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I can't seem to understand why, if there are s generalized coordinates, there end up being only 2s-1 integrals of the motion.
The solutions of Lagrange's equation will have 2s constants. Why couldn't one simply solve the 2s equations for the solutions q_i and dq_i/dt for the 2s constants?
For 1D motion of a free particle, for example, x = C_1*t + C_2 and dx/dt = C_1. Because the particle's velocity is constant, dC_1/dt = 0. Also, C_2 = x - dx/dt * t, so that dC_2/dt = dx/dt - dx/dt = 0. So C_1 and C_2 both appear to be constants of the motion. Integrals and constants of the motion are the same thing aren't they?
I'm thinking of the paragraph in Landau and Lifgarbagez's Mechanics where they say that that there are only 2s-1 integrals and that one of the solution constants can be taken as "an additive constant t_0 in the time". This statement also confuses me - the constants are determined by initial positions and velocities and end up involving these quantities. In what sense can you choose one of them as t_0?
The solutions of Lagrange's equation will have 2s constants. Why couldn't one simply solve the 2s equations for the solutions q_i and dq_i/dt for the 2s constants?
For 1D motion of a free particle, for example, x = C_1*t + C_2 and dx/dt = C_1. Because the particle's velocity is constant, dC_1/dt = 0. Also, C_2 = x - dx/dt * t, so that dC_2/dt = dx/dt - dx/dt = 0. So C_1 and C_2 both appear to be constants of the motion. Integrals and constants of the motion are the same thing aren't they?
I'm thinking of the paragraph in Landau and Lifgarbagez's Mechanics where they say that that there are only 2s-1 integrals and that one of the solution constants can be taken as "an additive constant t_0 in the time". This statement also confuses me - the constants are determined by initial positions and velocities and end up involving these quantities. In what sense can you choose one of them as t_0?