Question

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

Answer 1

@Hero @ganeshie8 @goformit100

Answer 2

Just so the smart people come and help you:P

Answer 3

haha thanks :)

Answer 4

Good luck(:

Answer 5

@Kira_Yamato

Answer 6

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

Answer 7

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

Answer 8

Where's Hero when you need him:P Lol

Answer 9

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.

Answer 10

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!

Answer 11

No problem! Happy to help!

Answer 12

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

Answer 13

what is the precise question?

Answer 14

i wanted to confirm the FVoA formula

Answer 15

post your formula, preferably with steps, to confirm

Answer 16

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

Answer 17

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 \(\beta=2\).

Answer 18

\[\frac{\partial x}{\partial t}=\beta\,\nu\]