Let $(M,g)$ be a Riemannian $n$-manifold, and $\varphi: M \to \Bbb R$ be a smooth map. Define another metric tensor by: $$\widetilde{g} = g - {\rm d}\varphi \otimes {\rm d}\varphi$$I know that $\widetilde{g}$ is Riemannian if $\|{\rm grad}_M\varphi\| < 1$ and Lorentzian if $\|{\rm grad}_M\varphi\| > 1$ $(\|\cdot\|$ induced by $g$, of course$)$.
To see the second case it is easy, it suffices to evaluate $\widetilde{g}$ in ${\rm grad}_M\varphi$ itself - it will be a timelike vector.
Is there an easy way to check that $\widetilde{g}$ will be positive definite? I don't quite know if there's a result about this, but I feel that computing the metric on ${\rm grad}_M\varphi$ should be enough. I'm not sure how to exactly compute $\det \widetilde{g}$ here... Thanks.
Take a vector field $v$, and compute $\bar g(v,v)$. You need to show this is strictly positive when $v \ne 0$. But $\bar g(v,v) = g(v,v) - (d\varphi(v))^2$. Isn't this enough to make it work?