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Given a Lie group $G$, whose most general transform depends on $\rho$ parameters, under the action of which an integral $I$ is invariant, there are $\rho$ linearly independent combinations of the Lagrange expressions which become divergences.

I know what they mean by a Lagrange expression: namely, $\frac{\partial L}{\partial \phi}-\sum \frac{d}{dx_i}\frac{\partial L}{\partial\frac{d\phi}{dx_i}}.$

  1. But what is the significance of the "linearly independent combinations of Lagrange expressions"?

  2. Also, I don't get how that expression is equivalent to $\frac{d\rho^0}{dt}+\nabla \cdot \vec{\rho}$ given that $\rho^{\mu}=[\rho^0,\vec{\rho}]$. Namely, what is $\rho$?

  3. Also, if my system follows Hamilton's principle, then how can that be a divergence? Namely, it should become zero, not infinity.

As you might have noticed, this is beyond my level. But I have to do a presentation on it, so please bear with me.

I've gotten my information from here: http://inside.mines.edu/~tohno/teaching/PH505_2011/Ryan_FinalPaperNoetherThm.pdf

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First of all, my opinion is that the paper on your link is full of notational inconsistencies and therefore causes a great amount of confusion for someone who struggles to understand Noether’s theorem. So, allow me to formulate the field-theoretic version of Noether’s theorem in a more, according to my taste, charming way.

Preliminaries 1: Lie Groups and Lie algebras

Each element of an $N$-dimensional lie Group $G$ can be parameterized by a point of a subspace of ${{\mathbf{R}}^{N}}$. In other words, an element $g\in G$ can be considered as a map: $g:{{\mathbf{R}}^{N}}\backepsilon \omega \mapsto {{g}_{\omega }}\in G$. Without loss of generality, we can assume that ${{g}_{0}}=e$ , the identity element of $G$. Now suppose that we have an $n$-dimensional real vector field:

$\varphi :\Omega \backepsilon x\mapsto \varphi \left( x \right)\in {{\mathbf{R}}^{n}}$

where $\Omega \subseteq {{\mathbf{R}}^{\left( 1,3 \right)}}$, with $n$ real components ${{\varphi }_{i}}\ ,\ i=1,..,n$, such that it obeys the boundary condition ${{\left. \varphi \right|}_{\partial \Omega }}=0$. Also, assume that a group element ${{g}_{\omega }}$ is implemented as an $n$-dimensional representation of $G$. We can now consider the linear transformation:

${{g}_{\omega }}:\varphi \to {{g}_{\omega }}\varphi \quad ,\quad {{\left( {{g}_{\omega }}\varphi \right)}_{i}}:=\sum\limits_{j=1}^{n}{{{\left( {{g}_{\omega }} \right)}_{ij}}{{\varphi }_{j}}}$

If $\varepsilon \in {{\mathbf{R}}^{N}}$ is a point infinitesimally close to $0$ then ${{g}_{\varepsilon }}$ will be a group element that is “infinitesimally close” to the identity $e$, in the sense that:

${{g}_{\varepsilon }}\varphi ={{g}_{0}}\varphi +\sum\limits_{I=1}^{N}{{{\varepsilon }_{I}}{{\left[ \tfrac{\partial }{\partial {{\omega }_{I}}}{{g}_{\omega }}\varphi \right]}_{\omega =0}}}+\mathcal{O}\left( {{\varepsilon }^{2}} \right)=\varphi +\sum\limits_{I=1}^{N}{{{\varepsilon }_{I}}{{T}_{I}}\varphi }+\mathcal{O}\left( {{\varepsilon }^{2}} \right)$

where ${{T}_{I}}\varphi :={{\left[ \tfrac{\partial }{\partial {{\omega }_{I}}}{{g}_{\omega }}\varphi \right]}_{\omega =0}}$. The vectors ${{T}_{I}}$ form a basis of the $n$-dimensional representation of the Lie Algebra $\mathrm{}$ of $G$ and they transform the field as:

${{T}_{I}}:\varphi \to {{T}_{I}}\varphi \quad ,\quad {{\left( {{T}_{I}}\varphi \right)}_{i}}=\sum\limits_{j=1}^{n}{{{\left( {{T}_{I}} \right)}_{ij}}{{\varphi }_{j}}}$

Preliminaries 2: Lagrangian formulation and Hamilton’s Principle

The dynamics of the field $\varphi \left( x \right)$ can be encoded in the “Lagrangian density” function $\mathsf{L}\left( \varphi ,{{\partial }_{\mu }}\varphi \right)$, which is a function such that the “Euler-Lagrange equations”:

$\frac{\partial \mathsf{L}}{\partial {{\varphi }_{i}}}-{{\partial }_{\mu }}\left( \frac{\partial \mathsf{L}}{\partial {{\varphi }_{i,\mu }}} \right)=0\quad ,\quad i=1,...,n$

are equivalent to the equations of motion of the field’s components. Having in our disposal $\mathsf{L}\left( \varphi ,{{\partial }_{\mu }}\varphi \right)$ we can restate the equations of motion as variational principle, the “Principle of Least Action” (“Hamilton’s principle”) as follows:

i) Define the “Action functional” as:

$S\left[ \varphi \right]:=\int\limits_{\Omega }{\mathsf{L}\left( \varphi ,{{\partial }_{\mu }}\varphi \right){{d}^{4}}x}$

ii) The functional derivative of $S\left[ \varphi \right]$ w.r.t. the field component ${{\varphi }_{i}}$ is a Lagrange expression, i.e.:

$\frac{\delta S}{\delta {{\varphi }_{i}}}=\frac{\partial \mathsf{L}}{\partial {{\varphi }_{i}}}-{{\partial }_{\mu }}\left( \frac{\partial \mathsf{L}}{\partial {{\varphi }_{i,\mu }}} \right)$

where for the derivation of this result the boundary condition ${{\left. \varphi \right|}_{\partial \Omega }}=0$ was been taken into account.

iii) By the above we deduce that if the field is a stationary point of $S\left[ \varphi \right]$, i.e. if:

$\frac{\delta S}{\delta {{\varphi }_{i}}}=0\quad ,\quad i=1,...,n$

then the Euler-Lagrange equations are satisfied and vice-versa. Concluding, the field’s equations of motion can be derived by a variational principle, the “Principle of Least Action”. This is “Hamilton’s principle”.

Definition of symmetry Group

The Lie group $G$ is a “symmetry group” for the theory of the field $\varphi $ if the action functional $S\left[ \varphi \right]$ is invariant (more generally, if it differences by a boundary term ${{S}_{\partial \Omega }}$) under the action of any ${{g}_{\omega }}\in G$, i.e. if: $S\left[ {{g}_{\omega }}\varphi \right]=S\left[ \varphi \right]$ For sake of simplicity, we will restrict to the case where the Lagrangian density is invariant:

$\mathsf{L}\left( {{g}_{\omega }}\varphi ,{{\partial }_{\mu }}\left( {{g}_{\omega }}\varphi \right) \right)=\mathsf{L}\left( \varphi ,{{\partial }_{\mu }}\varphi \right)$

which is a sufficient condition for the invariance of the action. Let’s investigate the case where ${{g}_{\omega }}$ is an infinitesimal transformation. Then the field components would vary as:

${{\delta }_{\varepsilon }}\varphi :={{g}_{\varepsilon }}\varphi -\varphi =\sum\limits_{I=1}^{N}{{{\varepsilon }_{I}}{{T}_{I}}\varphi }$

and their spatial derivatives as:

${{\delta }_{\varepsilon }}{{\varphi }_{,\mu }}:={{\partial }_{\mu }}\left( {{g}_{\varepsilon }}\varphi \right)-{{\partial }_{\mu }}\varphi ={{\partial }_{\mu }}\left( \varphi +\sum\limits_{I=1}^{N}{{{\varepsilon }_{I}}{{T}_{I}}\varphi } \right)-{{\partial }_{\mu }}\varphi ={{\partial }_{\mu }}\left( {{\delta }_{\varepsilon }}\varphi \right)$

Then the Lagrangian would vary as:

$\begin{align} & {{\delta }_{\varepsilon }}\mathsf{L}:=\mathsf{L}\left( {{g}_{\varepsilon }}\varphi ,{{\partial }_{\mu }}\left( {{g}_{\varepsilon }}\varphi \right) \right)-\mathsf{L}\left( \varphi ,{{\partial }_{\mu }}\varphi \right)=\sum\limits_{i=1}^{n}{\left[ \frac{\partial \mathsf{L}}{\partial {{\varphi }_{i}}}{{\delta }_{\varepsilon }}{{\varphi }_{i}}+\frac{\partial \mathsf{L}}{\partial {{\varphi }_{i,\mu }}}{{\delta }_{\varepsilon }}{{\varphi }_{i,\mu }} \right]} \\ & \quad \quad =\sum\limits_{i=1}^{n}{\left[ \frac{\partial \mathsf{L}}{\partial {{\varphi }_{i}}}\sum\limits_{I=1}^{N}{{{\varepsilon }_{I}}{{\left( {{T}_{I}}\varphi \right)}_{i}}}+\frac{\partial \mathsf{L}}{\partial {{\varphi }_{i,\mu }}}\sum\limits_{I=1}^{N}{{{\varepsilon }_{I}}{{\partial }_{\mu }}{{\left( {{T}_{I}}\varphi \right)}_{i}}} \right]} \\ & \quad \quad =\sum\limits_{I=1}^{N}{{{\varepsilon }_{I}}\sum\limits_{i=1}^{n}{\left[ \frac{\partial \mathsf{L}}{\partial {{\varphi }_{i}}}{{\left( {{T}_{I}}\varphi \right)}_{i}}+\frac{\partial \mathsf{L}}{\partial {{\varphi }_{i,\mu }}}{{\partial }_{\mu }}{{\left( {{T}_{I}}\varphi \right)}_{i}} \right]}} \\ & \quad \quad =\sum\limits_{I=1}^{N}{{{\varepsilon }_{I}}\sum\limits_{i=1}^{n}{\left[ \frac{\partial \mathsf{L}}{\partial {{\varphi }_{i}}}{{\left( {{T}_{I}}\varphi \right)}_{i}}+{{\partial }_{\mu }}\left[ \frac{\partial \mathsf{L}}{\partial {{\varphi }_{i,\mu }}}{{\left( {{T}_{I}}\varphi \right)}_{i}} \right]-{{\partial }_{\mu }}\frac{\partial \mathsf{L}}{\partial {{\varphi }_{i,\mu }}}{{\left( {{T}_{I}}\varphi \right)}_{i}} \right]}} \\ & \quad \quad =\sum\limits_{I=1}^{N}{{{\varepsilon }_{I}}\left( \sum\limits_{i=1}^{n}{\frac{\delta S}{\delta {{\varphi }_{i}}}{{\left( {{T}_{I}}\varphi \right)}_{i}}+{{\partial }_{\mu }}J_{I}^{\mu }} \right)} \\ \end{align}$

where:

$J_{I}^{\mu }:=\sum\limits_{i=1}^{n}{\frac{\partial \mathsf{L}}{\partial {{\varphi }_{i,\mu }}}{{\left( {{T}_{I}}\varphi \right)}_{i}}}=\sum\limits_{i,j=1}^{n}{\frac{\partial \mathsf{L}}{\partial {{\varphi }_{i,\mu }}}{{\left( {{T}_{I}} \right)}_{ij}}{{\varphi }_{j}}}$

So, the invariance of the Lagrangian density is expressed as:

${{\delta }_{\varepsilon }}\mathsf{L}=0\Rightarrow \sum\limits_{I=1}^{N}{{{\varepsilon }_{I}}\left( \sum\limits_{i=1}^{n}{\frac{\delta S}{\delta {{\varphi }_{i}}}{{\left( {{T}_{I}}\varphi \right)}_{i}}+{{\partial }_{\mu }}J_{I}^{\mu }} \right)=0}$

and since ${{\varepsilon }_{I}}$’s are considered independent parameters, then the above expression holds iff:

$\sum\limits_{i=1}^{n}{\frac{\delta S}{\delta {{\varphi }_{i}}}{{\left( {{T}_{I}}\varphi \right)}_{i}}}+{{\partial }_{\mu }}J_{I}^{\mu }=0\quad ,\quad I=1,..,N$

This is Noether’s first theorem! Now observe that if the fields are such that they obey the equations of motion (are “on-shell” in the physics jargon), i.e. if $\tfrac{\delta S}{\delta {{\varphi }_{i}}}=0$ then:

${{\partial }_{\mu }}J_{I}^{\mu }=0\quad ,\quad I=1,..,N$

i.e. that the currents $J_{I}^{\mu }$ are locally conserved. This is the essence of Noether’s theorem: it states that for any symmetry of a physical theory there is a corresponding conserved current; in my opinion it is one of the most beautiful theorems in physics. The above conservation law, which is in covariant form, can be re-expressed as:

${{\partial }_{\mu }}J_{I}^{\mu }=0\Rightarrow {{\partial }_{t}}J_{I}^{0}=-\nabla \cdot {{\mathbf{J}}_{I}}\Rightarrow {{\partial }_{t}}\int\limits_{{{\Omega }_{t}}}{J_{I}^{0}{{d}^{3}}x}=-\int\limits_{{{\Omega }_{t}}}{\left( \nabla \cdot {{\mathbf{J}}_{I}} \right){{d}^{3}}x}$

where ${{\Omega }_{t}}$ is an iso-temporal sub-surface of $\Omega $. Using Gauss’ divergence theorem we get:

${{\partial }_{t}}{{Q}_{I}}=-\int\limits_{\Sigma \equiv \partial {{\Omega }_{t}}}{{{\mathbf{J}}_{I}}\cdot d\,\mathbf{\Sigma }}$

where:

${{Q}_{I}}:=\int\limits_{{{\Omega }_{t}}}{J_{I}^{0}{{d}^{3}}x}$

Assuming a “Dirichlet boundary condition”:

${{\left. {{\mathbf{J}}_{I}} \right|}_{\partial {{\Omega }_{t}}}}=0$

or a “Neuman boundary condition”:

${{\mathbf{J}}_{I}}\left( x \right)\cdot \mathbf{n}\left( x \right)=0\ ,\ \forall x\in \partial {{\Omega }_{t}}\quad |\quad d\mathbf{\Sigma }=\mathbf{n}d\Sigma $

then:

$\int\limits_{\Sigma \equiv \partial {{\Omega }_{t}}}{{{\mathbf{J}}_{I}}\cdot d\mathbf{\Sigma }}=0$

so the ${{J}_{I}}$’s current conservation implies that the “charge” ${{Q}_{I}}$ is constant in time, i.e. that it is conserved.

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  • $\begingroup$ Surprisingly I understood almost everything in this post. I guess I'll look pretty smart today when giving my presentation :P $\endgroup$ – DLV Sep 25 '15 at 17:53
  • $\begingroup$ By the way are each $\phi_i$ different fields, or are they a component of one field? Also what do you mean by $R^{(1,3)}$ being contained in $\Omega$ ? $\endgroup$ – DLV Sep 25 '15 at 17:54
  • $\begingroup$ I am glad I could held. φi's are the different components of the field φ (e.g. the electric field E has three independend components E1, E2 and E3). And Ω is a subpsace of the Minkowski space, where the field is confined; it could as well be the entire Minkowski space $\endgroup$ – user3257624 Sep 25 '15 at 17:58
  • $\begingroup$ Hmm. Why use different subscripts $\mu$ and $i$ then? I mean, shouldnt $\partial_\mu \phi_i = 0$ if $\mu\neq i$? I guess I'll study this post a bit more then. $\endgroup$ – DLV Sep 25 '15 at 18:04
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    $\begingroup$ No! The index μ=0,1,2,3 refers to the space-time variable of the field while i=1,...,n to the field component. In our example of the electric field, the xμ would refer to the spacetime point where each field component Ei(x0,x1,x2,x3) is evaluated (here i=1,2,3) $\endgroup$ – user3257624 Sep 25 '15 at 18:30

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