# About Noether's theorem - question about how can I do such that I can study this theorem

My teacher asked me to try to write the main important steps which I have to learn such that I can understand the Noether's theorem.

Some few steps are:

• Lagrangian Formalism
• Hamiltonian Formalism
• Euler - Lagrange equation
• Poisson Brackets

Can you tell me another important steps which I have to learn? Or can you expand my steps?

Thanks

## 1 Answer

I'd say: Lagrangian $\rightarrow$ Hamitonian $\rightarrow$ Canonical transformations $\rightarrow$ Poisson brackets $\rightarrow$ Noether's theorem

Noether's theorem is an observation that for an infinitesimal canonical transformation $(q_1,\dots,q_n,p_1,\dots,p_n)\mapsto (Q_1,\dots,Q_n,P_1,\dots,P_n)$, and $|\epsilon|<<1$ , we have the generating function $$F_2(q_1,\dots,q_n,P_1,\dots,P_n)=\sum_i q_i P_i + \epsilon f_2(q_1,\dots,q_n,P_1,\dots,P_n,t),$$ which implies the following relationships between the old and new varables (and old and new Hamiltonians $H$ and $K$): $$Q_j = \frac{\partial F_2}{\partial P_j}= q_j + \epsilon \frac{\partial f_2}{\partial P_j} \\ p_j = \frac{\partial F_2}{\partial q_j}= P_j + \epsilon \frac{\partial f_2}{\partial q_j}\\ K = H + \epsilon \frac{\partial f_2}{\partial t}$$ We can write $f_2$ as a function of $p_j$ since $f_2(\dots,P_j,\dots)-f_2(\dots,p_j,\dots)=O(\epsilon^2)$, and define

$$\delta q_j := Q_j - q_j =\epsilon \frac{\partial f_2}{\partial p_j} \\ \delta p_j := P_j - q_j =- \epsilon \frac{\partial f_2}{\partial q_j}$$

And then for any function of the coordinates and momenta (and time), we have

$$\delta u = u(q_1+\delta q_1,\dots,p_1+\delta p_1,\dots, t + \delta t)- u(q_1,\dots,p_q,\dots,t)= \sum_i [\frac{\partial u}{\partial q_i}\delta q_i+\frac{\partial u}{\partial p_i}\delta p_i] + \frac{\partial u}{ \partial t}\delta t$$

and substituting the expressions from before, $$\delta u = \epsilon \sum_i [\frac{\partial u}{\partial q_i}\frac{\partial f_2}{\partial p_i}-\frac{\partial u}{\partial p_i}\frac{\partial f_2}{\partial q_i}] + \frac{\partial u}{ \partial t}\delta t = \epsilon\{u,f_2\}+ \frac{\partial u}{ \partial t}\delta t$$

This leads to Noether's theorem: $$\delta H = \epsilon\{H,f_2\}+ \frac{\partial H}{ \partial t}\delta t$$

which means that for a time independent Hamiltonian, every symmetry of the Hamiltonian ($\delta H = 0$) implies that there is a corresponding integral of motion $f_2$ (ie function for which $\{H,f_2\}=0$). Conversely, every integral of motion ($\{H,f_2\}=0$) generates a corresponding symmetry ($\delta H = 0$).