I think the vector case is a bit more subtle, for the following reason

first, note that \vec{F}(\vec{r}):=\vec{r}/r^3=\nabla (1/r)=\nabla G(\vec{r}) with G defined as in the last problem

so the fourier transform is just \vec{F}(\vec{k}) = \vec{k} G(\vec{k})=4pi \vec{k}/k^2 which is still rather nice

what makes things a little troublesome is that, unlike the second problem, the third component of \vec{F}(\vec{k}) is k_3-dependent and therefore the inverse Fourier transform won't be the same as in the first problem

this appears tractable for \vec{r}/r^3, but any higher z-derivatives seem to introduce some problems. so i'm wondering if you're interested in z- derivatives as well, or just the simpler x & y cases

i will be doing for sure the derivatives along x and y; for what concerns the z derivative i might have found a short-cut.

ok. to be clear, the issue i'm seeing is that z-derivatives lead to k_3 integrals of the form \int_-\infty^\infty k^n exp(i k_3 z) dk_3 /(k^2+k_3^2) which don't exist for n>1

@user3810266: what kind of shortcut did you have in mind?

i would only need a single z derivative anyway.

ok the short cut is the following

In practice i actually need to transform the following derivative: \frac{\partial}{\partial z} \frac{\vec{r}}{r^{3}} evaluated at z=h

which makes the function to transform to be h\frac{3x}{(x^2+y^2+h^2)^(5/2)}

which is the x component; otherwise the y component becomes:h\frac{3y}{(x^2+y^2+h^2)^(5/2)}

these two functions are respectively the derivative along x of \frac{x}{{(x^2+y^2+z^2)^(3/2)} evaluated at z=h and the derivative along y of of \frac{y}{{(x^2+y^2+z^2)^(3/2)}

for which the paper i'm following reports the transformation to be respectively \frac{ i k_{1}}{k} e^{-kz} and \frac{ i k_{2}}{k} e^{-kz}

well

awwww i miswrote

those two components should be fine, since you're essentially taking the planar fourier transform of partial_x partial_z and partial_y partial_z of 1/r

it's the z-component that's worrisome

true

actually the the functions i need to transform are the x and y derivatives of \frac{1}{(x^2+y^2+h^2)^(3/2)}

hmm. that's not quite the same as \vec{r}/r^3

numerator is different

i know

i both need the x and y derivatives of \vec{r}/r^3 and h\frac{3x}{(x^2+y^2+h^2)^(5/2)}

these are the functions that i have to transform

while for the first two you showed an elegant way to proceed

well, i'd suggest trying to frame it as much as possible in terms of derivatives

ok

which paper is this btw

h\frac{3x}{(x^2+y^2+h^2)^(5/2)} is actually the \frac{\partial^{2}}{\partial x \partial z} \frac{\vec{r}}{r^{3}} evaluated at z=h

wouldn't the second partial of something like \vec{r}/r^3 have denominator 1/r^7?

room topic changed to Semiclassical's methods: Conversations with Semiclassical (no tags)

(couldn't resist a small "semiclassical methods" pun)

true

my mistake again...i apologize

if it's partial_x partial_z of 1/r, that's actually not too bad

it is the \frac{\partial}{\partial x} \frac{z}{r^{3}} evaluated at z=h

ok, yeah, that should be doable

i found it to be -i \frac{k_{1}}{z} e^{-kz}

according to my shortcut

hmm

the i k_1 is definitely ok

the real question is just the FT of \vec{r}/r^3

taking x- or y- derivatives from there will just generate factors of i k_1 and i k_2 which are trivial

for \vec{r}/r^3 i'll take a moment to check if that works

sure!

ok: first, \vec{F}(\vec{r})=\vec{r}/r^3 = -\nabla (1/r) (said that above but missed the minus sign before by accident)

thanks again, i really appreciate your help...as engineer i'm not very good at this level of math

ok

no worries. i would be interested in seeing the paper itself though

so the same method as the second problem gives the 3D FT as \vec{F}(\vec{k}) = -\vec{k}*4\pi / k^2

sure, i can give you the title authors and year. and also the link

sure

let me now split up \vec{F} as \vec{F}_p (the xy-plane components) and the z-component F_z

fyi, the k^2 in the 3D FT is k_1^2+k_2^2+k_3^2 not just k_1^2+k_2^2. being a bit sloppy

ok, clear up now!+

for the planar part of \vec{F}(\vec{k}), the k_3 dependence is the same as that of 1/\vec{k}^2. thus we get the same inverse transform in z as before and so \vec{F}_p (k_1,k_2,z) = (FT of original problem) * (-i k_1,-i k_2) := (FT of original problem)*-i \vec{k}_p

so the planar part is pretty trivial

for the vertical/longitudinal component, we want the inverse transform in z of -i k_3*4\pi/k^2

which done directly requires us to compute the integral \int_{-\infty}^\infty dk_3 k_3 exp(-i k_3 z)/(k^2+k_3^2)

now, one way to handle this integral is to recognize it as i \partial_z acting on \int_{-\infty}^\infty dk_3 exp{- i k_3 z)/(k^2+k_3^2) = i partial_z pi exp(- |z| k)/k (same integral as in first problem)

ok clear

...i just realized something

rather head-smackingly obvious

if we want to get the z-component of the planar FT of \vec{r}/r^3

all we need to do is take a z-derivative of the planar FT of -1/r

ah true!

lol. that was rather silly

and since the planar FT of 1/r is 2 pi e^{-k |z|}/k

eyeballing the derivative, taking -\partial_z gives 2 pi sign(z) e^{-k |z|}

we get obviously the same result...but in a very easy way!

right. no sense in making our lives difficult!

though you can see from there what's a bit odd about that derivative, namely that at z=0 it doesn't make a ton of sense

but that's not an issue if you only care about z=h>0

yes exactly, i have h>0

is that result consistent with the paper (which i'd still like to see the link for)

the title is: Brownian motion near a partial-slip boundary: A local probe of the no-slip condition Authors : E. Lauga T.M. Squires Journal: Physics of Fluids Link : scitation.aip.org/content/aip/journal/pof2/17/10/10.1063/…

all these things are in the appendix

which is rather long

oh, so this is a fluid mechanics motivated problem?

yes, in Stokes regime

neat

where the equations are linear and prone to analytic solutions

right.

and squires is a pretty well known worker in the realm of fluid mechanics if memory serves

Yes he is he works at Berkeley

also, lauga includes the paper on his Cambridge webpage here

so no need to deal with paywalls

ehmmm what are paywalls?

(is there a way to save or download the conversation to be useful for the future?)

should be able to find this again, yes, since we're doing it on chat

probably the simplest way to keep track of it is to star/favorite this room (upper RH corner of the page)

and by paywalls i just mean that, if i want to look at the paper through that scitation link, i have to log in through my university's library in order to have access to it for free

but since lauga uploaded it to his own webpage, there's no need to deal with that

ahhhh clear I'm logged from my university atm so i didn't worried about that...

nod. i'm working off-campus at a coffee shop atm so i do

btw, if you look at the first equation in their appendix A

you'll see that they defined their Fourier transform using a factor of 1/2pi

which is what accounts for the factor of 2pi difference between the results I gave and those in their appendix

...actually, hmm. may have to take that back---I may have made an error myself

think my inverse fourier transform should've been 1/(2pi)^2 not 1/(2pi)

in which case the 2pi disappears anyways. hmmmm

anyways, i'd suggest that the best way to think about all the derivations in their appendix D

actually is was not caring too much about the coefficient in front of the transformations as it depends on how they defined it...but anyway i think itìs ok the factor

yes it is exactly what i meant to do with my short cut

well, it does matter a little

mostly for checking limiting values of the equation. it's nice to get rid of factors of 2pi discrepencies

entirely irrelevant for the functional form, though

anyways, what i was meaning to say was: if you can write f(r) in terms of Cartesian derivatives acting on 1/r

then taking the 2D fourier transform is easy: trade all x- and y- derivatives for i k_1 and i k_2 factors, and carry out any necessary z-derivatives. done.

some of those are a bit trickier than others, though, especially in the second part of the list. hmmm

yes \frac{xy}{r^{3}} for example...

yeah

luckily i don't have to transform that function

the fact that it has a factor of k^3 on the bottom is intriguing

hah, yes

(ugh, laptop decided it wanted to reboot for updates.)

one thing i do notice about some of those other examples is that they look like legendre polynomials

for instance, 1/r^3(1-3x^2/r^2) = -2/r^3 P_2(x/r)

so something interesting may be happening there, albeit perhaps irrelevant to your interests

anyways let me know if any other interesting questions come up from this

good luck!

of course I will! your help has been precious. I might come up with some other question about this paper, expecially regarding the integrals appearing in sections 2A and 2B that for the moment i'm integrating numerically...but that's another problem so thank you again!

Hi, coming a second back to the topic, i'm facing the usual two dimensional fourier transform of \frac{1}{r^{3}, i've found it to be \frac{1}{z} e^{-kz}; i took this fourier transform by considering the fourier transform of \frac{z}{r^{3} which is e^{-kz} and dividing it by z; is that correct?

Nice room title!