OpenStudy (anonymous):

Can anyone please explain to me the 'first variation of area' formula, when linked to a hyperspace?

4 years ago
OpenStudy (anonymous):

@Hero @ganeshie8 @goformit100

4 years ago
OpenStudy (anonymous):

4 years ago
OpenStudy (anonymous):

haha thanks :)

4 years ago
OpenStudy (anonymous):

Good luck(:

4 years ago
OpenStudy (anonymous):

@Kira_Yamato

4 years ago
OpenStudy (anonymous):

or if someone could at least confirm for me what the formula is? As i'm getting different formula from different sources Thanks :)

4 years ago
OpenStudy (kira_yamato):

Can't help on that problem... I'm not up to standard yet I guess... Sorry....

4 years ago
OpenStudy (anonymous):

Where's Hero when you need him:P Lol

4 years ago
OpenStudy (aakashsudhakar):

From what I understand (which, with Riemannian geometry, isn't much to be honest), isn't a First Variation of Area formula a formula that creates a relationship between a hypersurface's average curvature and the hypersurface's derivative (rate of change) as it 'expands' (so to speak) outwardly and normally? $\frac{ d }{ dt } dA = H dA$ I *think* that's it, but I'm relying on fragments of Riemannian geometry online databases that I've skimmed through, not actual coursework or anything. You may want to get confirmation from someone else.

4 years ago
OpenStudy (anonymous):

okay, no that's good! that's the formula that I have, just there was another one that contrasted elsewhere so i was a little bit wary of it! Thankyou so much!

4 years ago
OpenStudy (aakashsudhakar):

No problem! Happy to help!

4 years ago
OpenStudy (anonymous):

\begin{align} \mbox{area element: }{\rm d}\sigma&=\sqrt{{\rm det}g_{ij}}{\rm d}x^1\wedge\ldots\wedge{\rm d}x^{n-1}\\ \implies\frac{\partial}{\partial t}g_{ij}&=2\beta h_{ij}\\ \implies\frac{\partial}{\partial t}{\rm d}\sigma&=\beta\,H\,{\rm d}\sigma\\ {\rm Area}(S_t)&=\int_{S_1}\beta\,H\,{\rm d}\sigma\\ \end{align}

4 years ago
OpenStudy (anonymous):

what is the precise question?

4 years ago
OpenStudy (anonymous):

i wanted to confirm the FVoA formula

4 years ago
OpenStudy (anonymous):

post your formula, preferably with steps, to confirm

4 years ago
OpenStudy (anonymous):

Currently I have d/dt (dA)=HdA I'm only just approaching the concept of FVoA, as it's for my thesis. So I don't really know that much about it yet. I was just getting differing formula - one said 2H, one said H

4 years ago
OpenStudy (anonymous):

yes. that is true too. You see, the function $$\beta$$ and the factor "2" arise from the type of evolution of the hypersurface. "2H" is a specific solution when $$\beta=2$$.

4 years ago
OpenStudy (anonymous):

$\frac{\partial x}{\partial t}=\beta\,\nu$

4 years ago