Math4Teens - Path Integrals

Notes to a video lecture on //www.unizor.com

Path/Line/Curvilinear Integral

Path Integral of Scalar Function

Recall the definition of a definite integral (Riemann integral) of real function F(x) defined on interval [a,b]:
I = ∫_a^bF(x) dx = lim Σ_i∈[1,N] F(x_i)·Δx_i
where {x_i} is partitioning of segment [a,b] into N parts,
and we assume that N→∞, while the maximum width of intervals Δx_i=x_i−x_i−1 converges to zero.

This definition pertains to a function F(x) defined on a straight segment [a,b] and traditionally interpreted as an area under a graph of a function on Cartesian XY plane restricted on the left and right by x=a and x=b correspondingly.

Consider a function that is defined not on a straight segment of real numbers [a,b], but on some path (curve) in space we call γ.
Let's assume for definitiveness that our path is in three-dimensional Cartesian space and is defined by a set of three smooth functions
γ={x(s),y(s),z(s)}
of some real parameter s∈[0,S].

Generally speaking, parameterization of a curve in N-dimensional space is just a set of N real functions x_i(s) (i∈[1,N]) of real argument s such that
(a) all these functions have the same domain S - a finite or an infinite interval of real numbers;
(b) for each value of parameter s∈S a set of functions x_i(s) used as a set of N coordinates correspond to some point on a curve;
(c) the parameterization covers the entire curve: for each point on a curve there is at least one value of parameter s within domain S such that a set of functions x_i(s) represent the coordinates of this point.
The parameter s itself does not have any particular meaning. Depending on the problem, it can represent time, distance along the curve, arc length, or simply be an arbitrary convenient parameter.

For each position on this path a function F(x,y,z) is defined.

The path integral of this function F(x,y,z) along path γ is defined as
∫_γF(x,y,z)·dγ = ∫_[0,S]F(x(s),y(s),z(s))·ds
The expression above defines a new concept of path integral by connecting it to a regular integral based on a parameterization of a curved path.
Thus defined path integral is alternatively named line integral or curvilinear integral.

Obviously, this definition can be expanded to any N-dimensional space by replacing three-dimensional path γ={x(s),y(s),z(s)} with N-dimensional γ={x₁(s),x₂(s),...,x_N(s)}

Example
Consider a path γ in some mountainous area and a task of measuring the average barometric pressure along this path.
This is a typical problem with the path integral as a solution.
While path is three-dimensional, we can choose a length of a road from the beginning of the path s as a parameter and, for each value of this parameter (that is, for each point on a mountain path {x(s),y(s),z(s)}) we measure the barometric pressure H(s).
After integration of function H(s) by parameter s we can divide the result by the total length of a path to get the average barometric pressure.
So the most important issue in the path integral of a scalar function is to properly choose the parameter s to define the curved path as a function of a single parameter, thereby reducing the function defined for each point on a three-dimensional path to the function of a single parameter and reducing the path integral to a regular definite integral by a single variable.

Path Integral of Vector Function

Recall a concept of work in physics.
In a simple case of a straight road and a car moving along it from point A to point B, whose engine pulls the car with constant force F directed along the road, the work is a product of a force by a distance
W = F·d
where d is the length of segment AB.

In a more complicated case of a variable but still directed along the straight road force F=F(s) whose magnitude depends on position s on the road from A to B, the work can be approximated by dividing the segment AB into many small intervals, assuming that within each interval the force is practically the same, calculate the work on each interval as above, sum the work on all intervals and start increasing the number of intervals while decreasing their size.
In the limit, when the length of all intervals is infinitesimal, while their number increases to infinity, this calculation will result in an exact amount of work of an engine on a road AB expressed as an integral
W = ∫_[0,d]F(s)·ds

Consider now a case of a smooth but curved road from A to B with the force still directed along the road (that is, the direction of the force is always tangential to a road).
This case is no different than the above case where we introduced a path integral.
So, the work of an engine of a car moving along a curved road is just another example of application of path integral to physics.

The work evaluation in cases when the force is not directed along a path of a moving object, however, requires some other technique.
Simple algebraic product of the magnitude of force F(s) by ds and integrating by s would not produce the right result in these cases.

Let's generalize the definition of a path integral to accommodate the calculation of work performed by the force F(s) being a general vector-function not necessarily directed along the path of motion.

To come up with some meaningful formula, let's introduce a system of Cartesian X,Y,Z coordinates on a plane and define the curve by some vector-function r(s)={x(s),y(s),z(s)} parameterized by variable s∈[0,S] with r(0)=A and r(S)=B.

The force, as a vector, would be represented by a vector-function F(r) defined for each point on a curve by its three components
F(r) = {F_x(x,y,z),F_y(x,y,z),F_z(x,y,z)}

In physics the work performed by a constant vector of force F in case it's not colinear with a straight segment of movement s is defined as a product of a projection of the vector of force onto a direction of movement F_proj and the length |s| of the movement.

According to this rule for calculating the work, for the vector of force F and a segment s along which an object moves the work of this force is calculated as
W = |F|·|s|·cos(φ)
where φ is an angle between direction of the force and direction of the segment.

This formula can be written as a scalar (dot) product of two vectors:
W = F·s

Let's return to a task at hand - calculate the work performed by a generally directed force along a curved path.
Dividing, as before, curve AB into n small almost straight intervals, we see that on each interval the force is not directed along it, but makes some angle.
Therefore, for a small interval Δs_i and force F_i acting within this small interval the work performed by the force is
ΔW_i = F_i · Δs_i
where dot between two vectors means scalar product.

The total work for all n intervals is
W = Σ_i∈[1,n]ΔW_i
As the number of intervals increases to infinity and the size of each decreases to zero, this sum tens to some limit. This limit is called the path integral of vector F along path AB:
∫_AB[F(r)·dr]
where dot between two vectors means scalar product.

This definition can be rephrased by using the coordinate representation of vectors.
F(r) = F(x,y,z) =
= {F_x(x,y,z),F_y(x,y,z),F_z(x,y,z)}
dr = {dx,dy,dz}
Hence, infinitesimal amount of work our force performs on infinitesimal interval of a curved path equals to
dW = F(r)·dr =
= F_x(x,y,z)·dx+F_y(x,y,z)·dy+F_z(x,y,z)·dz

Considering our curve is defined parametrically as
r = r(s) = {x(s),y(s),z(s)}
we can define the path integral of a vector function above as
∫_AB[F(r)·dr] =
= ∫_[0,S]F_x(x(s),y(s),z(s))·dx(s) +
+ ∫_[0,S]F_y(x(s),y(s),z(s))·dy(s) +
+ ∫_[0,S]F_z(x(s),y(s),z(s))·dz(s) =
= ∫_[0,S]F_x(x(s),y(s),z(s))·(dx(s)/ds)·ds +
+ ∫_[0,S]F_y(x(s),y(s),z(s))·(dy(s)/ds)·ds +
+ ∫_[0,S]F_z(x(s),y(s),z(s))·(dz(s)/ds)·ds
thus reducing a path integral to a regular definite integral.

To bring our problem close to physics, we can use time t as a parameter instead of abstract s, having position of a moving object to be a function of time
x = x(t)
y = y(t)
z = z(t)
In this case dx(t)/dt is an X-component of the velocity of a moving object that we can denote as ẋ(t) and similar for other coordinates.

Using time as a parameter that defines the position of a moving object, the path integral that calculates the work would look like
∫_AB[F(r)·dr] =
= ∫_[0,S]F_x(x(t),y(t),z(t))·ẋ(t)·dt +
+ ∫_[0,S]F_y(x(t),y(t),z(t))·ẏ(t)·dt +
+ ∫_[0,S]F_z(x(t),y(t),z(t))·ż(t)·dt

Either way, the calculation of a path integral of a vector-function have been reduced to regular integration.

Math4Teens - Path Integral Properties

Notes to a video lecture on //www.unizor.com

Path Integral Properties

The properties of path integrals below follow from the definition of path integral in terms of regular integral using the parameterization of the path
γ = γ(s) = {x(s),y(s),z(s)} for s∈[0,S]
and corresponding properties of regular integrals.

Path Integrals Properties
of Scalar Functions

1. ∫_γ [F(x,y,z)+G(x,y,z)] dγ =
= ∫_γ F(x,y,z) dγ + ∫_γ G(x,y,z) dγ

With parameterization of path γ functions F and G become functions of one argument s
F(x(s),y(s),z(s)) = f(s)
G(x(s),y(s),z(s)) = g(s)
Hence,
∫_γ[F(x,y,z)+G(x,y,z)] dγ =
= ∫_[0,S][f(s)+g(s)] ds
and we can refer to the additive property of the regular integrals.


2. ∫_γ c·F(x,y,z) dγ = c·∫_γ F(x,y,z) dγ
for the same reasons as above.


3. Going along a path in an opposite direction changes the sign of an integral.
Indeed, assume that path γ is from A to B and is parameterized by s∈[0,S], which implies
γ(0)=A, γ(S)=B
Then
∫_AB F(x,y,z) dγ = ∫_[0,S]F(x(s),y(s),z(s))ds
∫_BA F(x,y,z) dγ = ∫_[S,0]F(x(s),y(s),z(s))ds
and, by properties of regular definite integrals, the latter is opposite in sign to the former.

4. Assume, path γ from A to B is divided by point M into two parts:
α from A to M and
β from M to B.
So, we can say γ=α+β.
Then we can state the following additive property of path integrals:
∫_γ F(x,y,z) dγ =
= ∫_α F(x,y,z) dα + ∫_β F(x,y,z) dβ
Here is the proof of this statement.
Consider the parameterization of path γ={x(s),y(s),z(s)}, where s∈[0,S].
Using this, point A corresponds to s=0, point B corresponds to s=S.
Assume, point M corresponds to s=s_M.
This allows to parameterize both parts of an original path:
α={x(s),y(s),z(s)}, where s∈[0,s_M]
β={x(s),y(s),z(s)}, where s∈[s_M,S]
Then, by definition of a path integral,
∫_γ F(x,y,z) dγ=∫_[0,S]F(x(s),y(s),z(s))ds
∫_α F(x,y,z) dα=∫_[0,sM]F(x(s),y(s),z(s))ds
∫_β F(x,y,z) dβ=∫_[sM,S]F(x(s),y(s),z(s))ds
From these definitions and the corresponding additive property of regular integrals follow the additive property of path integrals.


Path Integrals Properties
of Vector Functions

5. In physics we sometimes deal with gravitational, electrostatic and some other types of forces that have an important property called conservative.
Conservative force F={F_x,F_y,F_z}, as a vector, can be represented as partial derivatives of some scalar function f(x,y,z) by corresponding coordinates, which means:
F_x(x,y,z) = ∂f(x,y,z)/∂x
F_y(x,y,z) = ∂f(x,y,z)/∂y
F_z(x,y,z) = ∂f(x,y,z)/∂z
In physics concepts of potential and potential energy are based on such function.
Let's prove that path integral of a vector-function that represents a conservative force depends only on the endpoints of a path and does not depend on path itself between these endpoints.
In other words, no matter which path we choose between two fixed endpoints, path integral along it will be the same.
Given two endpoints in space A and B and a parametrically defined path AB that depends on parameter s∈[0,S]
r = r(s) = {x(s),y(s),z(s)}
with A={x(0),y(0),z(0)}
and B={x(S),y(S),z(S)}.
Our task is to integrate a vector-function that represents a conservative force F={F_x,F_y,F_z} along path AB.
This integral equals to
∫_AB[F(r)·dr] =
under integral we have a scalar (dot) product of two vectors
F={F_x,F_y,F_z} and dr={dx,dy,dz}
= ∫_[0,S][F_x(x(s),y(s),z(s))·dx(s) +
+ F_y(x(s),y(s),z(s))·dy(s) +
+ F_z(x(s),y(s),z(s))·dz(s)]
Assuming our vector-function represents a conservative force, that is
F_x = ∂f/∂x
F_y = ∂f/∂y
F_z = ∂f/∂z
we can rewrite this path integral as
∫_AB[F(r)·dr] =
= ∫_[0,S]{[∂f(x(s),y(s),z(s))/∂x]·dx(s) +
+ [∂f(x(s),y(s),z(s))/∂y]·dy(s) +
+ [∂f(x(s),y(s),z(s))/∂z·dz(s)]}
Notice that the expression in {...} is a full differential of function f(x,y,z).
Therefore,
∫_AB[F(r)·dr] = ∫_[0,S]df(x(s),y(s),z(s))
According to the rules of integration, this equals to
f(x(S),y(S),z(S))−f(x(0),y(0),z(0))
As we see, the path integral of a conservative vector-function depends only on the values of this scalar function f(x) at endpoints of a path.


6. If a vector-function F represents a conservative force, then the path integral along any closed loop path is zero.
Assume, path γ is a closed loop and is parameterized by s∈[0,S], which implied that A=s(0)=s(S)=B.
According to property #5 above, the values of a path integral along this path is
f(x(S),y(S),z(S))−f(x(0),y(0),z(0))
But points A=s(0)={x(0),y(0),z(0)} and B=s(S)={x(S),y(S),z(S)} coincide, which makes the expression above to be zero.
Alternative proof might be as follows. Let's choose any two points on a closed loop of this path, M and N.
Consider two different ways of moving from M to N along two different routes within path γ that we conditionally call "left" and "right".
Since path integral from M to N for conservative vector-function F is independent of path chosen, but only on the endpoints,
∫_[MleftN]F(γ)dγ = ∫_[MrightN]F(γ)dγ
Full loop along path γ can be represented as a sum of MleftN+NrightM.
But changing the direction of integration from MrightN to NrightM changes the sign of a path integral. ∫_[NrightM]F(γ)dγ = −∫_[MrightN]F(γ)dγ
Full path integral is equal to
∫_[MleftN]F(γ)dγ + ∫_[NrightM]F(γ)dγ =
= ∫_[MleftN]F(γ)dγ −∫_[MrightN]F(γ)dγ
As we see, we are subtracting a value from an equal value.
Therefore, integration along a full loop MleftNrightM results in zero.

Physics+ 4 All - Lagrangian for Non-Conservative Systems

Notes to a video lecture on UNIZOR.COM

Lagrangian for
Nonconservative System

All the advantages of Lagrangian Mechanics are best visible in its approach to the analysis of conservative mechanical system, where the forces acting on system's components are conservative and, as such, can be represented as gradients of a potential - a scalar field with values at each point in space depending only on the location.

Attempts to extend this approach to nonconservative systems encounter technical difficulties but are nevertheless worthwhile.
This lecture exemplifies this case.

Assume a simple system with forces depending not only on position but on velocity of an object as well.
Let a stone of mass m fall vertically into the water with no initial velocity.
The force of gravity will pull it down, but the water exerts a resistive force opposite to the direction of motion.
Assume the resistive force is proportional to the stone's speed v. Then the drag force has magnitude k·v, where k is the coefficient of resistance, and it always acts opposite to the direction of motion.


Our first task is to analyze the movement using Newtonian Second law.
Newton's Second Law produces the following differential equation of motion m·ẍ(t) = m·g − k·ẋ(t)
that can be simplified to a standard linear form
ẍ(t) + (k/m)·ẋ(t) − g = 0

The solution to this equation with initial conditions
x(0)=0 and x'(0)=0
is
x(t) =
= (mg/k)·t+(m²g/k²)·(e^−kt/m−1)
Condition x(0)=0 is satisfied.
The speed (time-derivative) is
x'(t) = (mg/k)·(1−e^−kt/m)
Condition x'(0)=0 is satisfied.
The acceleration is
x"(t) = g·e^−kt/m
Newton's differential equation of motion above is satisfied.

Two important conclusions:
1. Since the positive direction is downward, the acceleration remains positive but decreases exponentially toward zero.
2. Therefore, the speed is always increasing but asymptotically approaches its maximum mg/k.

Now we will try to approach this task from the Lagrangian viewpoint.
A simple Lagrangian in terms of a difference between kinetic and potential energy is not a good choice since the force of resistance to a movement is dependent on velocity and, therefore, is not conservative.

However, one possible Lagrangian, first proposed by Bateman and later developed by Caldirola and Kanai, is obtained as follows.
Consider again Newton's Second law equation for our system
ẍ(t) + (k/m)·ẋ(t) − g = 0
Suppose there exists a Lagrangian L(x(t),ẋ(t),t) suitable for this nonconservative system.
Then its Euler Lagrange equation
d/dt [∂L/∂ẋ] = ∂L/∂x
must be equivalent to Newton's Second equation above.
In other words, both equations must be transformable into each other.

1. Multiply Newton's Second law equation by e^kt/m.
e^kt/m[ẍ(t)+(k/m)·ẋ(t)−g] = 0
or
e^kt/m[ẍ(t)+(k/m)·ẋ(t)] = g·e^kt/m
2. Notice that
e^kt/m[ẍ(t)+(k/m)·ẋ(t)] =
= d/dt [e^kt/m·ẋ(t)]
3. Notice that an expression in [...] on the right side of the equation above can be written as a derivative by ẋ
e^kt/m·ẋ = ∂/∂ẋ[e^kt/m·½ẋ²]
4. Therefore, we have an equation
d/dt ∂/∂ẋ[e^kt/m·½ẋ²] = g·e^kt/m
5. The right side of the equation in item 4 equals to
∂/∂x [g·x·e^kt/m]
6. Our transformed Newton's Second law equation looks now as
d/dt ∂/∂ẋ[e^kt/m·½ẋ²] =
=∂/∂x [g·x·e^kt/m]
7. Adding to the expression in [...] on the left side of the equation any function of x won't change the partial derivative by ẋ. Adding to the expression in [...] on the right side of the equation any function of ẋ won't change the partial derivative by x.
Hence, we may replace both bracketed expressions by the single function
e^kt/m·½ẋ² + g·x·e^kt/m
8. Multiplying the function by the constant m and writing it in a more compact form does not change the Euler-Lagrange equation.
Hence, both expressions in [...] can be replaced by a function
L = e^kt/m·(½m·ẋ² + mg·x)
and we have the Euler-Lagrange equation
d/dt ∂L/∂ẋ = ∂L/∂x

Surprisingly or not, the Lagrangian in this case differs from the Lagrangian for a conservative system only by a multiplier e^kt/m since
½m·ẋ² is kinetic energy T of a falling stone,
−mg·x is its potential energy U and
½m·ẋ² + mg·x = T − U
and the Lagrangian for this nonconservative system is
L = e^kt/m·(T−U)

Unlike the Lagrangian of a conservative system, this Lagrangian depends explicitly on time through the exponential factor e^kt/m.
This explicit dependence reflects the continuous loss of mechanical energy caused by the resistive force. Therefore, although the motion can still be described by Euler-Lagrange equations, many of the elegant conservation properties of conservative systems no longer hold.

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