I know that there is a quantity called jerk (USA) or jolt (UK) which is the third-order derivative of position (i.e. the first derivative of acceleration). When we write down the second law of Newton, we write in the following generic form: \begin{equation} \sum\mathbf{F}(\mathbf{r}, \mathbf{\dot{{r}}},t) = m\ddot{\mathbf{r}} \end{equation} I am wondering what will be the consequences if we accept that there are forces acted upon the material point that depend upon higher derivatives of position. Will the initial value problem be well-defined? Can we solve it analytically? Thanks in advance.
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Possible duplicate of Why does newtons second law involve second derivative of position? – John Rennie Dec 09 '20 at 10:23
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1@JohnRennie I don't think that is the same thing. Asking why there aren't forces that depend on higher order derivative of position (this one) is not the same thing as asking why it is $F=m\dddot x$ (the link) – BioPhysicist Dec 09 '20 at 16:42
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I mean you can't even solve all second order diffy equations analytically afaik – tryst with freedom Dec 09 '20 at 16:45
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Dimitris, I suggest refining your question a bit. "What are the consequences" is pretty open ended, especially without specifying a specific scenario. Additionally, the question of the problem being well-defined and solvable analytically is more of a mathematics question. All of this combined leads to an unclear and unfocused post. For PSE I would suggest asking about the specific physics concept you are interested in. – BioPhysicist Dec 09 '20 at 16:53
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I am wondering what will be the consequences if we accept that there are forces acted upon the material point that depend upon higher derivatives of position. Will the initial value problem be well-defined? Can we solve it analytically?
First recapping what is True! (in certain regime),According to Newton's principle of determinacy all motions of a system are uniquely determined by their initial positions ($\mathbf{x}(t_0)\in \mathcal{R}^N$) and initial velocities ($\dot{\mathbf{x}}(t_0)\in \mathcal{R}^N$).
There is a function $\mathbf{F}:\mathcal{R}^N\times\mathcal{R}^N\times\mathcal{R}\rightarrow \mathcal{R}^N$ such that
$$\ddot{\mathbf{x}}=\mathbf{F}(\mathbf{x},\mathbf{\dot{x}},t)$$ It's called Newton's equation.
By the theorem of existence and uniqueness of solutions to ordinary differential equations, the function $\mathbf{F}$ and intial conditions uniquely determine a motion.
Galileo's principle and Newton's differential equation are basic experimental facts which lie at foundation of mechanics.
Now suppose You found (Very nice example is Dirac-Lorentz equation) that force depend on higher derivative so that Newton's differential equation now look like $$\mathbf{F}=\mathbf{F}(\mathbf{x},\mathbf{\dot{x}},\mathbf{\ddot{x}},\cdots,t)$$
Then by the theorem of existence and uniqueness (Picard–Lindelöf theorem) of solutions to ordinary differential equations, the function $\mathbf{F}$ and initial conditions $\mathbf{x}(t_0)$ $\dot{\mathbf{x}}(t_0)$, $\ddot{\mathbf{x}}(t_0)$ upto ($n-1$)th derivative of position.
Might be interested :
Why are there only derivatives to the first order in the Lagrangian?
Young Kindaichi
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