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

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.