Form a metric tensor from the kinetic energy A non-relativistic classical system can be described with the Lagrangian formalism. I have heard that one can construct a Riemannian manifold, using the kinetic energy to form the metric. The system then follows geodesics in the manifold. 
How does one construct the metric tensor? I'm looking for an equation here.
 A: It seems that OP is specifically asking about Jacobi's formulation of Maupertuis' principle for the abbreviated action
$$A[q, E] ~:=~\int \! p_i ~dq^i, \qquad p_i~:=~\frac{\partial L}{ \partial\dot{q}^i} ,\tag{1}$$ 
where we only consider (virtual) paths $q$ in the generalized position space ${\cal Q}$ with one and the same fixed energy $E$.
The (generalized, non-relativistic) kinetic energy 
$$ T~=~\frac{1}{2} M_{jk}(q) \dot{q}^j \dot{q}^k\tag{2}  $$
is often a quadratic function of the generalized velocities. Therefore we can build a Riemannian metric
$$ ds^2 ~=~M_{jk}(q) ~dq^j ~dq^k  \tag{3} $$
in the generalized position space ${\cal Q}$ with the help of the generalized mass matrix $M_{jk}(q)$. The abbreviated action (1) then takes a square root form:
$$ A[q, E] ~=~\int \sqrt{2(E-V(q)}ds.\tag{4} $$
If the potential energy $V(q)$ is a constant, then the square root in eq. (4) becomes a constant and the minimal value of the abbreviated action is achieved via geodesics in the generalized position space ${\cal Q}$.
References:


*

*H. Goldstein, Classical Mechanics; Section 8.6.

*L.D. Landau & E.M. Lifshitz, Mechanics, vol. 1, 1976; §44.
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A: The energy interval is
$$
(mc^2)^2~=~E^2~-~(pc)^2
$$
where the right hand side is the same as
$$
E^2~-~(pc)^2~=~m^2\left(\frac{dt}{ds}\right)^2~-~m^2\sum_{i=1}^3\left(\frac{dx_i}{ds}\right)^2~=~m^2\left(\frac{dx_\mu}{ds}\right)\left(\frac{dx^\mu}{ds}\right).
$$
The metric tensor is introduces as the "machine" that lowers and raises indices so that now
$$
1~=~g_{\mu\nu}\left(\frac{dx^\mu}{ds}\right)\left(\frac{dx^\nu}{ds}\right),
$$
and the line element is found by multiplying by $ds^2$
The kinetic energy is 
$$
K~=~g_{\mu\nu}\left(\frac{dx^\mu}{ds}\right)\left(\frac{dx^\nu}{ds}\right).
$$
We also have with $mc^2ds$ a term that is "energy-time." we may take this as the action
$$
mc^2\int ds~=~\int\sqrt{g_{\mu\nu}dx^\mu dx^\nu}.
$$
It is a typical upper class undergraduate to first year graduate school problem to perform the variation of this to find the geodesic equation.
