Baez taught the corresponding classical mechanics class last fall, but my teaching load was insane, and I couldn't manage to fit it into my schedule :\
first off, when $w=0$ it becomes diagonal. Moreover, the two operators along the diagonal are related by $\theta\mapsto \pi-\theta$. (I could do pi+theta but I've got reasons not to.)
@Secret In essence, it's a quantifier-swapping sort of dealie. It replaces "there exists an infinite thingy such that for every blah…" with "for every blah there exists a finite thingy such that…"
For example, compare $\exists n\forall k,n>k$ and $\forall k\exists n,n>k$
(and > should be $\geq$)
the strange thing is that $\exists n\forall k[n\geq k]$ seemed to have $n$ as a maximal element but $\forall k\exists n[n\geq k]$ does not seemed so (e.g. natural numbers fit that expression)
@0celo7 I suspect one could interpret that term rather literally as an elastic potential energy if, for instance, you took $u(x,y)$ to be the height of a surface above the xy-plane
with each part of the surface trying to go to $u=0$ in order to minimize elastic energy
for that matter, though, I think that the grad(u)^2 terms would also correspond to elastic energy, but now in regards to nearby points on the surface wanting to be at the same height
of course, for a diff geo presentation this is all irrelevant.
nerd gang nerd gang nerd gang (x57 more times), spent 10 $$$ on new game, amphetamine is better than cocaine, i read a paper forgot the author's name ...
@SoumyoB I was pushed by some insane advisor. That is, I had really bad advisor who used to always discourage me and make me feel bad about myself. Then, I changed him so in order to catch up in my master's I had to work really hard
@EricSilva What is the right name for the kind of manifold the path space $\Omega(M; p, q)$ of paths from $p$ to $q$ in $M$ is?
For $\gamma \in \Omega(M; p, q)$ the tangent space $T_\gamma \Omega(M; p, q)$ being the space of vector fields $X$ along $\gamma$ such that $X(p) = X(q) = 0$ (this is the vector field you variate $\gamma$ along by pushing $\gamma(t)$ along $X(\gamma(t))$)
@Daminark What's terrible about manifold with corners is that topologically they are the same thing as manifold with boundaries. $\{(x, y) \in \Bbb R^2 : x \geq 0, y \geq 0\}$ is homeomorphic to the upper half plane.
Corners are not topologically any different from boundaries. They are different smoothly!
@Mathein I will say there's a physical science requirement, and I'll be able to use mathematical logic, combinatorics, and something TBD that'll probably be basically math
@Daminark oh there's funny thing here, you have to take two courses that are not in the subject you're studying. You can't take "mathematical logic" for that, because that's a math course, but "mathematical logic for philosophers" is a philosophy course, so that's okay
I mean here we have logic in the philosophy department and it's rather different from that in the math department so maybe? That or lmao loopholes are the best
