Hello all readers.

I have recently thought what is the Lagrangian for a harmonic oscillator
with damping. For instance, consider the equation of motion

(A) x'' = (-k/m)x - ax'

Actually, the problem arises for a simple Newtonian body moving with
friction. If I'm right, many of the readers probabely know this, but the
conclusion I arrived at is that there is no Lagrangian. And that's why:
assume otherwise, i.e., that the above models may be formulated using a
Lagrangian, and therefore, eventually put into a Hamiltonian form. This will
automatically yield a conserved quantity (i.e. the energy) which is not
something trivial - otherwise, the dynamics would be trivial. Such a
conserved quantity simply does not exist for the above models. Another
argument comes from the Liouville theorem - damping results in an evident
reduction of the volume occupied by the possible states of the oscillator -
as these states reduce to x=0, v=0 with time. So, what is the common feature
of physically meaningful, but non-Hamiltonian - hence, non-quantizable -
models? There is another group of models which come to mind - physics living
on manifolds-with-boundary. For instance, a billiard exhibits
non-smouthness, and as a consequence, non-"Hamiltonianess". However, these
second type non-Hamiltonian models may be APPROXIMATED by Hamiltonian ones -
in the billiard case, those would be particles in a shraply rising potential
near the walls. Therefore, if we exclude the near-Hamiltonian models, only
one type seems to remain - models which describe non-adiabatic - i.e. with
heat transfer - procceses. Another example would be the billiard with energy
loss added in wall reflections. The Liouville argument shows the entropy
changes in these models what again implies them being characterized by
non-adiabaticity. This also relates to the possibility of quantizing, which
in the usual cases results in unitary, thus information preserving = entropy
conserving models. This is, therefore, not surprising that the definition of
quantum entropy S = -kTr(rho ln rho) leads to a conserved quantity, even
when the Hamiltonian changes with time. This is interesting, how may we
describe non-adiabatic models after all, and so, address the issue of
quantum entropy.

Best regards, squark.

How about:

  L = 1/2 (m(x')^2 - k x^2) + max't

(I used to think that only adiabatic systems have Lagrangians and
Hamiltonians)

This one fails. The canonical conjugate of x is (mx' + mat)

So the Hamiltonian:

  H = 1/2 (  m (x')^2 - k x^2 )

The time dependent term has canceled.
I think Hamilitonians indeed only work for energy conserving systems.
But Lagrangian systems can be more general.

I remember something about "the principle of minimum power", which is
general enough to include dissipation, contrary to the principle
of minimum energy.

[..]

The non adiabaticity arises when you let heat escape. If you include
the heat by modeling the heat explicitly as individual molecules,
it becomes adiabatic.
The problem is that a "deterministic" model has no entropy. You
would have to specify what you pretend you don't know about the
system. Then you have to count all systems that could fulfill
the requirements set by what you do know.

Gerard

Nah. If there is a Lagrangian formulation then there is a
Hamiltonian formulation and vice versa. The correspondence is
easiest when there are no constraints or the Lagrangian has a
non-degenerate Hessian. In that case the usual Legendre transform
suffices to move back and forth. In the more general case one can
use the methods of Dirac.

Conservation of energy is not required for a
Lagrangian/Hamiltonian formulation. For example, one can write
down a Lagrangian/Hamiltonian for a charged particle moving in a
prescribed time-varying electromagnetic field. The energy of the
particle is not, in general, conserved. (Better: there is no
conserved quantity that you would like to identify as energy.)

Given a set of DEs, when is there a Lagrangian (or Hamiltonian)?
This is a very old problem in mathematics called "the inverse
problem in the calculus of variations". A lot is known about it.
It goes back at least to Helmholtz!

See for example the paper by Anderson and Thompson,  in Memoirs
of the AMS, Volume 98, number 473, 1992.

Charles Torre

 P.S.

The true Lagrangian for an oscillator with damping is

(C) L = 1/2 (mx'^2-kx^2) exp(at)

It yields the conjugate momentum

(D) p = mx'exp(at)

and the equation of motion (A). The Hamiltonian is

(E) H = 1/2 (mx'^2+kx^2) exp(at)

to which the said before still applies. The exponent in (D) also solves
the problem with Liouvilles theorem.

Best regards, squark.

[Note for the moderator: it may be reasonable to link the three posts I
sent into a single one for the convenience of the readers]

Sent via Deja.com http://www.deja.com/
Before you buy.

I'm not sure this is true.  I can write down a Lagrangian with an
infinite number of time derivatives.  It's rather unclear how to turn
this into a Hamiltonian.  (This is a particularly perverse case, but
not completely without interest.)

--
======================================================================
Kevin Scaldeferri                       Calif. Institute of Technology
                      The INTJ's Prayer:
     Lord keep me open to others' ideas, WRONG though they may be.

In article <395242FD.E9448...@xs4all.nl>,
  Gerard Westendorp <west...@xs4all.nl> wrote:

Wait a second. This yields p = mx' + mat, thus the Euler-Lagrange
equation is mx'' = -kx - ma, NOT (A).

Best regards, squark.

Sent via Deja.com http://www.deja.com/
Before you buy.

In article <8j0e95$...@gap.cco.caltech.edu>, ke...@cco.caltech.edu (Kevin A. Scaldeferri) writes:

Well, ahem, hmmm. Okay. But you are stretching the definition of
"Lagrangian" a bit from its usual one ( I guess I thought
that the more or less standard classical mechanics kind of
Lagrangian was being discussed). As usual, most disagreements
are more about definitions than consequences of definitions. Normally
one defines a Lagrangian as a local function of the spacetime coordinates,
the dynamical variables and a finite number of their derivatives, which
is what I had in mind.

I guess I would call this kind of thing you are mentioning
a "non-local Lagrangian". It becomes tricky to say much about
these beasts in general except in a very formal, non-rigorous way.  
What are the Euler-Lagrange equations equations
for such a beast? A formal infinite series? How do you know when it
converges? Noether's theorem? Etc. I guess I don't see that many of the usual
nice tools of Lagrangian dynamics come easily into play when the Lagrangian is
non-local. In such cases where these things are
appearing, it is probably best to just forget Lagrangians and work at
the level of action functionals (that's just a gut feeling of mine).

I would be interested in hearing some more about
cases where it is useful to think in terms of Lagrangians with an
infinite number of derivatives. Oh. Maybe you are thinking about
effective actions in quantum theory? Fair enough. (Still,
even though an infinite number of derivatives may arise in a
derivative expansion of the "Lagrangian", one usually truncates
to a finite number of terms in any perturbative computation.
Then what I said still applies.)

Charles Torre

But in squarks example, a term in the Lagrangian in canceled
by the Legendre transformation. Thus, the Hamiltonian does
not include the term for damping.

So what Hamiltonian do you suggest for a damped oscillator?

Or even simpler, a first order system, like

  x' = ax

?

Gerard

In article <395242FD.E9448...@xs4all.nl>,
  Gerard Westendorp <west...@xs4all.nl> wrote:

Cool!

That's because you are using the wrong variables! Lets express the
Hamiltonian through the canonically conjugate x and p = mx' + mat:

(B) H = 1/2 ( (p - mat) / m^2 - k x^2 )

As you see, there is time dependance. When we use x and x', the
sympletic structure carries the time dependance. The funny thing is,
that we started with the time-translation invariant equation (A), and
got the time-translation-asymmetric Hamiltonian (B). That's where the
non-adiabatic nature reveals itself. This also explains where the
Noether theorem fails.

Hmm, my plan failed. The entropy -k Tr(rho ln rho) doesn't rise.

Best regards, squark.

Sent via Deja.com http://www.deja.com/
Before you buy.

Yes, sorry about that.

Gerard

Hey, that is cool.
It also allows you to interpret H as the total energy,
which decays with exp(at).

Still haven't found the Lagrangian for

  x' = ax

Any suggestions?

Gerard

No, I think I got it wrong.

Squark has already suggested a better Lagrangian.

Gerard

Why should there be one?

There is no Lagrangian, L=L(x,x'), such that the differential equation
x' = ax is the Euler-Lagrange equation of L. Likewise,
there is no corresponding Hamiltonian related by the Legendre dual
transformation.

How do I know?  If a system of DEs are Euler-Lagrange equations then
their linearization defines a formally self-adjoint differential operator
(this necessary condition is in fact locally sufficient).
This test for the existence of a Lagrangian constitutes the "Helmholtz
conditions". It is represents the tip of the iceberg in the study of the
inverse problem in the calculus of variations.

Since your equation is already linear it is pretty easy
to construct its linearization ;) . Since this linear operator is not
formally self-adjoint there ain't no Lagrangian.  

Some (many? most?) dynamical systems just don't admit variational principles.
I guess that means that they cannot be considered "quantizable" and hence
are somehow not "fundamental".  What's perhaps a little more amusing in
this regard are the systems that admit more than one Lagrangian since now
one has in priniciple different quantum theories with the same classical
limit and one has to decide which Lagrangian nature uses -  and why.

Charles Torre

Actually, the energy 1/2 (mx'^2+kx^2) decays with exp(-at).
And H stays constant.

Gerard

This was in reaction to the statement that adiabaticity is not required.
There may be other requirements, though.

[..]

What is formally self-adjoint?

Gerard

..

Could you explain what you mean by  "amusing", please, for us simple
folk who don't follow the nuance.   What's so funny and why?

--
Best regards,
Ralph E. Frost
Frost Low Energy Physics

"I go the way of all the earth. Be thou strong therefore, and show
thyself a man."  1 Kings 2:2

http://www.dcwi.com/~refrost/index.htm

This simple problem, amusingly, admits an explicitly time-dependent
Lagrangian and a corresponding time-dependent Hamiltonian.  The latter is
the energy of the oscillator (as it is usually defined) at t=0.  Naturally
it is conserved.

If you try this procedure with a multidimensional oscillator that does not
have the same exponential decay factors for all the modes you will typically
not be able to find a time-dependent Lagrangian such as the one above that
yields the equations of motion.  Instead you will need a second function F
known as Rayleigh's dissipation function which in combination with the usual
Lagrangian L yields the correct equations of motion:

  (d/dt)(d_L/d_x') - (d_L/d_x) + (d_F/d_x') = 0

(where x is one of the generalized coordinates, and d_ indicates a partial
derivative).  See Section 1.5 (pp. 23-24) of the second edition of Herbert
Goldstein's _Classical Mechanics_.

The dissipation function approach is not useful for certain fundamental
problems such as calculating entropy or deriving fluctuation-dissipation
theorems.  If that is your goal, I'd like to suggest the following exercise.

Take an electrical analogue of your damped oscillator, replacing the mass
and spring by and inductor and capacitor.  But do not use a resistor in
in place of the viscous damping pot;  instead, substitute a semi-infinite
transmission line.  This has the same terminal characteristics as a
resistor, but is an explicitly conservative system.  Now write the
Lagrangian or Hamiltonian for the combined system.  You will find, if you
work out the details, that the resistor is only part of the equivalent
circuit of the transmission line.  The other part is a source (voltage
source in series or current source in parallel).  If you consider an
initial value problem starting at t=0, then the resistor accounts for
the energy which flows from the resonant circuit into the transmission
line, while the source accounts for the energy initially stored in the
transmission line that flows into the resonant circuit.  If you consider
what has to be stored in the transmission line prior to t=0 to give rise
to specified initial conditions, you will find that the system is
time-reversal invariant (a T reversal swaps the roles of the resistor
and the source).  Next put the system in thermal equilibrium and determine
the power spectrum of that source.  You will have re-derived a
fluctuation-dissipation theorem first proved by H. Nyquist in 1928.

Best regards,

Mike Heard

You are right, this is a non-local Lagrangian.  To let the cat out of
the bag, what I am thinking of is the sort of Lagrangian encountered
in non-commutative field theories.

These cases are different from an effective action.  There is not a
momentum expansion the way there usually is in an effective theory.

OTOH, as I said, this case is a little perverse as the theories with
infinite numbers of time derivatives are not unitary.  The theories
with infinite space derivatives are okay, though.

--
======================================================================
Kevin Scaldeferri                       Calif. Institute of Technology
                      The INTJ's Prayer:
     Lord keep me open to others' ideas, WRONG though they may be.

In article <8ivp57$vh...@nnrp1.deja.com>,

Unfortunatelly, the quantization of this system doesn't make any sense
at all. The Shordinger equation generated my (E) may yield some kind of
damping, but (D) leads to a modification of the Heisnberg uncertainty,
which is altogether unreasonable: (delta x)(delta v) >= exp(-at)hbar/2m

Best regards, squark.

Sent via Deja.com http://www.deja.com/
Before you buy.

In article <3957EF3E.30451...@xs4all.nl>,
  Gerard Westendorp <west...@xs4all.nl> wrote:

If you're right, this means the Noether theorem is still valid in this
case - simply it yields a quantaty with EXPLICIT time dependance.

Best regards, squark.

Sent via Deja.com http://www.deja.com/
Before you buy.

Right. Well, there is one interesting exception. If 2
Lagrangians, say L_1 and L_2 have the same Euler-Lagrange
equations then their difference,

L_0 = L_1 - L_2

must have identically vanishing Euler-Lagrange equations. L_0 is
sometimes called a "null Lagrangian". Locally, null Lagrangians
can be expressed as a total derivative (or total divergence in a
field theory) just as you say, but this may not be true globally
if the configuration space of the theory has some topology. I
believe there is a theorem to the effect that, for a field
theory on an n-dimensional manifold (n=1 means mechanics), to
every representative of a degree n cohomology class on the
bundle of independent and dependent variables (i.e., the bundle
of fields) one can construct a Lagrangian that is not a total
divergence, but nevertheless has identically vanishing
Euler-Lagrange equations. To get an interesting example one
probably needs some cohomology in "field space" (rather than
just in spacetime). Probably I could cook up some examples if
you are perverse enough to really be interested in this
phenomenon. Anyway, this wasn't really what I was thinking of
when I made the comment about different Lagrangians and
quantization. As you say...

Excellent point. The more interesting possibility is that one
could have two Lagrangians whose Euler-Lagrange (EL) equations
are *equivalent* instead of identical. (I had inadvertently
drifted into this point of view when I made the comment about
inequivalent Lagrangians and quantum theory. Thanks for keeping
me honest.) This point of view gives a much more useful (and
much harder) form of the inverse problem in the calculus of
variations: when is there a Lagrangian whose EL equations are
*equivalent* (rather than identical) to a specified set of
equations. I say that this point of view is more useful since
one often times does not have equations expressed in just the
right form to be EL equations, even though there is an
underlying Lagrangian for the dynamical system of interest. For
example, the vacuum Einstein equations G_ab=0 (G is the Einstein
tensor) are not the EL equations of any Lagrangian. (Wait! Don't
shoot until after you read the next two sentences.) But they are
equivalent to a system of equations E_ab=0 which ARE EL
equations. Here E is the Einstein tensor multiplied by the
square root of the determinant of the metric.

The paper by Anderson and Thompson that I cited earlier in this
thread gives, I think, a pretty near state of the art treatment
of this more general type of inverse problem for ODEs. The PDE
version of this inverse problem is, I think, in a much more
primitive state. As I recall, in that paper you will find examples
of DEs which admit more than one Lagrangian such that the
various Lagrangians do not differ by a total derivative or
constant rescaling. These examples do NOT arise because of the
topological subtleties that I mentioned earlier, but rather
because of the freedom to choose alternative, but equivalent,
equations of motion. (Sorry. I don't have the paper available so
I can't whip out one of their examples. )

Charles Torre

Actually, here is one:

   L = 1/2 (y')^2 exp(-at)

This gives:

  y'' = ay'

Then, substitute x = y'

Gerard.

[Regarding the equation of motion x'=ax]

This is differential equation terminology. I guess the
quickest way to define it here would be to introduce a
scalar product (f,g), which is the just the integral (over,
say, t) of the product of f=f(t) and g=g(t) (over some
region). The linearization of a differential equation
defines a linear differential operator, call it L. Let's
denote by Lf the action of this operator on f. The formal
adjoint of L, denoted by L*, is computed using integration
by parts in the defining relation

               (f,Lg) = (L*f,g).

Just integrate by parts ignoring boundary terms to find out
what L* is. (See below for an alternative definition).
Exercise: show that if

               Lf = f' - af

then

              L*f = - f' - af.

You can think of formal self-adjointness of the linearized
equations as being just the statement that the second functional
derivative of the action integral is symmetric under interchange
of derivatives. So, this is
a functional analogue of the usual  sort of integrability
condition (dG=0) for an equation of the form dF = G. In this analogy
G represents the equations of motion, F is the Lagrangian and d is
the process of forming the Euler-Lagrange equations (the "functional
derivative").

*********************

Alternative definition: Given a linear differential
operator L, there exists a unique linear differential operator L*
such that, for any functions f and g

                 g Lf - f L*g = h',

for some h. All this can be generalized to more complicated
types of differential operators.

Charles Torre

Actually, the energy 1/2 (mx'^2+kx^2) decays with exp(-at).
And H stays constant.

Gerard

In article <8j39ms$...@gap.cco.caltech.edu>,
  ke...@cco.caltech.edu (Kevin A. Scaldeferri) wrote:

Are you referring to non-commutative geometry? How do this Lagrangians
arise there?

Best regards, squark.

Sent via Deja.com http://www.deja.com/
Before you buy.