Consider the half-plane $H^2 = \{(x,y)\in\mathbb{R}^2\ |\ y>0\}$, and suppose it models (using a little bit of fantasy, and being highly non-rigorous as far as the actual thing goes) the underground of the Earth, such that each layer of ground $y=\textrm{k}$ consists of harder material the more one nears the final layer $y=0$. This way, the deeper one goes the harder would it be for them to move, and hence it would be more expensive to drill the ground.
I suppose such a situation could be modelled (to an extent) through the well-known representation of the hyperbolic plane $(H^2,g_{H^2})$, where $$ g_{H^2}(x,y) = \begin{bmatrix}1/y^2 &0\\0 &1/y^2\end{bmatrix} $$ does not represent the metric which induces the proper length (for which we use the standard euclidean metric), but rather a "cost-length", such in a way that geodesics are the cost-minimising paths. This way, if one wishes to move from one point towards another, they could choose between a shorter but more expensive path (the line, which would be the standard geodesic of the plane) and a longer but "cheaper" path (geodesics of the hyperbolic plane).
First off, I am interested in knowing whether reasonings like the above have been used to describe (even with an high degree of approximation) some real situation or phenomenon.
If they have been, can you give me a reference?
Secondly, my example relies heavily on the fact that one has got this idea of difficulty of movement (which could be as well easiness of movement, if one has got a different metric), that in some way resembles some sort of friction.
Is there so at least one vector field (maybe a force field dependent on velocity) which describes this? Something like a friction, which grows larger the more one nears the bottom?
And viceversa, given some "friction", can one always find a geometry which models the situation?
If there is no vector field, what is there that can be useful to describe what happens?
I thought that, given the Levi-Civita connections of $(H^2,g_{\mathbb{R}^2})$ and $(H^2,g_{H^2})$ ($\nabla^{\mathbb{R}^2}$ and $\nabla^{H^2}$) and a path $c(t)$ in $H^2$, what I called "friction" could be $$\tilde{\nabla}_{c'(t)}c'(t) := 2\nabla^{\mathbb{R}^2}_{c'(t)}c'(t) - \nabla^{H^2}_{c'(t)}c'(t),$$ and $\tilde{\nabla}$ is a torsion-free connection. However, I have yet to find the proprieties of such connection and in which way it is related to the original situation. I would think (wild guess) that, if one can find a metric such that $\tilde{\nabla}$ is its Levi-Civita connection, then the geodesics would be those paths along which there is the best "bargain" between length and cost.
I myself have only recently started sudying riemannian geometry myself and I had no further education in physics than the one in high school (or better said what is the equivalent of high school in the country I live in), so my answer might not be very sophisticated.
What is a riemannian metric? It is a something which associates to every point an inner product, which then induces a norm on the corresponding tangent space. And what is a norm? Normally people say it's a way of measuring lenghts of vectors. Of course this is a good way to think about it, but what is a length? It is just a positive number! This is why I sometimes think of an inner product as something which associates to a vector a positive number (and not specifically a length). And there are a lot of positive things, for example a cost! or the amount of a force,etc. So of course you can think of the riemannian metric as associating different costs to a direction and a length (here "our usual length/our intuitive euclidean length"). So for example geodesics are locally cost minimizing curves.
To your question with the vector field; I don't think there would exist such a vector field, because vector fields associate a vector to every point, but here you're looking for something which associates a number to every point and direction (and "length")(e.g. it matters wether you go up or down at a certain point).
However my riemannian geometry course was taught by an analyst and I didn't have a geometer at my hand, so I don't know if experts think this is a good way to think about it.