We'll consider $(N, g)$ a Riamannian Manifold and $\overline{g} = e^{f}g$ a conformal metric. Let M be a hypersurface in N, $\overline{H}_M$ and $H_M$ the mean curvature of M with respect to the metrics $\overline{g}$ and g, respectively. I would like some help to prove that

$$ \overline{H}_M = e^{-f}( H_M -2g( \nabla f, \eta)) $$

where $\nabla$ is the gradient with respect to metric g and $\eta$ is a tangent field in M.

We'll consider $(N, g)$ a Riamannian Manifold and $\overline{g} = e^{2f}g$ a conformal metric. Let M be a hypersurface in N, $\overline{H}_M$ and $H_M$ the mean curvature of M with respect to the metrics $\overline{g}$ and g, respectively. I would like some help to prove that

$$ \overline{H}_M = e^{-f}( H_M -2g( \nabla f, \eta)) $$

where $\nabla$ is the gradient with respect to metric g and $\eta$ is a normal vector field in M.