Given a manifold $\mathcal{M}$ with coordinates $\psi : \mathcal{M} \rightarrow \mathbb{R^4}$ , $\Psi(p)= (r,\theta ,\phi,t)$ for $ p \in \mathcal{M}$
Suppose we have the active transformation $F : \mathcal{M} \rightarrow \mathcal{M}$ in coordinates given by
$$\Psi \circ F\circ \Psi^{-1}(r,\theta ,\phi,t)=(f(r),\theta ,\phi,t)$$
where is $f$ a function in $\mathbb{R}$
Given the Schwarzschild metric $$ \Psi \circ g\circ \Psi^{-1}(r,\theta ,\phi,t) =-\left(1-\frac{1}{ r}\right)dt^2+\left(1-\frac{1}{ r}\right)^{-1}dr^2+r^2\left(d \theta^2 +\sin^2 \theta d \phi^2\right) \;. $$ we can induce another metric $h$ by
$$\Psi \circ h\circ \Psi^{-1}(r,\theta ,\phi,t)=\Psi \circ F^*g\circ \Psi^{-1}(r,\theta ,\phi,t)=-\left(1-\frac{1}{ f(r)}\right)dt^2+\left(1-\frac{1}{ f(r)}\right)^{-1}f'(r)^2dr^2+f(r)^2\left(d \theta^2 +\sin^2 \theta d \phi^2\right) \;.$$
where $f'(r)=\frac{df(r)}{dr }$
Then for a curve $\gamma(\tau )=(2,0,0,\tau) \in \mathcal{M} $ we would have
then $h(\gamma'(\tau), \gamma'(\tau))=1-\frac{1}{ f(2)}$
so we have
$$\int_{[0, 1]} h(\gamma'(\tau), \gamma'(\tau))d\tau = -\int_{[0, 1]} \left(1-\frac{1}{ f(2)}\right)d\tau $$
where $\gamma'(\tau)=\frac{d\gamma}{d\tau }$
and $$\int_{[0, 1]} g(\gamma'(\tau), \gamma'(\tau))d\tau = \frac{1}{ 2} $$
Since they both satisfy the Enstein equation and they do not give the same line element, how to know which solution should we chose for a given problem.
