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

Physics Asked on July 10, 2021

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} = mddot{x}_{A} &= -k(x_{A} – x_{B}) – bdot{x}_{A},
F_{B} = mddot{x}_{B} &= -k(x_{B} – x_{A}) – bdot{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:
$$ vec{x}(t) = begin{bmatrix} A A end{bmatrix} + begin{bmatrix} B B end{bmatrix} e^{-2gamma t} $$
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^{-tsqrt{gamma^{2}-omega^{2}}} + begin{bmatrix} D -D end{bmatrix}e^{tsqrt{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(tsqrt{omega^{2} – gamma^{2}}) + begin{bmatrix} D -D end{bmatrix}sin(tsqrt{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 $E = frac{1}{2}mdot{x}_{A}^{2} + frac{1}{2}mdot{x}_{B}^{2} + frac{1}{2}k(x_{A}-x_{B})^{2}$ 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} = -bdot{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?

2 Answers

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 $ mddot{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)expleft(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.

Correct answer by SolubleFish on July 10, 2021

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.

Answered by Pavlo. B. on July 10, 2021

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