# Noether's theorem with or without boundary term in the classical mechanics

I'm recently reading both a variational calculus textbook and a classical mechanics textbook. I found that Noether's theorem is stated differently within those two backgrounds. I'm wondering about the origin of this kind of difference and how to form a consistent view.

## Variational calculus viewpoint [1]

Basically here we only talk about the invariance of the action $$S=\int L(q,\dot{q},t) d t \, .$$ The statement is that if under some general variation, $$t\rightarrow t^\prime \approx t+\epsilon \delta t \\ q_i^\prime\rightarrow q_i^\prime(t^\prime) \approx q_i(t) +\epsilon \delta q_i$$ the action doesn't change $$\int_{t_1^\prime}^{t_2^\prime} L(q^\prime,\dot{q}^\prime,t^\prime) d t^\prime = \int_{t_1}^{t_2} L(q,\dot{q},t) d t$$ then we have a constant of motion as $$\sum_{i} \frac{\partial L}{\partial \dot{q}_i} \delta q_i - (L-\sum_{i} \dot{q}_i \frac{\partial L}{\partial \dot{q}_i})\delta t$$ See Section 20, Eqs. (50, 51, 52) in Ref. [1] for detailed discussions. I have changed the notation for comparison with the following results.

## Classical mechanics viewpoint [2]

Here we talk about the invariance and quasi-invariance of the Lagrangian under a general variation with a constraint that the time doesn't change. If we follow the variational calculus viewpoint, we would then reach the conclusion that the constant of motion would be ($$\delta t =0$$) $$\sum_{i} \frac{\partial L}{\partial \dot{q}_i} \delta q_i$$ But, the final results of the constant of motion here actually have another term $$\sum_{i} \frac{\partial L}{\partial \dot{q}_i} \delta q_i - F(q,\dot{q},t)$$ And the additional $$F$$ term is related to the variation of the Lagrangian under the general variation $$\epsilon \frac{d}{dt} F(q,\dot{q},t) = L(q_i+\delta q_i,\dot{q}_i+\delta \dot{q}_i,t) - L(q_i,\dot{q}_i,t)$$ See Section Conservation Laws and the Action Principle (starting from pg. 25, Eqs. (14, 16, 20, 22, 25)) of Chapter 4 in Ref. [2] for more details.

## Which one of the above variations leave the EoM unchanged?

For physicists, an important feature of the symmetry/invariance property is that the equation of motion doesn't change after some variation. Or, one may says that the varied function is still the solution of the varied system if the original function is the solution of the original system.

Following this point, I can understand why the invariance of the action would be thought of as a symmetry/invariance property of the system. Since the value of the action after this variation is the same, then that the original function is a stationary point of the original action would imply that the varied function would also be a stationary point of the varied action. Then we can invoke the Hamilton's variational principle stating the equivalence between the stationary point of an action functional and the solution of that system.

But for the second viewpoint, I don't know how to argue that the Euler-Lagrangian equation stay the same with that additional $$F$$ term. An important difference here is that the variation of the velocity is not require to vanish at the endpoint, and thus we cannot say that the $$F(q,\dot{q},t)$$ term don't contribute under the general variation.

## References

1. Gelfand, I. M., & Fomin, S. V. (2012). Calculus of Variations. Dover Publications.
2. Sudarshan, E. C. G., & Mukunda, N. (2015). Classical Dynamics: A Modern Perspective. World Scientific Publishing

4. As for OP's very last point, note that the change of the action with $$\epsilon F$$ is only infinitesimally small, and that the proof of Noether's theorem only needs the EL equations for the unaltered/original action.