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Geodesics and potential function

MathOverflow Asked by Bruno Peixoto on October 29, 2020

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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