# Why do we only need the initial position and velocity to describe a system fully?

I understand this question has been asked many times and I have thoroughly looked at the previous questions so please read my question carefully before flagging it as a duplicate.

Landau & Lifshitz write in page one of their mechanics textbook:

If all the co-ordinates 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 co-ordinates $$q$$ and $$q˙$$ are given at some instant, the accelerations $$q¨$$ at that instant are uniquely defined.

Now I am a bit confused, is this saying that we only need to know the initial position and initial velocity to get the full time evolution of the system? which is due to the fact that the dynamics are governed by a second order differential equation?

I just want to give an example here and please tell me if I am correct in my thought process:

Take the example of a spring that obeys Hooke's law: $$F = m \ddot{x}= -kx$$ then a general solution can be: $$x(t) = A e^{i \sqrt{\frac{k}{m}}t}+ B e^{-i \sqrt{\frac{k}{m}}t}$$

so we do not need velocities or position for all time, for a given particle we just need initial conditions for velocity and position and then we are set?

my last question is why can we rearrange Newton's law in the form $$\ddot{x} = F(x,\dot{x},t)/m$$ How do we know for a fact that $$F$$ can generally depend $$x,\dot{x},t$$ and that is it?

Possible duplicate list:

Why are position and velocity enough for prediction and acceleration is unnecessary?

Why are coordinates and velocities sufficient to completely determine the state and determine the subsequent motion of a mechanical system?

Why forces are functions of time, position and velocity at most?

Why are there only derivatives to the first order in the Lagrangian?

Why are differential equations for fields in physics of order two?

• I think this question is a little unfocused. The question you raise in your example: for a given particle we just need initial conditions for velocity and position and then we are set? has a simple answer: yes. The last part of your question -- why do we assume the force is not a function of acceleration (or higher order derivatives of $x$) -- is a good question, but could easily be its own question (and probably should be). Part of the answer is Ostragradsky's theorem. In general, try to ask a single well-defined question per post. Apr 21, 2022 at 3:16
• @Andrew I think you misspoke and meant the acceleration initial condition. Obviously force is expressible as a function of acceleration (and almost always thus expressed).
– g s
Apr 21, 2022 at 4:50
• @gs Actually I meant to write what is there (although maybe reasonable people can disagree :)). What I had in mind (inspired by the OP's notation) is that Newton's laws are $\ddot{x} = F/m$, where $F$ is a function that has an independent definition. The classic examples of force laws (Hooke's law, Coloumb's law, Lorentz force, gradient of a potential) all involve the position and sometimes the velocity, but never the acceleration. Apr 21, 2022 at 4:55
• @Andrew Okay, but I'm going to be one of the reasonable people disagreeing, mostly because $x = f(y) \to y = f^{-1}(x)$ but also because $\ddot x = F/m$ iff $\dot m = 0$.
– g s
Apr 21, 2022 at 5:06
• @gs I'm not sure I fully understand your point. Newton's second law is not a definition of force. The force laws do not involve acceleration; this does not change if you express Newton's second law as $\dot{p}=F$. Maybe it would help to say that one of the motivations for my comment is if you write a Lagrangian for one degree of freedom $x(t)$ that leads to an equation of motion which is non-linear in $\ddot{x}$, there will also be higher order derivative terms like $\dddot{x}$, which in turn will lead to an Ostragradsky instability. Apr 21, 2022 at 5:39

How do we know for a fact that F can generally depend x,$$\dot x$$, t and that is it?
It's just usually the case. In fact, often we have $$F = -\nabla U$$, where U is only a function of x and t.