# Can Jacobi's formulation of Maupertuis' principle be derived in Riemannian geometry?

i want to arrive to hamilton-jacobi equation using the riemannian geometry.

So let $$\textbf{X}\in \mathfrak{X}(M)$$, where $$M$$ is Riemannian manifold whose metric is $$g:\textbf{T}M \times \textbf{T}M \longrightarrow \mathbb{R}$$. On the other hand, let suppose that $$\textbf{X}=grad f$$.

So $$g(\textbf{X},\textbf{X})=g(g^{ij}\frac{\partial f}{\partial x^{i}}\frac{\partial }{\partial x^{j}},g^{\mu \nu}\frac{\partial f}{\partial x^{\mu}}\frac{\partial }{\partial x^{\nu}})=g^{ij}g^{\mu\nu}\frac{\partial f}{\partial x^{i}}\frac{\partial f}{\partial x^{\mu}}g(\frac{\partial }{\partial x^{j}},\frac{\partial }{\partial x^{\nu}})=g^{ij}g_{j\nu}g^{\nu\mu}\frac{\partial f}{\partial x^{i}}\frac{\partial f}{\partial x^{\mu}}$$.

Finally, we get that: $$g(\textbf{X},\textbf{X})=g^{i\mu}\frac{\partial f}{\partial x^{i}}\frac{\partial f}{\partial x^{\mu}}$$.

Until here it is more or less the things that i know . The function that i defined before it is a lagrangian of free particle $$L=\frac{1}{2}g_{ij}\dot x^{i}\dot x^{j}$$, so we now that the curves of this lagrangian $$x^{k}(t)$$ are geodesic that is $$\nabla_{\dot x^{k}}\dot x^{k}=0$$. I do not know to prove with this information that $$g(\textbf{X},\textbf{X})$$ is a constant, where $$\textbf{X}$$ is the vector field generated by $$L$$

Finallly, i have another question of Jacobi's Principle. Let $$H\in C^{\infty}(\textbf{T}^{*}M)$$ be a hamiltonian function of the form $$H=\frac{1}{2}g^{ij}p_{i}p_{j}+V$$ where $$g$$ is the metric of the riemaniann manifold and $$V$$ is the potential function $$V:M\longrightarrow \mathbb{R}$$. The jacobi's principle said that the curves in $$M$$ are geodesic of metric $$(E-V)g_{ij}dx^{i}\otimes dx^{j}$$ and the new hamiltonian is $$h=\frac{1}{2}(\frac{g^{ij}}{E-V})p_{i}p_{j}$$. Again, the solution of the curves here are geodesic, but some of the properties of the geodisc are that the tangent curves are paralel to the curve and the aceleration is 0 so, if it is a potential, there is an aceleration. My questions are this geodeisc lies in a new manifold $$M_{2}?$$ If it is like this how can arrive to the original manifold?

Thanks a lot!!!!!

In a non-relativistic context the metric becomes like a position-dependent mass-tensor for the kinetic energy [1] $$T~=~\frac{1}{2}M_{jk}(q)\dot{q}^j\dot{q}^k, \qquad M_{jk}(q)~\equiv~g_{jk}(q), \tag{8.83}$$ where $$(ds)^2~=~g_{jk}(q)~\mathrm{d}q^j ~d\mathrm{d}q^k\tag{8.86}$$ so that $$T~=~\frac{1}{2}\left(\frac{ds}{dt}\right)^2.\tag{8.87}$$ The abbreviated action therefore becomes $$\int p_k\dot{q}^k ~\mathrm{d}t~=~\int 2T ~\mathrm{d}t~=~\int \sqrt{2T} ~\mathrm{d}s ~=~\int \sqrt{2(E-V(q))} ~\mathrm{d}s.\tag{8.89}$$ The latter is the sought-for Jacobi's formulation.

References:

1. Herbert Goldstein, Classical Mechanics, Section 8.6.