How can I determine the rotation of basis vectors in non-Euclidean space?

[latentcorpse]

Hi,

I have a 3d space with metric $$ds^2= -r^a dt^2 + r^bdr^2 +r^2 dy^2$$ and I need to construct an orthonormal frame.

The first of these three basis vectors is fixed, let's say as $$e_0=A \partial_t + B \partial_r + C \partial_y$$

To find the other two I set $$v_1=\partial_t, v_2=\partial_r$$ and then apply the Gram-Schmidt procedure. Ultimately I end up with

$$e_1=D \partial_t + E \partial_r + F \partial_y$$ and $$e_2=G \partial_t + H \partial_r + J \partial_y$$

Since I have exact (albeit rather lengthy and complicated expressions) for $$A,B,C$$ in terms of variables such as energy and momenta, the Gram-Schmidt procedure does give me expressions for all the other constants (although they are in terms of energy and momenta as well - importantly, they are not numbers!).

Now, I am trying to compare my result to the result in the literature where they explicitly construct such a basis with $$e_1$$ only pointing in the t and r directions. Since $$e_0$$ is fixed and I can't alter it at all, this is really a rotation of the other two orthonormal basis vectors i.e. I want to rotate $$e_1,e_2$$ such that $$e_1$$ no longer has a y component.

I tried to do this by multiplying as follows:
$$\begin{pvector} e'_1 \\ e'_2 \end{pvector} = \begin{array} \cos{\theta} & \sin{\theta} \\ -\sin{\theta} & \cos{\theta} \end{array} \begin{pvector} e_1 \\ e_2 \end{pvector}$$

(Does this even hold when we are no longer in Euclidean space?)

Imposing $$(e'_1)^y=0$$ (I use this notation to denote the y component), I can read off the following equation

$$0=\cos{\theta} (e_1)^y + \sin{\theta} (e_2)^y \Rightarrow \tan{\theta}=-\frac{(e_1)^y}{(e_2)^y}$$

However, this gives me an expression for $$\theta=\tan^{-1}{(-\frac{(e_1)^y}{(e_2)^y})}$$. This will be some ratio of the energy and momenta and not an actual number. So how can I use it to determine $$\sin{\theta}$$ and $$\cos{\theta}$$ which I will need in order to work out how the ohther components of $$e'_1,e'_2$$ look like?

Thanks very much for your help..

 

[fzero]

(Does this even hold when we are no longer in Euclidean space?)

One reason to introduce frame fields is that the Euclidean rotations act naturally on them.

Imposing $$(e'_1)^y=0$$ (I use this notation to denote the y component), I can read off the following equation

$$0=\cos{\theta} (e_1)^y + \sin{\theta} (e_2)^y \Rightarrow \tan{\theta}=-\frac{(e_1)^y}{(e_2)^y}$$

However, this gives me an expression for $$\theta=\tan^{-1}{(-\frac{(e_1)^y}{(e_2)^y})}$$. This will be some ratio of the energy and momenta and not an actual number. So how can I use it to determine $$\sin{\theta}$$ and $$\cos{\theta}$$ which I will need in order to work out how the ohther components of $$e'_1,e'_2$$ look like?

There are elementary trig identities like

$$ 1 + \tan^2\theta = \frac{1}{\cos^2\theta}$$

that you can use to express $\sin$ and $\cos$ in terms of $\tan$.