# Building a Lorentzian metric from an induced metric in Euclidean space

As an exercise, I wanted to apply the tool of taking some "time-evolving" surface embeddable in Euclidean space, defined parametrically as $$X_0(u, v, t), X_1(u, v, t), X_2(u, v, t)$$, and with the induced metric obtained as

$$g_{ab} = \frac{\partial X_{\mu}}{\partial a} \frac{\partial X_{\nu}}{\partial b} g_{\mu \nu}$$

Try to adapt the $$g_{ab}$$ to a Lorentzian metric description

So, my first instinct was to just take the $$g_{tt}$$ component and flip the sign, but I'm unconvinced this is a sound procedure, specially since the induced metric in its current coordinates is not even diagonal. However this gut instinct is grounded on the fact that the flip of sign will make the determinant of the metric negative, which is what it should be in a Lorentz metric.

Since I already picked up a special coordinate as the time, how can I relate the euclidean induced metric with the actual Lorentzian metric I want to work with?

The Riemannian metric is not sufficient. You need an additional line field $$v_{a}$$, and if you want your Lorentzian metric to be non-degenerate, this line field should be everywhere non-zero.

Choosing a couple $$(g_{ab}^{R}, v_{a})$$ give rise to the following Lorentzian metric

$$$$g_{ab}^{L} = g_{ab}^{R} - 2 \frac{v_{a}v_{b}}{g_{R}^{cd}v_{c}v_{d}}$$$$ where the sub(super)-scripts $$L,R$$ refers to Lorentzian and Riemannian respectively. Note that this is true in the other direction as well, i.e. given a Lorentzian metric, by choosing a Riemannian metric is it possible to diagonalize $$g^{L}_{ab}$$ with respect to $$g_{ab}^{R}$$ to obtain a line field, that is the corresponding eigenvector with negative eigenvalue.

References and additional details can be found in Ellis & Hawking book.

• suppose that I don't want to work on so-called "physical units" and want to see the $c$ terms on the metric. Is enough to just use $v_{a} = (c, 0, 0)$? Commented Jun 14 at 16:47
• actually is not, any units in the $v_a$ are cancelled, so it seems the Riemannian metric has to have the $c^2$ factors in the $g_{tt}$ and $c$ factors in the $g_{tu}, g_{tv}$ Commented Jun 14 at 18:18
• If you want the $c$ factor you should choose a priori which of the coordinates have a dimension of time.
– Pipe
Commented Jun 15 at 0:12