# state for a classical particle [duplicate]

In classical mechanics, if all the generalized coordinates and the generalized velocities are simultaneously specified, the state of the system is completely determined and one can calculate its subsequent motion. That also means if we know generalized coordinate $q$ and generalized velocity $\dot{q}$, then we can calculate acceleration $\ddot{q}$. How to prove this?

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## marked as duplicate by Qmechanic♦, Ron Maimon, Manishearth♦Dec 11 '12 at 10:56

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## 1 Answer

I believe the reason is that the classical equations of motion are second order differential equations in $q$ (Newton's equations or the Euler Lagrange equations). To solve a second order differential equation you need two boundary conditions. One choice of these boundary conditions could be the initial values of $q$ and $\dot q$, but they could equally well be the initial and final positions etc.

Now, I do not know a good reason why the equations of motion should be second order. They are the classical limit of quantum mechanics and relativity and I guess one really needs to start there with such reasoning...

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Hi Mistake Ink, thank you so much for your answer and pointing out the other similar question. – Timothy Sep 2 '12 at 19:28