﻿ Noether's Theorem for Fields
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Noether's Theorem for Fields

Emmy Noether's Beautiful Theorem extends to the continuous objects of fields. The extension just requires adding another layer to the mathematical formulation. The Lagrangian formalism for fields is covered elsewhere in Lagrangian Theory for Fields.

Let S be a region of three dimensional space and let ∂S be its boundary. T is an interval of time and R is the Cartesian product of S and T. The Lagrangian L for a field is the integral of the Lagrangian density L over all space; i.e.,

#### L = ∫SLd³x

The action U of a field is the time integral of the Lagrangian of the field

#### U = ∫TLdt but this can be represented as wellas U =∫RLd4x

For a local field the Lagrangian density depends only upon the field intensity and its derivatives at a point. Let φ denote the field intensity and ∂μφ its derivatives, where μ can take on values from 0 to 3, with the zeroeth component being the time component. The field intensity is defined over R.

Let δφ be an incremental variation in field intensity over R with δφ=0 on the boundary ∂R. Note that δ often indicates a functional variation rather than just a number. When the field intensity φ changes by a amount δφover R it produces a variation in the Lagrangian density δL given by

#### δL = (∂L/∂φ)δφ + (∂L/∂[∂μφ])δ(∂μφ)

where there is a summation over the repeated index μ.

The increment in action is given

Note that

Thus

#### δU = ∫R[(∂L/∂φ)δφ + (∂L/∂[∂μφ])δ(∂μδφ) ]d4x

Integration by parts (∫UdV=UV − ∫VdU) can be applied to the second term of the RHS of the above equation. The variation δφ is zero on the boundary of R so δU reduces to

#### δU = ∫R[(∂L/∂φ) − ∂μ(∂L/∂(∂μφ)]δφd4x

By the Principle of Least Action δU must be zero. Since the region of integration may vary this means the integrand must equal zero everywhere in space and time; i.e.,

#### (∂L/∂φ) − ∂μ(∂L/∂(∂μφ)) = 0 or, equivalently (∂L/∂φ) = ∂μ(∂L/∂(∂μφ))

This the Euler-Lagrange equation for a field.

The time-derivative of the field φ is ∂0φ. The canonical conjugate momentum associated with the field is then

#### ψ = (δL/δ(∂0φ)

The Hamiltonian density H is given by

## Noether's Theorem for Fields

As noted above

#### δL = (∂L/∂φ)δφ + (∂L/∂[∂μφ])δ(∂μφ)

From the Euler-Lagrange equation ∂μ(∂L/∂[∂μφ]) may be substituted for (∂L/∂φ) in the above expression to give

#### δL = ∂μ(∂L/∂(∂μφ))δφ + (∂L/∂[∂μφ])δ(∂μφ)

The RHS of the above has the form of the differential of a product; i.e.,

#### δL = ∂μ[(∂L/∂[∂μφ])δ]

The expression (∂L/∂[∂μφ])δ is called the conserved current of the system and is denoted as Jμ.

Thus

#### δL = ∂μ(Jμ)

Thus if a transformation of variables results in δL being the zero function then

#### ∂μJμ = 0 over all of R

This is Noether's Theorem extended to fields.

The integration over space of the time component of the conserved current J, ∫SJ0d³x, is called the conserved charge of the physical system. Momenta and energy are conserved charges.

(To be continued.)