Lagrangian and Hamiltonian EOM with dissipative force

Question

I am trying to write the Lagrangian and Hamiltonian for the forced Harmonic oscillator before quantizing it to get to the quantum picture. For EOM $$m\ddot{q}+\beta\dot{q}+kq=f(t),$$ I write the Lagrangian $$ L=\frac{1}{2}m\dot{q}^{2}-\frac{1}{2}kq^{2}+f(t)q$$ with Rayleigh dissipation function as $$ D=\frac{1}{2}\beta\dot{q}^{2}$$ to put in Lagrangian EOM $$0 = \frac{\mathrm{d}}{\mathrm{d}t} \left ( \frac {\partial L}{\partial \dot{q}_j} \right ) - \frac {\partial L}{\partial q_j} + \frac {\partial D}{\partial \dot{q}_j}. $$

On Legendre transform of $L$, I get $$H=\frac{1}{2m}{p}^{2}+\frac{1}{2}kq^{2}-f(t)q.$$

How do I include the dissipative term to get the correct EOM from the Hamiltonian's EOM?

Answer 1

Problem: Given Newton's second law

$$\begin{align} m\ddot{q}^j~=~&-\beta\dot{q}^j-\frac{\partial V(q,t)}{\partial q^j}, \cr j~\in~&\{1,\ldots, n\}, \end{align}\tag{1} $$

for a non-relativistic point particle in $n$ dimensions, subjected to a friction force, and also subjected to various forces that have a total potential $V(q,t)$, which may depend explicitly on time.

I) Conventional approach: There is a non-variational formulation of Lagrange equations

$$\begin{align} \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{q}^j}\right)-\frac{\partial L}{\partial q^j}~=~&Q_j, \cr j~\in~&\{1,\ldots, n\},\end{align}\tag{2} $$

where $Q_j$ are the generalized forces that do not have generalized potentials. In our case (1), the Lagrangian in eq. (2) is $L=T-V$, with $T=\frac{1}{2}m\dot{q}^2$; and the force

$$ Q_j~=~-\beta\dot{q}^j\tag{3} $$

is the friction force. It is shown in e.g. this Phys.SE post that the friction force (3) does not have a potential. As OP mentions, one may introduce the Rayleigh dissipative function, but this is not a genuine potential.

Conventionally, we additionally demand that the Lagrangian is of the form $L=T-U$, where $T=\frac{1}{2}m\dot{q}^2$ is related to the LHS of EOMs (1) (i.e. the kinematic side), while the potential $U$ is related to the RHS of EOMs (1) (i.e. the dynamical side).

With these additional requirements, the EOM (1) does not have a variational formulation of Lagrange equations

$$\begin{align} \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{q}^j}\right)-\frac{\partial L}{\partial q^j}~=~&0,\cr j~\in~&\{1,\ldots, n\},\end{align}\tag{4} $$

i.e. Euler-Lagrange equations. The Legendre transformation of eq. (2) to the Hamiltonian formulation is of the form

$$\begin{align} \frac{dq^j}{dt}~=~&\frac{\partial H}{\partial p_j},\cr \frac{dp_j}{dt} +\frac{\partial H}{\partial q^j}~=~&Q_j, \cr j~\in~&\{1,\ldots, n\},\end{align}\tag{5}$$

see this Phys.SE post for details.

II) Unconventional approaches:

1. Trick with exponential factor$^1$: Define for later convenience the function $$ e(t)~:=~\exp(\frac{\beta t}{m}). \tag{6}$$ A possible variational formulation (4) of Lagrange equations is then given by the Lagrangian $$\begin{align} L(q,\dot{q},t)~:=~&e(t)L_0(q,\dot{q},t), \cr L_0(q,\dot{q},t)~:=~&\frac{m}{2}\dot{q}^2-V(q,t).\end{align}\tag{7}$$ The corresponding Hamiltonian is $$ H(q,p,t)~:=~\frac{p^2}{2me(t)}+e(t)V(q,t).\tag{8}$$ One caveat is that the Hamiltonian (8) does not represent the traditional notion of total energy. Another caveat is that this unconventional approach cannot be generalized to the case where two coupled sectors of the theory require different factors (6), e.g. where each coordinate $q^j$ has individual friction-over-mass-ratios $\frac{\beta_j}{m_j}$, $j\in\{1, \ldots, n\}$. For this unconventional approach to work, it is crucial that the factor (6) is an overall common multiplicative factor for the Lagrangian (7). This is an unnatural requirement from a physics perspective.

2. Imposing EOMs via Lagrange multipliers $\lambda^j$: A variational principle for the EOMs (1) is $$\begin{align}L ~=~& m\sum_{j=1}^n\dot{q}^j\dot{\lambda}^j\cr &-\sum_{j=1}^n\left(\beta\dot{q}^j+\frac{\partial V(q,t)}{\partial q^j}\right)\lambda^j.\end{align}\tag{9}$$ (Here we have for convenience "integrated the kinetic term by parts" to avoid higher time derivatives.)

3. Classical Schwinger/Keldysh "in-in" formalism: The variables are doubled up. See e.g. eq. (20) in C.R. Galley, arXiv:1210.2745. Ignoring boundary terms$^2$ the Lagrangian reads $$\begin{align} \widetilde{L}(q_{\pm},\dot{q}_{\pm},t) ~=~&\left. L(q_1,\dot{q}_1,t)\right|_{q_1=q_+ + q_-/2}\cr ~-~&\left. L(q_2,\dot{q}_2,t)\right|_{q_2=q_+ - q_-/2}\cr ~+~&Q_j(q_+,\dot{q}_+,t)q^j_-\end{align}\tag{10}. $$ The initial conditions $$\left\{\begin{array}{rcl} q^j_+(t_i)&=&q^j_i,\cr\dot{q}^j_+(t_i)&=&\dot{q}^j_i\end{array}\right.\tag{11} $$ implement the system's underlying initial values. The final conditions $$\begin{align}\left\{\begin{array}{rcl} q^j_-(t_f)&=&0\cr \dot{q}^j_-(t_f)&=&0 \end{array}\right. & \cr\cr\qquad\Downarrow&\qquad\cr\cr \left.\frac{\partial \widetilde{L}}{\partial \dot{q}^j_+}\right|_{t=t_f}~=~&0 \end{align}\tag{12} $$ implement the physical limit solution $q_-^j= 0$. The doubling trick (10) is often effectively the same as introducing Lagrange multipliers (9).

4. Gurtin-Tonti bi-local method: See e.g. this Phys.SE post.

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$^1$ Hat tip: Valter Moretti.

$^2$ The variational problem (10)+(11)+(12) needs an appropriate initial term, which might not always exist! In particular, since we already imposed $4n$ boundary conditions (11)+(12), it would be too much to also impose the initial condition $$ q^j_-(t_i)~=~0. \qquad (\leftarrow\text{Wrong!})\tag{13}$$ Example: If
$$L~=~\frac{1}{2}m\dot{q}^2,\qquad Q~=~0,\tag{14}$$ then $$\widetilde{L}~\stackrel{(14)+(10)}{=}~m\dot{q}_+\dot{q}_-,\tag{15}$$ and one should add an initial term $\left. m\dot{q}_+q_-\right|_{t=t_i}$ to the action $$\widetilde{S}~=~\left. m\dot{q}_+q_-\right|_{t=t_i}+ \int_{t_i}^{t_f}\!dt~\widetilde{L} ,\tag{16}$$ so that an infinitesimal variation becomes $$ \delta \widetilde{S}~\stackrel{\rm IBP}{=}~\text{bulk terms} ~+~ \text{boundary terms},\tag{17}$$ where the boundary terms $$\begin{align}\text{boundary terms} ~=~&\left. mq_-\delta\dot{q}_+ \right|_{t=t_i} +\left. m\dot{q}_+\delta q_-\right|_{t=t_i} +\left[m\dot{q}_{\pm}\delta q_{\mp}\right]_{t=t_i}^{t=t_f}\cr ~=~&\left. mq_-\delta\dot{q}_+ \right|_{t=t_i} +\left. m\dot{q}_+\delta q_-\right|_{t=t_f} +\left[m\dot{q}_-\delta q_+\right]_{t=t_i}^{t=t_f}\cr ~\stackrel{(11)+(12)}{=}&~0.\end{align}\tag{18}$$ vanish, as they should.

Answer 2

Problem: solve the EOM

$$\ddot x + \beta \dot x + \omega^2 x = f(t)$$

As an approach we shall use, additionally to $x(t), \dot x(t)$, two new parameters $y(t), \dot y(t)$.

Let us, magically, introduce a Lagrangian for this auxiliary system

$$L(x, y, \dot x, \dot y, t) = \dot x \dot y - \beta \dot x y - \omega^2 x y - (x + y) f(t)$$

The important thing to notice is that the equations of motion for this system are

$$ \frac{d}{dt}\left( \frac{\partial L}{\partial \dot x} \right) - \frac{\partial L}{\partial x} = \ddot y - \beta \dot y + w^2 y - f(t) = 0\\ \frac{d}{dt}\left( \frac{\partial L}{\partial \dot y} \right) - \frac{\partial L}{\partial y} = \ddot x + \beta \dot x + w^2 x - f(t) = 0 $$

As one can see, we recover the equations of motion for our original system along with an auxiliary EOM.

From here on out, everything goes according to theory for Hamiltonian mechanics. We can find the generalized momenta:

$$ p_x = \frac{\partial L}{\partial \dot x} = \dot y - \beta y\\ p_y = \frac{\partial L}{\partial \dot y} = \dot x$$

And rewriting the Langrangian as a Hamiltonian

$$H(x, y, p_x, p_y, t) = p_x p_y + \omega^2 x y + \beta y p_y + (x + y) f(t)$$

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The method is a bit more general, see Conservative perturbation theory for nonconservative systems which introduced me to the idea of auxiliary parameters by the example of the Van der Pol oscillator.

As far as I can see, this method should play nicely even when $x \in \mathbb R^n$ in which case you would also choose $y \in \mathbb R^n$.

Answer 3

Another Lagrangian formulation of a dissipative force is to derive it from the interaction of many degrees of freedom (a bath) that are integrated out. In its quantum version this is the Caldeira-Leggett model. This is not just a mathematical trick but can be physically motivated. For example, this is how you derive radiation damping of particles by their interactions with the EM field.

With no loss of generality you can couple $q$ with many independent harmonic oscillators: $$ L=L_0+\sum \frac12(\dot x_i^2-\omega_i^2x_i^2) - x_iq $$ You can solve for the bath degrees of freedom using causal Green’s functions: $$ G_i(t)=H(t)\frac{\sin(\omega_it)}{\omega_i}\\ x_i=G_i*q $$ Substituting $x_i$ in the action you get an effective, (temporally) non local action in $q$. In terms of frequency, this adds a quadratic term: $$ S_B=-\int \left(\sum\frac1{(\omega+i\epsilon)^2-\omega_i^2}\right) \frac{|\hat q|^2}2\frac{d\omega}{2\pi}\quad \epsilon\to0^+\\ \hat q(\omega)=\int q(t)e^{i\omega t}dt \quad q(t)=\int \hat q(\omega)e^{-i\omega t}\frac{d\omega}{2\pi} $$ By spanning all the frequencies in the continuum limit, with $n(\omega)d\omega$ the number of bath's modes in the range $(\omega,\omega+d\omega)$ you get the parenthesised term: $$ \begin{align} G_0 &= \int\frac{n(f)}{(\omega+i\epsilon)^2-f^2}\frac{df}{2\pi} \\ &= \int PV\frac{n(f)}{\omega^2-f^2}\frac{df}{2\pi}-i\frac{n(\omega)}{2\omega} \end{align} $$ By choosing $n$ appropriately, you can reproduce essentially any linear dissipation you want. In your case, you can take: $$ n(\omega)=2\beta\omega^2 $$ The key ingredient is causality. Causality gives dissipation while anti-causality gives injection.

In a similar vein with radiation damping, you can motivate this model physically. The idea is to couple your degree of freedom to an elastic string. Consider a 1D scalar field $\phi(t,x)$ on the half line $x\geq 0$: $$ S = S_q+S_\phi+S_I\\ S_I= \int q\phi\delta'(x)dxdt = -\int q\partial_x\phi(x=0)dt \\ S_\phi=\int \frac12(\partial_t\phi^2-\partial_x\phi^2)dxdt $$ This translates the fact that the force applied on $q$ is the tension at the elastic string’s extremity. The field is given by: $$ \phi(x,t) = q(t-x) $$ So the force applied on $q$ is: $$ F_\phi = \partial_x\phi(0) = -\dot q-q\delta(0) $$ You recognise the desired damping term, and the second term is an extra recall term, which can be cancelled by the appropriate quadratic counterterm in $S_\phi$ (analogue of mass renormalisation in EM). This is why even in the original Leggett-Caldeira model, you typically also add a quadratic counter term to avoid this as well. This is just the mathematical translation of the fact that the string dissipates energy by generating waves.