# Special Riemannian metric on the product

Let $$M^2$$ be an orientable surface with boundary endowed with a Riemannian metric $$g$$. We know that if we put in the manifold $$M \times \mathbb{R}$$ the product metric $$\overline{g} = g + dt^2$$, then the slices $$M_t = M \times \{t\}$$ are totally geodesic and are free-boundary surfaces, which means that their boundary meet the boundary $$\partial (M \times \mathbb{R}) = \partial M \times \mathbb{R}$$ orthogonally.

Suppose now that we choose an angle $$\theta \in (0, \pi)$$. Is there a "natural" Riemannian metric $$g_\theta$$ on $$M \times \mathbb{R}$$ such that the slices $$M_t$$ are totally geodesic, meet the boundary of $$M \times \mathbb{R}$$ at a constant angle $$\theta$$ and $$g_{\pi/2}$$ agrees with the product metric $$\overline{g}$$? Is there an explicit formula?

Remark: Let $$N_t$$ be the unit normal to $$M_t$$ in the metric $$g_\theta$$, and $$\overline{N}$$ be the outward unit normal to $$\partial M \times \mathbb{R}$$ in this metric. We say that the slice $$M_t$$ meets the boundary $$\partial M \times \mathbb{R}$$ at an angle $$\theta$$ if $$g_\theta(N_t, \overline{N}) = \cos \theta$$ along $$\partial M_t$$.

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• what about $k\cdot t^2\cdot g+dt^2$? – Anton Petrunin Jun 5 '19 at 21:20
• @AntonPetrunin what is $k$ and what happens when $t=0$? – Eduardo Longa Jun 6 '19 at 17:43
• $k$ is a constant so the metric is defined only for $t>0$, but if you reparametrize it by $\ln t$ you get a metric on the product $M\times \mathbb{R}$. – Anton Petrunin Jun 7 '19 at 12:56
• Crossposted from math.stackexchange.com/questions/3237696/… – Nate Eldredge Jun 16 '19 at 3:50