Noether Theorem and
Conservation Laws
Background
Motion of a mechanical system is represented by a curve in extended configuration space with coordinates y={yi(x)} - a set of time-space coordinates describing a motion curve parameterized by independent variable x∈[a,b] for i∈[0,n].
These coordinates have physical meaning:
y0 is time t;
{yi} are a set of generalized coordinates {qi} for i∈[1,n];
The action functional
| Φ[t,q] = |
|
L(t,q,q')dt |
was re-parameterized as an integral by independent parameter x∈[a,b]
| Φ[t,q] = |
|
L(t,q,qx/tx)·tx·dx |
and expressed as
| Φ[y] = |
|
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 i≠0 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 i≠0
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
The conclusion of the previous lecture:
d/dx Σi𝓛yix·ζi = 0 for i∈[0,n] where
ζ={ζi}={dyi(ε)/dε|ε=0} is a set of generators for each time-space coordinate
and where
x in a subscript indicates a derivative of a corresponding function by parameter x:
tx=dt/dx and
{qix}={dqi/dx} for i∈[1,n]
Linear Momentum Conservation
Let's choose a single kth space coordinate (k∈[1,n]) and consider the following
t(ε) = t,
qk(ε) = qk + ε,
qi(ε) = qi for all i≠k.
This represents a uniform movement along the kth space coordinate qk.
The corresponding generators of this transformation are
ζk = dyk/dε|ε=0 = 1 and
ζi = 0 for i≠k.
The main result of the Noether's theorem was:
as long as the action functional is invariant under the transformation,
an expression
J = Σi𝓛yix·ζi
is a constant of motion along a trajectory and its x-derivative equals to zero for all x∈[a,b]:
dJ/dx = 0
Considering all ζi=0 for i∈[0,n] except ζk=1,
J = 𝓛ykx
and, therefore,
dJ/dx =
= d/dx 𝓛ykx({yi},{yix}) = 0
or
d/dx ∂/∂ykx 𝓛({yi},{yix}) = 0
To see what follows from this equation, let's return to the original Lagrangian
L(t,{qi},{qi'}),
where apostrophe at q indicates a time-derivative, using yk=qk equivalence for k≠0 and taking into account our definition of function 𝓛:
𝓛(y,yx) = L(t,{qi},{qix/tx})·tx
Notice that we represented qi'=dqi/dt (a generalized velocity) as (dqi/dx)/(dt/dx)=qix/tx.
Now
∂/∂ykx 𝓛({yi},{yix}) =
= ∂/∂qkx [L(t,{qi},{qix/tx})·tx]
As we indicated above, an expression qix/tx is a generalized velocity along ith space coordinate because
qix/tx = (dqi/dx)/(dt/dx) =
= dqi/dt = qi' = vi
Therefore, using the chain rule, we can write the expression above as
∂/∂qkx [L(t,{qi},{qix/tx})·tx] =
= ∂/∂qkx [L(t,{qi},{vi})·tx] =
= [∂/∂vk L(t,q,v)]·[∂vk/∂qkx]·tx =
[recall, vk=qkx/tx]
= [∂/∂vk L(t,q,v)]·(1/tx)·tx =
= ∂/∂vk L(t,q,v) =
= ∂/∂qk' L(t,q,q')
Since pk=∂/∂qk' L(t,q,q') is a definition of generalized momentum, we conclude that under the ε-transformation of a single space coordinate qk described above that leaves the action functional invariant or, in other words, possesses translational symmetry the generalized momentum is conserved.
The conservation of momentum pk could be derived directly from the Euler–Lagrange equations once we know that the Lagrangian is independent of a coordinate qk. Indeed, from the Euler-Lagrange equation
d/dt ∂L/∂qk' = ∂L/qk
follows that, if the right-hand side is zero (independence of Lagrangian L of coordinate qk), the left-hand side is zero as well, which means that generalized momentum
pk = ∂L/∂qk'
is constant (conserved).
Noether's theorem is important because it reveals that momentum conservation is a consequence of a continuous symmetry of the action and extends this principle to every continuous symmetry. Momentum conservation is therefore not an isolated fact but one example of a universal connection between symmetry and conservation laws.
CONCLUSION
Assuming the translation
qk ⟶ qk + ε
leaves action functional invariant,
the Noether conserved quantity would be
J = ∂𝓛/∂ykx = ∂L/∂qk' = pk
which is a generalized momentum along kth generalized coordinate.
Therefore,
dpk/dt = 0
which is the law of conservation of linear momentum.
Angular Momentum Conservation
Consider a rigid body rotating about a fixed axis with only two extended generalized coordinates describing its motion
y0 = t is time,
y1 = θ is an angle of rotation.
The ε-transformation (rotation) we would like to consider is the uniform rotation that can be expressed as
t(ε) = t,
θ(ε) = θ + ε.
Since the angle θ is simply a generalized coordinate, this situation is the same as in the previous one
t(ε) = t,
qk(ε) = qk + ε
with n=k=1,
generalized space coordinate being the angle of rotation q1=y1=θ
and the derivation of Noether's conserved quantity is identical to the derivation for linear momentum.
Therefore, everything we derived for linear momentum in the above case is valid for angular momentum
Assuming the rotation
θ(=y1) ⟶ θ + ε (=y1 + ε)
leaves action functional invariant,
the Noether conserved quantity would be
J = ∂𝓛/∂θx = ∂L/∂θ' = ℒ
which is a generalized angular momentum.
Therefore, the angular momentum is conserved
dℒ/dt = 0
which is the law of conservation of angular momentum.
Energy Conservation
The third important application of Noether's theorem is the law of conservation of energy, which follows from the invariance of the action under translations of time.
Let's choose a uniform translation of the time coordinate that does not affect any space coordinates:
y0(ε) = y0 + ε,
which is time transformation
t(ε) = t + ε
and
yi(ε) = yi for all i∈[1,n],
which means that all generalized coordinates remain unchanged
qi(ε) = qi for all i∈[1,n].
For this kind of transformation the corresponding generators are
ζ0 = dy0/dε|ε=0 = 1 and
ζi = 0 for i≠0.
According to Noether's theorem, the conserved quantity is
J = Σi𝓛yix·ζi = 𝓛y0x·ζ0 =
= ∂𝓛/∂tx = ∂/∂tx[L(t,q,qx/tx)·tx]
which is a constant of motion along a trajectory.
Let's perform all the required computations.
J = ∂/∂tx[L(t,q,qx/tx)·tx] =
[recall, vi=qix/tx=qi' - time derivative of a generalized coordinate]
= ∂/∂tx[L(t,{qi},{qix/tx})·tx] =
[using a formula of a derivative of a product of two functions]
= [∂/∂txL(t,{qi},{qix/tx})]·tx + L=
[apply the chain rule, taking into account that the dependence on tx enters only through the generalized velocities {qix/tx} = {vi} and using vi as a placeholder for qix/tx to shorter the notation]
= [Σi(∂L/∂vi)·(d(qix/tx)/dtx)]·tx + L =
= [Σi(∂L/∂vi)·(−qix/t²x)]·tx + L =
[substitute qix/tx²=vi·tx/tx²=vi/tx]
= −Σi(∂L/∂vi)·vi + L =
[recall, ∂L/∂vi=∂L/∂qi' is s generalized momentum pi]
= −Σi(pi·vi) + L
The final formula for a conserved quantity J is:
J = −Σi(pi·vi) + L
In all conservative mechanical systems considered in this course, the Lagrangian L has the form
L = T − U
where the kinetic energy T is a quadratic homogeneous function of generalized velocities and U is potential energy of a system.
Recall that
pi = ∂L/∂vi = ∂(T−U)/∂vi
Since potential energy U does not depend on generalized velocities,
pi = ∂T/∂vi
In classical mechanics the kinetic energy is a homogeneous quadratic (that is, of degree 2) function of the generalized velocities
T = Σi,jAijvivj
For any given quadratic homogeneous function
T = Σi,jAijvivj
the sum Σi[∂T/∂vi]·vi is equal to 2T as follows from the Euler theorem about homogeneous functions.
Here is a simple and elegant proof.
In case of quadratic homogeneous function
T(v1,...,vn) = Σi,jAijvivj
T(λ·v1,...,λ·vn) = λ²·T(v1,...,vn)
where λ - any real number.
Let's differentiate both sides by λ applying the chain rule for the left side
Σi[∂T/∂(λ·vi)]·vi = 2λ·T
Set λ=1 that results in λ·vi=vi, and the result is
Σi[∂T/∂vi]·vi = 2·T
Since generalized momentum is defined by
pi = ∂T/∂vi
and for ordinary mechanical systems L=T−U while U does not depend on the velocities, ∂L/∂vi = ∂T/∂vi
therefore,
Σi(pi·vi) = 2T
J = −2T + (T−U) = −(T+U)
which is a negative total energy of the system, whose conservation is equivalent to conservation of the total system's energy itself.
Therefore, the total energy is conserved under a time transformation that preserves the action functional, as described above.
