Manuscript Em02-E@1 · Em02-E-v1.tex · theorem · f3795050

Theorem: Complete relational lapse through second order

Complete classical lapse expansion › Composition with the relational map · lines 758–796 · proof included or linked

Em02-E@1#thm:lapse, Em02-E@1#eq:n1, Em02-E@1#eq:n2

The coefficients are

n_{1} n_{2} = N_{1} + N ˙_{0} u_{1}, = N_{2} + 2 u_{1} N ˙_{1} + N ˙_{0} u_{2} + N ¨_{0} u_{1}^{2} - 2 Y_{1}^{i} \partial_{i} n_{1} .

Every quantity on the right is given explicitly by (eq:u), (eq:q1), (eq:B), (eq:Y), and (eq:W1)–(eq:N).

The second spatial rod coefficient $Y_{2}$ does not occur in $n_{2}$ .

Proof. The composition rule gives

n_{2} = N_{2} + 2 α_{1}^{μ} \partial_{μ} N_{1} + α_{2}^{μ} \partial_{μ} N_{0} + α_{1}^{μ} α_{1}^{ν} \partial_{μ} \partial_{ν} N_{0} .

Since $N_{0}$ depends only on time, substitution of (eq:alpha) leaves

N_{2} + 2 u_{1} N ˙_{1} + 2 z_{1}^{i} \partial_{i} N_{1} + N ˙_{0} (u_{2} + 2 z_{1}^{i} \partial_{i} u_{1}) + N ¨_{0} u_{1}^{2} .

Use $z_{1} = - Y_{1}$ and $\partial_{i} (N_{1} + N ˙_{0} u_{1}) = \partial_{i} n_{1}$ . The spatial part of $α_{2}$ multiplies $\partial_{i} N_{0} = 0$ , which proves the last assertion.

Displayed formulas in this passage

Em02-E@1#eq:n1 · line 761
$n_{1} n_{2} = N_{1} + N ˙_{0} u_{1}, = N_{2} + 2 u_{1} N ˙_{1} + N ˙_{0} u_{2} + N ¨_{0} u_{1}^{2} - 2 Y_{1}^{i} \partial_{i} n_{1} .$
Em02-E@1#eq:n2 · line 761
$n_{1} n_{2} = N_{1} + N ˙_{0} u_{1}, = N_{2} + 2 u_{1} N ˙_{1} + N ˙_{0} u_{2} + N ¨_{0} u_{1}^{2} - 2 Y_{1}^{i} \partial_{i} n_{1} .$
unlabeled · line 779
$n_{2} = N_{2} + 2 α_{1}^{μ} \partial_{μ} N_{1} + α_{2}^{μ} \partial_{μ} N_{0} + α_{1}^{μ} α_{1}^{ν} \partial_{μ} \partial_{ν} N_{0} .$
unlabeled · line 788
$N_{2} + 2 u_{1} N ˙_{1} + 2 z_{1}^{i} \partial_{i} N_{1} + N ˙_{0} (u_{2} + 2 z_{1}^{i} \partial_{i} u_{1}) + N ¨_{0} u_{1}^{2} .$