Notes to a video lecture on UNIZOR.COM
General Approach
to Hamiltonian
In the previous introductory lecture we have examined the way to simplify the system of n Euler-Lagrange differential equations of second order into a system of 2n equations of the first order using coordinates q={qi} and momenta p={pi}.
For simple classical systems the Lagrangian Mechanics uses Euler-Lagrange equations
∂L/∂qi = d/dt ∂L/∂q̇i
where Lagrangian L=T−U,
T is the full kinetic energy of a system (independent of coordinates),
U is its potential energy (independent of velocities).
Instead, we constructed mathematically equivalent system of Hamiltonian equations
q̇i = ∂(T+U)/∂pi
−ṗi = ∂(T+U)/∂qi
and introduced a new function called Hamiltonian
H(q,p) = T(p) + U(q)
where q=(q1,...,qn) are coordinates
and p=(p1,...,pn) are momenta.
This transformation increased the number of variables from n to 2n, but reduced the complexity of differential equations to the first order and made them symmetric and esthetically more appealing.
However, this construction relied on special properties of the system:
kinetic energy T depended only on momenta,
potential energy U depended only on coordinates.
In general mechanical systems, the Lagrangian L(q,q̇,t) may depend on coordinates and velocities in a more complicated way, kinetic and potential energies might not be separable and the Lagrangian might not be written as a sum of a function of p and a function of q.
Therefore, the construction H=T+U is not applicable to the general case.
The Hamiltonian approach is attractive for these simple mechanical systems, so it is natural to ask whether it can be generalized.
More specifically, given a Lagrangian L(q,q̇,t), we'll construct a function H(q,p,t) whose partial derivatives generate the differential equations similar to above and equivalent to Euler-Lagrange equations of motion.
To accomplish this, we will reformulate the expression H=T+U to avoid explicitly using energies and, instead, express H in terms of coordinates, momenta and a Lagrangian.
Recall the calculations of the Hamiltonian for simple mechanical system in the previous introductory lecture:
T = Σi½mi·q̇i² = Σi½pi²/mi
Partial derivative of T by pi produces
∂T/∂pi = pi/mi = q̇i
Therefore,
q̇i = ∂T/∂pi
Our expression for H=T+U in that case can be represented as
H = 2T − (T−U) = 2T − L =
= Σimi·q̇i² − L =
= Σi(mi·q̇i)·q̇i − L =
= Σipi·q̇i − L
The above expression
H(q,p,t) = Σipi·q̇i − L(q,q̇,t)
formally seems to correspond to our task of constructing a function H(q,p,t) from coordinates, momenta and a given Lagrangian L(q,q̇,t).
It does not explicitly rely on kinetic or potential energy. Since it is written only in terms of the Lagrangian, generalized coordinates, generalized velocities, and generalized momenta, it is a natural candidate for the Hamiltonian in the general case.
It's very important to understand the change of a viewpoint on relationship between arguments, which is a key idea of the Hamiltonian Mechanics.
While in the Lagrangian formulation the independent variables are q, q̇ and t, in the Hamiltonian approach the independent variables are q, p and t with generalized velocities q̇ become dependent on them:
q̇i = q̇i(q,p,t)
Let's check if this expression corresponds to differential equations involving the Hamiltonian.
∂H(q,p,t)/∂pk =
[take into consideration that now qi and pi are independent variables, while generalized velocities q̇i are functions of these variables: q̇i=q̇i(q,p,t)]
= ∂/∂pk [Σipi·q̇i − L] =
[use chain rule]
= q̇k + Σipi·(∂q̇i/∂pk) −
− Σi(∂L/∂q̇i)·(∂q̇i/∂pk) =
= q̇k+Σi[pi−∂L/∂q̇i]·(∂q̇i/∂pk) =
[recall that pi=∂L/∂q̇i by definition]
= q̇k
as required.
∂H(q,p,t)/∂qk =
= Σipi·(∂q̇i/∂qk) − ∂L/∂qk −
− Σi(∂L/∂q̇i)·(∂q̇i/∂qk) =
= Σi[pi−∂L/∂q̇i]·(∂q̇i/∂qk) −
− ∂L/∂qk =
[recall that pi=∂L/∂q̇i by definition]
= − ∂L/∂qk =
[using Euler-Lagrange equation]
= − d/dt ∂L/∂q̇k =
[using the definition of generalized momentum]
= − d/dt pk = −ṗk
as required.
So, from the definitions of momenta as
pi = ∂L/∂q̇i
the Hamiltonian as
H(q,p,t) = Σipi·q̇i − L(q,q̇,t)
and Euler-Lagrange equation for L
∂L/∂qi = d/dt ∂L/∂q̇i
follows that both differential equations for H are satisfied:
∂H/∂pi = q̇i
∂H/∂qi = −ṗ
In reverse, assuming Hamiltonian equations are satisfied, we can derive Euler-Lagrang equations.
∂H(q,p,t)/∂qk =
= ∂/∂qk [Σipi·q̇i − L] =
= Σipi·(∂q̇i/∂qk) − ∂L/∂qk −
− Σi(∂L/∂q̇i)·(∂q̇i/∂qk) =
= Σi[pi−∂L/∂q̇i]·(∂q̇i/∂qk) −
− ∂L/∂qk =
[recall that pi=∂L/∂q̇i by definition]
= − ∂L/∂qk
Since we assumed that Hamiltonian differential equations are true,
∂H(q,p,t)/∂qk = −ṗk
Therefore,
− ∂L/∂qk = −ṗk
From the definition
ṗk = dpk/dt = d/dt ∂L/∂q̇k
follows:
∂L/∂qk = d/dt ∂L/∂q̇k
which is the Euler-Lagrange equation.
That completes the proof that a system of n second order Euler-Lagrange differential equations is equivalent to 2n first order Hamiltonian equations.
The construction of the Hamiltonian
H = Σipi·q̇i − L
from the Lagrangian L is called the Legendre transformation.
Thursday, July 16, 2026
Hamiltonian Introduction
Notes to a video lecture on UNIZOR.COM
Introduction to Hamiltonian
We assume you have a pretty good understanding of Lagrangian Mechanics. If not, the previous chapters of this course Lagrangian and Noether Theorem provide a description of its basic principles.
The Hamiltonian Mechanics is built upon Lagrangian Mechanics. The differences can be summarized as follows.
In Lagrangian Mechanics our main dynamic variables were generalized coordinatesq=(q1,...,qn) and their time derivatives - generalized velocities q̇=(q̇1,...,q̇n) (here we use Newtonian 'dot notation' to indicate a time derivative).
In Hamiltonian Mechanics it's the same generalized coordinates {qi} and, separately from coordinates, generalized momentap=(p1,...,pn) instead of velocities.
Generalized momenta of a mechanical system with the Lagrangian L(q,q̇,t) are defined as a set of components
pi=∂L/∂q̇i
This definition was already introduced in the lectureNoether p=m·v const of a previous chapter of this course.
Using this definition of generalized momentum, the Euler-Lagrange differential equation of the second degree
d/dt ∂L/∂q̇i = ∂L/∂qi (i∈[1,n])
would look simpler
d/dt pi = ∂L/∂qi or, shorter,
ṗi = ∂L/∂qi
In Lagrangian Mechanics we had n functions of time {qi(t)} (generalized coordinates) and a system of n Euler-Lagrange differential equations of the second order.
The number of unknowns was equal to the number of equations.
Instead, in this momentum-based approach, we have 2n functions of time {qi(t)} (generalized coordinates) and {pi(t)} (generalized momenta) with only n differential equations
(A) ṗi = ∂L/∂qi
We need n more equations to obtain a system of 2n first-order differential equations equivalent to n Euler-Lagrange equations of the second-order.
Consider a simple case of a conservative system of one point mass m in three-dimensional Euclidean space with Cartesian coordinates (q1,q2,q3), velocities q̇i and time-independent Lagrangian L equaled to a difference between kinetic T and potential U energies
L = T − U
Since kinetic energy T is independent of position qi, the same equation (A) above can be written as
ṗi = ∂(−U)/∂qi
or
−ṗi = ∂U/∂qi
In addition to these n differential equations, we can construct n more using the classical definition of the vector of momentumpi=m·q̇i and the kinetic energy expressed in terms of momenta p as
T = Σi½mi·q̇i² = Σi½pi²/mi
Partial derivative of T by pi produces
∂T/∂pi = pi/mi = q̇i
Therefore,
(B) q̇i = ∂T/∂pi
We have constructed n more differential equations to complete the system.
The differential equation for a time derivative of the generalized momentum, as we stated above, is
ṗi = ∂L/∂qi = ∂(T−U)/∂qi
Since kinetic energy is independent of coordinates, we can exclude it
(C) −ṗi = ∂U/∂qi
Equations (B) and (C) constitute 2n differential equations of the first order with 2n unknowns - coordinates and momenta
q̇i = ∂T/∂pi
−ṗi = ∂U/∂qi
The problem is, the first n equations depend on kinetic energy T, while the second group of n equations depends on potential energy U.
We would like a formulation in which both sets of equations are generated by a single function of the same variables (q,p), whose partial derivatives with respect to p give time derivatives of coordinates, that is velocities q̇, and with respect to q give time derivatives of the generalized momenta ṗ.
Recall that in our simple case the kinetic energy T is independent of positions qi (∂T/∂qi=0) and potential energy U is independent of momenta pi (∂U/∂pi=0).
Based on this property, we can add to partial differentiation of the first equation the potential energy U and add to partial differentiation of the second equation the kinetic energy T.
q̇i = ∂(T+U)/∂pi
−ṗi = ∂(T+U)/∂qi
Let's introduce a new function called Hamiltonian
H(q,p) = T(p) + U(q)
where q=(q1,...,qn) are coordinates
and p=(p1,...,pn) are momenta.
With this notation our system of 2n differential equations of the first degree looks quite symmetrical (some might say "beautiful")
q̇i = ∂H/∂pi
−ṗi = ∂H/∂qi
In this case of a simple mechanical system the Hamiltonian H=T+U represents a total (kinetic and potential) energy of our mechanical system, which makes our system of equation more related to real physical characteristics of a system.
In more general systems, however, the definition of the Hamiltonian is broader and not necessarily coincides with a total energy.
Looking ahead, let us state that the symmetric form of Hamilton's equations makes Hamiltonian Mechanics especially suitable for advanced topics such as canonical transformations, statistical mechanics and quantum mechanics.
Introduction to Hamiltonian
We assume you have a pretty good understanding of Lagrangian Mechanics. If not, the previous chapters of this course Lagrangian and Noether Theorem provide a description of its basic principles.
The Hamiltonian Mechanics is built upon Lagrangian Mechanics. The differences can be summarized as follows.
In Lagrangian Mechanics our main dynamic variables were generalized coordinates
In Hamiltonian Mechanics it's the same generalized coordinates {qi} and, separately from coordinates, generalized momenta
Generalized momenta of a mechanical system with the Lagrangian L(q,q̇,t) are defined as a set of components
pi=∂L/∂q̇i
This definition was already introduced in the lecture
Using this definition of generalized momentum, the Euler-Lagrange differential equation of the second degree
d/dt ∂L/∂q̇i = ∂L/∂qi (i∈[1,n])
would look simpler
d/dt pi = ∂L/∂qi or, shorter,
ṗi = ∂L/∂qi
In Lagrangian Mechanics we had n functions of time {qi(t)} (generalized coordinates) and a system of n Euler-Lagrange differential equations of the second order.
The number of unknowns was equal to the number of equations.
Instead, in this momentum-based approach, we have 2n functions of time {qi(t)} (generalized coordinates) and {pi(t)} (generalized momenta) with only n differential equations
(A) ṗi = ∂L/∂qi
We need n more equations to obtain a system of 2n first-order differential equations equivalent to n Euler-Lagrange equations of the second-order.
Consider a simple case of a conservative system of one point mass m in three-dimensional Euclidean space with Cartesian coordinates (q1,q2,q3), velocities q̇i and time-independent Lagrangian L equaled to a difference between kinetic T and potential U energies
L = T − U
Since kinetic energy T is independent of position qi, the same equation (A) above can be written as
ṗi = ∂(−U)/∂qi
or
−ṗi = ∂U/∂qi
In addition to these n differential equations, we can construct n more using the classical definition of the vector of momentum
T = Σi½mi·q̇i² = Σi½pi²/mi
Partial derivative of T by pi produces
∂T/∂pi = pi/mi = q̇i
Therefore,
(B) q̇i = ∂T/∂pi
We have constructed n more differential equations to complete the system.
The differential equation for a time derivative of the generalized momentum, as we stated above, is
ṗi = ∂L/∂qi = ∂(T−U)/∂qi
Since kinetic energy is independent of coordinates, we can exclude it
(C) −ṗi = ∂U/∂qi
Equations (B) and (C) constitute 2n differential equations of the first order with 2n unknowns - coordinates and momenta
q̇i = ∂T/∂pi
−ṗi = ∂U/∂qi
The problem is, the first n equations depend on kinetic energy T, while the second group of n equations depends on potential energy U.
We would like a formulation in which both sets of equations are generated by a single function of the same variables (q,p), whose partial derivatives with respect to p give time derivatives of coordinates, that is velocities q̇, and with respect to q give time derivatives of the generalized momenta ṗ.
Recall that in our simple case the kinetic energy T is independent of positions qi (∂T/∂qi=0) and potential energy U is independent of momenta pi (∂U/∂pi=0).
Based on this property, we can add to partial differentiation of the first equation the potential energy U and add to partial differentiation of the second equation the kinetic energy T.
q̇i = ∂(T+U)/∂pi
−ṗi = ∂(T+U)/∂qi
Let's introduce a new function called Hamiltonian
H(q,p) = T(p) + U(q)
where q=(q1,...,qn) are coordinates
and p=(p1,...,pn) are momenta.
With this notation our system of 2n differential equations of the first degree looks quite symmetrical (some might say "beautiful")
q̇i = ∂H/∂pi
−ṗi = ∂H/∂qi
In this case of a simple mechanical system the Hamiltonian H=T+U represents a total (kinetic and potential) energy of our mechanical system, which makes our system of equation more related to real physical characteristics of a system.
In more general systems, however, the definition of the Hamiltonian is broader and not necessarily coincides with a total energy.
Looking ahead, let us state that the symmetric form of Hamilton's equations makes Hamiltonian Mechanics especially suitable for advanced topics such as canonical transformations, statistical mechanics and quantum mechanics.
Friday, July 3, 2026
Noether's Theorem Conservation: UNIZOR.COM -> Physics+ 4 All -> Lagrangian -> Noether's Theorem -> Conservation
Notes to a video lecture on UNIZOR.COM
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
was re-parameterized as an integral by independent parameter x∈[a,b]
and expressed as
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 followingε-transformation of coordinates:
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
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
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.
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.
Subscribe to:
Posts (Atom)
