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.