Harmonic Osillator and Least Action

[tgupta2000@gmail.com]

Could someone elaborate on the steps to obtain the equation of motion
of a harmonic oscillator using the principle of least action, and then
extending it to a damped oscillator?
So, if L = T - V,
L = 0.5 m [ (xdot)^2 - (omega)^2 * (x)^2 ]

S = Integrate [ L dt ] ....

Is it essential to use trial functions?

How does one extend this to a damped oscillator?

Thanks in advance

[jambaugh]

You are asking for the derivation of the Euler-Lagrange equations which you can find in any text on Lagrangian mechanics. Here's the basic outline:

S = \int_{t_1}^{t_2} L dt
\delta S= \int \delta L dt = \int \left(\frac{\partial L}{\partial x}\delta x + \frac{\partial L}{\partial \dot{x}} \delta \dot{x}\right) dt
One argues that \delta \dot{x} = \frac{d}{dt}(\delta x) and integrate that term by parts.

\delta S =\int \left(\frac{\partial L}{\partial x}\delta x - \frac{d}{dt}\frac{\partial L}{\partial \dot{x}} \delta x\right) dt + \frac{\partial{L}}{\partial\dot{x}}\delta x|^{t_2}_{t_1}

One usually chooses the variations in the path \delta x(t) so that there is no variation at the end points so that the boundary term goes away.

\delta S =\int \left(\frac{\partial L}{\partial x} - \frac{d}{dt}\frac{\partial L}{\partial \dot{x}}\right)\delta x dt

One also assumes arbitrary variation in the path which means:
\delta S = 0 \Rightarrow \frac{\partial L}{\partial x} - \frac{d}{dt}\frac{\partial L}{\partial \dot{x}}=0
or
\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}=\frac{\partial L}{\partial x}
which are the famous Euler-Lagrange equations. Apply to the Lagrangian of your choice.

A dampened harmonic oscillator is an example of dissipative systems which aren't as easily described in Lagrangian terms. I'm not totally up to speed on Lagrangian methods for dissipative systems. There are lots of Google hits under "lagrangian dissipative" if you want to explore further.

[Igor Khavkine]

On Feb 2, 9:50 pm, tgupta2...@gmail.com wrote:
> Could someone elaborate on the steps to obtain the equation of motion
> of a harmonic oscillator using the principle of least action, and then
> extending it to a damped oscillator?

To find out more about the difficulties in describing damped/
dissipative systems with a Lagrangian formalism, you may find helpful
a section of Arnold Neumaier's excellent Theoretical Physics FAQ:

http://www.mat.univie.ac.at/~neum/physics-faq.txt
S1p. Dissipative dynamics and Lagrangians

Hope this helps.

Igor

[a student]

On Feb 3, 7:50 am, tgupta2...@gmail.com wrote:
> Could someone elaborate on the steps to obtain the equation of motion
> of a harmonic oscillator using the principle of least action, and then
> extending it to a damped oscillator?
> So, if L = T - V,
> L = 0.5 m [ (xdot)^2 - (omega)^2 * (x)^2 ]
>
> S = Integrate [ L dt ] ....
>
> Is it essential to use trial functions?
>
> How does one extend this to a damped oscillator?
>
> Thanks in advance

A Lagrangian for the damped oscillator was mentioned in an old
thread
( http://sci.tech-archive.net/Archive/sci.physics.research/2007-12/msg00065.html )
In particular, using explicit mass units, the Lagrangian is given by
L = (1/2) M [ (dx/dt)^2 - w^2 x^2 ]
where
M(t) := m exp[ \int_0^t ds b(s) ] .
One then obtains the corresponding equation of motion
d^2 x / t^2 + b dx/dt + w^2 x = 0
as desired.

Note if b is a constant, then
M(t) = m exp[ b t ].

[mpalenik (He-Jutsu)]

What's the justification for that redefinition of mass in the post you
linked to? The only justfication I can see is that it presupposes a
solution with exponential decay of amplitude, in otherwords, the mass
is redefined so that the answer matches a known solution. It doesn't
really seem like a method of FINDING the solution. Am I right in my
assessment, or is there something else going on that I missed?

On Feb 5, 1:55*pm, a student <of_1001_nig...@hotmail.com> wrote:
> On Feb 3, 7:50 am, tgupta2...@gmail.com wrote:
>
> > Could someone elaborate on the steps to obtain the equation of motion
> > of a harmonic oscillator using the principle of least action, and then
> > extending it to a damped oscillator?
> > So, if L = T - V,
> > L = 0.5 m [ (xdot)^2 - (omega)^2 * (x)^2 ]
>
> > S = Integrate [ L dt ] ....
>
> > Is it essential to use trial functions?
>
> > How does one extend this to a damped oscillator?
>
> > Thanks in advance
>
> A Lagrangian for the damped oscillator was mentioned in an old
> thread
> (http://sci.tech-archive.net/Archive/sci.physics.research/2007-12/msg0...)
> In particular, using explicit mass units, the Lagrangian is given by
> * * * L = (1/2) M [ (dx/dt)^2 - w^2 x^2 ]
> where
> * * *M(t) := m exp[ \int_0^t ds b(s) ] .
> One then obtains the corresponding equation of motion
> * * *d^2 x / t^2 + b dx/dt + w^2 x = 0
> as desired.
>
> Note if b is a constant, then
> * * M(t) = m exp[ b t ].

[a student]

On Feb 9, 10:38Â*am, "mpalenik (He-Jutsu)" <markpale...@gmail.com>
wrote:
> What's the justification for that redefinition of mass in the post you
> linked to? Â*The only justfication I can see is that it presupposes a
> solution with exponential decay of amplitude, in otherwords, the mass
> is redefined so that the answer matches a known solution. Â*It doesn't
> really seem like a method of FINDING the solution. Â*Am I right in my
> assessment, or is there something else going on that I missed?

What is presupposed in the thread
http://sci.tech-archive.net/Archive/sci.physics.research/2007-12/msg00065.html
is that the equation of motion for some physical
system parameterised by position x is given by
d^2 x / t^2 + b dx/dt + w^2 x = 0 ,
where b and w are given functions of time. The problem is then find
a
Lagrangian or Hamiltonian for the system which leads to that equation
of motion, via an action principle (and which reduces to the usual
oscillator case for b=0). Note that no actual solutions to the
equation of motion itself are presupposed, or anything about the
"mass" of the system.

The fact that the Lagrangian given in the thread has what looks like
a
time-dependent mass M(t) is a matter of interpretation, having no real
physical content (based on what the Lagrangian looks like in the
no-damping case b=0). Note that any desired positive function M(t)
is
possible, not just exponential functions, by choosing
b = (dM/dt) / M

Of course, even though the "correct" equation of motion follows, the
given Hamiltonian and Lagrangian are not physically fundamental, as
they do not model the physical causes of damping (even in the
classical
case) - i.e., dissipation via interaction with other systems.
However, I
suppose that if one did actually have a physical system with a
time-dependent mass, then one might take the Lagrangian to be
physically appropriate.

[John Polasek]

On Mon, 2 Feb 2009 20:50:25 +0000 (UTC), tgupta2000@gmail.com wrote:

>Could someone elaborate on the steps to obtain the equation of motion
>of a harmonic oscillator using the principle of least action, and then
>extending it to a damped oscillator?
>So, if L = T - V,
>L = 0.5 m [ (xdot)^2 - (omega)^2 * (x)^2 ]
>
>S = Integrate [ L dt ] ....
>
>Is it essential to use trial functions?
>
>How does one extend this to a damped oscillator?
>
>Thanks in advance
Goldstein discusses this briefly as Rayleigh's dissipation function,
which he inserts into the Lagrangian on page 22. The coverage
extendsto one trivial exercise, as linear or quadratic friction mucks
things up properly.
John Polasek

[Igor Khavkine]

On Feb 5, 7:55 pm, Igor Khavkine <igor...@gmail.com> wrote:
> On Feb 2, 9:50 pm, tgupta2...@gmail.com wrote:
>
> > Could someone elaborate on the steps to obtain the equation of motion
> > of a harmonic oscillator using the principle of least action, and then
> > extending it to a damped oscillator?
>
> To find out more about the difficulties in describing damped/
> dissipative systems with a Lagrangian formalism, you may find helpful
> a section of Arnold Neumaier's excellent Theoretical Physics FAQ:
>
> http://www.mat.univie.ac.at/~neum/physics-faq.txt
> S1p. Dissipative dynamics and Lagrangians

Having spent some time mulling it over, I'd like to add some details
to
Arnold's discussion. Here's a quote from the FAQ:

] Of course, in theory, a dissipative system is thought to be a
] contracted version of a bigger conservative system which includes
the
] envoironment, and in simple situations, this theoretical view can
] indeed be substantiated.
]
] If one models the dissipative environment explicitly, on gets a
] bigger conservative system, not a dissipative system. Of course,
this
] conservative system has a Hamiltonian or Lagrangian description, but
] it does not describe the dissipative system alone. When one
contracts
] it to the degrees of freedoms of the original system, one gets an
] integro-differential equation with memory, which is no longer
] described by a Hamiltonian or Lagrangian framework.

I would like to show in more detail how this contraction, from a large
conservative system to a small system coupled to a dissipative
environment, can be done. Then I'll open the question of whether the
resulting system is no longer in a Hamiltonian or Lagrangian
framework.

For simplicity, assume we are dealing with a linear system, where one
degree of freedom (x) is singled out as the "oscillator" and the rest
are labelled as the "environment". This coupled system has the
following Lagrangian and Hamiltonian:

L = 1/2 (m (x')^2 - k x^2) - x F + L_e ,

H = 1/2 (p^2/m + k x^2) + x Q + H_e .

m - oscillator mass
k - oscillator stiffness
Q - driving force due to the environment
L_e, H_e - environment Lagrangian and Hamiltonian, respectively

The equations of motion for the oscillator are

m x''(t) + k x(t) = -Q(t) .

The specific equations of motion for the environment are not
important,
so long as they are linear and inhomogeneous, with a driving force
proportional to x, just as the oscillator is driven by a force
proportional to Q (generalized Newton's third law in action).

Imposing initial conditions on the environment at t=0, but keeping the
motion of x(t) arbitrary for the moment, the equations of motion for
the environment can be solved uniquely. Specifically, the driving
force
can be expressed as

-Q(t) = int_0^t G(t-s) x(s) ds + F(t)
= int_0^t G(s) x(t-s) ds + F(t) ,

where G(t) depends on the specific dynamics of the environment, but is
otherwise fixed, while F(t) corresponds to the motion of the
environment with given initial conditions at t=0 if x(s) were set to
0.
This form is fairly general and deriving it is a simple exercise in
using Green functions to solve an inhomogeneous system of linear
equations. I recommend it to anyone interested in the matter. NB:
lower
limit of the integral is fixed because of the choice of t=0 for
specifying the boundary conditions. I will refer to G(t) as the memory
function and to F(t) as the noise.

As expected, the equation of motion for the oscillator becomes and
integro-differential equation

m x''(t) + k x(t) = int_0^t G(s) x(t-s) ds + F(t) .

If the environment is sufficiently large, the memory function G(t) may
take on a (perhaps not quite) arbitrary shape. Specifically, it may be
highly peaked about t=0, in which case the following expansion is
warranted:

int_0^t G(s) x(t-s) ds
~ int_0^t G(s) [x(t) - s x'(t) + s^2/2 x''(t) + ...] ds
~ dk x(t) - b x'(t) + dm x''(t) + ...

Truncating the expansion of x'(t-s) at some finite number of
derivatives is generally known as a Markov approximation. Note that if
G(t) is strongly peaked about t=0, then for t larger than the width of
that peak we can neglect the time dependence of the coefficients
dk, b and dm. The oscillator equation of motion now becomes

(m-dm) x''(t) + b x'(t) + (k-dk) x(t) = F(t) ,

which is none other than the sought equation of a damped harmonic
oscillator with a driving force as well as renormalized mass and
stiffness. An interesting feature of this derivation is the automatic
appearance of the noise F(t) along with the dissipative term b x'(t).
Both of these terms appear automatically when the environment degrees
of freedom are contracted. This is a well known and quite general
result which goes under the name of the Fluctuation-Dissipation
Theorem.

Now, here's where things get interesting. Even though I haven't
specified the dynamics of the environment explicitly, the total
oscillator + environment system is conservative and has a well defined
Hamiltonian description. Specifically, the total Hamiltonian H is
conserved and it also generates the canonical equations of motion

x' = {x,H} ,
p' = {p,H} ,

with the Poisson brackets defined as usual. At a fixed time t, the
oscillator variables Poisson commute with the environment variables

{x(t),Q(t)} = {x(t),H_e(t)} = 0 ,
{p(t),Q(t)} = {p(t),H_e(t)} = 0 .

When the environment driving force is decomposed into memory and
noise,
the following relations are implied:

{x(t), int G(s) x(t-s) ds} + {x(t),F(t)} = 0 and
{p(t), int G(s) x(t-s) ds} + {p(t),F(t)} = 0 .

A straightforward calculation also shows that

{F(t),F(s)} = G(s-t) .

In essence, the last three equations together with the Hamiltonian

H = 1/2 (p^2/m + k x^2) - x (int_0^t G(s) x(t-s) ds + F) + H_e,

where F(t) and H_e(t) are formal variables whose internal dynamics
are specified only through the memory function G(t).

In particular, we can set

G(t) = - b delta'(t) ,

then

int_0^t G(s) x(t-s) ds = - b x'(t) = - b p/m
H = 1/2 (p^2/m + k x^2) + x (b p/m - F) + H_e
x' = {x,H} = p/m ,
p' = {p,H} = -k x - b p/m + F
{x,F} = b/m
{p,F} = 0

and

m x'' + b x' + k x = F .

In a certain sense, the above equations define the damped harmonic
oscillator with noise as a Hamiltonian system.

Now, on to the Lagrangian. The total action is

S = int dt [1/2 (m (x')^2 - k^2 x^2) + x (G*x + F)] + S_e ,

where G*x denotes the convolution of the memory function with x(t),
which was previously written out in full. In principle, this is the
value of the action extremized over the environment degrees of
freedom,
keeping x(t) fixed. Extremizing again with respect to x(t) we should
recover the right equations of motion for the oscillator, with memory
and noise terms.

A subtle point here has to do with the boundary conditions on
environment variables. F(t) was defined as the value that Q(t) should
assume in the absence of coupling to the oscillator, and with fixed
boundary conditions at t=0. However, for the purposes of the
variational principle, fixed boundary conditions should be imposed at
two different times t1 and t2 and not at t=0. This is accomplished by
making the t=0 initial conditions defining F(t) depend on x(t). This
point is important because it requires F(t) to have non-trivial
variation with respect to x(t).

Let & denote the functional variation, then

&S = int dt [-&x (m x''+ k x - G*x - F) + x (G*&x + &F)] + &S_e ,

with &S_e + int dt x (G*&x + &F) = 0 .

Just as in the Hamiltonian case, a Lagrangian formulation of an
oscillator with memory and noise seems possible, as long as the memory
function G(t) is specified, along with the formal noise, F(t), and
environment action, S_e, are specified, whose variations with respect
to x(t) satisfy the last equation. Again, we could declare G(t) to be
singular, as before, and thus define a damped harmonic oscillator in a
Lagrangian framework.

The above exercise was rather lengthy, but I think instructive, at
least for me. I've not seen similar treatments of classical linear
systems with noise and memory elsewhere, however I'd think that these
properties would be well known by experts in some field. I have
however seen discussions of so called "quantum noise", which is very
similar to the Hamiltonian case above, though in a quantum setting.
Incidentally, I synthesized this discussion in the classical case from
section 17.4 of Mandel & Wolf's book on Quantum Optics, which treats
quantum noise.

It's not clear to me why the common discussions of the classical
Langevin equation, which precisely describes a system with dissipation
and noise, don't mention the fact that the Hamiltonian structure of
the
theory may be preserved, at least in a formal sense. Perhaps it's not
important in the same way as preserving the right commutation
relations
is in the case of quantum noise.

So, after this exercise, should it still be argued that classical
systems with dissipation and noise cannot be given a (perhaps only
formal) Hamiltonian or a Lagrangian description?

Igor

[mpalenik (He-Jutsu)]

What's the justification for that redefinition of mass in the post you
linked to? The only justfication I can see is that it presupposes a
solution with exponential decay of amplitude, in otherwords, the mass
is redefined so that the answer matches a known solution. It doesn't
really seem like a method of FINDING the solution. Am I right in my
assessment, or is there something else going on that I missed?

On Feb 5, 1:55*pm, a student <of_1001_nig...@hotmail.com> wrote:
> On Feb 3, 7:50 am, tgupta2...@gmail.com wrote:
>
> > Could someone elaborate on the steps to obtain the equation of motion
> > of a harmonic oscillator using the principle of least action, and then
> > extending it to a damped oscillator?
> > So, if L = T - V,
> > L = 0.5 m [ (xdot)^2 - (omega)^2 * (x)^2 ]
>
> > S = Integrate [ L dt ] ....
>
> > Is it essential to use trial functions?
>
> > How does one extend this to a damped oscillator?
>
> > Thanks in advance
>
> A Lagrangian for the damped oscillator was mentioned in an old
> thread
> (http://sci.tech-archive.net/Archive/sci.physics.research/2007-12/msg0...)
> In particular, using explicit mass units, the Lagrangian is given by
> * * * L = (1/2) M [ (dx/dt)^2 - w^2 x^2 ]
> where
> * * *M(t) := m exp[ \int_0^t ds b(s) ] .
> One then obtains the corresponding equation of motion
> * * *d^2 x / t^2 + b dx/dt + w^2 x = 0
> as desired.
>
> Note if b is a constant, then
> * * M(t) = m exp[ b t ].