# I Conflict of domain and endpoints in Noether's theorem

#### Van Ladmon

Summary
Conflicts arise on boundary when proving energy conservation using Noether's theorem. Different statement appear in Physics from Symmetry and Kleinert's Particles and Quantum Fields.
In the derivation of energy conservation, there is the transformation $q(t)\rightarrow q'(t)=q(t+\epsilon)$, whose end points are kind of fuzzy. The original path $q(t)$ is only defined from $t_1$ to $t_2$. If this transformation rule is imposed, $q'(t_2-\epsilon)=q(t_2)$ to $q'(t_2)=q(t_2+\epsilon)$ is not defined in the original path. Then how could the Lagrangian be integrated?

On P.98 of Jakob Schwichtenberg's book Physics from Symmetry, he stated that $\delta q(t_1)=\delta q(t_2)=0$ whereas Kleinert stated in his Particles and Quantum Fields $\delta q_s(t_a)$ and $\delta q_s(t_b)$ are not necessarily $0$. Who's correct?

This question is different from the endpoint questions since it is already clear that $q(t_2+\epsilon)\neq q(t_2)$.

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#### vanhees71

Science Advisor
Gold Member
You have to express everything in terms of the "old coordinates" including time. Then there's no problem with imposing the appropriate boundary conditions $\delta q(t_1)=\delta q(t_2)=0$ for Hamilton's principle. For time translations the symmetry condition is $\partial_t L=0$, i.e., $L=L(q,\dot{q})$ and the conserved quantity is $H=p \cdot q-L$ with $p=\partial L/\partial \dot{q}$.

#### Van Ladmon

You have to express everything in terms of the "old coordinates" including time. Then there's no problem with imposing the appropriate boundary conditions $\delta q(t_1)=\delta q(t_2)=0$ for Hamilton's principle. For time translations the symmetry condition is $\partial_t L=0$, i.e., $L=L(q,\dot{q})$ and the conserved quantity is $H=p \cdot q-L$ with $p=\partial L/\partial \dot{q}$.
But why Kleinert differentiates the two kinds of variations: $\delta q$ used in Hamilton's principle and $\delta_s q$ in Noether's theorem? He says that $\delta_s q$ need not be $0$ on the boundaries. Also, what do you mean by expressing everything in terms of "old coordinates"? Could you please give an example? Thanks.

#### vanhees71

Science Advisor
Gold Member
I don't know, why Kleinert does it this way.

I have my treatment of Noether's theorem unfortunately only in a German manuscript. I hope the formula density is high enough, so that you can understand the argument:

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