﻿ Noether's Theorem and Quantized Parameters

San José State University

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 Noether's Theoremand Quantized Parameters

What is generally known as Noether's Theorem states that if the Lagrangian function for a physical system is not affected by a continuous change (transformation) in the coordinate system used to describe it, then there will be a corresponding conservation law; i.e. there is a quantity that is constant. For example, if the Lagrangian is independent of the location of the origin then the system will preserve (or conserve) linear momentum. If it is independent of the base time then energy is conserved. If it is independent of the angle of measurement then angular momentum is conserved.

It has been perplexing since the publication of Noether's Theorem that no consequence has been found for discrete transformations, such as parity changes, which leave the Lagrangian of a system unchanged. These can be considered as transformations which are functions of a parameter that can take on only discrete values. At those discrete values the Lagrangian has the same value.

First the nature of the Lagrangian for a physical system must be explained.

## Lagrangian Dynamics

Let K be the kinetic energy of a system and V its potential energy. The Lagrangian L for the system is defined as the difference between the kinetic and the potential energy of the system; i.e., L = K − V.

A condition of a physical system is described by n coordinates {q1, q2, …, qn} and their time derivatives {v1, v2, …, vn}, where vi=dqi/dt for i=1 to n.

A physical system evolves over a period of time from 0 to T such that the integral of its Lagrangian

#### L = ∫0T L(q1(t),q2(t),…,qn(t);v1(t),v2(t),…,vn(t))dt

is a minimum.

The determination of the trajectory of the system, {qi(t) for i=1,…,n} is a problem in the calculus of variations. The result from the calculus of variations is that for the Lagrangian integral to be an extreme, a maximum or a minimum, at each instant of time

#### ∂L/∂qi = d(∂L/∂vi)/dt for all i=1,…,n

This is known as the Euler-Lagrange equation.

## The Conservation of Momenta

In general, the momentum pi associated with a coordinate qi is defined as

#### pi = (∂L/∂vi)

With this definition the Euler-Lagrange equation can be expressed as

#### ∂L/∂qi = dpi/dt

Thus if the Lagrangian is independent of qi; i.e., ∂L/∂qi=0; then dpi/dt=0; i.e., pi is constant over time and is said to be conserved.

Noether's Theorem is a generalization of the above. Suppose the coordinates {qi} are continuous functions of a parameter s. According to Noether's Theorem if the Lagrangian is independent of s then there is a quantity that is conserved.

## Proof of Noether'sTheorem

Consider a quantity (∂qi/∂s) and its product with the corresponding momentum pi. Call this product Hi; i.e.,

#### Hi = pi(∂qi/∂s)

Now consider the time derivative of Hi:

#### dHi/dt = (dpi/dt)(∂qi/∂s) + pid(∂qi/∂s))/dt

The order of the differentiations (by t and by s) in the second term on the right may be interchanged to give

Since

#### dpi/dt = ∂L/∂qiand pi = ∂L/∂vi

the time derivative of Hi reduces to

#### dHi/dt = (∂L/∂qi)(∂qi/∂s) + (∂L/∂vi)(∂vi/∂s)

But the right-hand-side of this equation is merely the rate of change of L having to do with the effect of a change in s. It is the change in L which occurs as a result of the effect of the change in s on qi and vi. A change in s may affect all of the coordinates and not just one. If everything is summed over the n coordinates the result is the total derivative of L with respect to s; i.e.,

#### Σ(dHi/dt) = dL/ds

The left-hand-side is just (dH/dt) where H=ΣHi. Thus if L is independent of s; i.e., dL/ds=0; then dH/dt=0 and thus H is constant over time; i.e., H is conserved.

## The Conservation of Energy

First note that the kinetic energy K depends only the vi's and does not depend the qi's. Furthermore the dependence of K upon the vi's is quadratic and such that if all of the vi's are multiplied by a factor λ the value of K is multiplied by λ². Thus the kinetic energy function K(v1,…,vn) is homogeneous of degree 2.

The potential energy V usually depends only upon the qi's.

Now consider the time derivative of the Lagrangian function; i.e.,

#### (dL/dt) = Σi[(∂L/∂qi)(dqi/dt) + (∂L/∂vi)(dvi/dt)]

By the Euler-Lagrange equation

#### ∂L/∂qi) = d(∂L/∂vi)/dt

This means the previous equation can be expressed as

#### dL/dt = Σi[(d(∂L/∂vi)/dt)vi + (∂L/∂vi)(dvi/dt)] which is the same as dL/dt = Σi[d(vi(∂L/∂vi)]/dt or, equivalently d[Σi[vi(∂L/∂vi)−L]/dt = 0

This means that the expression in the square brackets is constant over time; i.e., it is a conserved quantity. Let it be designated J and hence (dJ/dt)=0.

If potential energy function V is independent of the vi's and also independent of time t then

#### ∂L/∂vi = ∂K/∂viso J = Σvi(∂K/∂vi) − L

However K is a homogeneous function of degree 2 so by Euler's Theorem

Thus

#### J = 2K − L = 2K − (K−V) = K + V

So the conserved quantity J is total energy.

## The Independence of the Lagrangianto Discrete Changes in its System

Typical changes of the these sorts are parity changes and changes in the signs of the charges. A parity change where the signs of all values of a coordinate, say x, are reversed; i.e.

#### x => −x

Now suppose this parity change is represented as a continuous function of a parameter λ but λ is only allowed the values of ±1.

Such a function is

#### x = λx0

This is a contraction-dilation transformation. But not only does the x coordinat change but the velocity associated with it; i.e.,

#### v = (dx/dt) = λv0

Therefore, even if (∂L/∂x)=0, (∂L/∂λ) is not zero. The fact that L(λ=+1)=L(λ=−1) is of no significance. There is no conserved quantity for the parity transformation even though the Lagrangian has the same value at the two allowed values of the parity parameter.

Now consider a change the signs of the charges of a system. Suppose all chages are proportional to a unit charge e, say the charge of an electron. Then the transformation of charges can be represented as

#### e => −e

As in the case of a parity change the charge transformation can be represented as continuation of a contraction-dilation transformation but the allowed values of the parameter are only ±1.

#### e = λe0

As in the case of parity change not only do the magnitudes of the charges change but also their time rates of change and consequently also of the time rates of change of the electrical fields generated by the charges. But by Maxwell's equations if there is a rate change of the electrical\ field intensity the there is a magnetic field generated. Hence the physical behavior of the system would change over time and there would not be any conserved quantity. The fact that L(λ=+1)=L(λ=−1) is of no significance. There is no conserved quantity.

The fact that the Lagrangian is equal for two parameter values is not unique. If the Lagrangian function is not constant for all values of the pararmeter then there may be a whole continuum of Lagrangian values such that the Lagrangian is equal at two parameter values, as shown below. ## Conclusion

The fact that the Lagrangian of a system is unchanged by a discrete transformation of its coordinates is of no significance. There is no necessity for there to be a conserved quantity associated with the transformation.

(To be continued.)