I am confused about the vector and dual vector in the tetrad formalism. Start from Schwarzschild metric

$$\mathrm{d} s^2=\left(1-\frac{2m}{r}\right)dt^2 - \left(1-\frac{2m}{r}\right)^{-1}\mathrm{d} r^2-r^2(\mathrm{d}\theta^2+sin^2\theta \mathrm{d}\phi^2).$$ Change to the advanced Eddington-Finkelstein coordinate it becomes

$$\mathrm{d} s^{2}=\left (1-\frac{2m}{r} \right ) \mathrm{d} v^{2}-2 dv dr-r^{2}\left(\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{d} \phi^{2}\right).$$ Then I could use it to construct the following null tetrad.

$\begin{equation} \begin{array}{l} \ell ^{a}=(\frac{\partial}{\partial r})^a, \\ n^{a}=-(\frac{\partial}{\partial v})^a-\frac{1}{2}\left(1-\frac{2 m}{r}\right) (\frac{\partial}{\partial r})^a, \\ m^{a}=\frac{1}{\sqrt{2} r}\left((\frac{\partial}{\partial \theta})^a+\frac{i}{\sin \theta} (\frac{\partial}{\partial \phi})^a\right), \\ \bar{m}^{a}=\frac{1}{\sqrt{2} r}\left((\frac{\partial}{\partial \theta})^a-\frac{i}{\sin \theta} (\frac{\partial}{\partial \phi})^a\right). \end{array} \end{equation}$

The metric with respect to the tetrad is

$\begin{equation} g_{i j}=\left[\begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & -1 & 0 \end{array}\right]. \end{equation}$

What would be the form of dual vectors with respect to $\{v,r,\theta,\phi\}$?


1 Answer 1


Where did these equations come from? You have an index of the left hand sides of the equations but no on the right hand side... Such equations/definition are just plain wrong. What I think you should have written $$\begin{equation} \begin{array}{l} \ell=\frac{\partial}{\partial r}, \\ n=-\frac{\partial}{\partial v}-\frac{1}{2}\left(1-\frac{2 m}{r}\right) \frac{\partial}{\partial r}, \\ m=\frac{1}{\sqrt{2} r}\left(\frac{\partial}{\partial \theta}+\frac{i}{\sin \theta} \frac{\partial}{\partial \phi}\right), \\ \bar{m}=\frac{1}{\sqrt{2} r}\left(\frac{\partial}{\partial \theta}-\frac{i}{\sin \theta} \frac{\partial}{\partial \phi}\right). \end{array} \end{equation}$$ Now this perhaps does define some tetrad, if we further define the what are the coordinates (I'm going to assume $(x^0,x^1,x^2,x^3) = (r,v,\theta,\phi)$) and what is the metric. A general tensor $T$ can be written in a coordinate system as $T = T^a \frac{\partial}{\partial x^a}$, where we denote $\frac{\partial}{\partial x^a}$ as the (coordinate) basis.

From this it's obvious what you should have written as $l^a$ since $$ l = \frac{\partial}{\partial r } = l^r \frac{ \partial}{\partial r } $$ Thus your vector $l^\alpha = (1,0,0,0)$. Now if we want to lower these indices, we can, just as always do $$ l_a = g_{a m} l^m = g_{a r}l^r $$ Since $l^r$ is the only non-zero component. For the process for other tetrad vector is practically the same.

Edit: Now that we have the metric, we can finish our calculation of $l_a$. Since the only non-zero metric term with $r$ is $g_{v r}$ the only non-zero term of $l_a$ will be $l_v$

$$ l_v = g_{vr}l^r = -1 $$ So now $$ l_a = -1 \; dv_a $$

  • $\begingroup$ I edit the question, It should be more clear now. $\endgroup$
    – David Shaw
    Commented Mar 10, 2023 at 17:46
  • $\begingroup$ I have edited the answer $\endgroup$
    – Nitaa a
    Commented Mar 12, 2023 at 14:41
  • $\begingroup$ I think $g_{vr}$ should be one, beause the -2 is $g_{vr}+g_{rv}$. Actually, it still confused me. Why do we use metric in $\{v,r,\theta,\phi\}$ to lower the index, rather than using the metric of the tetrad to lower the index? $\endgroup$
    – David Shaw
    Commented Mar 12, 2023 at 22:08
  • $\begingroup$ Sorry you are right about the factor. $\endgroup$
    – Nitaa a
    Commented Mar 12, 2023 at 23:28
  • 1
    $\begingroup$ I already figure out this problem. Using abstract index a, and b to represent the vector index. $l_{a}=g_{ab}l^b$, as $l^b=(\partial/\partial r)^b$, $l_{a}=g_{ar}l^r=g_{vr}(dv)_a$. Thank you! $\endgroup$
    – David Shaw
    Commented Mar 14, 2023 at 15:24

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