# Why only 2 derivatives in classical mechanics? [duplicate]

The title conveys my true question, but for the sake of clarity I will now rephrase it in a more mathematical flavor using the Hamiltonian formalism of classical mechanics and the terminology of differential geometry, including differential forms and the exterior derivative $$d$$.

Consider a physical system with configuration space $$M$$, a manifold. Its phase space is the cotangent bundle $$T^*M$$, consisting of (position, momenta) pairs. There is a tautological $$1$$-form $$\theta$$ on $$T^*M$$ turning it into a symplectic manifold $$(T^*M,-d\theta)$$. Any function $$H\colon T^*M\to \mathbb R$$ may be regarded as a Hamiltonian. From $$H$$, the dynamics are given by the flow lines of the Hamiltonian vector field $$X_H$$, defined to be the unique vector field satisfying $$dH(Y)=d\theta(Y,X_H)$$ for all vector fields $$Y$$ on $$M$$. Moreover, the action is simply $$S=\theta(X_H)$$.

Since the dynamics are given by a vector field on the cotangent bundle, only two derivatives can appear in the equations of motion; since any classical system can be put in the Hamiltonian formalism, it appears that the equations of motion of any classical system can involve only the first two derivatives (equivalently, $$F=ma$$ where $$F$$ is a function of position and velocity).

However, there seems to be no mathematical reason for this limitation; certainly one can write down equations involving $$3$$ or more derivatives - so what makes them unable to appear in the laws of a classical system?

Phrased more mathematically, instead of starting with $$T^*M$$ (which can be thought of as the space of first order Taylor expansions to functions on $$M$$) we can start with the jet manifold $$J^r(M)$$, where $$J^0(M)=M$$ and $$J^1(M)=T^*M$$ and in general $$J^r(M)$$ consists of the $$r^{th}$$ order Taylor expansions to functions on $$M$$. For any $$r\geq 1$$, the jet manifold $$J^r(M)$$ carries a tautological $$1$$-form $$\theta$$ which is defined in a similar way as the tautological $$1$$-form on $$T^*M$$; in local coordinates $$\bigl(p^{(0)},p^{(1)},\ldots,p^{(r)}\bigr)$$ which describe curves $$f\colon \mathbb R\to M$$ with Taylor expansion $$f(t)=p^{(0)}+tp^{(1)}+\cdots+t^rp^{(r)}+O\bigl(p^{(r+1)}\bigr),$$ the tautological $$1$$-form on $$J^r(M)$$ is given by $$\theta\bigl(p^{(0)},p^{(1)},\ldots,p^{(r)}\bigr)=p^{(1)}_i dp^{(0)\ i}+\cdots+p^{(r)}_i dp^{(r-1)\ i}.$$ Exactly as before, we get a symplectic manifold $$\bigl(J^r(M),-d\theta\bigr)$$. Any function $$H\colon J^r(M)\to\mathbb R$$ may be thought of as a Hamiltonian, giving rise to a Hamiltonian vector field $$X_H$$ on $$J^r(M)$$ and thus to dynamics of a physical system.

Question. Why does classical mechanics stop at $$r=1$$ and have no need for higher order terms in the Taylor expansion? Alternatively, are there physical systems in which higher order derivatives do arise and thus require some modification (e.g. the one given above) to the Hamiltonian formalism?

## marked as duplicate by Qmechanic♦ classical-mechanics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 19 at 10:23

• Of course this is true, but I don't think it really hits at the heart of the question. Are there naturally occurring systems that truly require order N>2 differential equations, and as a result must be split into multiple simpler differential equations? (See the second sentence in my question above.) Note that this procedure, while "artificially" lowering the number of derivatives, increases the complexity of the system (by multiplying the dimension). Essentially it is an embedding of the jet bundle $J^r(M)$ in a tangent bundle $T^*(M^r)$. – pre-kidney Jan 19 at 9:56