One independent variable Consider the simplest case, a system with one independent variable, time. Suppose the dependent variables
q are such that the action integral 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 :\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 :\begin{align} 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{align} 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. : \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 : \begin{align} 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{align} which may be regarded as a function of
ε. Calculating the derivative at
ε = 0 and using
Leibniz's rule, we get : \begin{align} 0 = \frac{d I'}{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{align} Notice that the Euler–Lagrange equations imply : \begin{align} \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{align} Substituting this into the previous equation, one gets : \begin{align} 0 = \frac{d I'}{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{align} Again using the Euler–Lagrange equations we get : \frac{d}{d t} \left( \frac{\partial L}{\partial \dot{\mathbf{q}}} \frac{\partial \varphi}{\partial \varepsilon} \right) = \left( \frac{d}{d t} \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 : \begin{align} 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{align} From which one can see that :\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 \frac{\partial \varphi}{\partial \mathbf{q}} = 1 and so the conserved quantity simplifies to :\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.
Geometric derivation Noether's theorem can be seen as a consequence of the
fundamental theorem of calculus (known by various names in physics such as the
Generalized Stokes theorem or the
Gradient theorem): for a function S analytical in a domain {\cal {D}}, \int_{{\cal {\cal P}}}dS=0 where {\cal P} is a closed path in {\cal D}. Here, the
function S(\mathbf{q},t) is the action
function that is computed by the integration of the Lagrangian over optimal trajectories or equivalently obtained through the
Hamilton-Jacobi equation. As \partial S/\partial\mathbf{q}=\mathbf{p} (where \mathbf{p}is the momentum) and \partial S/\partial t=-H (where H is the Hamiltonian), the differential of this function is given by dS=\mathbf{p}d\mathbf{q}-Hdt. Using the geometrical approach, the conserved quantity for a symmetry in Noether's sense can be derived. The symmetry is expressed as an infinitesimal transformation:\begin{aligned} \mathbf{q'} & = & \mathbf{q}+\epsilon\phi_{\mathbf{q}}(\mathbf{q},t)\\ t' & = & t+\epsilon\phi_{t}(\mathbf{q},t) \end{aligned} Let {\cal C} be an optimal trajectory and {\cal C}' its image under the above transformation (\phi_{\mathbf{q}},\phi_{t})^{T} (which is also an optimal trajectory). The closed path {\cal P} of integration is chosen as ABB'A', where the branches AB and A'B' are given {\cal C} and {\cal C}' . By the hypothesis of Noether theorem, to the first order in \epsilon, \int_dS=\int_{{\cal C}'}dS therefore, \int_{A}^{A'}dS=\int_{B}^{B'}dS By definition, on the AA' branch we have d\mathbf{q}=\epsilon\phi_{\mathbf{q}}(\mathbf{q},t) and dt=\epsilon\phi_{t}(\mathbf{q},t). Therefore, to the first order in \epsilon, the quantity I=\mathbf{p}\phi_{\mathbf{q}}-H\phi_{t} is conserved along the trajectory.
Field-theoretic derivation Noether's theorem may also be derived for tensor fields \varphi^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
xμ 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 :x^\mu \rightarrow \xi^\mu = x^\mu + \delta x^\mu whereas the transformation of the field variables is expressed as :\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 \delta\varphi^A result from two factors: intrinsic changes in the field themselves and changes in coordinates, since the transformed field
αA depends on the transformed coordinates ξμ. To isolate the intrinsic changes, the field variation at a single point
xμ may be defined :\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 :\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. :{\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 : \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 : \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 : \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 : \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 :\begin{align} &\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{align} Thus, the change in the action can be written as : \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 : \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 :\begin{align} \delta x^{\mu} &= \varepsilon X^\mu \\ \delta \varphi^A &= \varepsilon \Psi^A = \bar{\delta} \varphi^A + \varepsilon \mathcal{L}_X \varphi^A \end{align} where \mathcal{L}_X \varphi^A is the
Lie derivative of \varphi^A in the
Xμ direction. When \varphi^A is a scalar or {X^\mu}_{,\nu} = 0 , :\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 :\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 :\frac{\partial}{\partial x^\sigma} j^\sigma = 0 where the conserved current equals : 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\,.
Manifold/fiber bundle derivation Suppose we have an
n-dimensional oriented
Riemannian or more generally
Lorentzian 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 T over
M.) Examples of this
M in physics include: • In
classical mechanics, in the
Hamiltonian formulation,
M is the one-dimensional manifold \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, \varphi_1,\ldots,\varphi_m, then the target manifold is \mathbb{R}^{m}. If the field is a real vector field, then the target manifold is
isomorphic to \mathbb{R}^{3}. Now suppose there is a
functional :\mathcal{S} \colon \mathcal{C}\rightarrow \mathbb{R}, called the
action. (It takes values in \mathbb{R}, rather than \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 \mathcal{S}[\varphi] is the
integral over
M of a function :\mathcal{L}(\varphi,\partial_\mu\varphi,x) called the
Lagrangian density, depending on \varphi, its
derivative and the position. In other words, for \varphi in \mathcal{C} : \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 \varphi at the
boundary if
M is
compact, or some limit on \varphi as
x approaches ∞. Then the
subspace of \mathcal{C} consisting of functions \varphi such that all
functional derivatives of \mathcal{S} at \varphi are zero, that is: :\frac{\delta \mathcal{S}[\varphi]}{\delta \varphi(x)} = 0 and that \varphi 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 \mathcal{L} \, \mathrm{d}^n x \right] = \int_{\partial N} f^\mu [\varphi(x),\partial\varphi,\partial\partial\varphi,\ldots] \, ds_\mu for all compact submanifolds
N of dimension 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}[\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. If the symmetry generated by
Q integrates to a continuous symmetry, we say that
Q is the generator of a
one parameter symmetry Lie group. Now, for any
N, because of the
Euler–Lagrange theorem, we have
on shell (and only on-shell) : \begin{align} Q\left[\int_N \mathcal{L} \, \mathrm{d}^nx \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}^nx + \int_{\partial N} \frac{\partial\mathcal{L}}{\partial(\partial_\mu\varphi)}Q[\varphi] \, \mathrm{d}s_\mu \\ & = \int_{\partial N} f^\mu \, \mathrm{d}s_\mu. \end{align} Since this is true for any
N, we have :\partial_\mu\left[\frac{\partial\mathcal{L}}{\partial(\partial_\mu\varphi)}Q[\varphi]-f^\mu\right] = 0 \; . But this is the
continuity equation for the current J^\mu defined by: :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).
Comments Noether's theorem is an
on shell theorem: it relies on the 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 : \int_{\partial N} J^\mu ds_{\mu} = 0 \; . The quantum analogs of Noether's theorem involving expectation values (e.g., \left\langle\int \partial \cdot \textbf{J}~d^{4}x \right\rangle = 0) probing
off shell quantities as well are the
Ward–Takahashi identities.
Generalization to Lie algebras 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 Q_1[\mathcal{L}] = \partial_\mu f_1^\mu and Q_2[\mathcal{L}] = \partial_\mu f_2^\mu 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 f^\mu_{12} = Q_1[f_2^\mu] - Q_2[f_1^\mu] . So, 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.
Generalization of the proof 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 \varphi and its first derivatives. Also, assume :Q[\mathcal{L}]\approx\partial_\mu f^\mu Then, : \begin{align} 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{align} for all \varepsilon. More generally, if the Lagrangian depends on higher derivatives, then : \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. == Examples ==