- #1
ngkamsengpeter
- 195
- 0
Given a Finsler geometry (M,L,F) and $$g_{ab}^L=\frac{1}{2} \frac{\partial^2 L}{\partial y^a \partial y^b}$$
$$g_{ab}^F=\frac{1}{2} \frac{\partial^2 F^2}{\partial y^a \partial y^b}$$
$$F(x,y)=|L(x,y)|^{1/r}$$
I manage to get the following form
$$g_{ab}^F=\frac{2|L|^{2/r}}{rL}( g_{ab}^L+\frac{2-r}{2rL} \frac{\partial L}{\partial y^a}\frac{\partial L}{\partial y^b})$$
However, how to obtain the inverse of the the metric? That is how to obtain the following form:
$$g^{Fab}=\frac{rL}{2|L|^{2/r}}( g^{Lab}-\frac{2(2-r)}{r(r-1)L} y^a y^b)$$
I am new to this so have no idea how to get the inverse. Any help will be appreciated. Thanks.
$$g_{ab}^F=\frac{1}{2} \frac{\partial^2 F^2}{\partial y^a \partial y^b}$$
$$F(x,y)=|L(x,y)|^{1/r}$$
I manage to get the following form
$$g_{ab}^F=\frac{2|L|^{2/r}}{rL}( g_{ab}^L+\frac{2-r}{2rL} \frac{\partial L}{\partial y^a}\frac{\partial L}{\partial y^b})$$
However, how to obtain the inverse of the the metric? That is how to obtain the following form:
$$g^{Fab}=\frac{rL}{2|L|^{2/r}}( g^{Lab}-\frac{2(2-r)}{r(r-1)L} y^a y^b)$$
I am new to this so have no idea how to get the inverse. Any help will be appreciated. Thanks.