Why Riemann distance better for PSD matrices than Euclidian distance? Can you explain to me why Riemann distance is better for positive semidefinite matrices (for example covariance matrices) than Euclidian distance?
Here is the riemannian distance:
$$
d\left(Σ_A,Σ_B\right)=\sqrt{ \sum_i{\ln^2⁡{λ_i (Σ_A,Σ_B)}}}
$$
Where $$\lambda_i(Σ_A,Σ_B)$$ is the generalized eigeinvalue of A and B (see https://en.wikipedia.org/wiki/Eigendecomposition_of_a_matrix#Generalized_eigenvalue_problem).

I do not have, or little knowledge in geodesics but it seems to be the reason why it is better.
Any clarification would be very helpful
Thank you very much!

I can see several advantages of Riemann distance over Euclidean distance for covariance matrix. 
First, Riemann distance is scaling free. if you change unit for both $\Sigma_A$ and $\Sigma_B$, you would expect the similarity should keep unchanged. i.e., $$d(\Sigma_A, \Sigma_B) = d(k\Sigma_A, k\Sigma_B) $$, which is not true for Euclidean distance.
Second, you would expect different impacts for discrepancies among diagonal elements vs. among off-diagonal elements, which is not true for Euclidean distance either.
Third, Riemann distance respects the positive semidefinite constraints. Considering the example that $\Sigma_A - dA$ is semidefinite while $\Sigma_A + dA$ is not, you would not expect they both have the same distance to $\Sigma_A$, which is the case of Euclidean distance. 
