Proof:Noeter theorem

This article presents the expanded proof of the Noeter theorem. Below, the indices that numerate the components of the field and those that numerate the coordinates are recovered.

Assumptions

Assume that the $M$-component field $u$ depends on the coordinates $x$ in the $N$-dimendionals space and, for some Lagrangian $L=L(x,u(x),u_\bullet(x))$, realizes the extremum of the action

$ (1) \displaystyle ~ ~ ~ ~ ~ ~ A=\int _{x\in \Omega} L~ \mathrm {d}^N x$

Assume, there exist some smooth transform $x\mapsto y$, $u(x)\mapsto v(y)$, dependent on the real parameter $\varepsilon$ such that at $\varepsilon\!=\!0$, the transform is identical, id est, $y\!=\!x$ and $v\!=\!u$; and in the linear approximation with respect to $\varepsilon$,

$ (2) \displaystyle ~ ~ ~ ~ ~ ~ y_j ~= ~x_j + \mathcal X_j(x) ~ \varepsilon ~=~ x_j + \chi_j(x)$

$ (3) \displaystyle ~ ~ ~ ~ ~ v_\alpha(y)=u_\alpha(x) + \Psi_\alpha (x) \varepsilon=u_\alpha(x) + \psi_j(x)$

Assume the variation takes place inside some domain $x\in \Omega$; assume $O$ is transform on $\Omega$.

Statement

Then, the variation of the action

$ (4) \displaystyle ~ ~ ~ ~ ~ ~ \delta A=

\int _{y \in O} L(y,v(y),v_\bullet(y)) ~\mathrm{d}^N y - \int _{x \in \Omega} L(x,u(x),u_\bullet(y)) ~\mathrm{d}^N x$

can be presented in the form of the integral of divergence:

$ (5) \displaystyle ~ ~ ~ ~ ~ ~ \delta A = \int

\left(

L \mathcal X_k+
\big( \Psi_\alpha -u_{\alpha,j} \mathcal X_{j} \big)

\frac

{\partial L}
{\partial u_{\alpha, k} }
\right)
_{\!\! ,k}
~\varepsilon~ \mathrm{d}^N x
+ \mathcal O (\varepsilon^2)

$

Proof

Lagrange-Euler

The Theorem by Lagrange-Euler states that the assumption of the stationary action leads to the Lagrange–Euler equation

$ (11) \displaystyle ~ ~ ~ ~ ~ ~

\frac {\partial L}{\partial u_\alpha} = \left( \frac {\partial L}{\partial u_{\alpha,j}} \right)_{\!\! , j}$

This equation is used below in the deduction,

Variation of the field derivative

From (2),(3) one may deduce the variation of the derivative of the field:

$ (21) \displaystyle ~ ~ ~ ~ ~ ~

v_{\alpha,j}(y)=\frac{ \partial v_\alpha(y)}{\partial y_j}= \frac {\partial (u_\alpha(y\!-\!\chi)+\psi_\alpha(x) )} {\partial y_j} + \mathcal{O}(\varepsilon^2)$

The term with derivative of $\psi$ is already of order of $\varepsilon$; so, here is no difference between $x$ and $y$ there. As for the first term, in the formal differentiation gives:

$ (22) \displaystyle ~ ~ ~ ~ ~ ~

v_{\alpha,j}(y)= u_{\alpha,j}(y\!-\!\chi)-u_{\alpha,k}(y\!-\!\chi) \chi_{k,j}+\psi_{\alpha,j} + \mathcal{O}(\varepsilon^2)$

While $y=x+\chi$, the derivatives of the transformed field can be espressed as functions of the old coordinates:

$ (23) \displaystyle ~ ~ ~ ~ ~ ~

v_{\alpha,j}(y)= u_{\alpha,j}(x)-u_{\alpha,k}(x) \chi_{k,j}+\psi_{\alpha,j} + \mathcal{O}(\varepsilon^2)$

With such a representation, in the first integral in (4), the variable of integration $y$ can be changed to $x$, allowing the combination of the two integrals and the simplification.

Change of variable of integration

In order to simplify (4), change the variable of integration from $y$ to $x$. In the first of order with respect to parameter $\varepsilon$ of the transformation, the only diagonal part of the determinant should be taken in to account;

$(31) \displaystyle ~ ~ ~ ~ ~ ~

\frac{\mathcal D y}{\mathcal D x}=\frac{\mathcal D (x+\chi(x))}{\mathcal D x}=1+\chi_{j,j}+ \mathcal O(\varepsilon^2)$

Then, the variation of action can be written as follows:

$ (32) \displaystyle ~ ~ ~ ~ ~ ~ \delta A=

\int _{x \in \Omega} L\big(x\!+\!\chi,\,u(x)\!+\!\psi,\,u_\bullet(x)-u_{\bullet}(x) \chi_{\bullet}+\psi_{\bullet}\big) (1\!+\!\chi_{j,j})~\mathrm{d}^N x +\mathcal O(\varepsilon^2)$

$ \phantom{(32)} \displaystyle ~ ~ ~ ~ ~ ~ \phantom{\delta A=}

- \int _{x \in \Omega} L\big(\,x,\,u(x),\,u_\bullet(x)\big) ~\mathrm{d}^N x+\mathcal O(\varepsilon^2)$

Represetnation of the integrand as divergence of current

Combining integrals in (32) and expanding the Lagrangian at $(x,u(x),u_\bullet(x))$, we get:

$ (32) \displaystyle ~ ~ ~ ~ ~ ~ \delta A=

\int _{x \in \Omega} \Big( L\big(x,u(x),u_\bullet(x)) \chi_{j,j} +L_{x_j}\chi_j $

$\displaystyle ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

+L_{u_\alpha}\psi_\alpha+L_{u_{\alpha,j}}\psi_{\alpha,j} - L_{u_{\alpha,j}}u_{\alpha,k} \chi_{k,j} \Big) \mathrm{d}^N x +\mathcal O(\varepsilon^2)$

In order to get the complefe divergence, add to and sustract from the integrand terms $L_{u_\alpha} u_{\alpha,j}\chi_j$ and $L_{u_{\alpha,k}} u_{\alpha,k,j}\chi_{j}$; then, the integrand $\mathcal A$ in (32) can be written as follows:

$(33) \displaystyle ~ ~ ~ ~ ~ ~ \mathcal{A}= L \chi_{j,j} +L_{x_j}\chi

+ L_{u_\alpha} u_{\alpha,j}\chi_j + L_{u_{\alpha,k}} u_{\alpha,k,j}\chi_{k,j}$

$\displaystyle ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

- L_{u_\alpha} u_{\alpha,j}\chi_j - L_{u_{\alpha,k}} u_{\alpha,k,j}\chi_{j} +L_{u_\alpha}\psi_\alpha+L_{u_{\alpha,j}}\psi_{\alpha,j} - L_{u_{\alpha,j}}u_{\alpha,k} \chi_{k,j} +\mathcal O(\varepsilon^2)$

The first line in (33) forms the divergence of $L \chi$; in the second line, the terms with $L_{u_\alpha}$ should be collected:

$(34) \displaystyle ~ ~ ~ ~ ~ ~ \mathcal{A}= L \chi_{j,j} +L_{,j}\chi_j +

$

$\displaystyle ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

+L_{u_\alpha} ( \psi_\alpha - u_{\alpha,j}\chi_j ) +L_{u_{\alpha,j}}(\psi_{\alpha,j}- u_{\alpha,k} \chi_{k,j} - u_{\alpha,k,j}\chi_{k}) +\mathcal O(\varepsilon^2) $

Using the Lagrange-Euler equation, $L_{u_\alpha} $ can be replaced to $\big(L_{u_{\alpha,k}} \big)_{,k}$. Then each or lines in (34) gives the complete divergence;

$(35) \displaystyle ~ ~ ~ ~ ~ ~ \mathcal{A}= L \chi_{j,j} +L_{,j}\chi_j +

$

$\displaystyle ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

+\big(L_{u_{\alpha,j}}\big)_{\!,j} ( \psi_\alpha - u_{\alpha,k}\chi_k ) +L_{u_{\alpha,j}}(\psi_{\alpha,j}- u_{\alpha,k} \chi_{k,j} - u_{\alpha,k,j}\chi_{k}) +\mathcal O(\varepsilon^2) $ Combination of last terms into the single derivative gives

$(36) \displaystyle ~ ~ ~ ~ ~ ~ \mathcal{A}= \Big( L \chi_{j}

+L_{u_{\alpha,j}} ( \psi_\alpha - u_{\alpha,k}\chi_k ) \Big)_{\!,j} +\mathcal O(\varepsilon^2) $

In such a way,

$ (37) \displaystyle ~ ~ ~ ~ ~ ~ \delta A= \int _{x \in \Omega}\! \mathcal{A}~ \mathrm{d}^N x =

\int _{x \in \Omega} \Big( L \chi_{j} +L_{u_{\alpha,j}} ( \psi_\alpha \! -\! u_{\alpha,k}\chi_k ) \Big)_{\!,j} ~ \mathrm{d}^N x +\mathcal O(\varepsilon^2) $

Due to the arbitrary domain $\Omega$ of integration, the invariance of the Action with respect to transform (2),(3) leads to the conservation of the Noeter current

$ (38) \displaystyle ~ ~ ~ ~ ~ ~ \Theta_j= L \mathcal X_{j} +L_{u_{\alpha,j}} ( \Psi_\alpha \! -\! u_{\alpha,k}\mathcal X_k )$

its divergence $\Theta_{j,j}=0$ .

(end of proof)

Conclusion

Some transforms of coordinates and fields keep the action invariant. These transforms correspond to the symmetry of the model. Each of continuous symmetries of the Action leads to the conservation of some quantity; this quantity is determined by the generators $\mathcal X$ and $\Psi$ of the transform through equation (37). In particular, in the "1+3" dimensional space, the thranslational, rotational and Galilean invariance lead to the conservation of 10 quantities: Energy-momentum (4 quantities), Angular momentum (3 quantities) and the Erenfestian momentum (3 quantities). The last three determine that the center of mass of any isolated system moves along the straight line with constant speed (Theorem of Erenfest); this is equivalent of the First Law of Newton; this symmetry prohibits the realization of inertioids and, in particular, the truters without discharge of mass (движители без расхода рабочего тела) that are "developed" in the Khrunichev State Research and Production Space Center in Russia.

The Noeter theorem applies not only to the Field theory but also to any system that can be described within the Lagrangian formalism. While such a system has any symmetries, id est, while the action remains at some transforms of the coordinates and/or, certain physical quantities conserve. In particular, this can be applied to the Ginzburg-Landau equation and to the Navier-Stokes equation.