Why not use higher derivatives than velocity for the state of a system?

Question

I understand that the state of a system at any instant of time is completely specified if we give the positions and velocities of all the particles, but why not also the accelerations and higher derivatives?

Answer 1

If you know the positions and velocities of all the particles, and the nature of the interactions between them, then you have completely specified a system.

This is because (in the Newtonian Language) the equation of motion is F(x,v) = ma, or in other words, F(x, x') = m x''.

This is a second-order DiffEQ, and as such requires two initial conditions: x and v (x and x') for each particle to obtain a unique solution. Conversely, specifying all the initial x's and v's reduces the family of solutions to the DiffEQ to a single, unique solution (in other words, knowing the initial x's and v's is necessary and sufficient to use F=ma to obtain the future histories of all the particles in the system).

Accelerations do not specify anything because the equation is second-order and requires initial conditions for x and v only. In other words, the acceleration is derivable because the force only depends on x and v: a = F(x,v)/m. If you know all the x's and v's, you know all the a's too.

Answer 2

The other answers are correct regarding the mathematics.

But ultimately your question goes deeper: why did nature chose equations of motion which are second order?

Well, you can dig deeper, if you like. You can learn about the re-formulation of classical mechanics with the Hamilton formalism. From there it is a straight forward way to quantum mechanics and its non-commutative operators. And that will take you to quantum field theory. In all these cases you will learn that nature seems to prefer the minimal self-consistent theory over every other. We don't completely understand why but we do observe it. (This comment does not imply that the minimal self-consistent theory is simple for humans... it usually is not.)

Personally I would venture an educated guess: it is all because of statistical mechanics. Once you throw a large amount of the same stuff on one heap and introduce a non-trivial interaction, you end up with phase transitions AKA critical phenomena. We have learned a very interesting thing about critical phenomena: they are universal. This means that no matter how they form and what the underlying details of the interaction are, at the critical point they all behave VERY much the same. You can describe them by a set of invariants. And I dare to bet that all of what we see in nature is the result of a giant phase transition. As such it will obey certain universal rules, one of which is the order of relevant perturbations around the critical point. And those perturbations are what leads to the the Hamiltonian brackets in classical mechanics, QM and QFT.

This is just a hunch. I can't prove it to you mathematically. But the strength of the argument rests on the fact that the universe is so enormous, that it is homogeneous and that its fundamental laws are all simple. Beat me if I am wrong. But I don't think so.

Answer 3

Indeed why not?

I think the state of a system can use more than location and velocity of its particles to be totally defined. For one thing, there are all kinds of state that can be specified. In fact, there are all kinds of energy state (thermal, gravitational, atomic) and each one of these has its own metrics for definition.

State simply means condition. Check this out:

1 a: mode or condition of being b (1): condition of mind or temperament (2): a condition of abnormal tension or excitement
2 a: a condition or stage in the physical being of something b: any of various conditions characterized by definite quantities (as of energy, angular momentum, or magnetic moment) in which an atomic system may exist [See source.]

I'm not clear your assertion is correct. Seems to me there are so many possible states that there must be more metrics than location and velocity needed to define them. And why not acceleration and jerk (rate of change of acceleration) for some of those metrics?

Answer 4

hmm ease?
when you integrate back to other things there's integration constants to be taken into account...
ie if you start, using position and acceleration, you have no idea what the starting velocity is... so you're missing information already.
if you use acceleration and velocity, you wont know the starting position...

Answer 5

technically i think you can. Its just a matter of taking derivitives of the velocity.

Essentially, velocity, accelration and the 'jerk' (the third derivitive) are all interchangeable, as long as you know initial conditions and the time intervals over which each of them operates.