# In Noether's theorem, what is a “classical solution of the equations of motion”?

I'm reading a book which states that:

for each generator of a global symmetry transformation, there is a current $j^{\mu}_{a}$ which, when evaluated on a classical solution of the equations of motion $\phi_{cl}$, is conserved. I.e. $\partial_{\mu}j^{\mu}_{a}\vert_{\phi=\phi_{cl}} =0$

I get the general principle, but I'm uncertain about the bit I have made italics. Can anyone shed some light on this?

-

2
It means $\phi_{cl}$ satisfies Euler-Lagrange equation. – Jia Yiyang Feb 8 at 14:23

## 2 Answers

1) The word classical in this context means $\hbar=0$.

2) In the context of an action principle, the Euler-Lagrange equations

$$\frac{\delta S}{\delta\phi^{\alpha}}~\approx~0$$

are often referred to as the (classical) equations of motion (eom), cf. comment by Jia Yiyang. Here the $\approx$ symbol means equality modulo eom.

Let on-shell (off-shell) refer to whether eom are satisfied (not necessarily satisfied), respectively.

3) In the context of a global continuous (off-shell) symmetry of an action, Noether's (first) theorem implies an off-shell Noether identity

$$d_{\mu} J^{\mu} ~\equiv~ - \frac{\delta S}{\delta\phi^{\alpha}} Y_0^{\alpha},$$

where $J^{\mu}$ is the full Noether current, and $Y_0^{\alpha}$ is a (vertical) symmetry generator.

This leads to an on-shell conservation law

$$d_{\mu} J^{\mu}~\approx~0.$$

-

The classical equation of motion can be find solving the Euler-Lagrange equation for the Lagrangian of the system $L=L(q,\dot{q})$, then EL equation states $$\frac{\partial L}{\partial q_i}-\frac{\text{d}}{\text{d}t}\frac{\partial L}{\partial\dot{q}_i},\text{ }i=1,\dots,n.$$ The conserved current $j_a^\mu$ come from the fact that if a system has a symmetry then something linked with the symmetry is kept by the system. If a system show a time symmetry then the Energy must be conserved, if a system show a traslational symmetry then linear momentum must be conserved and if a system has a rotational symmetry then angolar momentum must be conserved.

-

– joshphysics Feb 8 at 16:20
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8829483389854431, "perplexity_flag": "head"}