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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?

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  • $\begingroup$ Since $p=\frac{\partial L}{\partial \dot{q}}$ and you no longer have $\dot{q}$ how do you propose to define $p$? $\endgroup$ – ZeroTheHero Jun 18 '17 at 21:14
  • $\begingroup$ @ZeroTheHero Is it possible that $p=mq'$ still holds? $\endgroup$ – High GPA Jun 18 '17 at 21:17
  • $\begingroup$ 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 $\endgroup$ – user126422 Jun 19 '17 at 1:49
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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.

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

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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.

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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.

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