Talk:Noether's theorem/Proposed revision (May 24, 2006)

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.

Explanation

The word "symmetry" in the above statement refers more precisely to the covariance of the form that a physical law takes with respect to a one-dimensional Lie group of transformations satisfying certain technical criteria. The conservation law of a physical quantity is usually expressed as a continuity equation.

The formal statement of the theorem derives, from the condition of invariance alone, an expression for the current associated with a conserved physical quantity. The conserved quantity is called the Noether charge and the current, Noether current. The Noether current is defined up to a divergenceless vector field.

Applications

Application of Noether's theorem allows physicists to gain powerful insights into any general theory in physics, by just analyzing the various transformations that would make the form of the laws involved invariant. For example:

• the invariance of physical systems with respect to spatial translation (in other words, that the laws of physics do not vary with locations in space) gives the law of conservation of linear momentum;
• invariance with respect to rotation gives law of conservation of angular momentum;
• invariance with respect to time translation gives the well known law of conservation of energy

In quantum field theory, the analog to Noether's theorem, the Ward-Takahashi identities, yields further conservation laws, such as the conservation of electric charge from the invariance with respect to the gauge invariance of the electric potential and vector potential.

The Noether charge is also used in calculating the entropy of stationary black holes¹.

Proof

Suppose we have an n-dimensional manifold, M and a target manifold T. Let ${\mathcal {C}}$ 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 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,...,φ_m, then the target manifold is R. If the field is a real vector field, then the target manifold is isomorphic to R.

Now suppose there is a functional

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

called the action. (Note that it takes values into $\mathbb {R}$, rather than $\mathbb {C}$; 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

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

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

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

Suppose we are given boundary conditions, ie., 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 ${\mathcal {C}}$ consisting of functions φ such that all functional derivatives of S at φ are zero, that is:

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

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 ${\mathcal {C}}$, generated by a functional derivation, Q such that

$Q\left[\int _{N}d^{n}x{\mathcal {L}}\right]=\int _{\partial N}ds_{\mu }f^{\mu }[\phi (x),\partial \phi ,\partial \partial \phi ,...]$

for all compact submanifolds N or in other words,

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

for all x, where we set ${\mathcal {L}}(x)={\mathcal {L}}[\phi (x),\partial _{\mu }\phi (x),x]$.

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 one parameter symmetry Lie group.

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

$Q\left[\int _{N}d^{n}x{\mathcal {L}}\right]$	$=\int _{N}d^{n}x\left[{\frac {\partial {\mathcal {L}}}{\partial \phi }}-\partial _{\mu }{\frac {\partial {\mathcal {L}}}{\partial (\partial _{\mu }\phi )}}\right]Q[\phi ]+\int _{\partial N}ds_{\mu }{\frac {\partial {\mathcal {L}}}{\partial (\partial _{\mu }\phi )}}Q[\phi ]$
	$=\int _{\partial N}ds_{\mu }{\frac {\partial {\mathcal {L}}}{\partial (\partial _{\mu }\phi )}}Q[\phi ].$

Since this is true for any N, we have

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

But this is the continuity equation for the current

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

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).

Comments

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

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

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

Suppose 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

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

and

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

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

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

where f₁₂=Q₁[f₂^μ]-Q₂[f₁^μ]. So,

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

This shows we can (trivially) extend Noether's theorem to larger Lie algebras.

Generalization of the 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.

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

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

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

$q[\epsilon ][S]=\int d^{d}xq[\epsilon ][{\mathcal {L}}]$

$=\int d^{d}x\left({\frac {\partial }{\partial \phi }}{\mathcal {L}}\right)\epsilon Q[\phi ]+\left[{\frac {\partial }{\partial (\partial _{\mu }\phi )}}{\mathcal {L}}\right]\partial _{\mu }(\epsilon Q[\phi ])$

$=\int d^{d}x\epsilon \partial _{\mu }{\Bigg \{}f^{\mu }-\left[{\frac {\partial }{\partial (\partial _{\mu }\phi )}}{\mathcal {L}}\right]Q[\phi ]{\Bigg \}}$

for all ε.

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

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

Example 1: Conservation of Energy

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

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

(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, $Q[x(t)]={\dot {x}}(t)$. [Quantum field physicists would often put a factor of i on the right hand side.] Note that

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

This has the form of

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

so we can set

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

Then,

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

You might recognize the right hand side as the energy and Noether's theorem states that ${\dot {j}}=0$ (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

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

(called the energy) is conserved.

Example 2: Conservation of Linear Momentum

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

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

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

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

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

Note that

$Q_{i}[{\mathcal {L}}]=\sum _{\alpha }m_{\alpha }{\dot {x}}_{\alpha }^{i}-\sum _{\alpha <\beta }\partial _{i}V_{\alpha \beta }({\vec {x}}_{\beta }-{\vec {x}}_{\alpha })(t-t)$

$=\sum _{\alpha }m_{\alpha }{\dot {x}}_{\alpha }^{i}.$

This has the form of ${\frac {d}{dt}}\sum _{\alpha }m_{\alpha }x_{\alpha }^{i}$ so we can set

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

Then,

${\vec {j}}=\sum _{\alpha }\left({\frac {\partial }{\partial {\dot {\vec {x}}}_{\alpha }}}{\mathcal {L}}\right)\cdot {\vec {Q}}[{\vec {x}}_{\alpha }]-{\vec {f}}$

$=\sum _{\alpha }(m_{\alpha }{\dot {\vec {x}}}_{\alpha }t-m_{\alpha }{\vec {x}})$

$={\vec {P}}t-M{\vec {x}}_{CM}$

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

Example 3

Both examples above are over a one dimensional manifold (time). For an example involving spacetime, let's work out the case of a conformal transformation of a massless real scalar field with a quartic potential in (3 + 1)-Minkowski spacetime.

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

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

$Q[\phi (x)]=x^{\mu }\partial _{\mu }\phi (x)+\phi (x).$

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

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

This has the form of

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

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

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

Then,

$j^{\mu }=\left[{\frac {\partial }{\partial (\partial _{\mu }\phi )}}{\mathcal {L}}\right]Q[\phi ]-f^{\mu }$

$=\partial ^{\mu }\phi \left(x^{\nu }\partial _{\nu }\phi +\phi \right)-x^{\mu }\left({\frac {1}{2}}\partial ^{\nu }\phi \partial _{\nu }\phi -\lambda \phi ^{4}\right).$

Noether's theorem states that $\partial _{\mu }j^{\mu }=0$ (as one may explicitly check by substituting the Euler-Lagrange equations into the left hand side).

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