Astronomy before computers!

Sect. 2.7: Energy Function & Energy Conservation

• One more conservation theorem which we would expect to get from the Lagrange formalism is: CONSERVATION OF ENERGY.

• Consider a general Lagrangian L, a function of the coords qj, velocities qj, & time t:

L = L(qj,qj,t) (j = 1,…n)

• The total time derivative of L (chain rule): (dL/dt) = ∑j(∂L/∂qj)(dqj/dt) + ∑j(∂L/∂qj)(dqj/dt) + (∂L/∂t)

Or:

(dL/dt) = ∑j(∂L/∂qj)qj + ∑j(∂L/∂qj)qj + (∂L/∂t)

• Total time derivative of L:

(dL/dt) = ∑j(∂L/∂qj)qj + ∑j(∂L/∂qj)qj + (∂L/∂t) (1)

• Lagrange’s Eqtns: (d/dt)[(∂L/∂qj)] - (∂L/∂qj) = 0

Put into (1)

(dL/dt) = ∑j(d/dt)[(∂L/∂qj)]qj + ∑j(∂L/∂qj)qj + (∂L/∂t)

Identity: 1st 2 terms combine

(dL/dt) = ∑j(d/dt)[qj(∂L/∂qj)] + (∂L/∂t)

Or: (d/dt)[∑jqj(∂L/∂qj) - L] + (∂L/∂t) = 0 (2)

(d/dt)[∑jqj(∂L/∂qj) - L] + (∂L/∂t) = 0 (2)

• Define the Energy Function h:

h ∑jqj(∂L/∂qj) - L = h(q1,..qn;q1,..qn,t)

• (2) (dh/dt) = - (∂L/∂t) For a Lagrangian L which is not an explicit function of time (so that (∂L/∂t) =

0)

(dh/dt) = 0 & h = constant (conserved)• Energy Function h = h(q1,..qn;q1,..qn,t)

– Identical Physically to what we later will call the Hamiltonian H. However, here, h is a function of n indep coords qj & velocities qj. The Hamiltonian H is ALWAYS considered a function of 2n indep coords qj & momenta pj

• Energy Function h ∑jqj(∂L/∂qj) - L

• We had (dh/dt) = - (∂L/∂t)

For a Lagrangian for which (∂L/∂t) = 0

(dh/dt) = 0 & h = constant (conserved)

• For this to be useful, we need a

Physical Interpretation of h. – Will now show that, under certain circumstances, h = total

mechanical energy of the system.

Physical Interpretation of h• Energy Function h ∑jqj(∂L/∂qj) - L

• Recall (Sect. 1.6) that we can always write KE as:

T = M0 + ∑jMjqj + ∑jMjkqjqk

M0 (½)∑imi(∂ri/∂t)2 , Mj ∑imi(∂ri/∂t)(∂ri/∂qj)

Mjk ∑i mi(∂ri/∂qj)(∂ri/∂qk)

Or (schematically) T = T0(q) + T1(q,q) + T2(q,q)

– T0 M0 independent of generalized velocities

– T1 ∑jMjqj linear in generalized velocities

– T2 ∑jMjkqjqk quadratic in generalized velocities

• With almost complete generality, we can write (schematically) the Lagrangian for most problems of interest in mechanics as:

L = L0(q,t) + L1(q,q,t) + L2(q,q,t)

L0 independent of the generalized velocities

L1 linear in generalized the velocities

L2 quadratic in generalized the velocities

– For conservative forces, L has this form. Also does for some velocity dependent potentials, such as for EM fields.

L = L0(q,t) + L1(q,q,t) + L2(q,q,t) (1)

• Euler’s Theorem from mathematics:

If f = f(x1,x2,.. xN) = a homogeneous function of degree n of the variables xi, then

∑ixi(∂f/∂xi) = n f (2)

• Energy Function h ∑jqj(∂L/∂qj) - L (3)

• For L of form (1):

(2) h = 0L0 + 1L1+2L2 - [L0 + L1 + L2]

or h = L2 - L0

L = L0(q,t) + L1(q,q,t) + L2(q,q,t)

Energy function h ∑jqj(∂L/∂qj) - L = L2 - L0

• Special case (both conditions!):

a.) The transformation eqtns from Cartesian to

Generalized Coords are time indep.

In the KE, T0 = T1 = 0 T = T2

b.) V is velocity indep. L2 = T = T2 & L0 = -V

h = T + V = E Total Mechanical Energy• Under these conditions, if V does not depend on t, neither does L & thus

(∂L/∂t) = 0 = (dh/dt)

so h = E = constant (conserved)

Energy Conservation • Summary: Different Conditions:

Energy Function h ∑jqj(∂L/∂qj) - L ALWAYS: (dh/dt) = - (∂L/∂t)

SOMETIMES: L does not depend on t (∂L/∂t) = 0, (dh/dt) = 0 & h = const. (conserved)

USUALLY: L = L0(q,t) + L1(q,q,t) + L2(q,q,t)

h ∑jqj(∂L/∂qj) - L = L2 - L0

SOMETIMES: T = T2 = L2 AND L0 = -V h = T + V = E Total Mechanical Energy Conservation Theorem for Mechanical Energy: If h = E AND L does not depend on t, E is conserved!

• Clearly, the conditions for conservation of energy function h are DISTINCT from those which make it the total mechanical energy E.

Can have conditions in which:

1. h is conserved & = E

2. h is not conserved & = E

3. h is conserved & E

4. h is not conserved & E

Most common case in classical (& quantum) mech. is case 1.

• Stated another way: Two questions:

1. Does the energy function h = E for the system?

2. Is the mechanical energy E conserved for the system?

• Two aspects of the problem! DIFFERENT questions! – May have cases where h E, but E is conserved.– For example: A conservative system, using generalized coords in motion

with respect to fixed rectangular axes:

Transformation eqtns will contain the time

T will NOT be a homogeneous, quadratic function

of the generalized velocities!

h E, However, because the system is conservative, E is conserved! (This is a physical fact about the system, independent of coordinate choices!).

• It is also worth noting:

The Lagrangian L = T - U is independent of the choice of generalized coordinates.

The Energy function h ∑jqj(∂L/∂qj) - L depends on the choice of generalized coordinates.

• The most common case in classical (& quantum) mechanics is h = E and E is conserved.

Non-Conservative Forces • Consider a non-conservative system: Frictional forces obtained

from the dissipation function ₣ . Derivation with the energy function h becomes:

(dh/dt) + (∂L/∂t) = - ∑jqj(∂₣ /∂qj)

• Ch. 1: The formulation of ₣ shows it is a homogeneous, quadratic function of the q’s.

Use Euler’s theorem again: ∑jqj(∂₣ /∂qj) = 2₣

(dh/dt) = - (∂L/∂t) - 2₣ • If L is not an explicit function of time (∂L/∂t =0 ) AND h = E:

(dE/dt) = - 2₣ – That is, under these conditions, 2₣ = Energy dissipation rate.

Symmetry Properties & Conservation Laws (From Marion’s Book!)

• In general, in physical systems:

A Symmetry Property of the System

Conservation of Some Physical Quantity

Also: Conservation of Some Physical Quantity

A Symmetry Property of the System • Not just valid in classical mechanics! Valid in

quantum mechanics also! Forms the foundation of modern field theories (Quantum Field Theory, Elementary Particles,…)

We’ve seen in general that:

Conservation Theorem: If the Generalized Coord qj is cyclic or ignorable, the corresponding Generalized (or Conjugate) Momentum, pj (∂L/∂qj) is conserved.

• An underlying symmetry property of the system:

If qj is cyclic, the system is unchanged (invariant) under a translation (or rotation) in the “qj direction”.

pj is conserved

Linear Momentum Conservation

• Conservation of linear momentum:

If a component of the total force vanishes,

nF = 0, the corresponding component of total

linear momentum np = const (is conserved)• Underlying symmetry property of the system:

The system is unchanged (invariant) under a translation in the “n direction”.

np is conserved

Angular Momentum Conservation

• Conservation of angular momentum:

If a component of total the torque vanishes,

nN = 0, the corresponding component of total angular momentum nL = const (conserved)

• Underlying symmetry property of the system: The system is unchanged (invariant) under a rotation about the “n direction”.

nL is conserved

Energy Conservation

• Conservation of mechanical energy:

If all forces in the system are conservative, the total mechanical energy E = const (conserved)

• Underlying symmetry property of the system: (More subtle than the others!) The system is unchanged (invariant) under a time reversal.

(Changing t to -t in all eqtns of motion)

E is conserved

Summary: Conservation Laws

• Under the proper conditions, there can be up to 7 “Constants of the Motion” “1st Integrals of the Motion” Quantities which are Conserved (const in time):

Total Mechanical Energy (E)

3 vector components of Linear Momentum (p)

3 vector components of Angular Momentum (L)