1:04 AM
Why would $\vec r$ be a function of either of the angles on a sphere?

5 hours later…
5:37 AM
@Charlie As an example that it can be done, from $\mathbf{r} = x \hat{i} + y \hat{j} + z \hat{k}$, with $x = r \sin \theta \cos \phi$, $y = r \sin \theta \sin \phi$, and $z = r \cos \theta$, if we keep $r$ fixed and set $\phi = f(\theta)$ then you'd have it as a function of the angle $\theta$, but in general they can be independent right
@imbAF You should have $\mathbf{r}(r,\theta,\phi)$ not $\mathbf{r}(\theta,\phi)$, keeping $r$ fixed when you want to restrict to the surface of a sphere of some fixed radius $r$
If everything is parametrically restricted to a surface then you can set $r = r(u,v)$, $\theta = \theta(u,v)$, and $\phi = \phi(u,v)$, in which case $\vec{r} = \vec{r}(r(u,v),\theta(u,v),\phi(u,v)) := \vec{r}(u,v)$ and $\vec{A} = \vec{A}(r(u,v),\theta(u,v),\phi(u,v)) := \vec{A}(u,v)$ holds, obviously one can then use the notation $u = \theta$, $v=\phi$,
and from the context it's usually clear we're talking about 3D vectors so if they have two arguments $(\theta,\phi)$ it's clear we mean that 3 initial variables are expressed in terms of two parameters $\vec{r} = \vec{r}(\theta,\phi) = \vec{r}(x(\theta,\phi),y(\theta,\phi),z(\theta,\phi))$

5:59 AM
Can anybody help me?
The Green function is the solution of an equation to the input (i.e., source) $\delta(x'-x)$. Now for any other source $f(x)$ you have $f(x)=\int_{x'}f(x')\delta(x'-x)dx'$. You usually work with linear equations, so if $G(x',x)$ is the solution of your equation to $\delta(x'-x)$, then $\int_{x'}f(x')G(x',x)dx'$ is the solution of your equation to $f(x)$. — Newbie Jan 19 at 4:59

3 hours later…
9:09 AM
3

The current answers to the Academia HNQ Should I ask for permission to name a mathematical theorem? can be loosely paraphrased as "go for it" and "it would be bad if you didn't", e.g. There is no need to ask permission, and mostly likely they will be happy to have the theorem referred to with th...

5 hours later…
2:24 PM
@Slereah The mLab
2

AI generated?

2:36 PM
uses a parsing expression grammar
if that counts as AI

3:13 PM
makes as much sense to me as nlab sometimes does

2 hours later…
4:59 PM
What's the relationship between speed of light and gravity?
when things move at the speed of light near an object with a gravitational force, the pull should be weaker than the object that moves at a slower speed right?

2 hours later…
6:51 PM
@NiharKarve Provided the straightforward relative cohomologies are co-accessible, all co-reflectively co-perfect covers are co-fibrant. Historically, the notion of a lift of a L-accessible m-co-composable co-spectrum is an approximate solution to the problem of finding co-perfect arrows that satisfy ⋁rQQ=k(⨆p:G⇐sR) with respect to dendroidally P-co-reflective bundles over the chain complex of perfectly S-infinite ends.
I could totally buy that as legit

7:09 PM
That mLab stuff is simply incredible
Can't find a physics one yet
Title: Right sheaf
1. Idea
Sometimes, an analogous definition makes sense in the context of acyclic resolutions.
"In groupoid-like internal 0-cells (where by "extensional" we mean, of course, a lift of a ..." it messes up these bracket references all the time (e.g. here it didn't mention "extensional" before the brackets)
Title: Enriched pushout
1. Idea
Informally, if the evident diagram commutes, it is said that ...
Definition: In the context of co-homotopy theoretic morphisms, pure co-operads are simply co-pure bundles over endofunctors.
5. Related Concepts
Informally, an analogous definition makes sense in the context of co-framed co-structures ...
I'm learning more from m than n :\