Noether Theorem and Symmetry - Spacetime and Energy-Momentum Conservation
When discussing spacetime symmetry, we fix the field \(\phi \) such that it is unaffected by variations (\(\Psi=0\)). We are left with:
$$\partial _\mu \left[\left({\partial \mathcal{L}\over\partial \left(\partial _\mu \phi\right) } \partial _\alpha \phi - \delta ^\mu _\alpha L \right) ฮง^\alpha \right]=0$$
We define the energy-momentum tensor \(T^{\mu\nu}\):
$$T^{\mu\nu}\equiv {{\partial L}\over{\partial \left( \partial _\nu \phi \right)}} \partial ^\mu \phi -\eta ^{\mu\nu} L$$
The conservation current \(\partial _\nu T^{\mu\nu}=0\) represents:
$$\partial _\nu T^{0\nu}=0 \text{ Energy conservation}$$
$$\partial _\nu T^{i\nu}=0 \text{ Momentum conservation}$$
In relativity, the energy-momentum tensor for a free particle is:
$$T^{\mu\nu}\equiv mU^\mu U^\nu$$
For a classical particle, returning to non-relativistic terms, the trajectory \(q\) only depends on \(t\):
$$T^{\mu\nu}=T^{00}={\partial L\over \partial \dot{q} } \dot{q} -L=H$$
\(T^{00}\) is the Hamiltonian, representing energy conservation:
$$\partial _\nu T^{0\nu}=\partial _0 T^{00}={dH \over dt}=0$$
The energy does not change with time!
Classical Noether theorem can be stated as:
$$Classical Non-relativistic$$ | $$Relativistic$$ |
$$S=\int Ldt $$ | $$S=\int \mathcal{L}d^4 x$$ |
$$N\equiv {\partial L\over\partial \dot{q} _i } Q_i-\left({\partial L\over\partial \dot{q} _i } \dot{q} _i-L\right)T$$ | $$N^\mu \equiv {\partial \mathcal{L}\over\partial (\partial _\mu \alpha^\nu ) } \Psi^\nu-\left({\partial \mathcal{L}\over\partial (\partial _\mu \alpha^\nu ) } \partial _\alpha \alpha^\nu- \delta ^\mu _\alpha L\right) ฮง^\alpha $$ |
Originally written in Chinese by the author, these articles are translated into English to invite cross-language resonance.