Confusion with Noether's theorem: Time symmetric system with velocity-dependent terms?

Question

Noether's theorem says that any system that is time-translation-symmetric displays energy conservation, and vice-versa. However, I'm not sure if this is the case.

Suppose we have a two particles (of the same mass $m$) connected by a spring (of spring constant $k$) with a damping coefficent $b$: \begin{align*} F_{A} = m\ddot{x}_{A} &= -k(x_{A} - x_{B}) - b\dot{x}_{A}, \\ F_{B} = m\ddot{x}_{B} &= -k(x_{B} - x_{A}) - b\dot{x}_{B}. \end{align*} For reference, I'll write the solution to this below.

We'll write $\vec{x}(t) = \begin{bmatrix} x_{A}(t) \\ x_{B}(t) \end{bmatrix}$ along with $\gamma = \dfrac{b}{2m}$ and $\omega = \sqrt{\dfrac{2k}{m}}$ for succinctness. The general solution to the differential equations is a linear combination of the following solutions. First, we have the part where the two masses coincide: $$ \vec{x}(t) = \begin{bmatrix} A \\ A \end{bmatrix} + \begin{bmatrix} B \\ B \end{bmatrix} e^{-2\gamma t} $$ where $A, B$ are constants. Second, we have the oscillation part: \begin{align*} &\text{Overdamped } (\gamma > \omega): && \vec{x}(t) = e^{-\gamma t}\left( \begin{bmatrix} C \\ -C \end{bmatrix}e^{-t\sqrt{\gamma^{2}-\omega^{2}}} + \begin{bmatrix} D \\ -D \end{bmatrix}e^{t\sqrt{\gamma^{2}-\omega^{2}}} \right) \\[1.5ex] &\text{Critical damping } (\gamma = \omega): && \vec{x}(t) = \begin{bmatrix} C \\ -C \end{bmatrix} e^{-\gamma t} + \begin{bmatrix} D \\ -D \end{bmatrix} t e^{-\gamma t} \\[1.5ex] &\text{Underdamping } (\gamma < \omega): && \vec{x}(t) = e^{-\gamma t}\left( \begin{bmatrix} C \\ -C \end{bmatrix}\cos(t\sqrt{\omega^{2} - \gamma^{2}}) + \begin{bmatrix} D \\ -D \end{bmatrix}\sin(t\sqrt{\omega^{2} - \gamma^{2}}) \right) \end{align*} where $C, D$ are constants.

Clearly none of the forces are explicit functions of time, so the system is time-translation-symmetric. However, the energy $E = \frac{1}{2}m\dot{x}_{A}^{2} + \frac{1}{2}m\dot{x}_{B}^{2} + \frac{1}{2}k(x_{A}-x_{B})^{2}$ is clearly not conserved.

According to this and this, I can convert my system to a Lagrangian system with \begin{align*} L &= \frac{1}{2}m(\dot{x}_{A}^{2} + \dot{x}_{B}^{2}) - \frac{1}{2}k(x_{A} - x_{B})^{2}, \qquad Q_{k} = -b\dot{x}_{k} \qquad (k=A, B), \end{align*} and equations of motion \begin{align*} \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{x}_{k}} \right) - \frac{\partial L}{\partial x_{k}} = Q_{k} \qquad (k=A, B). \end{align*} This is again apparently a time-translation-symmetric system, but the usual definition of energy is not conserved.

How is this compatible with Noether's theorem?

Answer 1

Noether's theorem is usually proved assuming that the trajectories of the system belong to the extremum of the integral $$ \int_{t_1}^{t_2}L(q,\dot{q},t)dt. $$

Having generalized forces $Q_i$ in the right part of Lagrange equation violates the aforementioned assumption.

Answer 2

Noether's theorem applies to Lagrangian systems where no generalized forces are present. To have a mapping between conserved quantities and continuous symmetries of the system, you need to rewrite it with $Q_i =0$.

You can describe a damped oscillator $ m\ddot{x} = - k x - \lambda \dot{x}$ with the following lagrangian : \begin{equation}L(x,\dot{x},t) = \frac{1}{2}( m \dot{x}^2 - kx^2)\exp\left(\frac{\lambda t}{m}\right)\end{equation} I imagine you could write a similar lagrangian for two coupled oscillators.

However, it is then clear that you no longer have time-translation symmetry : energy is not conserved for the damped oscillator.