Hi All..

does $\Lambda^T\Lambda=I$ lead to $\Lambda\Lambda^T=I$? If so, how to prove it?

I think this should be the case if $\Lambda$ is an element of 0(3) because a group element should satisfy $\Lambda^{-1}\Lambda=I$ and $\Lambda\Lambda^{-1}=I$, but I don't know how to prove it.

@CaptainBohemian math.stackexchange.com/questions/3852/if-ab-i-then-ba-i

Thank you @Yly for your comments in the question/answers physics.stackexchange.com/questions/598114/… You do care about the truth, possibly as much as I do. +1000 for your comments.

Good morning

Consider a semi-Riemannian manifold $\zeta^{2,2}$ with metric, $g=\frac{dxdy}{xy}+\frac{dudv}{v-uv}.$ How could you define a 3-dimensional "slice" of $\zeta^{2,2}$? I realize the question is ill-posed as it stands, because I have to precisely define what I mean by "slice", but I'm struggling to precisely define it

in a Lorentzian setting you'd be looking for a foliation by Cauchy surfaces, but I have no idea why you're looking at signature (2,2) or what sort of "slice" you're loking for

okay. Also do you know how to express $\Bbb R^{3,1}$ as a product manifold?

I feel we've had this conversation before :P

$\mathbb{R}^{n,m} = \mathbb{R}^{p,q} \times \mathbb{R}^{s,t}$ where $p+s = n$ and $q +t = m$.

oh right, sorry. $\Bbb R^{1,0} \times \Bbb R^{1,0} \times \Bbb R^{1,1}$

is that correct^?

sure, that's one way to do it

Hey All..

I'm confused that this article says wool is a conductive material: scientificamerican.com/article/…

If it's conductive, why doesn't it behave like a metal and doesn't get back the electrons quickly?

Progress

Hello everyone

I've been stumped trying to prove the rolling without slipping condition for a particular special case. Anyone interested in helping me out?

@aderchox Wool is a pretty good insulator, but not as good as rubber. The article should probably say “electronegative,” but it’s for kids.

@rob Yeah guessed that might be the reason, but couldn't be sure about it. Thanks a lot.