# Finding the components of local Minkowski basis vectors in terms of the Schwarzchild coordinates

If I am an observer in a system with Schwarzchild geometry, I can construct a set of local basis vectors such that $\textbf{e}_{\alpha}\cdot\textbf{e}_{\beta} = \eta_{\alpha \beta}$ that are valid in my local region. How would I go about expressing the components of these basis vectors in terms of the global coordinates $(t,r,\theta,\phi)$? So what I'm wondering is how to calculate $(\textbf{e}_{\alpha})^{\mu}$ for the $\mu^{th}$ component of my basis vector.

In my lectures he quotes these formula without proof and I can't figure out how to derive them from the metric.

## 1 Answer

I am not totally sure I am understanding you well, but the usual Schwarzschild metric is $$ds^2=-\left(1-\frac{2GM}{r}\right)dt^2+\left(1-\frac{2GM}{r}\right)^{-1}dr^2+r^2(d\vartheta^2+\sin^2\vartheta d\varphi^2),$$ which means the matrix of the metric components is diagonal:

$$(g_{\mu\nu})=\text{diag}\left(1-\frac{2GM}{r},\left(1-\frac{2GM}{r}\right)^{-1},r^2,r^2\sin^2\vartheta\right).$$

Diagonality of the metric components mean that the coordinate system is orthogonal, so all coordinate bases are orthogonal. You can obtain an orthonormal frame then by dividing with the square root of the metric components because $g_{\mu\nu}=g(\partial_\mu,\partial_\nu)$ (where $\partial_\mu$ is the $\mu$-th coordinate basis vector), so the square of $\partial_\mu$'s length is $g(\partial_\mu,\partial_\mu)=g_{\mu\mu}$ (no sum implied), therefore $\mathbf{e}_\mu=\partial_\mu/\sqrt{|g_{\mu\mu}|}$.

What are the components of $\partial_\mu$? For any basis, the components of the $\mu$-th basis vector in that basis is given by $(0,0,...,1,...0)$ where the 1 is in the $\mu$-th spot, so $(\partial_\mu)^\nu=\delta_\mu^\nu$ (in this coordinate system only), therefore $$(\mathbf{e}_\alpha)^\mu=\frac{\delta_\alpha^\mu}{\sqrt{|g_{\alpha\alpha}|}}=\text{diag}\left(\sqrt{1-\frac{2GM}{r}}^{-1},\sqrt{1-\frac{2GM}{r}},\frac{1}{r},\frac{1}{r\sin\vartheta}\right).$$

Edit: If you are unfamiliar with invariant notation, consider that if $e^\mu_\alpha$ is the $\mu$th component of the $\alpha$th basis vector, then $e^\alpha_\mu$ is the inverse matrix of $e^\mu_\alpha$ (this inverse matrix is actually the dual frame).

The inverse matrices satisfy $$g_{\mu\nu}=\eta_{\alpha\beta}e^\alpha_\mu e^\beta_\nu$$, so

$$ds^2=g_{\mu\nu}dx^\mu dx^\nu=\eta_{\alpha\beta}e^\alpha_\mu e^\beta_\nu dx^\mu dx^\nu=\eta_{\alpha\beta}e^\alpha e^\beta,$$ where $e^\alpha=e^\alpha_\mu dx^\mu$, so for Schwarzschild specifically $$ds^2=-a(r)dt^2+a(r)^{-1} dr^2+r^2 d\vartheta^2+r^2\sin^2\vartheta d\varphi^2=-e^0e^0+e^1e^1+e^2e^2+e^3e^3,$$ from which you can read off $e^0=\sqrt{a(r)}dt$, $e^1=\sqrt{a(r)^{-1}}dr$, $e^2=rd\vartheta$ and $e^3=r\sin\vartheta d\varphi$ and since each of these expression contain only one differential, $e^0_0=\sqrt{a(r)}$ while $e^0_{\text{any other index}}=0$ etc., so $$e^\alpha_\mu=\text{diag}\left(\sqrt{a(r)},\sqrt{a(r)^{-1}},r,r\sin\vartheta\right),$$ but what you are looking for, $e^\mu_\alpha$ is the inverse of this matrix, which gives the same result I gave before the edit (with $a(r)=1-\frac{2GM}{r}$ obviously).