Hamiltonian Dynamics = Adiabatic Processes Only?
Squark
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
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
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
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
>
>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
Wait a second. This yields p = mx' + mat, thus the Euler-Lagrange
equation is mx'' = -kx - ma, NOT (A).
Best regards, squark.
Charles Torre
> 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
>
> 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.
>
> 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
> >
> > 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.
>
>
> 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
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
> 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
> 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
>
> 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
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
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
..
>
> 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
>
> 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)
>
> 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
>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:
>
>>>Hamiltonian formulation and vice versa.
>>
>> 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
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
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
> * 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
> >
> > Or even simpler, a first order system, like
> >
> > x' = ax
>
> Why should there be one?
>
> 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
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
> > 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
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
"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
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
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
> 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
>>
>> >
>> > So what Hamiltonian do you suggest for a damped oscillator?
>> >
>> > Or even simpler, a first order system, like
>> >
>> > x' = ax
>>
>>
>> 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
> 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
> > 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
>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
> 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
[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
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
>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
>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
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
>Try hep-th/9912072, Minwalla, Van Raansdonk, Seiberg, "Noncommutative
van Raamsdonk.
Aaron
--
Aaron Bergman
<http://www.princeton.edu/~abergman/>
Charles Torre
>
>
> 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
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
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
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
> 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
>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
Norbert Dragon <dra...@itp.uni-hannover.de> wrote:
>Charles Torre intoned:
>> 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.
>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.
The *action* will indeed be independent of the metric h on the
target space N, for the reason you cite. However, the *Lagrangian*
will depend on h. It's easiest to see exactly what's going on if
we take the simplest possible case, when both M and N are a circle.
By the way, we don't really need a metric on N to play this game;
all we really need is a closed but not exact n-form.
It's fun to try to quantize this sort of theory, by the way!
C. M. Heard
> 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.
But in fact the terminal impedance of a lossless semi-infinite
transmission line is purely real. There is no reactive component.
In an electrical circuit it behaves just like a resistor in that it
absorbs energy. That is not inconsistent with energy conservation,
however, because the energy that the transmission line absorbs is
stored in the electromagnetic fields within the transmission line.
If a voltage or current of finite duration is applied at the terminals
of the line the fields will take the form of travelling waves which
propagate away from the terminals. Because the line is semi-infinite
the travelling wave does not reflect back toward the terminals. That
is not true, by the way, for a line of finite length. The latter does
have a purely imaginary terminal impedance.
This is not so different from an ordinary resistor which dissipates
the energy it absorbs as heat. In that case the absorbed energy is
stored as the motion of atoms or molecules in the environment. Perhaps
an even better analogy is free space which absorbs any electromagentic
energy that is radiated by any source (e.g., an atom decaying from an
excited state to its ground state).
As I said in a previous post, I highly recommend analyzing the damped
LC circuit by substituting a transmission line for the damping resistor
and writing down the resultinh Lagrangian or Hamiltonian. This simple
model is exactly solvable and will teach you a great deal about the
physics of dissipative systems.
Mike
Jim Graber
to...@cc.usu.edu (Charles Torre) wrote:
{Snip previous problem}
> 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
>
>
I am interested in the inverse problem in General Relativity and its
alternatives. In particular a two step "inverse problem"
Step one: Going back from
a spherically symmetric solution(SSS) (preferably the general SSS) to a
set of field equations which it satisfies.
Step two: Going back from the field equations to an action. (I believe
this is the problem you have been discussing in a different context)
I understand that a unique answer is not to be expected to either step.
I was struck by your above discussion of the "boring" solution allowed
by introducing extra variables. I guess better a boring solution than no
solution at all. However, I wonder: Is there a "non-boring" solution to
the above problem?
Obviously the Einstein action or is it the Hilbert action solves the
inverse problem for the Schwarzschild solution. But what non boring
action or actions yield the general spherically symmetric solution?
Thanks Jim Graber
squ...@my-deja.com
dra...@itp.uni-hannover.de (Norbert Dragon) wrote:
> * 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.
An example would be a system of n-particles. Staright-forward
quatization leads to bosonic statistics, as such is the Hilbert space
of square-integrable functions on their configuration space. One way to
get different statistics is to choose a different line bundle over the
configuration space to make the state space. Another would be to change
the Lagrangean without changing the classical equations of motion. This
is achieved through adding a term of the form a(sum i>j)omega_ij where
omega_ij is the angular velocity of particle i with respect to particle
j. a should give a statistics determining number, evidently (a=0
corresponds to bosonic, a=1 to fermionic). Another example would be the
Wess-Zumino term which is, though a non-local term in the action.
Best regards, squark.
squ...@my-deja.com
to...@cc.usu.edu (Charles Torre) wrote:
> 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.?
Yes, but this is a much simpler case then usual gauge freedom. Firstly,
it is easy to gauge fix: just demand y(t_0)=y_0. Scondly, the
transformation is time-independant (and there's no issue of space at
all) so there's no real importance to identification of solutions
differing by a constant. The quantization of (G) is starightforward,
the action of the "gauge transformation" is plain y translation.
John Baez
Aaron Bergman <aber...@princeton.edu> wrote:
>In article <8jqgvc$ptm$1...@nnrp1.deja.com>, squ...@my-deja.com wrote:
>>Really? How do you describe it as a TQFT (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.
"Constructing the appropriate functor" really just amounts to
1) figuring out the Hilbert space associated to any compact
n-dimensional manifold representing "space" and 2) figuring
out the time evolution operator associated to any (n+1)-dimensional
cobordism representing "spacetime". These are eminently worthwhile
things to do if you're trying to understand a TQFT. They're not
just category-theoretic niceties: they're at the heart of the physics.
Let's think a minute about how it goes for the cohomological
field theories I described in my last post. For starters, let's
think about a *compact* (n+1)-dimensional manifold M representing
spacetime. This is a cobordism from the empty set to the empty
set - i.e., we've got a universe popping into existence from nothing,
living its merry life, and then winking back out of existence. We
thus know (from the TQFT axioms) that the corresponding time evolution
operator will be a mere number - the "partition function" of M.
Let's think about how to calculate this number. Remember, in our
cohomological field theory, a field is a map f: M -> X where X
is some space called the "target space". The Lagrangian is just
the pullback by f of some closed (but preferably not exact)
(n+1)-form w on X. In equations:
L(f) = f^*(w)
We then integrate this over M to get the action S(f). As we've
seen, this depends only on the homotopy class of f: M -> X.
Now let's do the path integral to compute the partition function!
For this, we want to integrate exp(iS(f)) over the space of all
fields f. Normally in field theory we start waving our hands madly
right at this point, because we don't really know how to do these
infinite-dimensional integrals. But in the present case exp(iS(f))
only depends on the homotopy class of f, so it's constant on each
connected component of the space of all fields. This sort of function
should be a lot easier to integrate than most.
This suggests a clever trick: to integrate exp(iS(f)), take the *sum*
of exp(iS(f)) over homotopy classes of maps from M to X. This
amounts to assuming that each connected component of the space
of maps has the same measure (which we normalize to 1). If you
have objections to this assumption, you are free to insert an extra
"weight" factor which reflects the different size of each connected
component.
What happens next depends a lot on whether there are *infinitely*
many homotopy classes of maps from M to X, or *finitely* many.
If there are infinitely many, our path integral may diverge - we
have to do the calculation to see if it does or not. If there
are finitely many, we are really in good luck: the path integral
is well-defined!
When will there be finitely many homotopy classes of maps from M
to X? The set of such homotopy classes is called [M,X]. Over in
the thread on geometric quantization, Toby has been calculating some
examples of this sort of thing - most of them worked out to be
infinite, but one worked out to be finite. Are there any good
criteria that ensure it's finite?
Sure! Topologists have been studying this stuff for eons, so
you just need to crack open a book and see what they say. For
example, suppose X is an Eilenberg-MacLane space K(G,m) where
is a *finite* group (abelian if m > 1). Then more or less by
definition,
[M,X] = H^m(X,G)
where the right-side means "the mth cohomology group of X with
coefficients in G". This is always a finite set when X is compact!
So Eilenberg-MacLane spaces of finite groups give very nice
examples of cohomological field theories where everything converges
and you can calculate everything to your heart's content. Not
just the partition function of a compact spacetime, but the
Hilbert spaces for manifolds representing space, and the time
evolution operators for manifolds representing spacetime.
This sort of example was pioneered by Dijkgraaf and Witten, and
then worked out in great detail by Freed and Quinn.
Chris Hillman
> But what non boring action or actions yield the general spherically
> symmetric solution?
"Solution" is a misnomer in this context. Do you mean to ask: what field
equations admit the Schwarzschild spacetime (M,g)? A static spherically
symmetric vacuum solution (M,g)? A dynamic spherically symmetric
nonvacuum solution (M,g)? Or what?
And what kind of theory do have in mind? A classical relativistic
gravitational field theory? A theory with a particular type of
Lagrangian, e.g. with specific additional gravitational field terms not
included in the Einstein-Hilbert Lagrangian for the gravitational field?
What about dimensions? Are you only interested in 3+1 spacetimes? Or are
you also interested in effective field theories derived from string
theory?
Note: the more general you make your question, the less likely anyone will
have an answer ready for you! So if you can focus on a very specific
question, the one which interests you most, you have a better chance of
getting a partial answer, an educated guess, or at least pointers to
related work.
Chris Hillman
Home Page: http://www.math.washington.edu/~hillman/personal.html
Jim Graber
ambiguous than I intended. i will tryto answer below.
In article
<Pine.OSF.4.21.000707...@goedel3.math.washington.edu>,
Chris Hillman <hil...@math.washington.edu> wrote:
> On 6 Jul 2000, Jim Graber wrote:
>
> > But what non boring action or actions yield the general spherically
> > symmetric solution?
>
> "Solution" is a misnomer in this context. Do you mean to ask: what
field
> equations admit the Schwarzschild spacetime (M,g)? A static
spherically
> symmetric vacuum solution (M,g)? A dynamic spherically symmetric
> nonvacuum solution (M,g)? Or what?
I would be interested in the answers to all of the above questions. The
one I originally had in mind was the second above. ie, the form in Wald
6.1.5 ds^2 = -f(r) dt^2 + h(r)dr^2 + r^2(do^2 + sin^2 o dp^2), where o
= theta and p = phi.
(Another alternative: PPN or Eddington type approximations to the above
exact form.)
>
> And what kind of theory do have in mind? A classical relativistic
> gravitational field theory? A theory with a particular type of
> Lagrangian, e.g. with specific additional gravitational field terms
not
> included in the Einstein-Hilbert Lagrangian for the gravitational
field?
> What about dimensions? Are you only interested in 3+1 spacetimes? Or
are
> you also interested in effective field theories derived from string
> theory?
General answer: As close to GR as possible without being identical or
equivalent to GR.
Several more specific answers:
My first choice: A scalar tensor alternative like in Damour &
Esposito-Farese gr-qc/9506063.
(However, if I understand the implications of what they say on page 32
in equation 5.7 this approach does not give allow the most general
spherically symmetric solution considered by other authors, because
the Eddington 2PN gamma parameter is not independent of the 1PN
parameters.)
My second choice: GR plus higher order terms such as the Gauss-bonnet
terms considered (in another context) in Deruelle & Dolezel
gr-qc/0004021.
Note that both these forms can be considered to be low energy limits of
string theory and might be viable candidates for a correct effective
theory of gravitation for some choices of parameters.
Obviously these papers (and others) give partial answers to my question,
but I am not aware of or else do not understand any paper which actually
gives an explicit general solution to the above inverse problem.
Oh yes dimensions: I would definitely prefer 3+1. If higher
dimensions are necessary, I need to understand how to reduce things
down to the effective theory in 3+1.(which I will need lots of help
with.)
>
> Note: the more general you make your question, the less likely anyone
will
> have an answer ready for you! So if you can focus on a very specific
> question, the one which interests you most, you have a better chance
of
> getting a partial answer, an educated guess, or at least pointers to
> related work.
I will take any help I can get. I am happy to start with simple cases
and work up to more general ones.
In the meantime, I have programmed some of the field equations derived
by Damour & Esposito-Farese (hereafter DEF) into Mathematica and am
trying to verify for myself (and understand) their statement referred to
above concerning the possible Eddington 2PN parameters accessible via
this approach. As I understand it, this theory or more
properly class of theories, eliminates some 2PN parameter combinations
which have not yet been eliminated experimentally.
>
> Chris Hillman
>
> Home Page: http://www.math.washington.edu/~hillman/personal.html
>
>
Thanks again for considering my question. Jim Graber
Gerard Westendorp
[..]
>
> 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.
>
I agree the mathematical structures this leads to are perhaps a
bit boring.
But I'm interested in the concept of a Lagrangian as a fundamental
principle in physics. It is interesting that some Lagrangians are
not interesting.
I have a much better intuitive understanding of Hamitonians
than of Lagrangians. What could be an intuitive picture of
a Lagrangian?
Gerard
alfred_...@my-deja.com
> > So what Hamiltonian do you suggest for a damped oscillator?
> >
> > Or even simpler, a first order system, like
> >
> > x' = ax
>From Charles Torre <to...@cc.usu.edu>
> Why should there be one?
>
> There is no Lagrangian,
[mathematical argument deleted]
> 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".
Now that you've said this, I'm going actually show how to include
such forces in quantum theory. This prescription is only being
stated here for the non-relativistic Schroedinger Equation, since
I'm not entirely sure how it generalizes into the relativistic
domain.
The Schroedinger equation, as it is generally stated, takes on the
form:
p^2/2m psi + V psi = E psi
where p = -i h-bar del, E = i h-bar d/dt and V is the potential for
a non-frictional force.
[Note: I'm using d/dt to denote the partial differential operator
which can't be easily rendered in ASCII].
It is often cited as a consequence that the density rho = (psi* psi)
satisfies the continuity equation
d rho/dt + div J = 0
where J is also defined in terms of psi. What is not as well known
is that if psi is written in polar form
psi = sqrt(rho) exp(i S / h-bar)
when J can be written as
J = rho v
where
v = grad S / m.
If you substitute this decomposition into the Schroedinger equation
in fact and separate out the real and imaginary parts, you will, in
fact, derive the continuity equation mentioned above -- but also
another equation:
E = 1/2 m v^2 + V + Q
where, here now, E is the "local energy" defined as E = -dS/dt, and
where Q is the term (first coined by Bohm) as the Quantum Potential,
defined as:
Q = -h-bar ^2 / 2m (del^2 r) / r
where r is the square root of rho. This reduction is known,
unofficially as the Bohm decomposition of the Schroedinger Equation.
However, this is NOT what I'm about to get at here. So we'll just
pass on by old Bohm and bid our adieu. What I'm about to get at is
the following question:
When can the original Schroedinger equation
be recovered from these two equations?
The answer is that the following two conditions must hold:
curl v = 0
grad E + d(mv)/dt = 0
Since v is a local velocity field for what essentially amounts to
a fluid whose pressure term is p = rho Q, it is only natural to try
and find an equation of state for p. To this end, the second
equation can be differentiated using the relation involving grad E
to eliminate E. Define the local force-per-unit-mass, b as
b = -rho/m grad V
Then, it follows that:
dv/dt + grad (v^2/2) - b/rho + grad(Q/m) = 0
or, after multiplying back the rho:
rho dv/dt + rho grad(v^2/2) + rho grad(Q/m) = b
The term rho dv/dt can be rewritten using the continuity equation:
rho dv/dt = d(rho v)/dt - v d(rho)/dt
= d(rho v)/dt + v div(rho v)
and using the fact that curl v = 0, we can write:
grad(v^2/2) = v.div(v)
Therefore, the first two terms can be expressed as:
d(rho v)/dt + div(rho v v)
where dyadic notation is used on the second term (i.e., the term
is the vector whose x component of div(rho v v) is div(rho v v_x).)
Now, as it turns out, the term rho grad(Q/m) is also a perfect
differential. I think it's of the form
rho grad(Q/m) = (h/2m)^2 div P
where P is the generalized pressure (i.e. stress tensor) which has
the form
P = (del rho)(del rho)/rho - 1/2 (del^2 rho)/rho
or something like that.
The result is a pair of equations of the form:
d rho/dt + div(rho v) = 0 (A)
d(rho v)/dt + div(rho v v + (h/2m)^2 P) = b (B)
combined with the constraint
curl v = 0. (C)
When b is defined as above (b = -rho/m grad V), the resulting equation
is equivalent to Schroedinger's equation, subject to the reverse
identification:
S is the momentum potential
grad S = m v
which defines S up to a space-constant function of time (this
function is clamped down by absorbing the constant of integration
after equation (B) is integrated); and
psi = sqrt(rho) exp(i S/h-bar).
However, the equations (A), (B) do not make any specific assumption
about the type of force that b corresponds to. It could represent
a frictional force Therefore, it is possible to use this system as
a generalization of Schroedinger's equation capable of handling
frictional forces.
squark
>There's nothing but quoted text here-- probably an accident...
>
>(I'm notifying you in case you thought you had posted something else.)
Of course I did! This deja is really going wild.
Okay, modified text:
>In article <396058AB...@xs4all.nl>,
> Gerard Westendorp <wes...@xs4all.nl> wrote:
>> Another Hamiltonian that gives x' = ax is:
>>
>> H = apx
>>
>> ->
>>
>> dp/dt = -ap
>> dx/dt = ax
Quantizing this is interesting, as there is an evident operator ordering ambiguity. We might try a symmetric prescritption, for instance:
^H = (-ia/2)(x d/dx + d/dx x)
^H = (-ia/2)(2 x d/dx + 1)
The eigenvalue equation is
2 x dpsi(x)/dx = (2iE/a - 1) psi(x)
dpsi(x)/dx = (iE/a - 1/2) psi(x)/x
This gives all the eigenvalue in the world and the eigenvectors
psi(x) = A x^(iE/a-1/2)
However, this leads to a model equivalent to the classical one as well. Actually, all quantizations of x' = ax do, because the later equation implies [x, x'] = 0.
Best regards, squark.
--== Sent via Deja.com http://www.deja.com/ ==--
Before you buy.
Gerard Westendorp
[...]
> Now that you've said this, I'm going actually show how to include
> such forces in quantum theory. This prescription is only being
> stated here for the non-relativistic Schroedinger Equation, since
> I'm not entirely sure how it generalizes into the relativistic
> domain.
This is probably not possible, because the time-space symmetry would
imply that time anisotropy gives space anisotropy.
(Friction = damping = "arrow of time")
But an arrow of space would be a bit weird. Physically that is,
mathematically it would be no problem.
[...rewriting the Schrodinger equation...]
> The result is a pair of equations of the form:
>
> d rho/dt + div(rho v) = 0 (A)
> d(rho v)/dt + div(rho v v + (h/2m)^2 P) = b (B)
>
> combined with the constraint
>
> curl v = 0. (C)
This looks a lot like fluid dynamics (a compresible, irrotational
fluid.)
However, I see that the equation has become non-linear, wheras the
original Schrodinger equation is linear. What is going on here?
> When b is defined as above (b = -rho/m grad V), the resulting equation
> is equivalent to Schroedinger's equation, subject to the reverse
> identification:
> S is the momentum potential
> grad S = m v
>
> which defines S up to a space-constant function of time (this
> function is clamped down by absorbing the constant of integration
> after equation (B) is integrated); and
>
> psi = sqrt(rho) exp(i S/h-bar).
>
> However, the equations (A), (B) do not make any specific assumption
> about the type of force that b corresponds to. It could represent
> a frictional force Therefore, it is possible to use this system as
> a generalization of Schroedinger's equation capable of handling
> frictional forces.
Another approach could be splitting the real and imaginary
part of psi:
psi = X + iY
You can get a 4th order equation:
(leaving out constants)
{(d/dt)^2 + (-(d/dx)^2 + V)^2 } X = 0
This look like a wave in a beam.
You should be able to put some friction in here also. But you
might not want to spoil the conservation of rho. This would
mean the particle is slowly going up in smoke. In your
equations, rho is explicitly conserved, but energy is not.
Gerard
Gerard Westendorp
> > Gerard Westendorp <wes...@xs4all.nl> wrote:
> >> Another Hamiltonian that gives x' = ax is:
[Vast wads of quoted text deleted]
> >> H = apx
> Quantizing this is interesting, as there is an evident operator
> ordering ambiguity.
Interesting that quantisation leads to an extra dimension.
We started off with one variable x(t) as a function of time
and ended up with a 1D field psi(x,t).
Gerard
Gerard Westendorp
Gerard Westendorp wrote:
>
[..]
>
>
> 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'.
>
I got confused here. You can use the normal procedure
for going from H to L:
L = px' - H
= (x'-ax)p
As Charles Torre explained, this is not a very useful
L. But stretching things helps to show the meaning
of these concepts.
Gerard
Charles Torre
all the threads, but I recently deduced the existence of this one...
Jim Graber wrote:
> I am interested in the inverse problem in General Relativity and
> its alternatives. In particular a two step "inverse
> problem" Step one: Going back from
> a spherically symmetric solution(SSS) (preferably the general
> SSS) to a set of field equations which it satisfies.
> Step two: Going back from the field equations to an action. (I
> believe this is the problem you have been discussing in a
> different context)
...
>
> Obviously the Einstein action or is it the Hilbert action solves
> the inverse problem for the Schwarzschild solution. But
> what non boring action or actions yield the general spherically
> symmetric solution?
I would expect that there are infinitely many actions that allow
for some spherically symmetric solutions. I am not sure what
*the* "general spherically symmetric solution" is. One can
demand spherical symmetry of, say, a spacetime metric. This
leads to a metric containing 4 free functions of two variables
(in 3 and higher dimensions). (Two of these functions can be
eliminated by adjusting some coordinates, but this is dangerous
if you are going to plug the resulting metric into an action and
vary it.) Anyway, so far, no field equations have been chosen, so
there are no "solutions" to speak of. Pick your favorite
Lagrangian and shove this metric in. Vary the 4 functions to get
the field equations. You get 4 PDEs whose solution (when it
exists) is the spherically symmetric solution for that
Lagrangian. I would expect that the solutions would depend upon
1 or more arbitrary constants depending upon how many
derivatives appear in the PDEs. Now, given such a solution
(e.g., Schwarzschild for the Einstein equations) you can ask how
many Lagrangians allow for that same solution. Is this what you
mean? I don't think this kind of problem has been explored much.
I can imagine that there are a large number of purely geometric
Lagrangians whose field equations, while not equivalent to the
(vacuum) Einstein equations, are satisfied by solutions to the
Einstein equations. For example, I believe that certain
"fourth-order gravity theories" include all Ricci-flat metrics
in their solution space. (One result along these lines that is
cute: the plane wave spacetimes solve virtually any covariant
field equations you can write down for a metric! In particular,
they even define exact solutions in string theory to all orders
in perturbation theory.)
Or do you want to take, say, the Schwarzschild solution and
(given some choice of fields) construct actions that give
solutions that approximate Schwarzschild? This latter problem
can usually be solved by modifying the Einstein-Hilbert action
by terms whose coefficients are sufficiently small. Is this the
kind of thing you have in mind?
Are you interested in keeping the same field content and
changing Lagrangians, or changing both the field content and the
Lagrangians? If the latter, I do not even know really what is the
Schwarzschild solution any longer since you have to tell me what
to do with the other fields, since there is the possibility of
field redefinitions, etc.
I see that Chris Hillman has been asking similar questions to
mine. Oh well, I'll just quit here.
Charles Torre
Jim Graber
to...@cc.usu.edu (Charles Torre) wrote:
> Between my leaky newsfeed and my summer schedule I am not seeing
> all the threads, but I recently deduced the existence of this one...
>
> Jim Graber wrote:
>
> > I am interested in the inverse problem in General Relativity and
> > its alternatives. In particular a two step "inverse
> > problem" Step one: Going back from
> > a spherically symmetric solution(SSS) (preferably the general
> > SSS) to a set of field equations which it satisfies.
> > Step two: Going back from the field equations to an action. (I
> > believe this is the problem you have been discussing in a
> > different context)
> ...
> >
> > Obviously the Einstein action or is it the Hilbert action solves
> > the inverse problem for the Schwarzschild solution. But
> > what non boring action or actions yield the general spherically
> > symmetric solution?
Thanks for deducing the existence of this thread and responding to it.
Your reply is already helpful to me.
I did post an answer to some of Chris's questions. Rather than repost
it, I will try to email it to you. (Actually written 7/8/00. Appeared
on newsgroup after moderation 7/10/00. Off of my news server now, but
still retrievable from deja)
>
> I would expect that there are infinitely many actions that allow
> for some spherically symmetric solutions. I am not sure what
> *the* "general spherically symmetric solution" is. One can
> demand spherical symmetry of, say, a spacetime metric. This
> leads to a metric containing 4 free functions of two variables
> (in 3 and higher dimensions).
This is exactly what I had in mind. Actually even the static solution
with only one variable will do for a start.
> (Two of these functions can be
> eliminated by adjusting some coordinates, but this is dangerous
> if you are going to plug the resulting metric into an action and
> vary it.)
This is exactly what I did. I don't understand why it is dangerous.
Can you explain?
> Anyway, so far, no field equations have been chosen, so
> there are no "solutions" to speak of.
Excuse my irreverent analogy, but perhaps you have heard the joke,
"Jesus is the answer. What is the question?"
These are the two inverse problems I have been trying to solve:
1. Schwarzschild is the answer. What is the question? ( This is just a
warm up for number 2)
2. General (static) spherically symmetric "solution" is the answer.
What are the field equations, and the Lagrangian?
So far, I have attempted to go to the field equations first and then
to the Lagrangian. I see you seem to think you should go to the
Lagrangian first. Why is this better?
> Pick your favorite
> Lagrangian and shove this metric in.
I don't have a favorite Lagrangian. Probably I need to get one fast.
In the meantime, my first choice is the scalar-tensor Lagrangian,
because a lot of work has been done with it, which I can draw upon, and
because it is supported as a low energy limit of string theory. (see my
reply to Chris for a little more detail). My second choice would be a
higher order gravity Lagrangian, particularly Gauss-Bonnet(again, low
energy limit of string theory. My third choice would be the general
metric affine gravity Lagrangian, because it is apparently one of the
most general forms considered so far.
(By the way, I do *not* understand string theory, and am not even very
comfortable with Lagrangians and variations.)
alfred_...@my-deja.com
> The result is a pair of equations of the form:
>
> d rho/dt + div(rho v) = 0 (A)
> d(rho v)/dt + div(rho v v + (h/2m)^2 P) = b (B)
>
> combined with the constraint
>
> curl v = 0. (C)
>From Gerart Westendorp <wes...@xs4all.nl>:
> This looks a lot like fluid dynamics (a compresible, irrotational
> fluid.)
That's exactly what it is -- except that the equation of state (i.e.,
the stress tensor) is quantum theoretic in origin. Were it not for the
additional term (h/2m)^2 P, it would be a classical fluid.
This stress tensor is the form (apart from sign, which I don't exactly
remember):
P^mn = rho u^m u^n - eta delta^mn div (rho u)
where
eta = h/2m
and
u = eta grad log rho
corresponds directly to the "osmotic velocity" of R. Nelson's Brownian
Motion interpretation of Schroedinger's Equation.
> However, I see that the equation has become non-linear, wheras the
> original Schrodinger equation is linear. What is going on here?
That just means that some systems of non-linear equations in fluid
dynamics can, by a suitable transformation, be turned into linear
systems.
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Before you buy.