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Hamiltonian Dynamics = Adiabatic Processes Only?

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Squark

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Jun 22, 2000, 3:00:00 AM6/22/00
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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.

Gerard Westendorp

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Jun 22, 2000, 3:00:00 AM6/22/00
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Squark wrote:
>
> 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.

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)

> 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 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.


[..]

> This is interesting, how may we
> describe non-adiabatic models after all, and so, address the issue of
> quantum entropy.


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


Charles Torre

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Jun 23, 2000, 3:00:00 AM6/23/00
to

Gerard Westendorp <wes...@xs4all.nl> writes:
>
> The time dependent term has canceled.
> I think Hamilitonians indeed only work for energy conserving systems.
> But Lagrangian systems can be more general.

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


squ...@my-deja.com

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Jun 23, 2000, 3:00:00 AM6/23/00
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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.


Kevin A. Scaldeferri

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Jun 23, 2000, 3:00:00 AM6/23/00
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In article <UFS$3V8Y...@cc.usu.edu>, Charles Torre <to...@cc.usu.edu> wrote:
>
>Gerard Westendorp <wes...@xs4all.nl> writes:
>>
>> The time dependent term has canceled.
>> I think Hamilitonians indeed only work for energy conserving systems.
>> But Lagrangian systems can be more general.
>
>Nah. If there is a Lagrangian formulation then there is a
>Hamiltonian formulation and vice versa.

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.


squ...@my-deja.com

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Jun 23, 2000, 3:00:00 AM6/23/00
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In article <395242FD...@xs4all.nl>,

Gerard Westendorp <wes...@xs4all.nl> wrote:
> How about:
>
> L = 1/2 (m(x')^2 - k x^2) + max't

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

Best regards, squark.

Charles Torre

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Jun 23, 2000, 3:00:00 AM6/23/00
to
In article <8j0e95$o...@gap.cco.caltech.edu>, ke...@cco.caltech.edu (Kevin A. Scaldeferri) writes:
> In article <UFS$3V8Y...@cc.usu.edu>, Charles Torre <to...@cc.usu.edu> wrote:

>>Nah. If there is a Lagrangian formulation then there is a
>>Hamiltonian formulation and vice versa.
>
> 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.)
>

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


Gerard Westendorp

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Jun 24, 2000, 3:00:00 AM6/24/00
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Charles Torre wrote:
>
> Gerard Westendorp <wes...@xs4all.nl> writes:
> >
> > The time dependent term has canceled.
> > I think Hamilitonians indeed only work for energy conserving systems.
> > But Lagrangian systems can be more general.
>
> 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.
>

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


squ...@my-deja.com

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Jun 24, 2000, 3:00:00 AM6/24/00
to
In article <395242FD...@xs4all.nl>,
Gerard Westendorp <wes...@xs4all.nl> wrote:
>
>
> Squark wrote:
> >
> > 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.
>
> How about:
>
> L = 1/2 (m(x')^2 - k x^2) + max't

Cool!

> > 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 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.

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.

> > This is interesting, how may we describe non-adiabatic models
> > after all, and so, address the issue of quantum entropy.

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

Gerard Westendorp

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Jun 24, 2000, 3:00:00 AM6/24/00
to

squ...@my-deja.com wrote:
>
> In article <395242FD...@xs4all.nl>,
> Gerard Westendorp <wes...@xs4all.nl> wrote:
> > How about:
> >
> > L = 1/2 (m(x')^2 - k x^2) + max't
>

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

Yes, sorry about that.

Gerard


Gerard Westendorp

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Jun 24, 2000, 3:00:00 AM6/24/00
to
squ...@my-deja.com wrote:

> 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)

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


Gerard Westendorp

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Jun 24, 2000, 3:00:00 AM6/24/00
to mtx...@coventry.ac.uk
mtx...@coventry.ac.uk wrote:

> In article <395242FD...@xs4all.nl> some poor uncited person wrote:

> >Yet another poor uncited guy wrote:

> >> For instance, consider the equation of motion

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

> >How about:


> >
> > L = 1/2 (m(x')^2 - k x^2) + max't

> I get dL/dx' = mx' + mat, so d/dt(dL/dx') = mx'' + ma.
> Then setting this equal to dL/dx=-kx just gives
> mx'' + ma = -kx
> or
> x'' = -(k/m)x - a
>
> rather than
>
> x'' = -(k/m)x - ax'
>
> Am I missing something here?

No, I think I got it wrong.

Squark has already suggested a better Lagrangian.

Gerard


Charles Torre

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Jun 26, 2000, 3:00:00 AM6/26/00
to
Gerard Westendorp <wes...@xs4all.nl> writes:

>
> So what Hamiltonian do you suggest for a damped oscillator?
>
> Or even simpler, a first order system, like
>
> x' = ax

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


Gerard Westendorp

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Jun 26, 2000, 3:00:00 AM6/26/00
to

Gerard Westendorp wrote:
>
> squ...@my-deja.com wrote:
> >
> > 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)
> >
>
> Hey, that is cool.
> It also allows you to interpret H as the total energy,
> which decays with exp(at).
>

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


Gerard


Gerard Westendorp

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Jun 26, 2000, 3:00:00 AM6/26/00
to

Charles Torre wrote:
>
> Gerard Westendorp <wes...@xs4all.nl> writes:
>
> >
> > So what Hamiltonian do you suggest for a damped oscillator?
> >
> > Or even simpler, a first order system, like
> >
> > x' = ax
>
> Why should there be one?
>

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


[..]

>
> 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.

What is formally self-adjoint?

Gerard


Ralph E. Frost

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Jun 27, 2000, 3:00:00 AM6/27/00
to
Charles Torre wrote:
>
> Gerard Westendorp <wes...@xs4all.nl> writes:

..


>
> 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.


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


C. M. Heard

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Jun 27, 2000, 3:00:00 AM6/27/00
to
"Squark" <squ...@mydeja.com> wrote:
>
> 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'
>
> [T]he problem arises for a simple Newtonian body moving with friction.
....
> The [ ... ] 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.

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


Kevin A. Scaldeferri

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Jun 28, 2000, 3:00:00 AM6/28/00
to
In article <4Q5ClM...@cc.usu.edu>, Charles Torre <to...@cc.usu.edu> wrote:
>In article <8j0e95$o...@gap.cco.caltech.edu>, ke...@cco.caltech.edu (Kevin A. Scaldeferri) writes:
>> In article <UFS$3V8Y...@cc.usu.edu>, Charles Torre <to...@cc.usu.edu> wrote:
>
>>>Nah. If there is a Lagrangian formulation then there is a
>>>Hamiltonian formulation and vice versa.
>>
>> 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.)
>>
>
>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.

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.

>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.)

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.

squ...@my-deja.com

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Jun 28, 2000, 3:00:00 AM6/28/00
to
In article <8ivp57$vho$1...@nnrp1.deja.com>,
squ...@my-deja.com wrote:

> 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)

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

squ...@my-deja.com

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Jun 28, 2000, 3:00:00 AM6/28/00
to
In article <3957EF3E...@xs4all.nl>,

Gerard Westendorp <wes...@xs4all.nl> wrote:
> Actually, the energy 1/2 (mx'^2+kx^2) decays with exp(-at).
> And H stays constant.

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

Charles Torre

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Jun 28, 2000, 3:00:00 AM6/28/00
to
Norbert Dragon <dra...@itp.uni-hannover.de writes:

> * Charles Torre to...@cc.usu.edu writes:
>
>> 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.
>

> The correspondence of Euler-Lagrange equations and Lagrangean is
> unique up to total derivatives.

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...

>
> However, it may turn out that different systems of equations have
> the same set of solutions, which poses the problem to find the
> functionals which become stationary exactly for a given set of
> functions.
>
> An example of two different, local functionals with the same set of
> stationary points is L_2 = a L_1 . Are there less trivial examples?

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


Gerard Westendorp

unread,
Jun 29, 2000, 3:00:00 AM6/29/00
to
Charles Torre wrote:
>
> Gerard Westendorp <wes...@xs4all.nl> writes:
>
> >
> > So what Hamiltonian do you suggest for a damped oscillator?
> >
> > Or even simpler, a first order system, like
> >
> > x' = ax
>
> 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.


Actually, here is one:


L = 1/2 (y')^2 exp(-at)

This gives:

y'' = ay'

Then, substitute x = y'


Gerard.


Charles Torre

unread,
Jun 29, 2000, 3:00:00 AM6/29/00
to
[Regarding the equation of motion x'=ax]

Gerard Westendorp <wes...@xs4all.nl> writes:


> Charles Torre wrote:
>
>>
>> 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.
>
> What is formally self-adjoint?
>

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


Gerard Westendorp

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Jun 29, 2000, 3:00:00 AM6/29/00
to
Gerard Westendorp wrote:

> squ...@my-deja.com wrote:

> > 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)

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

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


Gerard


squ...@my-deja.com

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Jun 29, 2000, 3:00:00 AM6/29/00
to
In article <8j39ms$1...@gap.cco.caltech.edu>,
ke...@cco.caltech.edu (Kevin A. Scaldeferri) wrote:
> ...

> 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.

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

squ...@my-deja.com

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Jun 29, 2000, 3:00:00 AM6/29/00
to
In article <eo565.456$0x.1...@nuq-read.news.verio.net>,

"C. M. Heard" <he...@vvnet.com> wrote:
> "Squark" <squ...@mydeja.com> wrote:
> >
> > I have recently thought what is the Lagrangian for a harmonic
> > oscillator with damping.
> > ...

> ...


> 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.

Sorry, but what is a transmission line?

squ...@my-deja.com

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Jun 29, 2000, 3:00:00 AM6/29/00
to
In article <3954864A...@xs4all.nl>,

Gerard Westendorp <wes...@xs4all.nl> wrote:
> Still haven't found the Lagrangian for
>
> x' = ax
>

Use the variable y satisfying y' = x. The equation of motion becomes

(F) y'' = ay'

Which corresponds to the Lagrangian

(G) L = m(y'^2)exp(-at)

This method may seem a bit unusual but it is perfectly analogous to
what is done with Maxwell theory, where we pass from the
electric/magnetic field to the 4-potential.

squ...@my-deja.com

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Jun 29, 2000, 3:00:00 AM6/29/00
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In article <8j84re$drn$1...@newsserver.rrzn.uni-hannover.de>,

dra...@itp.uni-hannover.de (Norbert Dragon) wrote:
> The correspondence of Euler-Lagrange equations and Lagrangean is
> unique up to total derivatives.

The problem is that the total derivative may change the quantum theory.
Example: topological Lagrangians. I may eleborate if you want.

Norbert Dragon

unread,
Jun 29, 2000, 3:00:00 AM6/29/00
to
* Charles Torre to...@cc.usu.edu writes:

> 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.

Seems that you are refering to Chern-polynomials.

They are most conveniently described by combining the Lagrangean
with the volume element d^n x and combining the fieldstrength
to two forms

F = 1/2 F_mn dx^m dx^n

F_mn = d_m A_n - d_n A_m - [A_m,A_n]

All Lagrangeans densities

L(A_n ,d_m A_n, d_k d_l A_n, ... ) d^n x,

which depend on A_n and its derivatives and which are gauge
invariant and have vanishing Euler-derivative
with respect to A are polynomials P(F) of degree n/2 in the
field strength two form F.

Each polynomial P(F) is a complete derivative of the Chern Simons
form, but not the derivative of a function of gauge covariant field
strengths.

This is the content of the covariant Poincare lemma.
If you know a proof which is simpler than in

Nucl. Phys. B 340, (1990) 187 (guess who one of the authors is)

I would be interested. I try to simplify and to teach the subject.

--

Norbert Dragon
dra...@itp.uni-hannover.de
http://www.itp.uni-hannover.de/~dragon


Charles Torre

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Jul 1, 2000, 3:00:00 AM7/1/00
to
Gerard Westendorp <wes...@xs4all.nl> writes:
> Charles Torre wrote:
>>
>> Gerard Westendorp <wes...@xs4all.nl> writes:
>>
>> >
>> > So what Hamiltonian do you suggest for a damped oscillator?
>> >
>> > Or even simpler, a first order system, like
>> >
>> > x' = ax
>>
>>
>> There is no Lagrangian, L=L(x,x'), such that the differential equation
>> x' = ax is the Euler-Lagrange equation of L.
>
>
> Actually, here is one:
>
>
> L = 1/2 (y')^2 exp(-at)
>
> This gives:
>
> y'' = ay'
>
> Then, substitute x = y'
>
>

Well, if I had said something like "there is no Lagrangian whose
Euler-Lagrange equations are *equivalent* to x'=ax *after a change of
variables*" you would have nailed me with this example.
But I was very careful how I worded
my claim ;) - which is correct as stated. You can't get x'=ax as
the Euler-Lagrange equations of a Lagrangian L=L(x,x').

Note that your change of variables x <-> y is not invertible (since
y=any-constant yields the same x=0). Also, your change of variables
is not purely local (since you need to integrate x to get at y).
Neither of these subtleties cause any real problems (I guess)
with such a simple example, but you can imagine that more complicated
kinds of examples could get pretty ugly.

Anyway, my predisposition to be a lawyer aside,
I agree that a standard strategy to avoid the kind of difficulty I was
highlighting is to make a (typically non-local, not invertible) change
of variables. I guess I see what kind of result you were actually
interested in. By the way, if you grant me non-invertible and non-local
changes of variables, equivalent equations, etc. you can probably make
a LOT of Lagrangians.


Charles Torre


Charles Torre

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Jul 1, 2000, 3:00:00 AM7/1/00
to
squ...@my-deja.com writes:
> In article <3954864A...@xs4all.nl>,
> Gerard Westendorp <wes...@xs4all.nl> wrote:
>> Still haven't found the Lagrangian for
>>
>> x' = ax
>>
>
> Use the variable y satisfying y' = x. The equation of motion becomes
>
> (F) y'' = ay'
>
> Which corresponds to the Lagrangian
>
> (G) L = m(y'^2)exp(-at)
>
> This method may seem a bit unusual but it is perfectly analogous to
> what is done with Maxwell theory, where we pass from the
> electric/magnetic field to the 4-potential.

Yes, this is a viable strategy for taking differential equations that,
strictly speaking, have no Lagrangian and finding an alternate set
of variables and equations that do admit a Lagrangian (perhaps
many!).

Your analogy is a good one. To stretch it a little further: Using the
y variables to formulate the model, you seem to have introduced
a miniature gauge invariance: since y and y+constant correspond to
the same x, it seems we should physically identify y(t) and y(t)+const.?

Charles Torre


C. M. Heard

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Jul 1, 2000, 3:00:00 AM7/1/00
to
squ...@my-deja.com wrote:
> "C. M. Heard" <he...@vvnet.com> wrote:
[ ... ]

> > 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.
>
> Sorry, but what is a transmission line?

There are many kinds, but perhaps the simplest example is a coaxial
transmission line. An idealized version would consist of two concentric
perfectly conducting cylinders. If you solve Maxwell's equations in the
region between the cylinders (assuming that they extend indefinitely
along the axis) you will find that the there exist propagating solutions
where the propagation direction is along the axis of the cylinders, with
E radial and B tangential. For such solutions the ratio of transverse
voltage (potential difference between the cylinders, i.e., line integral
of E) and the longitudinal current flowing parallel to the direction of
propagation (equal to the line integral of B) is a constant, called the
characteristic impedance, which is determined by the geometry. This is
the same terminal relation as that of a resistor. See exercise 8.1
of Jackson. Alternatively, look in an engineering E & M text such as
Ramo, Whinnery, and van Duzer.

Mike


Kevin A. Scaldeferri

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Jul 1, 2000, 3:00:00 AM7/1/00
to
In article <8jduti$2vc$1...@nnrp1.deja.com>, <squ...@my-deja.com> wrote:
>In article <8j39ms$1...@gap.cco.caltech.edu>,
> ke...@cco.caltech.edu (Kevin A. Scaldeferri) wrote:
>> ...
>> 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.
>
>Are you referring to non-commutative geometry? How do this Lagrangians
>arise there?

Yes, by non-commutative field theory, I mean field theory on a
non-commutative geometry.

Try hep-th/9912072, Minwalla, Van Raansdonk, Seiberg, "Noncommutative
Perturbative Dynamics" for a review. Trace the references back if you
want more basics.

Essentially, the effect of the non-commutativity of the geometry:

[x^m,x^n] = i theta^mn

is that the normal product of fields is replaced by the Moyal product

(f_1 * f_2)(x) = exp(i/2 theta^mn @/@y^m @/@z^n) f_1(y) f_2(z)

at y=z=x.

Voila, infinite numbers of derivatives. If you have space/time
non-commutivity, you have infinite numbers of time derivatives.

Norbert Dragon

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Jul 1, 2000, 3:00:00 AM7/1/00
to
* squ...@my-deja.com writes:

> The problem is that the total derivative may change the quantum theory.
> Example: topological Lagrangians. I may eleborate if you want.

You have at least one interested reader.

However, the name topological Lagrangeans is sometimes also used for
Chern-Simons forms which are not at all topological.

Considering path integrals it is not clear to me why the topological
contributions are restricted to the ones from topological Lagrangeans,
i.e. why do they have to derive from a _local_ action? I could equally
well imagine powers and more complicated functions of winding numbers
and other topological numbers to contribute to the action.

Charles Torre

unread,
Jul 1, 2000, 3:00:00 AM7/1/00
to
Norbert Dragon <dra...@itp.uni-hannover.de (Norbert Dragon)> writes:

[regarding Lagrangians whose Euler-Lagrange expression vanishes
identically and are nevertheless not total divergences]

> Seems that you are refering to Chern-polynomials.

...

>
> All Lagrangeans densities
>
> L(A_n ,d_m A_n, d_k d_l A_n, ... ) d^n x,
>
> which depend on A_n and its derivatives and which are gauge
> invariant and have vanishing Euler-derivative
> with respect to A are polynomials P(F) of degree n/2 in the
> field strength two form F.
>
> Each polynomial P(F) is a complete derivative of the Chern Simons
> form, but not the derivative of a function of gauge covariant field
> strengths.

> This is the content of the covariant Poincare lemma.
> If you know a proof which is simpler than in
>
> Nucl. Phys. B 340, (1990) 187 (guess who one of the authors is)
>
> I would be interested. I try to simplify and to teach the subject.
>


Hi Norbert. Nice paper. I don't know of anything off the top of
my head, although I would have thought that such results existed in
the math literature somewhere (perhaps in the body of literature related
to index theorems). I believe that analogous results for Lagrangians
built on the jet space of metrics have been obtained by Gilkey.

Your gauge theory example does indeed illustrate what I was
mentioning, but it is, perhaps, a little fancier than necessary
(given its essential use of gauge invariance). Here is a humbler
example (pointed out to me by Ian Anderson a long time ago).

Let M and N be compact manifolds of dimension n. Let h be a
Riemannian metric on N. The volume form v on N defined by h is
closed but not exact as a form on N. Let the dynamical fields be
maps u:M->N. (So the bundle of independent and
dependent variables is E=MxN and v represents an element of the
degree n cohomology of E.) Consider a generic map u; use it to
pull back v from N to M. This form on M can be used to define a
Lagrangian for u built from its 1-jet (that is, the Lagrangian
depends upon u and its first derivatives). The Lagrangian L is,
in local coordinates, just the square root of the determinant of
h, times the Jacobian determinant of the map u, times the
coordinate volume form. The action integral is, up to maybe a
numerical factor, the degree of the map u. The degree is an
integer, which implies that the Euler-Lagrange expression for L
vanishes identically, but of course L cannot be a total divergence or else
the degree would be zero for every map u (which it isn't).

-charlie

John Baez

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Jul 1, 2000, 3:00:00 AM7/1/00
to
Gerard Westendorp <wes...@xs4all.nl> wrote:

Charles Torre wrote:

>> There is no Lagrangian, L=L(x,x'), such that the differential equation
>> x' = ax is the Euler-Lagrange equation of L.

>Actually, here is one:

> L = 1/2 (y')^2 exp(-at)

No, that's not one.

>This gives:

> y'' = ay'

What Charles Torre said is that there's no Lagrangian of the form
L(x,x') which gives the equation x' = ax as its Euler-Lagrange equation.
You have written down an equation that is not of this form, which gives
some other equation as its Euler-Lagrange equation.

However, I agree that your answer might serve as a practical
workaround to solve an otherwise unsolvable problem by changing
the rules of the game a bit.

You might enjoy proving to yourself that Torre's claim is in fact
correct. Consider an arbitrary Lagrangian of the form L(x,x'), work
out the Euler-Lagrange equations, and show they can't be x' = ax.
The stuff he said about "formal self-adjointness" may sound complicated,
but if you consider this specific example you can sort of see what's
going on.

Toby Bartels

unread,
Jul 2, 2000, 3:00:00 AM7/2/00
to
Gerard Westendorp <wes...@xs4all.nl> wrote:

>Charles Torre wrote:

>>Gerard Westendorp <wes...@xs4all.nl> writes:

>>>So what Hamiltonian do you suggest for a damped oscillator?

>>>Or even simpler, a first order system, like x' = ax?

>>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.

>Actually, here is one: L = 1/2 (y')^2 exp(-at).
>This gives: y'' = ay'.
>Then, substitute x = y'.

If you intend to allow such shenanigans,
then the Hamiltonian is H = 1/2 p^2 exp(at).


-- Toby
to...@ugcs.caltech.edu


ba...@galaxy.ucr.edu

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Jul 2, 2000, 3:00:00 AM7/2/00
to
to...@cc.usu.edu (Charles Torre) wrote:

>Norbert Dragon <dra...@itp.uni-hannover.de (Norbert Dragon)> writes:

>[regarding Lagrangians whose Euler-Lagrange expression vanishes
>identically and are nevertheless not total divergences]

> Seems that you are refering to Chern-polynomials.

>Your gauge theory example does indeed illustrate what I was


>mentioning, but it is, perhaps, a little fancier than necessary
>(given its essential use of gauge invariance). Here is a humbler
>example (pointed out to me by Ian Anderson a long time ago).

>Let M and N be compact manifolds of dimension n. Let h be a
>Riemannian metric on N. The volume form v on N defined by h is
>closed but not exact as a form on N. Let the dynamical fields be
>maps u:M->N.

>The Lagrangian L is, in local coordinates, just the square root of the

>determinant of h, times the Jacobian determinant of the map u, times the
>coordinate volume form. The action integral is, up to maybe a
>numerical factor, the degree of the map u.

These two examples are both special cases of something, so let me
say what that something is.

Suppose our spacetime M is a compact n-dimensional manifold.

Consider a field theory where the field is a smooth map f: M -> N
from M to some manifold N of arbitrary dimension.

Suppose we can find a closed but not exact n-form w on N.

Then we get a field theory where the Lagrangian is the pullback of
w by the function f.

L = f^*(w)

Since w is closed, the integral of L over spacetime does not
change when we vary f slightly, so the Euler-Lagrange equations
for this field theory are trivial: every field solves the field
equations.

In other words, the action is *locally* constant on the space of
smooth fields f: M -> N.

However, the action need not be constant, since the space of fields
can have different connected components. In particular, the action
need not be zero, since w is not exact.

Charles Torre is basically considering the special case where N is
an n-dimensional manifold and w is a volume form on this manifold.
Norbert Dragon is basically considering the case where N = BG for
the group G = U(n); the Chern classes are closed but not exact
differential forms on BG. It's good to generalize this idea by
considering other groups G - we get different field theories from
different cohomology classes on BG. These cohomology classes are
called "characteristic classes", and they are well-understood for
lots of groups G, so one can list these field theories and study
them in great detail.

In general, the trick I'm talking about is called a "cohomological
field theory" - it's a special case of a topological field theory.


Gerard Westendorp

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Jul 3, 2000, 3:00:00 AM7/3/00
to

The Hamilton equations look more like solutions than
the original equation:

dp/dt = 0
dq/dt = p exp(at)


Another Hamiltonian that gives x' = ax is:

H = apx

->

dp/dt = -ap
dx/dt = ax

This Hamiltonian does not have a corresponding Lagrangian,
because there is no relation between p and x'.


Gerard


Aaron Bergman

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Jul 5, 2000, 3:00:00 AM7/5/00
to
In article <8jgckj$4...@gap.cco.caltech.edu>, Kevin A. Scaldeferri wrote:

>Try hep-th/9912072, Minwalla, Van Raansdonk, Seiberg, "Noncommutative

van Raamsdonk.

Aaron
--
Aaron Bergman
<http://www.princeton.edu/~abergman/>


Charles Torre

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Jul 5, 2000, 3:00:00 AM7/5/00
to
Gerard Westendorp <wes...@xs4all.nl> writes:
>
>
> Another Hamiltonian that gives x' = ax is:
>
> H = apx
>
> ->
>
> dp/dt = -ap
> dx/dt = ax
>
> This Hamiltonian does not have a corresponding Lagrangian,
> because there is no relation between p and x'.
>
>

You are allowing yourself additional variables and additional equations
(you are using x and p instead of just x). This leads to a particularly
boring version of the inverse problem of the calculus of variations.
I say this because, if you allow that, all equations have a Lagrangian:
Let your desired equation be D(x,x',...)=0. Introduce
a new variable p. Choose the Lagrangian L = pD. One of the
EL equations will be D=0.

Example: a Lagrangian whose EL equations give the equations
you gave above is

L(x,p,x',p') = p(x' - ax).

(By treating x and p as configuration variables I get a degenerate Lagrangian,
but this does not do any harm.)

Usually one is interested in taking the dynamical problem as given (e.g,
a variable x and an equation x'=ax) and seeing if there is a variational
principle for it without having to introduce any additional, ad hoc,
structure.

Charles Torre


squ...@my-deja.com

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Jul 5, 2000, 3:00:00 AM7/5/00
to
In article <2000070122...@math-cl-n02.ucr.edu>,
ba...@galaxy.ucr.edu wrote:
[A lot of cool stuff]

> In general, the trick I'm talking about is called a "cohomological
> field theory" - it's a special case of a topological field theory.

Really? How do you describe it as a TFT (i.e. construct the appropriate
functor)?

squ...@my-deja.com

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Jul 5, 2000, 3:00:00 AM7/5/00
to
P.S.

When we quantize the model, the observables in the x-sense must commute
with y-translations, i.e. we get that x-observables must be plain
functions of momentum. This makes the model kinda trivial. Actually,
there are no quantum uncertainties in it, so it's equivalent to the
classical model!

squ...@my-deja.com

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Jul 5, 2000, 3:00:00 AM7/5/00
to
The question is whether the transmittion line can have a non-imaginary
impedance. Because I doubt the last property is consistent with energy
conservation. If the impedance is imaginary, this is principally
different from the resistor.

Norbert Dragon

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Jul 5, 2000, 3:00:00 AM7/5/00
to
* Charles Torre to...@cc.usu.edu writes:

> Your gauge theory example does indeed illustrate what I was
> mentioning, but it is, perhaps, a little fancier than necessary
> (given its essential use of gauge invariance). Here is a humbler
> example (pointed out to me by Ian Anderson a long time ago).

> Let M and N be compact manifolds of dimension n. Let h be a
> Riemannian metric on N. The volume form v on N defined by h is
> closed but not exact as a form on N. Let the dynamical fields be

> maps u:M->N. (So the bundle of independent and
> dependent variables is E=MxN and v represents an element of the
> degree n cohomology of E.) Consider a generic map u; use it to
> pull back v from N to M. This form on M can be used to define a
> Lagrangian for u built from its 1-jet (that is, the Lagrangian

> depends upon u and its first derivatives). The Lagrangian L is,


> in local coordinates, just the square root of the determinant of
> h, times the Jacobian determinant of the map u, times the
> coordinate volume form. The action integral is, up to maybe a

> numerical factor, the degree of the map u. The degree is an
> integer, which implies that the Euler-Lagrange expression for L
> vanishes identically, but of course L cannot be a total divergence or else
> the degree would be zero for every map u (which it isn't).

Thank you for the simple example.
It is simpler than gauge theories with transformation

delta A_m = d_m Lambda

and rests on fields which transform as Goldsstone bosons

delta Phi = Lambda , where d_m Lambda = 0


The field strength 1-form of this transformation law is

D Phi = dx^m d_m Phi .

Your local Lagrangean seems to be the polynomial of order dim N
in this 1-form. I am slightly puzzled, however, about the
appearance of the metric h, it should drop out completely
because winding numbers can be defined without reference to a
metric.

Aaron Bergman

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Jul 5, 2000, 3:00:00 AM7/5/00
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In article <8jqgvc$ptm$1...@nnrp1.deja.com>, squ...@my-deja.com wrote:
>In article <2000070122...@math-cl-n02.ucr.edu>,
> ba...@galaxy.ucr.edu wrote:
>[A lot of cool stuff]
>> In general, the trick I'm talking about is called a "cohomological
>> field theory" - it's a special case of a topological field theory.
>
>Really? How do you describe it as a TFT (i.e. construct the appropriate
>functor)?

Do you really need all the category theory? It's certainly
pretty, but you can just as well define a TQFT as a field theory
whose observables are independent of the metric.

John Baez

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Jul 5, 2000, 3:00:00 AM7/5/00