Sunday, June 21, 2026

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 qi(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 {qi'(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)={qi(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
tt(ε)=T(ε,t,q)
qq(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=tx (so, dt=tx·dx) and
dq/dx={dqi/dx}={qix}=qx
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={qi} (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={yi} (i∈[0,n]) with
y0=t and
yi=qi for all i∈[1,n]).

Also, we replace derivatives
dt/dx=tx with dy0/dx=y0x
and
dq/dx={dqi/dx}={qix}=qx
for i≠0 with
dy/dx={dyi/dx}={yix}=yx.

So, a set of all derivatives {tx,qix} can be written as
dy/dx={dyi/dx}={yix}=yx
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)={yi(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 yi(x)=qi(x) for i0 and
yx(x)={yix(x)} (i∈[0,n]) signifies a set of all derivatives of time-space coordinates by parameter x with y0x(x)=tx(x), and yix(x)=qix(x) for i0
and a new function 𝓛() is defined for i∈[0,n] as
𝓛(y,yx) = 𝓛({yi},{yix}) =
= L(t,q,qx/tx)·tx =
= L(t,
{qi},{qix/tx})·tx

This representation of the same action functional is simpler because a new function 𝓛() under the integral symmetrically depends on n+1 functions {yi(x)} (i∈[0,n]) that encompass t(x) and all {qi(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={ζi}={dyi(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 ζx 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 yx are group variables representing all time-space coordinates in extended configuration space {yi} and {yix} 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
dyx(ε)/dε = d/dε[dy(ε)/dx] =
= d/dx
[dy(ε)/dε] = dζ/dx = ζx
=
b
a
(𝓛y·dy(ε)/dε+𝓛yx·dyx(ε)/dε)·dx
=
b
a
(𝓛y·ζ+𝓛yx·ζx)·dx

where group item 𝓛y·ζ represents
a sum Σi𝓛yi·ζi with i∈[0,n]
which in expanded form is
Σi𝓛/∂yi·[dyi(ε)/dε|ε=0]
and group item 𝓛yx·ζx represents
a sum Σi(𝓛yix·ζix) 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)[𝓛yx·ζ] =
= (d
𝓛yx/dx)·ζ + 𝓛yx·ζx

b
a
(d/dx)[𝓛yx·ζ]·dx =

=
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


[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
𝓛y·ζ represents Σi𝓛yi·ζi

getting
0 = [𝓛yx·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:
𝓛y = 𝓛/∂y = d/dx(∂𝓛/∂yx) =
= d
𝓛yx/dx
which nullifies an integral in the last identity.

Therefore,
[𝓛yx·ζ]|[a,b] = 0 or
𝓛yx(b)·ζ(b) − 𝓛yx(a)·ζ(a) = 0
where group parameter
𝓛yx·ζ represents Σi𝓛yix·ζi with the sum by i∈[0,n].

This means that the value of 𝓛yx·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[𝓛yx·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 = 𝓛yx·ζ
or, in expanded by coordinates format,
J = Σi𝓛yix·ζi with i∈[0,n]
Recall that
y={yi} is a set of time-space coordinates with i∈[0,n];
y0 is time t;
{yi} are a set of generalized coordinates {qi} with i∈[1,n];
ζ={ζi}={dyi(ε)/dε|ε=0} is a set of generators for each time-space coordinate;
𝓛(y,yx) = L(t,tx,q,qx/tx)·tx
where x in a subscript indicates a derivative of a corresponding function by parameter x:
tx=dt/dx and
{qix}={dqi/dx}

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

Sunday, April 12, 2026

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={qi}, 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,...,qn} and time parameter t.

Consider a family of transformation functions parameterized by variable ε
t(ε)=T(ε,t,q1,...,qn)=T(ε,t,q)
and (for each i∈[1,n])
qi(ε)=Qi(ε,t,q1,...,qn)=Qi(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 qi(0)=qi for each i∈[1,n].

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

Functions T(ε,t,q) and Qi(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 Qi(ε,t,q) to uniformly converge to T(0,t,q) and Qi(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
qi'(ε) = dqi(ε)/dt = qi(ε)'
Notice the order of ε-transformation and apostrophe that indicates a time derivative.
Thus, in this particular case
qi'(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
qi'(ε) = dqi(ε)/dt(ε)

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

The numerator, by chain rule, equals to
∂Qi/∂t+Σk(∂Qi/∂qk)·qk'
The denominator, by chain rule, equals to
∂T/∂t+Σk(∂T/∂qk)·qk'

Therefore,
qi'(ε) =
∂Qi /∂t+Σk(∂Qi /∂qk)·qk'
∂T/∂t+Σk(∂T/∂qk)·qk'
where
Qi = Qi(ε,t,q) = qi(ε)
T = T(ε,t,q) = t(ε)

As ε0, we can represent ε-transformed values {t(ε),qi(ε)} in terms of initial values {t(0)=t,qi(0)=qi} and infinitesimal increments like in the Taylor's formula:
t(ε)= t + ε·[∂t(ε)/∂ε|ε=0] +o(ε)
qi(ε)= qi + ε·[∂qi(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
ξi(t,q) = ∂qi(ε)/∂ε|ε=0 =
= ∂Qi(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,
qi(ε)= qi + ε·ξi(t,q) +o(ε)

Using the above representations, we can evaluate the terms of the formula for qi'(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)/∂qk =
= ∂t(ε)/∂qk =
= ∂
[t+ε·τ(t,q)+o(ε)]/∂qk
derivative of t by qk is 0;
ε is an independent of qk parameter

= ε·∂τ(t,q)/∂qk


∂Qi (ε,t,q)/∂t =
= ∂qi(ε)/∂t =
= ∂
[qi+ε·ξi(t,q)+o(ε)]/∂t
qi 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 ∂qi/dt=0;
ε is an independent of t parameter

= ε·∂ξi(t,q)/∂t


∂Qi (ε,t,q)/∂qk =
= ∂qi(ε)/∂qk =
= ∂
[qi+ε·ξi(t,q)+o(ε)]/∂qk
derivative of qi by qk is 1 for i=k and 0 otherwise ⇒ (using math symbol δ) derivative is equal to δik;
ε is an independent of t parameter

= δik + ε·∂ξi(t,q)/∂qk


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

Numerator =
= ε·∂ξi(t,q)/∂t +
+
Σk[δik+ε·∂ξi(t,q)/∂qk]·qk' =
= ε·∂ξi(t,q)/∂t +
+ qi' + ε·
Σk[∂ξi(t,q)/∂qk]·qk' =
= qi' +
+ε·
{∂ξi(t,q)/∂t+ Σk[∂ξi(t,q)/∂qk]·qk'}

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

Now the formula for a transformation of velocity looks like
qi'(ε) =
qi' + ε·A
1 + ε·B
where
A = ∂ξi(t,q)/∂t +
+
Σk[∂ξi(t,q)/∂qk]·qk'
and
B = ∂τ(t,q)/∂t +
+
Σk[∂τ(t,q)/∂qk]·qk'

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
qi'(ε) − qi' = ε·X + o(ε)
then we could say that
limε→0 (qi'(ε) − qi')/ε =
= d(qi'(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)·[qi' + ε·(A−B·qi')] =
= qi' + ε·A + o(ε)


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

Therefore,
qi'(ε) = qi' + ε·(A−B·qi') + o(ε)

Hence, the instantaneous rate of change of velocity at point {t,q} is
d(qi'(ε))/dε|ε=0 = A−B·qi' =
=
{∂ξi(t,q)/∂t +
+
Σk[∂ξi(t,q)/∂qk]·qk'}
− qi
{∂τ(t,q)/∂t +
+
Σk[∂τ(t,q)/∂qk]·qk'}

In the above expression we can recognize two full time derivatives
i(t,q)/dt = ∂ξi(t,q)/∂t +
+
Σk[∂ξi(t,q)/∂qk]·qk'
and
dτ(t,q)/dt = ∂τ(t,q)/∂t +
+
Σk[∂τ(t,q)/∂qk]·qk'
which makes our formula for a rate of change of velocity much simpler
d(qi'(ε))/dε|ε=0 =
= dξi(t,q)/dt − qi'·dτ(t,q)/dt

where
ξi(t,q)=∂qi(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 qi' (the second term in the above formula) represents an adjustment caused by ε-transformation of time.

Tuesday, March 17, 2026

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)={qi(t)}, where i∈[1,n] and n is the number of degrees of freedom.
Generalized velocities (time derivatives of generalized coordinates) q'(t)={qi'(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/∂qi' = ∂L/∂qi (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:
pi = ∂L/∂qi'
Previous lecture was dedicated to the proof of the Momentum Conservation law for each component pi, for which the Lagrangian is invariant (symmetrical) under a translation (shift) of the corresponding generalized cyclic coordinate qiqi:
∂L/∂qi=0 pi=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,...,qn,q1',...,qn',t)
is invariant under translation of qk by infinitesimal value ε
qkqk + ε
then the kth coordinate of the generalized momentum
pk = ∂L/∂qk'
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
θθ + 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 θθ+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 θθ+ε, but the Lagrangian of a system in a central field is invariant under the translation θθ+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
pθ = ∂L/∂θ'
is conserved.

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


Thus,
pθ = 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·rv

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·rv
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
ijk
r00
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'.

Sunday, March 15, 2026

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

Notes to a video lecture on UNIZOR.COM

Noether's Theorem
Symmetry and 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)={qi(t)}, where i∈[1,n] and n is the number of degrees of freedom.
Generalized velocities (time derivatives of generalized coordinates) q(t)={qi'(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
d/dt ∂L/∂qi' = ∂L/∂qi (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.

Symmetry and Momentum Conservation

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

More precisely, if the Lagrangian of a mechanical system is invariant under the translation of generalized coordinates, the generalized momentum of this system is invariant under this translation as well, which constitutes the Momentum Conservation law.

Let's introduce a concept of momentum in generalized coordinates.

We are familiar with a vector of momentum in Euclidean three-dimensional space with Cartesian coordinates (x,y,z) for a point-mass m, as a vector with three components
px(t) = m·x'(t)
py(t) = m·y'(t)
pz(t) = m·z'(t)

Another approach, that uses the kinetic energy of this object
T=½m·(x'²+y'²+z'²)
would be to define
px = ∂T/∂x'
py = ∂T/∂y'
pz = ∂T/∂z'

Both definitions are equivalent, but the latter leads us to the third definition using the Lagrangian
L=T−U
instead of just kinetic energy T, because potential energy U does not depend on velocity:
px = ∂L/∂x'
py = ∂L/∂y'
pz = ∂L/∂z'

Since the Lagrangian of a mechanical system is usable in both Cartesian and non-Cartesian (generalized) coordinates, we can define a generalized momentum
(p1,...,pn)
as a set of partial derivatives of the Lagrangian L by corresponding component of generalized velocity (∂L(...)/q1',...,∂L(...)/qn').

The time-dependent function
pk(t) = ∂L(...)/∂qk'
is called the kth component of the generalized momentum.

Consider a mechanical system with n degrees of freedom and its trajectory in generalized coordinates
q(t) = (q1(t),...,qn(t)).

Theorem
If the Lagrangian of this system
L(q1,...,qn,q1',...,qn',t)
is invariant under translation of qk by infinitesimal value ε
qkqk + ε
then the kth coordinate of the generalized momentum
pk = ∂L/∂qk'
is conserved.

Proof
The invariance of our Lagrangian under translation of qk means
L(...qk...,q') = L(...qk+ε...,q')
Let
ΔkL=L(...qk+ε...)−L(...qk...)=0
Then the partial derivative of the Lagrangian by qk is
∂L/∂qk = limε→0 ΔkL/ε = 0
When a partial derivative of the Lagrangian by a coordinate is zero, this coordinate is called cyclic.
The corresponding Euler-Lagrange equation for this coordinate is
d/dt ∂L/∂qk' = ∂L/∂qk

The right side is zero, as we stated above, which results in
d/dt ∂L/∂qk' = 0
from which, in turn, follows
∂L/∂qk' is constant.

According to a definition of a generalized momentum, we have proved the conservation of the kth component of a generalized momentum when the Lagrangian is invariant under the translation along the kth generalized coordinate.

∂L/∂qk=0 ∂L/∂qk'=pk=const
translation symmetry ⇒
⇒ conserved momentum

End of Proof.

Note 1:
In a more general case, if the Lagrangian is invariant under the translation of several generalized coordinates, all the corresponding components of the generalized momentum are conserved.

Note 2:
A deeper interpretation of this result is that momentum is the generator of spatial translations: an infinitesimal shift of a coordinate is produced by the corresponding component of momentum.
Symmetry produces conservation laws, and the conserved quantities generate the corresponding symmetry transformations.

Example
A stone is thrown in the air horizontally along X-axis with speed v from the height H.
Assume, the origin of Cartesian coordinates is on a ground levelimmediately under the initial position of a stone with X and Y axes are on the ground level with Z axis going vertically upward.

The coordinates of the stone trajectory are
x(t) = v·t
y(t) = 0
z(t) = H−½g·t²

The velocity components of the stone trajectory are
x'(t) = v
y'(t) = 0
z'(t) = −g·t

Kinetic and potential energies of the stone are
T = ½m·(x'²+y'²+z'²)
U = m·z

Lagrangian and its partial derivatives by coordinates areis
L=T−U
∂L/∂x = 0
∂L/∂y = 0
∂L/∂z = m ≠ 0

According to the theorem above, the linear momentum components are
px=const=∂L/∂x'=m·x'=m·v
py=const=∂L/∂y'=m·y'=0
pz≠const; ∂L/∂z'=m·z'=−g·t

Sunday, March 1, 2026

Noether's Theorem & Energy Conservation: UNIZOR.COM -> Physics+ 4 All -> Lagrangian - Noether E = T − U const

Notes to a video lecture on UNIZOR.COM

Noether's Theorem
Symmetry and Energy Conservation

History

In 1918, German mathematician Emmy Noether published the proof of a very important theorem named after her, for which Albert Einstein and other famous physicists called her one of the most significant mathematicians of her time.
The Noether Theorem itself was called "one of the most important mathematical theorems ever proved in guiding the development of modern physics".

This lecture is about a particular case of the Noether 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. We have introduced a concept of a functional Φ[f] that produces a numeric value for a function f (like a definite integral [a,b]f(x)dx gives an area under a curve that graphically represents a function). Basically, a functional can be considered a function of a point in a set of functions.

2. To find stationary points of a functional Φ[], like a function that brings a functional to a local minimum or maximum, we introduced a concept of the functional's variation δΦ. A solution f (a function) to an equation δΦ[f]=0 is such a stationary point.

3. Assume, we deal with a mechanical system in the n-dimensional configuration space and Cartesian coordinates or generalized coordinates transformable to and from Cartesian ones by one-to-one smooth time-independent transformation functions.
Many problems in Physics are related to finding a stationary point of a specific type of a functional known as action defined as
Φ[L(t)t∈[t1,t2]] =
= [t1,t2] L[q(t),q'(t),t]dt
where L is a known smooth real function called Lagrangian of a mechanical system,
q(t) is a set of generalized coordinates of a system's trajectory {q1(t),...,qn(t)},
q'(t) is a set of derivatives of these generalized coordinates by time {q'1(t),...,q'n(t)} and
t is an optional time parameter.

4. For this action functional from the equation δΦ[L(t)]=0 we have derived a differential equation known as Euler-Lagrange equation, solution to which are stationary points of the action functional:
d/dt ∂L/∂qi' = ∂L/∂qi (i∈[1,n])
We will shorten the name of this equation as "E-L".
Each such stationary point (that is, a trajectory q(t) that extremizes the action functional) is the real physical trajectory a mechanical system is moving along from point A in the configuration space at time t1 to point B at time t2.

5. For a classical conservative mechanical system in Cartesian coordinates we have defined its Lagrangian L as a difference between its total kinetic energy T (that depends only on the magnitudes of velocities q') and total potential energy U (that depends only on positions q) without explicit time dependency. Then we have proved that extremals of the action functional built with this Lagrangian are exactly the same as real trajectories obtained from solutions to Newton's Second law equations.

6. Clear solutions of Newton's Second law equations can be obtained only in Cartesian coordinates. If we use different generalized coordinates, the Euler-Lagrange equations can still deliver the trajectories, which makes the Lagrangian approach more universal and in some cases easier, especially in the presence of constraints, which can be avoided by properly choosing generalized coordinates.

Symmetry and Energy Conservation

Let's analyze the meaning of the Lagrangian's independence of the explicit time parameter t mentioned as "optional" in the introductory item 3 above.
In short, it means that the formula that expresses this Lagrangian includes only n coordinates qi and n their derivatives by time q'i (i∈[1,n]) and no explicit time parameter.
Consequently,
∂L(q,q',t)/∂t = 0
where q represents n coordinates qi and
q' represents their n time derivatives q'i (i∈[1,n]).

For our analysis we need to calculate a full time derivative of a Lagrangian of a real trajectory q(t) (that is, q(t) is a solution to E-L equation).
d/dt L[q,q',t] =
= Σi(∂L/∂qi)·(dqi/dt) +
+ Σi(∂L/∂qi')·(dqi'/dt) +
+ ∂L/∂t =
[dq/dt=q', dq'/dt=q"]
= Σi(∂L/∂qi)·qi' +
+ Σi(∂L/∂qi')·qi" +
+ ∂L/∂t =
[from E-L: ∂L/∂qi = d/dt ∂L/∂qi']
= Σi(d/dt ∂L/∂qi')·qi' +
+ Σi(∂L/∂qi')·qi" +
+ ∂L/∂t =
[combine two Σ's]
= Σi[(d/dt ∂L/∂qi')·qi' +
     + (∂L/∂qi')·qi"
] +
+ ∂L/∂t =
[recognize a full derivative of
(∂L/∂qi')·qi' in the [...] by time]

= Σi[d/dt ((∂L/∂qi')·qi')] +
+ ∂L/∂t =
[replace Σ d/dt with d/dt Σ]
= d/dt Σi[(∂L/∂qi')·qi'] +
+ ∂L/∂t

Therefore,
dL/dt = d/dt Σi[(∂L/∂qi')·qi'] +
+ ∂L/∂t
[combine two time derivatives]
d/dt i[(∂L/∂qi')·qi'] − L} =
= −∂L/∂t


Assign
E(q,q',t) = Σi[(∂L/∂qi')·qi'] − L
getting
d/dt E(q,q',t) = −∂L/∂t

As we mentioned above, we deal only with classical conservative mechanical system for which Lagrangian is independent of an explicit time parameter, that is
∂L(q,q',t)/∂t=0.
Therefore,
d/dt E(q,q',t) = 0
which means that
E(q,q',t) = Σi[(∂L/∂qi')·qi'] − L
is a constant of motion.

Our last stop is to interpret E(q,q',t) from the physical viewpoint based on Lagrangian's definition as a difference between kinetic (T) and potential (U) energies, that is L=T−U.

E(q,q',t) =
= Σi[(∂(T−U)/∂qi')·qi']−(T−U)
Since potential energy U does not depend on velocities qi', Σi[(∂(T−U)/∂qi')·qi'] =
= Σi[(∂T/∂qi')·qi']
Therefore,
E(q,q',t) =
= Σi[(∂T/∂qi')·qi']−(T−U)

If our coordinates {qj} are Cartesian,
T = Σj½mj·(qj')²
From this we easily derive
Σi[∂T/∂qi' · qi'] =
= Σi[∂/∂qi'j½mj·(qj')²] · qi'] =
only term #i from Σj that represents T contains qi', so partial derivative by qi' nullifies all terms except #i
= Σi[∂/∂qi'[½mi·(qi')²] · qi'] =
= Σi[mi·qi'·qi'] =
= Σi[mi·qi] =
= 2T
Thus,
E(q,q',t) = 2T−(T−U) = T+U
which is a total energy of a mechanical system,
and the constancy of E(q,q',t) established above means Energy Conservation law, which we have derived from the independence of the Lagrangian from time.

We have proven that for a conservative mechanical system in Cartesian coordinates the Conservation of Energy law is a consequence of the time-independence of its Lagrangian.
This time-independence is also called time symmetry, which means the Lagrangian is invariant under the transformation t→t+ε
So, time symmetryEnergy Conservation.


In generalized coordinates the result is the same, but the calculations are a bit more complex.

Assume, q(t) is a trajectory of a mechanical system in generalized coordinates and s(t) is the same trajectory in Cartesian coordinates.

Let Sj(q) be a one-to-one-smooth time-independent transformation functions from generalized to Cartesian coordinates:
sj = Sj(q1,...,qn) (j∈[1,n])

To get to the formula for kinetic energy in generalized coordinates, we will use the known formula for Cartesian coordinates and substitute Cartesian coordinates using the transformation function.
T = ½Σj∈[1,n] mj·[sj']² =
= ½Σj∈[1,n] mj·[d/dt Sj(q1,...,qn)]²

We will prove that kinetic energy in generalized coordinates is a homogenous quadratic form of generalized velocities that can be represented as
T = Σk,l [Bk,l(q1,...,qn)·qk'·ql']
where Bk,l(q1,...,qn) are some known functions of generalized coordinates.

Here is the proof.
1. Evaluating the value of
d/dt Sj(q1,...,qn)
in the above expression using the chain rule:
d/dt Sj(q1,...,qn) =
= Σk [∂/∂qkSj(q1,...,qn) · qk']

Let's drop (q1,...,qn) from Sj for brevity.
The expression for kinetic energy contains the square of d/dt Sj, that is
(d/dt Sj)² = Sj'² =
=
Σk,l [∂Sj/∂qk·∂Sj/∂ql · qk'·ql'] =
= Σk,l [A(j)k,l · qk'·ql']
where A(j)k,l = ∂Sj/qk·∂Sj/ql are known functions of generalized coordinates q.

2. Thus, the total kinetic energy in generalized coordinates is
T = ½Σj mj·Σk,l [A(j)k,l·qk'·ql'] =
= Σk,l [Bk,l·qk'·ql']
where Bk,l=½Σj mj·A(j)k,l

Knowing that kinetic energy in generalized coordinates is a homogenous quadratic form of generalized velocities with coefficients being functions of generalized coordinates
T = Σk,l [Bk,l·qk'·ql'] (k,l∈[1,n])
we can calculate
Σi[(∂T/∂qi')·qi']
that participates in the expression
E(q,q',t) =
= Σi[(∂T/∂qi')·qi']−(T−U)

We can use the Euler theorem that states
if T(q,q') is a homogenous form of q' then
Σi[(∂T/∂qi')·qi'] = 2T

For those who are not familiar with this theorem, here is the proof.

Since Bk,l are functions of only generalized coordinates q, they should be considered as constants when we partially differentiating T by generalized velocities qi'.

Expression
T = Σk,l [Bk,l·qk'·ql']
contains members with qi' and without it.
Those members that do not contain qi' will produce zero after partial derivation by qi'.
Those members that do contain qi' are
Σk [Bk,i·qk'·qi'] and
Σl [Bi,l·qi'·ql']
(notice, member Bi,i·qi'·qi' is included in both above sums, but it's correct because it has two qi')

Therefore,
∂T/∂qi' = ∂/∂qi' [Σk (Bk,i·qk'·qi')+
+ Σl (Bi,l·qi'·ql')] =
= Σk Bk,i·qk' + Σl Bi,l·ql'

Multiplying this by qi' and summarizing by i we get
Σi[(∂T/∂qi')·qi'] =
= ΣiΣk [Bk,i·qk'·qi'] +
+ ΣiΣl [Bi,l·ql'·qi'] =
= Σi,k [Bk,i·qk'·qi'] +
+
Σi,l [Bi,l·ql'·qi'] =
= T + T = 2T


Thus, we've got the same result in generalized coordinates as in Cartesian
Σi[(∂T/∂qi')·qi'] = 2T

Therefore,
E(q,q',t) = 2T − (T−U) =
= T + U = const

This is the same Conservation of total energy in generalized coordinates as we have proven it above in Cartesian coordinates.

We have proven that for a classical conservative mechanical system in generalized coordinates with the time-independent Lagrangian L equaled to the difference between kinetic (T) and potential (U) energies, the quantity T+U is conserved and equals to the total mechanical energy of a system, which means that the Conservation of Energy law is a consequence of the time-independence (time symmetry) of its Lagrangian.
Time SymmetryEnergy Conservation.

Monday, February 23, 2026

Euler-Lagrange Equations for Gravitation: UNIZOR.COM - Physics+ 4 All - Lagrangian

Notes to a video lecture on UNIZOR.COM

Euler-Lagrange Equations
for the Gravitational Field


The following is an illustration of using Lagrangian Mechanics to analyze the movement of a planet in a central gravitational field of the Sun.
It will also show how the Kepler's laws of planetary movements are derived from Euler-Lagrange equations.

Let the Sun be modeled as a point mass M and a planet be modeled as a point mass m.
In the lecture about a central field we proved that the orbit of a planet lies in a plane. That allows us to choose polar coordinates r(t) and θ(t) with the Sun at the origin as generalized coordinates.

To apply Euler-Lagrange equation, we have to express kinetic energy T and potential energy U in terms of generalized coordinates r and θ.

Kinetic energy depends on the square of the magnitude of velocity v. In polar coordinates this is expressed as a sum of squares of radial speed vr and perpendicular to it tangential speed vθ:
vr = r'
vθ = r·θ'
v² = vr²+vθ² = (r')²+(r·θ')²
T = ½mv² = ½m[(r')²+(r·θ')²]

Potential energy of a planet in the gravitational field (you can refer to lectures in
Physics 4 Teens → Energy → Energy of Gravitational Field) is
U = −G·M·m/r
where G is the universal Gravitational Constant

Lagrangian of a planet is
L = T − U =
= ½m
[(r')²+(r·θ')²] + G·M·m/r

The Euler-Lagrange equations for generalized coordinate are
∂L/∂r = d/dt ∂L/∂r'
∂L/∂θ = d/dt ∂L/∂θ'

From the equation for r:
∂L/∂r = m·r·(θ')² − G·M·m/r²
∂L/∂r' = m·r'
d/dt ∂L/∂r' = m·r"
Resulting equation is
m·r·(θ')² − G·M·m/r² = m·r"
Cancelling m produces
r·(θ')² − G·M/r² = r"

From the equation for θ:
∂L/∂θ = 0
(because L does not explicitly depend on θ)
∂L/∂θ' = m·r²·θ'
Resulting equation is
0 = d/dt m·r²·θ'
from which follows that
m·r²·θ' = const = |L|
The expression on the left side of the above equation is the magnitude of a angular momentum vector (usually it's denoted by symbol L, but we will use |L| to separate it from the Lagrangian).
Since the Lagrangian does not depend explicitly on θ, this coordinate is cyclic, and therefore the corresponding angular momentum is conserved.
So, this equation expresses the angular momentum conservation law.
At the same time the constancy of the expression r²·θ' means that equal areas are swept by the radius vector per unit of time. Thus, Kepler's Second Law follows directly from the Euler–Lagrange equations.
From this follows that r²·θ' is also a constant of motion (angular momentum per unit of mass), let's call this constant k. Now the second Euler-Lagrange equation looks like this
r²·θ' = |L|/m = k

The system of both Euler-Lagrange equations for two generalized coordinates is

r·(θ')² − G·M/r² = r"
r²·θ' = k

From the physical standpoint, the second equation determines how fast the planet moves along the orbit, while the first determines the shape of the orbit.

Let's use the expression for θ'=k/r² from the second Euler-Lagrange equation and substitute it into the first term r·(θ')² of the first equation
r·(θ')² = r·(k/r²)² = k²/r³
Now the first equation looks like
k²/r³ − G·M/r² = r"
This is exactly the equation we derived in the lecture Motion in Polar Coordinates of the Laws of Kepler part of this course, though the derivation in that part was much longer and involved a lot of vector manipulations in Cartesian coordinates.

Both Euler-Lagrange equations define r(t) and θ(t) as functions of time.
In a tradition of using polar coordinates, let's derive r as a function of θ.

Differentiation by time of an expression X can be done according to the following procedure
[A1] dX/dt = (dX/dθ)·(dθ/dt)
or, equivalently,
[A2] dX/dθ = (dX/dt)/(dθ/dt) = X'/θ'

Apply rule [A2] for X=r:
dr/dθ = r'/θ'
and, therefore,
r' = (dr/dθ)·θ' =
use the second Euler-Lagrange equation above resolving it for θ'=k/r²
= (dr/dθ)·k/r² = −k·d(1/r)/dθ

Consider the equality we've got
r' = −k·d(1/r)/dθ

Let's differentiate it by time. On the left side we will get r".

To differentiate d(1/r)/dθ on the right side by time, we will use the rule [A1] above getting
d/dt[d(1/r)/dθ] =
= d/dθ
[d(1/r)/dθ]·(dθ/dt) =
=
[d²(1/r)/dθ²]·θ' =
=
[d²(1/r)/dθ²]·k/r²

Equating derivatives of the left and the right sides, we get
r" = −(k²/r²)·[d²(1/r)/dθ²]

This expression for r" we substitute as the right side of the first Euler-Lagrange equation getting
r·(θ')² − G·M/r² =
= −(k²/r²)·
[d²(1/r)/dθ²]

Multiplying both sides by :
r³·(θ')² − G·M =
= −k²·
[d²(1/r)/dθ²]
Again substitute θ'=k/r² getting
r³·k²/r4 − G·M =
= −k²·
[d²(1/r)/dθ²]
or
(1/r) − G·M/k² = −[d²(1/r)/dθ²]

Assign x=1/r. In terms of x as a function of θ our equation becomes
d²x/dθ² + x = G·M/k²
This is a known type of differential equation, in a simplified form its solutions are
x(θ) = a + b·cos(θ)
More precisely, the general solution is
x(θ) = G·M/k² + C·cos(θ−θ0)
from which follows
r(θ) = 1/[G·M/k²+C·cos(θ−θ0)]

The last equation in polar coordinates represents second degree curves (called conic sections - circle, ellipse, parabola, hyperbola) with eccentricity e=C·k²/(G·M).
Geometrically,
if e is smaller than 1, a curve is an ellipse;
if e is equal to 1, a curve is a parabola;
if e is greater than 1, a curve is a hyperbola.

This equation describes an orbit of an object flying around the Sun along a trajectory that is a conic section. This is in compliance with Kepler's laws of planetary movement.

Thus, using the methodology of Lagrangian Mechanics, we were able to derive the Law of Conservation of Angular Momentum, Kepler's Second Law and the shape of a planetary orbit - a curve of the second degree (conic section).