Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 311 / 3 / i / Solution

For any nonzero $g$-causal vector $v^a$,For timelike $v$, the first term is negative. For null $v$, a nonzero null vector cannot be orthogonal to a timelike vector, so $(t\cdot v{)}^2>0$. Thusin both cases. Every $g$-causal direction is strictly timelike for $\tilde{g}$, so its light cone is strictly wider.