Noether's theorem says that if the following transformation is a symmetry of the Lagrangian
$$t \to t + \epsilon T$$
$$q \to q + \epsilon Q.$$
Then the following quantity is conserved
$$\left( \frac{\partial L}{\partial \dot{q}}\dot{q} – L \right) T – \frac{\partial L}{\partial \dot{q}} Q.$$
Suppose our Lagrangian is given by
$$ L = \frac{1}{2}m \dot{q}^2 – \ln t.$$
Then is not the Lagrangian invariant under the transformation given by
$$T = t$$
$$Q = 0~?$$
Making this transformation contributes only an additive constant to the Lagrangian, which will not affect the dynamics, and so we should conclude that such a transformation is indeed a symmetry of the Lagrangian. However, the quantity
$$ \left( \frac{\partial L}{\partial \dot{q}}\dot{q} – L \right) t = \left(\frac{1}{2}m\dot{q}^2 + \ln t \right)t$$
is clearly not conserved. The E-L equations imply that the kinetic energy is constant, and so this function is clearly an increasing function of time.
Where is my error?
Best Answer
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.
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.