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Noether's Theorem:
Its Explanation and Proof

In her 1918 article Invariante Variationsprobleme Emma Noether actually stated two theorems and their converses. Only the first of the four has gotten attention and the designation Noether's Theorem. Some comments will be made about the other three theorems once the first of them has been dealt with.

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.

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 vn=dqn/dt. For example, a body in a vertical, uniform gravitational field would be described by one dimension, its height z. Its kinetic energy would be T=½mv² where m is the body's mass. Its potential energy is V=mgz, where g is the acceleration due to gravity. Thus the Lagrangian function for the body is L(z,v)=½mv²−mgz.

A physical system evolves over some 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.

For the example of a body in a uniform gravitational field L(z,v)=½mv²−mgz. Thus the Euler-Lagrange equation is that

−mg = d(mv)/dt
which reduces to
dv/dt = −g
which has the solution
v(t) = v(0)−gt
and since v=dz/dt
z(t) = z(0) + v(0)t − ½gt²

Suppose there is no gravitational field; i.e., g=0. Then d(mv)/dt=0 and hence the momentum mv is constant. In other words, if the Lagrangian is independent of the height z, or more generally, location in a particular direction then momentum in that direction is conserved.

Consider next the more complex system of a particle of mass m at the end of a mass-less lever arm of length r which is free to rotate about an axis. Let θ be the angle of the lever arm with respect to the horizontal. The linear velocity is then v=r(dθ/dt)=rω. Thus the kinetic energy is K=½mr²ω². (The quantity mr² is called the moment of inertia and denoted I. Thus the kinetic energy could be expressed as K==½Iω². The potential energy is V=mgrsin(θ) where g is the acceleration due to gravity. Therefore the Lagrangian function is L(θ,ω)=½Iω²−mgrsin(θ). The Euler-Lagrange equation for this system is

mgrcos(θ) = d(mr²ω)/dt
which reduces to
dω/dt = (g/r)cos(θ)

To work toward a solution multiply each side of the above equation by the corresponding side of the equation ω=dθ/dt. The result is

ω(dω/dt) = (g/r) cos(θ)(dθ/dt)
which is equivalent to
d(½ω²)/dt = (g/r)(d(sin(θ))/dt)
which upon integration from 0 to t gives
½[(ω(t))² − (ω(0))²] = (g/r)[sin(θ(t)) − sin(θ(0))]

The above equation can be solved for ω(t)=dθ/dt and the result, in principle, integrated to obtain θ(t).

For a third example, consider an object with a moment of inertia of I spinning about its own axis. Its kinetic energy is then ½Iω² where ω is its angular velocity. There is no potential energy so the Lagrangian is independent of the angle of position θ. Thus the angular momentum (∂L/∂ω)=Iω is conserved.

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 a function of a continuous parameter s. According to Noether's Theorem if the Lagrangian is independent of s then there is a quantity that is conserved.

Proof:

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

d(∂qi/∂s))/dt = ∂(dqi/dt)/∂s = ∂vi/∂s

Since

dpi/dt = ∂L/∂qi
and
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.

Reconsider the case of a body moving in a uniform gravitational field. In that case L(z,v)=½mv²−mgz. Then p=∂L/∂v=mv. Now let the parameter s be just time t. Thus (∂z/∂s)=dz/dt=v. Therefore H=(mv)v=mv²=2K. In this case there is no reason to expect dL/dt to be zero and therefore kinetic energy per se would not be conserved.

The Conservation of Energy

First note that the kinetic energy K depends only the v's and does not depend the q's. Furthermore the dependence of K upon the v's is quadratic and such that if all of the v'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 q'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.

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

∂L/∂vi = ∂K/∂vi
so
J = Σvi(∂K/∂vi) − L

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

Σvi(∂K/∂vi) = 2K

Thus

J = 2K − L = 2K − (K−V) = K + V

So the conserved quantity J is total energy.

Hamiltonian Dynamics

The Hamiltonian H for a system is defined in terms of the Lagrangian as

H = Σipivi − L

The equations for the dynamics of the system are then

∂H/∂pi = dqi/dt = vi
∂H/∂qi = −dpi/dt + Fqi
∂H/∂t = ∂L/∂t

where Fqi represents nonconservative external forces or dissipative forces such as frictions that are not represented in terms of the gradients of the potential energy function V. If the kinetic energy function depends only upon products of the vi then H is just the total energy of the system. Then the dynamic equations reduce to

dqi/dt = ∂H/∂pi
and
 
dpi/dt = −∂H/∂qi

For such systems it immediately follows that if ∂L/∂qi=0 and hence ∂H/∂qi=0 that dpi/dt=0 and therefore pi is conserved. Also if ∂L/∂t=0 then total energy H is conserved.

More generally if

∂H/∂qi=0
then
dpi/dt = Fqi

That is to say, momenta changes only due to nonconservative external and dissipative forces.

The Hamiltonian approach to dynamics was generalized by the great Russian mathematician Lev Pontryagin in terms of Optimal Control Theory and the Maximum Principle.

The Nature of Noether's Theorem

In the above the illustrations of Noether's Theorem involved taking a specific change and showing the conserved quantity. Noether's Theorem is more general. It says that if a transformation of the coordinate system satisfies certain condition, namely being continuous, then necessarily there exist a quantity that is conserved. The exact quantity may not be known but nevertheless it is known that it exists. There may be other transformations, such as the inversion between right and left-handed coordinate systems, which leave the Lagrangian function unchanged but for which there are not conserved quantities.

To go further it is convenient to express the spatial, velocity and momenta coordinates as the vectors Q, V and P, respectively. Then the Lagrangian function is L(Q,V) and the Hamiltonian function is H(Q,P). Now the effect of transformation T on the coordinates is expressed as Q'=T(Q) with a corresponding transformation of the velocities V'=T(V) and of the momenta P'=P(T).

The transformation T could be linear translations or rotations. It could also be a scalar dilation or inflation in Q'=λQ where λ is a scalar. This would be of interest because the universe appears to be undergoing such an inflation.

(To be continued.)

Extensions of Noether's Theorem

Noether's Theorem applies for continuous transformations of the coordinate system such as linear translations, rotations or time shifts. There are no explicit quantities conserved as a result of the independence of the system to coordinate transformations such as parity transformation (transformations between right-handed and left-handed coordinate systems).

The Converse of Noether's Theorem

The converse is that if there is some continuous variable that is conserved by a physical system then there is some variable that the Lagrangian of the system is independent of. This is not true for a variable such as the charge of a system.

(To be continued.)


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