Noether's theorem

The essence of Noether's theorem is generalizing the notion of ignorable coordinates.

One can assume that the Lagrangian L defined above is invariant under small perturbations (warpings) of the time variable t and the generalized coordinates q. One may write

${\displaystyle {\begin{aligned}t&\rightarrow t^{\prime }=t+\delta t\\\mathbf {q} &\rightarrow \mathbf {q} ^{\prime }=\mathbf {q} +\delta \mathbf {q} ~,\end{aligned}}}$

where the perturbations δt and δq are both small, but variable. For generality, assume there are (say) N such symmetry transformations of the action, i.e. transformations leaving the action unchanged; labelled by an index r = 1, 2, 3, ..., N.

Then the resultant perturbation can be written as a linear sum of the individual types of perturbations,

${\displaystyle {\begin{aligned}\delta t&=\sum _{r}\varepsilon _{r}T_{r}\\\delta \mathbf {q} &=\sum _{r}\varepsilon _{r}\mathbf {Q} _{r}~,\end{aligned}}}$

where ε_r are infinitesimal parameter coefficients corresponding to each:

• generator T_r of time evolution, and
• generator Q_r of the generalized coordinates.

For translations, Q_r is a constant with units of length; for rotations, it is an expression linear in the components of q, and the parameters make up an angle.

Using these definitions, Noether showed that the N quantities

${\displaystyle \left({\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}\cdot {\dot {\mathbf {q} }}-L\right)T_{r}-{\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}\cdot \mathbf {Q} _{r}}$

are conserved (constants of motion).

I. Time invariance

For illustration, consider a Lagrangian that does not depend on time, i.e., that is invariant (symmetric) under changes t → t + δt, without any change in the coordinates q. In this case, N = 1, T = 1 and Q = 0; the corresponding conserved quantity is the total energy H[10]

${\displaystyle H={\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}\cdot {\dot {\mathbf {q} }}-L.}$

II. Translational invariance

Consider a Lagrangian which does not depend on an ("ignorable", as above) coordinate q_k; so it is invariant (symmetric) under changes q_k → q_k + δq_k. In that case, N = 1, T = 0, and Q_k = 1; the conserved quantity is the corresponding linear momentum p_k[11]

${\displaystyle p_{k}={\frac {\partial L}{\partial {\dot {q_{k}}}}}.}$

In special and general relativity, these apparently separate conservation laws are aspects of a single conservation law, that of the stress–energy tensor,[12] that is derived in the next section.

III. Rotational invariance

The conservation of the angular momentum L = r × p is analogous to its linear momentum counterpart.[13] It is assumed that the symmetry of the Lagrangian is rotational, i.e., that the Lagrangian does not depend on the absolute orientation of the physical system in space. For concreteness, assume that the Lagrangian does not change under small rotations of an angle δθ about an axis n; such a rotation transforms the Cartesian coordinates by the equation

${\displaystyle \mathbf {r} \rightarrow \mathbf {r} +\delta \theta \,\mathbf {n} \times \mathbf {r} .}$

Since time is not being transformed, T = 0, and N = 1. Taking δθ as the ε parameter and the Cartesian coordinates r as the generalized coordinates q, the corresponding Q variables are given by

${\displaystyle \mathbf {Q} =\mathbf {n} \times \mathbf {r} .}$

Then Noether's theorem states that the following quantity is conserved,

${\displaystyle {\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}\cdot \mathbf {Q} =\mathbf {p} \cdot \left(\mathbf {n} \times \mathbf {r} \right)=\mathbf {n} \cdot \left(\mathbf {r} \times \mathbf {p} \right)=\mathbf {n} \cdot \mathbf {L} .}$

In other words, the component of the angular momentum L along the n axis is conserved. And if n is arbitrary, i.e., if the system is insensitive to any rotation, then every component of L is conserved; in short, angular momentum is conserved.

Although useful in its own right, the version of Noether's theorem just given is a special case of the general version derived in 1915. To give the flavor of the general theorem, a version of Noether's theorem for continuous fields in four-dimensional space–time is now given. Since field theory problems are more common in modern physics than mechanics problems, this field theory version is the most commonly used (or most often implemented) version of Noether's theorem.

Let there be a set of differentiable fields ${\displaystyle \varphi }$ defined over all space and time; for example, the temperature ${\displaystyle T(\mathbf {x} ,t)}$ would be representative of such a field, being a number defined at every place and time. The principle of least action can be applied to such fields, but the action is now an integral over space and time

${\displaystyle {\mathcal {S}}=\int {\mathcal {L}}\left(\varphi ,\partial _{\mu }\varphi ,x^{\mu }\right)\,d^{4}x}$

(the theorem can be further generalized to the case where the Lagrangian depends on up to the n^th derivative, and can also be formulated using jet bundles).

A continuous transformation of the fields ${\displaystyle \varphi }$ can be written infinitesimally as

${\displaystyle \varphi \mapsto \varphi +\varepsilon \Psi ,}$

where ${\displaystyle \Psi }$ is in general a function that may depend on both ${\displaystyle x^{\mu }}$ and ${\displaystyle \varphi }$. The condition for ${\displaystyle \Psi }$ to generate a physical symmetry is that the action ${\displaystyle {\mathcal {S}}}$ is left invariant. This will certainly be true if the Lagrangian density ${\displaystyle {\mathcal {L}}}$ is left invariant, but it will also be true if the Lagrangian changes by a divergence,

${\displaystyle {\mathcal {L}}\mapsto {\mathcal {L}}+\varepsilon \partial _{\mu }\Lambda ^{\mu },}$

since the integral of a divergence becomes a boundary term according to the divergence theorem. A system described by a given action might have multiple independent symmetries of this type, indexed by ${\displaystyle r=1,2,\ldots ,N,}$ so the most general symmetry transformation would be written as

${\displaystyle \varphi \mapsto \varphi +\varepsilon _{r}\Psi _{r},}$

with the consequence

${\displaystyle {\mathcal {L}}\mapsto {\mathcal {L}}+\varepsilon _{r}\partial _{\mu }\Lambda _{r}^{\mu }.}$

For such systems, Noether's theorem states that there are ${\displaystyle N}$ conserved current densities

${\displaystyle j_{r}^{\nu }=\Lambda _{r}^{\nu }-{\frac {\partial {\mathcal {L}}}{\partial \varphi _{,\nu }}}\cdot \Psi _{r}}$

(where the dot product is understood to contract the field indices, not the ${\displaystyle \nu }$ index or ${\displaystyle r}$ index).

In such cases, the conservation law is expressed in a four-dimensional way

${\displaystyle \partial _{\nu }j^{\nu }=0,}$

which expresses the idea that the amount of a conserved quantity within a sphere cannot change unless some of it flows out of the sphere. For example, electric charge is conserved; the amount of charge within a sphere cannot change unless some of the charge leaves the sphere.

For illustration, consider a physical system of fields that behaves the same under translations in time and space, as considered above; in other words, ${\displaystyle L\left({\boldsymbol {\varphi }},\partial _{\mu }{\boldsymbol {\varphi }},x^{\mu }\right)}$ is constant in its third argument. In that case, N = 4, one for each dimension of space and time. An infinitesimal translation in space, ${\displaystyle x^{\mu }\mapsto x^{\mu }+\varepsilon _{r}\delta _{r}^{\mu }}$ (with ${\displaystyle \delta }$ denoting the Kronecker delta), affects the fields as ${\displaystyle \varphi (x^{\mu })\mapsto \varphi \left(x^{\mu }-\varepsilon _{r}\delta _{r}^{\mu }\right)}$: that is, relabelling the coordinates is equivalent to leaving the coordinates in place while translating the field itself, which in turn is equivalent to transforming the field by replacing its value at each point ${\displaystyle x^{\mu }}$ with the value at the point ${\displaystyle x^{\mu }-\varepsilon X^{\mu }}$ "behind" it which would be mapped onto ${\displaystyle x^{\mu }}$ by the infinitesimal displacement under consideration. Since this is infinitesimal, we may write this transformation as

${\displaystyle \Psi _{r}=-\delta _{r}^{\mu }\partial _{\mu }\varphi .}$

The Lagrangian density transforms in the same way, ${\displaystyle {\mathcal {L}}\left(x^{\mu }\right)\mapsto {\mathcal {L}}\left(x^{\mu }-\varepsilon _{r}\delta _{r}^{\mu }\right)}$, so

${\displaystyle \Lambda _{r}^{\mu }=-\delta _{r}^{\mu }{\mathcal {L}}}$

and thus Noether's theorem corresponds to the conservation law for the stress–energy tensor T_μ^ν,[12] where we have used ${\displaystyle \mu }$ in place of ${\displaystyle r}$. To wit, by using the expression given earlier, and collecting the four conserved currents (one for each ${\displaystyle \mu }$) into a tensor ${\displaystyle T}$, Noether's theorem gives

${\displaystyle T_{\mu }{}^{\nu }=-\delta _{\mu }^{\nu }{\mathcal {L}}+\delta _{\mu }^{\sigma }\partial _{\sigma }\varphi {\frac {\partial {\mathcal {L}}}{\partial \varphi _{,\nu }}}=\left({\frac {\partial {\mathcal {L}}}{\partial \varphi _{,\nu }}}\right)\cdot \varphi _{,\mu }-\delta _{\mu }^{\nu }{\mathcal {L}}}$

with

${\displaystyle T_{\mu }{}^{\nu }{}_{,\nu }=0}$

(we relabelled ${\displaystyle \mu }$ as ${\displaystyle \sigma }$ at an intermediate step to avoid conflict). (However, the ${\displaystyle T}$ obtained in this way may differ from the symmetric tensor used as the source term in general relativity; see Canonical stress–energy tensor.)

The conservation of electric charge, by contrast, can be derived by considering Ψ linear in the fields φ rather than in the derivatives.[14] In quantum mechanics, the probability amplitude ψ(x) of finding a particle at a point x is a complex field φ, because it ascribes a complex number to every point in space and time. The probability amplitude itself is physically unmeasurable; only the probability p = |ψ|² can be inferred from a set of measurements. Therefore, the system is invariant under transformations of the ψ field and its complex conjugate field ψ^* that leave |ψ|² unchanged, such as

${\displaystyle \psi \rightarrow e^{i\theta }\psi \ ,\ \psi ^{*}\rightarrow e^{-i\theta }\psi ^{*}~,}$

a complex rotation. In the limit when the phase θ becomes infinitesimally small, δθ, it may be taken as the parameter ε, while the Ψ are equal to iψ and −iψ*, respectively. A specific example is the Klein–Gordon equation, the relativistically correct version of the Schrödinger equation for spinless particles, which has the Lagrangian density

${\displaystyle L=\partial _{\nu }\psi \partial _{\mu }\psi ^{*}\eta ^{\nu \mu }+m^{2}\psi \psi ^{*}.}$

In this case, Noether's theorem states that the conserved (∂ ⋅ j = 0) current equals

${\displaystyle j^{\nu }=i\left({\frac {\partial \psi }{\partial x^{\mu }}}\psi ^{*}-{\frac {\partial \psi ^{*}}{\partial x^{\mu }}}\psi \right)\eta ^{\nu \mu }~,}$

which, when multiplied by the charge on that species of particle, equals the electric current density due to that type of particle. This "gauge invariance" was first noted by Hermann Weyl, and is one of the prototype gauge symmetries of physics.

Consider the simplest case, a system with one independent variable, time. Suppose the dependent variables q are such that the action integral

$${\displaystyle I=\int _{t_{1}}^{t_{2}}L[\mathbf {q} [t],{\dot {\mathbf {q} }}[t],t]\,dt}$$

is invariant under brief infinitesimal variations in the dependent variables. In other words, they satisfy the Euler–Lagrange equations

${\displaystyle {\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}[t]={\frac {\partial L}{\partial \mathbf {q} }}[t].}$

And suppose that the integral is invariant under a continuous symmetry. Mathematically such a symmetry is represented as a flow, φ, which acts on the variables as follows

${\displaystyle {\begin{aligned}t&\rightarrow t'=t+\varepsilon T\\\mathbf {q} [t]&\rightarrow \mathbf {q} '[t']=\varphi [\mathbf {q} [t],\varepsilon ]=\varphi [\mathbf {q} [t'-\varepsilon T],\varepsilon ]\end{aligned}}}$

where ε is a real variable indicating the amount of flow, and T is a real constant (which could be zero) indicating how much the flow shifts time.

${\displaystyle {\dot {\mathbf {q} }}[t]\rightarrow {\dot {\mathbf {q} }}'[t']={\frac {d}{dt}}\varphi [\mathbf {q} [t],\varepsilon ]={\frac {\partial \varphi }{\partial \mathbf {q} }}[\mathbf {q} [t'-\varepsilon T],\varepsilon ]{\dot {\mathbf {q} }}[t'-\varepsilon T].}$

The action integral flows to

${\displaystyle {\begin{aligned}I'[\varepsilon ]&=\int _{t_{1}+\varepsilon T}^{t_{2}+\varepsilon T}L[\mathbf {q} '[t'],{\dot {\mathbf {q} }}'[t'],t']\,dt'\\[6pt]&=\int _{t_{1}+\varepsilon T}^{t_{2}+\varepsilon T}L[\varphi [\mathbf {q} [t'-\varepsilon T],\varepsilon ],{\frac {\partial \varphi }{\partial \mathbf {q} }}[\mathbf {q} [t'-\varepsilon T],\varepsilon ]{\dot {\mathbf {q} }}[t'-\varepsilon T],t']\,dt'\end{aligned}}}$

which may be regarded as a function of ε. Calculating the derivative at ε' = 0 and using Leibniz's rule, we get

${\displaystyle {\begin{aligned}0={\frac {dI'}{d\varepsilon }}[0]={}&L[\mathbf {q} [t_{2}],{\dot {\mathbf {q} }}[t_{2}],t_{2}]T-L[\mathbf {q} [t_{1}],{\dot {\mathbf {q} }}[t_{1}],t_{1}]T\\[6pt]&{}+\int _{t_{1}}^{t_{2}}{\frac {\partial L}{\partial \mathbf {q} }}\left(-{\frac {\partial \varphi }{\partial \mathbf {q} }}{\dot {\mathbf {q} }}T+{\frac {\partial \varphi }{\partial \varepsilon }}\right)+{\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}\left(-{\frac {\partial ^{2}\varphi }{(\partial \mathbf {q} )^{2}}}{\dot {\mathbf {q} }}^{2}T+{\frac {\partial ^{2}\varphi }{\partial \varepsilon \partial \mathbf {q} }}{\dot {\mathbf {q} }}-{\frac {\partial \varphi }{\partial \mathbf {q} }}{\ddot {\mathbf {q} }}T\right)\,dt.\end{aligned}}}$

Notice that the Euler–Lagrange equations imply

${\displaystyle {\begin{aligned}{\frac {d}{dt}}\left({\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}{\frac {\partial \varphi }{\partial \mathbf {q} }}{\dot {\mathbf {q} }}T\right)&=\left({\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}\right){\frac {\partial \varphi }{\partial \mathbf {q} }}{\dot {\mathbf {q} }}T+{\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}\left({\frac {d}{dt}}{\frac {\partial \varphi }{\partial \mathbf {q} }}\right){\dot {\mathbf {q} }}T+{\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}{\frac {\partial \varphi }{\partial \mathbf {q} }}{\ddot {\mathbf {q} }}\,T\\[6pt]&={\frac {\partial L}{\partial \mathbf {q} }}{\frac {\partial \varphi }{\partial \mathbf {q} }}{\dot {\mathbf {q} }}T+{\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}\left({\frac {\partial ^{2}\varphi }{(\partial \mathbf {q} )^{2}}}{\dot {\mathbf {q} }}\right){\dot {\mathbf {q} }}T+{\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}{\frac {\partial \varphi }{\partial \mathbf {q} }}{\ddot {\mathbf {q} }}\,T.\end{aligned}}}$

Substituting this into the previous equation, one gets

${\displaystyle {\begin{aligned}0={\frac {dI'}{d\varepsilon }}[0]={}&L[\mathbf {q} [t_{2}],{\dot {\mathbf {q} }}[t_{2}],t_{2}]T-L[\mathbf {q} [t_{1}],{\dot {\mathbf {q} }}[t_{1}],t_{1}]T-{\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}{\frac {\partial \varphi }{\partial \mathbf {q} }}{\dot {\mathbf {q} }}[t_{2}]T+{\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}{\frac {\partial \varphi }{\partial \mathbf {q} }}{\dot {\mathbf {q} }}[t_{1}]T\\[6pt]&{}+\int _{t_{1}}^{t_{2}}{\frac {\partial L}{\partial \mathbf {q} }}{\frac {\partial \varphi }{\partial \varepsilon }}+{\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}{\frac {\partial ^{2}\varphi }{\partial \varepsilon \partial \mathbf {q} }}{\dot {\mathbf {q} }}\,dt.\end{aligned}}}$

Again using the Euler–Lagrange equations we get

${\displaystyle {\frac {d}{dt}}\left({\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}{\frac {\partial \varphi }{\partial \varepsilon }}\right)=\left({\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}\right){\frac {\partial \varphi }{\partial \varepsilon }}+{\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}{\frac {\partial ^{2}\varphi }{\partial \varepsilon \partial \mathbf {q} }}{\dot {\mathbf {q} }}={\frac {\partial L}{\partial \mathbf {q} }}{\frac {\partial \varphi }{\partial \varepsilon }}+{\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}{\frac {\partial ^{2}\varphi }{\partial \varepsilon \partial \mathbf {q} }}{\dot {\mathbf {q} }}.}$

Substituting this into the previous equation, one gets

${\displaystyle {\begin{aligned}0={}&L[\mathbf {q} [t_{2}],{\dot {\mathbf {q} }}[t_{2}],t_{2}]T-L[\mathbf {q} [t_{1}],{\dot {\mathbf {q} }}[t_{1}],t_{1}]T-{\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}{\frac {\partial \varphi }{\partial \mathbf {q} }}{\dot {\mathbf {q} }}[t_{2}]T+{\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}{\frac {\partial \varphi }{\partial \mathbf {q} }}{\dot {\mathbf {q} }}[t_{1}]T\\[6pt]&{}+{\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}{\frac {\partial \varphi }{\partial \varepsilon }}[t_{2}]-{\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}{\frac {\partial \varphi }{\partial \varepsilon }}[t_{1}].\end{aligned}}}$

From which one can see that

${\displaystyle \left({\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}{\frac {\partial \varphi }{\partial \mathbf {q} }}{\dot {\mathbf {q} }}-L\right)T-{\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}{\frac {\partial \varphi }{\partial \varepsilon }}}$

is a constant of the motion, i.e., it is a conserved quantity. Since φ[q, 0] = q, we get ${\displaystyle {\frac {\partial \varphi }{\partial \mathbf {q} }}=1}$ and so the conserved quantity simplifies to

${\displaystyle \left({\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}{\dot {\mathbf {q} }}-L\right)T-{\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}{\frac {\partial \varphi }{\partial \varepsilon }}.}$

To avoid excessive complication of the formulas, this derivation assumed that the flow does not change as time passes. The same result can be obtained in the more general case.

Noether's theorem may also be derived for tensor fields φ^A where the index A ranges over the various components of the various tensor fields. These field quantities are functions defined over a four-dimensional space whose points are labeled by coordinates x^μ where the index μ ranges over time (μ = 0) and three spatial dimensions (μ = 1, 2, 3). These four coordinates are the independent variables; and the values of the fields at each event are the dependent variables. Under an infinitesimal transformation, the variation in the coordinates is written

${\displaystyle x^{\mu }\rightarrow \xi ^{\mu }=x^{\mu }+\delta x^{\mu }}$

whereas the transformation of the field variables is expressed as

${\displaystyle \varphi ^{A}\rightarrow \alpha ^{A}\left(\xi ^{\mu }\right)=\varphi ^{A}\left(x^{\mu }\right)+\delta \varphi ^{A}\left(x^{\mu }\right)\,.}$

By this definition, the field variations δφ^A result from two factors: intrinsic changes in the field themselves and changes in coordinates, since the transformed field α^A depends on the transformed coordinates ξ^μ. To isolate the intrinsic changes, the field variation at a single point x^μ may be defined

${\displaystyle \alpha ^{A}\left(x^{\mu }\right)=\varphi ^{A}\left(x^{\mu }\right)+{\bar {\delta }}\varphi ^{A}\left(x^{\mu }\right)\,.}$

If the coordinates are changed, the boundary of the region of space–time over which the Lagrangian is being integrated also changes; the original boundary and its transformed version are denoted as Ω and Ω’, respectively.

Noether's theorem begins with the assumption that a specific transformation of the coordinates and field variables does not change the action, which is defined as the integral of the Lagrangian density over the given region of spacetime. Expressed mathematically, this assumption may be written as

${\displaystyle \int _{\Omega ^{\prime }}L\left(\alpha ^{A},{\alpha ^{A}}_{,\nu },\xi ^{\mu }\right)d^{4}\xi -\int _{\Omega }L\left(\varphi ^{A},{\varphi ^{A}}_{,\nu },x^{\mu }\right)d^{4}x=0}$

where the comma subscript indicates a partial derivative with respect to the coordinate(s) that follows the comma, e.g.

${\displaystyle {\varphi ^{A}}_{,\sigma }={\frac {\partial \varphi ^{A}}{\partial x^{\sigma }}}\,.}$

Since ξ is a dummy variable of integration, and since the change in the boundary Ω is infinitesimal by assumption, the two integrals may be combined using the four-dimensional version of the divergence theorem into the following form

${\displaystyle \int _{\Omega }\left\{\left[L\left(\alpha ^{A},{\alpha ^{A}}_{,\nu },x^{\mu }\right)-L\left(\varphi ^{A},{\varphi ^{A}}_{,\nu },x^{\mu }\right)\right]+{\frac {\partial }{\partial x^{\sigma }}}\left[L\left(\varphi ^{A},{\varphi ^{A}}_{,\nu },x^{\mu }\right)\delta x^{\sigma }\right]\right\}d^{4}x=0\,.}$

The difference in Lagrangians can be written to first-order in the infinitesimal variations as

${\displaystyle \left[L\left(\alpha ^{A},{\alpha ^{A}}_{,\nu },x^{\mu }\right)-L\left(\varphi ^{A},{\varphi ^{A}}_{,\nu },x^{\mu }\right)\right]={\frac {\partial L}{\partial \varphi ^{A}}}{\bar {\delta }}\varphi ^{A}+{\frac {\partial L}{\partial {\varphi ^{A}}_{,\sigma }}}{\bar {\delta }}{\varphi ^{A}}_{,\sigma }\,.}$

However, because the variations are defined at the same point as described above, the variation and the derivative can be done in reverse order; they commute

${\displaystyle {\bar {\delta }}{\varphi ^{A}}_{,\sigma }={\bar {\delta }}{\frac {\partial \varphi ^{A}}{\partial x^{\sigma }}}={\frac {\partial }{\partial x^{\sigma }}}\left({\bar {\delta }}\varphi ^{A}\right)\,.}$

Using the Euler–Lagrange field equations

${\displaystyle {\frac {\partial }{\partial x^{\sigma }}}\left({\frac {\partial L}{\partial {\varphi ^{A}}_{,\sigma }}}\right)={\frac {\partial L}{\partial \varphi ^{A}}}}$

the difference in Lagrangians can be written neatly as

${\displaystyle {\begin{aligned}&\left[L\left(\alpha ^{A},{\alpha ^{A}}_{,\nu },x^{\mu }\right)-L\left(\varphi ^{A},{\varphi ^{A}}_{,\nu },x^{\mu }\right)\right]\\[4pt]={}&{\frac {\partial }{\partial x^{\sigma }}}\left({\frac {\partial L}{\partial {\varphi ^{A}}_{,\sigma }}}\right){\bar {\delta }}\varphi ^{A}+{\frac {\partial L}{\partial {\varphi ^{A}}_{,\sigma }}}{\bar {\delta }}{\varphi ^{A}}_{,\sigma }={\frac {\partial }{\partial x^{\sigma }}}\left({\frac {\partial L}{\partial {\varphi ^{A}}_{,\sigma }}}{\bar {\delta }}\varphi ^{A}\right).\end{aligned}}}$

Thus, the change in the action can be written as

${\displaystyle \int _{\Omega }{\frac {\partial }{\partial x^{\sigma }}}\left\{{\frac {\partial L}{\partial {\varphi ^{A}}_{,\sigma }}}{\bar {\delta }}\varphi ^{A}+L\left(\varphi ^{A},{\varphi ^{A}}_{,\nu },x^{\mu }\right)\delta x^{\sigma }\right\}d^{4}x=0\,.}$

Since this holds for any region Ω, the integrand must be zero

${\displaystyle {\frac {\partial }{\partial x^{\sigma }}}\left\{{\frac {\partial L}{\partial {\varphi ^{A}}_{,\sigma }}}{\bar {\delta }}\varphi ^{A}+L\left(\varphi ^{A},{\varphi ^{A}}_{,\nu },x^{\mu }\right)\delta x^{\sigma }\right\}=0\,.}$

For any combination of the various symmetry transformations, the perturbation can be written

${\displaystyle {\begin{aligned}\delta x^{\mu }&=\varepsilon X^{\mu }\\\delta \varphi ^{A}&=\varepsilon \Psi ^{A}={\bar {\delta }}\varphi ^{A}+\varepsilon {\mathcal {L}}_{X}\varphi ^{A}\end{aligned}}}$

where ${\displaystyle {\mathcal {L}}_{X}\varphi ^{A}}$ is the Lie derivative of φ^A in the X^μ direction. When φ^A is a scalar or ${\displaystyle {X^{\mu }}_{,\nu }=0}$,

${\displaystyle {\mathcal {L}}_{X}\varphi ^{A}={\frac {\partial \varphi ^{A}}{\partial x^{\mu }}}X^{\mu }\,.}$

These equations imply that the field variation taken at one point equals

${\displaystyle {\bar {\delta }}\varphi ^{A}=\varepsilon \Psi ^{A}-\varepsilon {\mathcal {L}}_{X}\varphi ^{A}\,.}$

Differentiating the above divergence with respect to ε at ε = 0 and changing the sign yields the conservation law

${\displaystyle {\frac {\partial }{\partial x^{\sigma }}}j^{\sigma }=0}$

where the conserved current equals

${\displaystyle j^{\sigma }=\left[{\frac {\partial L}{\partial {\varphi ^{A}}_{,\sigma }}}{\mathcal {L}}_{X}\varphi ^{A}-L\,X^{\sigma }\right]-\left({\frac {\partial L}{\partial {\varphi ^{A}}_{,\sigma }}}\right)\Psi ^{A}\,.}$

Suppose we have an n-dimensional oriented Riemannian manifold, M and a target manifold T. Let ${\displaystyle {\mathcal {C}}}$ be the configuration space of smooth functions from M to T. (More generally, we can have smooth sections of a fiber bundle over M.)

Examples of this M in physics include:

• In classical mechanics, in the Hamiltonian formulation, M is the one-dimensional manifold ${\displaystyle \mathbb {R} }$, representing time and the target space is the cotangent bundle of space of generalized positions.
• In field theory, M is the spacetime manifold and the target space is the set of values the fields can take at any given point. For example, if there are m real-valued scalar fields, ${\displaystyle \varphi _{1},\ldots ,\varphi _{m}}$, then the target manifold is ${\displaystyle \mathbb {R} ^{m}}$. If the field is a real vector field, then the target manifold is isomorphic to ${\displaystyle \mathbb {R} ^{3}}$.

Now suppose there is a functional

${\displaystyle {\mathcal {S}}:{\mathcal {C}}\rightarrow \mathbb {R} ,}$

called the action. (It takes values into ${\displaystyle \mathbb {R} }$, rather than ${\displaystyle \mathbb {C} }$; this is for physical reasons, and is unimportant for this proof.)

To get to the usual version of Noether's theorem, we need additional restrictions on the action. We assume ${\displaystyle {\mathcal {S}}[\varphi ]}$ is the integral over M of a function

${\displaystyle {\mathcal {L}}(\varphi ,\partial _{\mu }\varphi ,x)}$

called the Lagrangian density, depending on φ, its derivative and the position. In other words, for φ in ${\displaystyle {\mathcal {C}}}$

${\displaystyle {\mathcal {S}}[\varphi ]\,=\,\int _{M}{\mathcal {L}}[\varphi (x),\partial _{\mu }\varphi (x),x]\,d^{n}x.}$

Suppose we are given boundary conditions, i.e., a specification of the value of φ at the boundary if M is compact, or some limit on φ as x approaches ∞. Then the subspace of ${\displaystyle {\mathcal {C}}}$ consisting of functions φ such that all functional derivatives of ${\displaystyle {\mathcal {S}}}$ at φ are zero, that is:

${\displaystyle {\frac {\delta {\mathcal {S}}[\varphi ]}{\delta \varphi (x)}}\approx 0}$

and that φ satisfies the given boundary conditions, is the subspace of on shell solutions. (See principle of stationary action)

Now, suppose we have an infinitesimal transformation on ${\displaystyle {\mathcal {C}}}$, generated by a functional derivation, Q such that

${\displaystyle Q\left[\int _{N}{\mathcal {L}}\,\mathrm {d} ^{n}x\right]\approx \int _{\partial N}f^{\mu }[\varphi (x),\partial \varphi ,\partial \partial \varphi ,\ldots ]\,ds_{\mu }}$

for all compact submanifolds N or in other words,

${\displaystyle Q[{\mathcal {L}}(x)]\approx \partial _{\mu }f^{\mu }(x)}$

for all x, where we set

${\displaystyle {\mathcal {L}}(x)={\mathcal {L}}[\varphi (x),\partial _{\mu }\varphi (x),x].}$

If this holds on shell and off shell, we say Q generates an off-shell symmetry. If this only holds on shell, we say Q generates an on-shell symmetry. Then, we say Q is a generator of a one parameter symmetry Lie group.

Now, for any N, because of the Euler–Lagrange theorem, on shell (and only on-shell), we have

${\displaystyle {\begin{aligned}Q\left[\int _{N}{\mathcal {L}}\,\mathrm {d} ^{n}x\right]&=\int _{N}\left[{\frac {\partial {\mathcal {L}}}{\partial \varphi }}-\partial _{\mu }{\frac {\partial {\mathcal {L}}}{\partial (\partial _{\mu }\varphi )}}\right]Q[\varphi ]\,\mathrm {d} ^{n}x+\int _{\partial N}{\frac {\partial {\mathcal {L}}}{\partial (\partial _{\mu }\varphi )}}Q[\varphi ]\,\mathrm {d} s_{\mu }\\&\approx \int _{\partial N}f^{\mu }\,\mathrm {d} s_{\mu }.\end{aligned}}}$

Since this is true for any N, we have

${\displaystyle \partial _{\mu }\left[{\frac {\partial {\mathcal {L}}}{\partial (\partial _{\mu }\varphi )}}Q[\varphi ]-f^{\mu }\right]\approx 0.}$

But this is the continuity equation for the current ${\displaystyle J^{\mu }}$ defined by:[15]

${\displaystyle J^{\mu }\,=\,{\frac {\partial {\mathcal {L}}}{\partial (\partial _{\mu }\varphi )}}Q[\varphi ]-f^{\mu },}$

which is called the Noether current associated with the symmetry. The continuity equation tells us that if we integrate this current over a space-like slice, we get a conserved quantity called the Noether charge (provided, of course, if M is noncompact, the currents fall off sufficiently fast at infinity).

Noether's theorem is an on shell theorem: it relies on use of the equations of motion—the classical path. It reflects the relation between the boundary conditions and the variational principle. Assuming no boundary terms in the action, Noether's theorem implies that

${\displaystyle \int _{\partial N}J^{\mu }ds_{\mu }\approx 0.}$

The quantum analogs of Noether's theorem involving expectation values (e.g., ${\textstyle \left\langle \int d^{4}x~\partial \cdot {\textbf {J}}\right\rangle =0}$) probing off shell quantities as well are the Ward–Takahashi identities.

Suppose we have two symmetry derivations Q₁ and Q₂. Then, [Q₁, Q₂] is also a symmetry derivation. Let us see this explicitly. Let us say

$${\displaystyle Q_{1}[{\mathcal {L}}]\approx \partial _{\mu }f_{1}^{\mu }}$$

and

$${\displaystyle Q_{2}[{\mathcal {L}}]\approx \partial _{\mu }f_{2}^{\mu }}$$

Then,

$${\displaystyle [Q_{1},Q_{2}][{\mathcal {L}}]=Q_{1}[Q_{2}[{\mathcal {L}}]]-Q_{2}[Q_{1}[{\mathcal {L}}]]\approx \partial _{\mu }f_{12}^{\mu }}$$

where f₁₂ = Q₁[f₂^μ] − Q₂[f₁^μ]. So,

$${\displaystyle j_{12}^{\mu }=\left({\frac {\partial }{\partial (\partial _{\mu }\varphi )}}{\mathcal {L}}\right)(Q_{1}[Q_{2}[\varphi ]]-Q_{2}[Q_{1}[\varphi ]])-f_{12}^{\mu }.}$$

This shows we can extend Noether's theorem to larger Lie algebras in a natural way.

This applies to any local symmetry derivation Q satisfying QS ≈ 0, and also to more general local functional differentiable actions, including ones where the Lagrangian depends on higher derivatives of the fields. Let ε be any arbitrary smooth function of the spacetime (or time) manifold such that the closure of its support is disjoint from the boundary. ε is a test function. Then, because of the variational principle (which does not apply to the boundary, by the way), the derivation distribution q generated by q[ε][Φ(x)] = ε(x)Q[Φ(x)] satisfies q[ε][S] ≈ 0 for every ε, or more compactly, q(x)[S] ≈ 0 for all x not on the boundary (but remember that q(x) is a shorthand for a derivation distribution, not a derivation parametrized by x in general). This is the generalization of Noether's theorem.

To see how the generalization is related to the version given above, assume that the action is the spacetime integral of a Lagrangian that only depends on φ and its first derivatives. Also, assume

${\displaystyle Q[{\mathcal {L}}]\approx \partial _{\mu }f^{\mu }}$

Then,

${\displaystyle {\begin{aligned}q[\varepsilon ][{\mathcal {S}}]&=\int q[\varepsilon ][{\mathcal {L}}]d^{n}x\\[6pt]&=\int \left\{\left({\frac {\partial }{\partial \varphi }}{\mathcal {L}}\right)\varepsilon Q[\varphi ]+\left[{\frac {\partial }{\partial (\partial _{\mu }\varphi )}}{\mathcal {L}}\right]\partial _{\mu }(\varepsilon Q[\varphi ])\right\}d^{n}x\\[6pt]&=\int \left\{\varepsilon Q[{\mathcal {L}}]+\partial _{\mu }\varepsilon \left[{\frac {\partial }{\partial \left(\partial _{\mu }\varphi \right)}}{\mathcal {L}}\right]Q[\varphi ]\right\}\,d^{n}x\\[6pt]&\approx \int \varepsilon \partial _{\mu }\left\{f^{\mu }-\left[{\frac {\partial }{\partial (\partial _{\mu }\varphi )}}{\mathcal {L}}\right]Q[\varphi ]\right\}\,d^{n}x\end{aligned}}}$

for all ${\displaystyle \varepsilon }$.

More generally, if the Lagrangian depends on higher derivatives, then

${\displaystyle \partial _{\mu }\left[f^{\mu }-\left[{\frac {\partial }{\partial (\partial _{\mu }\varphi )}}{\mathcal {L}}\right]Q[\varphi ]-2\left[{\frac {\partial }{\partial (\partial _{\mu }\partial _{\nu }\varphi )}}{\mathcal {L}}\right]\partial _{\nu }Q[\varphi ]+\partial _{\nu }\left[\left[{\frac {\partial }{\partial (\partial _{\mu }\partial _{\nu }\varphi )}}{\mathcal {L}}\right]Q[\varphi ]\right]-\,\dotsm \right]\approx 0.}$

Looking at the specific case of a Newtonian particle of mass m, coordinate x, moving under the influence of a potential V, coordinatized by time t. The action, S, is:

${\displaystyle {\begin{aligned}{\mathcal {S}}[x]&=\int L\left[x(t),{\dot {x}}(t)\right]\,dt\\&=\int \left({\frac {m}{2}}\sum _{i=1}^{3}{\dot {x}}_{i}^{2}-V(x(t))\right)\,dt.\end{aligned}}}$

The first term in the brackets is the kinetic energy of the particle, while the second is its potential energy. Consider the generator of time translations Q = d/dt. In other words, ${\displaystyle Q[x(t)]={\dot {x}}(t)}$. The coordinate x has an explicit dependence on time, whilst V does not; consequently:

${\displaystyle Q[L]={\frac {d}{dt}}\left[{\frac {m}{2}}\sum _{i}{\dot {x}}_{i}^{2}-V(x)\right]=m\sum _{i}{\dot {x}}_{i}{\ddot {x}}_{i}-\sum _{i}{\frac {\partial V(x)}{\partial x_{i}}}{\dot {x}}_{i}}$

so we can set

${\displaystyle L={\frac {m}{2}}\sum _{i}{\dot {x}}_{i}^{2}-V(x).}$

Then,

${\displaystyle {\begin{aligned}j&=\sum _{i=1}^{3}{\frac {\partial L}{\partial {\dot {x}}_{i}}}Q[x_{i}]-L\\&=m\sum _{i}{\dot {x}}_{i}^{2}-\left[{\frac {m}{2}}\sum _{i}{\dot {x}}_{i}^{2}-V(x)\right]\\[3pt]&={\frac {m}{2}}\sum _{i}{\dot {x}}_{i}^{2}+V(x).\end{aligned}}}$

The right hand side is the energy, and Noether's theorem states that ${\displaystyle dj/dt=0}$ (i.e. the principle of conservation of energy is a consequence of invariance under time translations).

More generally, if the Lagrangian does not depend explicitly on time, the quantity

${\displaystyle \sum _{i=1}^{3}{\frac {\partial L}{\partial {\dot {x}}_{i}}}{\dot {x_{i}}}-L}$

(called the Hamiltonian) is conserved.

Still considering 1-dimensional time, let

${\displaystyle {\begin{aligned}{\mathcal {S}}\left[{\vec {x}}\right]&=\int {\mathcal {L}}\left[{\vec {x}}(t),{\dot {\vec {x}}}(t)\right]dt\\[3pt]&=\int \left[\sum _{\alpha =1}^{N}{\frac {m_{\alpha }}{2}}\left({\dot {\vec {x}}}_{\alpha }\right)^{2}-\sum _{\alpha <\beta }V_{\alpha \beta }\left({\vec {x}}_{\beta }-{\vec {x}}_{\alpha }\right)\right]dt,\end{aligned}}}$

or ${\displaystyle N}$ Newtonian particles where the potential only depends pairwise upon the relative displacement.

For ${\displaystyle {\vec {Q}}}$, consider the generator of Galilean transformations (i.e. a change in the frame of reference). In other words,

${\displaystyle Q_{i}\left[x_{\alpha }^{j}(t)\right]=t\delta _{i}^{j}.}$