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Jun 22, 2000, 3:00:00 AM6/22/00

to

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.

Jun 22, 2000, 3:00:00 AM6/22/00

to

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

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

Jun 23, 2000, 3:00:00 AM6/23/00

to

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.

Jun 23, 2000, 3:00:00 AM6/23/00

to

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.

>

>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.

Jun 23, 2000, 3:00:00 AM6/23/00

to

In article <395242FD...@xs4all.nl>,

Wait a second. This yields p = mx' + mat, thus the Euler-Lagrange

equation is mx'' = -kx - ma, NOT (A).

Best regards, squark.

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:

> 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

Jun 24, 2000, 3:00:00 AM6/24/00

to

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.

>

>

> 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.

>

> 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

Jun 24, 2000, 3:00:00 AM6/24/00

to

> 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.

>

> >

> > 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

>

> 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.

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

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)

> 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

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:

> 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

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

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

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

Jun 27, 2000, 3:00:00 AM6/27/00

to

..

>

> 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

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>

> I have recently thought what is the Lagrangian for a harmonic oscillator

>

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

>

....

> 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.

> 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

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:

>

>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.

>>

>>>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.

>> 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.

Jun 28, 2000, 3:00:00 AM6/28/00

to

In article <8ivp57$vho$1...@nnrp1.deja.com>,

squ...@my-deja.com 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)

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

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.

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.

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.

>

> * 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

Jun 29, 2000, 3:00:00 AM6/29/00

to

> > 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?

>

> >

> > 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.

> 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.

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

Jun 29, 2000, 3:00:00 AM6/29/00

to

> > 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

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.

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?

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.

> > ...

"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?

Jun 29, 2000, 3:00:00 AM6/29/00

to

In article <3954864A...@xs4all.nl>,

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.

Jun 29, 2000, 3:00:00 AM6/29/00

to

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.

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.

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

Jul 1, 2000, 3:00:00 AM7/1/00

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

>>

>>

>>

>> >

>> > 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'

>

>

>> 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

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.

> 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

Jul 1, 2000, 3:00:00 AM7/1/00

to

[ ... ]

> > 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?

> > 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

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?

>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.

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.

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

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.

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

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.

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

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/>

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'.

>

>

>

>

> 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

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.

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)?

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!

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.

impedance. Because I doubt the last property is consistent with energy

conservation. If the impedance is imaginary, this is principally

different from the resistor.

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.

Jul 5, 2000, 3:00:00 AM7/5/00

to

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)?

>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.

Jul 5, 2000, 3:00:00 AM7/5/00