# Numerical Relativity: Components of the Lapse Function?

1. Oct 16, 2009

### Wallace

I have a technical question about numerical relativity, hopefully someone can help.

In the usual 3 + 1 decomposition in NR, the four gauge freedoms are expressed via the lapse, $$\alpha$$ and three shift components $$\beta^i$$. In a finite element numerical scheme, each grid point will have a value for these 4 components.

Now, looking at evolution equations on the other hand, I see terms that want to take the covariant derivative of the lapse function, which I can't understand since it is a scalar, and hence the covariant derivative reduces to the regular derivative.

To be explicit, take for instance the ADM evolution equation for the extrinsic curvature. The textbook I have ("Elements of Numerical Relativity" by Carles Bona and Carlos Palenzuela-Luque), write this down as

$$(\partial_t - L_{\beta} ) K_{ij} = -\alpha_{j;i} + \alpha [ ....]$$

I've left the rest of the equation out for simplicity (note that $$L_{\beta}$$ is the Lie Derivative, I couldn't work out how to make the nice curly L with the tex tags). See that both $$\alpha$$ and $$\alpha^i$$ appear which I don't understand. It would make sense if this book was using $$\alpha^i$$ to denote the shift vector, but as you can see from the LHS (and it made clear in the book) it uses $$\beta^i$$ for this.

Any ideas? I'm just replacing the Covariant derivative with the regular one for $$\alpha$$ but maybe the equation is actually telling me something very different?

2. Oct 16, 2009

### Wallace

I think I found the answer, comparing to other literature it looks like they were meaning

$$\alpha_i \equiv \partial_i \alpha$$

that makes sense, because the covariant derivative of that would not be 'trivial', i.e. reduce simply to the regular derivative.