# Orthogonal of tangent vector in Rindler coordinates [closed]

For 2D space time from $(t,x)$ to $(u,v)$ the transformation are $$t = u \sinh(v)$$$$x=u\cosh(v)$$

Asking to show that two families of curves $u = \textrm{constant}$ and $v = \textrm{constant}$ are orthogonal (i.e., that the scalar product of their tangent vectors is zero) by derive the components of the metric in the $(u,v)$ frame.

I get the metric in $(u,v)$ is $${ds}^2 = {du}^2 - u^2 {dv}^2$$

and say in $(u,v)$ there is

$$g_{ab}= \left[{\begin{matrix} 1 & 0 \\ 0 & -u^2 \\ \end{matrix}}\right]$$

the two tangent vectors are

$$R_{u}=(\sinh(v),\cosh(v))$$ $$R_{v}=(u \cosh(v),u\sinh(v))$$

and I cannot have the scalar product of them is zero ie:

$${R_u}{R_v}={g_{ab}}{R_u^a}{R_v^b}=u\sinh(v)\cosh(v) - u^3\sinh(v)\cosh(v) = (u-u^3)\sinh(v)\cosh(v) = ? 0$$

where does my error comes from? thanks a lot...

Once you have the metric in u,v coordinates, you notice that $g_{uv}=0$ and this shows that the basis vectors $e_u$ and $e_v$, respectively tangent to the v=const and u=const lines (no typo, you see why?) are orthogonal.