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I somehow could not find the answers to the question in Why are coordinates and velocities sufficient to completely determine the state and determine the subsequent motion of a mechanical system? to be satisfying and that's why I am asking the question again.

Suppose I have a 1 particle system. Suppose the $q_1$, $q_2$and $q_3$ coordinates are 3, 5 and 8 respectively. Suppose the velocities $\dot q_1$, $\dot q_2$, $\dot q_3$ are 5, 9 and 10. Now with this much of information I can predict the future motion of my system only if the the accelerations in the 3 directions are ZERO.

If a force is acting on the system then definitely acceleration in the 3 directions are not zero. In that case how can I calculate the future motion for the particle from the values of $q_1$, $q_2$, $q_3$, $v_1$, $v_2$ and $v_3$ alone. I think I should be given $\ddot q_1$ and $\ddot q_2$ and $\ddot q_3$.

I am quoting the paragraph written in the book of classical mechanics by Landau.

"If all coordinates and velocities are simultaneously specified, it is known from experience that the state of the system is completely determined and that its subsequent motion, can in principle be calculated. Mathematically, this means that, if all the coordinates $q$ and velocities $\dot q$ are given at some instant the accelerations $\ddot q$ at that instant are uniquely defined"

How is $\ddot q$ uniquely defined at any instant from $q$ and $\dot q$ of that instant. At Least $\dot q$ at 2 different instants should be know to calculate acceleration.

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The usual answer is that you need force laws that are functions of the positions and velocities of the objects. Then you need the masses of the objects. Then you get accelerations from Newton's second law. So initial positions and initial velocities and the force law give you the initial force. From that and the mass you get the initial acceleration. So an initial acceleration is possibly redundant (and possibly contradictory).

But that won't necessarily give you the motion since there can be multiple solutions to Newton's second law even when you specify the initial position and initial velocity. For instance, if the potential energy is $U(x)=-(1J)(x/\lambda)^{(4/3)}$ then there are multiple solutions to Newton's second law with $x(0)=0$ meters and $v(0)=0$ meters/second.

One solution is $x(t)=0$ meters. And if you let $\tau=\lambda \sqrt{3M/2J}$ (where $M$ is the mass of the particle) then, another solution is $x(t)=\lambda (t/\tau)^3$ and a third solution is $x(t)=-\lambda (t/\tau)^3$ and those are all solutions with constant jerk. But other solutions exist for instance if $t_1<0s<t_2$ then your solution could be $x(t)=\lambda ((t-t_1)/\tau)^3$ when $t<t_1$ and $x(t)=0$ meters, when $t_1\leq t\leq t_2$ and $x(t)=\lambda ((t-t_2)/\tau)^3$ when $t>t_2$ and those are just the ones where it is always going right except for a window of time. It could also halt forever, it could turn around when it starts moving again, it could have been sitting there forever and then start moving at any moment in any direction. All of those are perfectly fine solutions of Newton's second law that meet the same initial conditions.

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  • $\begingroup$ Nice example. The caveat is that the solution exist and is unique only if (=where) the derivative of the force with respect the position exist - and that is not the case for your potential at the zero - so there uniqueness breaks down. $\endgroup$
    – MiMo
    Feb 25, 2016 at 5:08
  • $\begingroup$ @MiMo Newton's first law of motion gives us a unique solution to those potentials, so we don't need caveats, except to claims that it's the second law (and initial position and initial velocity) alone that gives the solution. $\endgroup$
    – Timaeus
    Feb 25, 2016 at 5:20
  • $\begingroup$ Sorry, I meant caveat to the standard 'it is a second-order ordinary differential equation, so given position and velocity the solution is unique' $\endgroup$
    – MiMo
    Feb 25, 2016 at 20:43
  • $\begingroup$ ...is it always true that the first law gives us a unique solution? (it clearly does in your example) $\endgroup$
    – MiMo
    Feb 25, 2016 at 20:45
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By using Newton's Laws of Motion one can calculate all of the interparticle forces; knowing the initial positions and velocities gives you the starting point for your calculations.

When confronted with any system of differential equations one requires boundary conditions in order to obtain a specific solution. Initial conditions are agood starting point for physical problems.

So Landau is assuming you know the applicable forces if you are calculating, which gives the acceleration.

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  • $\begingroup$ Somehow you need masses too. Knowing the positions and the velocities might give you the forces (if you have the force laws). But you won't get acceleration from force if you don't have mass. $\endgroup$
    – Timaeus
    Feb 23, 2016 at 16:44
  • $\begingroup$ Presumably all other particle properties are already known, such as electric charge, magnetic response and mass; the OP only asked about position and velocity. $\endgroup$ Feb 24, 2016 at 0:10
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As other people have pointed out, strictly speaking, you need to know the forces and masses as well as the positions and velocities. You need the masses because Newton's second law for a particle is

$$ F_{net} = \dot{p} $$

Where you can easily find the initial momentum using the formula $p = mv$, which even works relativistically since the previous equation simply becomes

$$ p = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}} $$

So you can always find the future momentum (and hence velocity) by integrating the force against time, and you can always find the future position by integrating the velocity against time.

However, knowing the forces means you have to know the laws of physics; more importantly, it requires assuming that forces themselves only depend on positions, velocities, and perhaps particles of the properties themselves.

As it turns out, the forces we observe in classical mechanics follow these rules. So, when people say you just need the initial positions and velocities, what they really mean is that the laws of physics describe many possible ways for objects with given masses and velocities (and charges, spins, etc.) to move, but once you know their positions and velocities, there should in principle be only one way for them to move for the rest of time.

You can heuristically think of this of the fact that Newton's second law is a second order differential equation in $3 \times N$ variables and hence requires $2 \times 3 \times N$ independent constants, which are provided by positions and velocities, to specify a unique solution.

Strictly speaking, there are pathological examples of potentials (see Timaeus's answer for one of them) that do not demand unique solutions, but these tend to be completely unphysical since they usually require the system to be in a perfectly precise configuration occupying a subset of phase space with measure zero. These counterexamples also tend to involve very contrived potentials that do not occur physically. So, if you just want to understand what people like Landau mean when they say that positions and velocities specify solutions uniquely, you can safely ignore these pathological examples almost all of the time. By the time you get to physical phenomena that could conceivably behave pathologically, you will necessarily find that quantum effects become important anyway, and classical mechanics will no longer be directly useful.

That said, these examples perhaps point to intrinsic weaknesses in the mathematical and conceptual foundations of classical mechanics. So they are worth thinking about. But my understanding of your question was that you were wondering why $\ddot{q_i}$ are all specified by $\dot{q_i}$ and $q_i$, and the answer is that we have never seen a $\ddot{q_i}$ that arises spontaneously from nothing and that cannot be described in principle as coming from a force depending only on $\dot{q_i}$ and $q_i$.

All tests of classical mechanics (that are within the scope of validity of classical mechanics) reveal this to be true (to within experimental accuracy). You do need to know precisely what the positions and velocities of every particle are, so this isn't something that humans can use in practice, and it also doesn't describe reality perfectly because classical mechanics is inherently a low-velocity, large scale approximation of the laws of physics. But, to the extent that we can measure things precisely, we have always observed nature following this rule.

So, what this tells you is that our assumption that $F_{net}$ can only depend on unchanging properties of the particles themselves along with the velocities and positions of the other particles is consistent with experiment.

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  • $\begingroup$ Amazing how you claim you can find unique solutions after I already point out potentials for which there exist multiple solutions. And knowing the forces isn't the point. The point is to have force laws and have the force laws depend solely on positions and velocities. $\endgroup$
    – Timaeus
    Feb 24, 2016 at 7:12
  • $\begingroup$ Perhaps I am wrong, but your example doesn't seem to make sense. There is one solution in your example (assuming $\lambda$ constant), since $\dot{p} = F = -\frac{dU}{dx} = 0$ at $x = 0$, implying $v = 0$ and $x = 0$ for all time, which is undoubtedly a unique solution. If I am misunderstanding, please clarify your example. $\endgroup$ Feb 25, 2016 at 3:04
  • $\begingroup$ Thanks for pointing that out, I made a sign error. The potential needs to be negative to have multiple solutions for those initial conditions. $\endgroup$
    – Timaeus
    Feb 25, 2016 at 3:15
  • $\begingroup$ Do you mean a negative sign in front of the potential? In that case, physically, you would have an unstable solution, but it would still technically remain at $x = 0$, $v = 0$ in the absence of perturbation. It sounds like you're saying that any time you have an unstable solution (i.e. a local max in potential) it is mathematically possible to give a particle just enough energy to get to equilibrium, where it will stay forever, and that this leads to many past values of $x$ and $v$. I confess not to know how this edge case is treated philosophically, but in practice it has zero probability. $\endgroup$ Feb 25, 2016 at 3:26
  • $\begingroup$ I agree that the parameter space required to reach the top of the hill with zero kinetic energy to spare is a vanishingly small one. But if you go the route of "in practice" then only a small number of force laws actually come up in practice and classical mechanics is just an approximation anyway. The point is that it's an actual bald faced lie to claim that unique solutions exist for arbitrary potentials with well defined gradients. And someone thinking that could cause them to make errors in their analysis. Just like saying Hamilton-Jacobi is the same as Hamiltonian mechanics. $\endgroup$
    – Timaeus
    Feb 25, 2016 at 3:45

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