Calculating the minkowski metric in polar coordinates I'm trying to calculate the Minkowski metric on the positive hyperboloid in polar coordinates. I wrote a proof that looked fine to me, but re-reading it uncovered a point I don't understand and can't justify. Details following.
Definitions and notations:
Let h be the pseudo-rimannian metric on $\mathbb{R}^{n,1}$:
$$h=(dx^i)^2-dt^2$$
Let $\mathbb{H}^{n}_+$ be the hyperboloid, defined as following:
$$
\mathbb{H}^{n}_+=\{  (x,t)\in\mathbb{R}^{n,1}  \ \ \ :\ \ \  h(x,t)=-1\ ,\ t>0 \}
$$
Observe that $\phi: (0,\infty)\times S^{n-1}(1)\rightarrow \mathbb{H}^{n}_+$, defined as following, is a diffeomorphism:
$$
\phi (r,\theta )=(sinh(r)\theta ,cosh(r))
$$
Finally, let ${ds^{-1}}_n$ be the restriction of $h$ to $\mathbb{H}^{n}_+$
What I'd like to prove:
$$
\phi^*(ds^{-1}_n)=dr^2 +sinh^2(r){ds}_{n-1}
$$
Where $dr, ds_{n-1}$ are the euclidean metric restricted to $(0,\infty),S^{n-1}$ respectively.
How did I try to prove it
I've defined curves:
$$
\begin{aligned}
\gamma:I&\rightarrow S^{n-1}\ s.t. \\
\gamma(0)&=\theta \\
\gamma'(o)&=v \\
\beta:I&\rightarrow (0,\infty) \times S^{n-1}\ s.t.\\
\beta(t)&=(r,\gamma(t)) \\
\alpha:I&\rightarrow (0,\infty) \times S^{n-1}\ s.t.\\
\alpha(t)&=(r_0 +tr,\theta_0)
\end{aligned}
$$
Using those curves to calculate the differential at this vector gives us:
$$
\begin{aligned}
d\phi_{(r_0,\theta_0)}(0,v)&=(sinh(r_0)v,0)\\
d\phi_{(r_0,\theta_0)}(r,0)&=(rcosh(r_{0})\theta_{0},rsinh(r_{0}))
\end{aligned}
$$
Which provides me with:
$$
(\phi^{*}ds_{n}^{-1})_{(r_0,\theta_0)}((r_{1},0),(r_{2},0))=dr^{2}(r_{1},r_{2})
$$
And:
$$
(\phi^{*}ds_{n}^{-1})_{(r_0,\theta_0)}(((0,v),(0,w)))=sinh^{2}(r_0)ds_{n-1}(v,w)
$$
Which is almost what I want.
The problem:
The only thing left is to show:
$$
(\phi^{*}ds_{n}^{-1})_{(r_0,\theta_0)}(((r,0),(0,v)))=0
$$
Expanding the left-hand side of the equation, provides me with:
$$
\begin{aligned}
(\phi^{*}ds_{n}^{-1})_{(r_0,\theta_0)}(((r,0),(0,v)))=&ds_{n}^{-1}((rcosh(r_0)\theta_0,rsinh(r_{0})),(sinh(r_0)v,0))\\
=&rcosh(r_0)sinh(r_0)ds_{n-1}(\theta_0,v)
\end{aligned}
$$
My initial claim was that v was chosen as a tangent vector to the n-1 sphere at the point $\theta_0$ and thus $s_{n-1}(\theta_0,v)=0$, and we're done.
A closer look left me dumbfounded - $\theta_0$ is a point on the sphere, not a tangent vector. This makes my claim meaningless.
What am I missing here?
 A: Careful: the expression $ds_{n-1}(\theta_0,v)$ makes no sense since $\theta_0$ is not tangent to the sphere at $\theta_0$.
What makes sense is the euclidean product $\langle \theta_0,v\rangle$, which turns out to be zero, since $T_{\theta_0}S^{n-1}$ is precisely $\{\theta_0\}^{\perp}\subset \Bbb R^n$!
A way of seeing this is the following: consider $f\colon x \in \Bbb R^{n} \mapsto \|x\|^2-1 \in \Bbb R$.
For $x\in \Bbb R^n \setminus \{0\}$, $d_{x}f = 2\langle x, \cdot \rangle$ is surjective.
Then $S^{n-1} = f^{-1}(\{0\})$, $f$ is regular on $S^{n-1}$ and for $\theta\in S^{n-1}$, $T_{\theta}S^{n-1} = \ker d_{\theta}f = \{\theta\}^{\perp}$.
The result follows directly from your computations if you take care of the notations.
Below are these rigorous computations.

Let us fix $(r,\theta)\in \Bbb R^*_+\times S^{n-1}$.
The tangent space is given by
$$
T_{(r,\theta)}(\Bbb R^*_+\times S^{n-1}) = T_r\Bbb R^*_+\times T_{\theta}S^{n-1} = \Bbb R \times \{\theta\}^{\perp},
$$
where $\{\theta\}^{\perp}$ is the set of orthogonal vectors to $\theta$ in $\Bbb R^n$ with the euclidean metric.
For $(s,v)\in T_{(r,\theta)}(\Bbb R^*_+\times S^{n-1})$, we have
$$
d\phi_{(r,\theta)}(s,v) = \left(s\cosh(r) \theta + \sinh(r)v, s\sinh(r)\right).
$$
Denoting by $\langle\cdot,\cdot\rangle$ the euclidean metric of $\Bbb R^n$, the Minkowski metric is $\langle \cdot,\cdot\rangle - dt^2$, and we have
\begin{align}
(\phi^*ds_n^{-1})_{(r,\theta)}\left((s,v),(t,w)\right) &= {(ds^{-1}_n)}_{\phi(r,\theta)}\left(d\phi_{(r,\theta)}(s,v), d\phi_{(r,\theta)}(t,w) \right)\\
&= (ds^{-1}_n)_{\phi(r,\theta)}\left((s\cosh(r) \theta + \sinh(r)v, s\sinh(r)),(t\cosh(r) \theta + \sinh(r)w, t\sinh(r))\right)\\
&= \langle s\cosh(r)\theta + \sinh(r)v, t\cosh(r)\theta + \sinh(r)w  \rangle - s\sinh(r) t\sinh(r)\\
&= st\cosh^2(r)\langle \theta,\theta\rangle -st\sinh^2(r))\\
& \quad + s\cosh(r)\sinh(r)\langle \theta,w\rangle + t\cosh(r)\sinh(r)\langle v,\theta\rangle\\
& \quad + \sinh^2(r) \langle v,w\rangle.
\end{align}
Now, since $\langle \theta,\theta\rangle = 1$, $\cosh^2-\sinh^2 =1$, and $\langle\theta,w\rangle=\langle v,\theta\rangle = 0$, it follows that
$$
(\phi^*ds_n^{-1})_{(r,\theta)}\left((s,v),(t,w)\right) = st + \sinh^2(r)\langle v,w\rangle.
$$
The answer follows from the fact that $(ds_{n-1})_{\theta}(v,w)= \langle v,w\rangle$.
