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The general argument is as follows. By Newton's second law $\mathbf F=m\ddot{\mathbf{x}}$. Now it is said that this is a second-order ODE and hence requires $\mathbf x(0)$ and $\mathbf{\dot{x}}(0)$ as initial conditions to uniquely determine the trajectory $\mathbf x(t)$, and hence the state (which is anything that determines the trajectory uniquely given the dynamical law, here the functional form of $\mathbf F$) is completely specified by mentioning position and velocity (a total of 6 numbers in the 3-dimensional space) and hence a 6-dimensional state-space.

But the critical assumption was that the equation was a second-order ODE, which might not be the case if $\mathbf F$ also depended on say $\mathbf{\dddot{x}}$. (Araham-Lorentz force depends on $\mathbf{\dot{a}}$ for example.) Then the equation will no longer be second-order and will have a higher-dimensional state-space.

So is the frequently quoted statement that the state-space for a single particle is 6-dimensional wrong?

Edit: I’ve found that Newton’s principle of determinacy (that Arnol’d introduces in his epic book, Mathematical Methods of Classical Mechanics), provides an answer to my question, by basically postulating this.

Atom
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  • Possible duplicates: https://physics.stackexchange.com/q/18588/2451 , https://physics.stackexchange.com/q/4471/2451 , https://physics.stackexchange.com/q/52024/2451 , https://physics.stackexchange.com/q/160711/2451 and links therein. – Qmechanic Jun 22 '20 at 17:10
  • I think the answer to your question depends on the definition of the "phase space". If you define it to be the space of solutions to e.o.m., then yes, it will be 9-dimensional if you also have $\dddot{x}$ in them. – Prof. Legolasov Jun 22 '20 at 20:50
  • @Prof.Legolasov Yes, but I was, in essence, asking why are expressions of the forces observed in nature never go beyond the first derivative $\mathbf{\dot{x}}$. – Atom Jun 22 '20 at 20:57
  • And Arnol’d solves this by essentially postulating this. – Atom Jun 22 '20 at 20:58

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