Numerical Relativity: Components of the Lapse Function?

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In numerical relativity, the four gauge freedoms are represented by the lapse function, α, and three shift components, β^i. The discussion centers around the confusion regarding the covariant derivative of the lapse function, which is a scalar and typically reduces to a regular derivative. The ADM evolution equation for extrinsic curvature includes terms that suggest a more complex relationship involving both α and α^i, leading to questions about their definitions. The resolution appears to be that α_i is interpreted as the partial derivative of α, indicating that the covariant derivative is not trivial in this context. This clarification aligns with findings in other literature, enhancing the understanding of the equation's implications.
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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?
 
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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.
 
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