Noether Talk

7/21/2019 Noether Talk

1/25

Noethers Theorem

Rob Thompson

University of MinnesotaMath Club

February 3, 2011


2/25

Introduction

Noethers Theoremfirst appeared in

Invariante Variationsprobleme, Nachr. d. Konig. Gessellsch. d.Wiss. zu Gottingen, Math-phys. Klasse., 235-257 (1918).

Symmetries of variational problems yield conservation laws.


3/25

Conservation of momentum in mechanics

Often, a conservation lawmeans a quantity associated to a physicalsystem that remains unchanged as the system evolves in time.

Image taken from sparknotes.com

mcannon vcannon

| {z }cannon momentum

+ mball vball

| {z }ball momentum

= 0

Conservation ofmomentum in mechanics can be thought of as an

expression of Newtons second law, F = d

dt(mv).

Is energy conserved in this example?


4/25

Conservation of energy in mechanics

The energy of a particle is the sum of kinetic and potential energy:

E=T+ U

Umay be complicated, T =1

2x2

Suppose that dynamics are given by

x= f(x), f,x Rn,

where f=U

x for some U.

Then energy E is conserved.

dE

dt =

d

dt

1

2hx, xi+ U(x)

=hx, xi+hU

x, xi= hf+

U

x, xi= 0.

What ifUdepends on time?


5/25

Variational problems

Suppose we have a way to assign a weight to a given function.We search for an extremal of this weight.

(Well come back to mechanics in a minute!)

Example.

Length of graph off from (a, A) to (b, B) = Z b

a p1 + f0(x)2 dx,where f(a) =Aand f(b) =B .

-

6

!!!

!!!!!

a b x

f(x)

An extremal function fshould be a line, right?


6/25

EulerLagrange Equations

Shift ffrom a minimum value by adding v(x), where v(x) issome function satisfying v(a) =v(b) = 0 and a small parameter.

A family of variations

By ordinary calculus,d

d

=0

Z ba

p1 + (f0 +v0)2 dx=

Z ba

f0 +v0p1 + (f0 +v0)2

v0=0

dx

= Z b

a

f0

p1 + (f0)2v0 dx= 0.


7/25

EulerLagrange Equations

Integrating by parts we find

Z ba

f0p

1 + (f0)2v0 dx= f

0p1 + (f0)2

vb

a

Z ba

ddx

f0p1 + (f0)2

v dx

=

Z ba

d

dx

f0

p1 + (f0)2

v dx= 0.

Since our variation v is arbitrary this means d

dx

f0p1 + (f0)2

= 0.

With a little work, we see that fmust be a line.

The calculation we did works for any integrand of course, not justthe one for arclength.


8/25

EulerLagrange Equations in general

Theorem

For a function u to be an extremal of the integral

Z ba

L(t,u,ut) dt

u must satisfy the EulerLagrange equations

E(L) =L

u

d

dt

L

ut= 0

IfL depends on more than one variable u1, u2, . . . and its

derivatives, we get a system of EulerLagrange equations

E1(L) = L

u1

d

dt

L

u1t= 0, E2(L) =

L

u2

d

dt

L

u2t= 0, . . .

We can derive similar expressions ifL depends on higher

derivatives or more independent variables t1, t2, . . ..


9/25

Variational problems in optics

Optical path length = Z Q

P

n(x(s))ds, n the refractive index.

Image taken from wikipedia.org

The empirical fact that lights path minimizes the optical pathlength is Fermats Principle. It implies Snells Law:

sin(1)

sin(2)=

n1n2


10/25

Variational problems in mechanics

Actionof a mechanical system from time t1 to t2:

Z t2t1

(T U) dt

The empirical fact that the true path of the system minimizes

the action integral is called the Principle of Least Action.For most situations, kinetic energy depends only on the velocity: T =T(x) potential energy depends only on position: U=U(x)

This principle provides an alternative way to do mechanics. Thisviewpoint will help us to understand, via Noethers Theorem,where all the conservation laws in mechanics come from.


11/25

Newtons Equations as EulerLagrange equations

We can use the principle of least action to find Newtonsequations. Lets take the example of a 1-d harmonic oscillator.

Image taken from learner.org

T =1

2mx2 U=

1

2kx2

The EulerLagrange equations are

0 =(T U)

x

d

dt

(T U)

x=kx + mx

These are the same as Newtons equations!


12/25

Symmetries

A symmetry of a function is a transformation of its variables whichleaves the function unchanged.

Function Transformation: (x, u) goes to

x2 + u2 (x cos u sin, x sin+ u cos)u mx (x +, u + m)

u/x2 (x,2u)

Our symmetries come in groups, with parameters like ,,above.

Symmetries transform derivatives ux, uxx, . . . via the chain rule.

For example:

x7 x= x u7 u=2u = ux 7 ux= 1

xxux =ux


13/25

Symmetries and Infinitesimals

Given a symmetry group ofF, directional derivatives ofF in thedirection of symmetry are zero.

Example

For F =x2 + u2, we have the rotation symmetry

(x, u)7(x cos u sin, x sin+ u cos )

The direction of symmetry is v =u x

+ x u

.

v(F) =uF

x + x

F

u =2ux + 2xu= 0

@

@@Iv

F(x, u) = constantInfinitesimal symmetry v extends to derivatives:

v =u

x+ x

u = v(1) =u

x+ x

u+ (1 + u2x)

ux


14/25

Symmetries of Variational Problems

A symmetry of a variational problem is a transformation whichpreserves the value of the integral defining the problem.

Given an integral Z ba

L(t, u, ut) dt

we transform to the new variables

Z ba

L(t, u,ut) dt=

Z ba

L(t, u,ut)

dt

dtdt

To leave the value of the integral unchanged we must have

L(t, u, ut) =L(t, u,u

t)

dt

dt


15/25


If our infinitesimal symmetry is

t + u +

,t ut

then the condition of invariance takes the infinitesimal form

L

t+

L

u +

L

ut,t + L

d

dt = 0

| {z }change in Lagrangian

|{z}change in volume/measure

This infinitesimal formula provides the key to Noethers theorem.


16/25


Example

When measuring the length of curve, it doesnt matter if I rotate

the curve before measuring. Thus, the arclength variationalproblem

Zp1 + u2x dx should have rotational symmetry.

The symmetry generator of rotation is

u

x+ x

u+ (1 + u2x)

ux

We can check that

uLx

+ xLu

+ (1 + u2x)Lux+ L d(u)

dx

= (1 + u2x)L

ux+Lux = 0.

But you already knew that, right?


17/25

A Review

Variational problem: an integral we want to minimize

EulerLagrange equations: diff

erential equations givingpossible solutions to the variational problem

Conservation Law: A function (involving derivatives) which isconstant on solutions to the EulerLagrange equations

Variational symmetry: A symmetry which preserves theintegrand of the variational problem

Noethers Theorem.

Variational symmetries are in one to one correspondence withconservation laws for the associated EulerLagrange equations.


18/25

Noethers Theorem

The Theorem!

Suppose thatv =

t

+

u

+ ,t

u

t

generates a variational

symmetry for the variational problem

Z L(t, u, ut) dt. Then

L+ L

u

t

( ut)

is constant for solutions u(t) of the EulerLagrange equations.

You want proof?

The proof quite remarkable, but unfortunately wont fit in themargin of this slide....

P f f N h Th


19/25

Proof of Noethers Theorem

The proof of Noethers theorem at this point is a computation.

0 =L

t+

L

u +

L

ut,t + L

d

dt

=

L

t+

L

uut +

L

ututt

Lu

ut Lut

utt+ Lu

+ Lut

,t + L ddt

= d

dt(L)

L

uut

L

ututt+

L

u+

d

dt ut

d

dt

L

ut

Things collapse nicely into

= d

dt(L) +

L

u

ut

+

L

ut

d

dt

ut

P f f N h Th


20/25

Proof of Noethers Theorem

So we have

0 = d

dt (L) + L

u ut + Lut

d

dt ut .

The next step is to integrate by parts:

L

ut

d

dtut =d

dtL

utut +

d

dtL

utut ,

Thus,

0 = L

u

d

dt

L

u

t ut +

d

dtL

u

t ut +

d

dt(L)

| {z }EulerLagrange equations!

| {z }conservation law

QED?

C ti L i h i


21/25

Conservation Laws in mechanics

Using our new found tools, we can rediscover the conservation lawsof mechanics without much work.

Kinetic energy: T(x) =12

X

mx

Potential energy: U(t,x) depends on problem...

Newtons equations are the EulerLagrange equations:

mx =

U

x

A variational symmetry v =

t+

x+ ,t

x satisfies

v(T U) + (T U) ddt

= 0.

For each such v we have a corresponding conservation law:

K=m

x (T+ U).

C ti L i h i


22/25

Conservation Laws in mechanics

Symmetry condition: v(T U) + (T U)d

dt

= 0

Conservation law: K=m

x (T+ U).

Energy.

U is independent oft iffour variational problem has time

translation symmetryv =

t . Thus the total energy T+ U isconserved.

Momentum.

Consider spatial translation x 7x +a, which is generated by

v =P a x . This is a variational symmetry iffU istranslationally invariant, i.e. v(U) = 0. This produces theconservation of linear momentum

P

ma x.

Angular momentum.

Th B hist h


23/25

The Brachistochrone

In 1696, Johann Bernoulli posed the Brachistochrone problem:

Given two points A and B in a vertical plane, what is

the curve traced out by a point acted on only by gravity,which starts at A and reaches B in the shortest time?

Image taken from storyofmathematics.com

According to Conduitt, Newton solved the problem in an eveningafter learning of it, working from 4pm to 4am.

The Brachistochrone


24/25

The Brachistochrone

Lets see how long it takes you to solve it, Noether style.

Exercise.

Step 1.

Write down an integral which gives the total time to travel from Ato B along the curve u(x).

Step 2.Look for symmetries/conservation laws.

Step 3.

Use a conservation law to help parametrize your curve u(x).

References


25/25

References

Peter Olver, Applications of Lie Groups to DifferentialEquations, GTM 107.

Emmy Noether, trans. M.A. Tavel, Invariant VariationProblems, arXiv:physics/0503066v1

Nina Beyers, E. Noethers Discovery of the Deep ConnectionBetween Symmetries and Conservation Laws,arXiv:physics/9807044v2

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