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I am doing a course in dynamical systems, and the term "time invariant" when it comes to systems is quite not clear.

I understand if time is not an explicit variable in the equation then it's a time invariant system, but again aren't all dynamical systems from time dependent?

I mean, a pendulum location and speed is dependent on its time, and yes the laws of physics governing it are the same regardless of the time, but this won't affect that it's still dependent on time in its motion

So what am I missing or have misunderstood? Thanks alot!

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  • $\begingroup$ Have you seen a Lagrangian that describes a dynamical system yet? $\endgroup$ – edm Feb 16 at 12:25
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A dynamical system is described by a Lagrangian function $L$. It is a function that takes time, position and velocity as inputs, so we write it as $L(t,q,\dot q)$, where $t$ is time variable, $q$ is position variable, $\dot q$ is velocity variable.

To say that a dynamical system is time-invariant means $\frac{\partial L}{\partial t}=0$.

Physically, time-invariant means the following:

You do an experiment observing the movement of a pendulum that experiences no friction. You give the pendulum an initial push, or some initial height, or a mixture of both. You record the whole motion of the pendulum for a while. You do the same experiment again some time later, with the same initial conditions. If the system is time-invariant, the whole motion is the same as before.

If the system is time-dependent, for example, the pendulum is charged, and someone messes with your lab by applying some electric field with varying field strength over time, you may observe different motions of pendulum if you start your experiment at different times.

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  • $\begingroup$ Interesting, I have a question regarding the system you have proposed. Imagine the same Pendulum experiment being influenced under an electric field which is a function of time; if this will be a system which under the same condition at another point in time is observed, would be also time-invariant regarding that the conditions of the experiment are the same and the resulting motion the same. How would this be expressed in Lagrangian? will the electric force be similar to the way the gravitational force is expressed? $\endgroup$ – QUTADAH Feb 16 at 14:27
  • $\begingroup$ @QUTADAH The case where there is an electric field as a function of time is not time-invariant. The Lagrangian will be expressed as $$L=\frac{1}{2}m\dot\theta^2-U(\theta)-V(t,\theta)$$ where $U$ is gravitational potential energy and $V$ is electric potential energy. $\endgroup$ – edm Feb 16 at 14:41
  • $\begingroup$ Thanks alot, i get it now. That was a big help! $\endgroup$ – QUTADAH Feb 16 at 15:01
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In classical mechanics, the Lagrangian $L$ and Hamiltonian $H$ satisfy$$\frac{dH}{dt}=\frac{\partial H}{\partial t}=-\frac{\partial L}{\partial t},$$which is not in general $-\frac{dL}{dt}$. A time-invariant system has $\frac{\partial L}{\partial t}=0$, i.e. $H$ (a popular definition of energy) is conserved.

Let's discuss the damped harmonic oscillator. The Lagrangian is $L=\frac12me^{\gamma t}(\dot{x}^2-\omega^2x^2)$, so$$0=\frac{1}{me^{\gamma t}}\left[\frac{d}{dt}\frac{\partial L}{\partial\dot{x}}-\frac{\partial L}{\partial x}\right]=\ddot{x}+\gamma\dot{x}+\omega^2x.$$The undamped oscillator has $\gamma=0$ so $\partial_tL=0$, but more generally $\partial_tL=\gamma L$. The damped oscillator loses energy over time, i.e. $\dot{H}<0$.

A physicist would no doubt note this doesn't reduce the total energy of the universe; the oscillator is dissipative, gifting energy to its environment, which the above Lagrangian doesn't talk about. However, something analogous to the above calculation applies to an expanding universe. The whole universe's Lagrangian density ends up proportional to $\sqrt{|g|}$, so energy isn't conserved, at least not if you define it as the total Hamiltonian. In particular, we get an $e^{\gamma t}$ factor appearing for de Sitter space, $\gamma$ being its Hubble parameter.

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