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

Notes to a video lecture on UNIZOR.COM

Noether's Theorem
Angular Momentum Conservation

History

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

Background

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

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

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

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

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

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

Angular Momentum Conservation

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

i	j	k
r	0	0
r'	r·θ'	0

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

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

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

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

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

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
p_x(t) = m·x'(t)
p_y(t) = m·y'(t)
p_z(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
p_x = ∂T/∂x'
p_y = ∂T/∂y'
p_z = ∂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:
p_x = ∂L/∂x'
p_y = ∂L/∂y'
p_z = ∂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,...,p_n)
as a set of partial derivatives of the Lagrangian L by corresponding component of generalized velocity (∂L(...)/q₁',...,∂L(...)/q_n').

The time-dependent function
p_k(t) = ∂L(...)/∂q_k'
is called the k^th component of the generalized momentum.

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

Theorem
If the Lagrangian of this system
L(q₁,...,q_n,q₁',...,q_n',t)
is invariant under translation of q_k by infinitesimal value ε
q_k → q_k + ε
then the k^th coordinate of the generalized momentum
p_k = ∂L/∂q_k'
is conserved.

Proof
The invariance of our Lagrangian under translation of q_k means
L(...q_k...,q') = L(...q_k+ε...,q')
Let
Δ_kL=L(...q_k+ε...)−L(...q_k...)=0
Then the partial derivative of the Lagrangian by q_k is
∂L/∂q_k = lim_ε→0 Δ_kL/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/∂q_k' = ∂L/∂q_k

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

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

∂L/∂q_k=0 ⇒ ∂L/∂q_k'=p_k=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
p_x=const=∂L/∂x'=m·x'=m·v
p_y=const=∂L/∂y'=m·y'=0
p_z≠const; ∂L/∂z'=m·z'=−g·t

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),...,q_n(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/∂q_i' = ∂L/∂q_i (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 q_i 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 q_i 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] =
= Σ_i(∂L/∂q_i)·(dq_i/dt) +
+ Σ_i(∂L/∂q_i')·(dq_i'/dt) +
+ ∂L/∂t =
[dq/dt=q', dq'/dt=q"]
= Σ_i(∂L/∂q_i)·q_i' +
+ Σ_i(∂L/∂q_i')·q_i" +
+ ∂L/∂t =
[from E-L: ∂L/∂q_i = d/dt ∂L/∂q_i']
= Σ_i(d/dt ∂L/∂q_i')·q_i' +
+ Σ_i(∂L/∂q_i')·q_i" +
+ ∂L/∂t =
[combine two Σ's]
= Σ_i[(d/dt ∂L/∂q_i')·q_i' +
     + (∂L/∂q_i')·q_i"] +
+ ∂L/∂t =
[recognize a full derivative of
(∂L/∂q_i')·q_i' in the [...] by time]
= Σ_i[d/dt ((∂L/∂q_i')·q_i')] +
+ ∂L/∂t =
[replace Σ d/dt with d/dt Σ]
= d/dt Σ_i[(∂L/∂q_i')·q_i'] +
+ ∂L/∂t

Therefore,
dL/dt = d/dt Σ_i[(∂L/∂q_i')·q_i'] +
+ ∂L/∂t
[combine two time derivatives]
d/dt {Σ_i[(∂L/∂q_i')·q_i'] − L} =
= −∂L/∂t

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

If our coordinates {q_j} are Cartesian,
T = Σ_j½m_j·(q_j')²
From this we easily derive
Σ_i[∂T/∂q_i' · q_i'] =
= Σ_i[∂/∂q_i'[Σ_j½m_j·(q_j')²] · q_i'] =
only term #i from Σ_j that represents T contains q_i', so partial derivative by q_i' nullifies all terms except #i
= Σ_i[∂/∂q_i'[½m_i·(q_i')²] · q_i'] =
= Σ_i[m_i·q_i'·q_i'] =
= Σ_i[m_i·q_i'²] =
= 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 symmetry ⇒ Energy 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 S_j(q) be a one-to-one-smooth time-independent transformation functions from generalized to Cartesian coordinates:
s_j = S_j(q₁,...,q_n) (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 = ½Σ_j∈[1,n] m_j·[s_j']² =
= ½Σ_j∈[1,n] m_j·[d/dt S_j(q₁,...,q_n)]²

We will prove that kinetic energy in generalized coordinates is a homogenous quadratic form of generalized velocities that can be represented as
T = Σ_k,l [B_k,l(q₁,...,q_n)·q_k'·q_l']
where B_k,l(q₁,...,q_n) are some known functions of generalized coordinates.

Here is the proof.
1. Evaluating the value of
d/dt S_j(q₁,...,q_n)
in the above expression using the chain rule:
d/dt S_j(q₁,...,q_n) =
= Σ_k [∂/∂q_kS_j(q₁,...,q_n) · q_k']

Let's drop (q₁,...,q_n) from S_j for brevity.
The expression for kinetic energy contains the square of d/dt S_j, that is
(d/dt S_j)² = S_j'² =
= Σ_k,l [∂S_j/∂q_k·∂S_j/∂q_l · q_k'·q_l'] =
= Σ_k,l [A^(j)_k,l · q_k'·q_l']
where A^(j)_k,l = ∂S_j/q_k·∂S_j/q_l are known functions of generalized coordinates q.

2. Thus, the total kinetic energy in generalized coordinates is
T = ½Σ_j m_j·Σ_k,l [A^(j)_k,l·q_k'·q_l'] =
= Σ_k,l [B_k,l·q_k'·q_l']
where B_k,l=½Σ_j m_j·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 = Σ_k,l [B_k,l·q_k'·q_l'] (k,l∈[1,n])
we can calculate
Σ_i[(∂T/∂q_i')·q_i']
that participates in the expression
E(q,q',t) =
= Σ_i[(∂T/∂q_i')·q_i']−(T−U)

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

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

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

Expression
T = Σ_k,l [B_k,l·q_k'·q_l']
contains members with q_i' and without it.
Those members that do not contain q_i' will produce zero after partial derivation by q_i'.
Those members that do contain q_i' are
Σ_k [B_k,i·q_k'·q_i'] and
Σ_l [B_i,l·q_i'·q_l']
(notice, member B_i,i·q_i'·q_i' is included in both above sums, but it's correct because it has two q_i')

Therefore,
∂T/∂q_i' = ∂/∂q_i' [Σ_k (B_k,i·q_k'·q_i')+
+ Σ_l (B_i,l·q_i'·q_l')] =
= Σ_k B_k,i·q_k' + Σ_l B_i,l·q_l'

Multiplying this by q_i' and summarizing by i we get
Σ_i[(∂T/∂q_i')·q_i'] =
= Σ_iΣ_k [B_k,i·q_k'·q_i'] +
+ Σ_iΣ_l [B_i,l·q_l'·q_i'] =
= Σ_i,k [B_k,i·q_k'·q_i'] +
+ Σ_i,l [B_i,l·q_l'·q_i'] =
= T + T = 2T

Thus, we've got the same result in generalized coordinates as in Cartesian
Σ_i[(∂T/∂q_i')·q_i'] = 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 Symmetry ⇒ Energy Conservation.

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 v_r and perpendicular to it tangential speed v_θ:
v_r = r'
v_θ = r·θ'
v² = v_r²+v_θ² = (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 r²:
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).

Lagrangian and Constraints: UNIZOR.COM - Physics+ 4 All - Lagrangian

Notes to a video lecture on UNIZOR.COM

Constraints

As we have learned by now, Euler-Lagrange equations in Cartesian coordinates are equivalent to Newton's Second Law, provided the forces are derivable from a potential and constraints are ideal (do no virtual work).
At the same time Euler-Lagrange equations are applicable not only in Cartesian coordinates, but in generalized as well.
In simple cases of conservative mechanical systems in Cartesian coordinates Newton's Second law might be even a preferable choice, but in complex cases of multiple objects moving and acting upon each other the complication of using vectors of force and acceleration might prompt us to choose Euler-Lagrange equations in Cartesian or generalized coordinates as the main tool to obtain trajectories of moving objects.

There is, however, another reason why Euler-Lagrange equations might be a better choice.
In previous lectures we discussed a few times a simple case of a pendulum in a gravitational field.

The specifics of this example is that the movement of a pendulum is constrained by a rigid rod, which necessitates introduction of constraints into a system of equations based on the Newton's Second law.

In the lecture Pendulum of this chapter we approached the task of finding the formula of this pendulum's movement using an angle of deviation of its rod from a vertical. We started analyzing the problem using Cartesian coordinates, but gave up noting the complexity of the problem.
This was not really the application of Newton's Second law. Actually, we used the non-Cartesian parameter (angle) to solve the problem, and we expressed the movement in terms of this angle, not the Cartesian coordinates of the pendulum's bob.

Let's try to move with Cartesian coordinates a little further to make sure that this is not a good way to approach this problem.
Here is how this task is supposed to be approached using the classical Newtonian approach.
Let x(t) and y(t) be time-dependent coordinates of a point-mass bob at the end of a rod moving along a circular trajectory. Assume, positive direction of the X-axis on a picture above goes to the right and positive direction of the Y-axis goes down.
Then an angle α of a deviation of a rod from a vertical is defined by
cos(α(t)) = y(t)/L
sin(α(t)) = x(t)/L

Two forces act on a bob of a pendulum:
(a) constant weight W directed vertically down along Y-axis, its projections on coordinate axes are:
W_x = 0
W_y = m·g
(b) tension of a rod T(t) that holds the bob on its end, its unknown magnitude is T(t) and its projections on coordinate axes are:
T_x(t) = −T(t)·sin(α) = −T·(x/L)
T_y(t) = −T(t)·cos(α) = −T·(y/L)

These parameters allow to construct differential equations of motion based on Newton's Second law
m·x"(t) = −T(t)·(x/L)
m·y"(t) = m·g − T(t)·(y/L)
In addition, we have to satisfy the constraint
x²(t) + y²(t) = L²

We have three equations with three unknowns x(t), y(t) and T(t).
We will skip (t) for brevity.

First, let's eliminate T by invariant transformations of the above equations.
(a) transfer m·g in the second equation to the left
m·x" = −T·(x/L)
m·y" − m·g = −T·(y/L)
(b) multiply the first equation by x, the second - by y and add them together
m·x·x" + m·y·y" − m·y·g =
= −T·(x²+y²)/L
(c) note that x²(t)+y²(t)=L², which allows to substitute it into the above equation and resolve it for T
m·x·x" + m·y·y" − m·y·g =
= −T·L
T = −(m/L)·(x·x" + y·y" − y·g)
(d) by differentiating an identity x²+y²=L² by time twice we get another identity:
2x·x' + 2y·y' = 0
x·x' + y·y' = 0
x'² + x·x" + y'² + y·y" = 0
x·x" + y·y" = −(x'² + y'²)
(e) use this in the expression for T above
T = (m/L)·(x'² + y'² + y·g)
(f) we can substitute T with its expression from (e) into original equations getting two equations with only two unknown functions x(t) and y(t) (mass m cancels out)
x"(t) = −(x/L²)·(x'²+y'²+y·g)
y"(t) = g−(y/L²)·(x'²+y'²+y·g)

The solution to the above system of differential equations is lengthy, it does not resolve into any elementary functions and the obtained integral as a solution is not pretty (it involves elliptical functions).

Let's forget about Cartesian coordinates and switch back to an angle as the main parameter.
It has two main advantages:
(a) there is only one parameter, not two like in Cartesian coordinates;
(b) there is no need to involve a constraint because it is implicitly built into an approach we took.

In the simplest form for the above example, two Cartesian parameters (x and y) minus one constraint (x²(t)+y²(t)=L²) results in one unconstrained parameter (angle α in this case) that is one degree of freedom.

The above can be generalized to the following.

Consider a mechanical system moving in n-dimensional configuration space and described by n time-dependent coordinates:
s₁,...,s_n.
Assume, there are m constraints that affect the motion of this system
f⁽¹⁾(s₁,...,s_n)=0,
...
f^(m)(s₁,...,s_n)=0.

Then under certain conditions (see below) the movement of this system can be described by n−m generalized parameters, which means that the system has n−m degrees of freedom.
These conditions include:
(a) constraints must be holonomic (expressed as equations that contain functions of only coordinates and time, not the velocities or other non-positional parameters);
(b) constraints must be independent (not derived from one another) which for holonomic constraints would follow from the requirement of linear independency of their gradients.

It should be noted that each individual constraint, provided its gradient is nonzero, defines a surface (a smooth manifold of dimension n−1) in a configuration space.
A trajectory of a system constrained by m holonomic constraints with linearly independent gradients belongs to an intersection of all such surfaces (which is a smooth manifold of dimension n−m) defined by all m constraints.
The system's velocity at any point on its trajectory must be tangential to this intersection of surfaces and, therefore, perpendicular to each constraint's gradient.
With m linearly independent gradients that together span an m-dimensional space, the tangent space where all trajectories belong to span orthogonal to it (n−m)-dimensional manifold, and generalized coordinates are just coordinates in this constrained manifold.

The rigorous mathematical proof of the above statement and exact specification of conditions of its applicability are beyond the scope if this course.

However, some geometric interpretation might be helpful.
Imagine the n-dimensional Euclidean space Rⁿ and a single constraint in it
f(s₁,...,s_n) = 0
For example, for a movement of a single point-mass we use n=3, our own three-dimensional space R³ with Cartesian coordinates x, y and z in it and a constraint - an equation of a sphere of radius r
x²+y²+z²=r².

The constraint means that the movement of our mechanical system expressed as coordinate functions of time s_i(t) (i∈[1,n]) is such that
f(s₁(t),...,s_n(t)) = 0
for any moment of time t.
In our example it means that a point-mass must be always on a surface of a sphere, which can be parameterized by only two parameters.

In general, under conditions described above, one constraint reduces the number of degrees of freedom of a mechanical system by one.

Transverse Waves: UNIZOR.COM - Physics4Teens - Waves

Notes to a video lecture on http://www.unizor.com Transverse Wave We are familiar with longitudinal waves, like sound waves in the air. Their defining characteristic is that molecules of air (the medium) are oscillating along the direction of the wave propagation, which, in turn, causing oscillation of pressure at any point along the direction of wave propagation. Consider a different type of waves. Take a long rope by one end, stretching its length on the floor. Make a quick up and down movement of the rope's end that you hold. The result will be a wave propagating along the rope, but the elements of rope will move up and down, perpendicularly to the direction of waves propagation. (open this picture in a new tab of your browser by clicking the right button of a mouse to better see details) These waves, when the elements of medium (a rope in our example) are moving perpendicularly to a direction of waves propagation, are called transverse. Such elementary characteristics of transverse waves as crest, trough, wavelength and amplitude are clearly defined on the picture above. Some other examples of transverse waves are strings of violin or any other string musical instrument. Interesting waves are those on the surface of water. They seem to be transverse, but, actually, the movement of water molecules is more complex and constitutes an elliptical kind of motion in two directions - up and down perpendicularly to waves propagation and back and forth along this direction. Our first problem in analyzing transverse waves is to come up with a model that resembles the real thing (like waves on a rope), but yielding to some analytical approach. Let's model a rope as a set of very small elements that have certain mass and connected by very short weightless links - sort of a long necklace of beads. Every bead on this necklace is a point-object of mass m, every link between beads is a solid weightless rod of length r. Both mass of an individual bead m and length of each link r are, presumably, very small. In theory, it would be appropriate to assume them to be infinitesimally small. Even this model is too complex to analyze. Let's start with a simpler case of only two beads linked by a solid weightless rod. Our purpose is not to present a complete analytical picture of waves, using this model, but to demonstrate that waves exist and that they propagate. Consider the following details of this model. Two identical point-objects α and β of mass m each on the coordinate plane with no friction are connected with a solid weightless rod of length r. Let's assume that α, initially, is at coordinates (0,−A), where A is some positive number and β is at coordinates (a, −A). So, both are at level y=−A, separated by a horizontal rod of length r from x=0 for α to x=a for β. We will analyze what happens if we move the point-object α up and down along the vertical Y-axis (perpendicular to X-axis), according to some periodic oscillations, like yα(t) = −A·cos(ω·t) where A is an amplitude of oscillations, ω is angular frequency, t is time. Incidentally, for an angular frequency ω the period of oscillation is T=2π/ω. So, the oscillations of point α can be described as yα(t) = −A·cos(2π·t / T) Point α in this model moves along the Y-axis between y=−A and y=A. It's speed is y'α(t) = A·ω·sin(ω·t) = = A·(2π/T)·sin(2π·t/T) The period of oscillations is, as we noted above T = 2π/ω During the first quarter of a period α moves up, increasing its speed from v=0 at y=−A to v=A·ω at the point y=0, then during the next quarter of a period it continues going up, but its speed will decrease from v=A·ω to v=0 at the top most point y=A. During the third quarter of a period α moves down from y=A, increasing absolute value of its speed in the negative direction of the Y-axis from v=0 to the same v=A·ω at y=0, then during the fourth quarter of a period it continues going down, but the absolute value of its speed will decrease from v=A·ω to v=0 at the bottom y=−A. Positions of objects α and β during the first quarter of a period of motion of α at three consecutive moments in time are presented below. As the leading object α starts moving up along Y-axis, the led by it object β follows it, as seen on a picture above. Let's analyze the forces acting on each object in this model. Object β is moved by two forces: tension from the solid rod Tβ(t), directed along the rod towards variable position of α, and constant weight Pβ. Object α experiences the tension force Tα(t), which is exact opposite to Tβ(t), the constant weight Pα and pulling force Fα(t) that moves an entire system up and down. There is a very important detail that can be inferred from analyzing these forces. The tension force Tα(t), acting on object α, has vertical and horizontal components from which follows that pulling force Fα(t) cannot be strictly vertical to move object α along the Y-axis, it must have a horizontal component to neutralize the horizontal component of Tα(t). This can be achieved by having some railing along the Y-axis that prevents α to deviate from the vertical path. The reaction of railing will always neutralize the horizontal component of the Tα(t). Without this railing the pulling force Fα(t) must have a horizontal component to keep α on the vertical path along the Y-axis. If α and β are the first and the second beads on a necklace, we can arrange the railing for α. But, if we continue our model and analyze the movement of the third bead γ attached to β, there can be no railing and the horizontal component of the tension force Tβ(t) will exist and will get involved on some small scale. All-in-all, transverse motion is not just movement of components up and down perpendicularly to the wave propagation, it is also a longitudinal motion of these components, though not very significant in comparison with transverse motion and often ignored in textbooks. The really obvious reason for transverse motion to involve a minor horizontal movement in addition to a major vertical one is that you cannot lift up a part of a horizontally stretched rope without a little horizontal shift of its parts as well, because a straight line is always shorter than a curve. Let's now follow the motion of β as α periodically moves up and down, starting at point (0,−A), according to a formula yα(t) = −A·cos(ω·t) with its X-coordinate always being equal to zero. Since position of α is specified as a function of time and the length of a rod connecting α and β is fixed and equal to r, position of β can be expressed in term of a single variable - the angle φ(t) from a vector parallel to a rod directed from α to β and the positive direction of the Y-axis. At initial position, when the rod is horizontal, φ(0)=π/2. Then coordinates of β are: xβ(t) = r·sin(φ(t)) yβ(t) = yα(t) + r·cos(φ(t)) During the first quarter of a period α moves up from y=−A to y=0, gradually increasing its speed and pulling β by the rod upwards. While β follows α up, it also moves closer to the Y-axis. The reason for this is that the only forces acting on β are the tension force along the rod and its weight. Weight is vertical force, while tension acts along the rod and it has vertical (up) and horizontal (left) components. Assuming vertical pull is sufficient to overcome the weight, β will be pulled up and to the left, closer to the Y-axis. If this process of constant acceleration of α continued indefinitely, β would be pulled up and asymptotically close to the Y-axis. Eventually, it will just follow α along almost the same vertical trajectory upward. The angle φ(t) in this process would gradually approach π (180°). With a periodic movement of α up and down the Y-axis, the trajectory of β is much more complex. Let's analyze the second quarter of the period of α's oscillations, when it moves from y=0 level, crossing the X-axis, to y=A. After α crosses the X-axis it starts to slow down, decelerate, while still moving up the Y-axis. As α decelerates in its vertical motion up, the composition of forces changes. Now β will continue following α upwards, but, instead of being pulled by the rod, it will push the rod since α slows down. That will result in the force Tα(t), with which a rod pushes α, to be directed upwards and a little to the left, as presented on the above picture. Also Tβ(t), the reaction of the rod onto β, will act opposite to a tension during the first quarter of a period. Now it has a vertical component down to decelerate β's upward motion and a horizontal component to the right. The latter will cause β to start moving away from the Y-axis, while still going up for some time, slowing this upward movement because of opposite force of reaction of the rod. When α reaches level y=A at the end of the second quarter of its period, it momentarily stops. The behavior of β at that time depends on many factors - its mass m, amplitude A and angular frequency ω of α's oscillations, the length of the rod r. Fast moving leading object α will usually result in a longer trajectory for β, while slow α will cause β to stop sooner. During the third quarter of its period α starts moving downward, which will cause β to follow it, but not immediately because of inertia. Depending on factors described above, β might go up even higher than α. In any case, the delay in β's reaching its maximum after α has completed the second quarter of a period is

UNIZOR.COM - Physics 4 Teens - Electromagnetism - Magnetic Field - Problems 1

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

Magnetism - Lorentz Force - Problems 1

Problem 1a
Consider the experiment pictured below.

A copper wire (yellow) of resistance R is connected to a battery with voltage U and is swinging on two connecting wires (green) in a magnetic field of a permanent magnet.
All green connections are assumed light and their weight can be ignored. Also ignored should be their electric resistance. Assume the uniformity of the magnetic field of a magnet with magnetic field lines directed vertically and perpendicularly to a copper wire.
The mass of a copper wire is M and its length is L.
The experiment is conducted in the gravitational field with a free fall acceleration g.
The magnetic field exerts the Lorentz force onto a wire pushing it horizontally out from the field space, so green vertical connectors to a copper wire make angle φ with vertical.
What is the intensity of a magnetic field B?

Solution

T·cos(φ) = M·g
T = M·g/cos(φ)
F = T·sin(φ) = M·g·tan(φ)
I = U/R
F = I·L·B = U·L·B/R
M·g·tan(φ) = U·L·B/R
B = M·R·g·tan(φ)/(U·L)

Problem 1b
An electric point-charge q travels with a speed v along a wire of length L.
What is the value of the equivalent direct electric current I in the wire that moves the same amount of electricity per unit of time?
What is the Lorentz force exerted onto a charge q, if it moves in a uniform magnetic field of intensity B perpendicularly to the field lines with a speed v.

Solution
Let T be the time of traveling from the beginning to the end of a wire.
T = L/v
I = q/T = q·v/L
F = I·L·B = q·v·B
Notice, the Lorentz force onto a wire in case of only a point-charge running through it does not depend on the length of a wire, as it is applied only locally to a point-charge, not an entire wire. Would be the same if a particle travels in vacuum with a magnetic field present.

Problem 1c
An electric point-charge q of mass m enters a uniform magnetic field of intensity B perpendicularly to the field lines with a speed v.

Suggest some reasoning (rigorous proof is difficult) that the trajectory of this charge should be a circle and determine the radius of this circle.

Solution
The Lorentz force exerted on a point-charge q, moving with speed v perpendicularly to force lines of a permanent magnetic field of intensity B, is directed always perpendicularly to a trajectory of a charge and equals to F=q·v·B (see previous problem).

Since the Lorentz force is always perpendicular to trajectory, the linear speed v of a point-charge remains constant, while its direction always curves toward the direction of the force. Constant linear speed v means that the magnitude of the Lorentz force is also constant and only direction changes to be perpendicular to a trajectory of a charge.

According to the Newton's Second Law, this force causes acceleration a=F/m, which is a vector of constant magnitude, since the Lorentz force has constant magnitude and always perpendicular to a trajectory, since the force causing this acceleration is always perpendicular to a trajectory.
So, the charge moves along a trajectory with constant linear speed and constant acceleration always directed perpendicularly to a trajectory.

Every smooth curve at any point on an infinitesimal segment around this point can be approximated by a small circular arc of some radius (radius of curvature) with a center at some point (center of curvature). If a curve of a trajectory on an infinitesimal segment is approximated by a circle of some radius R, the relationship between a radius, linear speed and acceleration towards a center of this circle (centripetal acceleration), according to kinematics of rotational motion, is
a = v²/R
Therefore, R = v²/a

Since v and a are constant, the radius of a curvature R is constant, which is a good reason towards locally circular character of the motion of a charge. It remains to be proven that the center of the locally circular motion does not change its location, but this is a more difficult task, which we will omit.
Hence,
R = v²/a = m·v²/F =
= m·v²/q·v·B = m·v/q·B