The intuition behind the definition of geodesics on a Riemannian manifold. (A non-technical question) In the text I'm studying, the idea behind the definition of a geodesic on a Riemannian manifold was sketched via paths in $\mathbb{R}^n$. I have trouble understanding some aspects of it.
Let $\gamma: I \to \mathbb{R}^n$ be a path. It is a geodesic if $$ \ddot{\gamma}_T \equiv 0, $$ where $\ddot{\gamma}_T$ denotes the tangential component of the total acceleration $\ddot{\gamma}$.
A geodesic is a "straight as possible" path on a manifold. Clearly, a line $$\gamma(t) = vt+p, \quad v,p \in \mathbb{R}^n$$ satisfies the given characterisation by acceleration in $\mathbb{R}^n$. But wouldn't a perfectly circular path also do? Have I misunderstood something or is this some property of the geodesics that the heuristic description fails to convey.
 A: I guess that you are confused because your definition of ‘geodesic’ requires you to isometrically embed the given Riemannian manifold $ M $ into an ambient Euclidean space $ \mathbb{R}^{N} $ for $ N \in \mathbb{N} $ large enough. As is well known, we need $ N $ to be sufficiently large to perform such an embedding (by the Nash Embedding Theorem, it suffices to choose $ N = \dfrac{m(m + 1)(3m + 11)}{2} $, where $ m = \dim(M) $). Once you have isometrically embedded $ M $ into $ \mathbb{R}^{N} $, then distances on $ M $ are preserved and any vector in $ \mathbb{R}^{N} $ whose tail is attached to a point in $ M $ will have well-defined tangential and normal components with respect to $ M $.
If $ \gamma $ is a smooth path constrained to a smooth sub-manifold $ M $ of $ \mathbb{R}^{n} $, then $ \gamma $ is a geodesic of $ M $ if and only if the acceleration vector $ \ddot{\gamma}(t) $ attached to the point $ \gamma(t) $ has no component along $ M $.
If $ \gamma $ is a smooth path in $ \mathbb{R}^{n} $, then the so-called ‘tangential component’ of the acceleration vector $ \ddot{\gamma}(t) $ attached to the point $ \gamma(t) $ is simply $ \ddot{\gamma}(t) $ itself. This is because we are assuming our manifold to be $ \mathbb{R}^{n} $, which is its own ambient Euclidean space. Hence, in this rather peculiar case, we are asking if $ \gamma $ is a geodesic of $ \mathbb{R}^{n} $ and not if $ \gamma $ is a geodesic of some smooth sub-manifold of $ \mathbb{R}^{n} $, as we have not specified any smooth sub-manifold in the first place.
As any particle moving along a circular path $ \gamma $ in $ \mathbb{R}^{n} $ exhibits centripetal acceleration in $ \mathbb{R}^{n} $, we necessarily have $ \ddot{\gamma}_{T} = \ddot{\gamma} \not\equiv \mathbf{0}_{\mathbb{R}^{n}} $, which shows that $ \gamma $ is not a geodesic of $ \mathbb{R}^{n} $. The geodesics of $ \mathbb{R}^{n} $ are necessarily straight lines, as mentioned in the OP.
The geodesics of a smooth sub-manifold of $ \mathbb{R}^{n} $, however, are generally curved in $ \mathbb{R}^{n} $. For example, the geodesics of a $ 2 $-sphere are its great circles, which are clearly circles in $ \mathbb{R}^{3} $.
