Action principle, Lagrangian mechanics, Hamiltonian mechanics, and conservation laws when assuming Aristotelian mechanics $F=mv$

Define a physical system when Aristotelian mechanics $$F=mv$$ instead of Newtonian mechanics $$F=ma$$.

Then we could have action $$I=\int L(q,t)dx$$ rather than $$\int L(q',q,t)dx$$.

• Is there an action principle?

• Will the formula $$I=\int p d q$$ still hold?

• What will be the Hamiltonian and conservation laws look like in this case?

• Since $p=\frac{\partial L}{\partial \dot{q}}$ and you no longer have $\dot{q}$ how do you propose to define $p$? – ZeroTheHero Jun 18 '17 at 21:14
• @ZeroTheHero Is it possible that $p=mq'$ still holds? – High GPA Jun 18 '17 at 21:17
• If you define p as mv and the third law is still valid (f1=-f2), then you can easily verify that for a constant force I is a function of t – user126422 Jun 19 '17 at 1:49

Unlike Newtonian mechanics $$m\ddot{q}^i~=~-\frac{\partial V(q)}{\partial q^i}, \tag{N}$$ the Aristotelian mechanics $$m\dot{q}^i~=~-\frac{\partial V(q)}{\partial q^i} \tag{A}$$ is always dissipative and has no conventional stationary action principle. (See also this related Phys.SE post.) This implies that any attempt to define corresponding Aristotelian notions of Lagrangian & Hamiltonian mechanics, Noether current & conservation laws, are severely crippled from the onset.

• By saying "dissipative", did you pre-assume that kinetic energy is $(1/2)mv^2$ and energy is conserved? – High GPA Jun 19 '17 at 1:43
• $\uparrow$ No. – Qmechanic Jun 19 '17 at 4:11

To start you do not have $\vec F~=~m\vec v$ just on dimensional reasons. Instead of mass $m$ we might use $\vec F_v~=~\sigma\vec v$. This is a force due to viscosity or friction. We may then write a potential energy by using the work-energy theorem $\int\vec F_v\cdot d\vec r$ $=~\sigma\int\vec v\cdot d\vec r$. Because $\vec v~=~d\vec r/dt$ we have $W~=~\sigma\int\vec v\cdot\vec v dt$. Now we define a type of potential $U~=~-W$.

All looks good, but there is a problem. Let's go back to $\int\vec F_v\cdot d\vec r$ and consider the integration around a loop of constant radius $R$. The integration variable is the angle $\theta$ so that $d\vec r~=~R\vec\theta d\theta$. Clearly then $\vec v\cdot d\vec r$ $=~R^2\omega d\theta$ for the velocity $\vec v~=~R\omega\vec\theta$ for a constant $\omega$ angular velocity. This means the closed integration is $$\oint\vec F_v\cdot d\vec r~=~2\pi\sigma R^2\omega.$$ For a conservative force this is zero. There is then not a potential $U$ that conserves energy with kinetic energy $K~=~\frac{1}{2}mv^2$. The action $I~=~\oint\vec p\cdot d\vec q$ for the momentum $\vec p~=~\int \vec F dt$ is then not an invariant.

The force $\vec F_v~=~\sigma\vec v$ is not conservative and it indicates that energy and action or angular momentum are being taken away, for $\sigma~<~0$, or for this positive it means there is some source of energy or a "torque" introducing angular momentum (action) into the system.

Aristotelian mechanics with conservative "forces" can be written as $m\dot{\vec{x}}+\vec{\nabla}L=0$, where I have denoted the potential $L$ instead of $V$ because its dimension is that of angular momentum, and I don't want people saying "you can't do that because of dimensional analysis". First-order Euler-Lagrange equations are achievable by introducing an auxiliary variable, viz. $L=\vec{y}\cdot\left(m\dot{\vec{x}}+\vec{\nabla}L\right)$. It is worth shifting this by a total derivative to make $\vec{y}$ dynamical, viz. $L=\vec{y}\cdot\vec{\nabla}L-m\vec{x}\cdot\dot{\vec{y}}$. (The Schrödinger equation can be obtained from a Lagrangian in which the "auxiliary variable" is $\psi^\ast$, because complex numbers allow such a "don't invent anything new" trick. Shifting by a total derivative is in that case justified by a desire for Hermiticity.)

Varying $\vec{y}$ gives us the ELE we want. (As a matter of completeness, varying $x_i$ gives us $\sum_jy_j\partial_i\partial_j L-m\dot{y}_i=0$ with $\partial_i:=\frac{\partial}{\partial x_i}$, i.e. $m\dot{\vec{y}}-\vec{\nabla}\left(y\cdot\vec{\nabla}L\right)=0$.) I'll leave as an exercise the addition of terms for non-conservative forces, in analogy for how this achieves a Lagrangian formulation of Newtonian mechanics with non-conservative forces.

Eugene Wigner discussed the symmetries and conservation laws of Aristotelian physics in the very short paper Conservation Laws in Classical and Quantum Physics. The usual conservation laws do not hold.