# A particular case when Lagrange equation is equivalent to equation of motion on a Riemannian manifold

Suppose a particle is moving on a surface of a sphere,then it contains a holonomic constraint and so the three Cartesian co-ordinates are available with a constraint equation(equation of surface in Cartesian co-ordinate system). Then there must exist two generalized co-ordinates in form of which Lagrange equation can be constructed.Then is it true to say that the generalized co-ordinate system in this particular case represent Riemannian manifold and so Lagrange equation is equation of motion on this Riemannian manifold?I should made one more important specification that work done by forces of constraints are zero.

-
Could you clarify your question? Do you want to know if the coordinates of the generalized coordinate system that you want to set up is a Reimannian manifold? OR,do you want to know if Lagrange's Equations are equations of motion or not? If its the second then yeah! obviously Lagrange's Equations do give you equations of motion. – Debangshu Nov 27 '12 at 22:00
No,I mean that whether the generalized co-ordinate system in this particular case will represent Reinmmanian manifold?And if yes then Lagrange equations are expressing motion on a Reinmmanian manifold. – Abhinav Anand Nov 28 '12 at 14:37

If you look at a particle constrained to move on the surface of a sphere, and the motion is frictionless, then you can use the usual geometric formalism of classical mechanics to describe the situation. The sphere constitutes the configuration space for the system (traditionally denoted $Q$), and in this case is two dimensional. The Lagrangian $L$ is a function which is dependent, not only on the two coordinates of the configuration space ${q_i}$, but also on the two velocities ${v_i}$. So in this case $L$ is a function on the tangent bundle of the sphere.
The motions of the particle are given by the Euler-Lagrange equations, in turn given by extrema of the action $$I=\int L(q_i,v_i)dt$$
So in this picture, you can work intrinsically on the sphere and forget that it is embedded in $\mathbb{R^3}$, and that it arose as a constraint surface.