Why doesn't Newton's Second Law include higher-order mass?

I suspect this has been asked here before, but I didn't find anything using Search.

Why is Newton's second law only second-order in position? For instance, could there exist higher-order masses $m_i$ with

$$F(x) = m\ddot{x} + \sum_{i=3}^{\infty} m_i x^{(i)}?$$

Are there theoretical reasons why $m_i$ must be exactly zero for $i>2$? If not, if these masses existed but were extremely small, would we be able to tell experimentally (e.g. by observing galactic motion)?

I think Newton's first law, linearity of motion, and the time-reversal symmetry strongly constrain the form of Newton's second law.

Generically, the linear relation between the position and the force is given by

$$F = \beta_0 x+ \beta_1 \frac{dx}{dt} + \beta_2 \frac{d^2x}{dt^2} + \beta_3 \frac{d^3x}{dt^3}+\cdots.$$

The time-reversal symmetry demands that all odd derivatives vanish, $\beta_{2k+1} = 0$ for $k = 0,1,2\cdots$. The spatial homogeneity demands $\beta_0 = 0$. Therefore,

$$F = \beta_2 \frac{d^2x}{dt^2} + \beta_4 \frac{d^4x}{dt^4} + \cdots = \beta_2 a + \beta_4 \frac{d^2a}{dt^2}+\cdots.$$

Here $a$ is the acceleration. The first law demands that, when $F=0$, the only solution is the trivial solution $a=0$. We see that, if all higher coefficients $\beta_4, \beta_6, \cdots$ are not zero, there will be other solutions. Thus, the only possible form is

$$F = \beta_2 \frac{d^2x}{dt^2}.$$

I think that in our present framework, we would lump additional terms on the right hand side of your equation with force, i.e., move them over to the left. For exaple, the Abraham-Lorentz force.

As is the case with the above-mentioned force, the more derivatives of $x$ you involve, the more non-local in time you get. If a particle's position function as a function of time, $x(t)$, can be Taylor expanded as a function of $t$, then knowing sufficiently many derivatives at $t=t_0$ allows you to calculate it's motion at some later time $t\gt t_0$. So if you mix $\ddot{x}$ with higher derivatives, then the acceleration of the particle would be determined in part by it's position in the future. This violates causality and causes all sorts of nasty things.

• Do you have a reference where this is explained more in detail ? – Mathusalem Jan 14 '14 at 17:23
• What precisely? Are you familiar with Taylor series? – lionelbrits Jan 14 '14 at 17:33

Why is Newton's second law only second-order in position?

Simple answer: it is the simplest that works well enough, and in fact it works very well. If the equation of motion had higher derivatives of $x$ than second, this could lead to un-physical motion. For example, there were attempts by Lorentz, Abraham and Dirac to use the equation of motion

$$m\ddot{x} = C\,\dddot{x},$$

for charged particle, but it does not work well. This equation has a solution $x(t) = x_0 e^{\frac{m}{C}t}$ which does not conform to known behaviour of charged particles.

• I believe "works well enough" here means that it satisfies experiments. – BMS Jan 14 '14 at 19:17

One avenue of approach is to consider what this would imply for the simple harmonic oscillator -- that is, if Newton's Second Law was $F = m x^{\left(n\right)}$ : $$m x^{\left(n\right)} = - k x \ \ \ \Rightarrow \ \ \ x^{\left(n\right)} + \omega^2 x = 0,$$ where $\omega = \sqrt{k / m}$. Assuming a solution $$x = A \exp\left(\lambda t\right),$$ the equation of motion gives $$\lambda^n + \omega^2 = 0 \ \ \ \Rightarrow \ \ \ \lambda_j = \omega^{2/n} \exp\left[i \pi \left(\frac{1+2j}{n}\right)\right], \ \ \ j = 0, 1, ..., n-1.$$ For $n=1$, $\lambda_0 = -\omega^2$, and $x \propto \exp\left(-k t / m\right)$, which decays exponentially.

For $n=2$, $\lambda_0 = \pm i \omega$, so $x \propto \exp\left(\pm i \omega t\right)$ as desired/observed in nature.

For $n>2$, $\mathcal{Re}\left(\lambda_0\right) = \omega^{2/n} \cos\left(\pi /n\right) > 0$, so $x$ blows up exponentially. You can't just throw these solutions away, because there have to be $n$ unknown $A_j$'s. So, even though there exist solutions like the $n=2$ case for some $n>2$ -- namely, when $\left(1+2j\right)/n = 1/2, 3/2$ -- they will always be accompanied by solutions that aren't seen in nature.