# Conservation of energy for a class of Lagrangians with explicit time dependence

I have read in my book that if $\frac{\partial L}{\partial t}=0$, then the quantity $L-\frac{\partial L}{\partial \dot{q}} \dot{q}$ is conserved, and we call it the energy of the system.

But if this is the energy of the system there is something I misunderstand with the fact that adding a term like $\frac{d}{dt} f(q,t)$ doesn't change the equations of motion.

Indeed, we could write :

$$L'=L+\alpha t$$

And $L'$ and $L$ describe the same physics of the system because $t$ can be written as $\frac{d (\frac{t^2}{2})}{dt}$.

This second lagrangian $L'$ explicitly depends on time and then the energy is not conserved, if we do the calculation we would have this equality :

$$L'-\frac{\partial L'}{\partial \dot{q}} \dot{q}=\alpha t+b$$

So it looks like the energy is not conserved in time for the new lagrangian.

So there is something I misunderstand...

• $b$ is conserved and that is defined as the energy. The LHS is not the energy, since $L'$ does not satisfy to the condition. This is a special case of Noether's theorem. – Phoenix87 Dec 27 '16 at 14:23
• What do you call "LHS" ? Im a beginner in lagrangian mechanics. If $b$ is not the energy, how can we define it in a general case ? In the book I read they define the energy as $-L+\frac{\partial L}{\partial \dot{q}}\dot{q}$ so they are wrong in a general case ? – StarBucK Dec 27 '16 at 14:45

1. No explicit time dependence of the Lagrangian is just a simplifying sufficient condition for Noether's theorem. It is not necessary. Recall that Noether's Theorem (in its original form) concerns a quasi-symmetry of the action $S$, cf. this Phys.SE answer.
2. If the Lagrangian is of the form $$\tag{1} L(q,\dot{q},t)~=~ L_0(q,\dot{q}) +\frac{dF(q,t)}{dt},$$ where $L_0(q,\dot{q})$ has no explicit time dependence, then a notion of energy is $$\tag{2} \dot{q}^i\frac{\partial L_0}{\partial \dot{q}^i }-L_0.$$ It is possible to view the energy (2) as a Noether charge associated with time translations by modifying the techniques of this Phys.SE post.