4-vector and 4-covector, the answer in there seems to use relativity while saying it's not using relativity, and the whole answer comes from relativity alone, however the claim is supposed to be non-relativity and QM gives what relativity gives, so idk
 
as the dude says it's not rly a QM relation
since you can also get it from the classical Poisson brackets
Feynman has given the physicists many of their Bad Math
Especially the differentiating under the integral sign trick
 
2:14 PM
Yeah, but the fact the force takes the form it does, partially composed of a curl, the rest a normal vector, $F_{\mu \nu} = (\mathbf{E},\mathbf{B})$, why doesn't this hold for every force based off this derivation
 
While he didn't invent it, it was a fairly obscure trick before
 
He found diff under the integral sign in that book everybody wants to read because he mentions it in his bio
 
@bolbteppa Well, what relativistic force doesn't?
Relativistic force on a point particle, that is
 
Who is using relativity here, that's the point
 
This just in: electrons are a societal construct
 
2:17 PM
I mean, if you strip this of its relativistic context, all you have is just that the forcre is conservative
Wait, I think the relativity comes in as $p^2 = m^2$
and $\dot m = 0$
 
That's why the answer is bad, he says after that 'we haven't used relativity yet' even though he used it there, but the link in the OP to the Feynman article doesn't use any relativity at all, the guy was just trying to explain what Feynman was doing in relativistic notation so is mostly fine
Even Feynman was confused so...
 
2:33 PM
I have a question regarding Gauss's law
For calculating the electric field produce d by a infinitely charged plane sheet I considered the Gaussian surface to be a cylinder
 
why
pillbox integration, my man
 
The electric field through both the circular ends is equal and opposite and yet the total flux was the sum of of the fluxs thru both ends should they cancel each other out
they have the same Areas, same Electric field intensity ( but in opposite direction)
EA - EA = 0
I am following my textbook it says nothing about pillbox
like the fig in the link
 
@Semiclassical you around?
 
2:49 PM
just arrived
what's up? @EmilioPisanty
@susanJ note that the flux is not $\int |E|\,da$ but $\int \vec{E}\cdot d\vec{a}$
 
@Semiclassical care to take a critical eye to last week's slides and tomorrow's work-in-progress?
 
so you need to take care about the direction of the normal vector in $d\vec{a}$
sure, I'll take a gander
fun fact, my advisor sent me a puzzle over the weekend which you might appreciate
If you use Stirling's formula, you get $\Gamma(i z)\Gamma(-i z)\sim \frac{2\pi}{z} e^{-\pi z}$ as $z\to\infty$
by comparison, the reflection formula yields the exact answer as $$\Gamma(iz)\Gamma(-i z)= \frac{2\pi}{z} e^{-\pi z} (1 - e^{-2\pi z})^{-1}$$
These are obviously compatible, but if you expand the denominator of the latter answer then it seems like you're getting a countable sequence of corrections to Stirling's formula (which is just the saddle point approximation on $\Gamma(z)=\int_{t=0}^\infty e^{z\ln t-t}\,dt$).
@EmilioPisanty typo on first bullet of slide 8: continuosusly->continuously
 
@Semiclassical wait, which slide 8?
first one or second one?
 
oh. lecture one, sorry
bit late for that, i guess
 
page 8 as marked or on the pdf?
 
2:59 PM
as listed at the bottom right of the slide
 
@Semiclassical yeah, a bit ;-)
I'll make them public at some point, so it's not wasted either, though
 
right
 
Any book for elementary excitations in condensed matter ?
 
0
Q: Causally benign spacetimes and the Minkowski torus

SlereahA notion encountered in field theory on non-globally hyperbolic manifolds is the notion of a spacetime being causally benign with respect to some field $\phi$, which is defined thusly : A neighbourhood $U$ is causally regular if $\bar U$ has a neighbourhood $U_1$ such that for every solution $...

plz halp
 
@EmilioPisanty my favorite example of a function with natural boundary, btw, is $1+z+z^2+z^4+z^8+\cdots=\sum_{n=0}^\infty z^{2^n}$
 
3:03 PM
These days Yurtsever works at some big data company
Sad!
 
...oh.
that's basically your example as well
deeeeeerp
 
0
A: Why is chemical potential equivalent to a true potential?

user187456Chemical potential is the difference in free energy to add a particle to the system. Consider two systems (1,2) with a contact with the reservoir at temperature $T$, then total free energy is \begin{equation} G = G_{1}+G_{2} \end{equation} But since the total number of particles is conserved \beg...

 
lacunary shite
 
3:04 PM
still publishing papers, tho
Livin' the dream
 
@Semiclassical that's a good 'un indeed
 
clocks in at nine pdf pages, so I obviously agree ;-P
 
Ä° know yurtseven
 
3:06 PM
but the appearance of the natural boundary on something that ougher be "physical" is way scarier
@Semiclassical btw that series is wrong
you're off by 1
There are two hard things in computer science: cache invalidation, naming things, and off-by-one errors.
 
0
Q: Can an "incomplete" comment be considered an unhelpful comment?

Niamh O'SSObvious unhelpful comments usually consist of abusive and negative comments as well as completely irrelevant ones and conversational ones. Those kind of comments are easy to spot. But what about "incomplete" comments? Ones where someone is trying to make a valid argument but is incomplete and se...

 
feh. including 1 vs. not including 1 can't make a difference on whether there's a natural boundary
 
@Semiclassical oh, I agree
 
not including 1 has the benefit of 0 -> 0, though
 
n = 0
 
3:08 PM
but it's either $1+z+z^2+z^4+z^8+\cdots$ or $\sum_{n=0}^\infty z^{2^n}$ though
 
oh, blah
yeah
 
@Semiclassical =P
 
thought I'd made that work
 
I'm just relaying my quandary from last week
keep the 1 and make the resemblance to the geometric series better?
 
oh lol, you did the same?
ahh
 
3:09 PM
or ditch the one and make the series cleaner?
 
do any of you physicists understand the hartogs phenomenon
in a physical way
 
right
I actually asked a MO question about that series a while back
11
Q: Distribution of zeroes of lacunary functions

SemiclassicalIn a recent Math Stack Exchange question I asked about the function $$f(z)=\sum_{n=0}^\infty z^{2^n},$$ and was informed of its status is a canonical example of a lacunary series with natural boundary at $|z|=1$. A phenomenon observed by the accepted answer was that this function has a multitude ...

 
Photons can be vacumm lattice phonons ?
 
@Semiclassical oh, are we showing off lacunary-series-questions-on-MO pedigrees now?
15
Q: Why are lacunary series so badly behaved?

Emilio PisantyHi! I just came across the Ostroski-Hadamard gap theorem, and while I can understand the proofs as well as the principle that the series $\sum_{n=0}^\infty z^{2^n}$ ought to have a singularity at every $2^n$-th root of unity for every $n$, I feel I'm missing some intuition into what exactly is g...

 
loool
 
3:12 PM
I blame the parents. — Will Jagy Mar 5 '12 at 1:26
^ lolz
 
hah, nice
 
Anonymous
@user187456 They are different
 
You could say that it's my first question on MO.
 
Any proof? Or explanation why not ?
 
(given that we've established that it's OK to make off-by-one errors.)
 
3:14 PM
I think the same is true for mine.
Two things I wonder about re: the propagator example
 
@Blue why not
 
the first is what happens if I modify the boundary conditions to $\psi(\theta+2\pi)=e^{i\eta}\psi(\theta)$ with $\eta\in[0,2\pi)$
 
@Semiclassical nothing much, I should expect
 
yeah, I think you're right
 
[high school level stuff] I am surprised that we cannot talk about the slope of the tangent of a polar curve $r = f(\theta)$ without reference to x and y
 
3:16 PM
It may or may not map exactly into the original example
21
A: The position-representation matrix elements of the propagator for a particle in a ring

Emilio PisantyYou wouldn't think it, from how easy it is to pose this question, but it is ridiculously nontrivial. As it happens, it is entirely impossible to find the position-basis matrix elements of this propagator. So far you've done good, and the identification $$ U\left( t_{2},t_{1}\right) =e^{\frac {-...

 
only sorta interesting is that the dependence should be periodic with respect to $\eta$
 
http://tutorial.math.lamar.edu/Classes/CalcII/PolarTangents.aspx
The derivation $\tan \psi = r \frac{d\theta}{dr}$ is absolutely tedious
 
but the propagator will still be a series in $e^{i a n^2}$
 
but that's not so interesting
right
what I'm more interested in is how that plot of the behavior on the unit disk would change if you include an actual potential
...wait
 
@Semiclassical That plot on the unit disk is just for illustration
 
3:19 PM
$\hat{H}=\omega\frac{\partial^2}{\partial\theta^2}$?
That'd make sense to me when $\omega<0$.
and actually in that case you'd have $\hat{H}e^{i n\theta}=-\omega n^2 e^{i n\theta}$
 
So $E_n=-\omega n^2$.
 
that... will be fixed in the public version ;-)
 
lol
the joy of minus signs
What I'm mostly curious about is whether that natural boundary is a symptom of having $V=0$ here
i.e. would you still have that if you had a potential like $V=\alpha \cos\theta$
Once you have a potential, I guess the point is that the solutions in momentum space are actual functions rather than just distributions
@EmilioPisanty Another way one could study this, I think, would be to consider a loop of N sites and examine how the propagator in that case will behave as $N\to\infty$
in that scenario I presume there would be finitely many poles on the boundary which would accumulate to a natural boundary as $N\to\infty$
 
@Semiclassical I shouldn't think so
 
3:29 PM
Hmm
 
for any bounded potential, at sufficiently high $n$, it'll still look like $E_n \sim n^2$ ish
 
True.
And then the gap theorem gets you back to expecting a natural boundary
 
exactly
 
to the extent that the low energy states matter, then, it's for the behavior near the center of the unit circle?
 
3:42 PM
@Semiclassical have a look at slide 27 on the new one
@Semiclassical what unit circle?
 
|q|=1
 
the unit circle in $q=e^{-i\omega t}$?
 
right
 
I guess you could say that?
the $|q|\to 0$ limit is the limit of $\mathrm{Im}(t)$ large and negative
 
3:44 PM
so I imagine there's some connection there to low energies
 
kk
My intuition is mostly that, if the gap theorem tells you that the boundary behavior is dictated by high energy states
then the only natural place to look for the signature of low energy states is away from the boundary
I guess the point for your strong-field approximation is that the low-energy states should be pretty irrelevant tho?
 
Hey @ACuriousMind
 
@ACuriousMind are you really here
I have a doubt
 
He was briefly
and then disappeared
I too have a doubt
Do you have any wise words on this issue @0celo7
1
Q: Causally benign spacetimes and the Minkowski torus

SlereahA notion encountered in field theory on non-globally hyperbolic manifolds is the notion of a spacetime being causally benign with respect to some field $\phi$, which is defined thusly : A neighbourhood $U$ is causally regular if $\bar U$ has a neighbourhood $U_1$ such that for every solution $...

 
@EmilioPisanty I like the chainsaw, lol
 
3:53 PM
@Semiclassical yeah, the excited bound states don't do that much when all the action is in high-energy continuum states
@Semiclassical yeah, I'm pretty proud of the chainsaw
took inordinately long to get working
there's two hard problems in computer science: we only have one joke and it's not funny.
lolz
 
Not wrong
 
@EmilioPisanty conversely, that suggests why the natural boundary is usually not relevant when you're doing stuff that involves low energies
 
@Semiclassical wait, what?
 
I mean I'm sure the joke was funny in 1930 for a few days
when the foundations of computer science were established
 
I think we started mixing SFA and the particle-in-a-ring at some point
 
3:55 PM
But its time has gone
 
well, I mean, if the natural boundary is a symptom of the high energy states
 
@Slereah cache invalidation was a thing in the thirties?
 
@EmilioPisanty No, but binary was
 
then presumably when you do low-energy stuff you're in a scenario where the natural boundary isn't relevant
 
@Semiclassical maybe? I don't know how to make that work rigorously though
 
3:57 PM
yeah, fair
 
@EmilioPisanty I thought it was the usual binary joke
 
also the Jeff Atwood tweet above
There are two hard things in computer science: cache invalidation, naming things, and off-by-one errors.
 
indeed that joke isn't funny
 
ah, c'mon, it definitely gets at least a 0.5 on the chuckle-o-meter
 
3:59 PM
I think what I partly have in mind is that, if I were to use a particle on N sites rather than on a ring, then there's only a finite number of states and therefore a finite number of terms in the propogator
 
@Semiclassical that's a tricky limit to do, though
 
oh, sur.e
 
if you start with $N$ sites with periodic boundary conditions and you keep adding sites, do you end up with a continuous position in a circle, or with an infinite chain of discrete sites?
 
pretty sure it goes to the former.
 
(real question, btw. A friend a couple of doors down is doing similar stuff.)
@Semiclassical why?
 
4:01 PM
hmm. actually
 
note that the two alternatives are Fourier duals.
$$
\begin{array}{ccc}
\begin{pmatrix}
\text{Fourier transform}\\
t\color{blue}{\text{ unbounded}}\text{ and }\color{blue}{\text{continuous}}\\
\omega\color{blue}{\text{ unbounded}}\text{ and }\color{blue}{\text{continuous}}
\end{pmatrix}
& &
\begin{pmatrix}
\text{Fourier series}\\
t\color{green}{\text{ bounded}}\text{ and }\color{blue}{\text{continuous}}\\
\omega\color{blue}{\text{ unbounded}}\text{ and }\color{green}{\text{discrete}}
\end{pmatrix}
\\ & & \\
\begin{pmatrix}
\text{Time series}\\
 
lol
nice
 
i.e. you start at DFT on the lower right and you add more and more sites
 
right.
 
does that take you up, or does that take you to the left?
 
4:03 PM
Not sure.
I don't have a lot of time-dependent intuition
 
particularly given that those two are equivalent, if you re-map what you label as $t$ and what you label as $\omega$
 
Though I'm not sure why in my scenario $t$ would be bounded/discrete to begin with
 
@Semiclassical $t$ is $x$ is $x_n$ is your discrete sites
 
5 mins ago, by Semiclassical
I think what I partly have in mind is that, if I were to use a particle on N sites rather than on a ring, then there's only a finite number of states and therefore a finite number of terms in the propogator
"a particle on N sites" is just discrete $t$ as far as that diagram is concerned
swap out $t, \omega$ for $x,p$ if you want to
 
4:06 PM
hmm. alright.
 
4:43 PM
Page 6 & 7 Dirac explains his logic for his gamma matrices, hmm
 
@JohnRennie Hey ! Good afternoon :)
 
Hi :-)
 
http://www.smallperturbation.com/physics-proof
"Those Physicists And Their "Physics Proofs"
$$\{ \gamma^{\mu} , \gamma^{\nu} \} = 2 \eta^{\mu \nu} I$$
How big do our matrices have to be in order to satisfy this? They obviously cannot be 1x1 matrices because these are just numbers that commute. It turns out that they have to be at least 4x4 but all published sources I have seen fail at explaining why. I will go through the physics proof that is often given and then set the record straight by writing a real proof. If it appears nowhere else, let it appear here!"
 
@bolbteppa : the evidence is what's really important. Evidence like optical clocks go slower when they're lower. IMHO people like Kip Thorne don't pay enough attention to the evidence.
 
You are just using bad equations (authors) then, in your language :p
 
4:54 PM
@JohnRennie What's up ? Lol I'm really tired of looking at the books all day long , how do you spend your time when you're not working (also when you're not here) ?
 
I just mess around really.
 
Omg you're Messi ?
 
At the moment I'm dismantling and reconstructing some ebooks to change the fomatting
 
Anonymous
@Tanuj cringey joke :P
 
@Blue my bad , that was really poor.
 
4:57 PM
No-one ever formats e-books exactly the way I want them, so I generally reformat them before loading them on my tablet. With .epubs that's fairly easy to do.
 
Anonymous
The kindle format is pretty good
 
omg putting Dirac matrices in Jordan canonical form
 
Anonymous
They spread out the text and it's easy on the eyes
 
I've only encountered .epub once in my life , and didn't even know what it was before I googled it
 
@Blue The MOBI format is proprietary and in fact not publically documented
 
4:59 PM
@bolbteppa : I'm not. Have a look at the GR section of Is The Speed of Light Everywhere the Same? written by the PhysicsFAQ editor Don Koks. When you use bad equations or authors, you end up with wormholes and time machines and grandfather paradoxes et cetera.
 
Anonymous
@JohnRennie Yeah, I know....wish it were open source
 
@Blue use epub and a reader like Aldiko
 
Anonymous
Does it have a Desktop app? (Aldiko)
 
You can use the app Calibre to convert MOBI to EPUB
@Blue yes, Adobe Digital Editions.
 
@JohnRennie is there anything you don't know ?
 
5:02 PM
@Tanuj anything that is non-nerdy I am hopelessly clueless about
 
@JohnRennie you're the coolest nerd I've known though ! :)
 
Cool nerd is a tautology :-)
 
Lol :') I beg to differ.
 
@Tanuj careful, your cool points are slipping away :-)
 
@Blue Advice needed
 
Anonymous
5:04 PM
Okay?
 
@Blue In these last three weeks , should I go for all out NCERT for Phy and Chem ?
 
Anonymous
@DanielSank @Mithrandir24601 @heather The private beta of the QC site is up. You can access it now!
 
Anonymous
Check your mails
 
Anonymous
@Tanuj Yeah, along with as many past papers as possible
 
Anonymous
Try to memorize inorganic and organic fully
 
5:06 PM
@Blue I just noticed!!! :D
 
I mean for chapters like heat , thermo , semi conductors , and magnetism , will NCERT be sufficient ?
oops
 
Anonymous
@Mithrandir24601 I'll ask a couple of questions, there today. I'm thinking :)
 
Anonymous
There are already two questions
 
Anonymous
Check them out
 
Anonymous
5:08 PM
"How can quantum computing help to develop a strong AI"
 
Anonymous
"Could a Turing Machine simulate a quantum computer?"
 
Anonymous
Ugh, both the questions don't show much research effort :/
 
Anonymous
I'm not sure we'd like to have those kinds there
 
Anonymous
@Tanuj Do all the questions at the back and previous years'
 
Anonymous
The Turing machine question is too common. They would have found an answer if they googled a bit
 
5:11 PM
@Blue it'll take a couple of hours to get good questions, I expect. The one on Turning machines is basic, but I feel OK with having it. Can we set up a QC chat?
 
Anonymous
Hmm, the chat is missing
 
Anonymous
They haven't set it up yet
 
Anonymous
Should I mail Robert?
 
Anonymous
Or perhaps let's just wait
 
Anonymous
A couple of days
 
5:12 PM
@Blue I did read a meta post somewhere about this being acceptable in general though
 
@Blue Got it ! Wish me luck :)
 
I've got a rehearsal now though, so I'll be properly on in a couple of hours
 
Anonymous
Added a couple of comments:
 
Anonymous
"Please add your research efforts in the way of trying to find an answer. It's important that the first few questions set a good example of how we want the site to be in the long term."
 
Anonymous
@Mithrandir24601 Cya!
 
5:22 PM
@Blue Yeah, I'm there.
 
I just googled "causally benign spacetime" and my post is the first result
I'm doooooomed
I might have to shoot Yurtsever an email
Wait, Friedman says that the subset has to be spacelike
which sounds like a more reasonable condition, although I'm not 100% sure either
 
6:14 PM
@DanielSank sigh, these things keep cropping up
 
6:40 PM
@EmilioPisanty Don't worry too much about it, bro.
Also, gimme dat sweet, sweet music.
 
@DanielSank ooooohhh, lemme see
@DanielSank how well do you know Dusminguet and/or La Troba Kung Fu?
 
The whozeewhat?
 
as starting points
roughly the same people, if I remember correctly
or at least the same lead singer, Joan Garriga
 
Anonymous
Ah, cool
 
7:38 PM
@EmilioPisanty fun follow up to that question I mentioned in passing earlier: there's a 1991 paper by Berry which talks about it (and mentions Christopher Howls in the acknowledgements, heh)
 
@Semiclassical reference?
 
The main point of interest for me is the final paragraph prior to the acknowledgements on the last page
 
@Semiclassical oooofff
 
heh
I mostly skipped the stuff in the middle tbh
 
@Semiclassical check out the newest version of the second file btw
boy is this shit slow
36 slides
 
7:40 PM
you've added a good deal of stuff to the end, yeah
 
I'm beginning to fear that there's no way in hell I'll be able to get to the end of the second hour
I'll just run out of slides midway through hour no. 2
 
hrm
btw, i came across the following notes when I was trolling for some sources earlier
you might find it interesting reading
main thing that's nice is that it's front-loaded with a lot of examples
 
@Semiclassical goodness
resurgence
yes, but not for today
 
right
 
hello condensed matter theorists i have a question.. .basically in a crystal, hk is the crystal momentum, not the electron momentum (they dont even have momentum I think.). it is claimed that the mean value of the speed of electrons is the derivative of E_nk with respect to k. How come they are treated as if they had a momentum worth hk?
 
7:45 PM
::glances around nervously:: Condensed matter theorists? Where?!
3
 
Pieter is... oh wait he isn't here
@Pieter ah he's here
 
@EmilioPisanty I'm not sure where they even start on the formal resurgence stuff
most of it just seems to be motivational examples
loooots of saddle point analysis stuff
 
Hm
Yurtsever's last known email is at Raytheon
But he doesn't work there anymore
should I contact him via Linkedin
apparently he's working in quantum computing these days
oh wait, his resume has an email
 
8:03 PM
@no_choice99 "Crystal momentum" means the momentum of electrons in a crystal lattice.
 
8:17 PM
I thought it was the momentum of a crystal when you threw it at someone
 
8:29 PM
Email sent to Yurtsever
Hopefully he will answer
 
Hey, do any of you guys know what sympathetic sideband cooling is?
 
Is that some kind of 80's bandana
 
ahaha, it's a relatively new thing I guess
@DanielSank you work in quantum computing right? do you know how they cool in ion traps past the doppler cooling phase?
 
You misspelt "hell" :P
3
 
8:43 PM
@Pieter I did not know that Pieter. when one do p psi on the psi of Bloch electrons, one does NOT get hbar k and hbar k is the crystal momentum
so i wouldn't think that the crystal moemntum, hbar k, is associated to any electron
 
gosh heckies
 
@no_choice99 The dispersion relations in the band structure give the energies of Bloch states as a function of crystal momentum in the first Brillouin zone.
The standing waves at the zone boundary have zero group velocity. And a gap to the next higher energy branch of the band.
 
hmm I see @Pieter ok so far
 
9:51 PM
The hover-over text from today's xkcd:
is brilliant.
It's &%#@ sure the case that neither the manufacturer nor google are looking out for my (android) phone anymore.
I'm going to have to root the thing just to clear the unused mandatory apps to make room for the enormous updates to other mandatory apps that have appeared lately.
 
10:12 PM
Quick question: When measuring the dynamic viscosity by dropping a sphere in a fluid and measuring the terminal velocity, why would the dynamic viscosity be a function of the radius of the sphere? Shouldn't the viscosity be constant?
 
@Kelthar I do not work with ions. I work with superconducting qubits. Unfortunately, I don't know the answer to your question.
 
10:23 PM
@JannikPitt Errr ... the viscosity doesn't have to be a function of size for that measurement to work (in fact it is easier if it isn't). It's just that the terminal velocity should depend on on the size in a manner that lets you extract the viscosity from your data.
If we use Stokes drag $D = 6 \pi \eta R v$, then plotting terminal velocity as a function of inverse-radius (for constant $m$ and $\eta$) should give you line of slope $\frac{mg}{6 \pi \eta}$ (and zero intercept).
Hm. A better assumption would be constant density which yields $$v_t = \frac{2\rho g}{9 \eta}R^2 \;.$$
So you plot $v_t$ against $R^2$ and slope is $\frac{2\rho g}{9 \eta}$.
Or you can ditch all this old-school stuff and just use a fitting package.
 
@dmckee In our lab class we had to measure the dynamic viscosity by measuring the time it takes 3 spheres of different radii to fall to the bottom of the fluid. Then we calculated the viscosity by Stokes' law and plotted the different values. (Shouldn't in theory all three values be the same?) Then we extrapolated the "real viscosity" by a linear regression taking the value of the obtained function at r=0 as the final value of the measurement. I really don't undertstand what's going on
Why would eta(r) be a linear function and why is eta(0) the undistorted value of the dynamic viscosity?
 
The Stoke's drag rule depends on very low Reynolds number. If the sizes you had were just fairly low, then the extrpolation would make sense.
You are collecting data in an accessible but not ideal regime and then projecting the data into the ideal regime.
That is a step up in sophistication from the simple analysis I was suggesting.
 
10:39 PM
But how do I know that eta(r) is linear? You could use any function to extrapolate r->0
 
Some old notes of mine say that the Stokes rule in dominate for $\mathrm{Re} < 1$, and you can use $\mathrm{Re} = 2 v \rho R/\eta$.
So you may be assuming that the Reynolds Number is linear in $R$ with $\eta$ approximately constant.
 
Okay I understand that.
 
But I think you should take a question like "Why are we making this extrapolation?" to someone who really understands the lab. Because I'm just guessing.
 
10:55 PM
Yeah my supervisor said something along the lines eta(r) can't be linear because there's a factor of r^3 in the equation haha
 

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