Noether's theorem and time-dependent Lagrangians

The Lagrangian (and the action as a whole)

$$ L = \frac{1}{2}m \dot{q}^2 - \ln t$$

is not invariant under the transformation given by $$T = t, \qquad Q = 0.$$

The rescaling of $t$ by factor $1+\epsilon$ also modifies the time derivatives: $ \delta \dot{q} = -\epsilon \dot{q}$ (and the measure of integration $dt$), so the suggested quantity is not conserved.

Where is my error?

So, the error is in choosing the wrong Lagrangian / transformation.

Noether's theorem works just fine for explicitly time dependent Lagrangians. Here is another example of Lagrangian with explicit time-dependence: $$ L = \frac{m \dot{q}^2}{2} e^{\alpha t}. $$ Such type of Lagrangian could arise for example, if we try to obtain the equations of motions for dissipative system.

The Euler-Lagrange equation for this system after omitting common factor reads as $$ \ddot{q} = -\alpha \dot{q}. $$

This Lagrangian is invariant under the infinitesimal transformation: $$ t\to t' = t + \epsilon, \qquad q \to q' = q -\epsilon\frac{\alpha q}{2}. $$

Substituting these $T$ and $Q$ in the Noether's theorem we obtain the quantity $$ A=\frac{m }{2} e^{\alpha t}\cdot (\dot{q}^2 + \alpha \dot{q} q). $$

Its time derivative is $$ \dot{A} = \frac{m }{2} e^{\alpha t}\cdot (\alpha \dot{q}^2 + \alpha^2 \dot{q} q +2 \ddot{q}\dot{q} + \alpha \ddot{q} q + \alpha \dot{q}^2) , $$ If we use E-L equation to eliminate $\ddot{q}$ we obtain $\dot{A}=0$, so the quantity is conserved as it should be.

Let's see how we can play with Noether's theorem in the "conservation of energy" example.

First of all, we apply a time-independent variation to the Lagrangian, \begin{equation} t \rightarrow t+ \epsilon \end{equation} For a general Lagrangian, it will change \begin{equation} L(x(t), \dot{x}(t), t)\rightarrow L'(x(t), \dot{x}(t), t ) = L(x(t), \dot{x}(t), t-\epsilon ) \end{equation} what we require in this case, is that the Lagrangian is invariant(not just covariant) as a scalar, \begin{equation} L=L' \implies \frac{\partial L}{\partial t} = 0 \end{equation} That's a property true alone any path.

Then we use an induced time-dependent variation, \begin{equation} t \rightarrow t+ \epsilon(t) \quad x(t) \rightarrow x(t) + \epsilon(t) \dot{x(t)} \end{equation} The change of the action in this case is \begin{equation} \delta S = \int_0 ^T \frac{\partial L }{\partial x} \dot{x(t)} \epsilon(t) + \frac{\partial L } {\partial \dot{x} } \frac{d}{dt}(\epsilon(t) \dot{x}(t)) = \int_0^T dt \epsilon(t) ( \frac{\partial L }{\partial x} \dot{x} + \frac{\partial L } {\partial \dot{x} } \ddot{x} ) + \frac{\partial L } {\partial \dot{x} } \dot{x} \dot{\epsilon} \end{equation} Now adding $\frac{\partial L}{\partial t}=0$ to the first two terms in the parentheses to make it a total derivative, since this is true for all paths in the configuration space . For the second term, we integrate by part to get(variation has zero boundary condtions), \begin{equation} \delta S = \int_0^T dt \epsilon(t) \frac{d}{dt}(L - \frac{\partial L } {\partial \dot{x} } \dot{x}) \end{equation} Alone the classical path, any variation will extremize the action; in particular our special variation $\epsilon(t)$ induced by the symmetry action will. We get the energy conservation equation, \begin{equation} \frac{d}{dt}(L - \frac{\partial L } {\partial \dot{x} } \dot{x}) = 0 \quad \text{along classical path} \end{equation}

In other words, you have to choose $T$ and $Q$ to be constant(independent of time) to dig out the symmetries of the Lagrangian. Then you can apply the time-dependent variation to get a special "equation of motion" along the classical path: conservation law.

I need to elaborate the symmetry equation I have derived.

Under a time-independent variation, we say Lagrangian actually changes \begin{equation} L' = L(x(t-\epsilon), \dot{x}(t-\epsilon), t-\epsilon ) \end{equation} This is due to an identity, \begin{equation} L(x(t),\dot{x}(t),t) = L'(t') \quad t'=t+\epsilon \end{equation} That means the value of Lagrangian of a point in configuration space is independent of the time coordinate describing it. It's like you can use different time zones to describe an event. They only differ by a constant.

However, this is not true if we apply a time dependent variation, \begin{equation} \frac{d}{dt}(x(t-\epsilon)) = \frac{d}{dt}( x(t) - \epsilon \dot{x}(t) ) = \dot{x}(t) - \dot{\epsilon}\dot{x(t)} - \epsilon \ddot{x}(t) \ne \dot{x}(t) - \epsilon \ddot{x}(t) = \dot{x}(\tau)|_{\tau = t-\epsilon} \end{equation} the time dependent variation forces the velocity to change in a another way different from what we like. It's like time is not flowing uniformly, such that even if you displace your velocity vector to the original point before the translation, it still changes, because the rate will depend on how time flows.

In the technical aspect, you variation can't lead to \begin{equation} \frac{\partial L}{\partial t} = 0 \end{equation} which is equivalent to the conservation law.

1. Noether's theorem also works for OP's time-dependent Lagrangian $$ L~=~T-V,\qquad T ~=~\frac{1}{2}m \dot{q}^2, \qquad V(t)~=~\ln t. \tag{A} $$

2. The time-dependent potential $V(t)=\ln t$ can be viewed as changing conventions for the zero-level of the potential energy. Nevertheless, the kinetic energy $T$ is a constant of motion.

3. To prove kinetic energy conservation, OP essentially makes the same mistake as OP in this Phys.SE question: Perhaps counter-intuitively, the relevant infinitesimal transformation is not a pure time translation. The key infinitesimal transformation is instead $$ \delta q~=~\varepsilon\dot{q} .\tag{B}$$ Time-translation can be included or excluded as explained in my Phys.SE answer here. Let us for simplicity exclude time-translation $$ \delta t ~=~0. \tag{C}$$ Then the transformation $\delta$ can no longer "feel" the potential term $V$. We are therefore back in a standard application of Noether's theorem.

4. The infinitesimal transformation of the Lagrangian $$ \delta L ~\stackrel{(A)+(B)+(C)}{=}~\ldots~=~\varepsilon \frac{dT}{dt} \tag{D} $$ is a total derivative, i.e. the transformation (B) & (C) is a quasisymmetry of the Lagrangian (A).

5. The bare Noether charge is momentum times generator. That is: $p\dot{q}$. The full Noether charge $$p\dot{q}-T~=~T\tag{E}$$ is unsurprisingly the kinetic energy.