# Noethers theorem

Noether's theorem is a central result in theoretical physics that expresses the one-to-one correspondence between the symmetries and the conservation laws. This exact equivalence holds for all physical laws based upon the action principle defined over a symplectic space. It is named after the early 20th century mathematician Emmy Noether.

## Explanation

The word "symmetry" in the previous paragraph really means the covariance of the form that a physical law takes with respect to a one-dimensional Lie group of transformations which satisfies certain technical criteria. The conservation law of a physical quantity is usually expressed as a continuity equation.

The most important examples of the theorem are the following:

• the energy is conserved if and only if the physical laws are invariant under time translations (if their form does not depend on time)
• the momentum is conserved iff the physical laws are invariant under spatial translations (if the laws do not depend on the position)
• the angular momentum is conserved iff the physical laws are invariant under rotations (if the laws do not care about the orientation); if only some rotations are allowed, only the corresponding components of the angular momentum vector are conserved

A Noether charge is a physical quantity conserved as an effect of a continuous symmetry of the underlying system. One theoretical use of the Noether charge is in calculating the entropy of stationary black holesTemplate:Fn.

## Mathematical statement of the theorem

Informally, Noether's theorem can be stated as (technical fine points aside):

To every differentiable symmetry generated by local actions, there corresponds a conserved current.

The vice versa part is actually harder to prove and the proof of it is omitted in this article but the main idea is very simple: consider the conservating value as a new Hamiltonian; the evolution generated by this Hamiltonian will be the symmetry transformation.

## Applications

The formal statement of the theorem derives an expression for the physical quantity that is conserved -- and hence also defines it (actually, its current) -- from the condition of invariance alone. Actually, this conserved current is not uniquely defined. In the formulation given in the proof below, for example, fμ is only defined up to a divergenceless vector field. But if you think about it, any two conserved currents differ by a divergenceless vector field - for example:

When it comes to quantum field theory, the invariance with respect to general gauge transformations also gives the law of conservation of quantities such as electric charge, though there are some subtleties here; the conservation law here is based on the Ward-Takahashi identities for the BRST symmetry. Thus, the result is a very important contribution to physics in general, as it helps to provide powerful insights into any general theory in physics, by just analyzing the various transformations that would make the form of the laws involved invariant.

## Proof

Suppose we have an n-dimensional 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)

Before we go on, let's give some examples:

• In classical mechanics, in the Hamiltonian formulation, M is the one-dimensional manifold 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, φ1,...,φm, then the target manifold is R. If the field is a real vector field, then the target manifold is isomorphic to R. There's actually a much more elegant way using tangent bundles over M, but for the purposes of this proof, we'd just stick to this version.

Now suppose there is a functional

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

called the action. (Note that it takes values into $\displaystyle \mathbb{R}$ , rather than $\displaystyle \mathbb{C}$ ; this is for physical reasons, and doesn't really matter for this proof.)

To get to the usual version of Noether's theorem, we need additional restrictions on the action. We assume S[φ] is the integral over M of a function

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

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

$\displaystyle S[\varphi]\equiv\int_M d^nx \mathcal{L}[\varphi(x),\partial_\mu\varphi(x),x].$

Suppose given boundary conditions, which are basically a specification of the value of φ at the boundary if M is compact, or some limit on φ as x approaches ∞ (this will help in doing integration by parts). The subspace of $\displaystyle \mathcal{C}$ consisting of functions, φ such that all functional derivatives of S at φ are zero, that is:

$\displaystyle \frac{\delta}{\delta \phi(x)}S[\phi]=0$

and 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

$\displaystyle Q\left[\int_N d^nx\mathcal{L}\right]=\int_{\partial N}ds_\mu f^\mu[\phi(x),\partial\phi,\partial\partial\phi,...]$

for all compact submanifolds N or in other words,

$\displaystyle Q[\mathcal{L}(x)]=\partial_\mu f^\mu(x)$

for all x, where in the usual manner of physicists (i.e. mathematicians won't do this) we set :$\displaystyle \mathcal{L}(x)=\mathcal{L}[\phi(x), \partial_\mu \phi(x),x]$ .

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 1-parameter symmetry Lie group.

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

 $\displaystyle Q\left[\int_Nd^nx\mathcal{L}\right]$ $\displaystyle =\int_Nd^nx\left[\frac{\partial\mathcal{L}}{\partial\phi}- \partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right]Q[\phi]+ \int_{\partial N}ds_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}Q[\phi]$ $\displaystyle =\int_{\partial N}ds_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}Q[\phi].$

Since this is true for any N, we have

$\displaystyle \partial_\mu\left[\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}Q[\phi]-f^\mu\right]=0.$

You might immediately recognize this as the continuity equation for the current

$\displaystyle J^\mu\equiv\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}Q[\phi]-f^\mu$

which is called the Noether current associated with the symmetry. The continuity equation tells us 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).

This is not generally well-known, but Noether's theorem is really a reflection of the relation between the boundary conditions and the variational principle. Assuming no boundary terms in the action, Noether's theorem implies that

$\displaystyle \int_{\partial N}ds_\mu J^\mu=0.$

Noether's theorem is an on shell theorem. The quantum analog of Noether's theorem are the Ward-Takahashi identities.

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

$\displaystyle Q_1[\mathcal{L}]=\partial_\mu f_1^\mu$

and

$\displaystyle Q_2[\mathcal{L}]=\partial_\mu f_2^\mu$

(it doesn't matter if this holds off shell or only on shell). Then,

$\displaystyle [Q_1,Q_2][\mathcal{L}]=Q_1[Q_2[\mathcal{L}]]-Q_2[Q_1[\mathcal{L}]]=\partial_\mu f_{12}^\mu$

where f12=Q1[f2μ]-Q2[f1μ]. So,

$\displaystyle j_{12}^\mu=\left(\frac{\partial}{\partial (\partial_\mu\phi)}\mathcal{L}\right)(Q_1[Q_2[\phi]]-Q_2[Q_1[\phi]])-f_{12}^\mu.$

This means we can (trivially) extend Noether's theorem to larger Lie algebras. Suppose L is a Lie algebra and there is a realization of it as symmetry derivations. Then, Noether's theorem would apply as well.

### A more general and elegant proof

This applies to any derivation Q, not just symmetry derivations and also to more general functional differentiable actions, including ones where the Lagrangian depends on higher derivatives of the fields and nonlocal actions. 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 any ε on shell, or more compactly, q(x)[S] 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.

How is this related to the version given above? Simple. Assume the action is the spacetime integral of a Lagrangian which only depends on φ and its first derivatives. Also, assume

$\displaystyle Q[\mathcal{L}]=\partial_\mu f^\mu$

(either off-shell or only on-shell is fine). Then,

$\displaystyle q[\epsilon][S]=\int d^dx q[\epsilon][\mathcal{L}]$
$\displaystyle =\int d^dx \left(\frac{\partial}{\partial \phi}\mathcal{L}\right) \epsilon Q[\phi]+ \left[\frac{\partial}{\partial (\partial_\mu \phi)}\mathcal{L}\right]\partial_\mu(\epsilon Q[\phi])$
$\displaystyle =\int d^dx \epsilon \partial_\mu \Bigg\{f^\mu-\left[\frac{\partial}{\partial (\partial_\mu\phi)}\mathcal{L}\right]Q[\phi]\Bigg\}$

for all ε.

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

$\displaystyle \partial_\mu\left[f^\mu-\left[\frac{\partial}{\partial (\partial_\mu\phi)}\mathcal{L}\right]Q[\phi]-2\left[\frac{\partial}{\partial (\partial_\mu \partial_\nu \phi)}\right]\partial_\nu Q[\phi]+\partial_\nu\left[\left[\frac{\partial}{\partial (\partial_\mu \partial_\nu \phi)}\mathcal{L}\right] Q[\phi]\right]-\,\cdots\right]=0.$

### Example 1

OK, that was a general proof. Let's look at a specific case. We work with a 1-dimensional manifold with the topology of R (time) coordinatized by t. We assume

 $\displaystyle S[x]\,$ $\displaystyle =\int dt \mathcal{L}[x(t),\dot{x}(t)]$ $\displaystyle =\int dt \left\{\frac{m}{2}g_{ij}\dot{x}^i(t)\dot{x}^j(t)-V[x(t)]\right\}$

(i.e. a Newtonian particle of mass m moving in a curved Riemannian space (but not curved spacetime!) of metric g with a potential of V).

For Q, consider the generator of time translations. In other words, $\displaystyle Q[x(t)]=\dot{x}(t)$ . (Quantum field) physicists would often put a factor of i on the right hand side. Note that

$\displaystyle Q[\mathcal{L}]=m g_{ij}\dot{x}^i\ddot{x}^j-\frac{\partial}{\partial x^i}V(x)\dot{x}^i.$

This has the form of

$\displaystyle \frac{d}{dt}\left[\frac{m}{2} g_{ij}\dot{x}^i\dot{x}^j-V(x)\right]$

so we can set

$\displaystyle f=\frac{m}{2} g_{ij}\dot{x}^i\dot{x}^j-V(x).$

Then,

 $\displaystyle j\,$ $\displaystyle =\left(\frac{\partial}{\partial \dot{x}^i}\mathcal{L}\right)Q[x]-f$ $\displaystyle =m g_{ij}\dot{x}^j\dot{x}^i-\left[\frac{m}{2} g_{ij}\dot{x}^i\dot{x}^j-V(x)\right]$ $\displaystyle =\frac{m}{2}g_{ij}\dot{x}^i\dot{x}^j+V(x).$

You might recognize the right hand side as the energy and Noether's theorem states that $\displaystyle \dot{j}=0$ (i.e. the conservation of energy is a consequence of invariance under time translations).

More generally, if the Lagrangian does not depend explicitly on time, the quantity (called the energy)

$\displaystyle \sum_i \left (\frac{\partial}{\partial \dot{x}^i}\mathcal{L}\right )\dot{x^i}-\mathcal{L}$

is conserved.

### Example 2

Let's still work with one dimensional time. This time, let

 $\displaystyle S[\vec{x}]\,$ $\displaystyle =\int dt \mathcal{L}[\vec{x}(t),\dot{\vec{x}}(t)]$ $\displaystyle =\int dt \left [\sum^N_{\alpha=1} \frac{m_\alpha}{2}(\dot{\vec{x}}_\alpha)^2 -\sum_{\alpha<\beta} V_{\alpha\beta}(\vec{x}_\beta-\vec{x}_\alpha)\right]$

i.e. N Newtonian particles where the potential only depends pairwise upon the relative displacement.

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

$\displaystyle Q_i[x^j_\alpha(t)]=t \delta^j_i.$

Note that

$\displaystyle Q_i[\mathcal{L}]=\sum_\alpha m_\alpha \dot{x}_\alpha^i-\sum_{\alpha<\beta}\partial_i V_{\alpha\beta}(\vec{x}_\beta-\vec{x}_\alpha)(t-t)$
$\displaystyle =\sum_\alpha m_\alpha \dot{x}_\alpha^i.$

This has the form of $\displaystyle \frac{d}{dt}\sum_\alpha m_\alpha x^i_\alpha$ so we can set

$\displaystyle \vec{f}=\sum_\alpha m_\alpha \vec{x}_\alpha.$

Then,

$\displaystyle \vec{j}=\sum_\alpha \left(\frac{\partial}{\partial \dot{\vec{x}}_\alpha}\mathcal{L}\right)\cdot\vec{Q}[\vec{x}_\alpha]-\vec{f}$
$\displaystyle =\sum_\alpha (m_\alpha \dot{\vec{x}}_\alpha t-m_\alpha \vec{x})$
$\displaystyle =\vec{P}t-M\vec{x}_{CM}$

where $\displaystyle \vec{P}$ is the total momentum, M is the total mass and $\displaystyle \vec{x}_{CM}$ is the center of mass. Noether's theorem states that $\displaystyle \dot{\vec{j}}=0$ (i.e. $\displaystyle \vec{P}=M\dot{\vec{x}}_{CM}$ ).

### Example 3

Both examples above are over a one dimensional manifold (time). How about spacetime? Well, we'd have Noether currents. Let's see how this goes for the case of a conformal transformation of a massless real scalar field with a quartic potential in (3 + 1)-Minkowski spacetime.

 $\displaystyle S[\phi]\,$ $\displaystyle =\int d^4x \mathcal{L}[\phi (x),\partial_\mu \phi (x)]$ $\displaystyle =\int d^4x \left( \frac{1}{2}\partial^\mu \phi \partial_\mu \phi -\lambda \phi^4\right )$

For Q, let's consider the generator of a spacetime rescaling. In other words,

$\displaystyle Q[\phi(x)]=x^\mu\partial_\mu \phi(x)+\phi(x).$

The second term on the right hand side is due to the "conformal weight" of φ. Note that

$\displaystyle Q[\mathcal{L}]=\partial^\mu\phi\left(\partial_\mu\phi+x^\nu\partial_\mu\partial_\nu\phi+\partial_\mu\phi\right)-4\lambda\phi^3\left(x^\mu\partial_\mu\phi+\phi\right).$

This has the form of

$\displaystyle \partial_\mu\left[\frac{1}{2}x^\mu\partial^\nu\phi\partial_\nu\phi-\lambda x^\mu\phi^4\right]=\partial_\mu\left(x^\mu\mathcal{L}\right)$

(where we have performed a change of dummy indices) so we can set

$\displaystyle f^\mu=x^\mu\mathcal{L}.\,$

Then,

$\displaystyle j^\mu=\left[\frac{\partial}{\partial (\partial_\mu\phi)}\mathcal{L}\right]Q[\phi]-f^\mu$
$\displaystyle =\partial^\mu\phi\left(x^\nu\partial_\nu\phi+\phi\right)-x^\mu\left(\frac{1}{2}\partial^\nu\phi\partial_\nu\phi-\lambda\phi^4\right).$

Noether's theorem states that $\displaystyle \partial_\mu j^\mu=0$ (as you may explicitly check by substituting the EL equations into the left hand side).

If you try to find the Ward-Takahashi analog of this equation, you'd run into a problem because of anomalies.