Noether's theorem

In order to discuss the relation between symmetry and conservation laws, we start from eq. (4.33). We do not assume that the variations of the fields vanish on $\partial S$, but we take into account the field equations, so that the last two integrals vanish. We obtain

$$\int_{\partial S} \theta = \int_S \delta \lambda, \qquad \t... ... \Psi^U \pi_U + \delta \omega^{\alpha} \wedge \sigma_{\alpha}.$$ (4.39)

Since $S$ is arbitrary, we have

$$d \theta = \delta \lambda$$ (4.40)

and if $\lambda$ is invariant, namely $\delta \lambda = 0$, the 3-form $\theta$ is conserved, namely $d \theta = 0$. This equation can be considered as a formulation of the Noether theorem.

We consider two different kinds of infinitesimal symmetry transformations. If we use eq. (2.11), that describes a symmetry property of the Lagrangian, for any element $\kappa$ of the Lie algebra of the symmetry group $\mathcal{F}$, we obtain the conserved quantity

$$\theta(\kappa) = \theta^G(\kappa) + \theta^M(\kappa),$$

$$\theta^G(\kappa) = C^{\alpha}{}_{\beta}(\kappa) \omega^{\beta} \wedge \sigma_{\alpha}, \qquad \theta^M(\kappa) = S^U{}_V(\kappa) \Psi^V \pi_U.$$ (4.41)

If we consider an infinitesimal diffeomorphism generated by the infinitesimal vector field $B$, we have, since $\lambda$, does not depend on $s$ explicitly,

$$\delta \Psi^U = B \Psi^U, \qquad \delta \omega^{\alpha} = L(B) \omega^{\alpha},$$

$$\delta \lambda = L(B) \lambda = i(B) d \lambda + d i(B) \lambda.$$ (4.42)

If we put

$$\tau(B) = \theta - i(B) \lambda = B \Psi^U \pi_U + L(B) \omega^{\alpha} \wedge \sigma_{\alpha} - i(B) \lambda,$$ (4.43)

from eq. (4.40) we obtain

$$d \tau(B) = i(B) d \lambda.$$ (4.44)

For $B = - A_{\alpha}$, we have $\tau(- A_{\alpha}) = \tau_{\alpha}$ and these quantities are interpreted as the density and the current density of (10+n)-momentum with respect to the moving frame $s$. We have introduced a minus sign because we are considering active transformations of the fields, defined, for scalar fields, by

$$s \to s', \qquad \Psi(s) \to \Psi'(s), \qquad \Psi'(s') = \Psi(s).$$ (4.45)

One has to remember that the energy density is described by $\tau^0 = - \tau_0$. In fact, even in elementary mechanics, the Hamiltonian is the generator of the passive time translations.

From eq. (4.38) we obtain

$$i_{\alpha} d \lambda = - d \tau_{\alpha} = 0$$ (4.46)

and therefore

$$d \lambda = 0.$$ (4.47)

This formula, that follows from the field equations and the invariance of the Lagrangian under diffeomorphisms, implies that the action integral depends only on $\partial S$ and not on the details of $S$. It is trivially satisfied in a theory based on the 4-dimensional spacetime.

In conclusion, for any choice of the vector field $B$, as a consequence of the invariance under diffeomorphisms, we have the conservation law

$$d \tau(B) = 0.$$ (4.48)

Note that the conservation laws can be easily derived from the field equations. We have derived them from the action integral for pedagogical reasons and to make more clear their connection with the symmetry properties.

In order to show that eq. (4.48) follows from a generalized Gauss law, we introduce the 2-form

$$\sigma(B) = - b^{\alpha} \sigma_{\alpha}, \qquad B = b^{\alpha} A_{\alpha}$$ (4.49)

and, by means of the formula

$$L(B) \omega^{\alpha} = i(B) d \omega^{\alpha} + d i(B) \omeg... ...pha} = b^{\beta} L(A_{\beta}) \omega^{\alpha} + d b^{\alpha},$$ (4.50)

we see that

$$d \sigma (B) = - d b^{\alpha} \wedge \sigma_{\alpha} + b^{\alpha} \tau_{\alpha} = - \tau(B).$$ (4.51)

In the following Sections, we consider Lagrangian forms invariant with respect to the infinitesimal Lorentz transformations (2.13) and from eq. (4.41) we obtain the conserved quantities

$$\theta_{[ik]} = \hat F_{[ik] \beta}^{\alpha} \omega^{\beta} \wedge \sigma_{\alpha} + \Sigma_{[ik]}{}^U{}_V \Psi^V \pi_U.$$ (4.52)

Their conservation does not depend on the geometry of $\mathcal{S}$. If $\mathcal{S}$ is a bundle of frames, as in Chapter 1, we have

$$\tau_{[ik]} = \theta_{[ik]} + i_{[ik]} \lambda.$$ (4.53)

Note that

$$d i_{[ik]} \lambda = L(A_{[ik]}) \lambda - i_{[ik]} d \lambda = 0$$ (4.54)

and the three forms that appear in eq. (4.53) are all conserved.