Hamiltonian - General Approach

Notes to a video lecture on UNIZOR.COM

General Approach
to Hamiltonian

In the previous introductory lecture we have examined the way to simplify the system of n Euler-Lagrange differential equations of second order into a system of 2n equations of the first order using coordinates q={q_i} and momenta p={p_i}.

For simple classical systems the Lagrangian Mechanics uses Euler-Lagrange equations
∂L/∂q_i = d/dt ∂L/∂q̇_i
where Lagrangian L=T−U,
T is the full kinetic energy of a system (independent of coordinates),
U is its potential energy (independent of velocities).

Instead, we constructed mathematically equivalent system of Hamiltonian equations
q̇_i = ∂(T+U)/∂p_i
−ṗ_i = ∂(T+U)/∂q_i
and introduced a new function called Hamiltonian
H(q,p) = T(p) + U(q)
where q=(q₁,...,q_n) are coordinates
and p=(p₁,...,p_n) are momenta.

This transformation increased the number of variables from n to 2n, but reduced the complexity of differential equations to the first order and made them symmetric and esthetically more appealing.

However, this construction relied on special properties of the system:
kinetic energy T depended only on momenta,
potential energy U depended only on coordinates.

In general mechanical systems, the Lagrangian L(q,q̇,t) may depend on coordinates and velocities in a more complicated way, kinetic and potential energies might not be separable and the Lagrangian might not be written as a sum of a function of p and a function of q.
Therefore, the construction H=T+U is not applicable to the general case.

The Hamiltonian approach is attractive for these simple mechanical systems, so it is natural to ask whether it can be generalized.
More specifically, given a Lagrangian L(q,q̇,t), we'll construct a function H(q,p,t) whose partial derivatives generate the differential equations similar to above and equivalent to Euler-Lagrange equations of motion.

To accomplish this, we will reformulate the expression H=T+U to avoid explicitly using energies and, instead, express H in terms of coordinates, momenta and a Lagrangian.

Recall the calculations of the Hamiltonian for simple mechanical system in the previous introductory lecture:
T = Σ_i½m_i·q̇_i² = Σ_i½p_i²/m_i

Partial derivative of T by p_i produces
∂T/∂p_i = p_i/m_i = q̇_i
Therefore,
q̇_i = ∂T/∂p_i
Our expression for H=T+U in that case can be represented as
H = 2T − (T−U) = 2T − L =
= Σ_im_i·q̇_i² − L =
= Σ_i(m_i·q̇_i)·q̇_i − L =
= Σ_ip_i·q̇_i − L

The above expression
H(q,p,t) = Σ_ip_i·q̇_i − L(q,q̇,t)
formally seems to correspond to our task of constructing a function H(q,p,t) from coordinates, momenta and a given Lagrangian L(q,q̇,t).
It does not explicitly rely on kinetic or potential energy. Since it is written only in terms of the Lagrangian, generalized coordinates, generalized velocities, and generalized momenta, it is a natural candidate for the Hamiltonian in the general case.

It's very important to understand the change of a viewpoint on relationship between arguments, which is a key idea of the Hamiltonian Mechanics.
While in the Lagrangian formulation the independent variables are q, q̇ and t, in the Hamiltonian approach the independent variables are q, p and t with generalized velocities q̇ become dependent on them:
q̇_i = q̇_i(q,p,t)

Let's check if this expression corresponds to differential equations involving the Hamiltonian.
∂H(q,p,t)/∂p_k =
[take into consideration that now q_i and p_i are independent variables, while generalized velocities q̇_i are functions of these variables: q̇_i=q̇_i(q,p,t)]
= ∂/∂p_k [Σ_ip_i·q̇_i − L] =
[use chain rule]
= q̇_k + Σ_ip_i·(∂q̇_i/∂p_k) −
− Σ_i(∂L/∂q̇_i)·(∂q̇_i/∂p_k) =
= q̇_k+Σ_i[p_i−∂L/∂q̇_i]·(∂q̇_i/∂p_k) =
[recall that p_i=∂L/∂q̇_i by definition]
= q̇_k
as required.

∂H(q,p,t)/∂q_k =
= Σ_ip_i·(∂q̇_i/∂q_k) − ∂L/∂q_k −
− Σ_i(∂L/∂q̇_i)·(∂q̇_i/∂q_k) =
= Σ_i[p_i−∂L/∂q̇_i]·(∂q̇_i/∂q_k) −
− ∂L/∂q_k =
[recall that p_i=∂L/∂q̇_i by definition]
= − ∂L/∂q_k =
[using Euler-Lagrange equation]
= − d/dt ∂L/∂q̇_k =
[using the definition of generalized momentum]
= − d/dt p_k = −ṗ_k
as required.

So, from the definitions of momenta as
p_i = ∂L/∂q̇_i
the Hamiltonian as
H(q,p,t) = Σ_ip_i·q̇_i − L(q,q̇,t)
and Euler-Lagrange equation for L
∂L/∂q_i = d/dt ∂L/∂q̇_i
follows that both differential equations for H are satisfied:
∂H/∂p_i = q̇_i
∂H/∂q_i = −ṗ

In reverse, assuming Hamiltonian equations are satisfied, we can derive Euler-Lagrang equations.
∂H(q,p,t)/∂q_k =
= ∂/∂q_k [Σ_ip_i·q̇_i − L] =
= Σ_ip_i·(∂q̇_i/∂q_k) − ∂L/∂q_k −
− Σ_i(∂L/∂q̇_i)·(∂q̇_i/∂q_k) =
= Σ_i[p_i−∂L/∂q̇_i]·(∂q̇_i/∂q_k) −
− ∂L/∂q_k =
[recall that p_i=∂L/∂q̇_i by definition]
= − ∂L/∂q_k
Since we assumed that Hamiltonian differential equations are true,
∂H(q,p,t)/∂q_k = −ṗ_k
Therefore,
− ∂L/∂q_k = −ṗ_k

From the definition
ṗ_k = dp_k/dt = d/dt ∂L/∂q̇_k
follows:
∂L/∂q_k = d/dt ∂L/∂q̇_k
which is the Euler-Lagrange equation.
That completes the proof that a system of n second order Euler-Lagrange differential equations is equivalent to 2n first order Hamiltonian equations.

The construction of the Hamiltonian
H = Σ_ip_i·q̇_i − L
from the Lagrangian L is called the Legendre transformation.

Hamiltonian Introduction

Notes to a video lecture on UNIZOR.COM

Introduction to Hamiltonian

We assume you have a pretty good understanding of Lagrangian Mechanics. If not, the previous chapters of this course Lagrangian and Noether Theorem provide a description of its basic principles.

The Hamiltonian Mechanics is built upon Lagrangian Mechanics. The differences can be summarized as follows.

In Lagrangian Mechanics our main dynamic variables were generalized coordinates q=(q₁,...,q_n) and their time derivatives - generalized velocities q̇=(q̇₁,...,q̇_n) (here we use Newtonian 'dot notation' to indicate a time derivative).
In Hamiltonian Mechanics it's the same generalized coordinates {q_i} and, separately from coordinates, generalized momenta p=(p₁,...,p_n) instead of velocities.

Generalized momenta of a mechanical system with the Lagrangian L(q,q̇,t) are defined as a set of components
p_i=∂L/∂q̇_i
This definition was already introduced in the lecture Noether p=m·v const of a previous chapter of this course.

Using this definition of generalized momentum, the Euler-Lagrange differential equation of the second degree
d/dt ∂L/∂q̇_i = ∂L/∂q_i (i∈[1,n])
would look simpler
d/dt p_i = ∂L/∂q_i or, shorter,
ṗ_i = ∂L/∂q_i

In Lagrangian Mechanics we had n functions of time {q_i(t)} (generalized coordinates) and a system of n Euler-Lagrange differential equations of the second order.
The number of unknowns was equal to the number of equations.

Instead, in this momentum-based approach, we have 2n functions of time {q_i(t)} (generalized coordinates) and {p_i(t)} (generalized momenta) with only n differential equations
(A) ṗ_i = ∂L/∂q_i
We need n more equations to obtain a system of 2n first-order differential equations equivalent to n Euler-Lagrange equations of the second-order.

Consider a simple case of a conservative system of one point mass m in three-dimensional Euclidean space with Cartesian coordinates (q₁,q₂,q₃), velocities q̇_i and time-independent Lagrangian L equaled to a difference between kinetic T and potential U energies
L = T − U

Since kinetic energy T is independent of position q_i, the same equation (A) above can be written as
ṗ_i = ∂(−U)/∂q_i
or
−ṗ_i = ∂U/∂q_i

In addition to these n differential equations, we can construct n more using the classical definition of the vector of momentum p_i=m·q̇_i and the kinetic energy expressed in terms of momenta p as
T = Σ_i½m_i·q̇_i² = Σ_i½p_i²/m_i

Partial derivative of T by p_i produces
∂T/∂p_i = p_i/m_i = q̇_i
Therefore,
(B) q̇_i = ∂T/∂p_i
We have constructed n more differential equations to complete the system.

The differential equation for a time derivative of the generalized momentum, as we stated above, is
ṗ_i = ∂L/∂q_i = ∂(T−U)/∂q_i
Since kinetic energy is independent of coordinates, we can exclude it
(C) −ṗ_i = ∂U/∂q_i

Equations (B) and (C) constitute 2n differential equations of the first order with 2n unknowns - coordinates and momenta
q̇_i = ∂T/∂p_i
−ṗ_i = ∂U/∂q_i
The problem is, the first n equations depend on kinetic energy T, while the second group of n equations depends on potential energy U.

We would like a formulation in which both sets of equations are generated by a single function of the same variables (q,p), whose partial derivatives with respect to p give time derivatives of coordinates, that is velocities q̇, and with respect to q give time derivatives of the generalized momenta ṗ.

Recall that in our simple case the kinetic energy T is independent of positions q_i (∂T/∂q_i=0) and potential energy U is independent of momenta p_i (∂U/∂p_i=0).
Based on this property, we can add to partial differentiation of the first equation the potential energy U and add to partial differentiation of the second equation the kinetic energy T.
q̇_i = ∂(T+U)/∂p_i
−ṗ_i = ∂(T+U)/∂q_i

Let's introduce a new function called Hamiltonian
H(q,p) = T(p) + U(q)
where q=(q₁,...,q_n) are coordinates
and p=(p₁,...,p_n) are momenta.
With this notation our system of 2n differential equations of the first degree looks quite symmetrical (some might say "beautiful")
q̇_i = ∂H/∂p_i
−ṗ_i = ∂H/∂q_i

In this case of a simple mechanical system the Hamiltonian H=T+U represents a total (kinetic and potential) energy of our mechanical system, which makes our system of equation more related to real physical characteristics of a system.

In more general systems, however, the definition of the Hamiltonian is broader and not necessarily coincides with a total energy.

Looking ahead, let us state that the symmetric form of Hamilton's equations makes Hamiltonian Mechanics especially suitable for advanced topics such as canonical transformations, statistical mechanics and quantum mechanics.

Noether's Theorem Conservation: UNIZOR.COM -> Physics+ 4 All -> Lagrangian -> Noether's Theorem -> Conservation

Notes to a video lecture on UNIZOR.COM

Noether Theorem and
Conservation Laws


Background

Motion of a mechanical system is represented by a curve in extended configuration space with coordinates y={y_i(x)} - a set of time-space coordinates describing a motion curve parameterized by independent variable x∈[a,b] for i∈[0,n].

These coordinates have physical meaning:
y₀ is time t;
{y_i} are a set of generalized coordinates {q_i} for i∈[1,n];

The action functional

Φ[t,q] =

t(b) ∫ t(a)

L(t,q,q')dt

was re-parameterized as an integral by independent parameter x∈[a,b]

Φ[t,q] =

b ∫ a

L(t,q,qx/tx)·tx·dx

and expressed as

Φ[y] =

b ∫ a

𝓛(y,yx)·dx

where
y(x)={y_i(x)} (i∈[0,n]) signifies a set of all time-space coordinates parameterized by x∈[a,b], that is a trajectory in extended configuration space, with y₀(x)=t(x), and y_i(x)=q_i(x) for i≠0 and
y_x(x)={y_ix(x)} (i∈[0,n]) signifies a set of all derivatives of time-space coordinates by parameter x with y_0x(x)=t_x(x), and y_ix(x)=q_ix(x) for i≠0
and a new function 𝓛() is defined for i∈[0,n] as
𝓛(y,y_x) = 𝓛({y_i},{y_ix}) =
= L(t,q,q_x/t_x)·t_x =
= L(t,{q_i},{q_ix/t_x})·t_x

The conclusion of the previous lecture:
d/dx Σ_i𝓛_yix·ζ_i = 0 for i∈[0,n] where
ζ={ζ_i}={dy_i^(ε)/dε|_ε=0} is a set of generators for each time-space coordinate
and where
x in a subscript indicates a derivative of a corresponding function by parameter x:
t_x=dt/dx and
{q_ix}={dq_i/dx} for i∈[1,n]

Linear Momentum Conservation

Let's choose a single k^th space coordinate (k∈[1,n]) and consider the following ε-transformation of coordinates:
t^(ε) = t,
q_k^(ε) = q_k + ε,
q_i^(ε) = q_i for all i≠k.
This represents a uniform movement along the k^th space coordinate q_k.

The corresponding generators of this transformation are
ζ_k = dy_k/dε|_ε=0 = 1 and
ζ_i = 0 for i≠k.

The main result of the Noether's theorem was:
as long as the action functional is invariant under the transformation,
an expression
J = Σ_i𝓛_yix·ζ_i
is a constant of motion along a trajectory and its x-derivative equals to zero for all x∈[a,b]:
dJ/dx = 0

Considering all ζ_i=0 for i∈[0,n] except ζ_k=1,
J = 𝓛_ykx
and, therefore,
dJ/dx =
= d/dx 𝓛_ykx({y_i},{y_ix}) = 0
or
d/dx ∂/∂y_kx 𝓛({y_i},{y_ix}) = 0

To see what follows from this equation, let's return to the original Lagrangian
L(t,{q_i},{q_i'}),
where apostrophe at q indicates a time-derivative, using y_k=q_k equivalence for k≠0 and taking into account our definition of function 𝓛:
𝓛(y,y_x) = L(t,{q_i},{q_ix/t_x})·t_x
Notice that we represented q_i'=dq_i/dt (a generalized velocity) as (dq_i/dx)/(dt/dx)=q_ix/t_x.

Now
∂/∂y_kx 𝓛({y_i},{y_ix}) =
= ∂/∂q_kx [L(t,{q_i},{q_ix/t_x})·t_x]

As we indicated above, an expression q_ix/t_x is a generalized velocity along i^th space coordinate because
q_ix/t_x = (dq_i/dx)/(dt/dx) =
= dq_i/dt = q_i' = v_i

Therefore, using the chain rule, we can write the expression above as
∂/∂q_kx [L(t,{q_i},{q_ix/t_x})·t_x] =
= ∂/∂q_kx [L(t,{q_i},{v_i})·t_x] =
= [∂/∂v_k L(t,q,v)]·[∂v_k/∂q_kx]·t_x =
[recall, v_k=q_kx/t_x]
= [∂/∂v_k L(t,q,v)]·(1/t_x)·t_x =
= ∂/∂v_k L(t,q,v) =
= ∂/∂q_k' L(t,q,q')

Since p_k=∂/∂q_k' L(t,q,q') is a definition of generalized momentum, we conclude that under the ε-transformation of a single space coordinate q_k described above that leaves the action functional invariant or, in other words, possesses translational symmetry the generalized momentum is conserved.

The conservation of momentum p_k could be derived directly from the Euler–Lagrange equations once we know that the Lagrangian is independent of a coordinate q_k. Indeed, from the Euler-Lagrange equation
d/dt ∂L/∂q_k' = ∂L/q_k
follows that, if the right-hand side is zero (independence of Lagrangian L of coordinate q_k), the left-hand side is zero as well, which means that generalized momentum
p_k = ∂L/∂q_k'
is constant (conserved).

Noether's theorem is important because it reveals that momentum conservation is a consequence of a continuous symmetry of the action and extends this principle to every continuous symmetry. Momentum conservation is therefore not an isolated fact but one example of a universal connection between symmetry and conservation laws.

CONCLUSION

Assuming the translation
q_k ⟶ q_k + ε
leaves action functional invariant,
the Noether conserved quantity would be
J = ∂𝓛/∂y_kx = ∂L/∂q_k' = p_k
which is a generalized momentum along k^th generalized coordinate. Therefore,
dp_k/dt = 0
which is the law of conservation of linear momentum.

Angular Momentum Conservation

Consider a rigid body rotating about a fixed axis with only two extended generalized coordinates describing its motion
y₀ = t is time,
y₁ = θ is an angle of rotation.

The ε-transformation (rotation) we would like to consider is the uniform rotation that can be expressed as
t^(ε) = t,
θ^(ε) = θ + ε.

Since the angle θ is simply a generalized coordinate, this situation is the same as in the previous one
t^(ε) = t,
q_k^(ε) = q_k + ε
with n=k=1,
generalized space coordinate being the angle of rotation q₁=y₁=θ
and the derivation of Noether's conserved quantity is identical to the derivation for linear momentum.

Therefore, everything we derived for linear momentum in the above case is valid for angular momentum

Assuming the rotation
θ(=y₁) ⟶ θ + ε (=y₁ + ε)
leaves action functional invariant,
the Noether conserved quantity would be
J = ∂𝓛/∂θ_x = ∂L/∂θ' = ℒ
which is a generalized angular momentum.
Therefore, the angular momentum is conserved
dℒ/dt = 0
which is the law of conservation of angular momentum.

Energy Conservation

The third important application of Noether's theorem is the law of conservation of energy, which follows from the invariance of the action under translations of time.

Let's choose a uniform translation of the time coordinate that does not affect any space coordinates:
y₀^(ε) = y₀ + ε,
which is time transformation
t^(ε) = t + ε
and
y_i^(ε) = y_i for all i∈[1,n],
which means that all generalized coordinates remain unchanged
q_i^(ε) = q_i for all i∈[1,n].

For this kind of transformation the corresponding generators are
ζ₀ = dy₀/dε|_ε=0 = 1 and
ζ_i = 0 for i≠0.

According to Noether's theorem, the conserved quantity is
J = Σ_i𝓛_yix·ζ_i = 𝓛_y0x·ζ₀ =
= ∂𝓛/∂t_x = ∂/∂t_x[L(t,q,q_x/t_x)·t_x]
which is a constant of motion along a trajectory.

Let's perform all the required computations.
J = ∂/∂t_x[L(t,q,q_x/t_x)·t_x] =
[recall, v_i=q_ix/t_x=q_i' - time derivative of a generalized coordinate]
= ∂/∂t_x[L(t,{q_i},{q_ix/t_x})·t_x] =
[using a formula of a derivative of a product of two functions]
= [∂/∂t_xL(t,{q_i},{q_ix/t_x})]·t_x + L=
[apply the chain rule, taking into account that the dependence on t_x enters only through the generalized velocities {q_ix/t_x} = {v_i} and using v_i as a placeholder for q_ix/t_x to shorter the notation]
= [Σ_i(∂L/∂v_i)·(d(q_ix/t_x)/dt_x)]·t_x + L =
= [Σ_i(∂L/∂v_i)·(−q_ix/t²_x)]·t_x + L =
[substitute q_ix/t_x²=v_i·t_x/t_x²=v_i/t_x]
= −Σ_i(∂L/∂v_i)·v_i + L =
[recall, ∂L/∂v_i=∂L/∂q_i' is s generalized momentum p_i]
= −Σ_i(p_i·v_i) + L

The final formula for a conserved quantity J is:
J = −Σ_i(p_i·v_i) + L

In all conservative mechanical systems considered in this course, the Lagrangian L has the form
L = T − U
where the kinetic energy T is a quadratic homogeneous function of generalized velocities and U is potential energy of a system.

Recall that
p_i = ∂L/∂v_i = ∂(T−U)/∂v_i
Since potential energy U does not depend on generalized velocities,
p_i = ∂T/∂v_i

In classical mechanics the kinetic energy is a homogeneous quadratic (that is, of degree 2) function of the generalized velocities
T = Σ_i,jA_ijv_iv_j

For any given quadratic homogeneous function
T = Σ_i,jA_ijv_iv_j
the sum Σ_i[∂T/∂v_i]·v_i is equal to 2T as follows from the Euler theorem about homogeneous functions.
Here is a simple and elegant proof.
In case of quadratic homogeneous function
T(v₁,...,v_n) = Σ_i,jA_ijv_iv_j
T(λ·v₁,...,λ·v_n) = λ²·T(v₁,...,v_n)
where λ - any real number.
Let's differentiate both sides by λ applying the chain rule for the left side
Σ_i[∂T/∂(λ·v_i)]·v_i = 2λ·T
Set λ=1 that results in λ·v_i=v_i, and the result is
Σ_i[∂T/∂v_i]·v_i = 2·T

Since generalized momentum is defined by
p_i = ∂T/∂v_i
and for ordinary mechanical systems L=T−U while U does not depend on the velocities, ∂L/∂v_i = ∂T/∂v_i
therefore,
Σ_i(p_i·v_i) = 2T
J = −2T + (T−U) = −(T+U)
which is a negative total energy of the system, whose conservation is equivalent to conservation of the total system's energy itself.

Therefore, the total energy is conserved under a time transformation that preserves the action functional, as described above.

Noether's Theorem Derivation: UNIZOR.COM -> Physics+ 4 All -> Lagrangian -> Noether's Theorem -> Derivation

Notes to a video lecture on UNIZOR.COM

Derivation of Noether Theorem


Background

The previous lectures of the Noether Theorem part of the course Physics+ 4 All have introduced the concepts of symmetry, parameterized group of continuous transformations and a concept of an extended configuration space that combines time and generalized coordinates into one set of coordinates.

We suggested that the symmetries relevant to the laws of motion are transformations of extended time-space coordinates that leave the action functional invariant.

This lecture is about mathematical derivation of certain conservation laws as logical consequences from the symmetries of transformations.


Summary of Assumptions

(1) Let us consider an extended configuration space of a mechanical system with coordinates {t,q}, where t is time and q is a set of generalized coordinates q_i(t) (i∈[1,n]).

(2) This system is described by its Lagrangian L(t,q(t),q'(t)) where q'(t) is a set of generalized velocities {q_i'(t)} (time derivatives of generalized coordinates).

(3) The trajectory of the movement of this system, a curve in the extended configuration space, is described by parameterized functions
t(x) and
q(x)={q_i(x)}, i∈[1,n]
where x is an abstract parameter changing from real value x=a to x=b with {t(a),q(a)} being the start and {t(b),q(b)} being the finish point of the movement.

(4) Given a group of continuous transformations of the extended configuration space parameterized by ε
t⟶t^(ε)=T(ε,t,q)
q⟶q^(ε)=Q(ε,t,q)
with ε=0 causing a transformation to be the identity transformation, that is t⁽⁰⁾=t and q⁽⁰⁾=q.
We assume, the transformation functions T(ε,t,q) and Q(ε,t,q) are sufficiently differentiable.

(5) This transformation of points (t,q)⟶(t^(ε),q^(ε))
induces the transformation of every trajectory
{t(x),q(x)}⟶{t^(ε)(x),q^(ε)(x)}
where x∈[a,b].
We further assume that the transformed trajectory
{t^(ε)(x),q^(ε)(x)}
belongs to the same class of physically admissible trajectories of a mechanical system defined by its properties and the laws of physics.

(6) Let's assume that the action functional of the movement of this mechanical system

Φ[t,q] =

t(b) ∫ t(a)

L(t,q,q')dt

is invariant under this induced transformation of trajectories as parameter of transformation ε is infinitesimally changing from zero.
This assumption can be expressed as
(d/dε)Φ[t^(ε),q^(ε)]|_ε=0 = 0
and we assume sufficient differentiability of the action functional by parameter ε.


Derivation of Noether Theorem

The problem with the above representation of the action functional is that not only an expression under an integral is transformed, but the limits of integration t(a) and t(b) change as well, which significantly complicates the analysis of the behavior of the action functional under ε-transformation of time and generalized coordinates.
The road to simplification is the parameterized representation of a trajectory {t(x),q(x)}.
Using this, we can replace
(a) dt = (dt/dx)·dx
(b) q' = dq/dt = (dq/dx)/(dt/dx)
(c) integration by t on [t(a),t(b)] can be now replaced with an integration by x - the parameter changing on a fixed segment [a,b].

Let's rewrite the action functional as the integral by x using abbreviations
dt/dx=t_x (so, dt=t_x·dx) and
dq/dx={dq_i/dx}={q_ix}=q_x
for brevity

Φ[t,q] =

b ∫ a

L(t,q,qx/tx)·tx·dx

At this point we would like to bring some time-space uniformity.
Since both time t and generalized space coordinates q={q_i} (i∈[1,n]) are all functions of one parameter x (x∈[a,b]) and all have equal standing as coordinates in an extended time-space configuration space, it makes sense to use a single letter
y={y_i} (i∈[0,n]) with
y₀=t and
y_i=q_i for all i∈[1,n]).

Also, we replace derivatives
dt/dx=t_x with dy₀/dx=y_0x
and
dq/dx={dq_i/dx}={q_ix}=q_x
for i≠0 with
dy/dx={dy_i/dx}={y_ix}=y_x.

So, a set of all derivatives {t_x,q_ix} can be written as
dy/dx={dy_i/dx}={y_ix}=y_x
where i∈[0,n].

Now we can simplify the formula for action functional by replacing the Lagrangian under the integration with a function that treats all time-space coordinates equally:

Φ[y] =

b ∫ a

𝓛(y,yx)·dx

where
y(x)={y_i(x)} (i∈[0,n]) signifies a set of all time-space coordinates parameterized by x∈[a,b], that is a trajectory in extended configuration space, with y₀(x)=t(x), and y_i(x)=q_i(x) for i≠0 and
y_x(x)={y_ix(x)} (i∈[0,n]) signifies a set of all derivatives of time-space coordinates by parameter x with y_0x(x)=t_x(x), and y_ix(x)=q_ix(x) for i≠0
and a new function 𝓛() is defined for i∈[0,n] as
𝓛(y,y_x) = 𝓛({y_i},{y_ix}) =
= L(t,q,q_x/t_x)·t_x =
= L(t,{q_i},{q_ix/t_x})·t_x

This representation of the same action functional is simpler because a new function 𝓛() under the integral symmetrically depends on n+1 functions {y_i(x)} (i∈[0,n]) that encompass t(x) and all {q_i(x)} (i∈[1,n]) functions of parameter x and n+1 derivatives of these functions by x, and the parameter x is not a subject of ε-transformation.

In addition, the limits of integration by x are from a to b which are constant and not affected by the ε-transformation.

The latter form of function 𝓛() under an integral allows to express the assumption about the invariance of the action functional under ε-transformations
y⟶y^(ε)
which means t⟶t^(ε), q⟶q^(ε)),
as
(d/dε)Φ[y^(ε)]|_ε=0 = 0
in a symmetrical way relative to all time-space coordinates in extended configuration space and use the known apparatus of Calculus to perform all the required operations.

Since our transformations are continuous, ε-transformations with infinitesimal ε are infinitesimal, that is the increments in coordinates
Δy^(ε)=y^(ε)−y
are infinitesimal as well.
At the same time, the ε-derivatives of the changing coordinates characterize the speed of their change by a transformation.
Expressions
ζ={ζ_i}={dy_i^(ε)/dε|_ε=0}
are called generators of the transformation.
We will use them below.

Also,
d/dε[dy^(ε)/dx]|_ε=0 =
= d/dx[dy^(ε)/dε]|_ε=0 = dζ/dx

Let's apply our assumption about invariance of the action functional under ε-transformation and equate the ε-derivative of the action functional at ε=0 to zero and do the calculations.
We'll abbreviate derivatives with subscriptors for brevity (like ζ_x for dζ/dx) and omit the |_ε=0 to shorten the formulas:
0 = (d/dε)Φ[y^(ε)] =

= (d/dε)

b ∫ a

𝓛(y(ε),yx(ε))·dx =

where y and y_x are group variables representing all time-space coordinates in extended configuration space {y_i} and {y_ix} with i∈[0,n].
We can change the order of differentiation by ε and integration by x because we assumed that the function under the integral is sufficiently smooth, so the convergence theorem holds.

=

b ∫ a

(d/dε)

𝓛(y(ε),yx(ε))·dx =

use the rules for differentiation of multi-variable functions, subscriptions to indicate the corresponding derivative, definition ζ=dy^(ε)/dε|_ε=0 and the rule for interchanging the differentiation by two different variables
dy_x^(ε)/dε = d/dε[dy^(ε)/dx] =
= d/dx[dy^(ε)/dε] = dζ/dx = ζ_x

=

b ∫ a

(𝓛y·dy(ε)/dε+𝓛yx·dyx(ε)/dε)·dx

=

b ∫ a

(𝓛y·ζ+𝓛yx·ζx)·dx

where group item 𝓛_y·ζ represents
a sum Σ_i𝓛_yi·ζ_i with i∈[0,n]
which in expanded form is
Σ_i∂𝓛/∂y_i·[dy_i^(ε)/dε|_ε=0]
and group item 𝓛_yx·ζ_x represents
a sum Σ_i(𝓛_yix·ζ_ix) with i∈[0,n]
which expands analogously with a subscript x indicating a derivative by x

We have derived with a fundamental identity

b ∫ a

[Σ(i𝓛yi·ζi)+Σi(𝓛yix·ζix)]·dx = 0

where summation by i is for i∈[0,n].

Recall the following rules for integration by parts.

(d/dx)[𝓛_yx·ζ] =
= (d𝓛_yx/dx)·ζ + 𝓛_yx·ζ_x
⇒

b ∫ a

(d/dx)[𝓛yx·ζ]·dx =

=

b ∫ a

(d𝓛yx/dx)·ζ·dx +

b ∫ a

𝓛yx·ζx·dx

⇒
[𝓛_yx·ζ]|_[a,b] =

=

b ∫ a

(d𝓛yx/dx)·ζ·dx +

b ∫ a

𝓛yx·ζx·dx

⇒

[𝓛yx·ζ]|[a,b] −

b ∫ a

(d𝓛yx/dx)·ζ·dx =

=

b ∫ a

𝓛yx·ζx·dx

The above expression for

b ∫ a

𝓛yx·ζx·dx

can be substituted into the fundamental identity presented above

b ∫ a

(𝓛y·ζ+𝓛yx·ζx)·dx = 0

where group item
𝓛_y·ζ represents Σ_i𝓛_yi·ζ_i

getting
0 = [𝓛_yx·ζ]|_[a,b] +

+

b ∫ a

[𝓛y − d/dx(𝓛yx)]·ζ·dx

Since the ε-transformation leaves the action functional invariant, it preserves the set of stationary trajectories of the action mapping one stationary trajectory onto itself (shift along a trajectory) or to another one (jump to another trajectory).
Since all physical trajectories are stationary for the action functional and, therefore, are the solutions of the Euler–Lagrange equations, the transformed trajectory, being stationary as well, also satisfies the Euler–Lagrange equation:
𝓛_y = ∂𝓛/∂y = d/dx(∂𝓛/∂y_x) =
= d𝓛_yx/dx
which nullifies an integral in the last identity.

Therefore,
[𝓛_yx·ζ]|_[a,b] = 0 or
𝓛_yx(b)·ζ(b) − 𝓛_yx(a)·ζ(a) = 0
where group parameter
𝓛_yx·ζ represents Σ_i𝓛_yix·ζ_i with the sum by i∈[0,n].

This means that the value of 𝓛_yx·ζ is the same at x=a and x=b.
This holds for any subinterval [a,b] of any physically admissable trajectory.
Since the endpoints can be chosen arbitrarily along the trajectory, the above expression is constant along the trajectory and
d/dx[𝓛_yx·ζ] = 0

Noether Theorem

Every continuous symmetry of the action functional (every ε-transformation of the extended configuration space that leaves the action functional invariant) is associated with a conserved quantity along physical trajectories.
In extended configuration space, the conserved quantity is
J = 𝓛_yx·ζ
or, in expanded by coordinates format,
J = Σ_i𝓛_yix·ζ_i with i∈[0,n]
Recall that
y={y_i} is a set of time-space coordinates with i∈[0,n];
y₀ is time t;
{y_i} are a set of generalized coordinates {q_i} with i∈[1,n];
ζ={ζ_i}={dy_i^(ε)/dε|_ε=0} is a set of generators for each time-space coordinate;
𝓛(y,y_x) = L(t,t_x,q,q_x/t_x)·t_x
where x in a subscript indicates a derivative of a corresponding function by parameter x:
t_x=dt/dx and
{q_ix}={dq_i/dx}

We have just proven that
J is a constant of motion and, therefore,
dJ/dx = 0

Noether's Theorem Math 1: UNIZOR.COM -> Physics+ 4 All -> Lagrangian -> Noether's Theorem Math 1: ε−Transformation

Notes to a video lecture on UNIZOR.COM

Noether Math 1:
ε−Transformation

In 1918 Emmy Noether rigorously proved that the conservation laws are consequences of certain continuous symmetries of time, space and other symmetries appearing in classical mechanics, field theory, quantum mechanics, general relativity and other parts of theoretical physics.

In a very simplified form she proved that, if the laws of physics are the same today and tomorrow, or here and there, then some physical characteristics of a mechanical system must remain invariant, that is unchanged.

Symmetry

Imagine a physicist performs an experiment in a box.
We then rotate the box. If the physicist repeats this experiment with exactly the same initial configurations and finds no difference in the results, we say the laws governing the experiment are rotationally symmetric.

Similarly, if there are no differences in experiments under spatial or temporal translation, then the laws describing the experiment are spatially or temporally invariant.
Spatial translation is when the physicist repeats the experiment at a different location.
Temporal translation is when the physicist repeats the same experiment after a certain waiting period.

The mechanical laws are represented by the equations of motion.
So, "the same results of an experiment" means that the laws (that is, the equations of motion) are the same (invariant) relative to a corresponding transformation.

In Lagrangian Mechanics, this idea of symmetry takes a precise formulation.

A symmetry is a continuous transformation of time t, space coordinates q={q_i}, or other dynamical variables under which the Lagrangian L describing the mechanical system remains unchanged or changes only by a total time derivative dF(t,q(t),q'(t))/dt of some smooth function F(t,q(t),q'(t)), where q'(t) signifies a time-derivative of a coordinate q, that is velocity.

A nuance related to a total time derivative of some smooth function F(t,q(t),q'(t)) might need an explanation.
If the Lagrangian will change by dF(t,q(t),q'(t))/dt, the action functional
Φ[L(t)] = ∫_ab[L(t,q,q')+dF/dt]dt
defined on a trajectory with endpoints q(a)=A and q(b)=B fixed will change by the boundary term F(b)−F(a).
Considering the end points of a trajectory are fixed as initial conditions, the extremals will be preserved, leading to the same Euler–Lagrange equations and the same physical trajectories.

Let's rigorously define the symmetry as a continuous transformation we are talking about.
Consider a configuration space of n degrees of freedom with generalized coordinates q={q₁,...,q_n} and time parameter t.

Consider a family of transformation functions parameterized by variable ε
t^(ε)=T(ε,t,q₁,...,q_n)=T(ε,t,q)
and (for each i∈[1,n])
q_i^(ε)=Q_i(ε,t,q₁,...,q_n)=Q_i(ε,t,q)
define a transformations from original space-time coordinates {t,q} to the new ones {t^(ε),q^(ε)} if t⁽⁰⁾=t and q_i⁽⁰⁾=q_i for each i∈[1,n].

Sometimes for brevity we will use
q^(ε)=Q(ε,t,q)
implying q_i^(ε) on the left and Q_i(ε,t,q) on the right side for all i from 1 to n.

Functions T(ε,t,q) and Q_i(ε,t,q) are assumed to be continuous with respect to the first argument ε for any t and q.

Usually, we will require even a stronger requirements for transformational functions T(ε,t,q) and Q_i(ε,t,q) to uniformly converge to T(0,t,q) and Q_i(0,t,q) as ε→0.

In addition, to define it more rigorously, we require that all transformations make up a group (more precisely, the Lie group) in a sense that transformations can be combined:
{t,q}→{t^(α),q^(α)}→{t^(β),q^(β)}
should be equivalent to
{t,q}→{t^(α+β),q^(α+β)}

In particular, any (ε)-transformation is reversible using (−ε)-transformation because of the additivity mentioned above:
(ε)+(−ε)=(0).
and, as we stated above,
t = t⁽⁰⁾ = T(0,t,q)
q = q⁽⁰⁾ = Q(0,t,q)

Functions T(ε,t,q) and Q(ε,t,q) define the parameterized by ε transformation of time and position.
Since, ultimately, we need to check the Lagrangian for invariance to transformations, we also need to know how velocities are changing with transformation because Lagrangian depends on them.

The derivation of a formula that defines this change is rather complex and the rest of this lecture is dedicated precisely to dealing with transformation of velocity.

If transformation does not involve time then the transformed velocity is just a time-derivative of the transformed position
q_i'^(ε) = dq_i^(ε)/dt = q_i^(ε)'
Notice the order of ε-transformation and apostrophe that indicates a time derivative.
Thus, in this particular case
q_i'^(ε) = dQ(ε,t,q)/dt

The complication in our case is, not only coordinates are changing, but time as well.
So, the correct definition of velocity transformation is
q_i'^(ε) = dq_i^(ε)/dt^(ε)

Let's start with some simple manipulation.
dq_i^(ε)/dt^(ε) =
= [dq_i^(ε)/dt] / [dt^(ε)/dt] =
= [dQ_i(ε,t,q)/dt] / [dT(ε,t,q)/dt]

The numerator, by chain rule, equals to
∂Q_i/∂t+Σ_k(∂Q_i/∂q_k)·q_k'
The denominator, by chain rule, equals to
∂T/∂t+Σ_k(∂T/∂q_k)·q_k'

Therefore,

qi'(ε) =

∂Qi /∂t+Σk(∂Qi /∂qk)·qk' ∂T/∂t+Σk(∂T/∂qk)·qk'

where
Q_i = Q_i(ε,t,q) = q_i^(ε)
T = T(ε,t,q) = t^(ε)

As ε→0, we can represent ε-transformed values {t^(ε),q_i^(ε)} in terms of initial values {t⁽⁰⁾=t,q_i⁽⁰⁾=q_i} and infinitesimal increments like in the Taylor's formula:
t^(ε)= t + ε·[∂t^(ε)/∂ε|_ε=0] +o(ε)
q_i^(ε)= q_i + ε·[∂q_i^(ε)/∂ε|_ε=0] +o(ε)

An expression
τ(t,q) = ∂t^(ε)/∂ε|_ε=0 =
= ∂T(ε,t,q)/∂ε|_ε=0
is a derivative of the time transformation function by parameter of transformation ε at the initial point of transformation ε=0 and is called the infinitesimal generator of time transformation.
So, in terms of this generator,
t^(ε)= t + ε·τ(t,q) +o(ε)

An expression
ξ_i(t,q) = ∂q_i^(ε)/∂ε|_ε=0 =
= ∂Q_i(ε,t,q)/∂ε|_ε=0
is a derivative of the space transformation function by parameter of transformation ε at the initial point of transformation ε=0 and is called the infinitesimal generator of space transformation.
So, in terms of this generator,
q_i^(ε)= q_i + ε·ξ_i(t,q) +o(ε)

Using the above representations, we can evaluate the terms of the formula for q_i'^(ε).

∂T(ε,t,q)/∂t =
= ∂t^(ε)/∂t =
= ∂[t+ε·τ(t,q)+o(ε)]/∂t
derivative of t by t is 1;
ε is an independent of t parameter
= 1 + ε·∂τ(t,q)/∂t

∂T(ε,t,q)/∂q_k =
= ∂t^(ε)/∂q_k =
= ∂[t+ε·τ(t,q)+o(ε)]/∂q_k
derivative of t by q_k is 0;
ε is an independent of q_k parameter
= ε·∂τ(t,q)/∂q_k

∂Q_i (ε,t,q)/∂t =
= ∂q_i^(ε)/∂t =
= ∂[q_i+ε·ξ_i(t,q)+o(ε)]/∂t
q_i is the initial position in space (before transformation is applied) and, when used in partial derivatives of function that depends on it and other parameters, like time t, is treated as independent of other parameters, thus derivative ∂q_i/dt=0;
ε is an independent of t parameter
= ε·∂ξ_i(t,q)/∂t

∂Q_i (ε,t,q)/∂q_k =
= ∂q_i^(ε)/∂q_k =
= ∂[q_i+ε·ξ_i(t,q)+o(ε)]/∂q_k
derivative of q_i by q_k is 1 for i=k and 0 otherwise ⇒ (using math symbol δ) derivative is equal to δ_ik;
ε is an independent of t parameter
= δ_ik + ε·∂ξ_i(t,q)/∂q_k

We are ready to evaluate numerator and denominator of the above expression for q_i'^(ε).

Numerator =
= ε·∂ξ_i(t,q)/∂t +
+ Σ_k[δ_ik+ε·∂ξ_i(t,q)/∂q_k]·q_k' =
= ε·∂ξ_i(t,q)/∂t +
+ q_i' + ε·Σ_k[∂ξ_i(t,q)/∂q_k]·q_k' =
= q_i' +
+ε·{∂ξ_i(t,q)/∂t+ Σ_k[∂ξ_i(t,q)/∂q_k]·q_k'}

Denominator =
= 1 + ε·∂τ(t,q)/∂t +
+ Σ_k[ε·∂τ(t,q)/∂q_k]·q_k' =
= 1 +
+ε·{∂τ(t,q)/∂t+ Σ_k[∂τ(t,q)/∂q_k]·q_k'}

Now the formula for a transformation of velocity looks like

qi'(ε) =

qi' + ε·A 1 + ε·B

where
A = ∂ξ_i(t,q)/∂t +
+ Σ_k[∂ξ_i(t,q)/∂q_k]·q_k'
and
B = ∂τ(t,q)/∂t +
+ Σ_k[∂τ(t,q)/∂q_k]·q_k'

Let's not forget our goal - to establish how the velocity is transformed by ε-transformation, which implies differentiation of the transformed velocity by a transformation parameter ε at point ε=0 to find instantaneous velocity rate of change at the initial position {t,q}.

If the above expression would be of the form
q_i'^(ε) − q_i' = ε·X + o(ε)
then we could say that
lim_ε→0 (q_i'^(ε) − q_i')/ε =
= d(q_i'^(ε))/dε|_ε=0 = X

To achieve this goal, we will convert our formula

qi'(ε) =

qi' + ε·A 1 + ε·B

into more desirable form using this easily verifiable identity
(1+ε·B)·[q_i' + ε·(A−B·q_i')] =
= q_i' + ε·A + o(ε)

Dividing both parts by (1+ε·B) will result in
q_i' + ε·(A−B·q_i') =
= [q_i' + ε·A + o(ε)] / (1+ε·B) =
= [q_i' + ε·A] / (1+ε·B) + o(ε) =
= q_i'^(ε) + o(ε)
Therefore,
q_i'^(ε) = q_i' + ε·(A−B·q_i') + o(ε)

Hence, the instantaneous rate of change of velocity at point {t,q} is
d(q_i'^(ε))/dε|_ε=0 = A−B·q_i' =
= {∂ξ_i(t,q)/∂t +
+ Σ_k[∂ξ_i(t,q)/∂q_k]·q_k'} −
− q_i'·{∂τ(t,q)/∂t +
+ Σ_k[∂τ(t,q)/∂q_k]·q_k'}

In the above expression we can recognize two full time derivatives
dξ_i(t,q)/dt = ∂ξ_i(t,q)/∂t +
+ Σ_k[∂ξ_i(t,q)/∂q_k]·q_k'
and
dτ(t,q)/dt = ∂τ(t,q)/∂t +
+ Σ_k[∂τ(t,q)/∂q_k]·q_k'
which makes our formula for a rate of change of velocity much simpler
d(q_i'^(ε))/dε|_ε=0 =
= dξ_i(t,q)/dt − q_i'·dτ(t,q)/dt
where
ξ_i(t,q)=∂q_i^(ε)/∂ε|_ε=0
represents the rate of change of position by infinitesimal ε-transformation, whose time derivative (the first term in the above formula) is the rate of change of velocity without taking into consideration the transformation of time
and
τ(t,q)=∂t^(ε)/∂ε|_ε=0
whose time derivative multiplied by q_i' (the second term in the above formula) represents an adjustment caused by ε-transformation of time.

Noether's Theorem and Angular Momentum Conservation:UNIZOR.COM -> Physics+ 4 All -> Lagrangian - Noether l = m·r⨯v const

Notes to a video lecture on UNIZOR.COM

Noether's Theorem
Angular Momentum Conservation

History

In 1918, German mathematician Emmy Noether published the proof of a very important theorem named after her.
Albert Einstein and other physicists considered her one of the most significant mathematicians of her time.
Noether's theorem has been called one of the most important mathematical theorems guiding the development of modern physics.
This lecture is about a particular case of Noether's theorem as it applies to classical mechanics using the Lagrangian approach.

Background

Recall these important items from the prior lectures of this course Physics+ 4 All on UNIZOR.COM.

1. Configuration space with generalized coordinates q(t)={q_i(t)}, where i∈[1,n] and n is the number of degrees of freedom.
Generalized velocities (time derivatives of generalized coordinates) q'(t)={q_i'(t)}.

2. Lagrangian
L(q,q') = T(q,q') − U(q)
- the difference between the total kinetic T and total potential U energies of a system.
Optionally, the Lagrangian might be explicitly dependent on time, but in this and subsequent lectures we assume that energies and, therefore, the Lagrangian do not explicitly depend on time, but only on positions and velocities. This makes the Lagrangian depend on time implicitly through the motion q(t).
Systems with the Lagrangian explicitly depending on time (like in case of variable gravitational field, driven oscillations or time-dependent electromagnetic field) are not considered here.

3. Action functional
Φ[L(q(t),q'(t))_t∈[t1,t2]] =
= ∫_[t1,t2] L(q(t),q'(t))dt

4. Euler-Lagrange equations
These equations are established for each generalized coordinate
d/dt ∂L/∂q_i' = ∂L/∂q_i (i∈[1,n])
These are differential equations with unknown position functions q(t).
Their solutions are the extremals of the action functional above and, at the same time, the only candidates for real physical trajectories of a mechanical system in its configuration space with generalized coordinates.

5. Generalized Momentum
By definition, the generalized momentum is a set of partial derivatives of the Lagrangian by generalized velocities:
p_i = ∂L/∂q_i'
Previous lecture was dedicated to the proof of the Momentum Conservation law for each component p_i, for which the Lagrangian is invariant (symmetrical) under a translation (shift) of the corresponding generalized cyclic coordinate q_i→q_i+ε:
∂L/∂q_i=0 ⇒ p_i=const
This momentum conservation law is essential for this lecture about angular momentum.

Angular Momentum Conservation

Our purpose is to prove a particular case of Noether's theorem about the law of conservation of angular momentum of a mechanical system.

But instead of directly proving this for a particular case of rotational symmetry, we will use the already proven in the previous lecture result for a transformation of generalized coordinates.

Recall the general theorem proven in the previous lecture:
If the Lagrangian
L(q₁,...,q_n,q₁',...,q_n',t)
is invariant under translation of q_k by infinitesimal value ε
q_k → q_k + ε
then the k^th coordinate of the generalized momentum
p_k = ∂L/∂q_k'
is conserved.

Let's use this theorem for a special case of a conservative planar system (two-dimensional system on the Euclidean plane) moving in a central field (the one with a potential depending only on a distance from a central point, like gravitational field) with polar coordinates:
radius r that we will interpret as a generalized coordinate q₁ and
angle θ that we will interpret as a generalized coordinate q₂.

A rotation of a physical system in the plane corresponds to a translation of the angular coordinate
θ → θ + ε

Let's express the Lagrangian L(r,θ,r',θ') in terms of these variables.
L = T − U
is the difference between kinetic and potential energies.

Kinetic energy
T(r,θ,r',θ') = ½m·v²
depends on the mass m and the magnitude of its speed v that in polar coordinates can be calculated using Cartesian coordinates (x,y) as follows:
x = r·cos(θ)
y = r·sin(θ)
x' = r'·cos(θ)−r·sin(θ)·θ'
y' = r'·sin(θ)+r·cos(θ)·θ'
v² = (x')² + (y')²
(x')² = (r')²·cos²(θ) −
− 2r'·cos(θ)·r·sin(θ)·θ' +
+ r²·sin²(θ)·(θ')²
(y')² = (r')²·sin²(θ) +
+ 2r'·sin(θ)·r·cos(θ)·θ' +
+ r²·cos²(θ)·(θ')²

After cancelling plus and minus of the middle term in a sum of two last expressions above and taking into consideration that
sin²(θ) + cos²(θ) = 1
we obtain
v² = (x')² + (y')² =
= (r')² + r²·(θ')²

Therefore, kinetic energy T is
T = ½m·[(r')² + r²·(θ')²]
which is independent of angle θ and invariant to its translation θ→θ+ε.

Potential energy U of a central field is, as we mentioned above, depends only on the distance from the central point. If our polar system of coordinates has an origin in that point, we can express the potential energy as U(r) - also independent of angle θ and, therefore, invariant to its translation.

A rotation of a physical system in the plane corresponds to a translation of the angular coordinate θ→θ+ε, but the Lagrangian of a system in a central field is invariant under the translation θ→θ+ε, that is angle θ is cyclic and
∂L/∂θ = 0

From this and the generalized Momentum Conservation law proven in the previous lecture follows that the corresponding momentum
p_θ = ∂L/∂θ'
is conserved.

Let's express the momentum p_θ in term of polar coordinates, taking into account that potential energy U does not depend on velocities.
p_θ = ∂L/∂θ' = ∂(T−U)/∂θ' =
= ∂T/∂θ' =
= ∂/∂θ' {½m·[(r')² + r²·(θ')²]} =
= m·r²·θ'

Thus,
p_θ = m·r²·θ'
is a constant of motion of a conservative planar system in a central field.

The expression m·r²·θ' is exactly the magnitude of the angular momentum vector of a point-mass moving about the origin of polar coordinates (see Note 1 below), which in classical mechanics is defined as the mass multiplied by a vector product of radius and velocity vectors and directed perpendicular to a plane of rotation
𝓁 = m·r⨯v

Thus the conserved generalized momentum corresponding to rotational symmetry is the angular momentum.

As a conclusion for this and the previous lecture, the translational symmetry leads to conservation of linear momentum, while rotational symmetry leads to conservation of angular momentum
_________
Note 1
The magnitude of the angular momentum vector can be calculated from its definition
𝓁 = m·r⨯v
by expressing vectors in their Cartesian coordinate form, directing the X-axis along vector r, Y-axis to be perpendicular to it within a plane of motion and Z-axis to be perpendicular to other two axes and, therefore, perpendicular to an entire plane of motion.
Then
r = (r,0,0)
v = (r',r·θ',0)
Their vector product is the determinant of a matrix with i, j and k being unit vectors along the corresponding axes

i	j	k
r	0	0
r'	r·θ'	0

This determinat equals to
i·0 + j·0 + k·r·r·θ'
Which means that the angular momentum vector is perpendicular to the plane of motion and its magnitude equals to r²·θ'.