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I try to assemble concepts of differential geometry for my own comprehension of the subject. I understand a manifold is a higher dimensional surface. It has a metric which perform inner product in the tangent space. A curve on the respective manifold has a covariant derivative, which remains on the tangent bundle. A geodesic is a manner to comprehend straight lines on a manifold. They might be closed like on a sphere. They are the main tool of physicists to comprehend the universe through the lagrangian and hamiltonian framework. On the lagrangian framework, there is a kinetic $K = g_{ij} \dot{x}_i \dot{x}_j$ and potential $U = V(x)$ energies which computes the lagrangian $L = K - U$. The ausence of potential energy coincide with the geodesic equation. The Einstein notation is in force here.

In mathematical terms, I do not comprehend the role of the potential or dissipative term on the geodesic equation and further explanations on similar manner as I will explain shortly.

As far I comprehend, the geodesic statement is: given two points A and B, the geodesic which binds both points on a simply connected non-compact smooth manifold is the solution to the boundary value problem of former equation below.

The same statement but for the latter equation entertains other interpretation, no longer a geodesic at strict sense. I comprehend from physical perspective that given the manifold endowed by a metric, the second and third are relative to the actuation of forces on the motion particle. But for this, one defines the particle, which is merely an abstract conception, a trick to better comprehend intuitively.

\begin{equation} \ddot{x}^j + \Gamma^j_{i k} \dot{x}^i \dot{x}^k = 0 \end{equation}

\begin{equation} \ddot{x}^j + \Gamma^j_{i k} \dot{x}^i \dot{x}^k + g^{ji} \frac{\partial V}{\partial x^i} = 0 \end{equation}

\begin{equation} \ddot{x}^j + \Gamma^j_{i k} \dot{x}^i \dot{x}^k + g^{ji} \frac{\partial V}{\partial x^i} + g^{ji} \frac{\partial R}{\partial \dot{x}^i} = 0 \end{equation}

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  • $\begingroup$ What is your question here? From the looks of it, $R$ is just another potential term just like $V$, that could be absorbed into it. $\endgroup$ Commented Aug 23, 2020 at 11:35
  • $\begingroup$ The questions are: What is the geodesic equation when the boudary values are not points A and B but initial position and velocity? What is the influence of the potential term on the geodesic equation when employed as a Point A B condition? The same for with Rayleigh i.e . dissipative function? $\endgroup$ Commented Aug 23, 2020 at 12:10
  • $\begingroup$ The role of those terms is the same as their role in Newtonian classical mechanics, where the Riemannian manifold is flat Euclidean space. So you should ask a physicist or look in a classical mechanics textbook to see how such terms appear. Your questions have no mathematical answer, only an answer in physical intuition. $\endgroup$
    – Ben McKay
    Commented Aug 23, 2020 at 12:14
  • $\begingroup$ Why doesn't it have a mathematical answer? The physical statement relies on the mathematical toolset to perform its calculation. $\endgroup$ Commented Aug 23, 2020 at 12:16
  • $\begingroup$ @BrunoPeixoto I'm not sure what the question actually is. If I re-interpret it as asking for a geometrical set-up to discuss lagrangian/hamiltonian systems with potentials, then I have two comments: (1) you may be able to rewrite the equation as a geodesic equatoin for an affine connection which is not necessarily metric (cf. Newton-Cartan theory) and (2) you may be able to view it as a geodesic equation in an auxiliary space relative to the Levi-Civita connection (cf. the Eisenhart lift). $\endgroup$ Commented Aug 23, 2020 at 12:50

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