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Suppose I have a simplified pendulum (massless string of length $l$, ball of mass $m$ and some non-zero volume). I want to derive the equations of motion, but I want to take air resistance under account. I know how to do it by applying Newton's 2nd law, but is there any way I can do that using lagrangian or hamiltonian? What would be the form of lagrangian/hamiltonian in such case?

Diracology
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gabe
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3 Answers3

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For a general damping force there is no Least Action Principle. However you can obtain the equations of motions from d'Alembert Principle. The procedure is explained here and the equations of motions are $$\frac{d}{dt}\frac{\partial L}{\partial \dot q_i}-\frac{\partial L}{\partial q_i}=Q_i^p,$$ where $L=T-V$ is the Lagrange function and $$Q_i^p\equiv\sum_\alpha\vec F_\alpha\cdot\frac{\partial \vec r_\alpha}{\partial q_i},$$ is the so-called generalized force which depends on the damping force $\vec F_\alpha$ acting of particle $\alpha$.

Since for general damping we cannot formulate a variational principle, $L$ itself is not enough determine the dynamic evolution of the system. Therefore there is no Hamiltonian $H$ satisfying $$\delta S=\delta\int_{t_1}^{t_2}\left[\sum_i p_i\dot q_i-H(q,p,t)\right]dt=0.$$ Recall that the canonical equations can also be obtained from the variational principle above so there is no function $H$ satisfying the canonical equations and the equations of motion of the system (which shall take the damping into account) simultaneously.

In the case of linear damping, you can follow the Bateman idea which is mentioned in ZeroTheHero's answer or use the Rayleigh dissipation function

Diracology
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It depends on the form of the damping, but if you write $$ L=e^{\alpha t}\left(\textstyle\frac{1}{2}m\dot{q}^2 -V(q)\right) $$ then you get an equation of motion of the form $$ e^{\alpha t}\left( m\ddot{q}+m\alpha \dot{q}+\frac{dV}{dq}\right)=0 $$ which may account for the friction depending on your friction model.

The Hamiltonian would be computed in the usual way but clearly $L$ depends explicitly on $t$ so the energy is not conserved in your system.

Edit: changed $\alpha\to -\alpha$ after a correct comment.

ZeroTheHero
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  • Why $e^{-at}$? I don't see where it came from. I also forgot to mention that I'd like the damping to depend on the square of velocity. Well, the reason I'm asking such question is that ultimately I want to simulate double pendulum with air resistance and deriving equations from lagrangian seems to be easier than force method. – gabe Jul 07 '17 at 22:32
  • I think you want the opposite sign for $\alpha $. – J.G. Jul 07 '17 at 23:16
  • @J.G. Indeed. I have just made the correction. – ZeroTheHero Jul 07 '17 at 23:31
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    @G.Fil As far as I know, this $e^{\alpha t}$ factor is phenomenological, i.e. it is observed to produce the appropriate damping term. I do not know of any trick that captures a damping quadratic in $\dot{q}$: you might have to use the generalized equations of motion, where the term that does not come from a potential is added "by hand". – ZeroTheHero Jul 07 '17 at 23:33
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    It's not phenomenological; there's a deep reason for the choice. This is a bit of overkill, but the $e^{\alpha t}$ factor is analogous to the $\sqrt{|g|}$ factor in general relativity if we regard time as a $1$-dimensional Riemannian manifold. Then $q$ is a field on that manifold, with the equation of motion's $\dot{q}$ term proportional to the Christoffel term $\Gamma_{00}^0=\frac{1}{2}g^{00}\partial_0 g_{00}=\partial_0\ln\sqrt{g}$, from which the $g=e^{2\alpha t}$ choice follows. (Effectively it's a de Sitter manifold.) – J.G. Jul 08 '17 at 06:43
  • @J.G.wow! I just learned something! But why would you want to think of time in this way, and does this provide additional insight that would help handle additional (more complicated) cases? – ZeroTheHero Jul 08 '17 at 06:56
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In a sense yes; you'll need to write the force of air resistance as if it can be derived from a velocity dependent potential.

A common example of this is a particle in a magnetic field. The force of the magnetic field on the particle depends on its velocity, so the potential is written as $\vec{A} \cdot \vec{v}$.

Señor O
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