Let V be vectorial space.
Define a metric $g$ that comes from an inner product $\langle ,\rangle$ ; and another metric $g’(r)$ also coming from an inner product $\langle ,\rangle’$ depending on a parameter $r\in \mathbb{R^+}$.
Is there any way to determine the relation between the angle that forms $v$,$w\in V$ with respect each of the metrics?
I mean, we can determine that
$$\cos(\Theta_{vw})=\frac{\langle v,w\rangle}{\sqrt{\langle v,v\rangle \langle w,w\rangle }} $$
$$\cos(\Theta_{vw}’)=\frac{\langle v,w\rangle ’}{\sqrt{\langle v,v\rangle’\langle w,w\rangle’}} $$
So, there exist any correspondence between $\Theta_{vw}$ and $\Theta_{vw} ’$?
In my specific case I have a metric defined by the matrix $g=\mathrm{diag}(1,1)$ and the metric $g=\mathrm{diag}(1,\frac{r^2}{1-\alpha r^2})$ where $\alpha>0$ and $0<r<\sqrt{\alpha}$. I have solved the question asked on the exercise but I started thinking about a more general case!
Sorry if this is a dumb question.
A partial answer:
I'll stick to finite dimensional vector spaces. Without loss of generality, suppose that $v$ and $w$ are unit vectors (relative to $\langle \cdot, \cdot \rangle$). There exists a positive definite map $B:V \to V$ such that $\langle v,w \rangle ' = \langle v,Bw \rangle$. From my previous work in this post, we can deduce in the particular case that $v,w$ are orthogonal (i.e. $\cos(\Theta_{u,v}) = 0$) that $$ \langle v,w \rangle' \leq \frac{\kappa(B) - 1}{\kappa(B) + 1} \sqrt{\langle v,v\rangle' \langle w,w\rangle'}, $$ which is to say that $\cos(\Theta_{v,w}') \leq \frac{\kappa(B) - 1}{\kappa(B) + 1}$ (where $\kappa(B)$ denotes the condition number of $B$, i.e. the ratio of its largest and smallest eigenvalues).
I believe that with some work, this upper bound can be generalized to arbitrary vectors $v,w$ to give an upper bound to $\cos(\Theta'_{v,w})$ in terms of $\cos(\Theta_{v,w})$.