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Symmetries and Conservation Laws On Noether’s First Theorem and Clarifying the Presentation of Hamilton’s Principle Jason Saroni August 10, 2016 The relationship between symmetries and conservation laws captured in Noether’s Theorem is insightful. It shows how mathematical deductive reasoning on matters of physical elegance reaches necessarily true conclusions as fundametal as the laws of physics. Viewing physics through the preservation or breaking of symmetries is a powerful probe into the goal of the field as a general analysis of the physical universe. In the modern literature in high energy physics, many explorations are done as such. One of the most successful is the 1964 paper by Peter Higgs “Broken Symmetries and the Masses of Gauge Bosons” deductively reasoning for the existence of a particle that gives matter mass, later discovered at CERN. Because it is such an important physical insight, in this paper we explore the basics: the relationship between symmetries and conservation laws for infinitesimal transformations that can be extended to large ones. The goal is nothing less than a comprehensive approach from accessible basics to far reaching conclusions. Along the way we clarify Hamilton’s principle and present what we think is a statement most in line with the correct interpretation.

Fundamental Ideas on the Calculus of Variations The journey begins with the calculus of variations. The idea of functions and integration is available online. 1

Lemma 0.1. (Fundamental Lemma of the Calculus of Variations) If b

Z

A(t)η(t)dt = 0 a

where A(t) is a continuous function in [a, b], η(t) is an arbitrary continuous function in [a, b], and η(a) = η(b) = 0. Then A(t) = 0 throughout [a, b]. Proof. This is a proof by contradiction. Suppose that A(c) 6= 0 for some a < c < b and the integral vanishes. For definiteness assume that A(c) > 0 (the same reasoning can be used with A(t) < 0). Because of the continuity of A(t), an interval (t1 , t2 ) exists in which A(t) > 0. Since η(t) is arbitrary, we can design it such that it is non-zero in the interval (t1 , t2 ). For example, η(t) = (t − t1 )(t2 − t) for t1 < t < t2 η(t) = 0 for t < t1 and t > t2 The integral does not vanish with this choice of η(t) contradicting the supposition that it vanishes. Therefore A(t) = 0 throughout [a, b]. Q.E.D. Definition 0.1. A functional is a mapping from a well defined set of functions to real numbers R where the functions map some independent variable to some number in R. The corresponding action is a definite integral of the functional with respect to the independent variable.  For example given the distance element ds2 = dx2 + dy 2 and the function y(x) in rectangular coordinates, we can construct the following distance funcitonal, Z s2 D= ds s1

Z

(x2 ,y2 )

D=

p

(dx2 + dy 2 )

(x1 ,y1 )

Z

x2

D=

p (1 + (y 0 (x))2 )dx

x1

2

p f [y(x), y 0 (x); x] = (1 + (y 0 (x))2 ) is a functional and D is the corresponding action. D here is just the distance between two points s1 and s2 in rectangular coordinates. It can be seen that an action is also a functional since it meets the required criteria in the definition. Suppose we construct an action, Z b   µ µ J(ε) = f q(ε) (t), q˙(ε) (t); t dt a

as follows, The function q µ (t) with µ = 1, 2, 3 is a generalized position function for an orthogonal coordinate system in 3 dimensions. For example, for the Cartesian coordinate system, q 1 ≡ x, q 2 ≡ y, q 3 ≡ z. For the cylindrical polar coordinate system q 1 ≡ r, q 2 ≡ z, q 3 ≡ θ. We will use these later. The position function q µ (t) specifies a trajectory from point (a, q µ (a)) to point (b, q µ (b)) that makes J an extremal (either maximum or minimum). We embed the extremal q µ (t) in a family of infinite other paths (η µ (t) is arbitrary) of the action from point (a, q µ (a)) to point (b, q µ (b)) and represent them as, µ q(ε) (t) = q µ (t) + εη µ (t)

where ε is a variable continuous parameter and η(a)  = η(b) = 0. From µ µ the paths we create a one-dimensional functional f q(ε) (t), q˙(ε) (t); t . The corresponding action with respect to t defined as J is a function of ε since ε is variable. Z J(ε) =

b

f a





µ µ q(ε) (t), q˙(ε) (t); t

dt

is well defined. Theorem 0.2. (Euler-Lagrange Equation) Consider an action J whose funcµ tional f depends on the dependent variable q(ε) (t) and its first derivative µ µ q˙(ε) (t), where µ = 1, 2, 3. Suppose that q(ε) (t) = q µ (t) + εη µ (t) where ε is 3

a continuous parameter, q µ (t) make J an extremal, and t is an independent variable: Z b   µ µ f q(ε) (t), q˙(ε) (t); t dt J(ε) = a

µ

The q (t) that makes J an extremal is a the solution of the Euler-Lagrange equation ∂f d ∂f = , ∂q µ dt ∂ q˙µ given that

∂f ∂q µ

and

d ∂f dt ∂ q˙µ

µ = 1, 2, 3

are continuous in t.

Proof. J is extremized by q µ (t) which occurs when ε = 0. From calculus, we have that an extremum occurs when dJ/dε = 0 so,   dJ =0 dε ε=0 Using the chain rule again from calculus, # Z b" µ µ ∂f ∂q(ε) dJ ∂f ∂ q˙(ε) = + µ dt µ dε ∂q(ε) ∂ε ∂ q˙(ε) ∂ε a 

dJ dε

Z b

 = ε=0

a

 ∂f µ ∂f µ η + µ η˙ dt = 0 ∂q µ ∂ q˙

Using the product rule from calculus,   d ∂f µ ∂f d ∂f η = µ η˙ µ + η µ µ dt ∂ q˙ ∂ q˙ dt ∂ q˙µ   ∂f µ d ∂f µ d ∂f η˙ = η − ηµ µ µ ∂ q˙ dt ∂ q˙ dt ∂ q˙µ Substituting into the integrand, Z b a

  b ∂f d ∂f ∂f µ µ − η dt + η =0 ∂q µ dt ∂ q˙µ ∂ q˙µ a 4

∂f The integrated term vanishes because η(a) = η(b) = 0. Suppose that ∂q µ and d ∂f are continuous in t, then the integrated term in brackets is continuous dt ∂ q˙µ in t by the combination theorem for continuous functions. Because η µ (t) is an arbitrary function in t throughout [a, b] and the integrand quantity in brackets is continuous in t throughout [a, b] all the conditions for the ∂f d ∂f Fundamental Theorem of the Calculus of Variations are met and ∂q µ − dt ∂ q˙µ = 0 throughout [a, b].

∂f d ∂f = µ ∂q dt ∂ q˙µ Q.E.D. Taking, b

Z J(ε) = a

  µ µ f q(ε) (t), q˙(ε) (t); t dt

We define the unique functional f for which ε = 0 as the Lagrangian L. That is, the Lagrangian is the functional constructed from functions or trajectories that extremize the associated action with respect to t. The Lagrangian is a unique functional. d ∂L ∂L = µ ∂q dt ∂ q˙µ

(1)

for the Lagrangian and this is called the Euler-Lagrange equation. Suppose that the extremum function q µ (t) represents the trajectory of a particle in a classical field e.g. the trajectory of a stone in a gravitational field U . From empirical data, the trajectory the particle follows is a solution to the cause and effect differential equation, −

∂U d2 q µ = m ∂q µ dt2

(2)

also known as Newton’s second law (the cause is on the left and the effect on the right side of the equal sign) where U = − GMr m with G the gravitational constant, M the mass of the earth, m the mass of an individual stone,

5

and r its distance from the earth’s center. For (1) to have physical meaning we apply inspection to the case under consideration and find L such that (1) is equivalent to the physically observed cause and effect equation (2). The Lagrangian determined by inspection for (1) to coincide with (2)  µ 2 1 µ µ is L (q (t), q˙ (t); t) = 2 m dqdt(t) − U (q µ (t)) = T (q˙µ (t)) − U (q µ (t)) where T and U are the kinetic and potential energies. The Lagrangian is the functional for which the action J(ε) is minimized with respect to the continuous µ parameter ε that gives different trajectories q(ε) (t). The presentation of Hamilton’s principle is subject to wrong interpretations and an example is that in the undergraduate physics textbook Classical Dynamics of Particles and Systems. Hamilton’s Principle is stated as follows: “Of all the possible paths along which a dynamical system may move from one point to another within a specified time interval (consistent with any constraints), the actual path followed is that which minimizes the time integral of the difference between the kinetic and potential energies.” Rb The book states that what is minimized is a (T − U )dt but in fact this is the minimum of the functional J(ε) in the case of classical mechanics.

It is worth defining two quantities in terms of the Lagrangian L as we will see. The first is the canonical momentum pµ conjugate to q µ , pµ ≡

∂L ∂ q˙µ

(3)

The second is the Hamiltonian H (q µ (t), q˙µ (t); t) of the particle, H (q µ (t), q˙µ (t); t) ≡ pµ · q˙µ − L For the classical case particle in a gravitational field discussed above,  µof the 2 plugging L = 21 m dqdt(t) − U (q µ (t)) in (3) gives, pµ ≡

∂L dq µ (t) = mq˙µ (t) = m ∂ q˙µ dt 6

The canonical momentum equals the physical momentum mq˙µ . The Hamiltonian becomes, µ

µ 2

µ

H (q (t), q˙ (t); t) = m (q˙ ) −



1 m (q˙µ )2 − U 2



1 = m (q˙µ )2 + U = T + U 2

The result is the total energy of the particle. Because pµ ≡ d ∂L , dt ∂ q˙µ ∂L = p˙µ ∂q µ

∂L ∂ q˙µ

and

∂L ∂q µ

=

(4)

In addition, because the Lagrangian L is a function of q µ , q˙µ , and t, taking the full derivative of L with respect to t we use the chain rule and get, dL ∂L dt ∂L dq µ ∂L dq˙µ = + µ + µ dt ∂t dt ∂q dt ∂ q˙ dt dL ∂L ∂L ∂L = + µ q˙µ + µ q¨µ dt ∂t ∂q ∂ q˙ but pµ ≡

∂L ∂ q˙µ

and

∂L ∂q µ

= p˙µ so dL ∂L = + p˙µ q˙µ + pµ q¨µ dt ∂t

∂L dL − p˙µ q˙µ − pµ q¨µ = dt ∂t using the product rule we factorize the left hand side, d ∂L (L − pµ q˙µ ) = dt ∂t µ µ µ because H (q (t), q˙ (t); t) ≡ pµ · q˙ − L we get, ∂L −H˙ = (5) ∂t where H and L are functions of variables q µ , q˙µ , and t. We now look at the concept of invariance. 7

Invariance Invariance should not be confused with conservation law. When we say a quantity is invariant, we mean that when it is viewed under different frames of reference, it is the same. Conversely when we say a quantity is conserved, we mean that when it is viewed under one reference frame, it is never changes but it may not be the same as measured in another reference frame. The quantity we will analyze for invariance is the action J (q µ (t), q˙µ (t); t) shorthand J where L is the Lagrangian,  Z b  dq µ (t) µ ; t dt L q (t), J= dt a The transformation from one reference frame to another that we consider is the following infinitesimal linear transformation, t0 (t, q µ ) = t + ετ (t, q µ ) q µ0 (t, q ν ) = q µ + εζ µ (t, q ν ) where τ (t, q µ ), and ζ µ (t, q ν ) are the generators of the transformation and ε is infinitesimal. The vector components are represented by µ which ranges from 1 to 3 where ν ranges from 1 to 3 for each µ. Next we state a formal definition of invariance and then state a final definition that implies the former. Definition 0.2. The functional  Z b  dq µ (t) µ ; t dt J= L q (t), dt a is said to be invariant under the infinitesimal transformation t0 (t, q µ ) = t + ετ (t, q µ ) q µ0 (t, q ν ) = q µ + εζ µ (t, q ν ) if and only if J 0 − J = εs ,

where s > 1 and s is an integer

 Notice that because ε is infinitesimal, εs is always equal to 0 by definition of an infinitesimal. Then J 0 = J is the invariance. 8

Definition 0.3. The functional  Z b  dq µ (t) µ J= L q (t), ; t dt dt a is said to be invariant under the infinitesimal transformation t0 (t, q µ ) = t + ετ (t, q µ ) q µ0 (t, q ν ) = q µ + εζ µ (t, q ν ) if and only if L0

dt0 − L = εs , dt

where s > 1 and s is an integer

 Proof. We prove that the latter definition implies the former. Consider the functional  Z b  dq µ (t) µ J= L q (t), ; t dt dt a make the infinitesimal transformation t0 (t, q µ ) = t + ετ (t, q µ ) q µ0 (t, q ν ) = q µ + εζ µ (t, q ν ) then b0

  µ0 0 µ0 0 dq (t ) 0 ; t dt0 J = L q (t ), 0 dt 0 a  0 Z b0  µ0 0 dt 0 µ0 0 dq (t ) 0 J = L q (t ), ;t dt 0 dt dt a0 0

Z

integration is now with respect to t so we change the bounds of integration appropriately. Using shorthand for the primed Lagrangian we get, Z b dt0 0 J = L0 dt dt a it follows in shorthand 9

Z b

0

J −J =

0 0 dt

L a

dt

 − L dt

dt0

0

If L0 dt − L = εs in the finite range from a to b then L0 dt − L = 0 hence dt 0 J = J as required. Theorem 0.3. (The Rund-Trautman Identity) If the functional  Z b  dq µ (t) µ J= L q (t), ; t dt dt a is invariant under the infinitesimal transformation t0 (t, q µ ) = t + ετ (t, q µ ) q µ0 (t, q ν ) = q µ + εζ µ (t, q ν ) then the following identity holds ∂L ∂L τ + µ ζ µ − H τ˙ + pµ ζ˙ µ = 0 ∂t ∂q Proof. If the functional is invariant under the infinitesimal transformation, 0 then, L0 dt − L = εs where s > 1. We differentiate the left and right sides dt 0 with respect to ε and set ε = 0. L0 and dt are functions of ε but L is not dt and so we use only the product rule and get,    0 d dt0 dL L + =0 (6) dε dt ε=0 dε ε=0 because t0 (t, q µ ) = t + ετ (t, q µ ), to t. Then

dt0 dt

d dε

= 1 + ετ˙ , where the dot is with respect



dt0 dt

 = τ˙

for the second term we begin by using the chain rule for the primed Laµ0 grangian function L0 of variables q µ0 , dqdt0 , and t0

10



dq µ0 dt0

dL0 ∂L0 dt0 ∂L0 dq ∂L0 d = 0 + µ0 +  µ0  dε ∂t dε ∂q dε dε ∂ dqdt0 µ0



using our infinitesimal transformations 

=

dq µ0 dt0

∂L0 ∂L0 µ ∂L0 d   τ + ζ + 0 dq µ0 ∂t0 dε ∂q µ ∂ dt0



we then plug this in (4) and set ε = 0. In doing this we note that the primed partial derivatives simply become unprimed. The Lagrangian we consider is a polynomial or quotient function of its variables hence taking the partial derivative of the primed Lagrangian with respect to a certain primed variable we get an expression of a form that corresponds to the partial derivative of the unprimed Lagrangian with respect to the corresponding unprimed variable. Setting ε = 0 we simply get the partial derivative of the unprimed Lagrangian with respect to the corresponding unprimed variable. Because the last expression contains a full derivative with respect to ε of a quantity that contains ε, we first have to express the quantity in terms of ε, take the full derivative with respect to ε, then only set ε = 0.   µ0   dq ∂L µ ∂L  d dt0  ∂L τ + µζ + µ =0 (7) Lτ˙ + ∂t ∂q ∂ q˙ dε 0

from the last term, we expand

dq µ0 dt0

using differential calculus

dq µ + εdζ µ dq µ 0 q˙µ + εζ˙ µ = = dt0 dt + εdτ 1 + ετ˙ then   µ0   " d dqdt0   = d dε dε 0

q˙ + εζ˙ µ 1 + ετ˙ µ

   ζ˙ µ (1 + ετ˙ ) − q˙µ + εζ˙ µ τ˙  = ζ˙ µ −q˙µ τ˙ = (1 + ετ˙ )2 

!# 0

0

(5) becomes 11

Lτ˙ +

 ∂L ∂L ∂L  τ + µ ζ µ + µ ζ˙ µ − q˙µ τ˙ = 0 ∂t ∂q ∂ q˙

rearranging   ∂L µ ∂L ∂L ∂L µ τ + µ ζ + L − µ q˙ τ˙ + µ ζ˙ µ = 0 ∂t ∂q ∂ q˙ ∂ q˙ because pµ ≡ identity

∂L ∂ q˙µ

and H (q µ (t), q˙µ (t); t) ≡ pµ · q˙µ −L we get a Rund-Trautman ∂L ∂L τ + µ ζ µ − H τ˙ + pµ ζ˙ µ = 0 ∂t ∂q

Q.E.D. ∂L Because ∂q ˙µ , and ∂L = −H˙ following the Euler-Lagrange Equation µ = p ∂t (holds if the functional J is an extremal), we substitute and get,

˙ + p˙µ ζ µ − H τ˙ + pµ ζ˙ µ = 0 −Hτ ˙ − H τ˙ + p˙µ ζ µ + pµ ζ˙ µ = 0 −Hτ using the product rule we factorize, d (pµ ζ µ − Hτ ) = 0 dt integrating both sides, pµ ζ µ − Hτ = const. where const. is a constant of integration. We present these results formally as Noether’s First Theorem. Theorem 0.4. (Noether’s First Theorem) Suppose we have an extremal functional Z b J= L(q µ , q˙µ ; t)dt a

12

If under the infinitesimal transformation t0 (t, q µ ) = t + ετ (t, q µ ) q µ0 (t, q ν ) = q µ + εζ µ (t, q ν ) the functional is invariant according to the definition dt0 − L = εs , where s > 1 and s is an integer dt then the following conservation law holds: L0

pµ ζ µ − Hτ = const. This is a simplified and specific representation to the physical world of a more general theorem by Emmy Noether.

References [1] Dwight E. Neuenschwander Emmy Noether’s Wonderful Theorem. 2011. [2] Yvette Kosmann-Schwarzbach The Noether Theorems. July 2010. [3] James Stewart Calculus. 2012. [4] Stephen T. Thornton and Jerry B. Marion Classical Dynamics of Particles and Systems. 2004 [5] Peter W. Higgs Broken Symmetries and the Masses of Gauge Bosons. Phys. Rev. Letters, Vol 13, No. 16, October 1964.

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